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IL Ranker Series Chemistry for NEET Grade 12 Module 1
ISBN 978-81-985044-3-2 [SECOND
EDITION]
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Preface
The IL Ranker Series for NEET is a comprehensive series of books designed to help students prepare for one of the most important exams on their path to becoming doctors or medical professionals. The National Eligibility cum Entrance Test (NEET) is a critical step for students aiming to enter the medical field, testing not only knowledge but also their ability to think and understand the application of the concepts. Recognising the skills needed to crack the NEET and the challenges the students face, this series has been crafted to serve as a guide, mentor, and companion on their path to achieving their dream rank in NEET.
The IL Ranker Series stands out because it's built keeping NEET aspirants in mind. We've included a variety of features to cater to their learning needs and to prepare them thoroughly for the exam.
This meticulously designed book series offers a comprehensive exploration of key concepts. Organised into clear topics and subtopics, it facilitates efficient learning. Practice questions and checkpoints solidify understanding, while illustrations and tables enhance visualisation and comprehension. Chapter reviews provide a quick revision tool, ensuring knowledge retention.
The comprehensive coverage of the NCERT syllabus is supplemented by advanced questions, providing holistic preparation for the NEET exam. It incorporates diverse question types, including matching, statement, assertion & reason, and brain teasers to enhance critical thinking and application skills. The resource includes expert tips, theory-based questions, and a mix of difficulty levels, aiming to improve conceptual understanding, cross-topic synthesis, and retention of key concepts.
This second edition of the IL Ranker Series for NEET is more than just a set of books. It's a commitment to help you learn, grow, and succeed. We've designed it to ignite your passion for the medical profession and to foster a lifelong love of learning. With every page you turn, you're moving one step closer to your dream of cracking NEET with a good rank.
Key Features of the Book
Chapter Outline
1.1 Types of Solutions
1.2 Methods of Concentration
1.3 Solubility
This outlines topics or learning outcomes students can gain from studying the chapter. It sets a framework for study and a roadmap for learning.
Specific problems are presented along with their solutions, explaining the application of principles covered in the textbook. Solved Examples
1. What is the molality of a solution of H2SO4 having 9.8% by mass of the acid?
Sol. 9.8% by mass of H2SO4 contains 9.8 g of H2SO4 per 100 g of solution.
Therefore, if mass of solution = 100 g, mass of solute, H2SO4 = 9.8 g,
Try yourself:
1. In a solution of H 2 SO 4 and water, mole fraction of H2SO4 is 0.9. How many grams of H2SO4 is present per 100 g of the solution?
Ans: 98
Try Yourself enables the student to practice the concept learned immediately.
This comprehensive set of questions enables students to assess their learning. It helps them to identify areas for improvement and consolidate their mastery of the topic through active recall and practical application.
CHAPTER REVIEW
Types of Solutions
■ A solution is a homogeneous mixture of two or more non–reacting components. Formation of solution is a physical process.
TEST YOURSELF
1. The mole fraction of a solvent in aqueous solution of a solute is 0.6. The molality of the aqueous solution is (1) 83.25 (2) 13.88 (3) 37 (4) 73
It offers a concise overview of the chapter’s key points, acting as a quick revision tool before tests.
This is a focused practice with topic-wise questions based on NCERT textbook content. It is designed to enhance students’ success in NEET by aligning with recent exam trends.
Exercises
NEET DRILL FURTHER EXPLORATION
Holding substantial weightage in the NEET Biology paper, these questions improve analytical judgement of statements.
Known for their low scoring rate and high weightage in recent NEET exams, these questions play a crucial role in improving students’ critical thinking skills to assess the logical relationship between the assertion and the reason.
MATCHING TYPE QUESTIONS
STATEMENT TYPE QUESTIONS
ASSERTION AND REASON TYPE QUESTIONS
BRAIN TEASERS
FLASHBACK
CHAPTER TEST
Modelled after the NEET exam format, this test is based on a specific chapter. It serves as a tool for students to evaluate their time management skills and gauge their mastery level in a particular chapter.
This section comprises questions that extend beyond the NCERT content yet remain relevant to NEET, preparing students for additional and pertinent challenges beyond the textbook.
These include questions for practising the correlation of information across different topics. A significant number of matching questions appear in the NEET Biology paper and are easy to score.
These complex questions that combine fun and critical thinking are aimed at fostering higher order thinking skills and encourage analytical reasoning.
Hand-picked questions from previous years NEET offer an insight into the types of questions and the important topics that are probable to appear in NEET.
Chapter Outline
1.1 Types of Solutions
1.2 Methods of Concentration
1.3 Solubility
1.4 Vapour Pressure
1.5 Solution of a Solid in a Liquid
1.6 Colligative Properties
1.7. Abnormal Molar Masses
Pure substances are very rare. In normal life we come across mixtures containing two or more substances.
For example, their usage depends on their composition. The properties of german silver (mixture of copper with nickel and zinc) are different from that of bronze (mixture of copper with tin). Almost all processes in body occur in some kind of liquid solution. A homogeneous mixture of two or more substances is called a solution.
SOLUTIONS CHAPTER 1
1.1 TYPES OF SOLUTIONS
T he composition and the properties are uniform throughout the homogeneous mixture. The component that is present in a larger quantity is generally called solvent and the component present in a minor quantity is called solute. A binary solution contains only two components. Solvent determines physical state in which the solution exists. Based on the physical state, solutions are classified into three types: 1. solid solutions 2. liquid solutions 3. gaseous solutions. Each type of solution is further classified into 3 more types. Types of solutions and their examples are listed in Table 1.1
Note that the absorption of H 2 gas on palladium is known as occlusion.
Among different types of solutions, solid in liquid type solutions are most frequently studied.
Ethanol
Sucrose
Solution of hydrogen in palladium
Amalgam of mercury with sodium
Copper dissolved in gold
Table 1.1 Types of solutions and common examples
1.1.1 Aqueous and Non-Aqueous Solutions
Solution prepared by using water as solvent is called aqueous solution. Alcoholic solutions contain ethyl alcohol as solvent. Non-aqueous solutions have benzene, chloroform, ether, carbon tetrachloride, etc., as solvents.
Concentration is the term used to express the amount of solute present in a definite amount of solution. It is also called strength of the solution. The concentration of a solution can be expressed quantitatively in several ways.
1.2 METHODS OF CONCENTRATION
Concentration of a solution represents the composition of the solution. It expresses the relative quantities of solute and solvent. The following methods are used to express the concentration of a solution.
1.2.1 Weight Percent (or) Mass Percent (w/w)
Weight in grams of a solute present in 100 g of a solution is called its weight percent (w/w).
Weight percent of solute =
Weightofsolute 100
Weightofsolution ×
Weight volume percent or percentage of solute (w/v):
Weight in grams of a solute present in 100 mL of a solution is called its weight volume percent (w/v).
Weight volume percent of solute =
Weightofsolute 100
Volumeofsolution ×
1.2.2 Volume Percent (V/V)
Volume in millilitres of a solute present in 100 mL of a solution is called its volume percent.
Volume percent of solute =
Volumeofsolute 100
Volumeof solution ×
1.2.3 Molarity
It is defined as the number of gram moles of the solute present in one litre of solution. It is denoted by ‘M’. Units of molarity are mol L–1. It is dependent on temperature. As temperature increases, the volume of a solution increases and molarity decreases. Molarity (M) is given as,
Number of moles of solute () M
Volume in litres of solution (V) = n (or)
1000 w M
GMWV =× w n GMW =
Here, w is mass of solute in grams and V is volume of solution in millilitres. GMW is gram molecular weight of the solute.
A molar solution is the one in which one gram mole of solute is present in one litre of solution. One millimole of solute present in one millilitre solution is also called one molar solution. Molar (1M), semimolar (0.5M), decimolar (0.1M), centimolar (0.01M), millimolar (0.001M), etc., are in common use.
The mass of solute (w) in grams, present in V litres of a solution, can be calculated from molarity of the solution (M) as, w = M × V × GMW
Molarity also indicates the number of millimoles of the solute dissolved in one millilitre of the solution.
(1 mol = 1000 millimoles)
Some useful relations:
1. Number of millimoles of the solute present in V mL of the solution = M × V
2. Number of moles of solute present in V mL of the solution MV 1000 × =
3. Number of moles of solute present in V L of solution = MV
4. When a solution is diluted,its molarity decreases, but number of moles of solute before and after dilution remains constant.
M1V1 = M2V2
V1 = Volume of the solution before dilution
M 1 = Molarity of the solution before dilution
V2 = Volume of the solution after dilution
M2 = Molarity of the solution after dilution
5. Volume of water added to get a solution of known molarity,
10. When two solutions are titrated against each other as per the following equation: 1212 nAnBmCmD +→+
At the equivalence point of the titration, The molarities of the two solutions are related by the equation 1122 12 MVMV nn =
1.2.4 Concepts of Equivalent Weight
The equivalent weight of a substance is weight in grams of substance which furnishes or combines with one gram of hydrogen ions (H+ion) in a solution
Equivalent Weight of an Acid
6. Molarity = 1000 w GMWV ×
w = weight of the solute in grams
V = volume of the solution in millilitres
GMW = molecular weight of the solute in grams
7. When weight/volume percentage is given, () 10/% wV M GMW × =
8. When weight percentage of solution and specific gravity or density are given, then molarity is given by 10specificgravity(/)% M GMW ×× = ww
10d(/)% M GMW ×× = ww
where (w/w)% = percentage by weight
d = Density in gram/mL
9. For a mixture of two solutions of different molarities, the molarity of the resulting solution is given by
Molecularweightofacid E Basicityofacid =
Number of replaceable hydrogens present in one molecule of acid is called its basicity.
(Its structure is OH | HPOH || O ; it furnishes only two H+ ions)
Equivalent Weight of Base Base
Molecular weight of base E Acidity of base =
Number of replaceable hydroxyl groups present in one molecule of a base is called its acidity.
Example: 1. Ba(OH)2 Molecularweight 171.33 E 85.67 22 === 2. NaOH Molecularweight 40 E 40 11 ===
3. ()2 CaOH Molecularweight 74 E 37 22 ===
Equivalent Weight of a Salt
Formula weight of the salt ESalt Total number of units of charge on cation or anion of the salt =
Example:
1. NaCO23
Formulaweight 106 E 53 22 ===
2. NaCl
Formulaweight 58.5 E 58.5 11 ===
3. ()24 3 AlSO
Formulaweight 342 E 57 66 ===
Equivalent Weight of an Oxidising Agent (or) Oxidant
The weight in grams of a substance that gains one mole of electrons is called gram equivalent weight of oxidant.
Formula weight of oxidant E
oxidant
Decrease in oxidation state of oxidant per a formula unit =
Electrons gained by oxidant (or)
KMnO4 as an oxidant:
i. In acidic medium
MnO4– + 8H+ + 5e–→Mn+2 + 4H2O
Change in oxidation number = 7–2 = 5 or
Number of electrons gained = 5
Formulaweight 158 E 31.6 55 ===
KMnO4
ii. In dilute alkaline medium or in neutral medium
MnO4– +2H2O + 3e–→MnO2 + 4OH–
Change in oxidation number = 7 – 4 = 3
(or) Number of electrons gained = 3 KMnO4
Formulaweight 158 E 52.6 33 ∴===
iii. In strongly alkaline medium: 2 44 MnOeMnO +→
Change in the oxidation state is one unit.
Number of electrons gained is 1.
Potassium Dichromate as an Oxidant:
Cr2O7–2+14H+ + 6e– →2Cr+3 + 7H2O
Change in oxidation number for one ‘Cr’ atom = 3
Change in oxidation number for two ‘Cr’ atoms = 6
Number of electrons gained = 6
Formulaweight 294 E 49 66 ===
KCrO227
Equivalent Weight of Reducing Agent (or) Reductant
Formulaweightofreductant E
The weight in grams of substance that loses one mole of electrons is called gram equivalent weight of reductant. ()
Electronslostbyreducantor
Increaseinoxidationstateofreducant per a formula unit = Mohr’s salt is ferrous ammonium sulphate. Its formula is FeSO4 (NH4)2 SO4. 6H2O. In its reactions in acidic medium, ferrous (Fe+2) ion gets oxidised to ferric (Fe +3); 23 FeFee ++− →+ so, EMohr’s salt
Formulaweight 392 11 ==
1.2.5 Normality
It is defined as the number of gram equivalents of the solute present in one litre of a solution. It is denoted by ‘N’. Unit of normality is eq L –1. It is dependent on temperature. As the temperature increases, the volume increases and normality decreases.
Normality (N) is given as,
Numberofequivalentsofsolute N (or)
Volumeofsolution inlitres =
Weightofsolute 1000 N GramequivalentweightVolume of solutionin mL =×
Some useful relations:
1) Number of equivalents of solute = N×V; volume V is in litres.
2) Number of milli equivalents of solute= N×V; volume V is in millilitres.
3) The mass of a solute (w) in grams, present in V litres of a solution, can be calculated from normality of the solution (N) as,
w = N × V × GEW
Here,GEW is the gram equivalent weight of solute.
4) Normality of the mixture when two solutions of same solute are mixed;
1122 12 N = NVNV VV + +
Here V 1 + V 2 = Total volume o f the solution.
5) When V a mL of a strong acid of normality N a is mixed with Vb mL of a strong base of normality Nb;
i) If N a V a = NbVb , the solution is neutral.
ii) If N a V a > NbVb, the solution is acidic.
Normality w.r.t H+ = aabb ab NVNV VV +
iii) If N a V a < NbVb, the solution is alkaline.
Normality w.r.t OH– = bbaa ab NVNV VV +
6) Normality–Molarity interrelation:
i) For acids:
Normality = Molarity × basicity of acid
ii) For bases:
Normality = Molarity × acidity of base
iii) For salts:
Normality = Molarity × total number of units of +ve charge or –ve charge of the salt ion
iv) For oxidising (or) reducing agents:
Normality = Molarity × total change in oxidation state per mole of oxidant or reductant
7) For exact neutralisation of a strong acid with a strong base: aa
Weightofbase NV(mL)
GEWofbase1000 =
WeightofacidNV(mL)
bb
GEWofacid1000 =
8) In case of dilution: N1V1 = N2V2
The volume of the solution before and after dilution are V1 and V2, and normalities are N1 and N2 respectively.
1.2.6 Molality
Molality is defined as the number of gram moles of the solute present in one kilogram of solvent. It is denoted by ‘ m ’. Units of molality are mol kg –1. It is independent of temperature. Molality is the most accurate and,theoretically,the best method of expressing concentration.
Molality (m) is given as () () Numberofmolesofsoluten m NumberofkgofsolventW =
w1000 m GMWW =× w n GMW =
Here w and W are masses of solute and solvent respectively in grams. GMW is the gram molecular mass of solute.
Parts per Million
Trace quantities of solute in a solution is conveniently expressed in parts per million (ppm).
ppm = 6 Numberofpartsofcomponent 10
Total numberof parts of solution ×
Some useful relations:
1. 10Solubility m Grammolecularweight × = (solubility in g of solute 100 g solvent)
2. ()MolarityVinL m Weightofsolventinkg × =
3. () 1000Molarity m 1000specificgravityMGMW × = ×−× (here, M = Molarity)
4. Mole fraction of solute = m 1000 m Molecularweightofsolvent
1.2.7 Mole Fraction
It is the ratio of number of moles of a component to the total number of moles of the solution. It is denoted by the symbol ‘X’. If number of moles of component substances A and B in a solution are, respectively, nA and nB, Mole fraction of component,
A = () A A AB n X nn = +
Mole fraction of component, B = = + B B AB n X (nn)
In a binary solution of A and B, XA+XB = 1. The sum of the mole fractions of all components in a solution is unity. One hundred times mole fraction is called mole percentage. Mole fraction and mole percentage have no units. They do not vary with a change in temperature of solution.
1. What is the molality of a solution of H 2SO4 having 9.8% by mass of the acid?
Sol. 9.8% by mass of H2SO4 contains 9.8 g of H2SO4 per 100 g of solution.
Therefore, if mass of solution = 100 g, mass of solute, H2SO4 = 9.8 g, and mass of solvent = 100 – 9.8, = 90.2 g
Molality = 24 Number of moles of HSO Mass of solvent in kg
9.81000 9890.2 =× =1.1 mol kg–1
Try yourself:
1. In a solution of H 2 SO 4 and water, mole fraction of H2SO4 is 0.9. How many grams of H2SO4 is present per 100 g of the solution?
Answer: 98
TEST YOURSELF
1. Calculate normality of 0.98% W V
H2SO4 solution.
(1) 0.1 N (2) 1 N
(3) 0.2 N (4) 2 N
2. The mole fraction of a solvent in aqueous solution of a solute is 0.6. The molality of the aqueous solution is
(1) 83.25 (2) 13.88
(3) 37 (4) 73
3. Mole fraction of solute in 4.5 molal aqueous solution is (1) 0.05 (2) 0.025 (3) 0.0375 (4) 0.075
4. 14.3 g of Na2CO3 xH2O completely neutralizes 100 mL of 1 N H2SO4 solutions. Value of x is
(1) 3 (2) 10 (3) 5 (4) 2
5. The molarity of 1.5 N H3PO4 solution is (1) 1.3 M (2) 0.75 M
(3) 0.5 M (4) 4.5 M
6. The number of millimoles of H2SO4 present in 5 L of 0.2 N H2SO4 solution is (1) 500 (2) 1000 (3) 250 (4) 0.5 × 10–3
7. The volume of water that must be added to a mixture to 250 mL of 6 M HCl and 650 mL of 3 M to obtain 3 M HCl solution is (1) 75 mL (2) 150 mL (3) 300 mL (4) 250 mL
8. Molarity of 1 m aqueous NaOH solution (density of the solution is 1.02 g/mL) (1) 1 M (2) 1.02 M
(3) 1.2 M (4) 0.98
9. Equivalent weight of Baeyer’s reagent is (M = molecular weight)
Solubility of a solute in a given solvent represents the maximum quantity of the solute that can be present in dissolved state in the saturated solution at a given temperature.
1.3.1 Solubility of a Solid in a Liquid
When a solid solute is added to a solvent, some solute dissolves and its concentration increases in the solution. This process is known
as dissolution. When the solute is continuously added with constant shaking,some solute particles in solution collide with the solid solute particles and get separated out of solution. This process is known as crystallisation. A stage is reached when the two processes occur at the same rate. Under such conditions, number of solute particles going into solution will be equal to the solute particles separating out and a state of dynamic equilibrium is reached.
Solute + Solvent Solution
The solution at this stage is said to be saturated solution. (An unsaturated solution is the one in which more solute can be dissolved at the same temperature.) The concentration of the saturated solution is called the ‘solubility’. Thus, solubility may be defined as follows:
The solubility of a solid in a liquid at any temperature is defined as the maximum amount of the solid (solute) in grams which can be dissolved in 100 g of the liquid (solvent) to form the saturated solution at that particular temperature. Molar concentration of saturated solution is called its molar solubility.
Factors Affecting the Solubility of a Solid in a Liquid
The factors on which the solubility of a solid in a liquid depends are:
i) nature of the solute and the solvent
ii) temperature
iii) pressure
Nature of the solute and the solvent: In general, a solid dissolves in a liquid which is chemically similar to it. This is expressed by saying “Like dissolves like”. This statement implies that ionic (polar) compounds, like KCl, dissolve in polar solvents like, water. They are very much less soluble or almost insoluble in non-polar solvents, like benzene, ether, etc. Similarly, non-polar compounds,like naphthalene, anthracene, etc., are soluble in
non-polar solvents,like benzene, ether, carbon tetrachloride, etc. They are very less soluble in water.
Temperature: The effect of temperature on solubility depends on heat of solution and it can be explained based on Le Chatelier’s principle.
Heat of solution = Lattice energy + Hydration energy
The solubility of solids which dissolve in liquid solvents with absorption of heat ( D H = + ve) increase with increase of temperature.
The solubility of solids which dissolve in liquid solvents with release of heat ( D H = –ve) decreases with increase of temperature.
In a nearly saturated solution, if the dissolution process is endothermic, the solubility of a substance should increase with rise in temperature.
For salts whose lattice energy is greater than hydration energy, solubility increases with increase in temperature.
In a nearly saturated solution, if the dissolution process is exothermic,The solubility of a sub-stance should decrease with rise in temperature. For such salts hydration energy exceeds lattice energy.
Examples: Ce 2 (SO 4 ) 3 , Na 2 SO 4 , CaCl 2 , Li2SO4, CuSO4 etc.
For NaCl lattice energy(184 kcal) is nearly equal to hydration energy (182.8 kcal). Hence, its solubility does not vary much with temperature.
Pressure: Pressure has a very little effect on the solubility of a solid in a liquid because solids and liquids are highly incompressible.
1.3.2 Solubility of a Gas in a Liquid
Almost all gases are soluble in water, though to different extents. The existence of aquatic life in lakes, rivers, sea, etc. is due to dissolution of oxygen gas of the air in water.
Solubility of a gas in a liquid at a particular temperature is also expressed in terms of molarity (moles of the gas dissolved per litre of the solution to form the saturated solution, i.e., in terms of mol L–1) or in terms of mole fraction (X) of the gas.
Factors Affecting the Solubility of a Gas in a Liquid
The important factors on which the solubility of a gas in a liquid depends are:
i) nature of the gas and the solvent
ii) temperature
iii) pressure
Nature of the Gas and the Solvent
Gases like hydrogen, oxygen, nitrogen, etc. dissolve in water only to a small extent, whereas gases like HCl, NH3, etc. are highly soluble. The greater solubility of the later gases is due to their reaction with the solvent, water.
Oxygen, nitrogen, and carbon dioxide are much more soluble in ethyl alcohol than in water at the same temperature and pressure, while H2S and NH3 are more soluble in water than in ethyl alcohol. Evidently, the greater solubility of a gas in a solvent is again due to the chemical similarity between the gas and the solvent.
Effect of Temperature
The solubility of a gas decreases with increase in temperature. This is expected because the dissolution of a gas in a liquid is always an exothermic process, i.e., it is accompanied by evolution of heat.
Gas + Solvent Solution + Heat
Applying Le Chatelier’s principle, it is evident that increase in temperature would shift the equilibrium in the backward direction, i.e., the solubility would decrease.
Effect of Pressure – Henry’s Law
For the solution of a gas in a liquid, consider a system, as shown in Fig 1.1 (a). The lower part is the solution and the upper part is gaseous system at a pressure P and temperature T. Suppose the system is in dynamic equilibrium, i.e., rate of gaseous particles entering and leaving the solution is the same, which means that rate of dissolution = rate of evaporation. Now, on increasing the pressure over the system, as shown in Fig 1.1 (b) , the gas gets compressed to a smaller volume. Hence, the number of gaseous particles per unit volume increases. As a result, the number gaseous particles striking the surface of the solution and, hence, entering into it also increases, till a new equilibrium is reestablished. Thus, on increasing the pressure of the gas above the solution, the solubility increases.
depends upon the nature of the gas, the nature of the solvent, and the temperature.
Henry’s law may also be stated as follows:
The solubility of a gas in a liquid at a particular temperature is directly proportional to partial pressure of the gas in equilibrium with the liquid at that temperate.
Dalton, during the same period, had concluded independently that if a mixture of gases are simultaneously in equilibrium with the liquid at a particular temperature, the solubility of any gas in the mixture is directly proportional to the partial pressure of that gas in the mixture.
For a gas A, Henry’s law can be written as:
XA = A Kp ' . . . (ii),
where XA is the mole fraction of the gas in the solution, pA is the partial pressure of the gas above the solution, and K’ is proportionality constant whose value depends upon the nature of the gas, nature of the solvent, and the temperature.
Quantitatively, the effect of pressure on the solubility of a gas in a liquid was studied by Henry and is called Henry’s law. It is stated as follows:
The mass of gas dissolved in a given volume of the liquid at constant temperature is directly proportional to the pressure of the gas present in equilibrium with liquid.
Mathematically, m ∝ p or m = K p . . . . .(i), where m = mass of the gas dissolved in a unit volume of the solvent, p = pressure of the gas in equilibrium with the solvent, K = constant of proportionality whose value
For example, the solubility of pure N2 in water at 298 K and a partial pressure 0.78 atm (which is the partial pressure of N2 in air at 1.0 atm) is 5.3 × 10–3 mol L–1 (using molarities in place of mole fractions). If the partial pressure is doubled to 1.56 atm, the solubility of N 2 is doubled to 1.06 × 10–2 M.
From equations (ii) AA 1 Px K' = or PA = KH XA . . . (iii), where H 1 K K' = , is called Henry’s constant.
This is the most commonly used form of Henry’s law and may be defined as follows:
The partial pressure of a gas (p) is directly proportional to the mole fraction (X) of the gas in the solution.
Fig. 1.1 Pressure increases the solubility of gas
Using this expression, the unit K H will be atm or bar (or kbar).
This is a convenient expression for testing the validity of Henry’s law. Plotting equilibrium pressures pA, versus corresponding mole fractions XA, a straight line plot passing through the origin is obtained with slope = K H (in atm or bar), as shown in Fig. 1.2 , for solubility of HCl gas in cyclohexane at 293 K.
and the nature of the solvent at the same temperature. The KH values of some gases in different solvents at 298 K are given in Table 1.2
Table 1.2
Values of Henry’s constant for some gases in different solvents at 298 K
Fig.1.2 Solubility of HCl in cyclohexane at 293 K
Different gases have different values of K H at the same temperature and in the same solvent. From these values, the following results may be drawn:
i. Henry’s constant, KH, is a function of the nature of the gas.
ii. Greater the value of K H , lower is the solubility of the gas at the same partial pressure [according to eq n . (iii), at a particular temperature].
iii. The value of KH increases with increase in temperature, implying that the solubility decreases with increase in temperature at the same pressure.
It is for this reason that aquatic species feel more comfortable in cold water (in which O2 gas dissolved is more) than in warm water (in which O2 dissolved is less).
Further, as already mentioned, the value of K H depends upon the nature of the gas
Limitations of Henry’s law: Henry’s law is applicable only if the following conditions are satisfied:
i) The pressure should be low and the temperature should be high, i.e., the gas should behave like an ideal gas.
ii) The gas should not undergo compound formation or dissociation in the solvent
2. Oxygen gas is bubbled through water at 293 K, exerting a partial pressure of 0.98 bars. Find the solubility of oxygen in g L–1 (given: KH of oxygen gas is 34 kbar).
Sol. Henry’s law: p = KH ()
Compared to the number of moles of water, number of moles of dissolved O
Mass of water = 1000 g)
is negligible.
p = KH O2 X
0.98 bar = 34 × 1000 bars × WO2 1000 32 18
Therefore, mass of O2 dissolved per 1000 mL of water,
2 1 O 0.9832 W 0.05gL 3418 × == ×
Try yourself:
2. How many grams of dissolved CO 2 is expected to dissolve in 500 g of soda water when packed under 2.5 atm of CO2 at 298 K
(KH of CO2 = 1.67 × 108 Pa)?
TEST YOURSELF
Ans: 1.86
4. Which of the following gases will greatly deviate from Henry’s law in water?
(1) H2 (2) N2
(3) NH3 (4) CH4
5. Low concentration of oxygen in the blood and tissues of people living at high altitude is due to
(1) Low temperature
(2) Low atmospheric pressure
(3) High atmospheric pressure
(4) Both low temperature and high atmospheric pressure
6. K H of N 2 is 1×10 5 atm. The moles of N 2 dissolved in 10 mole of water is? (Given: Pressure of air is 5 atm and mole fraction of N2 in air is 0.8)
(1) 4 × 10–4 (2) 4 × 10–5
(3) 5 × 10–2 (4) 5 × 10–4
1. The Henry’s law constant for O 2 dissolved i n water is 4.34 × 10 4 atm at a certain temperature. If the partial pressure of O 2 in a gas mixture that is in equilibrium with water is 0.434 atm, what is the mole fraction of O2 is solution?
(1) 1 × 10–5 (2) 1 × 10–4
(3) 2 × 10–5 (4) 2 × 10–6
2. Two gases ‘A’ and ‘B’ have Henry’s constant (KH) values as 44 kPa and 66 kPa at 293 K. At 293 K, if ‘A’ and ‘B’ are dissolved in water, where the partial pressure ratio of A to B is 1 : 3, then the ratio of solubility of A to B would be
(1) 1 : 2 (2) 9 : 2
(3) 2 : 9 (4) 2 : 1
3. H2S,a toxic gas with rotten egg-like smell,is used for qualitative analysis. If the solubility of H2S in water at STP is 0.195 mol kg–1, then the Henry’s law constant is
(1) 285.6 bar
(2) 324.8 bar
(3) 462.9 bar
(4) 534.8 bar
Answer Key
(1) 1 (2) 1 (3) 1 (4) 3
(5) 2 (6) 1
1.4 VAPOUR PRESSURE
The pressure exerted by the vapour molecules of a liquid when they are in equilibrium with the liquid at a given temperature is called vapour pressure of the liquid at that temperature.
Vapour pressure of liquids is measured by barometric method.
Liquids having lower boiling points show higher vapour pressure.
Vapour pressure of a liquid depends upon the following factors:
1) Nature of liquid
2) Temperature
Vapour pressure of a liquid at a given temperature is independent of the quantity of liquid, the surface area, and on the shape of the vessel in which it is kept.
Volatile liquids and non-volatile liquids: Liquids having weaker intermolecular attractions have higher vapour pressures. Such liquids are called volatile liquids. Acetaldehyde, diethylether, pentane, etc., are examples of volatile liquids. On the other hand, intermolecular forces are strong in a metallic liquid, like mercury. Vapour pressures of such liquids are lower. They are commonly called non-volatile liquids.
Vapour pressure and temperature: With the increase of temperature, vapour pressure of a liquid increases exponentially.
When a liquid is heated, its average kinetic energy of molecules increases. Large number of molecules possess sufficient energy to overcome the intermolecular attractions and escape into the vapour state. Therefore, the rate of evaporation increases and the vapour pressure increases exponentially. Variation of vapour pressures of some liquids with temperature is given graphically in Fig.1.3
1.4.1 Vapour Pressure of Liquid–Liquid Solution
For a solution of liquid in liquid, as both the components of the solutions are volatile, each component will form vapour above the solution. When equilibrium is reached, each component will exert a vapour pressure, called its partial vapour pressure, whose value depends upon the mole fraction of the component in the solution and the vapour pressure of the component in its pure state. These studies were made by a French chemist, F.M. Raoult, and he put forward the following result known after him as Raoult’s law:
In a solution, the vapour pressure of a component at a given temperature is equal to the mole fraction of that component in the solution multiplied by the vapour presure of that component in its pure state.
Let us consider a mixture of two completely miscible volatile liquids A and B, having the mole fractions XA and XB. Suppose at a certain temperature, their partial vapour pressures are pA and pB and the vapour pressures in the pure state are p° A and p°B. According to Roult’s law, oo
pxpandpxp == … (1)
pppxpxp =+=+ . . . (2) (or)
p1xpxpppxp =++=−+
Boiling point: Vapour pressure of a liquid increases with increase in temperature, till its value approaches atmospheric pressure. The temperature at which the vapour pressure of the liquid becomes equal to the atmospheric pressure is called boiling point of the liquid (Tb).
p1xpxpppxp =++=−+ … (3)
Evaporation occurs at all temperatures but boiling occurs only at boiling point. The boiling point of a liquid changes with a change in external pressure and also by addition of volatile or non-volatile substances.
Since p° A and p° B are constant at a particular temperature, it is evident from equation (3) that the total vapour pressure is a linear function of the mole fraction X B (or X A as XA = 1 – XB). Thus, a straight line should be obtained, when P is plotted against XA or XB. Such a plot is shown in Fig. 1.4 . The lines (I) and (II) give the plots of partial pressure versus mole fraction and the line (III) that of the total pressure versus mole fraction.
Fig.1.3 Vapour pressure curves of some liquids
Fig.1.4 Vapour pressure versus mole fraction of components
When X A = 1, i.e., the liquid is pure A, p = p° A, and when XB = 1, i.e., the liquid is pure B, p = p° B
The composition of vapour phase in equilibrium with the solution is determined by the partial pressures of the components. If y A and y B are the mole fractions of the components, ‘A’ and ‘B’ respectively in the vapour phase, from Dalton’s law of partial pressures, or (ptotal=Total vapour pressure of the system)
pA = yA Ptotal; pB = yB Ptotal
In general pi = yi Ptotal
1.4.2 Ideal and Non-ideal Solutions
Liquid in liquid solutions can be classified as ideal solutions and non-ideal solutions.
Ideal Solutions
An ideal solution is the solution in which each component obeys Raoult’s law under all conditions of temperatures and concentrations.
An ideal solution will satisfy the following conditions:
i. There will be no change in volume on mixing the two components,
i.e., D V mixing = 0
ii. There will be no change in enthalpy (i.e., no heat is evolved or absorbed) when the two components are mixed, i.e, D H mixing = 0
For example, when we mix 50 cm 3 of benzene with 50 cm3 of toluene, the volume of the solution is found to be exactly 100 cm3, i.e, D V mixing = 0
Hence, the solution obtained is ideal.
An ideal solution may be defined as the solution in which no volume change and no enthalpy change take place on mixing the solute and the solvent in any proportion.
At the molecular level, an ideal solution may be defined as follows:
An ideal solution of the components A and B is defined as the solution in which the intermolecular interactions between the components (A–B inter molecular attractions) are of the same magnitude as the intermolecular interactions found in the pure components (A–A inter molecular attractions and B–B intermolecular interactions found in the pure components).
A few examples of ideal solutions are given below:
i. Benzene + Toluene
ii. n-Hexane + n-Heptane
iii. Enthyl bromide + Ethyl chloride
iv. Chlorobenzene + Bromobenzene
Non-ideal Solutions
A solution which does not obey Roult’s law at all compositions is called a non-ideal solution.
For such solutions, D Vmixing≠0, D H mixing ≠0. For example,when we mix sulphuric acid (solute) with water (solvent), the amount of heat generated is very large and change in volume is also observed. This is due to formation of a non-ideal solution.
In terms of molecular interactions, a nonideal solution may be defined as follows:
A non-ideal solution is the solution in which solute and solvent molecules interact with one another with a different force than the forces of interaction between the molecules of the pure components.
Types of Non-ideal solutions are divided into two types as explained below.
Non-ideal Solutions Showing Positive Deviations
When a component B is added to another component A, sometimes the partial pressure of a component A is found to be more than expected on the basis of Raoult’s law. A similar effect is observed for the other component B in the reversed mixing. The total vapour pressure for any solution is thus, greater than that corresponding to an ideal solution of the same composition. Such behaviour of solutions is described as a positive deviation from Roult’s law Fig.1.5
Vapour pressure of solution
Vapour pressure
fraction (X)
Fig.1.5 Vapour pressure of binary solutions Showing positive deviation
The boiling points of such solutions are relatively lower as compared to those of the pure components (because higher the vapour pressure, lower is the boiling point). For one intermediate composition, the total vapour pressure of such a solution will be the highest and the boiling point will be the lowest. This
solution acquires the property of boiling at a constant temperature and its composition remains unchanged. Liquid mixtures which distill without any change in composition are called azeotropes or azeotropic mixtures. In case of solutions showing positive deviations, we get minimum boiling (point) azetropes.
The positive deviations are exhibited by liquid pairs for which the A–B molecular interaction forces are lower than the A–A or the B–B molecular interaction forces. For example, mixtures of ethanol and cyclohexane (or acetone) show positive deviation. In pure ethanol, a very high fraction of the molecules are hydrogen bonded, as shown below:
On adding cyclohexane (or acetone), its molecules get in between the molecules of ethanol, thus breaking the hydogen bonds and reducing ethanol–ethanol attractions considerably.
Thus, for a non-ideal solution showing positive deviation,
(i) p
(ii) D Hmixing=+ve (iii) D V mixing = +ve
A few more examples of non-ideal solutions showing positive deviations are given below:
i. Acetone + Carbon disulphide
ii. Acetone + Ethyl alcohol
iii. Acetone + Benzene
iv. Methyl alcohol + Water
v. Ethyl alcohol + Water
vi. Carbon tetrachloride + Chloroform
vii. Carbon tetrachloride + Benzene
viii. Carbon tetrachloride + Toluene.
Non-ideal Solutions Showing Negative Deviations
If, for the two components A and B, the forces of interaction between the A and B molecules
are stronger than the A–A and B–B forces of interaction, the escaping tendency of A and B types of molecules from the solution becomes less than those in the pure liquids. In other words, for any composition of the solution, the partial vapour pressure of each component will be less and the total vapour pressure of the solution will also be less than that expected from Raoult’s law Fig. 1.6.
Vapour pressure of solution
Fig.1.6 Vapour pressure of binary solutions showing negative deviation
These solutions are said to show negative deviations from Raoult’s law. Such solutions have relatively higher boiling points as compared to those of the pure components (because lower the vapour pressure, higher is the boiling point). For one intermediate composition, the total vapour pressure of the solution will be the least and the boiling point will be the highest. Such a solution will also distil without any change in composition, and it provides an example of another kind of azeotrope. We call it the maximum boiling azeotrope.
For example, negative deviation from Raoult’s law is exhibited by a mixture of chloroform (CHCl3) and acetone, (CH3)2CO. When these are mixed, the hydrogen bonding takes place between the two molecular species, as shown below, due to which the escaping
tendency of either of the liquid molecules is decreased. Consequently, the boiling point of solution increases.
Hydrogen bonding between chloroform and acetone
In case of solution showing negative deviations, a slight decrease in volume and evolution of heat takes place on mixing, as excepted (i.e., D V and D H both are negative).
Thus, for a non-ideal solution showing negative deviation,
(i) pA < X
(ii) D H mixing = –ve
(iii) D V mixing = –ve
A few more examples of non-ideal solutions showing negative deviations are given below:
1. Chloroform + Benzene
2. Chloroform + Diethyl ether
3. Acetone + Aniline
4. HCl + Water
5. HNO3 + Water
6. Acetic acid + Pyridine
1.4.3 Azeotropic Mixtures
A solution which distils without a change in composition at a particular composition is called azeotropic mixture or azeotrope. Azeotrope is a binary mixture of a particular composition, which cannot be resolved by distillation.
Azeotropic mixtures are non-ideal solutions. They have same composition in liquid phase and vapour phase.
They are of two types:
(i) minimum boiling point azeotropes
(ii) maximum boiling point azeotropes
The solutions which show a large positive deviation from Raoult’s law form minimum boiling azeotropes at a specific composition. Rectified spirit obtained from fermented sugars contains about 95.6% ethanol by volume and is an example of minimum boiling azeotrope.
3. At 88 °C ,vapour pressure of benzene is 900 mm Hg, and vapour pressure of toluene is 360 mm Hg. What is the mole fraction of benzene in the mixture of benzene and toluene that will boil at 88°C?
Sol. oo
TotalAABB Ppxpx =+
Here, say liquid (A) is benzene and liquid (B) is toluene. At boiling point, vapour pressure becomes equal to atmospheric pressure, 760 mm Hg.
From this equation, XA = mole fraction of benzene = 0.74
Try yourself:
3. At 80°C, the vapour pressure of liquid A is 520 mm Hg and that of B is 1000 mm Hg. If a mixture of solution of A and B boils at 80°C and 1.0 atm, what will be the mole fraction of A in the solution?
Ans: 0.5
TEST YOURSELF
1 Which statement about the composition of vapour over an ideal solution of 1:1 molar mixture of benzene and toluene is correct?
Assume the temperature is constant (25°C)
Vapour pressure data at (25°C):
Benzene = 75 mm Hg
Toluene = 22 mm Hg
(1) The vapour will contain higher percentage of benzene.
(2) The vapour will contain higher percentage of toluene.
(3) The vapour will contain equal amounts of benzene and toluene.
(4) Not enough information is given to make a prediction.
2. The vapour pressure of two liquids P and Q are 80 torr and 60 torr, respectively. The total vapour pressure obtained by mixing 3 moles of P and 2 moles of Q would be (1) 68 torr (2) 20 torr
(3) 140 torr (4) 72 torr
3. The vapour pressure of two pure isomeric liquids X and Y are 200 torr and 100 torr, respectively, at a given temperature. Assuming that a solution of these components obeys Raoult’s law, the mole fraction of component X in vapour phase in equilibrium with the solution containing equal amounts of X and Y, at the same temperature, is
(1) 0.33 (2) 0.50 (3) 0.66 (4) 0.80
4. The vapour pressure of a pure liquid A is 60 mm, at 25 °C. It forms an ideal solution with another liquid B. The mole fraction of B is 0.6, and total pressure is 64 mm. The vapour pressure of B at 25°C, in mm, is
(1) 75 (2) 66.6
(3) 52 (4) 120
5. The system that forms maximum boiling azeotrope is
(1) Acetone + chloroform
(2) n – hexane + n –heptane
(3) Benzene + Toluene
(4) Carbon disulphide + Acetone
6. The boiling point of C6H6, CH3OH, C6H5NH2 and C6H 5NO 2 are 80°C, 65°C, 184°C and 210.9° C. Which will show highest vapour pressure at room temperature?
(1) C6H6 (2) CH3OH
(3) C6H5NH2 (4) C6H5NO2
7. The vapour pressure of water at 300 K in a closed container is 0.4 atm. If the volume of the container is doubled, its vapour pressure at 300 K will be
(1) 0.8 atm (2) 0.2 atm
(3) 0.4 atm (4) 0.6 atm
8. A solution contains hydrocarbons (A) and (B) in the ratio of 2: 3. The vapour pressures of pure hydrocarbons at 25°C are 300 and
100 mm of Hg, respectively. Mole fraction of ‘B’ in vapour phase is (1) 2/3 (2) 3/4 (3) 1/3 (4) 3/2
Answer Key
(1) 1 (2) 4 (3) 3 (4) 2
(5) 1 (6) 2 (7) 3 (8) 3
1.5 SOLUTION OF A SOLID IN A LIQUID
Consider a solution of cane sug ar or glucose or urea or a salt in water and a solution of sulphur or iodine or naphthalene dissolved in carbon disulphide. These are solutions of a solid in a liquid type. Some physical properties of such solutions are quite different from those of pure solvent.
In a pure liquid, the entire surface is occupied by the molecules of the liquid, as shown in Fig.1.7(a) . If a non-volatile solute is added to a solvent, a homogeneous solution is formed. The vapour pressure of the solution is from the solvent alone. The number of solvent molecules are relatively less at the surface in a solution, as shown in Fig.1.7(b).
The ratio of lowering of vapour pressure of solution to the vapour pressure of pure solvent is called relative lowering of vapour pressure of solution. s PP P °− ° is rela tive lowering of vapour pressure of solution (RLVP ).
1.5.1 Raoult’s Law
The law states that the relative lowering of vapour pressure of a solution is equal to the mole fraction of its solute.
According to Raoult’s law, at a given temperature, the vapour pressure of solution (P) is directly proportional to mole fraction of solvent.
Let a solution contain nA moles of solvent and n B moles of solute. Then,according to Raoult’s law,
AAAA PX,PkX α= ,
When pure solvent is taken, i.e., X A = 1, then k = PA 0 = vapour pressure of pure solvent.
On substituting the value of k in the above equation, we get,
0 AAA PX P = A A 0 A P X P = = 1 – XB
(a) Pure solvent (b) Solution of a solid in liquid solvent
Fig.1.7 Illustration of decrease in vapour pressure
Consequently, the number of solvent molecules escaping from the surface is correspondingly reduced. The vapour pressure is also reduced. The decrease in vapour pressure is directly proportional to quantity of solute.
The difference between the vapour pressure of the pure solvent (P 0) and vapour pressure of solution (Ps) is called lowering of vapour pressure (DP) of solvent in a solution.
Subtracting the above two quantities from (1), we get A B 0 A P 11X P −=− AB XX1∴+= 0 AA B 0 A PP X P = = mole fraction of solute.
0 AAB 0 AB A PPn nn P = +
In dilute solutions, n A >>> nB; nnnABA∴+≈
0 AAB 0 A A PPn n P = or o As B o A A pp n n p =
0 AABA 0 BA A PPwM MW P =×
(Here, pA = ps = vapour pressure of solution)
It is called simplified Raoult’s law, where WB, WA are the weights of solute and solvent, M B, M A are the molecular weights of solute and solvent respectively.
If a solution obeys Raoult’s law for all concentrations, its vapour pressure would vary linearly from zero to the vapour pressure of pure solvent. A plot of vapour pressure is linear with mole fraction of the solvent, as shown in Fig.1.8. Vapour pressure of pure solvent
Vapour pressure
0 1 Mole fraction of solvent
Fig.1.8 Vapour pressure as a function of mole fraction
RLVP is generally used to calculate molecular weight of non-volatile solute.
Limitations of Raoult’s law: It is applicable to 1. dilute solutions only;
2. if the solute is non-volatile and is in molecular state
3. solutions containing solut es, which undergo neither dissociation nor association
1.6 COLLIGATIVE PROPERTIES
The p hysical and chemical properties of aqueous solutions containing solutes in general depends on the nature and also on the structure of solutes. But there are some properties which
depend on number of solute particles (ions or molecules) but not on their nature. Such properties are called colligative properties. Properties of solutions which depend on the number of particles of solute, irrespective of their nature, are called colligative properties. There are four colligative properties. They are:
1. relative lowering of vapour pressure of solution;
2. elevation of boiling point of solution;
3. depression of freezing point of solution;
4. osmotic pressure of solution;
Colligative properties are generally used to determine molecular weights of non-volatile solutes.
1.6.1 Relative Lowering of Vapour Pressure
Vapour pressure of a solvent in solution is less than that of the pure solvent. The lowering of vapour pressure depends only on the concentration of solute particles and is independent of their size or molecular weight. If po is vapour pressure of pure solvent, X1 and X2 are mole fractions of solvent and solute in a solution, respectively, the vapour pressure of solution ‘p’ is gi ven by Raoult’s law as: ps = po X1 .............................................(1)
The lowering of the vapour pressure of solvent ( D p) is given as: D p = p° – ps = p° – p°X1 = p°(1–X1) ...(2)
But the sum of mole fractions of both solvent and solute in a solution is unity. Hence, D p = p° X2...........................................(3)
(4) solute solutesolvent n nn = +
For dilute solution, nsolute < < nsolvent.
Therefore, ssolute solvent ppn pn °− = °
solutesolvent
solutesolvent WM MW =× ... (5)
T hus, R LVP is directly proportional to number of particles; so, it is a colligative property.
If solute undergoes association or ionisation, the above equation (5) cannot be applied. The lowering of vapour pressure is almost twice for a solution containing 58.5 g (one mole) of sodium chloride in one litre aqueous solution than for a solution containing 342 g (one mole) of sucrose in one litre water. This is because the number of solute particles in 342 g of sucrose is No, (molecules) but the number in 58.5 g of sodium chloride is 2N o , (ions) as sodium chloride ionises almost completely in aqueous solutions (Here, N o = Avogadro’s number).
4. The vapour pressure of an aqueous solution of sucrose at 373 K is found to be 750 mm Hg. What is the molarity of the solution?
Sol. Raoult’s law,
This can be simplified as ssolute ssolvent
Try yourself:
4. Find the mass of non-volatile solute (molecular mass = 40) that should be dissolved in 57 g of octane (Molecular mass = 114) to decrease its vapour pressure to 80%
Ans: 4
1.6.2 Elevation of Boiling Point
The boiling point of a liquid is the temperature at which the vapour pressure of the liquid becomes equal to the atmospheric pressure. The boiling point of a solution containing non–volatile solute is always greater than that of pure solvent. This increase in boiling point is called the elevation in boiling point. The elevation in boiling point of solution is due to lowering of vapour pressure of solution. Hence, the solution has to be heated more to make the vapour pressure equal to the atmospheric pressure.
Alternatively, the elevation in boiling point may be explained on the basis of plots of vapour pressure versus temperature, as shown in Fig.1.9.
Boiling point of solvent
Atmospheric pressure SolventSolution E Vapour pressure
Boiling point of solution
Temperature (K)
Fig. 1.9 Elevation of boiling point
As at any temperature, vapour pressure of the solution is less than that of the solvent, the curve for the solution lies below that of the solvent, as shown by the curve CD. The temperatures at which the vapour pressure of the solvent and the solution become equal
to the atmospheric pressure are T b 0 and T b , respectively. Obviously, T b > T b 0 . The difference, called the elevation in boiling point, is given by = Tb – Tb 0 .
It is evident that greater the lowering in vapour pressure (Dp), higher is the elevation in boiling point (DTb), i.e., according to Raoult’s law, the Δp is directly proportional to the mole fraction of the solute in the solution.
Hence, D Tb ∝ X2 (or) D Tb = kX2 where k is a constant of proportionality.
But 2 2 12 n X nn = + if the solution is dilute 2 2 1 n X n = 22 21 11 1 b nn
If the mass of solvent, w 1 = 1 kg, then evidently, 2 1 n m w = , molality of the solution.
Also, for a given solvent, its molecular mass
M 1 is constant so that kM 1 = K b , an other constant. Hence, the above result reduces to Tbb = Km ∆
As molality is the number of moles of the solute dissolved per 1000 g of the solvent, if w2 grams of the solute of molecular mass M 2 are dissolved in w1 grams of the solvent,
2 21 w 1000 m Mw =×
Hence, the above formula becomes 2 bb 21 w 1000 TK Mw ∆=××
w2 = weight of solute
M2 = molecular weight of solute
w1 = weight of solvent in grams
o 2 bbbb 21 w 1000 TTTK Mw ∆=−=××
This formula is often used for the calculation of molecular masses of non–ionic solutes (i.e., non–electrolytes) where K b is called the boiling point elevation constant or ebullioscopic constant, and ‘m’ is the molality of the solution. If m = 1, then D Tb = Kb.
Molal elevation constant may be defined as the elevation in boiling point when the molality of the solution is unity, (i.e, 1 mol of the solute is dissolved in 1 kg (1000 g) of the solvent). The units of kb are degree/molality or K/m or 0C/m or K kg mol –1. Calculation of molal elevation constant from enthalpy of vapourisation is as follows. 2
where L v = enthalpy vapourisation per gram of the solvent
D H vap = enthalpy vapourisation per mole of the solvent
M1 = molecular mass of the solvent
R = 8.314 K –1 mol –1, if L v or D H vap are in joule.
R = 2 cal deg–1mol–1, if L v or D H vap are in calorie.
The value of K b depends only upon the nature of solvent.
5. An aqueous solution containing one gram of urea (molecular mass = 60) boils at 100.25 °C. Calculate the boiling point of the aqueous solution containing 3 g of glucose (molecular mass = 180) in the same volume of the solution.
Sol. bb TKm∆= In urea solution: b T0.25C∆=° () solute solute solventinKg W 1g m MW60W == ××
Therefore, b 1 0.25K 60W =× × . . . . . (1)
In glucose solution: bb 3 TK 180W ∆=× × . . .(2) Fr om the above equation D T b of glucose solution = 0.25°C.
Therefore, boiling point of the given aqueous solution of glucose = 100°C + 0.25°C = 100.25°C.
Try yourself:
5. When 0.4 g of a solute is dissolved in 40 g of diethyl ether (k b = 2.16 K kg mol–1), its boiling point is increased by 0.17 K. Then find the molar mass of the solute.
Ans: 127
1.6.3
Depression in Freezing Point
Freezing point of a substance is the temperature at which the solid and the liquid forms of the liquid are in equilibrium with each other, i.e., the solid and liquid forms of a substance have the same vapour pressure, e.g., ice and water at 0°C have the same vapour pressure.
The freezing point of a solution containing non- volatile solute is always less than that of pure solvent. This decrease is called the depression in freezing point. The depression in freezing point of solution is due to lowering of vapour pressure of solution.
The depression in freezing point may be explained on the basis of plots of vapour pressure versus temperature, as shown in Fig.1.10.
On cooling, the vapour pressure of the liquid solvent decreases along the curve AB. At the point B, the solid starts appearing and the vapour pressure decreases steeply along the path BC (because solids have lower vapour pressure). At B, the liquid and the solid solvent are in equilibrium and have the same vapour pressure. Thus, B is corresponding to the freezing point Tf 0 of the pure solvent. As the
vapour pressure of solution is less than that of the solvent, the curve for the solution lies below that of the solvent. On cooling, it will follow the path DE. At E, the solid appears. Hence, E is corresponding the freezing point Tf of the solution. Obviously, Tf is less than Tf0. The difference is called the depression in freezing point, D Tf
It is given by DTf = Tf ° –Tf = Kf m, where Kf is a constant, known as freezing point depression constant or cryoscopic constant of the solvent, and ‘m’ is the molality of the solution, i.e., the number of moles of the solute dissolved in 1000 grams (1 kg) of the solvent.
If m = 1, DTf = Kf. Hence, molal depression constant may be defined as the depression in freezing point when the molality of the solution is unity, i.e, one mole of the solute is dissolved in 1000 g (1 kg) of the solvent.
The units of K f are degrees/molality, i.e, K/m or K kg mol–1; the value Kf depends only upon the nature of solvent.
2 ff 21 w 1000 TK Mw ∆=×× where w2 = weight of the solute w1 = weight of the solvent M2 = molecular weight of solute
Fig. 1.10 Depression in Freezing Point
Calculation of molal depression constant from enthalpy of fusion is as follows:
1.6.4 Osmotic Pressure
where D Hf = enthalpy of fusion per mole
T0 = freezing point of the liquid (pure solvent)
Lf = latent heat of fusion per gram of the solvent
M1 = molecular mass of the solvent
R = 8.314JK–1mol–1 if Lf or are in joule = 2 cal deg–1mol–1 if Lf or are in calorie
6. When 36 g of a solute is dissolved in 1.2 kg of water, the solution freezes at –0.93°C. What is the molecular mass of the solute?(K f of water = 1.86 K kg mol–1)
Sol. Freezing point of water = 0°C
So, the depression in freezing point of the solution = D Tf = 0 – (–0.93) = 0.93°C.
D Tf = kf m 36
0.931.86 M1.2 =× × 1.8636 M 60
0.931.2 × == ×
Hence, molecular mass of the solute = 60 amu
Try yourself:
6. The temperature of a city was –93°C. A car was used wherein radiator was filled with 5 L of water. What weight of ethyleneglycol (molecular mass = 62) were added to the water of the radiator in order the use the car for travelling (kf of water = 1.86 kg molal–1)
Ans: 15.5 kg
When a dilute solution of a solute is separated from its solvent by a semipermeable membrane, the solvent flows into the solution through the membrane slowly. Such a phenomenon also occurs when two solutions of the same solute with different concentrations are separated by a semipermeable membrane.
The membrane which allows only the molecules of solvent to pass through it, but not the solute molecules, is called a semipermeable membrane.
Parchment paper, cellophane paper, pig’s bladder, some animal membranes, and inorganic precipitate membranes (copper ferrocyanide) are commonly used as semipermeable membrane, in osmosis.
The process of solvent flowing into the solution when the solution and solvent are separated by a membrane is called osmosis. In other words, osmosis is inflow of solvent from dilute solution to concentrated solution.
The hydrostatic pressure developed on the diluted aqueous solution at equilibrium state due to inflow of water when the solution is separated from the water by a semipermeable membrane is also called osmotic pressure.
Osmosis is also called endosmosis because of the inflow of solvent molecules. The osmosis process can be illustrated, as shown in Fig.1.11
Solution h π=hρg
Solvent
Fig. 1.11 Phenomenon of osmosis
Semipermeable membrane
A semipermeable membrane (parchment paper) is tied to a thistle funnel.
Thistle funnel
The funnel is filled with a dilute solution of a solute. This is placed in a beaker containing water. The liquid level in the stem of the funnel gradually increases for some time due to entry of water present in the beaker into the funnel by osmosis. After some time the level of solution in the funnel remains constant as it reaches an equilibrium state. As the water level increases in the stem, the pressure exerted by the column of water (hydrostatic pressure) also increases. This acts on the solution in the direction opposite to that of the direction of the flow of water (osmotic pressure) from the beaker into the funnel through the membrane.
The flow of water into the funnel is due to the osmotic pressure of the solution. The pressure that just stops the flow of solvent is called osmotic pressure of the solution.
The osmotic pressure depends on the concentration of the solution. Osmotic pressure is illustrated in Fig.1.12
Osmotic pressure method is used for deter-mining molecular masses of proteins, polymers, and other macromolecules.
Solutions having same osmotic pressure at a given temperature are called isotonic solutions. Osmosis does not occurs when such solutions are separated. Osmotic pressure associated with the fluid inside the blood cell is equivalent to that of 0.9% (w/v) sodium chloride solution, called normal saline solution.
If we place cells in a solution containing more than 0.9% ( w / v ) sodium chloride, water will flow out of the cells and cells would shrink. Such a solution is called hypertonic. Hypertonic solutions have higher concentration and possess greater osmotic pressure. If the salt concentration is less than 0.9% (w/v), the solution is called hypotonic. In this case, water will flow into the cells and they would swell. Hypotonic solutions have less concentration and possess lesser osmotic pressure compared to concentrated solutions. Considering 1M glucose solution and 2M glucose solution, 1M glucose solution is hypotonic with respect to 2M glucose solution and 2M glucose solution is hypertonic with respect to 1M glucose solution.
7. A 6% aqueous glucose (molecular mass = 180) solution and 2% aqueous solution of an unknown non-electrolytic and non-volatile compounds are isotonic. Find the molecular mass of the unknown compound.
Fig.1.12 Illustration of osmotic pressure
Molar mass of the solute can be calculated from the experimental determination of osmotic pressure ( π ). RT ., π==CRTw MV where M is molecular weight of solute and w is weight of solute .
V = volume of solution in litres.
Sol. Since, the given solution are isotonic, their osmotic pressures must be the same, π1 = π2. The solutes are non-electrolytes and non-volatile; hence, their particles concentrations must be the same.
Therefore, C1 = C2
62 180M = M = 60
The molecular mass of the unknown solute = 60 amu.
Try yourself:
7. A solution of urea (molecular mass = 60) of strength, 8.6 g L –1 is isotonic with 5% (W/V %) solution of a non-volatile organic compound. Find the molecular mass of the organic compound.
Ans: 348.8
1.6.5 Reverse Osmosis
If pressure higher than the osmotic pressure is applied to the solution side, solvent will flow out of the solution. This process is called reverse osmosis. Reverse osmosis is used in desalination of sea water. A schematic set up is shown in Fig.1.13. for desalination.
salted meat or candied fruit loses water, shrivels, and dies.
vii) If the osmotic pressure of the contents of the living cell is not equal to that of the contents surrounding it outside, two phenomena take place. They are haemolysis and plasmolysis.
Haemolysis is a process of entering of contents into the cell. The cell bulges and, finally, bursts.
Plasmolysis is a process of losing contents from the cell. The cell collapses.
TEST YOURSELF
1. Which of the following is colligative property?
(1) vapour pressure
(2) boiling point
(3) freezing point
(4) osmotic pressure
Fig.1.13 Desalination Using Reverse Osmosis
The following phenomena take place due to osmosis:
i) A raw mango placed in concentrated salt solution shrivels.
ii) Wilted flowers revive when placed in fresh water.
iii) A carrot becomes limp when placed in salt water.
iv) People taking a lot of salt or salty food experience water retention in tissue cells and intercellular spaces,result in puffiness (or) swelling,called edema.
v) Water movement from soil into plant roots and subsequently into upper portion of the plant
vi) The preservation of meat by salting and fruits by adding sugar protects against bacterial action because bacterium on
2. Relative lowering in vapour pressure of a solution containing 1 mole urea in 54 g H2O is
(1) 1/55
(2) 3/55
(3) 3/4
(4) 1/4
3. Relative lowering of vapourpressure is maximum for
(1) 0.1 m urea
(2) 0.1 m NaCl
(3) 0.1 m Al2(SO4)3
(4) 0.1 m MgCl2
4. The correct order of freezing point for 10 g aqueous solution each of urea, glucose and sucrose is
(1) ()()() fglucosefureafsucrose TTT >>
(2) ()()() fsucrosefglucosefurea TTT >>
(3) ()()() fureafglucosefsucrose TTT >>
(4) ()()() fsucrosefureafglucose TTT>>>
5. How many grams of methyl alcohol should be added to 10 L tank of water to prevent its freezing at 268 K?
(Kf of H2O = 1.86 K–kg/mol)
(1) 880.07 g
(2) 899.04 g
(3) 860.21 g
(4) 878.06 g
6. Two moles of glycol is dissolved in 891.24 g of water. The solution is cooled to –4.8°C. How many gram of ice is formed in the cooling process?
[Kf of water = 1.86 K kg mol –1]
(1) 106.24
(2) 116.24
(3) 126.24
(4) 136.24
7. At room temperature, a dilute solution of urea is prepared by dissolving 0.60 g of urea in 360 g of water. If the vapour pressure of pure water at this temperature is 35 mmHg, lowering of vapour pressure will be: ( molar mass of urea =60 g mol–1)
(1) 0.027 mm Hg
(2) 0.031 mm Hg
(3) 0.017 mm Hg
(4) 0.028 mm Hg
8. The molal depression constant depends upon (1) nature of the solute
(2) nature of the solvent
(3) heat of solution of the solute in the solvent
(4) vapour pressure of the solution
9. An aqueous solution containing nonvolatile, non -electrolyte freezes at –0.186°C
(Kf = 1.86 K kg mol–1, kb = 0.512 K mol–1). The elevation of boiling point of the same solution is
(1) 0.186
(2) 0.152
(3) 0.512 1.86
(4) 0.0512
10. Equimolal dilute solutions containing different non-volatile, non-electrolyte solutes in the same solvent have
(1) same boiling point but different freezing point
(2) same freezing point but different boiling point
(3) same boiling and same freezing points
(4) different boiling and different freezing points
11. Yg of non – volatile organic substance of molecular mass M is dissolved in 250 g benzene. Molal elevation constant of benzene is Kb. Elevation in its boiling point is given by (1) b M KY (2) 4 b KY M
(3) 4 b KY M
(4) b KY M
12. The relationship between osmotic pressure(P1) at 273 K when 10 g glucose, 10 g urea (P2) and 10 g sucrose (P3) are dissolved in 250 ml of water is
(1) P1 > P2 > P 3
(2) P 3 > P1 > P2
(3) P2 > P1 > P 3
(4) P2 > P 3 > P1
Answer Key
(1) 4 (2) 4 (3) 3 (4) 2
(5) 3 (6) 2 (7) 3 (8) 2
(9) 4 (10) 3 (11) 2 (12) 3
1.7 ABNORMAL MOLAR MASSES
Ionic compounds when dissolved in water, d issociate into cations and anions. If we dissolve one formula weight of sodium chloride, (58.5 g) in water, we expect one mole each of Na+ and Cl– ions to be released in the solution. If this happens, there would be two moles of solute particles in the solution.
If interionic attractions are ignored, one mole of sodium chloride in one kg of water would be expected to increase the boiling point by 1.04 K instead of 0.52 K. If we assume that NaCl is completely dissociated in water and increase in boiling point of water is 1.04 K, the molar mass of NaCl would be 29.25 g mol –1 .
This conclusion brings to light that experimentally determined molar mass is always lower than the true value when there is dissociation of solute into ions.
Similarly, some solutes, like acetic acid, when dissolved in solvents, may associate to form dimers due to hydrogen bonding. This happens in solvents with low dielectric constant. Therefore, the number of solute particles decreases. Molar masses of solute determined from such solutions will be higher than that of true mass.
Molar mass that is determined either lower or higher than the expected value is called abnormal molar mass.
A factor ‘i’, known as the Van’t Hoff factor was introduced to account for the extent of dissociation or association. This (i) is defined as follows:
van’t Hoff’s factor, Normalmolarmass i Abnormalmolarmass =
Calculated molar mass of solute
Observed molar mass of solute i = (or)
Totalnumberofmolesofparticles
afterdissociationorassociation
Numberofmolesofparticles
beforedissociationorassociation i =
1.7.1 Solute Dissociation or Ionisation
If a solute is dissociated or ionised in solutions to give ‘n’ ions and ‘ a ’ is the degree of ionisation, then:
n nAA
Initial moles 1 0
Number of moles after dissociation 1 a n a
Number of particles before ionisation = 1
Number of particles after ionisation = 1n1(n1) −α+α=+−α
van’t Hoff factor, 1(n1) i 1 +−α =
Degree of ionisation, i1 n1 α=
Range of ‘i’ for dissociation: 1 < i < n
1.7.2 Solute Association
If a solute is associated in solutions, n molecules associate and is the degree of association,
n nAA
Initial moles 1 0
Number of moles after dissociation 1– aa/ n
Number of particles before association = 1
Number of particles after association = 1(/n) −α+α
van’t Hoff factor, 1(/n) i 1 −α+α =
α=α=
Degree of association, () 1ii1 or 1 1(1/n) 1 n
Range of ‘i’ for association: 1 n < i < 1
1.7.3 Colligative Properties with Van’t Hoff Factor
When the van’t Hoff factor is included in the mathematical equations of colligative properties and used for molecular mass calculations, correct molecular masses of solutes are obtained. The modified equations are as follows.
Relative lowering of vapour pressure of solvent, o s solute o pp iX p =
Depression of freezing point, ff TiKm∆=
Elevation of boiling point, bb TiKm∆=
Osmotic pressure of solution, iCST π=
In case of dissociation of solute, i > 1
In case of association of solute, i < 1
W hen there is no dissociation or association, i = 1
The van’t Hoff factor increases upon dilution and approaches 2 for electrolytes like NaCl and MgSO 4. It gets close to 3 for K2SO4 and 5 for K4[Fe(CN)6]. The van’t Hoff factor for some electrolytes are listed in Table 1.3.
Table 1.3 The van’t Hoff factor, (i) for some electrolytes
of i van’t Hoff factor, i for
Salt
8. 75.2 g of phenol is dissolved in 1.0 kg of a solvent (kf = 14 K molal–1). If depression in freezing point of the solution is 7 K, what percentage of phenol that dimerises in the solvent > (Molecular mass of phenol = 94 )?
Sol. D Tf = i kf m f f T 794 i 00514 km1475.2 ∆× === ××
At equilibrium, 1 – aa /2 i1 2 α =−α+ 0.6251 2 α =− a = 0.75
Therefore, phenol undergoes 75% dimersation.
Try yourself:
8. In 0.01 M aqueous solution, if BaCl 2 undergoes 49% dissociation, find its van’t Hoff factor.
Ans: 1.98
TEST YOURSELF
1. What is the van’t Hoff factor of ferric sulphate (Assume 100% ionisation)? (1) 3 (2) 4 (3) 5 (4) 2
2. van’t Hoff factor is highest for _____ molal K2SO4
(1) 1
(2) 0.001
(3) 0.1
(4) 0.01
3. The van’t Hoff factor for 0.1 m KCl aqueous solution is 1.85. The degree of ionisation of KCl solution is
(1) 0.15
(2) 0.17
(3) 0.85
(4) 0.3
4. The van’t Hoff factor of aq. K2SO4 at infinite dilution has value equal to (1) 1
(2) 2
(3) 3
(4) Between 2 and 3
5. The degree of dissociation (α) of a weak electrolyte, Ax B y is related to van’t Hoff factor (i) by the expression
(1) i1 xy1 α= ++
(2) xy1 i1 +− α=
(3) xy1 i1 ++ α=
(4) () i1 xy1 α= +−
6. If BaCl 2 ionises to an extent of 80% in aqueous solution, the value of van’t Hoff factor is
(1) 0.4
(2) 2.6
(3) 0.8
(4) 2.4
7. 54% of a solute is dimerized in a solution. van’t Hoff factor of the solute in the solution is
(1) 0.73
(2) 0.63
(3) 0.42
(4) 0.84
8. The freezing point of benzene decreases by 0.45°C,when 0.2 g of acetic acid is added to 20 g of benzene. If acetic acid associates to form a dimer in benzene, the percentage association of acetic acid in benzene will be:
(Kf for benzene = 5.12 K kg mol −1)
(1) 74.6%
(2) 94.6%
(3) 64.6%
(4) 80.4%
9. 0.004 M Na 2 SO 4 is isotonic with 0.01 M glucose. The degree of dissociation of Na2SO4 is
(1) 75%
(2) 50%
(3) 25%
(4) 85%
10. Which of the following aqueous solutions has the highest boiling point?
(1) 0.1 M KNO3
(2) 0.1 M Na3PO4
(3) 0.1 M BaCl2
(4) 0.1 M K2SO4
Answer Key
(1) 3 (2) 2 (3) 3 (4) 3
(5) 4 (6) 2 (7) 1 (8) 2
(9) 1 (10) 2
CHAPTER REVIEW
Types of Solutions
■ A solution is a homogeneous mixture of two or more non–reacting components. Formation of solution is a physical process.
■ A solution of a solid in another solid is known as a solid solution. Many alloys are solid solutions.
■ An alloy of a metal with mercury is called an amalgam.
■ Aqueous solutions are those prepared using water. Non–aqueous solutions have other solvents.
■ The components do not lose their identity during the formation of a solution.
■ Dynamic equilibrium is established between dissolved solute and undissolved solute in a solution.
■ During formation of solution entropy of system increases due to greater disorder.
Methods of Concentration
■ A solution whose molar concentration is definitely known is a standard solution.
■ The most ideal method of expressing concen-tration is the molality ( m).
■ The commonly used method of expressing concentration is molarity ( M).
■ Mass by volume percentage (w/v) indicates the mass of a solute in 100 mL of solution.
■ Mass by weight (w/w) indicates the mass of a solute in 100 g of solution.
■ Molarity indicates the number of moles of the solute dissolved in one litre of the solution or the number of millimoles of the solute dissolved in one millilitre of the solution.
■ Number of millimoles of the solute present in V mL of the solution is given as product
of volume and molarity. It is given as V×M
■ Number of moles of solute present in V litres of solution is given as MV.
■ When a solution is diluted, its molarity decreases.
■ V1M1 = V2M2, where V1 = Volume of the solution before dilution, M 1 = Molarity of the solution before dilution, V 2 = Volume of the solution after dilution and M2 = Molarity of the solution after dilution
■ Molarity, w1000 GMWV ×
■ w = weight of the solute in grams and
■ V = volume of the solution in millilitres
■ M = 10(%w/v) GMW (or) M = () () ()10d%w/w GMW , where d is density in grams per mL.
■ Equivalent weight of a substance expressed in grams is known as gram-equivalent weight or gram-equivalent or equivalent.
■ Number of gram-equivalents = Weight GEW
■ Equivalent weight = Molecular weight n , where n is acidity of a base or basicity of an acid or valency or charge of ion or number of electrons transfered or number of faradays.
■ Normality indicates the number of gram–equivalents of solute present in one litre of the solution.
■ Number of gram equivalents of solute in V litres of solution is given as, NV
■ A normal solution means 1N solution.A decinormal solution is N/10 or 0.1 N solution. A centinormal solution is N/100 or 0.01 N solution.
■ Normality, N = w1000 GEWV ×
■ = N10(%w/v) GEW (or) N = () () ()10d%w/w GEW
■ Normality is given as, N = Molarity × n.
■ When a solution is diluted, its normality decreases.
■ V 1 N 1 (before dilution) = V 2 N 2 (after dilution)
■ Molarity and normality decrease with increase in temperature, except from 0 °C to 4°C.
■ Normality of the mixture when two solutions of same solute are mixed.
N = 1122 12 NVNV VV + +
■ Molality ( m ) indicates the number of moles of a solute dissolved in 1000 g or one kilogram of the solvent.
■ Molality (m) = w1000 GMWW × × where w is weight of the solvent in grams.
■ Mole fraction of the solute = 1 solute
12 X = + n nn Mole fraction of the solvent = Xsolvent
2 12 = + n nn , where n1 and n2 are number of moles of solute and solvent.
■ For a binary solution, Xsolute + Xsolvent = 1.
■ Weight percentage, molality, and mole fraction are independent of temperature.
Solubility
■ Solubility represents the number of moles of solute present in a litre of saturated solution, at a given temperature.
■ In a solution of gas in liquid, the solubility of gas increases with decrease of temperature.
■ In the solution of gas in liquid, the solubility of gas increases with increase of its partial pressure.
■ Henry’s law states that the mass of a gas dissolves in a solvent is directly proportional to its pressure.
pgas = KH m or pgas = KH X gas m = masses of gas, X = mole fraction of gas in solution.
Vapour Pressure
■ The process of escape of liquid molecules into space is called evaporation or vapourisation.
■ Rate of evaporation depends on nature of liquid, surface area of liquid, temperature, and flow of air current over the surface.
■ Rapid evaporation results in decrease in temperature, leading to intense cooling.
■ When a liquid and its vapour are in equilibrium with each other, the pressure exerted by the vapour over the liquid surface is known as vapour pressure of the liquid.
■ Volatile liquids with high vapour pressure at a given temperature have low boiling points. Less volatile liquids have high boiling points.
■ The vapour molecules have a higher potential energy than the liquid molecules at the same temperature.
■ The vapour pressure of a liquid depends upon the nature of the liquid and temperature.
■ With an increase in temperature, the vapour pressure of a liquid increases exponentially.
■ A liquid boils at that temperature when its vapour pressure becomes equal to the atmos-pheric pressure.
■ When the external pressure is decreased, the boiling point of liquid decreases.
■ The vapour pressure of a solution of a non-volatile solute is less than the vapour pressure of the pure solvent at the same temperature.
■ If P ° is the vapour pressure of the pure solvent and P S is the vapour pressure of the solution at the same temperature, then, lowering of vapour pressure is Po – P s
■ The vapour pressure of a solution of a nonvolatile solute,is directly proportional to the mole fraction of the solvent in the solution.
PS = Po Xsolvent
Solution of a Solid in Liquid
■ Lowering of vapour pressure is directly proportional to mole fraction of solute
■ Raoul’t law states that the relative lowering of vapour pressure of a solution is equal to the mole fraction of the solute in the solution.
■ Raoult’s law equation is given as, °
° = + PPn s PNn
■ Relative lowering of vapour pressure is independent of temperature.
■ An ideal solution is one which obey Raoult’s law at all concentrations and temperatures.
■ In an ideal solution the volume changes are additive.
■ In the formation of an ideal solution no heat is evolved or absorbed.
■ Chemically similar liquids form ideal solutions.
■ An ideal solution can be separated into two pure components by fractional distillation.
■ Raoults law is applicable to dilute solutions only.
■ When the solute is non-volatile, if the solute does not undergo either ionisation (or) association, Raoult’s law is good to apply.
■ Relative lowering of vapour pressure is determined by Ostwald and Walkers method.
■ In a dilute solution n is very small compared to N, 0S 0 PP w/mw.M
PW/MW.m ==
w = weight of the solute,
m = molecular weight of the solute,
W = weight of the solvent and
M = molecular weight of the solvent
Colligative Properties
■ The properties which depend on the number of particles of the solute but not on the nature of solute are called colligative properties.
■ Lowering of vapour pressure, elevation of boiling point, depression of freezing point and osmotic pressure are the four colligative properties.
■ All the colligative properties can be used to determine the molecular weight of non–volatile solute. The best method is using Osmotic pressure.
■ When a non-volatile solute is dissolved in the pure solvent its boiling point increases. e.g. sea water boils at greater than 100°C.
■ The difference between boiling points of solution containing a non-volatile solute ( T b ) and the pure solvent T o is called elevation of boiling point D Tb =Tb –T o
■ Liquids with high boiling point have low vapour pressures and are less volatile.
■ Elevation of boiling point is directly proportional to molality of the solution.
■ b bb w1000 TK.mK
where K b is molal elevation constant of the solvent or ebullioscopic constant
■ K b value changes from one solvent to another solvent. Kb = 0.52 degree kg/mol for water
■ 2 b b v RT K 1000 . L = where, L v is Latent heat of vapourisation.
■ The temperature at which the pressure of a liquid is equal to that of the solid is called freezing point.
■ When a non - volatile solute is dissolved in a solvent, the freezing point of the solvent decreases. eg., Sea water freezes below 0°C.
■ Water taken in automobile radiators is mixed with glycerol (or) glycol to decrease its Freezing point (F.P.) to prevent the formation of ice when surroungding temperature falls.
■ The depression in Freezing point is given by Tf∆ is T o –T s, where T o is F.P. of pure solvent and T s is F.P. of solution.
■ Depression in freezing point is directly propor-tional to molality of the solution.
■ Cellophane paper, cell walls, pig’s gall blader, Copper ferrocyanide Cu2 [Fe(CN)6], etc., are examples of membranes.
■ Egg free from outer shell placed in distilled water enlarges due to endosmosis and when placed in NaCl solution shrinks (plasmolysis) due to exosmosis respectively.
■ Pressure developed due to osmosis is called osmotic pressure.
■ Osmotic pressure is the excess pressure which must be applied to a solution to prevent the flow of the solvent into the solution.
■ Osmotic pressure is directly related to molar concentration and temperature.
■ Osmotic pressure ( π ) in terms of concentration (C) is, π = CRT
■ Molecular weight of solute in terms of osmatic pressure ( π ) is given as, wRT Vπ
■ Solutions having same osmotic pressure are known isotonic solutions.
■ Isotonic solutions generally have the same molar concentrations at a given temperature.
■ The solution having lower osmotic pressure is known as hypotonic solution and that having higher osmotic pressure is known as hypertonic solution.
where Kf is molal depression constant of the solvent or cryoscopic constant
■ 2 f f f RT K 1000L = where, Lf is Latent heat of fusion.
■ The passage of solvent molecules from a solution of low concentration into a solution of higher concentration through a semi-permeable membrane is known as osmosis.
■ Semipermeable membrane is one which allows the solvent molecules to pass through but not solute particles.
■ Osmotic pressure method is widely used to determine molecular masses of proteins, polymers and other macro molecules.
■ The colligative properties of solutions depend on the total number of solute particles present in solution.
■ For different molar concentrations of the same solute, the colligative property has greater value for the more concentrated solution.
Abnormal Molecular Masses
■ For different solutes of same molar concentration, the colligative properties
have the greater value for the solution which gives more number of particles on ionisation.
■ Certain solutes in solution are found to associate, leads to a decrease in the number of particles in the solutions. Thus, it results in a decrease in the values of colligative properties.
■ The colligative properties are inversely related, in general to the molecular mass of solute.
■ Electrolytes dissociate in solution to give two or more ions. Such solutions exhibit higher values of colligative properties. The molecular masses of such substances as calculated from colligative properties will be less than their normal values.
■ Certain solutes that undergo dissociation (or) association in solution are found to show abnormal molecular mass. The extent of dissociation (or) association of solutes in solution is determined by Van’t Hoff factor.
■ Van’t Hoff factor ( i) is given as, i = Normal molar mass
Observed molar mass
Theoritically,
i Observed colligative property
Normal colligative property =
■ For ideal solutions with no association or dissociation of solute the Van’t Hoff factor i = 1. For solutes showing association, i < 1 and showing dissociation, i > 1.
■ If a molecule of solute on dissociation gives ‘ n’ ions and ais the degree of dissociation,
i = 1 + a (n – 1)
Degree of dissociation ( a) = 1 1 i n
■ If a solute froms associated molecules (A) n and is the degree of association, i = 1 11 −α−
n
Degree of association
■ In terms of Van’t Hoff factor ( i)
Elevation of boiling point. D Tb = iKbm
Depression of freezing point. D Tf = iKfm
Osmotic pressure freezing point a = iCRT
NEET DRILL
Level - I
Types of Solutions
1. Occlusion of Hydrogen on Palladium is an example for ..... type solution
(1) gas in solid (2) solid in gas (3) gas in liquid (4) liquid in gas
2. Math the List-I with List-II.
List-I List-II
(A) Gaseous solutions (i) German silver
(B) Liquid solution (ii) Milk
(C) Solid solution (iii) Sand in water
(D) Colloidal solution (iv) Aqueous alcoholic solution (v) Air
(A) (B) (C) (D)
(1) v iv i ii
(2) i iii ii iv
(3) iv ii v i
(4) ii iii i iv
3. The characteristic property of solution is (1) formation of solution is physical change
(2) solute and solvent in the solution can be separated by filtration
(3) solute and solvent in the solution can be separated by decantation (4) solution can be represented with a chemical formula
4. Which among the following is a physical change?
(1) Burning of coal
(2) Burning of sulphur
(3) Dissolution of glucose in water
(4) Burning of white phosphorus
5. If air is taken as a binary solution, the solvent is
(1) N2 (2) O2 (4) CO2 (4) H2
Methods of Concentration
6. To halve the molarity of a solution the following should be adopted (1) weight of the solute to be doubled (2) weight of the solvent to be doubled (3) volume of the solvent to be doubled (4) volume of the solution to be doubled
7. Molarity of the liquid HCl if density of the solution is 1.17 g/cc is
(1) 36.5 (2) 18.25
(3) 32.5 (4) 42.10
8. Density of a 2.05 M solution of acetic acid in water is 1.02 g/mL .The molality of the solution is (1) 2.28 mol kg–1 (2) 0.44 mol kg–1 (3) 0.14 mol kg–1 (4) 3.28 mol kg–1
9. The units of Normality are (1) mole, lit –1 (2) mole, kg –1 (3) equivalent lit –1 (4) equivalent kg –1
10. 5 mL of 1 N HCl, 20 mL of N/2 H2SO4 and 30 mL of N/3 HNO3 are mixed together and the volume made to one litre. The normality of the resulting solution is (1) N/5 (2) N/10 (3) N/20 (4) N/40
11. ‘M’ is the molecular weight of KMnO4. The equivalent weight of KMnO4 when it reacts according to the equation 2KMnO4 + 3H2SO4 + 5H 2C 2O 4 → K 2SO 4 + 2MnSO 4 + 8H 2O + 10CO2
(1) M/2 (2) M/3
(3) M (4) M/5
12. The number of millimoles of H2SO4 present in 5 litres of 0.2N H2SO4 solution is (1) 500 (2) 1000 (3) 250 (4) 0.5 × 10–3
13. A one molal solution is one that contains (1) 1 g of the solute in 1000 g. of solvent
(2) 1 g mole of solute in 1000 mL of solution
(3) 1 g mole of solute in 22.4 lits of solution
(4) 1 g mole of solute in 1000 g. of solvent
14. The molarity of a 9.8% w w SO H 24 density 1.1 g/cc is
(1) 1 M (2) 0.55 M
(3) 0.1 M (4) 1.1 M
15. 6 g of Urea is dissolved in 90 g of water. The mole fraction of solute is
(1) 1/5 (2) 1/50
(3) 1/51 (4) 1/501
16. 100 mL of ethyl alcohol [d = 0.92 g/ml] and 900 mL of water [d = 1 g/ml] are mixed to form 1 lit solution. The Molarity and molality of the resulting solution are
(1) 2M and 2m (2) 2M and 2.22m (3) 2.2M and 1.1 m (4) 2M and 1m
17. Which of the following solution is more concentrated
18. A gaseous mixture contain four gases A, B, C and D. The mole fraction of ‘B’ is 0.5. The mole fraction of “A” is
(1) 0.525 (2) 0.375
(3) 0.625 (4) 0.732
19. The mole fraction of solvent in 0.1 molal aqueous solution is
(1) 0.9982 (2) 0.017
(3) 0.017 (4) 0.17
Solubility
20. Incorrect statement is (K H = Henry’s constant)
(1) KH is characteristic constant for a given gas - solvent system
(2) Higher is the value of K H , Lower is solubility of gas for a given partial pressure of gas
(3) KH has temperature dependence (4) KH decreases with temperature
21. Which of the following statement is correct about Henry’s law?
(1) It is applicable only when pressure is high
(2) It is applicable only when temperature is very high
(3) It is applicable only when gas reacts with the solvent
(4) KH (Henry’s constant) is a function of the nature of the gas
22. If 0.05 mole of gas are dissolved in 500 grams of water under 1 atm. pressure, 0.1 moles will be dissolved if the pressure is 2atm. It illustrates
(1) Graham’s Law
(2) Dalton’s Law
(3) Henry’s Law
(4) Boyle’s Law
23. In a nearly saturated solution, the solubility increases with increase in temperature, then ∆ is
(1) + Ve (2) –Ve
(3) + Ve (or) – Ve (4) 0
24. Among the following gases which gas has the highest Henry’s law constant, K H value in water at the same temperature?
(1) O2 (2) N2 (3) H2 (4) He
Vapour Pressure
25. Which of the following liquid pairs shows a positive deviation from Raoult’s law?
(1) Water – hydrochloric acid
(2) Benzene – Methanol
(3) Water – Nitric acid
(4) Acetone – chloroform
26. If liquid A and B form ideal solutions, then (1) the enthalpy of mixing is zero
(2) the entropy of mixing is zero
(3) the free energy of mixing is zero
(4) the free energy as well as entropy of mixing are each zero.
27. Among the following mixtures, dipole–dipole as major interactions is present in (1) KCl and water
(2) Benzene and carbon tetrachloride
(3) Benzene and ethanol
(4) Acetonitrile and acetone
28. Which has greater lowering of vapour pressure?
(1) 0.1m Glucose
(2) 0.1m Glucose
(3) 0.1m Sucrose
(4) equal in all cases
29. Relative lowering of vapour pressure is maximum for
(1) 0.1 m urea
(2) 0.1 m NaCl
(3) 0.1 m MgCl2
(4) 0.1 m Al2(SO4)3
30. Boiling point is least for (1) 0.1 m urea
(2) 0.2 m urea
(3) 0.1 m NaCl
(4) 0.2 m MgCl2
31. A solution that obeys Raoult’s law is called
(1) normal solution
(2) non-ideal solution
(3) ideal solution
(4) saturated solution
32. For an ideal solution of two components A and B, If xA and yA are mole fractions of component ‘A’ in solution and vapour phase respectively, then the slope of linear line in the graph drawn between 1/ xA and 1/yA is
(1) PPAB 00 + (2) PPAB 00 /
(3) PPBA 00 + (4) PPAB 00
33. A liquid is in equilibrium with its vapour at its boiling point, on the average, the molecules in the two phases have equal (1) Inter molecular forces
(2) Potential energy
(3) Temperature
(4) Kinetic energy
34. A solution has 1:4 molar ratio of pentane and hexane. The vapour pressure of pure hydrocarbons at 20ºC are 440 mm Hg for pentane and 120 mm of Hg for hexane. The mole fraction of pentane in vapour phase would be
(1) 0.200 (2) 0.478
(3) 0.547 (4) 0.786
35. The vapour pressure of pure water at 25°C is 30 mm. The vapour pressure of 10% (W/W) glucose solution at 25°C is
(1) 31.5 mm (2) 30.6 mm
(3) 29.67 mm (4) 26.56 mm
36. The vapour pressure of a pure liquid ‘A’ is 60mm, at 25°C. It forms an ideal solution with another liquid ‘B’. The mole fraction of ‘B’ is 0.6 and the total Pressure is 64mm. Then the vapour pressure of ‘B’ at 25°C is
(1) 66.6mm (2) 75 mm
(3) 52mm (4) 120mm
Solution of Solid in a Liquid
37. At a given temperature
a) Vapour pressure of a solution containing nonvolatile solute is proportional to mole fraction of solvent
b) Lowering of vapour pressure of solution containing nonvolatile solute is proportional to mole fraction of solute
c) Relative lowering of vapour pressure is equal to mole fraction of solute The correct combination is
(1) a only (2) a, b only (3) a, b and c only (4) b, c only
38. At 20°C, the vapour pressure of diethyl ether is 442mm. When 6.4 g of a non-volatile solute is dissolved in 50 g of ether, the vapour pressure falls to 410 mm. The Molecular weight of the solute is
(1) 150 (2) 130.832
(3) 160 (4) 180
Colligative Properties
39. If acetone (BP = 329 K) and CS2 (BP = 320 K) are mixed in a definite composition, so that the mixture of the two behave like pure liquid and boils at 312 K then it is (1) not an Azeotrope
(2) maximum boiling azeotrope
(3) minimum boiling azeotrope
(4) rectified spirit
40. Which one of the following is correct for solutions of components of A and B to follow Raoult’s law?
(1) A-B attractive force is greater than A-A and B-B attractive forces
(2) A-B attractive force is less than A-A and B-B attractive forces
(3) A-B attractive force remains same as A-A and B-B attractive forces
(4) volume of solution is different from sum of volumes of A and B components
41. When mercuric iodide is added to the aqueous solution of potassium iodide, (1) freezing point is raised (2) freezing point is lowered (3) freezing point does not change (4) boiling point does not change
42. During depression of freezing point in a solution the following are in equilibrium
(1) liquid solvent, Solid solvent
(2) liquid solvent, Solid solute
(3) liquid solute, Solid Solute
(4) liquid solute, Solid solvent
43. Solution A contains 7 g/L MgCl 2 and solution B contains 7 g/L of NaCl. At room temperature, the osmotic pressure of (1) solution A is greater than B
(2) both have same osmotic pressure (3) solution B is greater than A (4) can’t determine
44. The molal freezing point constant of water is 1.86 C/M. Therefore the freezing point of
45. A 5% solution of cane sugar (mol. wt. = 342) is isotonic with 1% solution of substance X. The molecular weight of X is (1) 34.2 (2) 171.2
(3) 68.4 (4) 136.8
46. The properties of solutions which depend only on the number of particles of solute (or the number of moles of solute) but independent of the nature of the solute are called
(1) extensive properties
(2) intensive properties
(3) colloidal properties
(4) colligative properties
47. The relationship between the value of Osmotic pressure (π) of the solution obtained by dissolving 6 g.L -1 of acetic acid (π1) and 7.45 g.L-1 of KCl (π2) is
48. The degree of dissociation (α) of a weak electrolyte Ax B y is related to Van’t Hoff factor (i) by the expression
(1) i xy 1 1 (2) i xy 1 1
(3) –xy i 1 1 (4) xy i 1 1
49. For a weak monobasic acid, if pKa = 4, then at a concentration of 0.01 M of the acid solution, the van’t Hoff factor is (1) 1.01 (2) 1.02 (3) 1.10 (4) 1.20
Abnormal Molecular Masses
50. The Vant Hoff’s factor ‘i’ accounts for (1) extent of solubility of solute (2) extent of mobility of solute
(3) extent of dissolution of solute
(4) extent of dissociation of solute
51. Relative lowering of vapour pressure is maximum for
(1) 0.1m urea (2) 0.1 m NaCl
(3) 0.1m MgCl2 (4) 0.1m Al2(SO4)3
52. Which of the following salt will have same value of Van’t Hoff’s factor (i) as that of K4[Fe(CN)6]
(1) Al(NO3)3 (2) NaCl (3) Na2SO4 (4) Al2(SO4)3
53. The freezing point of a solution containing 8.1 gms of HBr in 100 gms of water assuming acid to be 90% ionised (K f for water = 1.86 K. Kg/mol)
(1) 0°C (2) –0.35°C (3) –3.53°C (4) 0.85°C
54. The value of Kf for water is 1.86°, calculated from Glucose solution. The value of K f for water calculated from NaCl solution will be (1) 3.72 (2) 0.93 (3) Zero (4) 1.86
55. A solution of 0.816 g of a compound ‘A’ dissolved in 7.5g of Benzene freezes at 1.59°C. The molecular weight of compound ‘A’ if Kf of benzene is 4.9 and melting point of Benzene is 5.51°C (1) 120 (2) 136 (3) 32 (4) 180
56. The molal elevation constant of water is 0.51. The boiling point of 0.1 molal aqueous NaCl solution is nearly:
(1) 100.05°C (2) 100.1°C (3) 100.2°C (4) 101.0°C
57. The Osmotic pressure of solution containing 4.0 g of solute (molar mass 246) per litre at 27°C is (R =0.082 L atm K –1 mol–1).
(1) 0.1 atm (2) 0.2 atm (3) 0.4 atm (4) 0.8 atm
58. A 5% solution of cane sugar is isotonic with 0.5% of X. The molecular weight of substance X is
(1) 34.2 (2) 119.96 (3) 95.58 (4) 126.98
59. The osmotic pressure of the solution obtained by mixing 200 cm3 of 2% (massvolume) solution of urea with 200 cm3 of 3.42% solution of sucrose at 20°C is (1) 4 bar (2) 1.2 bar (3) 2.62 bar (4) 15.4 bar
60. 0.004 M solution of Na 2 SO 4 is isotonic with 0.01 M solution of glucose at same temperature. The approximate degree of dissociation of Na2SO4 is (1) 50% (2) 25% (3) 75% (4) 85%
61. Molal depression constant for water is 1.86 K Kg mole–1. The freezing point of a 0.05 molal solution of a non electrolyte in water is
63. The depression in freezing point of 0.01 m aqueous solution of urea, sodium chloride and sodium sulphate is in the ratio (1) 1 : 1 : 1 (2) 1 : 2 : 3 (3) 1 : 2 : 4 (4) 2 : 2 : 3
64. Which of the following salt will have same value of Van’t Hoff’s factor [i] as that of K4[Fe(CN)6]?
(1) Al2(SO4)3 (2) NaCl
(3) Al(NO3)3 (4) Na2 (SO4)
65. The van’t Hoff factor for BaCl 2 at 0.01 M concentration is 1.98. The percentage dissociation of BaCl2 at this concentration is
(1) 49 (2) 69 (3) 89 (4) 98
66. The freezing point of equimolal aqueous solution will be highest for (1) C6H5NH3Cl (2) Ca(NO3)2 (3) La(NO3)3 (4) C6H12O6
67. If 0.1 M solution of glucose and 0.1 M solution of urea are placed on two sides of a semipermeable membrane to equal heights, then it will be correct to say that (1) glucose will flow towards urea solution (2) there will be no net movement across membrane (3) urea will flow towards glucose solution (4) water will flow from urea solution to glucose
68. An aqueous solution freezes at–0.186°C (K f = 1.806; K b = 0.512°). What is the elevation on boiling point?
(1) 0.186 (2) 0.512 (3) 0.86 (4) 0.0512
69. If α is the degree of dissociation of Na 2SO4, the Vant Hoff factor (i) used for calculating the molecular mass is (1) 1 + a (2) 1 – a (3) 1 + 2 a (4) 1 – 2 a
Level - II
Methods of Concentration
1. The mole fraction of the solute in 1 molal aqueous solution is:
(1) 0.009 (2) 0.018 (3) 0.027 (4) 0.036
2. The Normality of 0.98 % (w/v) H 2 SO 4 solution is
(1) 0.1N (2) 0.2N (3) 0.4N (4) 1N
3. 100 mL of 0.1N FeSO 4 solution will be completely oxidised by ‘ x’ gms of K2Cr2O7 in acidic medium (Mol. wt = 294). The value of ‘x’ is
(1) 4.9 (2) 2.94 (3) 0.49 (4) 1.47
4. The normality of which solution is highest?
(1) 1 M HCl (2) 1 M H3PO4
(3) 1 M H2SO4 (4) 1 M C2H2O4
5. Which of the following solution is more concentrated?
(1) 0.3% H3PO4
(2) 0.3 MH3PO4
(3) 0.3 m H3PO4
(4) 0.3 NH3PO4
6. The correct relationship between molarity (M) and molality (m) is (d = density of the solution, in KgL–1, M2 = molar mass of the solute in kg mol–1)
(1) M md mM 1 2
(2) M m mMd 1 2
(3) M mM md 1 2
(4) M md mM 1 2
7. Regarding molarity, which of the following statements are correct?
a) Units of molarity mol kg –1
b) Molarity of dibasic acid is half of its normality
c) Normality × GEW / GMW
d) Molarity always equals to its molality
(1) a,b (2) a,b,c
(3) b,c (4) a,c
8. The density of 20% (w/v) aqueous NaOH solution is 1.20 g mL –1. What is the mole fraction of water?
(molar mass of NaOH = 40 g.mol –1)
(1) 0.95 (2) 0.917
(3) 0.97 (4) 0.94
9. X is a non-volatile solute and Y is a volatile solvent. The following vapour pressures are observed by dissolving X in Y
X/mol L–1 Y/mm of Hg
0.10 P 1
0.25 P 2
0.01 P 3
The correct order of vapour pressure is
(1) P1 < P2 < P 3
(2) P 3 < P2 < P1
(3) P 3 < P1 < P2
(4) P2 < P1 < P 3
10. 11.1 g. of CaCl 2 is present in 100 mL of the aqueous solution. The chloride ion concentration is (1) 1 M (2) 2 M (3) 0.5 M (4) 0.2 M
11. 100 mL each of 1 M AgNO3 and 1 M NaCl are mixed. The nitrate ion concentration in the resulting solution is (1) 1 M (2) 0.5 M (3) 0.75 M (4) 0.25 M
12. 100 mL of 1M HCl, 200 mL 2M HCl and 300 mL 3M HCl are mixed. The Molarity of the resulting solution is (1) 1M (2) 2.66M (3) 2.33M (4) 4.25 M
13. 100 mL of 1M HCl, 200 mL of 2M HCl and 300 mL of 3M HCl are mixed with enough water to get 1M solution. The volume of water to be added is (in mL) (1) 600 (2) 700 (3) 800 (4) 125
14. The volume of 0.025M Ca(OH) 2 solution which can neutralise 100 mL of 10–4 M H3PO4 is
(1) 10 mL (2) 60 mL
(3) 0.6 mL (4) 2.8 mL
Solubility
15. O 2 is bubbled through water at 293K, assuming that O2 exerts a partial pressure of 0.98 bar, the solubility of O 2 in gm.L–1 is
(Henry’s law constant = 34 k bar)
(1) 0.025 (2) 0.05
(3) 0.1 (4) 0.2
16. H 2S a toxic gas with rotten egg like smell is used for the qualitative analysis, if the solubility of H 2S in water, at STP is 0.195 mole. kg–1, the Henry’s law constant is (1) 285.6 bar (2) 324.8 bar
(3) 462.9 bar (4) 534.8 bar
17. Air contains O2 and N2 in the ratio 0.2 : 0.8. If Henry law constant for O 2 and N 2 are 3.3 × 107 torr and 6.6 × 107 torr respectively, then the ratio of mole fractions of O 2 to N2 dissolved in water at 1 bar pressure is
(1) 1 : 1 (2) 2 : 1
(3) 1 : 2 (4) 1 : 3
18. The quantity of CO2 in 500ml of soda water when packed under 2.5 atm CO2 pressure at 298 k is ....... gm (Henry’s law constant 1.67 × 108 pa at 298K)
(1) 0.64 (2) 1.86
(3) 6.4 (4) 18.6
19. Henry law constant for the solubility of methane in benzene at 298K is 4.27×10 5 mm Hg then the solubility of methane in benzene at 298K under 760 mm Hg is
(1) 7.8 × 10–2 mole/kg
(2) 7.8 × 10–3 mole/kg
(3) 7.8 × 10–4 mole/kg
(4) 1.78 × 10–3 mole/kg
20. Which one of the following gases has the lowest value of Henry’s Law constant?
(1) N2 (2) He
(3) CO2 (4) O2
21. The solubility of N2(g) in water exposed to the atmosphere, when the partial pressure of 593 mm is 5.3 × 10–4 M. Its solubility at 760 mm and at the same temperature is:
(1) 4.1 × 10–4 M
(2) 6.8 × 10–4 M
(3) 1500M
(4) 2400M
22. As temperature increases, vapour pressure of a liquid (1) increases linearly (2) decreases linearly (3) increases exponentially (4) decreases exponentially
Vapour Pressure
23. The plot of 1/xA versus 1/yA (where xA and yA are the mole fraction of A in liquid and vapour phases, respectively) is linear with slope and intercept respectively are given as (y-axis = 1/yA, x-axis = 1/xA)
(1) PPPPP ABABB 00 00 0 /, /
(2) PPPPP ABBAB 00 00 0 /, /
(3) PPPPP BAABA 00 00 0 /, /
(4) PPPPP BABAB 00 00 0 /, /
24. The vapour pressure of pure benzene and toluene are 160 Torr and 60 Torr respectively. The mole fraction of benzene in vapour phase in contact with equimolar solution of benzene and toluene would be
(1) 0.50 (2) 8/3
(3) 0.73 (4) 0.27
25. A non-volatile solute (A) is dissolved in a volatile solvent (B). The vapour pressure of resultant solution is Ps. The vapour pressure of pure solvent is P0B. If “X” is mole fraction, which of the following is correct?
(1) PPXSBA 0
(2) PPXBSB 0 =
(3) PPX SBB 0
(4) PPXBSA 0 =
26. If x1 and x2 represent the mole fraction of a component A in the vapour phase and liquid mixture respectively and p A° and pB°
represent vapours pressures of pure A and pure B, then total vapour pressure of the liquid mixture is
(1) px x A 0 1 2 (2) px
27. Two liquids X and Y form an ideal solution. The mixture has a vapour pressure of 400 mm at 300 K when mixed in the molar ratio of 1 : 1 and a vapour pressure of 350 mm when mixed in the molar ratio of 1 : 2 at the same temperature. The vapour pressures of the two pure liquids X and Y respectively are
(1) 250 mm, 550 mm
(2) 350 mm, 450 mm
(3) 350 mm, 700mm
(4) 550 mm, 250 mm
28. 15 cm3 of liquid ‘x’ and 20 cm3 of liquid ‘y’ are mixed at 20°C and the volume of solution was measured to be 35.1 cm3 then correct reaction is
(1) H mix 0 solution shows +ve deviation
(2) H mix 0 solution shows +ve deviation
(3) H mix 0 solution shows –ve deviation
(4) H mix 0 solution shows –ve deviation
29. Which of the following statements are correct
a) the boiling point of a solution is greater than pure solvent
b) the temperature where the vapour pressure of liquid equals to atmospheric pressure is called its boiling point
c) the vapour pressure of pure solvent is less than the vapour pressure of solution containing non volatile solute.
d) the temperature of liquid remained in the container after evaporation is more than before the evaporation
(1) a,b (2) b,c (3) c,d (4) a,d
30. The graph obtained by taking vapour pressure (P) of a liquid on y-axis and temperature (T) on x-axis will be (1)
31. Which graph of the following represents the graph between log p (on Y - axis) and 1/T (on X - axis)?
(1) log P
32. Which of the following solutions will have the lowest vapour pressure
(1) 0.1M Glucose
(2) 0.1M NaCl
(3) 0.1 M BaCl2
(4) 0.1 M Al2(SO4)3
33. In the depression of freezing point experiment, it is found that
a) The vapour pressure of the solution is less than that of pure solvent
b) The vapour pressure of the solution is more than that of pure solvent
c) Only solute molecules solidify at the freezing point
d) Only solvent molecules solidify at the freezing point
(1) a,b (2) b,c
(3) a,d (4) a,b,c
34. For a dilute solution, Raoult’s law states that:
(1) The relative lowering of vapour pressure is equal to the mole fraction of solute.
(2) The relative lowering of vapour pressure is equal to the mole fraction of solvent
(3) The relative lowering of vapour pressure is proportional to the amount of solute in solution
(4) The vapour pressure of the solution is equal to the mole fraction of solvent
35. Which of the following pair shows a positive deviation from Raoults law?
(1) H2O – HCl
(2) C6H6 – CH2OH
(3) H2O – HNO3
(4) CH3COCH3 – CHCl3
36. The vapour pressure of water at 20°C is 17.54mm Hg. then the vapour pressure of the water in the apparatus shown after the piston is lowered, decreasing the volume of the gas above the liquid to one half of its initial volume (temp. constant) is
(1) 8.77 mm Hg (2) 17.54 mm Hg
(3) 35.08 mm Hg
(4) between 8.77 and 17.54 mm Hg
37. An aq. solution containing 64% by weight of volatile liquid ‘A’ (molecular mass 128) has pressure of 145mm, then vapour pressure of ‘A’ is (V.P. of H2O is 155 mm)
(1) 150mm (2) 145mm
(3) 105mm (4) 21mm
Solution of Solid in a Liquid
38. The relative lowering of vapour pressure of 0.2 molal solution in which the solvent is Benzene is
(1) 15.6 × 10–4 (2) 15.6 × 10–3
(3) 15.6 × 10–1 (4) 0.05
39. The amount of Glucose to be dissolved in 500g of water so as to produce the same lowering in vapour pressure as that of 0.2 molal aqueous urea solution
(1) 9 g. (2) 18 g
(3) 36 g (4) 1.8 g
Colligative Properties
40. A Current of dry air was first passed through the bulb containing solution of ‘A’ in water and then through the bulb containing pure water. The loss in mass of a solution bulb is 1.92 g. Where as that in pure water bulb is 0.08 g, then mole fraction of ‘A’ is
(1) 0.86 (2) 0.2
(3) 0.96 (4) 0.04
41. RLVP of an aqueous solution of non volatile solute is 0.4. Molality of the solution is (1) 15 m (2) 37 m
(3) 45 m (4) 10 m
42. Which of the following is colligative property?
(1) vapour pressure
(2) boiling point
(3) freezing point
(4) osmotic pressure
43. Vapour pressure is highest for (1) 0.4 m urea
(2) 0.3 m sucrose
(3) 0.2 m fructose
(4) 0.1 m glucose
44. Among the following the azeotropic mixture is
(1) CCl4 +CHCl3
(2) C6H14 + C7H16
(3) C2H5Br + C2H5Cl (4) Br Cl +
45. An azeotropic solution of two liquids has a boiling point lower than either of them when it
(1) shows negative deviation from Raoult’s law
(2) shows no deviation from Raoult’s law
(3) shows positive deviation from Raoult’s law
(4) is saturated
46. The ratio of the value of any colligative property for KCl solution to that of sugar solution is
(1) 1 : 1 (2) 1 : 2
(3) 2 : 1 (4) 4 : 1
47. Which one of the following statements is false?
(1) Two sucrose solutions of same molality prepared in different solvents will have the same freezing point depression.
(2) The osmotic pressure ( π) of a solution containing non-volatile solute is given by the equation π = MRT where M is the molarity of the solution
(3) Raoult’s law states that the vapor pressure of a component over a solution is proportional to its mole fraction in the solution.
(4) The correct order of osmotic pressure for 0.01M aqueous solution of each compound is BaCl2 > KCl > CH3 COOH > sucrose
48. Molal elevation in boiling point constant depends on
(1) nature of solvent
(2) nature of solute
(3) heat of solution of the solute in the solvent
(4) vapour pressure of solution
Abnormal Molar Mass
49. Y gm of a non volatile solute of molar mass M is dissolved in 250 g of benzene. If K b is molal elevation constant, the value of DTb is given by:
(1) 4 M KY b (2) 4 KY M b
(3) K Y bM 4 (4) K Y bM
50. 139.18 g of glucose is added to 178.2 g of water the vapour pressure of water for this aqueous solution at 1000 C is
(1) 700.7 torr
(2) 759 torr
(3) 7.6 torr
(4) 76 torr
51. The relative lowering of vapor pressure of 0.2 molal solution in which the solvent is Benzene is
(1) 15.6 × 10–4 (2) 15.6 × 10–3
(3) 15.6 × 10–1 (4) 0.05
52. Which will form maximum boiling azeotrope?
(1) C6H6+C6H5CH3 solution
(2) HNO3+H2O solution
(3) C2H5OH+H2O solution
(4) n – hexane + n - heptanes
53. Which among the following pair of aqueous solutions have same freezing point?
(1) 0.1 M KCl & 0.1 M CaCl2
(2) 0.5 M MgCl2 & 1.5 M Glucose
(3) 1 M Al2 (SO4)3 & 4 M urea
(4) 2M Sucrose & 1 M K2SO4
54. A decimolar solution of K4[Fe(CN)6] at 300 K is 50% dissociated, then its osmotic pressure is nearly_______
(1) 7.5 atm (2) 3.61 atm
(3) 21.34 atm (4) 12.32 atm
55. Which has minimum osmotic pressure?
(1) 200 mL of 2 M NaCl solution
(2) 200 mL of 1 M glucose solution
(3) 200 mL of 2 M Urea solution
(4) none of these
56. The solution containing 6.8g of non-ionic solute in 100g of water was found to freeze at –0.93°C. If Kf for water is 1.86, the molar Mass of solute is
(1) 13.6 (2) 68
(3) 34 (4) 136
57. The Osmotic pressure of solution containing 4.0g of solute (molar mass 246) per litre at 270C is (R = 0.082L atm k-1 mol-1)
(1) 0.1 (2) 0.2
(3) 0.4 (4) 0.8
58. 0.004M Na 2 SO 4 is isotonic with 0.01M glucose. Degree of dissociation of Na2SO4 is:
(1) 75% (2) 50% (3) 25% (4) 85%
59. The relationship between osmotic pressure at 273K when 10g glucose (P1), 10g urea (P2) and 10 g sucrose (P3) are dissolved in 250 mL of water is
(1) P1 > P2 > P 3 (2) P 3 > P1 > P2
(3) P2 > P1 > P 3 (4) P2 > P 3 > P1
60. Which of the following statement is false?
(1) Raoult’s law states that vapour pressure of a component over a solution is proportional to its mole fraction.
(2) Osmotic pressure is given by the expression π = MRT where, M is molarity.
(3) The correct order of osmotic pressures of 0.01 M aqueous solution of each compound is BaCl2 > KCl> CH3COOH > Sucrose.
(4) Two sucrose solutions of same molarity prepared in different solvents will have same freezing point depressions.
61. Blood is isotonic with
(1) 0.16 M NaCl (2) Conc.NaCl
(3) 50% NaCl (4) 30% NaCl
62. FeCl 3 on reaction with K4[Fe(CN) 6] in aq solution gives blue colour and these are separated by a semi permeable membrane Due to osmosis there is (1) blue colour formation in the solution of FeCl3
FURTHER EXPLORATION
1. 29.2% (w/W) HCl stock solution has density of 1.25 g per ml The volume (mL) of stock solution required to prepare a 200 mL solution 0.4 M HCl is (1) 5 (2) 6 (3) 8 (4) 15
(2) blue colour formation in the solution of K4[Fe(CN)6]
(3) blue colour formation in both the solutions
(4) no blue colour formation
63. 1.0 molal aqueous solution of an electrolyte X 3Y2 is 25% ionized. The Boiling point of the solution is (kb of H2O = 0.52 K kg mol–1) (1) 375.5 K (2) 374.04 K (3) 377.12 K (4) 373.25 K
64. The osmotic pressure of a phenol solution in an organic solvent is determined to be 20% less than expected, it is due to
(1) Phenol is 20% ionised
(2) Phenol is 20% dimerised (3) Phenol is 40% dimerised (4) Phenol is 80% dimerised
65. Vant Hoff factor is highest for _____ molal K2SO4
(1) 1 (2) 0.001
(3) 0.1 (4) 0.01
66. 75% of solute is trimerised in a solution. Van’t Hoff factor of the solute in solution is
(1) 0.5 (2) 0.005 (3) 0.0005 (4) 0.05
67. If a is the degree of dissociation of Na2SO4, the van’t Hoff s factor (i) used for calculating the molecular mass is:
(1) 1 + a (2) 1 – a
(3) 1 + 2 a (4) 1 – 2 a
2. Two liquids A and B at a particular temperature (T) exert a pressure of 280 mm and 168 mm respectively. The solution of A and B with the mole fraction of A equal to 0.32 at the same temperature exerts a total vapour pressure of 376 mm. Which one of the following is correct based on the above data?
(1) Partial pressure of A is 190.4 mm.
(2) Partial pressure of B is 53.76 mm.
(3) For the above solution to be ideal, total vapour pressure should be 203.84 mm.
(4) Above solution is non-ideal solution with negative deviation 0.01.
3 The vapour pressures of pure liquids A and B are 400 and 600 mmHg, respectively at 298 K. On mixing the two liquids, the sum of their initial volumes is equal to the volume of the final mixture. The mole fraction of liquid B is 0.5 in the mixture. The vapour pressure of the final solution, the mole fractions of components A and B in vapour phase, respectively are:
(1) 500 mmHg, 0.4, 0.6
(2) 450 mmHg, 0.5, 0.5
(3) 500 mmHg, 0.5, 0.5
(4) 450 mmHg, 0.4, 0.6
4. The correct statement(s) among the following is?
I) The values of Vant-Hoff factor(i) at Various concentrations for K2SO4(incomplete dissociation) increases with dilution as, 2.32 at 0.1m, 2.70 at 0.01 m, 2.84 at 0.001m but it will be less than 3
II) 0.155 M NaCl solution is isotonic with the fluid inside the blood cell
III) Celluose acetate placed over a suitable support is used as a semi – permaeable membrane in reverse osmosis as it can bare quite high pressure which is required for reverse osmosis
IV) Osmotic pressure method is used to determine molar masses of proteins, polymers and other macromolecules
(1) 1 (2) 0.001
(3) 0.1 (4) 0.01
Matching Type Questions
1. Match List-I with List-II.
List-I List-II
(A) n-hexane, n-heptane (i) Dimerisation
(B) CO2, H2O (ii) ionisation
(C) C6H5COOH, Benzene (iii) Raoults law (obey)
(D) C6H5COOH, H2O (iv) Henrys law (obeys)
(a) (b) (c) (d)
(1) iii iv i ii
(2) iv i ii iii
(3) i ii iii iv
(4) ii iii iv i
2. Match List-I with List-II.
List - I List-II
(A) Lowering of vapour pressure (i) PP P o o
(B) Relative lowering of vapour pressure (ii) PP P w m M W o o
(C) Raoult’s law (iii) P0 –P
(D) Ideal solution (iv) Obeying Raoults law
(v) Boiling point
(a) (b) (c) (d)
(1) iii i ii iv
(2) iv i ii iii
(3) i ii iii iv
(4) ii iii iv i
3. Using V.P vs temp graph match the following
1 3 2 3 ,
List-I List-II
(A) P at 90° (i) 0.23
(B) X chlorobenzene in vapur phase (ii) 0.73
(C) X benzene in vapour phase at 130°C (iii) 733.33
(D) P at 130°C (iv) 433.33
(a) (b) (c) (d)
(1) iii iv i ii
(2) iv i ii iii
(3) i ii iii iv
(4) ii iii iv i
4. Match List-I with List-II.
List - I List-II
(A) x y (i) Vapour pressure vs. composition
(B) y x (ii) Boiling point of non-ideal mixture vs. composition
(C) y x (iii) log (V.P.) vs. (temperature)–1
(D) y x (iv) Vapour pressure vs. temperature.
(a) (b) (c) (d)
(1) iii iv i ii
(2) iv i ii iii
(3) i ii iii iv
(4) iv iii ii i
5. Match List-I with List-II.
List - I List-II
(A) Al2(SO4)3; α = 0.7 (i) i=3.4
(B) Na3PO4; α=0.91 (ii) i=2.8
(C) CaCl2; α=0.9 (iii) i=3.8
(D) FeCl3; α=0.8 (iv) i=3.7
(a) (b) (c) (d)
(1) iii iv ii i
(2) iv i ii iii
(3) i ii iii iv
(4) ii iii iv i
6. Match List-I with List-II.
List - I List-II
(A) Benzene+ Toluene (i) Does not form azeotrope
(B) CO2, H2O (ii) n-hexane + n-heptane
(C) chloroform + acetone (iii) Form maximum boiling azeotrope
(D) Ethanol + Water (iv) can be separated by fractional distillation
(a) (b) (c) (d)
(1) iii i,iv ii i,iv
(2) i, iv i,iv iii ii
(3) i i,iv i,iv ii
(4) iii ii i,iv i,iv
7. Match List-I with List-II.
List - I List-II
(A) Mole fraction (i) No. of g equivalents in 1000 ml of solution
(B) Molarity (ii) Always less than one
(C) Normality (iii) Greater than or equal to molarity
(D) Molality (iv) No. of g moles present in 1000 ml solution
(v) No. of g moles of solute 1 kg of solvent
(a) (b) (c) (d)
(1) i ii iii iv
(2) ii iii i v
(3) i iv ii iii
(4) ii iv iii v
8. Match List-I with List-II. List-I List-II
(A) Lowering of vapour pressure (i) PP P
(B) Relative lowering of vapour pressure (ii) PP P W m M W
(C) Raoult’s law (iii) P° – P
(D) Ideal solution (iv) Obeying Raoults law
(v) Boiling point
(A) (B) (C) (D)
(1) iii ii i iv
(2) iv i ii iii
(3) iii i ii iv
(4) i iii iv ii
9. Match List-I with List-II.
List-I List-II
(A) C6H14+C7H16 (i) ∆Hmix > 0
(B) C2H5OH + CH3COCH3 (ii) ∆Hmix < 0
(C) CHCl3 + CH3COCH3 (iii) ∆Hmix = 0
(iv) ∆Smix = 0
(v) No. of g moles of solute 1kg of solvent
(a) (b) (c)
(1) iv i ii
(2) i iv ii
(3) iii i ii
(4) iii i iv
Statement Type Questions
Directions for following questions
Each question has two statements. Statement I and statement II. Mark the correct answer as
(1) if both statement I and statement II are correct,
(2) if both statement I and statement II are incorrect,
(3) if statement I correct but statement II is incorrect,
(4) if statement I incorrect but statement II is correct.
1. S-I : Ethanol + water solution shows positive deviation from Raoult’s law.
S-II : Acetone + chloroform solution shows positive deviation from Raoult’s law
2. S-I : ∆Vmix and ∆Smix for an ideal solution is zero.
S-II : A – B interactions in an ideal solution are nearly equal to that of A – A and B – B interactions.
3. S-I : Addition of Hgl2 to aqueous solution of Kl increases the freezing point
S-II : A complex K2[Hgl4] is formed
4. S-I : Total vapor pressure over the solution can be related to mole fraction of any one component.
S-II : Depending on the vapor pressures of the pure components 1 and 2, total vapor pressure over the solutions decreases or increases with the increase of the mole fraction of component 1
5. S-I : 0.1 molal NaCl solution in water(Kf = 1.86°molal –1) have a freezing point of –1.86.
S-II : Van’t Hoff factor of NaCl solution isequal to 2 for 100% ionization.
S-II : Acetone + Aniline solution shows positive deviation from Raoult’s law
Assertion and Reason Type Questions
In each of the following questions, a statement of Assertion (A) is given, followed by a corresponding statement of Reason (R). Mark the correct answer as
(1) if both (A) and (R) are true and (R) is the correct explanation of (A),
(2) if both (A) and (R) are true but (R) is not the correct explanation of (A),
(3) if (A) is true but (R) is false,
(4) if both (A) and (R) are false.
1. (A) : Addition of ethylene glycol (non volatile) to water lowers the freezing point of water, hence used as antifreeze.
(R) : Addition of ethylene glycol (non volatile) to water lowers the freezing point of water, hence used as antifreeze.
2. (A) : The molecular weight of acetic acid in benzene is more than the actual value of the solute..
(R) : Molecules of acetic acid dimerize in benzene due to hydrogen bonding. In the light of the given statements, choose the correct answer from the options given.
3. (A) : The solubility of the gas in a liquid increases with increases of pressure at constant temperature..
(R) : The solubility of the gas in a liquid depends on pressure at constant temperature.
4. (A) : The vapour pressure of a liquid deceases if some non-volatile solute is dissolved in it.
(R) : The relative lowering of vapour pressure of a solution containing a non volatile solute is equal to the mole fraction of the solute in the solution.
5. (A) : The vapour pressure of 0.1 M sugar solution is more than that of 0.1 M KCl solution.
(R) : Lowering of vapour pressure is directly proportional to the number of species present in the solution.
6. (A) : Reverse osmosis is used in the desalination of sea water.
(R) : When the applied pressure is more than the osmotic pressure, pure water is squeezed out of sea water through the membrane.
7. (A) : If a liquid solute is added to a solvent which is more volatile than solvent, then vapour pressure of solution may increase, that is Ps > P0
(R) : In the pressure of a more volatile liquid solute, only the solute will form the vapours and solvent will not
8. (A) : The freezing point of 0.1 M sugar and 0.1 M KCl are equal.
(R) : Lowering of vapour pressure is does not depend upon the number of species present in solution..
9. (A) : Cooking time is reduced in a pressure cooker.
(R) : Boiling point of water inside is elevated
BRAIN TEASERS
1. Pure water freezes at 273 K and 1 bar. The addition of 34.5 gm of ethanol to 500 g of water changes the freezing point of the solution. Use the freezing point depression constant of water as 2K mol–1. The figure show below represent plot of vapour pressure (V.P) g mol–1).
10. (A) : The solubility of solid in a liquid depends upon lattice energy and hydration energy.
(R) : If hydration energy is greater than lattice energy, the solid dissolves in liquid.
11. (A) : Molarity of 0.02 N HNO3 is 0.02 M.
(R) : Molarity and normality of a solution are always same.
12. (A) : Relative lowering of vapour pressure is independent of temperature.
(R) : Relative lowering of vapour pressure is nothing but the mole fraction of solute.
13. (A) : Raoult’s law is applicable to ideal solutions only.
(R) : In ideal solutions association or dissociation does not takes place.
14. (A) : Mercuric iodide dissolves in potassium iodide solution forming K2HgI4 and the freezing point gets elevated..
(R) : The number of particles in K 2Hgl4 is same as those in Hgl2.
2. Mixture of volatile components A and B has total vapour pressure (in torr): p = 254 – 119XA where xA is mole fraction of A in mixture Hence, pA 0 and pB 0 are (in torr) (1) 254, 119 (2) 135, 254 (3) 119, 254 (4) 154, 119
3. The molar volume of X (I) d = 0.9g / mL increases by a factor of 3000 as it vaporizes at27°C and the Y(I) d = 0.88g /mL increases by a factor of 8000 at27°C. A miscible liquid solution of X and Y at 27°C has a vapour pressure of 50 torr. The mole fraction of Y in solution is : (Given: R = 0.082 atm/L/ mol/K; Molar mass of X = 75; Molar mass of Y = 88)
(1) 0.48 (2) 0.52 (3) 0.62 (4) 0.247
FLASHBACK
1. If 8 g of a non-electrolyte solute is dissolved in 114 g of n-octane to reduce its vapour pressure to 80%, the molar mass (in g mol–1) of the solute is ________ . If solution is assumed as dilute solution. [Given that molar mass of n-octane is 114 g mol –1] (2020-II)
(1) 20 (2) 40
(3) 60 (4) 80
2. The mixture which shows positive deviation from Raoult’s law is: (2020-I)
(1) Benzene + Toluene
(2) Acetone + Chloroform
(3) Chloroethane + Bromoethane
(4) Ethanol + Acetone
3. The freezing point depression constant (Kf) of benzene is 5.12 K kg mol–1. The freezing point depression for the solution of molality 0.078 m containing a non-electrolyte solute in benzene is (rounded off up to two decimal places): (2020-I)
(1) 0.80 K (2) 0.40 K
(3) 0.60 K (4) 0.20 K
4. Isotonic solutions have same (2020-II)
(1) Boiling temperature
(2) Vapour pressure
(3) Freezing temperature
(4) Osmotic pressure
5. Which of the following statements is correct regarding a solution of two components A and B exhibiting positive deviation from ideal behaviour? (2019-Odisa)
(1) Intermolecular attractive forces between A-A and B-B are equal to those between A-B.
(2) Intermolecular attractive forces between A-A and B-B are stronger than those between A-B.
(3) ∆ mix H =0 at constant T and P.
(4) ∆ mix V = 0 at constant T and P.
6. Toluene in the vapour phase is in equilibrium with a solution of benzene and toluene
having mole fraction of toluene 0.50. If vapour pressure of pure benzene is 119 torr and that of toluene is 37.0 torr at the same temperature, mole fraction of toluene in vapour phase will be: (2017)
(1) 0.462 (2) 0.237 (3) 0.506 (4) 0.325
7. Consider the following liquid-vapour equilibrium. Liquid Vapur
Which of the following relation is correct? (2016-I)
(1) dP dT H T v ln 22 (2) dP dT H RT v ln 2
(3) dG dT H RT v ln 22 (4) dP dT H RT v ln
8. Which of the following statements about the composition of the vapour over an ideal 1:1 molar mixture of benzene and toluene is correct? Assume that the temperature is constant at 25°C (Given, vapour pressure data at 25°C benzene =12.8 kPa, toluene =3.85 kP a (2016-I)
(1) The vapour will contain equal amounts of benzene and toluene.
(2) Not enough information is given to make a prediction.
(3) The vapour will contain a higher percentage of benzene.
(4) The vapour will contain a higher percentage of toluene.
9. Which one of the following is incorrect for ideal solution? (2016-I)
(1) ∆Hmix = 0 (2) ∆Umix = 0
(3) ∆P = Pobs –P calculated Roult’s law = 0
(4) ∆Gmix = 0
10. At 100oC, the vapour pressure of a solution of 6.5g a solute in100g water is 732mm. If kb= 0.52 the, boiling point of this solution will be (2016-II)
(1) 102oC (2) 103oC
(3) 101oC (4) 100oC
11. The van’t Hoff factor (i) for a dilute aqueous solution of the strong electrolyte barium hydroxide is (2016-II)
(1) 0 (2) 1
(3) 2 (4) 3
12. What is the mole fraction of the solute in a 1.00 m aqueous solution? (2015-Reexam)
(1) 1.770 (2) 0.0354
(3) 0.0177 (4) 0.177
13. Which of them is not equal to zero for an ideal solution? (2015-cancelled)
(1) ∆Vmix
(2) ∆P=Pobserved-PRaoult
(3) ∆Hmix
(4) ∆Smix
14. The boiling point of 0.2 mol kg-1 solution of X in water is greater than equimolal solution of Y in water. Which one of the following
CHAPTER TEST
1. Given below are two statements. one is labelled Assertion (A) and the other is labelled Reason (R).
(A) : Copper dissolved in gold is a solid solution
(R) : Solute determines the physical state in which solution exists.
(1) Both (A) and (R) are true and (R) is the correct explanation of (A)
(2) Both (A) and (R) are true and (R) is not the correct explanation of (A)
(3) (A) is true and (R) is false
(4) (A) and (R) are false
2. Match List-I with List-II.
List - I List-II
(A) Gaseous solution (i) German silver
(B) Liquid solution (ii) Milk
statements is true in this case? (2015-cancelled)
(1) Molecular mass of X is less than the molecular mass of Y
(2) Y is undergoing dissociation in water while X undergoes no change.
(3) X is undergoing dissociation in water.
(4) Molecular mass of X is greater than the molecular mass of Y
15. Which one of the following electrolytes has the same value of van’t Hoff factor (i) as that of Al2(SO4)3 (if all are 100 % ionized)? (2015-cancelled)
(1) Al(NO3)3
(2) K4[Fe(CN6)]
(3) K2SO4
(4) K3[Fe(CN6)]
(C) Solid solution (iii) Sand in water
(D) Colloidal solution (iv) Aqueous Alcoholic solution (v) Air
(A) (B) (C) (D)
(1) v iv i ii
(2) i iii ii iv
(3) iv ii v ii
(4) ii iii i iv
3. Which combination of the following terms is matched correctly?
I) Vapour pressure
II) Intermolecular forces
III) Latent heat of vaporization
(1) I-high, II-weak, III-small
(2) I-high, II-strong, III-large
(3) I-low, II-weak, III-large
(4) I-low, II-strong, III-small
4. Occlusion of Hydrogen on Palladium is an example for ___ type solution
(1) gas in solid (2) solid in gas (3) gas in liquid (4) liquid in gas
5. How much solid oxalic acid (Molecular weight 126) has to be weight to prepare 100 mL exactly 0.1 (N) oxalic solution in water?
(1) 1.26 g (2) 0.126 g (3) 0.63 g (4) 0.063 g
6. In one molal solution contains 0.5 mole of a solute, there is (1) 1000 g of solvent (2) 500 mL of solvent (3) 500 g of solvent (4) 100 mL of solvent
7. Liquid ‘M’ and liquid ‘N’ form an ideal solution. The vapour pressures of pure liquids ‘M’ and ‘N’ are 450 and 700 m Hg, respectively, at the same temperature. Then correct statement is:
(xM= Mole fraction of ‘M’ in solution; (xN= Mole fraction of ‘N’ in solution; (yM= Mole fraction of ‘M’ in vapour phase; (yN= Mole fraction of ‘N’ in vapour phase;
(1) x x y y M N M N =
(2) x x y y M N M N >
(3) x x y y M N M N < (4) (xM – yM) < (xN – yN)
8. A solution of sodium sulfate contains 92 g of Na+ ions per kilogram of water. The molality of Na+ ions in that solution in mol kg is: (1) 4 (2) 8 (3) 12 (4) 16
9. Which of the following is dependent on temperature?
(1) Molarity
(2) Mole fraction
(3) Weight percentage
(4) Molality
10. Mole fraction of solute in 4.5 molal aqueous solution is (1) 0.05 (2) 0.025 (3) 0.0375 (4) 0.075
11. The volume of 0.025M Ca(OH) 2 solution which can neutralise 100ml of 10 -4 M H3PO4 is (1) 10ml (2) 60ml
(3) 0.6ml (4) 2.8ml
12. The number of millimoles of H2SO4 present in 5 liters of 0.2 N H2SO4 solution is (1) 500 (2) 1000 (3) 250 (4) 05 10 3
13. Density of a 2.05 M solution of acetic acid in water is 1.02g/mL. The molality of the solution is
(1) 2.28 mol kg-1 (2) 0.44 mol kg (3) 1.14 mol kg-1 (4) 3.28 mol kg
14. 18 g of glucose is dissolved in 180 grams of water. Vapour pressure of solution is 178.2 mm . Vapour pressure of water at the same temperature will be (1) 180 mm (2) 180.9 mm
(3) 181.8 mm
(4) 182.7 mm
15. A solute has 2:3 molar ratio of toluene to benzene. The vapour pressure of benzene and toluene at 25 o C are 28 and 95 bar respectively. The mole fraction of toluene vapour is (1) 0.693 (2) 0.326 (3) 0.548 (4) 0.855
16. 6g. of a non volatile solute is dissolved in 90g of water, such that the lowering in vapour pressure is 2%. The molecular weight of the solute is (1) 65 (2) 92 (3) 60 (4) 80
17. Solubility of a gas in water at STP is 0.2m. Henry’s constant of the gas is equal to
(1) 278 bar (2) 556 bar
(3) 369 bar (4) 171 bar
18. The normal boiling point of water is 373 K (at 760 mm). Vapour pressure of water at 298 K is 23 mm. If enthalpy of vapourisation is 40.656 KJ/mol, the boiling point of water at 23 mm atmospheric pressure will be
(1) 250 K (2) 51.6 K
(3) 294 K (4) 12.5 K
19. Given below are two statements. one is labelled Assertion (A) and the other is labelled Reason (R).
(A) : A non volatile solute is added in liquid solvent the freezing point of mixture decreases.
(R) : Vapour pressure decreases by addition of non volatile solute, so equilibrium point where V.P. of solid and V.P. of liquid are equal can reach at lower temp
(1) Both (A) and (R) are true and (R) is the correct explanation of (A)
(2) Both (A) and (R) are true and (R) is not the correct explanation of (A)
(3) (A) is true and (R) is false
(4) (A) and (R) are false
20. An aqueous solution of 6.3g oxalic acid dihydrate is made upto 250 mL. the volume of 0.1 N NaOH required to completely neutralise 10 mL of this solution is
(1) 40 mL (2) 20 mL
(3) 10 mL (4) 4 mL
21. Which of the following solutions of complex compounds have maximum vapour pressure?
(1) [Co(NH3)6]Cl3
(2) [Co(NH3)5Cl]Cl2
(3) [Co(NH3)4Cl2]Cl
(4) [Co(NH3)3Cl3]
22. The phase diagram for the pure solvent (solid lines) and the solution (non volatilenon Electrolyte solute) (dashed lines) are recorded below: 1 atm P T L
The quantity indicated by L in the figure is (1) ∆P (2) ∆Tf
(3) Kfm (4) Kb. m
23. Henry’s law constant for the molality of methane in benzene at 298 K is 4.27 × 10 5 mm Hg. Calculate the solubility of methane in benzene at 298 K under 760mm Hg
(1) 1.78 × 10–4 (2) 1.78 × 10–3
(3) 1.78 × 104 (4) 1.78 × 10–6
24. The percentage composition by weight of an aqueous solution of a solute (molecular mass 150) which boils at 373.26k (kb= – 0.52) (approximately)
(1) 5.24 (2) 15 (3) 7.12 (4) 10.3
25. If two substances A and B have PP A o B o :: = 12 and also X A : XB in solutions as 1 : 2, then mole fraction of A in vapours is (1) 0.33 (2) 0.25 (3) 0.52 (4) 0.2
26. A 5% solution of cane sugar is isotonic with 0.5% of x the molecular weight of substance x is.
(1) 34.2 (2) 119.96
(3) 95.58 (4) 126.98
27. Which of the following aqueous solution has the least boiling point?
(1) 0.1 M KNO3
(2) 0.1 M Na3PO4
(3) 0.1 M BaCl2
(4) 0.1 M K2SO4
28. When 36.0 g of a solute having the empirical formula CH2O is dissolved in 0.8 kg of water, the solution freezes at -0.93 0C. What is the formula of the solute? (kf = 1.860C kg mol–1)
(1) C6H12O6 (2) C3H6O3
(3) CH2O (4) C2H4O2
29. A complex is represented as COCl 3 XNH 3 Its 0.1 molal solution in water shows ∆Tf = 0.558K, Kf for H2O is 1.86K molarity–1 Assuming 100% ionization of complex and co-ordination number of Co, is six, Calculate formula of complex.
(1) [CO(NH3)6]Cl3
(2) [CO(NH3)5 Cl]Cl2
(3) [CO(NH3)4 Cl2]Cl
(4) [COCl6]NH3
30. Correct statement about the following solutions (Assuming one mole of each solute is dissolved in 1 L solution)
(2) Boiling point : NaCl <(NH2)2 CO > Ca3 (PO4)2 < BaCl2
(3) Freezing point : NaCl >(NH2)2 CO > Ca3 (PO4)2 < BaCl2
(4) Freezing point : NH 2) 2 CO > NaCl < Ca3(PO4)2 < BaCl2
31. FeCl 3 on reaction with K 4 [F e (CN) 6 ] in aq. Solution gives blue colour. These are separated by a semipermeable membrane PQ as shown. Due to osmosis there is:
(4) No blue colour formation
32. Match the following List-I List-II
(A) Van’t Hoff (i) Cryoscopic constant factor
(B) Kf (ii) Isotonic solutions
(C) Solutions with same Osmotic pressure (iii) Normal molar mass / Abnormal molar mass
(D) Azeotropes (iv) Solutions with same composition in vapor and liquid phase
Choose the correct answer from the options given below
(A) (B) (C) (D)
(1) iii i ii iv
(2) iii ii i iv
(3) iii i iv ii
(4) i iii ii iv
33. At a certain temperature, the value of the slope of the plot of osmotic pressure( π ) against concentration C in moL–1 of a certain polymer solution is 291 R. The temperature at which osmotic pressure is measured, is (R is gas constant)
(1) 271°C (2) 18°C
(3) 564 K (4) 18 K
34. Which of the following solution has the highest freezing point?
(1) 0.1 mol KCl in 1 Kg water
(1) Blue colour formation in side X
(2) Blue colour formation in side Y
(3) Blue colour formation in both of the sides X and Y
(2) 0.1 mol K2SO4 in 1 Kg water
(3) 0.1 mol Urea in 1 Kg water
(4) 30 Kg of glucose in 1 Kg water
35. The units of the quantity of the factor DT K f f are
(1) Moles / Lit
(2) Moles / Kg
(3) Equivalents / Lit
(4) Equivalents / Kg
36. Solubility curve of Na2SO4.10H2O in water with temperature is given as Temp 34oC
Solutility (gl -1 )
From the Figure, we can say that
(1) Solution process is exothermic
(2) Solution process is exothermic till 34°C and endothermic after 34°C
(3) Solution process is endothermic till 34°C and exothermic thereafter
(4) Solubility process is endothermic
37. X is a non- volatile solute and Y is a volatile solvent. The following vapour pressures are observed by dissolving X in Y
X/ mol–1 Y/mm of Hg
0.10 P1
0.25 P2
0.01 P 3
(1) P1 < P2 < P 3
(2) P 3 < P2 < P1
(3) P 3 < P1 < P2
(4) P2 < P1 < P 3
38. A solute has 2:3 molar ratio of toluene to benzene. The vapour pressure of benzene and toluene at 25°C are 28 and 95 bar respectively. The mole fraction of Toluene vapour is
(1) 0.658 (2) 0.326
(3) 0.548 (4) 0.855
39. The total vapour pressure of a 4 mole% solution of NH3 in water at 293k is 50.0torr. The vapour pressure of pure water is 17.0 torr at this temperature. Applying Henry’a law and Raoult’s laws, calculate the total vapour pressure for 5mole% solution
(1) 58.25 torr (2) 33 torr
(3) 42.1 torr (4) 52.25 torr
40. The vapour pressure of water at 20°C is 17.5mm Hg. If 18 gms of glucose is added to 178.2 gms of H 2O at 20°C, the vapour pressure of resulting solution is
(1) 17.325 mm Hg
(2) 15.750 mm Hg
(3) 16.5 mm Hg
(4) 17.675 mm Hg
41. Two 5 molal solutions are prepared by dissolving non-electrolyte, non-volatile solute separately in the solvents X and Y. The molecular weights of solvents are Mx and M y respectively where Mx= y 4 M 3 . The relative lowering of vapour pressure of the solution in “X” is “m” times that of the solution in”Y”. Given that the number of moles of solute is very small in comparison to that of solvent, the value of “m” is:
(1) 4/3 (2) 3/4
(3) 1/2 (4) 1/4
42. Ratio of O2 and N2 in the air is 1:4. Find out the ratio of their solubilities in terms of mole fractions of O2 and N2 dissolved in water at atmospheric pressure and room temperature.
KOtorr
KNtorr H H 2 7 2 7 33010 66010 . .
(1) 1 : 2
(2) 2 : 1
(3) 1 : 1
(4) None of t these
43. Two liquids A and B are mixed at temperature T in a certain ratio to form an ideal solution. It is found that the partial pressure of A, i.e.,P A is equal to P B the pressure of B for liquid mixture. What is the total pressure of the liquid mixture in terms of PP AB 00 and ?
44. Temperature Henr’s constant Pressure
273 K 600 atm 0.30 atm
333 K 3400 atm p2
If solutions of CO2 in H2O is heated from 273 to 333K, pressure (p 2) needed to keep CO2 in the solution is
ANSWER KEY
NEET
(1) 0.108 atm (2) 1.7 atm
(3) 0.212 atm (4) 0.468 atm
45. Two liquids A and B at a particular temperature (T) exert a pressure of 280 mm and 168 mm respectively. The solution of A and B with the mole fraction of A equal to 0.32 at the same temperature exerts a total vapour pressure of 376 mm. Which one of the following is correct based on the above data?
(1) Partial pressure of A is 190.4 mm.
(2) Partial pressure of B is 53.76 mm.
(3) For the above solution to be ideal, total vapour pressure should be 203.84 mm
(4) Above solution is non-ideal solution with negative deviation
Electrical current flows in metals without changes occurring in metals. Solutions of acids, alkalies and salts conduct electricity, but material changes occur during conduction. Electrochemistry is mainly concerned with the relation between electrical energy and chemical energy.
2.1 ELECTRICAL CONDUCTIVITY
The substance that does not allow the flow of electricity through it is called an insulator.
S.No
1. Conductivity is due to mobility of free electrons from negative to positive end
Diamond is a familiar example of an insulator. Rubber, glass, plastics, etc., are other examples of insulators.
Electric current is considered in general as the flow of electric charges through the conducting medium. This medium is termed as an electrical conductor. Electrical conductor is defined as the substance that allows electric current to pass through it. Electrical conductors may be solid metals, fused metals, molten salts, or aqueous solutions of acids, bases, and salts. Electrical conductors are broadly classified into two types.
Electronic conductors: Metals, alloys, graphite, and salts like cadmium sulphide, copper sulphide, etc., are examples of electronic conductors.
Electrolytic conductors: Fused salts and aqueous solutions of acids, alkalies, salts, etc., are examples of electrolytic conductors. The main difference between the two types of electrical conductors are listed in Table 2.1
Conductivity is due to mobility of free ions towards oppositely charged electrodes
2. During conduction, matter is not transferred During conduction, matter is transferred in the form of ions
3. Passage of current may bring only physical changes.
4. The conductivity power is high
5. An increase in temperature decreases conductivity due to increased thermal vibration of particles or resistance increases
Passage of current brings physical as well as chemical changes.
The conductivity power is relatively less.
With an increase in temperature, conductivity increases due to the increase in the degree of ionisation and increase in the mobility of ions.
Table 2.1 Differences in the properties of two types of conductors
In electrolytic conductors, the flow of current is due to the movement of ions in the solution under the influence of an applied field. The mobility of ions under applied voltage is called electrical migration or mobility or conductivity of ions.
Electronic conduction is also called metallic conduction. It depends upon nature, structure, density, temperature, and valency of the metal. It also depends upon the composition of the alloy formed between metals.
2.1.1 Electrolytic Dissociation
Electrolyte is defined as a substance that conducts electric current in molten or in solution state and is simultaneously decomposed. There are substances that do not conduct electricity even in the solution state. Such substances are called non-electrolytes. Cane sugar, urea, glucose, etc., are examples of non-electrolytes. The important points of the theory of electrolytes are:
1. An electrolyte in aqueous solution breaks up into two oppositely charged ions and the ions are in equilibrium with the unionised electrolyte, BA B+ + A–
2. The process of splitting of the electrolyte into ions is called ionisation or dissociation. The fraction of total number of molecules present as ions in the solution is called degree of dissociation. It is den oted by ‘ a ’.
Numberofmoles ofelectrolyte asions
Totalnumberofmoles of electrolyte α=
3. The degree of dissociation depends upon nature of the electrolyte, concentration of the electrolyte and temperature.
4. The equilibrium constant for the dissociation of electrolyte (K) is related to the concentration (C) as K = Ca2/(1–a).
5. When an electric current is passed through the solution of electrolyte, the ions move
towards electrodes of opposite charge and are discharged.
6. The degree of dissociation of electrolyte increases with decrease in concentration. It approaches unity at infinite dilution. So, at infinite dilution, all electrolytes are said to be completely ionised.
7. Degree of ionisation increases with increase in temperature.
8. Theory of electrolytes is applicable for weak electrolytes only.
2.2 ELECTROLYTIC CONDUCTANCE
Conductivity of electrolytic solution depends on the nature of eletrolyte, concentration, and temperature of the solution.
Electrolytic solutions also obey Ohm’s law in a manner similar to electronic conductors. A conductivity cell is shown in Fig.2.1.} Platinum wires Black Platinum plates
Electrolyte Area, a Distance, l
Fig. 2.1 Conductivity cell
The resistance ‘R’ to the flow of current offered by the solution of electrolyte in the cell is inversely proportional to the area of cross section (a) of the electrode, and it is directly proportional to the distance of separation (l).
Resistance offered by the solution, () or ∝=ρ l RR a l a
Here, ρ is the specific resistance, also called resistivity. The ratio of length ( l) and area of
cross section (a) is called cell constant. Unit of cell constant is cm –1 or m–1.
Once the cell constant is determined, we can use it for measuring the resistance or conductivity of any solution. The set up for the measurement of the resistance is shown in Fig.2.2 .
Conductivity cell containing electrolyte
Fig. 2.2 Arrangement for measurement of resistance of a solution of an electrolyte
It consists of two resistances R 3 and R 4 , a variable resistance R 1 and the conductivity cell having the unknown resistance R 2. The Wheatstone bridge is fed by an oscillator O (a source of A.C. power in the audio frequency range 550 to 5000 cycles per second). P is a suitable detector (a headphone or other electronic device) and the bridge is balanced when no current passes through the detector. Under these conditions, unknown resistance
14 3 2 RR R R =
These days, inexpensive conductivity meters are available that can directly read the conductance or resistance of the solution in the conductivity cell. Once the cell constant and the resistance of the solution in the cell is determined, the conductivity of the solution is given by the equation * Cellconstant G k= = RR
The reciprocal of resistance is known as conductance (C).
1 11 / C Rla ==× ρ 1 Ck(or)kC (/a) a ==× l l
Here, k (kappa) is specific conductance, also called conductivity. It is the reciprocal of specific resistance. Resistance is expressed in ohm. The symbol of ohm is ‘Ω’. Conductance is expressed in ohm–1 or mho. In the SI system, ohm –1 is siemens. The symbol of siemen is ‘S’. Units of specific conductance are ohm –1cm–1 in CGS system and S m–1 in SI system. Specific conductance is the conductance of the electrolyte in unit volume of the solution. Conductance and specific conductance are one and the same if cell constant is unity.
The conductance of an electrolytic solution can be expressed as:
1. Specific conductance ( k)
2. Molar conductance ( Λm)
3. Equivalent condutance ( Λeq)
2.2.1 Specific Conductance or Conductivity (k)
The specific conductance of an electrolyte is defined as the conductance of the solution enclosed between two parallel electrodes of unit area of cross section separated by a unit distance or the conductance of the electrolyte in the solution of volume of 1 cm3 (CGS units) or the conductance of the electrolyte in the solution of volume of 1 m 3 (SI units).
In practice, the concentration of solutions of acids or bases or salts is expressed in molarity or normality. Conductance of electrolytic solution with reference to the concentration is of two types: molar conductance ( Λm) and equivalent conductance ( Λeq).
Molar Conductivity (Λm)
Molar conductivity of a solution at a dilution V is the conductance of all the ions produced from one mole of the electrolyte dissolved in V cm3 of the solution when the electrodes are
one cm apart and the area of the electrodes is so large that the whole of the solution is contained between them. It is usually represented by Λ m .
Relationship between Molar Conductivity and Conductivity
The molar conductivity of a solution is usually not found directly but is calculated from the specific conductivity. The relationship between molar conductivity and specific conductivity may be obtained as follows:
Consider a rectangular vessel with its two opposite walls one cm apart and made of some metal sheet so that these act as the electrodes.
Case-I : Suppose 1 cm 3 of the solution containing 1 mole of the electrolyte is taken in the vessel, the conductance of this solution will then be its conductivity. Further, if 1 cm3 of the solution taken contains one mole of the electrolyte, the conductance of the solution will be its molar conductivity. Thus, when 1 cm 3 of the solution containing one mole of the electrolyte is considered, the molar conductivity is equal to its conductivity.
Case-II : Suppose that 4 cm 3 of the solution containing one mole of the electrolyte is taken. The conductance of the solution will be still equal to its molar conductivity at this dilution but, now, there will be four cubes each of volume 1 cm3, as shown in Fig.2.3
molar conductivity, is four times the specific conductivity.
Therefore, the molar conductivity is related to the specific conductivity as follows: 10001000 Molarity mVmcc kVorkk c Λ=×Λ=×=×
where k is the conductivity and V is the volume of the solution containing one mole of the electrolyte and c is the molar concentration, i.e., mol L–1 (or mol dm–3).
Unit: The units of Λ m will be ohm–1 cm2 mol–1 or S cm2 mol–1 or Ω–1 cm2 mol–1 or S m2 mol–1 in SI units.
1 S m2 mol–1 = 104 S cm2 mol–1
Therefore, the formula, along with units of each quantity, may be written as
However, if molarity is expressed in mol L–1, the above formula becomes ()() 1 21 m 31 Sm Smmol 1000LmMolarity(molL) κ ∧= ×
Equivalent Conductivity (Λeq)
The conductance of each 1 cm 3 of the solution is equal to its conductivity so that the total conductance of the solution, i.e.,
Equivalent conductivity of solution at a dilution V is defined as the conductance of all the ions produced from one gram equivalent of the electrolyte dissolved in V cm3 of the solution when the distance between the electrodes is one cm and the area of the electrodes is so large that the whole solution is contained between them. Its symbol is Λ eq Equivalent conductivity = specific conductance × V Λ eq = k× V
Fig. 2.3 Molar conductivity – specific conductance relationship
If the solution has a concentration of c gram equivalents per litre, i.e., c gram equivalents are present in 1000 cm3 of the solution, then the volume of solution containing one gram
equivalent will be 1000 , c i.e., 1000 V c = .
Variation of Conductance with Concentration
As the ions are current conducting species, conductivity of an electrolyte depends on the concentration or number of ions present in a unit volume of the solution. Therefore, conductivity of solution increases with increase in concentration of solution. Equivalent conductance or molar conductance decreases with increase in concentration, as shown in Fig.2.4
increases. The increase in the conductivity of a weak electrolyte at lower concentrations is attributed to increase in the extent of dissociation. The variation in equivalent conductance with concentration is not significant in the case of strong electrolytes because strong electrolytes are almost completely ionised at all concentrations, and increase in eq Λ with dilution is only due to decrease in the inter-ionic forces. In case of weak electrolytes, the decrease in eq Λ is significant with increase in concentration.
Λ eq increases on dilution and reaches a maximum value at infinite dilution because degree of dissociation of weak electrolyte becomes nearly equal to 1.
The variation of equivalent conductance of an electrolyte with concentration depends on the type (valencies of ions) of electrolyte than on its nature.
■ For strong univalent electrolytes (KCl), the decrease in equivalent conductance () eq Λ with increase in concentration is not very large.
■ The decrease in equivalent conductance () eq Λ with concentration is more as the valency of the ions increases.
■ Weak electrolytes like CH3COOH, NH4OH exhibit an apparently different behaviour due to incomplete ionisation.
Fig. 2.4 Variations of equivalent conductivity with concentration
As the concentration decreases for any electrolyte, the equivalent conductance
The ratio of equivalent conductance at any concentration () C Λ and at infinite dilution () 0 Λ is called conductance ratio.
Table 2.2 Conductivity and molar conductivity of KCl solutions at 298.15 K
Degree of dissociation (or) conductance ratio c m o m Λ α= Λ or c m m . α Λ Λ For a weak electrolyte, the conductance ratio is equal to de gree of dissociation.
a is higher for 0.01 M CH 3COOH when compared to that for 0.1 M CH 3COOH.
For weak electrolytes, limiting equivalent conductance cannot be calculated from eq Λ vs graph, because, for weak electrolytes, at low concentrations, eq Λ increases sharply with decrease in the concentration.
For weak electrolytes, Λ 0 is indirectly calculated from the Kohlrausch law.
On the basis of k (specific conductance) value, electrolytes are of two types.
Equivalent conductance of an electrolytic solution at very low concentration is known as equivalent conductance at infinite dilution or at zero concentration or limiting equivalent conductance () o eq Λ . Direct determination of conductance at zero concentration of solution is not practically possible. The work of Debye and Huckel on the quantitative treatment of equivalent conductance marked the commencement of a new era in electrochemistry.
For strong electrolytes, Λ m vs c graph is a straight line with a negative slope as shown in Fig.2.5.
Intercept is Δ0 Slope=-veStrongelectrolyteWeakelectrolyte
c
Fig. 2.5 Variation of equivalent conductance with square root of concentration
For strong electrolytes, limiting equivalent conductance can be known by extrapolating m Λ vs c graph to zero concentration.
m0 Kc Λ=Λ−
In this equation, K is a constant. It depends on the type of electrolyte (1 : 1 or MX etc.)
1) Weak electrolytes: These electrolytes have low k value, e.g., carboxylic acids.
ClCH 2 COOH has higher k value than CH3COOH.
2) Strong electrolytes : These electrolytes have high k value. Example: salts, mineral acids, alkalies.
The conductance of an electrolyte is due to its ionisation. The extent of ionisation is maximum for weak electrolytes as dilution reaches to maximum.
Equivalent conductance of an electrolyte can be determined experimentally by the following method:
i. Determination of conductance by using conductance meter
ii. Calculation of specific conductance from conductance l ck a ×=
iii. Calculation of equivalent conductance
1000 × Λ= eq k N
1. Specific conductance is 0.0382 ohm–1 cm–1 for a solution that is 0.1 M in KCl and 0.2 M in MCl (here, M is a metal radical). If the ionic conductances of K+ and Cl– ions in the solution are 74 and 76 S cm2 mol–1, find the ionic conductance of M+ ion.
Sol. () o m KCl Λ = 74 + 76 = 150 S cm2 mol–1 = m 1000 0.1 × Λ= k 3 21 KCl
1. Water with conductivity, K = 7.5 × 10–8 Ω–1 cm –1 is used to make 0.01 M solution of acetic acid. Calculate the molar conductivity of acetic acid, when its conductivity is 1.465 × 10–4 Ω–1 cm–1 at this concentration. Ans: 14.64 S cm2 –1mol
TEST YOURSELF
1. Which of the following is a good conductor of electricity?
(1) Diamond (2) Graphite (3) Solid NaCl (4) Wood
2. Which of the following solutions of NaCl has the higher specific conductance?
(1) 0.001 N (2) 0.01 N (3) 0.1 N (4) 0.5 N
3. The specific conductance of 0.1 N KCl solution at 23ºC is 0.012 ohm–1 cm –1. The resistance of cell containing the solution at the same temperature was found to be 55 ohm. The cell constant will be (1) 0.142 cm–1 (2) 0.66 cm–1 (3) 0.918 cm–1 (4) 1.12 cm–1
4. The conductance of 0.1 M HCl solution is greater than that of 0.1 M NaCl. This is because
(1) HCl is more ionised than NaCl
(2) HCl is an acid whereas NaCl solution is neutral
(3) H+ ions have greater mobility than Na + ions
(4) H+ ions have lesser mobility than Na + ions
Answer Key
(1) 2 (2) 4 (3) 2 (4) 3
2.3 KOHLRAUSCH LAW
Kohlrausch studied the molar conductivities at infinite dilution Λ o m for a number of pairs of strong electrolytes, each pair having a common cation or anion. Then, he calculated the difference of these Λ o m values for each pair. It is observed that the difference between Λ o m values for each pair of sodium and potassium salts having a common anion was same, irrespective of what this anion was. Similarly, the difference in the Λ o m values for each pair of salts having the different anions and a common cation are the same, irrespective of what this cation was.
Table 2.3 Conductances of some ions at infinite dilution and 25°C
Kohlrausch concluded that each ion makes a definite contribution to the total molar conductivity of an electrolyte at infinite dilution, irrespective of the nature of the other ion of the electrolyte. This individual contribution of an ion towards the total molar conductivity of the electrolyte is called molar ionic conductivity. As a result of these studies, Kohlrausch, in 1876, put forward a generalisation, known after him as ‘Kohlrausch’s law’.
The limiting molar conductivity of an electrolyte (i.e., molar conductivity at infinite dilution) is the sum of the limiting ionic conductivities of the cation and the anion each multiplied with the number of ions present in one formula unit of the electrolyte.
Mathematically, o oo m forAB +−Λ=λ+λ xyxy
where, Λ o m is the limiting molar conductivity of the electrolyte, ο + λ and ο λ are the limiting molar conductivities of the cation (A y+) and the anion (Bx–), respectively.
For example,
ο Λ m for NaCl = +− οολ+λ NaCl
ο Λ m for BaCl2 = 2 2 +− οολ+λ BaCl
ο Λ m for Al2(SO4)3 = 32 4 23+− οολ+λ AlSO
In terms of equivalent conductivities, Kohlrausch’s law is defined as follows:-
ο Λ=λ+λoo eqca
where, ο λ c and λ o a are called the limiting ionic conductivities for the cation and the anion, respectively.
Relationship between molar conductivity and equivalent conductivity of electrolytes: Since normality = nf × molarity, eq m f 1 n Λ=Λ ()()() o o3o2 m24m m4 3 AlSO2Al3SO Λ+− =λ+λ ()() o m4 3 o eq24 3 f AlSO AlSO n Λ
() o m24 3 AlSO 6 Λ =
( n-factor of Al2SO4=6)
o m λ = molar ionic conductivity of the ion at infinite dilution or at zero concentration.
o o3o2 eq24 eq eq4 3 AlSOAlSO Λ+− =λ+λ
()()()
o eq λ = equivalent ionic conductivity of the ion at zero concentration or at infinite dilution.
2. The molar conductivities at infinite dilution of Ba(OH) 2 , BaCl 2 , and NH 4 Cl solutions are 457.6, 240.6, and 129.8 S cm 2 mol –1 , respectively. Find the value of ο Λ eq and ο Λ m of NH4OH solution.
2. The equivalent conductivities at infinite dilution of Ba(CN)2, HCl, and BaCl2 are 100, 400, and 120 S cm2 eq –1. If the equivalent conductivity of 0.1 M HCN solution is 3.8 S cm2 eq–1, find the degree of dissociation of HCN at this concentration.
Ans: 0.001
TEST YOURSELF
1. The molar conductivities NaOAc Λ and HCl Λ at infinite dilution in water at 25°C are 91.0 and 426.2 S cm2/mol, respectively. To calculate HOAc Λ , the additional value required is
(1) 2 HO Λ (2) KCl Λ
(3) NaOH Λ (4) NaCl Λ
2. Specific conductivity of 0.01 M solution of HX is 10-8 mho cm–1. Molar conductance of the solution is equal to X mho cm2 mole–1. Value of ‘X’ equal to
(1) 0.1 (2) 10
(3) 0.001 (4) 1
3. Conductance of Al+3 is ‘x’ Sm2 geq–1 and that -2 4 SO is ‘y’ Sm2 geq–1 Molar conductance of aluminium sulphate is ___ Sm 2 mol–1
(1) 6 + xy (2) x+y
(3) 6(x+y) (4) 2 x + 3 y
4. The conductivity of a 0.8 M solution of an electrolyte is 200 Ω –1 cm –1 . Its molar conductivity (in Ω–1 cm2 mol–1) is
(1) 2.5 × 104 (2) 2.5 × 105 (3) 0.25 (4) 2.5
5. Resistance of 0.2 M solution of an electrolyte is 50 Ω. The specific conductance of the solution is 1.4 m–1. The resistance of 0.5 M solution of the same electrolyte is 280 Ω. The molar conductivity of 0.5 M solution of the electrolyte in S m 2 mol–1 is (1) 5 × 103 (2) 5 × 102 (3) 5 × 10–4 (4) 5 × 10–3
6. Molar conductance of 0.01M weak monoprotic acid (HA) is 40 mho cm 2 mol–1. pH of the solution is (limiting molar conductance of HA is 400 mho cm 2 mol–1).
(1) 2 (2) 4 (3) 2.3 (4) 3
7. At 300 K, the conductivity of 0.01 mol dm-3 aqueous solution of acetic acid is 19.5 × 10–5 mho cm–1 and limiting molar conductivity of acetic acid at the same temperature is 390 mho cm2 mol–1. The degree of dissociation of acetic acid is
8. Molar ionic conductance of Ca2+ and 3 4 PO at infinite dilution are ‘x ’ mho cm 2 mol –1 and ‘y’ mho cm2 mol–1 respectively. Molar conductance of calcium phosphate at infinite dilution is ___ mho cm2 mol–1
(1) (x + y) (2) 3 x + 2 y
(3) 6 (x + y) (4) 32 6 xy +
Answer Key
(1) 4 (2) 3 (3) 3 (4) 2 (5) 3 (6) 4 (7) 2 (8) 2
2.4 ELECTROLYSIS
Sodium combines with chlorine directly to form sodium chloride. The reaction involves oxidation of Na to Na+ and reduction of Cl to Cl–. Thus, the reaction between a metal and a non-metal is described as a redox reaction.
The formation of sodium chloride from its elements is an example of spontaneous change. The process of getting back the elements from an electrolyte involves a non-spontaneous redox reaction. Such a process is made possible using electrical energy.
Electrolyte: A substance that is in the molten state or in the dissolved state, containing ions and functioning as an electrically conducting medium, is called an electrolyte.
Electrolytic conductors are called electrolytes. Salts in molten state, like NaCl, KCl, fused hydrides and halides of Ag, Ba, Pb. a-form of Ag2S, solutions of NaCl, KCl, K2SO4, CuSO4, HCl, H2SO4, HNO3, NaOH, KOH etc. in water are electrolytic conductors. In general,
all fused salts or that which contain free ions are electrolytes.
Non-electrolyte: A substance that does not conduct electric current either in the molten state or in aqueous solutions is called as a non-electrolyte.
Examples: Urea solution, aqueous solution of sugar, non polar covalent carbon compounds
Electrolysis: Electrolysis is a non-spontaneous redox reaction which occurs under the influence of an applied electromotive force (EMF)
Electrolysis is performed in a device called the electrolytic cell. So, in this cell, electrical energy is used to carry out a redox reaction. Electrons enter into the cell at cathode and, hence, it is negatively charged. Electrons flow in the external circuit from anode to cathode.
Electrons Anode Cathode Electrolyte
Fig. 2.6 Experimental setup of electrolysis
Positively charged ions migrate towards the negative electrode and negatively charged ions migrate towards the positive electrode. A common setup of electrolysis is shown in Fig.2.6.
During electrolysis, the cations move towards the negative electrode (cathode) and the anions move towards the positive electrode (anode). They get discharged and get deposited or liberated.
Let BA be an electrolyte, containing B + and A – ions. During electrolysis, B + ions move towards cathode, whereas A– ions move towards anode.
At cathode: BeB +−+→ (reduction)
At cathode, gain of electrons or electronation or reduction takes place.
At anode: AAe →+ (oxidation)
At anode, loss of electrons or deelectronation or oxidation takes place.
During electrolys the electrons enter into the electrolyte at cathode, and they are taken up by the cations. The electrons lost by the anions leave the electrolyte at anode. As a result, ions are converted into atoms or molecules and get deposited or liberated at respective electrodes.
The minimum potential that is required for performing electrolysis and subsequent discharge of ions at their respective electrodes is called discharge potential. It varies from one ion to another ion.
When a solution contains two or more anions and cations, then the ion with the lower discharge potential will get discharged first.
Discharge potentials of anion:
Order of deposition of anions:
Decreasing order of discharge potentials for various cations:
The process is illustrated by taking the electrolysis of molten sodium chloride as an example. When electric current is passed through molten sodium chloride, the Na+ ions move towards the cathode and the Cl– ions move towards the anode.
2NaCl2Na2Cl+−→+
At cathode (reduction): 2Na2e2Na +−+→
Sodium ions are reduced and sodium metal (Na) is deposited at cathode.
At anode (oxidation): 2 2ClCl2e →+
Chloride ions are oxidised and Cl 2 gas is liberated at anode.
Examples and Applications
Electrolysis of fused and aqueous solutions with inert electrodes is used for producing generally metal at cathode and non-metal at anode. Though hydrogen is a non-metal, it is electrolytically collected at cathode from aqueous solutions. However, hydrogen is liberated at anode during the electrolysis of fused metal hydrides.
The electrolysis of aqueous sodium chloride solution differs from that of molten sodium chloride. Ionisation of aqueous salt solution produces Na+ and Cl– ions. Na+ ions travel to the cathode but cannot undergo reduction to metal at platinum cathode in aqueous solutions. Water preferentially undergoes reduction to give hydrogen gas.
22 2HO2eH2OH +→+
(Reduction or electronation at cathode)
Chloride ions travel to the anode and undergo oxidation to give chlorine gas.
2 2Cl2eCl →+
(Oxidation or de-electronation at anode)
∴ In electrolysis, concentrated aqueous NaCl, H2 gas is obtained at cathode and Cl2 is obtained at anode. But Na+ and OH– ions remain in the solution. So, the pH increases during electrolysis.
However, on prolonged electrolysis, when most of the chloride ions are oxidised, water is oxidised finally to liberate oxygen gas at anode.
Electrolysis of water, using platinum electrodes liberates hydrogen at cathode and oxygen at anode. Water is a very
poor electrolyte. In order to increase the conductivity, a small amount of an electrolyte is dissolved in water during electrolysis.
At cathode: 22 4HO4e2H4OH +→+
At anode: 2H2O → 4e– + O2 + 4H+
6H2O→ 2H2 + O2 + 4H2O
The overall reactions is:
→+
Thus, for the electrolysis of 2 moles of water,
4 faradays of electricity are required.
The products of electrolysis at inert electrodes are the same using dilute sulphuric acid solution, using dilute caustic soda solution or using an aqueous potassium sulphate solution as an electrolyte.
Electrodes that do not involve in chemical reaction during electrolysis are called inert electrodes.
Example: Graphite, platinum, etc.
Active electrodes involve in chemical reaction during electrolysis. In general, anode metal dissolves in solution. Example: Cu anode in aqueous CuSO4
When the aqueous solutions of salts of alkali metals, alkaline earth metals, Al +3, Zn+2, Fe+2 and Ni+2 etc., are electrolysed by using inert electrodes, at cathode, H2 is obtained by the reduction of either H + or H 2O molecules in preference to metal ions.
22 2HO2eH2OH +→+ or 2(g) 2H2eH +−+→
When the aqueous solutions of SO 4 –2 , NO 3 – , and PO 4 –3 are electrolysed by using inert electrodes at anode, H 2 O molecules undergo oxidation in preference to oxoanions to give O2.
Pt 22 2HOO4H4e+−→++
2.4.1 Faraday’s Laws
NOnooxidation
Pt 3
Pt 2 4
SOnooxidation
Faraday put forward two laws regarding the quantity of electricity passed through electrolytes and the amount of substance deposited or liberated at electrodes.
POnooxidation
Pt 3 4
A metal salt solution is electrolysed by using the same metal electrodes.
At cathode, metal is deposited due to reduction of metal ion and at anode metal ions are obtained due to the oxidation of anodic metal.
Example: Electrolysis of CuSO 4 with Cu electrodes
() 22 4 4aq CuSOCuSO+−→+
At cathode: 2 Cu2eCu +−+→
At anode: 2 CuCu2e+−→+
Some examples of electrolysis reactions and the products obtained at the respective electrodes are summarised in Table 2.3.
These laws are also known as fundamental laws of electrochemistry and they give quantitative relationships of electrolysis. These laws are applicable to all electrolytes at all the temperatures, pressures, and other conditions.
First Law
The amount of substance deposited or liberated or dissolved at an electrode during electrolysis is directly proportional to the quantity of electricity passing through the electrolyte.
The first law is expressed as m ∝ q m=eq . . . . . (1) where m = weight of the substance in grams q = quantity of charge in coulombs q = c × t
2HO2eH2OH +→+ Fused NaOH Pt 22 4OH2HOO4e →++
NaeNa +−+→ Aq. NaOH Pt 22 4OH2HOO4e →++
22 2HO2eH2OH +→+
Aq. HCl Pt 2 2ClCl2e →+ Pt 2 2H2eH +−+→
Aq. H2SO4 Pt 22 2HOO4e4H−+→++ Pt 2 2H2eH +−+→
50% H2SO4 Pt 4228 2HSOHSO2e →+ Pb 2 2H2eH +−+→
Aq. K2SO4 Pt 22 2HOO4e4H−+→++
Fused CuCl2 Pt 2 2ClCl2e →+
Aq.CuCl2 Pt 2 2ClCl2e →+
Aq.CuSO4 Pt 22 2HOO4e4H−+→++
Fused NaH Pt 2 2HH2e →+
Aq.AgNO3 Pt 22 2HOO4e4H−+→++
Pt 22 2HO2eH2OH +→+
Pt 2 Cu2eCu +−+→
Pt 2 Cu2eCu +−+→
Pt 2 Cu2eCu +−+→
Pt NaeNa +−+→
Pt AgeAg +−+→
Table 2.3 Electrolysis of some substances and the respective products
c = current strength in amperes
t = time in seconds
m ∝ c×t (or) m = e × c × t ......(2)
where ‘e’ is a constant known as electrochemical equivalent, and it is the characteristic of the substance deposited. If q = 1 coulomb, then e = m.
Electrochemical equivalent may be defined as the mass of substance deposited by passing one coulomb of electricity through the electrolyte. It is also defined as the mass of substance deposited by passing a current strength of one ampere for one second. The units of electrochemical equivalent are gram/ amperes or gram /coulomb. The SI unit of electrochemical equivalent is kg coulomb –1 .
Second Law
If the same quantity of electricity is passed through different electrolytes, then the weight of substances deposited at the respective electrodes are in the ratio of their chemical equivalents. The second law is mathematically expressed as, w ∝E, where ‘E’ is the equivalent weight of the substance.
If the same quantity of electricity is passed through two or more different electrolytes, the second law is applied. Let the weight of the substance ‘A’ be wA and equivalent weight of ‘A’ be EA. Let the weight of the substance ‘B’ be wB and equivalent weight of ‘B’ be EB. Then, according to the second law,
(3)
All the above 1, 2 and 3 equations are valid only when current efficiency is 100% i.e., η = 1
current efficiency
Actual mass 100 Theoreticalmass η=×
Faraday is defined as the quantity of electricity required to deposit or dissolve one gram equivalent weight of a substance.
From first law m e q =
From second law E e F =
Therefore, mE e qF ==
The total charge present on one mole of electrons is known as faraday.
Both coulomb and faraday are the units of quantity of charge. Coulomb represents a smaller quantity of charge and faraday represents a larger quantity of charge.
One faraday is usually taken as 96500 coulomb. Units of faraday are coulomb/mole.
Electrochemical equivalent (e) is
Equivalentweight e 96,500 = (or)
Atomicweight e n96,500 = ×
Here, n is the valency of the element. Faraday’s laws of electrolysis are useful for the determination of the amounts of substances deposited, the quantity of electricity, the number of electrons passing through the electrolyte, equivalent weights, atomic weights and valency.
3. Electrolysis of dilute aqueous solution of NaCl was carried out by passing 0.01 Amperes of current. How much times is required to liberate 0.01 moles of H 2 gas at cathode in the electrolysis.
Sol. –22H2eH ++→
One mole of H2 gas is liberated with 2F charge. The charge required for 0.01 moles H2 = 0.01 × 2F
q = 0.01 × 2 × 96,500 C
q = i × t = 0.01 × 2 × 96,500 C
0.01296500C t 296500C 0.01 ×× ==×
Try yourself:
3. For converting 54 g of water into H 2 and O2 gases, how long a steady current of 96.5 amperes should be passed into the water? (Report the answer in minutes)
Ans: 100 minutes
TEST YOURSELF
1. When an electric current is passed through acidified water, 112 mL of hydrogen gas at N.T.P was collected at the cathode in 965 seconds. The current passed in ampere is (1) 1.0 (2) 0.5 (3) 0.1 (4) 2.0
2. Which among the following has highest electrochemical equivalent?
(1) Ca (2) Na (3) Mg (4) Al
3. The electric charge for electrode deposition of one equivalent of the substance is (1) one ampere per second
(2) 965000 C per second
(3) one ampere per hour
(4) charge on 1 mole of electrons
4. How many electrons are delivered at the cathode during electrolysis by a current of 1 amp in 60 seconds?
(1) 3.74 × 1020 (2) 6.0 × 1023
(3) 7.48 × 1021 (4) 6.0 × 10020
5. Electric charge on 1 g ion of N 3– is (1) 4.8 ×10–19 C
(2) 10 × 1.6 × 10–19 C (3) 1.6 × 10–19 C (4) 2.89 × 105 C
6. A silver cup is plated with silver by passing 965 coulombs of electricity. The amount of silver deposited is (1) 9.89 g (2) 107.87 g (3) 1.0787 g (4) 1.002 g
7. One coulomb of charge passes through solutions of AgNO3 and CuSO4. The ratio of the amounts of silver and copper deposited on platinum electrodes used for electrolysis is
8. Dilute nitric acid on electrolysis using platinum electrodes yields (1) both oxygen and hydrogen at cathode (2) both oxygen and hydrogen at anode (3) H2 at cathode and O2 at anode (4) oxygen at cathode and H2 at anode
9. When 0.1 mole 2 4 MnO is oxidised, the quantity of electricity required to completely oxidize 2 4 MnO and 4 MnO is (1) 96500 C (2) 2 × 96500 C (3) 9650 C (4) 96.50 C
10. One faraday of electricity will liberate one gram atom of the metal from the solution of (1) CuCl2 (2) CuSO4 (3) AgNO3 (4) AuCl3
Electrochemical cell is a device in which electrical energy is produced at the expense of chemical energy.
2.5.1 Voltaic Cell
When plates of two dissimilar metals are placed in a conducting liquid, such as an aqueous solution of a salt, the resulting system becomes a source of electricity. When a piece of zinc is kept in copper sulphate solution for some time, the zinc piece turns red. This is due to the deposition of copper on zinc. The reaction is given as
Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)
In this redox reaction, two electrons are transferred from zinc atom to cupric ion. However, these electrons cannot be trapped to obtain any small quantity of electrical energy because they are directly transfering from zinc to Cu+2 within the vessel. When the reactants are separated from each other and given an indirect contact, the electrical energy can be trapped.
A device in which electrical energy is generated by performing a redox reaction is called galvanic cell. this cell is also called voltaic cell or electrochemical cell.
So, chemical energy is converted into electrical energy in a galvanic cell. A galvanic cell can be reversed to function as an electrolytic cell. A typical galvanic cell is shown in Fig.2.7
Zinc anode
Copper cathode
Zinc sulphate solution Coper sulphate solution
Fig. 2.7 Galvanic cell
Comparison between the two types of cells are given in the Table 2.4.
Electrolytic cell
A galvanic cell consists of two half cells or electrodes. One electrode contains zinc metal plate dipped in ZnSO4 solution and the other is copper plate in CuSO 4 solution. The two half cells are joined by a salt bridge. In Daniell cell, salt bridge is replaced by a porous pot.
When the Zn and Cu electrodes are joined externally by a wire, the following observations are made:
1. There is a flow of electric current through the external circuit.
2. ‘Zn’ rod loses its mass, while ‘Cu’ rod gains the mass.
3. The concentration of ZnSO 4 solution increases, while the concentration of CuSO4 solution decreases.
4. The solutions in both the compartments remain electrically neutral.
The standard emf of a Daniell cell is 1.1 V. If an external potential of 1.1 V is applied on this cell, there will be no chemical reaction and no flow of electrons. If the external potential is less than 1.1 V, the spontaneous redox reaction is continuous. If the external potential is more than 1.1 V, the cell functions as an electrolytic cell and the electrical energy is used to carry out non-spontaneous chemical reactions. Functioning of Daniell cell with applied external voltage is shown Fig.2.8(a) Fig.2.8(b)
Electrochemical cell
1. Electrical energy is converted into chemical energy. Chemical energy is converted to electrical energy
2. Anode is positive electrode. Cathode is a positive electrode.
3. Cathode is negative electrode. Anode is a negative electrode.
4. Oxidation occurs at anode. Oxidation occurs at anode.
5. Reduction occurs at cathode. Reduction occurs at cathode.
6. Electricity flows from the circuit to the cell. Electricity flows from the cell to the circuit.
7. Irreversible and non spontaneous. Reversible and spontaneous at irreversible.
8. Ions are discharged at both the electrodes. Ions are discharged only on cathode.
Table 2.4 Comparison between the two types of cells
Fig. 2.8 Daniell cell with applied external voltage (a) less than 1.1 V and (b) more than 1.1 V
Chemical Reactions in a Galvanic Cell
Zinc goes into solution as Zn2+ ions in the half cell in the left hand side.
2 Zn(s)Zn(aq)2e +−→+
Since the reaction involves the loss of electrons, it is known as oxidation half reaction. This electrode is known as anode. The electrons from the Zn electrode flow through the external wire to the Cu electrode. Cu+2 ions of the CuSO4 solution take up these electrons to form Cu atoms.
Cu2+(aq) + 2e– → Cu(s)
Since this reaction involves gain of electrons, it is known as reduction half reaction. This electrode is known as cathode.
At anode: 2 ZnZn2e+−→+ (oxidation)
At cathode: 2 Cu2eCu +− +→ (reduction)
The net cell reactions is Zn + Cu+2 → Zn+2 + Cu
Voltmeter is used to measure the potential difference between the two electrodes. In the Daniel cell, the salt bridge is replaced with a porous pot.
The electrical neutrality of the Daniell cell is maintained by the porous vessel.
Salt Bridge
Salt bridge is a U-shaped table. It is filled with a strong electrolyte, such as KCl, NaCl, KNO3 etc.
It must be inert towards the chemicals used in the electrochemical cells. The cations and anions of the electrolyte of the salt bridge must have same migratory speed.
In the salt bridge the electrolytes are held as a gel like agar-agar.
In the absence of salt bridge, ions of electrolytes of galvanic cell are accumulated in the half cells. Then, electrons flow is stopped and the cell stops functioning.
When salts bridge is connected, it supplies its ions to the half cells of galvanic cell and maintain electrical neutrality.
Important functions of salts bridge are: (1) It completes the electrical circuit (2) It maintains electrical neutrality (3) It minimises the liquid – liquid junction potential.
Line Notation for Electrochemical Cells
1. The line notation for the anodic half cell is written on the left and that for the cathodic half cell on the right.
Representation of Daniell cell is Zn(s)/Zn2+(C1)||Cu+2(C2)|Cu(s) C1 and C2 are the molarities of the Zn +2 and Cu+2 in the electrolytic solutions.
2. A single vertical line or comma indicates a change in state or phase.
3. A double vertical line is used to indicate the elimination of liquid junction potential (using salt bridge) between the half cells.
4. When gases are involved, their pressures are written in parentheses.
Representation of hydrogen – oxygen cells is
Pt, H2 (P1) |HCl(C)| O2 (P2), Pt.
This indicates that HCl solution with concentration, C acts as an electrolytic in both H2-half cell and O2-half cell. P1 and P2 are pressures of H2 and O2 respectively.
4. What is the line representation of the cell for the cell reaction? (The molarity of the solution is 1.0 M and pressure of the gas is 1.0 gas)
2AgClH2HCl2Ag +→+
()()()() s2g aq s
Sol. In the given cell H2 is oxidised to H+ ions. AgCl is reduced to Ag(s) and Cl–. Platinum is required for electrical constant.
4. What are the half cell reactions possible for the electrochemical cell: Pt (s)| () Br2 |Br–(0.01M)||H+(0.03M|H2(g)(1.0 bar)|Pt(s)|H2(g) (1.0 bar)?
In this half cell, inert electrode is dipped in the solution containing ions of a species having two different oxidation states.
Examples: i. Pt/Fe+3, Fe+2
Electrode reaction: Fe +3 Fe+2
ii. Quin hydrone electrode, Pt/Q, QH 2
Electrode reaction:
Q + 2H+ 2e QH2
Here,
Q = Quinone, QH2 = hydroquinone
Single Electrode Potential
In each half cell, there is a tendency of a metal to go into the solution as ions, leaving behind electrons at the electrode. At the same time, metal ions of the solution show tendency to deposit on the metal electrode. At equilibrium, there is a separation of charges between metal and the solution. As a result, a potential difference exists between them. The electrical potential difference set up between the metal and its ions in the solution is called electrode potential (i.e., single electrode potential).
If the potential is developed due to the tendency of electrode to undergo oxidation, it is called its oxidation potential. If the potential is developed due to the tendency of the of electrode to undergo reduction, it is called its reduction potential. For a given electrode, if oxidation potential is (V) volts, its reduction potential will be (–V) volts. Electrode potential is an intensive property.
Electrode potential depends upon the
1. nature of the metal (or non-metal)
2. concentration of the electrolyte
3. temperature of the electrolyte solution
4. pressure in case of gas electrode
5. number of electrons involved per atom.
2.5.3 Standard Electrode Potential (E°)
The potential of an electrode measured at standard conditions of 25°C temperature and 1 bar pressure, and using ions at unit concentration, is known as standard electrode potential. It is denoted by ‘E°’.
For any half cell or single electrode, standard reduction potential = – (standard oxidation potential) For example, 2 Zn2eZn; +−+→ the standard reduction potential is – 0.76 V. 2 ZnZn2e; +−→+ the standard oxidation potential is + 0.76 V.
2.5.4 Measurement of Electrode Potentials
The absolute value of the electrode potential of a single electrode (i.e., single electrode potential or half cell potential) cannot be determined. Because no net reaction occurs in a half cell. It can be measured using some electrode as a reference electrode.
The hydrogen electrode is primary reference electrode for measuring single electrode potential. The construction of standard hydrogen electrode is shown in Fig.2.9.
The standard hydrogen electrode (SHE) consists of a platinum plate dipped into a solution of protons with one molar concentration at 25°C. Hydrogen gas is passed over the Pt plate at one atm. The half cell reaction is written as 2H + (aq) + 2e– → H2(g)
H2(g) (at 1 atm)
Pt black electrode
H3O+(aq) (1M)
2.9 Standard hydrogen electrode
The half cell is represented as, Pt, H +(1M)/ H2(1atm), E°= 0.0 volts. The potential of SHE or NHE is arbitrarily taken as zero.
It is a reversible electrode with respect to H+ ion, i.e., it can act like a cathode or anode.
1 2 2 HHe+− →+− when acting like anode
2 1 HeH 2 +−+→− when acting like cathode
Fig.
Secondary reference electrodes like quinhydrone electrode and calomel electrode are also used now-a-days. Single electrode potentials are measured in a potentiometer using reference electrodes (for SHE, [H+]=1M).
For the determinations of electrode potential of any electrode, an electrochemical cell is set up using the electrode or the half cell under study and standard hydrogen electrode. The emf of the cell is measured. It gives electrode potential of the electrode under study.
For example, the determination of standard electrode potential of Zn|Zn+2(1M) and SHE is set up. Then, voltage is recorded (it comes out to be 0.76 volts). In this cell, one can observe that the concentration of Zn+2 increases, while that of H+ decreases. Thus, the cell reaction is Zn + 2H+ → Zn+2 + H2. Hence, zinc electrode acts as anode.
EMF of cell = (standard reduction potential of cathode)
—(Standard reduction potential of anode)
0.76 = 2 H/HZn/Zn EE++ οο
0.76 = 2 ZnZn 0E + ο
Therefore, 2 Zn/Zn E + ο = – 0.76 volts.
Similarly, when standard potential of copper electrode is to be measured, a cell is set up with copper half cell and SHE. Then, the measured voltage is expected to be 0.34 volt. In the cell, one can observe that the concentration of Cu+2 decreases while the concentration of H+ increases. Thus, the cell reaction is
Cu+2 + H2 → Cu + 2H+
Hence, the coper electrode acts a cathode.
EMF of the cell, 0.34 V = 2 Cu/CuH/H2 EE++ οο
Therefore, 2 Cu/Cu E0 + ο = + 0.34 V
In electrode – SHE couple, if the electrode undergoes oxidation, the standard reduction potential of the electrode is negative.
In this manner, the standard electrode potentials of a large number of electrodes were determined using SHE as a reference electrode. These electrode potentials are relative standard half cell potentials.
TEST YOURSELF
1. In Daniell cell, the process that occurs at anode is
(1) ()() 2 aq s Zn2eZn +−+→
(2) ()() 2 CusCuaq2e +−→+
(3) ()() 22 ZnCuaqZnaqCu ++ +→+
(4) ()() 2 ZnsZnaq2e +−→+
2. At 298 K, reduction potential of H-electrode is –0.118 V. Hydroxyl ion concentration of the solution in which the H-electrode is immersed is ____M
(1) 10–10 (2) 10–9 (3) 10–12 (4) 10–11
3. Saturated solution of KNO 3 is used to construct ‘salt – bridge’ because (1) velocity of K + is greater than that of NO3
(2) velocity of NO3 is greater than that of K+
(3) velocities of both K+ and NO3 are nearly the same
(4) KNO3 is highly soluble in water
4. Which of the following statements is correct w.r.t. both electrolytic cell and galvanic cell?
(1) In both cells, anode is shown by +ve sign.
(2) In both cells, cathode is shown by –ve sign.
(3) In both cells, reduction reaction takes place at the cathode.
(4) In both cells, oxidation reaction takes place at the cathode.
Answer Key
(1) 4 (2) 3 (3) 3 (4) 3
2.6 ELECTROCHEMICAL SERIES
The arrangement of electrodes in the order of increasing standard reduction potentials is called electrochemical series. This is also called activity series.
The negative sign of standard reduction potential indicates that when the electrode is connected with standard hydrogen electrode, it acts as the anode and oxidation takes place. A positive sign of standard reduction potential indicates that the electrode acts as cathode when connected with a standard hydrogen electrode. The activity series containing important half-cells is listed in Table 2.5.
The values of standard electrode potentials for some selected half-cell reduction reactions are given. It can be seen that the standard electrode potentials for fluorine is the highest, indicating that fluorine gas (F 2 ) has the maximum tendency to get reduced to fluoride ions (F–). Therefore, fluorine is the strongest oxidising agent and fluoride ion is the weakest reducing agent.
Lithium has the lowest electrode potential, indicating that lithium ion is the weakest oxidising agent, while lithium metal is the most powerful reducing agent in an aqueous solution. It may be seen that as we go from top to bottom in Table 2.5, the standard electrode potential increase.
Thus, the tendency to undergo the corresponding reaction (reduction) decreases, the reducing power of the particles decreases.
2.6.1 Applications of Electrochemical Series
With the help of electrochemical series, one can understand the spontaneity of certain reactions and one can predict the occurrence of metal-salt solution reactions. Using these series reducing and oxidising abilities of elements can be estimated.
Chemical Reactivity of Metals
The metal that has high negative value of standard reduction potential readily loses electrons and it is converted into cations. Such a metal is said to be chemically very active. The chemical reactivity of metals decreases from top to bottom in the activity series.
Relative Strength of Oxidising and Reducing Agents
The reducing ability decreases from top to bottom in the electrochemical series. The oxidation ability increases with increase in the reduction potential.
Oxidising power ∝ reduction potential
Reducing power ∝ oxidation potential
Fluorine, with highest reduction potential, is the powerful oxidising agent. Lithium, with least reduction potential, is the powerful reducing agent in aqueous medium based on the potentials.
A given metal in activity series can reduce all the metal ions present below it. A given metal ion can oxidise all metal atoms present above it. A metal can reduce all the non-metals present below it in the activity series.
For storing a salt solution a metal container with smaller SRP value should not be used. Instead a metal container with higher SRP can be used.
Examples: FeSO4 solution () 2 o Fe/Fe E0.44v + =− cannot be stored in Zn container () 2 o Zn/Zn E0.76v + =− but it can be stored in a Cu container () 2 o Cu/Cu E0.34v +=+
A non-metal can oxidise all metal atoms present above it in the activity series. It can also oxidise non-metal anions present above it.
Table 2.5 Comparison between the two types of cells
S.No.
1. Li(aq)eLi(s) +−+→
2. K(aq)eK(s) +−+→ K+ / K – 2.93
3. 2 Ca(aq)2eCa(s) +−+→
4. Na(aq)eNa(s) +−+→
5. 2 Mg(aq)2eMg(s) +−+→ Mg2+/Mg – 2.37
6. 3 Al(aq)3eAl(s) +−+→ Al3+/Al – 1.66
7. 2 Zn(aq)2eZn(s) +−+→ Zn2+/Zn – 0.76
8. 2 Fe(aq)2eFe(s) +−+→
9. 2 Co(aq)2eCo(s) +−+→
– 0.28
10. 2 Ni(aq)2eNi(s) +−+→ Ni2+/Ni – 0.25
11. 2 Sn(aq)2eSn(s) +−+→ Sn2+/Sn – 0.14
12. 2 2H(aq)2eH(g) +−+→ Pt, H+/H2 0.00
13. 2 Cu(aq)2eCu(s) +−+→ Cu2+/Cu + 0.34
14. 2 I(s)2e2I(aq) +→ Pt, I2/I– + 0.54
15. Ag(aq)eAg(s) +−+→ Ag+/Ag + 0.80
16. 2 2 Hg(aq)2e2Hg(l) +−+→ Hg+/Hg + 0.85
17. 2 Br(l)2e2Br(aq) +→
18. 2 Cl(g)2e2Cl(aq) +→
19. 3 22 O(g)2H(aq)2eO(g)HO(l) +− ++→+
20. 2 F(g)2e2F(aq) +→
Displacement of Hydrogen from Dilute Acids by Metals
A metal with negative SRP can displace H+ from dilute acids to liberate H 2 gas.
A metal with positive SRP like Cu, Ag, Pt etc., cannot displace H+ from dilute acids to liberate H2.
Standard EMF of Cell
Br2/Br– + 1.07
Cl2/Cl– + 1.36
The sum of standard oxidation potential of anode and reduction potential of cathode is called standard cell emf or the difference in the reduction potentials of cathode and anode is also called cell standard emf of the cell, cell E ο ο=+ celloxred (anode)(cathode) EEE
But oxidation potential = –reduction potential
οο =− o cellcathodeanodeEEE
=− ooo cellrightleftEEE
In a similar way, the cell potential or emf of cells (Ecell) = Ecathode – Eanode (i.e., difference of reduction potentials)
5. Can PbO2 oxides Mn+2 according to the reaction: PbO2(s) + Mn+2 + 4H2O → Pb+2 + MnO4– + H+ Assume that all the reactants and products are in their standard states
(Given 2 Mn/MnO4 E1.51V +− ο =− and
2 2 PbO/Pb1.47V E + ο =+
Sol. For the given reaction, cellca EEE οοο =−
22 24 PbO/PbMnO/Mn EE+−+ οο =−
(difference of reduction potentials)
= + 1.47 – (+1.51) = – 0.04 V
Since, the potential of the cell reaction is negative, the given reaction is not spontaneous. In other words, the PbO2 cannot oxidise Mn+2
Try yourself:
5. Find the standard emf of the cell reaction, 22 34 27 2 CrO3Sn14H2Cr3Sn7HO −++++ ++→++
1. The standard reduction potentials for the following half-reactions are given against each at 298 K.
()() 2 aq s Zn2eZn;0.762 +−+=−
()() 3 aq s Cr3eCr;0.74V +−+=−
2H2eH;0.0V +−+=
()2g
() 32 maq aq FeeFe;0.77V +−++=+
Which is the strongest oxidising agent?
(1) Zn(s) (2) Cr(s) (3) H2(g) (4) Fe3+(aq)
2. The standard electrode potentials M/M E + ο of four metals A, B, C and D are –1.2 V, 0.6 V, 0.85 V, and –0.76 V, respectively. The sequence of deposition of metals on applying potential is
(1) A,B,C,D (2) B,D,C,A (3) C,B,D,A (4) D,A,B,C
3. Which of the following has least tendency to liberate H2 from mineral acids
(1) Cu (2) Mn
(3) Ni (4) Zn
4. Which of the following is true for electrochemical cell made with SHE and Cu electrode?
(Given 2 Cu/Cu E0.36V + ο=+ )
(1) H2 is cathode and Cu is anode
(2) H2 is anode and Cu is cathode
(3) Reduction occurs at H2 electrode
(4) Oxidation occurs at Cu electrode
5. Regarding the standard hydrogen electrode, the incorrect statement is (1) H2 gas is bubbled at 1 bar
(2) 2 M HCl is electrolyte
(3) SRP is arbitrarily taken as Zero. (4) It is a primary reference electrode.
6. Zn gives H2 gas with H2SO4 and HCl but not with HNO3 because (1) Zn acts as oxidising agent when reacts with HNO3.
(2) HNO3 is weaker acid than H2SO4 and HCl.
(3) In electrochemical series Zn is above H2. (4) NO3 is reduced in preference to H3O+
7. The standard reduction potentials of Cu +2 , Ag + , Hg +2 , and Mg +2 are +0.34 V, +0.80 V + 0.79 V, and –2.37 V respectively. With increasing voltage, the sequence of deposition of metals on the cathode from a molten mixture containing all those ions is (1) Ag, Hg, Mg, Cu (2) Cu, Hg, Ag, Mg (3) Ag, Hg, Cu, Mg (4) Cu, Hg, Mg, Ag
8. The chemical reaction
()()()() s2g aq s 2AgClH2HCl2Ag +→+ taking place in a galvanic cell is represented by the notation.
(1) ()()()()() s2g aq PtH1bar1MKClAgClAgss ,
PtH1bar1MHC1 , lMAgAg +
(2) ()()()()() s2g aq aqs
(3) ()()()()() s2g aq PtH1bar1MHClAgClAgss ,
(4) ()()()()() s2g aqs s PtH1bar1MHClAgAgCl , Answer Key
(1) 4 (2) 3 (3) 1 (4) 2
(5) 2 (6) 4 (7) 3 (8) 3
2.7 NERNST EQUATION (FOR ELECTRODE POTENTIALS)
Effect of concentration and temperature on electrode potentials is explained by Nernst equation.
The electrode potential required under different conditions (i.e., if the electrolyte ion is not with 1.0 M concentration or not with 1.0 bar pressure and temperature is not 298 K), can be obtained using Nernst equation.
2.7.1 For a Metal Electrode (or Half Cell)
The Nernst equation for the electrode reaction,
()() n aq s MneM +−+→
The electrode potential, ++
Here, n M/M E + ο = Standard reduction potential
R = gas constant (8.314 J K –1 mol–1)
T = Absolute temperature
n = number of electrons involved in the reaction
F = one Faraday = 96,500 C
For pure liquids or solids or gases at 1.0 atmospheric pressure, the molar concentration is taken as unity.
Therefore, [M] = 1 M/MM/M M/M RT1 EE F M 2.303RT1 E log F M ++ + ο + ο
nn n n n n n n at 298 K M/MM/M 0.0591 EE log M ++ ο + =−
For a gas electrode
If the electrode reaction is X2(g) + 2e – →2X–(aq)
Electrode potential is 22 2 // 2 2.303RT EE log F ο
6. What is the half cell potential of the half cell, Pt, Cl2 (2 atm)//Cl– (0.1 M). standard reduction potential of chlorine electrode is 1.36 volts.
Sol. For the given half cell, the reduction half reaction is ()()2g aq
Cl2e2Cl +
Since, temperature is not mentioned, the potential can be estimated at 298 K.
For the reaction
0.1 0.01 Q P22
0.059 1.3620.301 2
2 Cl/Cl E1.428Volts∴=+
Try yourself:
6. Estimate the reduction potential value of magnesium half cell containing 0.1 M solution of Mg+2 () 2 Mg/Mg E0.2363V + ο =−
Ans: –0.266 volts
2.7.2 Nernst Equation for a Galvanic Cell
For a cell reaction: ne aAbBcCdD +→+ cd
CD 2.303RT EE log nF AB
cellcell ab
at 298 K
cellcell ab CD 0.059 EElog nF AB
For a Daniell cell reaction:
2e 22 ZnCuZnCu ++ +→+ 2 cellcell 2 Zn 2.303RT EE log 2F Cu + ο +
(here cellcathodeanode EEE οοο =− )
7. Calculate the emf of the cell at 25°C, ()() 32 Cr/Cr0.1M//Fe0.01MFe ++
Given : 3 Cr/Cr E0.74V + ο =− and 2 Fe E0.44V + ο =−
Sol. In the given cell, the cell reaction is 6e 23 2Cr3Fe2Cr3Fe ++ ++
Q = 104
cellCa EEE οοο =− = (–0.44) – (–0.74 = 0.3 Volts
EMF of Cell, Ecell = cell 0.059 ElogQ n ο 4 cell 0.059 Elog10 6 ο () 0.059 0.34 6 =− cell E0.26volts∴=
Try yourself:
7. Estimate the cell potential at 298 K of the cell
Ag/Ag+(0.01 M)//Cl–(0.2M)/Cl2 (0.5 atm)/Pt
Given : Ag/Ag E0.8V + ο=+ and 2 Cl/Cl E1.36V ο=+ Ans: 0.17 volts
2.7.3 Equilibrium Constant from Nernst Equation
The cell reaction of the Daniell cell is ()()() 22 saq aq ZnCuZnCu ++ +→+
If the cell is functioning, the concentration of Zn +2 keeps on increasing, while the concentration Cu+2 keeps on decreasing. At the same time voltage of the cell slowly decreases. After some time, voltmeter gives zero reading, as there is no change in concentrations of both Zn +2 and Cu +2. This indicates that the cell reaction attains equation state.
2 cellcell 2 Zn 2.303RT EE log nF Cu + ο + =−
at equilibrium Ecell = 0 and 2 cc 2 Zn QK Cu + +
cell C 2.303RT 0E logK 2 ο =− (or)
cell C 2.303RT E logK 2 ο =
at 298 K, cell C 0.059 ElogK 2 ο =
In general, for any cell at equilibrium, cell C 2.303RT E logK nF ο = at 298 K cell C 0.059 ElogK n ο = or cell C nE logK 0.059 ο =
8. Find the log K value (here K = equilibrium constant) for the disproportionation reaction:
()()() 2 aq saq 2CuCuCu ++ + at 298 K.
(Given: 2 Cu/Cu Cu/Cu E0.52V,E0.15V +++ οο=+=+ )
Sol. For the reaction at equilibrium, E cell = 0
2 C Cu/CuCu/Cu 0.059 EE logK 1 +++ οο =−−
C 0.059 0.520.15logK 1 =−− = 0.37 – 0.059 log KC C 0.37 logK6.27 0.59 ==
Try yourself:
8. Calculate log KC for the reaction, Fe+2 + Ce+4 Fe+3 + Ce+3
The maximum reversible work done by a galvanic cell is given by
ΔG = – nF Ecell
Here, Ecell = emf of the cell
nF = charge of n moles of electrons passed
For the cell reaction: ()()() 2e 22 (aq) s aqs ZnCuZnCu ++ +→+
ΔG = – 2F Ecell
When coefficients are changed in the reaction
()()()() 4e 22 s aq aq s 2Zn2Cu2Zn2Cu ++ +→+
ΔG = – 4FEcell
If the concentration of all the reaction species is unity at 298 K,
ΔG° = – n F E° cell and ΔG° = – 2.303 RT log K
(K = equilibrium constant)
Thermodynamic Data of Galvanic Cell
1. For enthalpy change of reaction (ΔH): Gibbs equation: ΔG = ΔH – TΔS …… (1) or ΔG = ΔH + () P G T T ∂∆ ∂ ………… (2)
nn ( ΔG = –nFE)
(3)
2. Temperature coefficient of emf cell:
EHFE TFT ∂∆+ = ∂ n n . . . . . . . (4)
3. Entropy change of reaction:
(from equations (1) and (2))
(or)
Hence for each of the above half reactions, ΔG is calculated.
Hence,
2.7.5 Electrode Potential and Solubility Product
Let us consider silver electrode dipped in a solution of AgNO3. If excess of NaI is added to this half cell, the equilibrium established is AgI + e Ag + I–. This is corresponding to the electrode, I–/AgI/Ag. This is metal –insoluble salt–soluble anion electrode.
Nernst equations for the half cell potentials are as follows: o Ag/AgAg/Ag 0.0591 EE log 1 Ag ++ + =−
(for the reaction Ag+ + e Ag)
I/AgI/AgI/AgI/Ag
(For the reaction AgI + e Ag + I–)
Since, Ag/Ag
I/AgI/Ag
EE0.059logAgI −+ οο+− =+
I/AgI/AgAg/Ag
οοο −=−−− 332211 nFEnFEnFE
()() 3 20.340210.522 E 1 ο +− =
3 E0.1584volts ο=+
Try yourself:
9. Calculate the half cell potentials
3 Au3eAu,E1.5V ++→°=
u AeAu,E1.7V ++→°=
Then find the standard half cell potential for the reaction, Au+3 + 2e → Au+
Ans: 1.4 volts
TEST YOURSELF
1. Reduction potential of iron electrode is minimum when it is in contact with ________Fe2+ solution.
(1) 10–3 M (2) 10–3 M (3) 10–2 M (4) 1 M
or sp
EE0.059logK −+ οο=+
I/AgI/AgAg/Ag
9. If standard reduction potentials of Cu +/Cu, Cu +2/Cu are 0.522 volts and 0.3402 volts, respectively, calculate the standard half cell potential for the electrodes, Cu +2/Cu+
Sol. n1
CueCu,E0.522V +=+→°= (1) n2
2 Cu2eCu,E0.3402V +=+→°=+ ...(2)
2 CueCu,E? +++→°= .
(3)
Electrode potentials are not additive, but ΔG values are additive.
2. Find the emf of the cell in which the following reaction takes place at 298 K. ()()()()()() 0.001M 0.001M 2 s aq aq s Ni2AgNi2Ag ++ +→+ (Given that 0 cell 2.303RT E1.05V,0.059 at 298K ) F == (1) 1.0385 V (2) 1.385 V (3) 0.9615 V (4) 1.05 V
3. Give that zn/Zn22Cd/Cd E0.763VandE0.403V ++ °=−°=− the emf of the cell () () 22 ZnZn//CdCd a0.004a0.2 ++ == will be given by (1) E0.36log0.0590.004 22
=−+
4. Consider the following four electrodes:
A = Cu2+ (0.0001M) / Cu(s)
B = Cu2+(0.1 M) / Cu(s)
C = Cu2+ (0.01 M) / Cu(s)
D = Cu2+ (0.001 M) / Cu(s)
If the standard reduction potential of Cu +2/ Cu is +0.34 V, the reduction potentials (in volts) of the above electrodes follow the order
(1) A > D > C > B
(2) B > C > D > A
(3) C > D > B > A
(4) A > B > C > D
5. The standard electrode potential for the following reaction is +1.33 V. What would be its potential when the pH is 2.0?
(1) +1.820 V (2) +1.990 V (3) +1.608 V (4) + 1.0542 V
6. For a cell reaction involving two electron transfer the standard emf of the cell is found to be 0.295 V at 25°C. At the same temperature, equilibrium constant will be (1) 10 (2) 1 × 1010 (3) 1 × 10–10 (4) 29.5 × 10–2
7. If the cell E ο for a given reaction has a negative value, then which of the following gives the correct relationships for the values of ΔG° and K eq ?
(1) eq G0;K1∆>< (2) eq G0;K1∆>>
(3) eq 1 0; G K ∆<> (4) eq G0;K1∆<<
8. The standard electrode potential of a Daniell cell is 1.1 V. Find the standard Gibbs energy for the reaction
Battery is an example of an electrochemical c ell. It consists of two or more voltaic cells connected in series. The voltage of a battery should not vary much during its use. Mainly two types of cells are in use: primary cells and secondary cells.
2.8.1
Primary Cells
Primary cell is a galvanic cell that cannot be recharged once the reactants are consumed. This is because the electrode reactions cannot reversed to regenerate reactants from the products of the cell reaction.
The most familiar example of primary cell is the dry cell, commonly called Leclanche cell, which is used in clocks, transistors, and torches.
Dry Cell
Dry cell consists of a zinc container that acts as anode and graphite as cathode rod, surrounded by powdered manganese dioxide and carbon, as shown in Fig.2.10.
Carbon rod (cathode)
Fig. 2.10 Primary Cells: Common Dry Cell
The zinc electrode is in contact with paste of ammonium chloride and zinc chloride. The paste, C+MnO2 and NH4Cl + ZnCl2 are separated by a porous sheet. The zinc vessel is surrounded by a cardboard. The zinc vessel is sealed with pitch. The reactions that occur in dry cell are complex, but approximately shown as,
At cathode (reduction):
3 MnONHeMnO(OH)NH +− ++→+
24
(or) 22 MnOHOeMnO(OH)OH ++→+
MnO2 acts as cathodic depolariser.
At anode (oxidation): 2 Zn(s)Zn2e+−→+
The secondary reactions taking place in dry cell are
432 NHClOHNHClHO +→++
2 3 Zn2NH2ClZnCl.2NH23 +−++→
322 or[Zn(NH)Cl]
Ammonia produced in the reaction forms a complex, 2 34 [Zn(NH)] + . The potential of the cell is nearly 1.5 V.
Mercury Cell
Mercury cell is another dry cell suitable for low current devices. It consists of a paste of potash and zinc oxide acting as electrolyte. A paste of carbon and mercuric oxide acts as cathode and zinc amalgam as anode. The electrode reactions are,
At cathode (reduction):
2 HgOHO2eHg(l)2OH ++→+
At anode (oxidation): (s)2
ZnHg2OHZnOHOHg2e −+→+++
The overall reaction is, (s) ZnHgHgOZnO2Hg(s)() −+→+ l
This cell does not involve any ion, changing its concentration. That is why this cell can give almost constant voltage.
The potential of mercury dry cell is approximately 1.35 V.
Anode Cap
Anode
2.8.2
Secondary Cells
The secondary cells are rechargeable. Their chemical reactions can be reversed by supplying electrical energy to the cells. These are also known as storage cells.
Most common reversible storage cell is lead accumulator, used in inverters and automobile vehicles. It is called acid storage cell. Alkali storage cell is called Edison battery.
Anode
Cathode
Negative plates: lead grids filled with spongy lead
Lead storage battery consists of two lead electrodes, and both are reversible. Sponge lead is used as anode and a grid of lead packed with plumbic oxide is used as cathode. A 38% aqueous solution of sulphuric acid is used as electrolyte.
It is represented as:
Pb|PbSO4(s), H2SO4(aq) PbSO4(s), PbO2(s)/ Pb.
In a simple way, it can be represented as Pb/ H2SO4. PbO2
When the battery is in use, i.e., during discharge, the cell reactions are:
The reaction at cathode is 2 2(s)4(aq)(aq) PbOSO4H2e −+− +++→
PbSO2HO4(s)2(l) +
The reaction at anode is ()()() 2 44 s aq s PbSOPbSO2e +→+
The overall cell reaction is: discharge
PbPbO2HSO(s)2(s)24(aq)++→
2PbSO2HO4(s)2(l) +
During the charging of the battery, the electrode reactions are reversed as follows:
At anode: PbSO4(s)+2H2O(l) →
PbO2(s) + SO4–2(aq) + 4H+(aq) + 2e–
At cathode
PbSO4(s) + 2e– → Pb(s) + SO4–2(aq)
Lead sulphate is accumulated at cathode and anode is converted into lead and lead dioxide, respectively.
Charging and discharging of lead storage battery is known as double sulphation. The voltage increases with the concentration of electrolyte. The voltage varies from 1.88 V (5% H2SO4) to 2.15 V (40% H2SO4).
Current is generated when lead accumulator undergoes the process of discharging. During this process, sulphuric acid is consumed and its concentration decreases. When the concentration of sulphuric acid decreases to a certain low value, the accumulator should be charged. During charging, reverse reaction takes place and H2SO4 is generated resulting in an increase in its concentration or specific gravity. Hence, the cell is tested by determining the specific gravity of H 2SO4
Nickel–Cadmium Cell
Another important secondary cell is nickel–cadmium cell.
Positive plate
Separator
Negative plate
2.13 Rechargeable Ni–Cd Cell
The Ni-Cd cell is more expensive to manufacturer, but it has longer life than lead storage battery. The overall reaction during discharge of nickel–cadmium cell is,
A galvanic cell that directly converts chemical energy into electrical energy is called a fuel cell.
The fuel cell also has two electrodes and one electrolyte. The fuel is oxidised at anode and oxidant is reduced at cathode. Fuel and oxidant are continuously and separately supplied towards the two electrodes of the cell at which they undergo reaction. Fuel cells are primary cells and are capable of supplying
Fig.
2: Electrochemistry
current as long as they are provided with fuel and oxidant.
Fuel cells are now used in automobiles on an experimental basis. Fuel cells have silent operation and high efficiency, and they do not cause pollution.
The absence of moving parts in fuel cells eliminates wear and tear problems. They were used as power sources in spacecrafts Gemini and Apollo. The water formed in a hydrogen–oxygen cell is pure, and hence it is used for drinking purpose in spaceships.
TEST YOURSELF
1. The overall reaction of a hydrogen–oxygen fuel cell is
(1) ()()() 22 g eq O2HO4e4OH ++→
Fig. 2.14 Hydrogen–oxygen fuel cell
Hydrogen–oxygen, (as shown in Fig.2.14.) fuel cell consists of porous electrodes suspended in concentrated caustic soda solution.
Fuel and oxidant gases are bubbled at the surface of porous carbon electrodes embedded with suitable catalysts. Catalysts like finely divided platinum or palladium are incorporated into the electrode for increasing the rate of electrode reactions.
The electrode reactions are,
At cathode (reduction):
O2HO4e4OH ++→
2(g)2() (aq)
At anode (oxidation):
H2OH2e2HO2() +→+
2(g) (aq)
The overall reaction is given as
2(g)2(g)2() 2HO2HO +→
The heat of combustion is directly converted into electrical energy. The efficiency of energy conversion in fuel cells is generally more than 70%.
Methane–oxygen fuel cell consists of platinum electrodes suspended in phosphoric acid, which acts as an electrolyte.
(2) ()()() 22 g aq 2H4OH4HO4e +→+
(3) ()()() 222 gg 2HO2HO +→
(4) ()() 2 aq 4OH4e2HO +→
2. Select the incorrect statement.
(1) Primary batteries are non-rechargeable, so can be used only once
(2) Secondary batteries are rechargeable
(3) Primary batteries may act as electrolytic cells
(4) Secondary batteries may act as electrolytic cells
3. During the discharge of lead storage battery 1F of electricity is generated. Amount of PbSO 4 formed at cathode would be (MW of PbSO4 = 304)
(1) 304 g (2) 152 g
(3) 608 g (4) 456 g
4. During the charging of lead storage battery, the reaction occurring at cathode is represented by
5. A depolariser used in dry cell batteries is (1) manganese dioxide (2) manganese dioxide
(3) potassium hydroxide (4) sodium phosphate
6. Which of the following reactions occurs at the cathode of a common dry cell?
(1) Mn → Mn2+ + 2e–
(2) () 24 3 MnONHeMnOOHNH +− ++→+
(3) 2 3 23 Zn2NH2ClZnCl2NH +−++→⋅
(4) 2 ZnZn2e+−→+
Answer Key
(1) 3 (2) 3 (3) 2 (4) 4
(5) 2 (6) 2
2.9 CORROSION
Natural tendency of conversion of a metal into its undesirable compound on interaction with the environment is known as corrosion. It is an oxidative deterioration of metals.
Rusting of iron (Fe 2 O 3 .xH 2 O), tarnishing of silver (Ag 2S), and development of green coating on copper (CuCO3(OH)2) are some important examples of corrosion. Metallic corrosion causes enormous damage to bridges, buildings, ships, etc.
2.9.1 Mechanism of Corrosion
Process of corrosion may be chemical or electrochemical. Electrochemical corrosion is considered as the anodic dissolution of metal undergoing corrosion. Anodic dissolution involves oxidation of metal. M →Mn+ + ne–
Electrochemical corrosion occurs if the environmental conditions of metal favour the formation of a galvanic cell with the metal acting as anode.
If oxygen is not uniformly distributed on the surface of the metal, metal corrodes at those points when concentration of oxygen is less.
The portion of metal with access to high concentration of oxygen functions as cathode and with low oxygen as anode.
Metal with differential oxygenation functions as a galvanic cell. The portion of metal access to lower oxygen concentration acts as anode and undergoes anodic dissolution, as shown in Fig.2.15.
Fig. 2.15 Corrosion of iron
Rusting of iron is a common example of differential of oxygenation type corrosion. The mechanism corrosion is quite complex. At a particular spot of object of iron, it behaves as anode, at which oxidation takes place and metal dissolves.
At anode (oxidation),
Fe(s)→Fe2+ + 2e–; E° = – 0.44 V
Electrons released at anodic spot move through the metal and go to another spot and reduce oxygen in the presence of proton. The spot behaves as cathode, where oxygen is reduced.
At cathode (reduction)
O2(g) + 4H+(aq) + 4e– →2H2O(ℓ); E° = 1.23V. The overall reaction is
Since E cell is positive, the free energy change is negative, suggesting the corrosion is spontaneous. Further, ferrous iron is oxidised to ferric iron in the atmospheric oxygen and forms rust, which is chemically hydrated ferric oxide, Fe2O3 . xH2O.
Hence, for the rusting process to occur, the iron must be exposed to both water and oxygen.
2.9.2 Preventive Methods of Corrosion
The main principle underlying the method of prevention of corrosion is to separate the metal from the environment. Some of the methods are:
1. Painting the metal object protects the surface of the metal from the impact of the environment.
Metal surface can also be covered by using chemicals like bisphenol.
2. Prevention of metal surface from electrical conducting media.
3. Alloying the metal with more anodic metal, e.g., coating of iron with zinc (galvanisation of iron).
Iron pole in sea water undergoes corrosion faster than iron on beach, because iron in sea water suffers high salt concentration because dissolved ions make water a better conductor of electric current.
CHAPTER REVIEW
Electrical Conductivity
■ The substance that allows the flow of electricity through it is called electrical conductor.
■ Electrical conductors are of two types electronic and electrolytic.
TEST YOURSELF
1. Metal can be prevented from rusting by (1) Connecting iron to more electropositive metal – a case of cathodic protection
(2) Connecting iron to more electropositive metal –a case of anodic protection
(3) Connecting iron to less electropositive metal –a case of anodic protection
(4) Connecting iron to less electropositive metal –a case of cathodic protection
2. The composition of rust is (1) Fe2O3xH2O (2) Fe2O3.2H2O (3) Fe2O36H2O (4) Fe2O3
3. Galvanization is applying a coating of (1) Cr (2) Cu (3) Zn (4) Pb
4. Rusting of iron is catalysed by which of the following?
(1) Fe (2) Zn (3) N2 (4) H+
5. The Zn acts as sacrificial or cathodic protection to prevent rusting of iron because (1) OP OP E of ZnE of Fe οο <
(2) OP OP E of ZnE of Fe οο > (3) OP OP E of ZnE of Fe οο =
(4) Zn is cheaper than Fe Answer Key
(1) 1 (2) 1 (3) 3 (4) 4 (5) 2
■ Electronic conductors conduct electricity due to mobility of free electrons. Examples: Graphite, gas carbon, petroleum coke, metals, alloys and solid salts like CdS, CuS, etc.,
■ Electrolytic conductors conduct electricity due to mobility of free ions. Examples: Aqueous solutions of salts, acids and bases and fused salts like NaCl, KCl, etc.
■ Solutions of alkali metals in liquid ammonia are mixed conductors. They contain solvated cations and solvated electrons.
■ Substances that do not conduct electric current in the molten state or in solution are called non-electrolytes. Examples: sugar, glucose, urea, fructose, etc.
■ Non-electrolytes are non-polar covalent substances. They do not undergo ionisation.
■ All electrolytes spontaneously dissociate into charged particles when dissolved in water.
MA M+ + A–
■ The degree of ionisation (a) is the fraction of the total number of ionised molecules in a solution.
■ The extent of ionisation is different for different electrolytes, and it depends on the nature of the electrolyte, concentration of electrolyte, and temperature.
■ The degree of ionisation increases with increase in dilution, and at infinite dilution, it approaches unity.
■ Strong electrolytes ionise to a greater extent. Examples: Strong acids, strong bases, and soluble salts
■ Weak electrolytes ionise to a lesser extent. Examples: Weak acids, weak bases and sparingly soluble salts
■ Degree of ionisation of electrolytes increases with increase in temperature.
■ Arrhenius theory is applicable to only weak electrolytes.
Electrolytic Conductance
■ Resistance to the flow of electricity in a solution is directly proportional to the length and inversely proportional to the area of the tube in which the solution is taken.
■ Resistivity is the resistance of a conductor of 1 m length and 1 m2 area of cross section.
■ Units of specific resistance is ohm. metre (Ω.m)
■ For a given electrolytic cell, the quantity a is constant, which is called cell constant
■ Unit of cell constant is taken as m–1 or cm–1
■ Conductance is the ease of flow of electric current through the conductor. Conductance is the reciprocal of resistance.
■ Unit of conductance is ohm –1 or Ω–1 (or) mho. In SI system, it is seimen (S).
■ Specific conductance or conductivity (K, kappa) is the conductance of a conductor of 1 m length and 1m2 area of cross section.
■ Conductance taken as the reciprocal of resistivity.
■ Units of specific conductance is ohm–1.m–1 (or) S.m–1.
■ Conductance of all the ions produced by one gram equivalent weight of an electrolyte in V mL of solution is called equivalent conductance,
Λ eq = κ .V
■ Λ eq = 1000 N κ× , where N is normality .
■ Units of equivalent conductivity are Ω–1 . cm2. equiv–1 (or) S.cm2. eq–1.
■ Conductance of all the ions produced by one gram mole of an electrolyte in Vml of solution is called molar conductivity, Λ m = κ .V
■ Λ m = 1000 M κ× , where M is molarity.
■ Units of Λ m are Ω–1.cm2.mol–1 or S.m2.mol–1
■ The conductance of increase in all electrolytes increases with temperature.
■ With decrease in concentration of solution or increase in dilution, Λ m increases, Λ m increases and κ decreases.
■ At infinite dilution, concentration approaches zero. Molar conductance at this dilution is called limiting molar conductivity m 0.
■ Equivalent conductance at this dilution is called limiting equivalent conductivity ^0.
■ In case of strong electrolytes, Λ m or m increases to less extent with dilution.
■ Strong electrolytes are almost completely ionised at all concentrations. Increase in Λ m or m , with dilution is only due to decrease in the interionic forces.
■ For strong electrolytes, variation of conductance with dilution is given by Huckel Onsagar equation.
C0 KC ∧=∧−
Kohlrausch Law
■ Kohlrausch law states that, the limiting equivalent conductance of an electrolyte is the sum of equivalent conductivities of cations and anions.
() 0 00mxyABxy+−Λ=λ+λ
Electrolysis
■ The chemical decomposition of an electrolyte by the influence of electric current is called electrolysis.
■ In an electrolytic cell, electrical energy is converted into chemical energy.
■ During electrolysis, reduction (electronation) takes place at cathode and oxidation (de-electronation) takes place at anode.
■ During electrolysis, electrons flow from anode to cathode in the external circuit.
■ When a solution contains different kinds of cations and anions, during electrolysis, the cation of the least el ectropositive element
and the anion of the least electronegative element are discharged first.
■ Electrodes that do not involve in chemical reaction during electrolysis are called inert electrodes. Examples: graphite, platinum, gold, etc.
■ Active electrodes involve in chemical reaction during electrolysis. In general, anodic metal dissolves in solution. Example: Cu anode in aqueous CuSO 4.
■ Quantitative relationship of electrolysis was given by Faraday. There are two Faraday’s laws.
■ The mass of the substance evolved or deposited or dissolved at an electrode is directly proportional to the quantity of electricity passed through the electrolyte.
■ Mass, m = eQ, where e is electrochemical equivalent
■ Quantity of electricity ( Q) is taken as the product of current strength in amperes (A) and time in seconds ( t).
■ Q = it. C is A s.
■ If same quantity of electricity is passed into different electrolytes (connected in series), the ratio of the amount of substances deposited at the respective electrode is equal to the ratio of their equivalent weights.
■ 11 22 mE mE = Here, m1, m2 are masses of elements and E1, E2 are chemical equivalents
Electrochemical Cells
■ The cell in which chemical energy is converted to electrical energy is called as galvanic or voltaic cell.
■ In galvanic cell, oxidation takes place at anode and electrons are released, and reduction takes place at cathode and electrons are used up.
■ Electrons travel along external circuit from anode to cathode. Cathode is +ve electrode and anode is –ve electrode. These are connected by salt bridge.
■ Daniel cell is an example of galvanic cell. It is constructed using Zn and Cu electrodes and a porous partition.
■ The representation of the Daniel cell is Zn(s) / Zn2+(aq) // Cu2+(aq) / Cu(s)
■ The standard emf of a Daniel cell is given as 1.1 volt.
■ Single electrode potential is the potential developed at metal and metal ion interface or non-metal and non-metal ion interface.
■ Non-metallic element is generally gas. Single electrode of non-metal / nonmetal ion needs an inert metal rod like Pt. Example: Pt, H2(g) / H+(aq)
■ The magnitude of potential developed by a single electrode depends on the nature of metal or non-metal, number of electrons involved in half-cell reaction, concentration of ions in solution and temperature.
Electrochemical Series
■ Normal hydrogen electrode, NHE or SHE is represented as: Pt, H2(g)(1atm) /H+(1M).
■ The potential of SHE is assumed to be ‘zero’ volts.
■ A saturated calomel electrode is now used as secondary reference electrode. It is represented as Hg/ Hg 2Cl 2 (s), KCl (saturated).
■ Potential of saturated calomel electrode is taken as –0.2422 volts.
■ Standard electrode potential (E o ) is the potential developed by single electrode if concentration of ions is 1 M at 298 K and 1 atmosphere pressure.
■ The potentials are standard reduction potentials only (SRP).
■ Standard oxidation potential (SOP) is equal to SRP in magnitude, but with opposite sign.
■ Electrochemical series is the arrangement of various electrode systems in the ascending order of their SRP values. It is also called activity series.
■ All the metals that are present above hydrogen in the electrochemical series are called active metals. They have –ve SRP values.
■ Active metals liberate hydrogen from dilute mineral acids.
■ All the elements that are present below hydrogen in the activity series have +ve SRP values. They do not liberate H2 from acid.
■ A metal with less SRP has high tendency to lose electrons, undergoes oxidation and acts as reducing agent. It is represented on LHS in galvanic cell and replaces other metal ions that have more SRP from their salt solution.
■ Metal with more SRP has high tendency to gain electrons. It undergoes reduction and acts as an oxidant. It is represented on RHS in galvanic cell and is replaced by other metals that have less SRP from its salt solution.
■ An element with a lower reduction potential is a more powerful reducing agent.
■ Reduction power of some metals is in the order
Li > K > Ca > Zn > Cu
■ An element with a higher reduction potential is a more powerful oxidising agent.
■ The order of oxdising ability of halogens is F2 > Cl2 > Br2 > I2
■ The potential difference between the two electrodes when no current is drawn from the cell is known as the emf of the cell.
■ If the emf of the cell is negative, the cell reaction is non-spontaneous and that cell cannot be constructed.
Nernst Equation
■ If the salt bridge is removed, the cell reaction stops, the emf of the cell becomes zero and ions move randomly.
■ Nernst equation gives the dependence of the electrode potential on the concentration of ion.
■ Nernst equation holds good for reversible electrodes.
■ For metal electrodes, the half-cell reaction is Mn++ne– M.
The Nernst equation is:
0 n M 2.303RT EElog nF M +
■ Simplified form of Nernst equation for metal electrodes at 298 K is
0 0.059
EElogC n =+
■ For non-metal electrodes, the half cell reaction is A + ne – A n–. The Nernst equation is: n 0 EElog2.303RT[A] nF[A] =−
■ Simplified form of Nernst equation for non-metal electrodes at 298 K is:
0 0.059 EElogC n =−
■ For hydrogen electrode, E = – 0.059 pH
■ Concentration cell is a galvanic cell in which both the electrodes are of same type but the electrolytes have different concentrations.
Cu|CuSO4(C1)|| CuSO4(C2)|Cu
■ The cell potential of the concentration cell can be calculated using the equations:
Ecell= 2.303RT nF log 2 1 C C = 0.0591 n = log 2 1 C C
■ When the cell reaction is in equilibrium, the cell emf is zero. The Nernst equation is E0 = 2.303RT nF logKc (or) E0 = 0.0591 n logKc
■ The change in free energy is represented by ΔG. For spontaneous reaction, ΔG will have a negative value (ΔG = –ve)
■ Electrical work done in one second is the product of electrical potential and total charge passed.
■ To obtain maximum work from a galvanic cell, the charge has to be passed reversibly.
■ The reversible work done by a galvanic cell is the decrease in its free energy.
W max = – ΔG.
■ When no current is drawn from the cell, the emf of the cell is E, the amount of charge passed is nF (n is the number of electrons and F is faraday).
ΔG = – nFE
■ From standard free energy, the equilibrium constant can be calculated
ΔG° = –RT lnkc; 0 c G2.303RTlogK∆=−
Batteries
■ Battery is a single galvanic cell or more than one cell connected in series. Commercial batteries are of two types.
■ Primary cell becomes dead after some time and cannot be used again. Examples: Dry cell, mercury cell
■ Secondary battery can be recharged and can be used again. Examples: Lead storage battery, nickel–cadmium cell, alkali storage cell (Edison’s battery).
■ Dry cell is also called Leclanche cell. It is a Zn container. Cathode is a graphite rod, surrounded by powdered MnO 2 and carbon. Gap between cathode and anode is filled with moist paste of NH4Cl and ZnCl2, which acts as electrolyte. Cell potential is 1.5 volt.
■ Mercury cell is used for low current devices. Anode is ZnHg. Cathode is a paste of HgO and carbon. Electrolyte is a paste of KOH and ZnO. Cell potential is 1.35 volts.
■ Lead storage battery is used in automobiles and inverters. Anode is lead grids filled with spongy lead. Cathode is lead grids filled with PbO2.
■ Cathode and anode in lead storage battery arranged alternately and separated by thin fiber glass sheets. Electrolyte is 38% (by weight) solution of H 2SO4.
■ Overall cell reaction in lead storage battery
Pb(s) + PbO2(s) + 2H2SO4(aq) discharging charging
2PbSO4(s) + 2H2O(l)
■ During the working of storage cell, H2SO4 is used up (during discharge). The cell reactions are reversed. Each set of anode and cathode produces a potential of 2 volts.
■ Nickel–Cadmium cell has longer life than lead storage battery. Anode is Cd(s), cathode is Ni(OH) 3(s) and electrolyte is moist NaOH or KOH. Cd acts as reductant and Ni(OH)3 as oxidant.
Corrosion
■ Corrosion is the process of gradual destruction of metal by the environment
■ Rusting of iron, tarnishing of silver and green coating on copper are examples of corrosion.
■ Corrosion is due to the formation of oxide or other salts on the metal. Corrosion is an electrochemical phenomenon.
■ During rusting of iron, at a particular place, iron undergoes oxidation, forming hydrated ferric oxide (rust).
Fe2O3 + xH2O → Fe2O3.xH2O
■ Prevention of corrosion of iron can be prevented by painting iron surface, by electroplating the iron with corrosion resistant metals like chromium and connecting with sacrificial anodes.
■ Coating of iron surface with zinc metal is called galvanization.
■ Metals used as sacrificial anode are more electropositive elements like Zn, Mg, Al and their alloys.
Exercises
Level - I
Electrical Conductivity
1. Which one of the following materials conducts electricity ?
(1) diamond
(2) barium sulphate
(3) crystalline sodium chloride
(4) fused potassium chloride
2. Which of the following is conductor of electricity
(1) diamond
(2) graphite
(3) carborundum
(4) silica
3. In metallic conductor the current is conducted by flow of (1) electrons (2) atoms
(3) ions (4) molecules
4. Choose the wrong statement
(1) Electrical conductance of an electrolytic conductor increases with increase in temperature.
(2) Electrical conductance of a metallic conductor increases with increase in temperature.
(3) Electrical conductance of a metallic conductor decreases with increase in temperature.
(4) Degree of dissociation of an electrolyte increases with dilution.
5. Which of the following is not a conductor of electricity?
(1) Solid NaCl
(2) Cu
(3) Fused NaCl
(4) Brine solution
6. Which of the following is an example of weak electrolyte
(1) H3BO3 (2) H2SO4
(3) HNO3 (4) HCIO3
Electrolytic Conductance
7. Conductivity (units Siemen’S) is directly proportional to the area of the vessel and the concentration of the solution in it and is inversely proportional to the length of the vessel, then the unit of constant of proportionality is
(1) S m mol–1
(2) Sm2 mol–1
(3) S–2 m2 mol
(4) S2 m2 mol-2
8. If the specific conductance and conductance of a solution is same, then its cell constant is equal to
(1) 1 (2) 2
(3) 10 (4) 100
9. The distance between two electrodes of a cell is 2.5 cm and area of each electrode is 5 cm2 the cell constant (in cm–1) is (1) 2 (2) 12.5
(3) 7.5 (4) 0.5
10. Which of the following solutions of NaCl has the higher specific conductance ?
(1) 0.001N (2) 0.01N
(3) 0.1 N (4) 1 N
11. Molar conductivity of a solution is 1.26 × 102 Ω–1cm2mol–1 Its molarity is 0.01M. Its specific conductivity will be
(1) 1.26 × 10–5
(2) 1.26 × 10–3
(3) 1.26 × 10–4
(4) 0.0063
12. The unit of specific conductivity is
(1) Ohms cm–1
(2) Ohms cm–2
(3) Ohm–1 cm
(4) Ohm–1 cm–1
13. Effect of dilution on conduction is as follows
(1) Specific conductance increases, molar conductance decreases
(2) Specific conductance decreases, molar conductance increases
(3) Both increase with dilution
(4) Both decrease with dilution
14. The value of molar conductivity of HCl is greater than that of NaCl at a particular temperature because
(1) Molecular mass of HCl is greater than that of NaCl
(2) Mobility of H+ ions is more than that of Na+ ions
(3) HCl is strongly acidic
(4) Ionisation of HCl is larger than that of NaCl
15. The unit of equivalent conductivity is
(1) Ohm cm
(2) Ohm–1cm–2(g equivalent)–1
(3) S cm–2
(4) Ohm cm2 (g equivalent)
Kholrausch Law
16. According to Kohlrausch law, the limiting value of molar conductivity of an electrolyte A2B is (1)
(1) 100 m m K C λ=
(2) 000 m vv +− +−λ=λ+λ
(3) 1000 eq eq K C λ=
(4) 0 mca λ=λ+λ
18. If the molar conductance values of Ca2+ and Cl– at infinite dilution are respectively 118.88 × 10–4 m2 mho mol–1 and 77.33×10–4 m2 mho mol–1 then that of CaCl2 is (in m2 mho mol–1)
(1) 118.88 × 10-4
(2) 154.66 × 10-4
(3) 273.54 × 10-4
(4) 196.21 × 10-4
19. The molar conductivities 00 NaOAcHCl and ΛΛ at infinite dilution in water at 25°C and 91.0 and 426.2 S cm2 / mol respectively. To caculate the additional value required is (1) 0 NaCl Λ (2) 2 0 HO Λ (3) 0 KCl Λ (4) 0 NaOH Λ
20. The equivalent conductances of two strong electrolytes at infinite dilution in H2O (where ions move freely through a solution) at 25°C are given below 3 2 CHCOONa 91.0Scm/ equiv Λ= o 2 HCl 426.2Scm/ equiv Λ= o
What additional information / quantity one need to calculate 0 Λ of an aqueous solution of acetic acid ?
(1) 3 of CHCOOK Λ o
17. The equation representing Kohlrausch law from the following is(v+ are number of cations and v- are number of anions )
(2) + Ë of H o
(3) 0 2 of ClCHCOOH Λ
(4) 0 of NaCl Λ
21. Calculate HOAc ∞ Λ using appropriate molar conductances of the electrolytes listed above at infinite dilution in H 2 O at 25°C
22. The difference in molar conductances between KCl and KNO 3 at infinite dilution at 298K is 34.9. What is the difference in molar conductances between KNO3 and LiNO3 at infinite dilution in the same units at 298K
(1) 34.9 (2) 68.8 (3) 3.49 (4) 6.88
23. Which of the following equation gives correct relation between molar conductance of strong electrolyte, and it’s concentration
(1) MKC ∞ λ=λ+
(2) ()MKC ∞ λ=λ−
(3) ()MKC ∞ λ=λ−
(4) MKC ∞ λ=λ−
24. The ionic equivalent conductivity of -2 24 , COK + and Na + ions are X, Y and Z Scm 2 Eq –1 respectively. The 0 eq Λ of (NaOOC–COOK) based on the data given is
y xz
++
(1) Scmeq2-1 2
z xy
++
(2) Scmeq2-1 2
(3) () Scmeq2-1 ++ xyz
25. Kohlrausch law is useful in the calculation of
(A) pH of weak electrolyte (B) degree of dissociation of weak electrolyte
(C) K w of water
(D) K sp of sparingly soluble salt
(1) A, C and D only
(2) A and C only (3) A, B and C only (4) A, B, C and D
Electrolysis
26. Three Faradays of electricity was passed through on aqueous solution of iron (II) bromide. The mass of Iron metal (at mass 56) deposited at the cathode is (1) 56 g (2) 84 g (3) 112 g (4) 168 g
27. What is the amount of Al deposited on the electrolysis of moltenAl 2 O 3 when a current of 9.65 A is passed for 10.0 seconds
(1) 0.9 g (2) 0.009 g (3) 9 g (4) 0.09 g
28. Product at anode in electrolysis of dilute sulphuric acid is (1) H2S (2) H2 3) SO2 (4) O2
29. Process which occur at anode in the electrolysis of fused sodium chloride is (1) Na+ is oxidised (2) Cl– is oxidised (3) Cl is reduced (4) Na is reduced
xyz
(4) Scmeq2-1 22
++
30. Electro chemical equivalent is highest for (1) Sodium (2) Hydrogen (3) Aluminium (4) Magnesium
31. How many coulombs of electricity are required for the reduction of 1 mol of 2 4 MnOtoMn−+ ?
(1) 96500 C
(2) 1.93 ×105C
(3) 4.83×105C
(4) 9.65×106C
32. 1 faraday can deposit/liberate one gram atom of metal at cathode during the electrolysis of (inert electrodes are used)
(1) aq. AgNO3
(2) aq. CuSO4
(3) aq. NaCl
(4) aq. ZnCl2
33. If three faraday of electricity are passed through the solutions of AgNO3,CuSO4, and AuCl3 the molar ratio of the cations deposited at the cathodes will be
(1) 1 : 1: 1 (2) 1 : 2 : 3
(3) 3 : 2 : 1 (4) 6 : 3 : 2
34. If a current of 1.5 ampere flows through a metallic wire for 3 hours, then how many electrons would flow through the wire?
(1) 2.25×1022 electrons
(2) 1.13×1023 electrons
(3) 1.01×1023 electrons
(4) 4.5×1023 electrons
35. 810 g of acidulated water is subjected to electrolysis using 9.65 × 104 amperes. Time required for electrolysis is (1) 45 seconds (2) 90 seconds
(3) 180 seconds (4) 360 seconds
Electro Chemical Cells
36. For cell reaction,Zn+Cu 2+→ Zn 2++Cu, cell representation is:
(1) Zn/ Zn2+||Cu2+/Cu
(2) Cu/Cu2+||Zn2+/Zn
(3) Cu|Zn2+||Zn|Cu2+ (4) Cu2+|Zn||Zn2+|Cu
37. Which of the following is the cell reaction that occurs when the following half-cells are combined?
I2+2e–→2I–(1M); E°=+0.54V
Br2+2e–→2Br–(1M); E°=+1.09V
(1) 2Br–+I2→Br2+2I–
(2) I2+Br–→ 2I– +2Br–
(3) 2I–+Br2→I2+2Br–
(4) 2I–+2Br2→I2+Br2
38. For the galvanic cell,Cu|Cu2+||Ag+|Ag.
Which of the following observations is not correct?
(1) Cu acts as anode and Ag acts as cathode
(2) Ag electrode loses mass and Cu electrode gains mass.
(3) Reaction at anode Cu→Cu2++2e–
(4) Copper is more reactive than silver
39. In the cell, Zn| Zn2+||Cu2+|Cu, the negative terminal is
(1) Cu (2) Cu2+
(3) Zn (4) Zn2+
40. The cell reaction of the galvanic cell : ()()()() 2+2+ saqaql CuCuHgHg is (1) Hg+Cu2+→Hg2++Cu
(2) Hg+Cu2+→Cu++Hg+
(3) Cu+Hg→CuHg
(4) Cu+Hg2+→Cu2++Hg
41. Which of the following is the correct cell representation for the given cell reaction?
Zn+H2SO4→ ZnSO4+H2
(1) Zn| Zn2+||H+|H2
(2) Zn| Zn2+||H+,H2|Pt
(3) Zn|ZnSO4|H2SO4|Zn
(4) Zn|H2SO4| ZnSO4|H2
42. Following reactions are taking place in a Galvanic cell, Zn→ Zn2++2e–;Ag++e–→Ag
Which of the given representations is the correct method of depicting the cell?
(1) Zn(s)|Zn2+(aq)||Ag+(aq)|Ag(s)
(2) Zn2+|Zn||Ag|Ag+
(3) Zn(aq)|Zn2+(s)||Ag+(s)|Ag(aq)
(4) Zn(s)|Ag+(aq)||Zn2+(aq)|Ag(s)
43. Regarding the standard hydrogen electrode, the incorrect statement is
(1) H2 gas is bubbled at 1 bar
(2) 2 M HCl is electrolyte
(3) SRP is arbitrarily taken as Zero. (4) It is a primary reference electrode.
44. Which cell will measure the standard electrode potential of copper electrode?
45. Reduction potential of a hydrogen electrode at pH = 1 is:
(1) 0.059 volt (2) 0 volt
(3) –0.059 volt (4) 0.59 volt
Electrochemical Series
46. Given that 22 O/HO E=+1.23V
2-2284 SO/SO E=2.05V2 Br/Br E=+1.09V
3+ Au/Au E=+1.4V
The strongest oxidizing agent is:
(1) 2-SO28 (2) Au3+ (3) Br2 (4) O2
47. Which of the following is not an application of electro chemical series?
(1) To compare the relative oxidizing and reducing power of substances.
(2) To predict evolution of hydrogen gas on reaction of metal with acid.
(3) To predict spontaneity of a redox reaction.
(4) To calculate the amount of metal deposited on cathode.
48. Zn gives hydrogen with H2SO4 and HCl but not with HNO3 because
(1) Zn acts as oxidizing agent when reacts with HNO3
(2) HNO 3 is weaker acid than H 2SO 4 and HCl
(3) Zn is above the hydrogen in electrochemical series
(4)NO3 is reduced in preference toH+ ion.
49. Which of the following is the correct order in which metals displace each other from the salt solution of their salts?
(1) Zn,Al,Mg,Fe,Cu
(2) Cu,Fe,Mg,Al, Zn
(3) Mg,Al, Zn,Fe,Cu
(4) Al,Mg,Fe,Cu,Zn
50. The metal that cannot displace hydrogen from dilute HCl is:
(1) Al (2) Fe
(3) Cu (4) Zn
51. E° of Mg2+/Mg,Zn2+/Zn and Fe2+/Fe are are –2.37V, –0.76V and –0.44V respectively. Which of the following statements is correct
(1) Zn will reduce Fe2+
(2) Zn will reduce Mg2+
(3) Mg oxidises Fe
(4) Zn oxidises Fe
52. Four metals A, B, C & D are having respectively standard electrode potential are –3.05, –1.66, –0.40 & 0.80 V. Which one will be the strong reducing agent
(1) A (2) B
(3) C (4) D
53. E° for Fe 2+ +2e→Fe is–0.44 V; E° Zn2++2e→Zn is –0.76V Then
(1) Zn is more electropositive than Fe
(2) Fe is more electropositive than Zn
(3) Zn is more electronegative
(4) No reaction takes place
54. The correct statement among the following, about electrochemical series (S.R.P. increasing order) is
(1) the metals occupying top positions in the series do not liberate hydrogen with dilute acids
(2) the substances which are stronger reducing agents and stronger oxidising agents are placed below & top respectively
(3) a metal higher in the series will displace the metal from its solution which is lower in the series
(4) various electrodes are arranged in a series in the descending order of their potentials
55. The E0 values for
Al+|Al=+0.55 V and Tl+|Tl =–0.34V
Al+3|Al = -1.66V and Tl+3|Tl=+1.26V
identify the incorrect statement.
(1) Al is more electropositive than Al
(2) Tl+3 is a good reducing agent than Tl+
(3) Al+ is unstable in solution
(4) Tl can be easily oxidized to Tl+ than Al+3
Nernst Equation
56. Emf of the cell Zn(s)/Zn+2(C1)//Fe2+(C2)/ Fe(s) can be increased by
(1) increasing the size of Iron electrode
(2) increasing [Fe2+]
(3) increasing [Zn2+]
(4) All of these
57. The emf of the cell in which of the following reaction, Zn (s)+Ni 2+(0.1M)→ Zn 2+ (1.0 M)+Ni (s) occurs is found to 0.5105 V at 298 K. The standard emf of the cell is
(1) 0.4810 V (2) 0.5696 V
(3) –0.5105 V (4) 0.5400 V
58. The Nernst equation giving dependence of electrode potential on concentration is
(1) 0 n+ M 2.303RT E=E+log nF M
(2)
0 M 2.303RT E=E+lognFM
(3) n+ 0 M 2.303RT E=E-lognFM
(4) 0 n+ 2.303RT E=E-logM nF
59. A copper electrode is in contact with copper sulphate solution. The reduction potential of copper electrode when copper sulphate solution is diluted by 10 times will be
(1) decreases by 0.0295 V
(2) increases by 0.0295 V
(3) decreases by 0.059 V
(4) increases by 0.059 V
60. The standard EMF of Daniel cell is 1.10 volt. The maximum electrical work obtained from the Daniel cell
(1) 212.3 kJ
(2) 175.4 kJ
(3) 106.15 kJ
(4) 53.07 kJ
61. Reduction potential of Chlorine electrode in contact with 0.001 M HCl solution is2 1 Pt,Cl/Cl 2 E=+1.36V
(1) 1.54 V (2) 1.24 V (3) 1.39 V (4) 1.45 V
62. The relationship between free energy and electrode potential is
(1) D G=–nFE (2) D G=nFE
(3) nFE G R ∆= (4) H G nFE ∆ ∆=
63. The potential of hydrogen electrode depends on
(A) Concentration of acid in which the platinum foil is immersed
(B) Temperature
(C) Pressure at which hydrogen gas is bubbled through the solution
(1) A, B & C
(2) A & B only
(3) A only
(4) A & C only
64. Reduction potential of Pt,Cl 2 (1atm)/ HCl(0.1M) is () 2 0 / 1.36 ClCl EV =+ (1) + 1.36V (2) + 1.39V (3) + 1.45V (4) + 1.42V
65. The correct Nernst equation for the given cell
Pt(s)|Br2(l)|Br–(M)||H+(M)|H2(g)(1bar)|Pt(s) is (1)
66. The pressure of H2 required to make the potential of H 2 electrode zero in pure water at 298 K is:
68. What will be value of E cell for reaction ()()() -3 + +2 s 0.01M 10M Mg+2AgMg+2Ag given 0 cellE is 3.17 V (1) 2.67 (2) 2.13 (3) 3.27 (4) 3.051
69. Cu electrode is dipped in 0.1M of its solution at 298K. Reduction potential of electrode is +2 0 Cu/Cu E=+0.34V (1) +0.301V (2) +0.34V (3) +0.31V (4) +0.37V
70. SRP values of Ni, Cl2 electrodes are -0.25 & 1.36 V respectively. Then the EMF of the following cell is Ni/Ni+2(0.01M)||Cl–(0.01M)/Cl2,Pt. (1) 1.79 V (2) –1.79 V (3) 1.17 V (4) 1.11 V
Batteries
71. Which of the following reactions occurs at the cathode of a common dry cell
(1) Mn→Mn2++2e–
(2) MnO2+NH+4+e–→MnO(OH)+NH3
(3) Zn+2+2NH3+2Cl–→ZnCl2.2NH3
(4) Zn→Zn2++2e–
72. Dry cell is commonly used in transistors and clocks. Identify correct statement about dry cell.
(1) Cathode is zinc
(2) Anode is carbon
(3) zinc is oxidised
(4) carbon is oxidised
73. The reaction taking place at cathode in the lead storage battery during its discharging is
(1) Pb+SO4–2→ PbSO4 + 2e–
(2) PbO2+4H++SO4–2+2e–→PbSO4+2H2O
(3) PbSO 4 +2H 2 O → PbO 2 + 4H + +SO 4 –2 +2e–
(4) PbSO4+2e–→ Pb+SO4–2
74. A secondary cell is one
(1) can be recharged by itself
(2) can be recharged by passing current through it in the same direction
(3) can be recharged by passing current through it in the opposite direction (4) cannot recharged.
75. The overall reaction of a hydrogen –oxygen fuel cell is:
76. With respect to fuel cell prepared from H2 and O2 gases, the false statement is (1) It is free from pollution
(2) This is more efficient than conventional method of generating electricity
(3) The reaction occuring at anode is O2(g)+2H2O+4e–→4OH–
(4) These take little time to go into operation.
Level - II
Electrical Conductivity
1. Which of the following is not a semiconductor
(1) CuO
(2) Silicon
(3) Ge
(4) Mixed metal oxides at 150 K
2 Which of the following is not true ?
(1) Metallic conduction is due to the movement of electrons in the metal
(2) Electrolytic conduction is due to mobility of ions in solution
(3) The current carrying ions are not necessarily discharged at the electrodes
(4) The metallic conduction increases with increase in temperature, whereas electrolytic conductions decreases with increase of temperature
3. Amongst the following, the form of water with the lowest ionic conductance at 298 K is:
(1) sea water
(2) water from well
(3) saline water used for intravenous injection
(4) distilled water
4. Aqueous solution of which of the following compounds is the best conductor of electric current?
(1) Ammonia NH3
(2) Fructose C6H12O6
(3) Acetic acid, C2H4O2
(4) Hydrochloric acid, HCl
5. The correct trend of conductivity (in S m–1) at 298.15 K is represented in
(1) Ag > Cu > Au > 0.1M HCl
(2) Ag >Au > Cu > 0.1M HCl
(3) 0.1 M HC l> Ag> Au > Cu
(4) 0.1 M HCl> Ag > Cu > Au
6. The decrease in electrical conductivity of metals with increase in temperature is due to increase in
(1) the velocity of electrons
(2) the resistance of the metal (3) the number of electrons
(4) the number of metal atoms
Electrolytic Conductance
7. Which of the following solutions of KCl has higher specific conductance than others?
(1) 1N (2) 0.1 N
(3) 0.01 N (4) 0.001 N
8. Molar conductivity of a solution is 1.26×10 2 Ω –1 cm 2 mol –1 . Its molarity is 0.01M. Its specific conductivity will be
(1) 1.26×10–5 (2) 1.26×10–3
(3) 1.26×10–4 (4) 0.0063
9. The conductivity of 0.001 M acetic acid is 5 × 10–5 S cm–1 and Λ° is 390.5 S cm2 mol–1 then the calculated value of dissociation constant of acetic acid would be
(1) 81.78 × 10–4
(2) 81.78 × 10–5
(3) 18.78 × 10–6
(4) 18.78 × 10–7
10. The specific conductance of salt of 0.01M concentration is 1.06×10 –4 mho.cm –1 . Molar conductance of same solution in mho.cm2.mol–1 is
(1) 106.1 (2) 10.61 (3) 1.061 (4) 1.061×10–4
11. Molar conductance of an electrolyte increase with dilution according to the
equation: O mm Ac Λ=Λ−
Which of the following statements are true?
(A) This equation applies to both strong and weak electrolytes.
(B) Value of the constant A depends upon the nature of the solvent.
(C) Value of constant A is same for both BaCl2 and MgSO4
(D) Value of constant A is same for both BaCl2 and Mg(OH)2
Choose the most appropriate answer from the options given below:
(1) (A) and (B) only
(2) (A), (B) and (C) only (3) (B) and (C) only
(4) (B) and (D) only
12. The conductivity of a solution containing 2.08 g of anhydrous barium chloride in 200 mL solution is 6×10–3 Ohm.cm–1. The molar conductivity of the solution (in Ohm cm–1 mol–1)is x × 102. The value of x is (Atomic mass of Ba = 137, Cl = 35.5)
(1) 1.2 (2) 2.4 (3) 3.6 (4) 3.0
13. specific conductivity of 0.04 M “AB” is ______S m–1. (Λ in mho cm2 mol–1 and c in mole/litre)
(1) 100 (2) 1 (3) 10–2 (4) 10–1
14. Molar conductance of 0.2 M weak mono protic acid HA is 4.95 mho cm 2 mol −1 . Molar conductance of 0.002 M HA at the same temperature would be nearly (‘α‘ is too small compared to 1)
(1) 495 mho cm2 mol-1
(2) 4.95 mho cm2 mol-1
(3) 49.5 mho cm2 mol-1
(4) 0.0495 moh cm2 mol-1
Kohlrausch Law
15. m ° Λ for NaCl, HCl and NaA are 126.4, 425.9 and 100.5 S cm2 mol–1 respectively. If the conductivity of 0.001M HA is 5×10–5 S cm–1, degree of dissociation of HA is
(1) 0.25 (2) 0.50
(3) 0.75 (4) 0.125
16. At 25°C, the molar conductances at infinite dilution for the strong electrolytes NaOH, NaCl and BaCl 2 are 248 × 10 –4 , 126 × 10 –4 and 280 × 10 –4 S m 2 mol –1 respectively. m λ Ba(OH)2 in Sm2 mol–1 is.
(1) 52.4 × 10–4
(2) 524 × 10–4
(3) 402 × 10–4
(4) 262 × 10–4
17. The limiting molar conductivities (Λ°) for NaCl, KBr an KCl are 126, 152 and 150 S cm2 mol–1 respectively. Then Λ° for NaBr is
(1) 128 S cm2 mol–1
(2) 302 S cm2 mol–1
(3) 278 S cm2 mol–1
(4) 288 S cm2 mol–1
18. The equivalent conductances of two strong electrolytes at infinite dilution in H 2 (where ions move freely through a solution) at 25°C are given below:
3 2 91.0S cmeq Λ= CHCOONa HCl =426.2Scm/ eq Ë
What additional information/quantity one needs to calculate Λ°of an aqueous solution of acetic acid?
(1) Λ° of NaCl
(2) Λ° of CH3COOK
(3) The limiting equivalent conductance of () H Λ+
(4) Λ° of chloroacetic acid (ClCH2COOH)
19. Molar conductivity of BaCl 2 ,KCl and KOH is x,y,z mho cm2 mol–1 respectively. Equivalent conductance of Ba(OH)2 is
(1) x+2z–2y
(2) x + z – y
(3) x2z2y 2 +−
(4) 2(x+2z–2y)
20. Molar ionic conductance of Ca+2 is x Sm2 mol –1 and that of PO 4 –3 is y S m 2 mol –1 equivalent conductance of calcium phosphate is....S m2 eq–1
(1) x + y (2) (3x + 2y)
(3) 6(3x + 2y) (4) 3x2y 6 +
21. Kohlrausch’s law is applicable
(1) for concentrated solution (2) at infinite dilution
(3) for concentrated as well as dilute solutions
(4) none
22. The conductivity of saturated solution of AgCl is found to 1.86 × 10 –6 ohm –1 cm –1 and that of water is 6×10 –8 ohm –1 cm–1.If λ°AgCl is 138 ohm–1 cm2 eq–1 the solubility product of AgCl is
(1) 1.3 × 10–5 (2) 1.69 × 10–10
(3) 2 × 10–10 (4) 2.7 × 10–10
23. Molar ionic conductivities of a bivalent electrolyte are 57 and 73. The molar conductivity of the solution will be (1) 130 S cm2 mol–1
(2) 65 S cm2 mol–1
(3) 260 S cm2 mol–1
(4) 187 S cm2 mol–1
24. At 25°C, the ionic mobility of CH3COO–,H+ respectively 4.1×10–4, 3.63×10–3 cm/s. The conductivity of 0.001M CH3COOH is 5×10–5 S cm1.Dissociation constant of CH3COOH is cm.
(1) 1.53 × 10–5 (2) 3 × 10-4
(3) 3 × 10-5 (4) 3 × 10-6
25. At 298 K, molar conductance of 0.1 M weak monoacidic base, BOH is 39.6 mho. cm2.mol–1. Molar conductance of BOH at infinite dilution is 396 mho.cm 2.mol –1 . pH of centimolar solution of BOH is
(1) 11 (2) 10.699 (3) 12 (4) 9.301
Electrolysis
26. During the electrolysis of acidulated water, the mass of hydrogen obtained is ‘ x’ times that of O2 and the volume of H2 is ‘y’ times that of O2. The ratio of ‘y’ and ‘ x’ is a
(1) 16 (2) 8
(3) 0.125 (4) 0.25
27. One coulomb of charge passes through solutions of AgNO 3 and CuSO 4 . The ratio of the amounts of silver and copper deposited on platinum electrodes used for electrolysis is
(1) 108 : 63.5 (2) 54 : 31.75
(3) 108 : 31.75 (4) 215.8 : 31.75
28. How many coulombs of electricity are required for the oxidation of one mole of water to dioxygen?
(1) 9.65 × 104 C (2) 1.93 × 104 C
(3) 1.93 × 105 C (4) 19.3 × 105 C
29. What is the time (in sec) required for depositing all the silver present in 125 mL of 1M AgNO3 solution by passing a current of 241.25 amperes ? (1F = 96500 coulombs)
(1) 10 (2) 50
(3) 1000 (4) 100
30. A const ant current was flown for one minute through a solution of KI. At the end of experiment, liberated I2 consumed 150 mL of 0.01Msolution of Na 2 S 2 O 3 . Average rate of current flow is
(1) 9.65A (2) 4.825A
(3) 2.4125A (4) 1.206A
31. When 6 ×1022 electrons are used in the electrolysis of a metallic salt, 1.9 gm of the metal is deposited at the cathode. The atomic weight of that metal is 57. So, oxidation state of the metal in the salt is
(1) +2 (2) +3
(3) +1 (4) +4
32. The density of copper is 8 g/cc. Number of coulombs required to plate an area of 10 cm × 10 cm on both sides to a thickness of 10 –2 cm using CuSO 4 solution as electrolyte is
(1) 48,250 (2) 24,314 (3) 96,500 (4) 10,000
33. When 3.86 amperes current are passed through an electrolyte for 50 minutes, 2.4 g of a divalent metal is deposited. The gram atomic weight of the metal (in grams) is (1) 24 (2) 12 (3) 64 (4) 40
34. When 965 amp current is passed through aqueous solution of salt X using platinum electrodes for 10 sec, the volume of gases liberated at the respective electrodes is in 1:1 ratio. Then X is (1) MgSO4 (2) AgCl (3) MgCl2 (4) KNO3
35. When electricity is passed through molten AlCl3, 13.5 g of Al is deposited. The number of Faradays must be (1) 0.5 (2) 1.0 (3) 1.5 (4) 2.0
36. 965 amp current is passed through molten metal chloride for one minute and 40 seconds during electrolysis. The mass of metal deposited is 9 gm at the cathode. The valency of metal atom (at.wt = 27) is
(1) 4 (2) 3
(3) 2 (4) 1
37. On passing current through molten KCl, 19.5 g of K is deposited. The amount of Al deposited by the same quantity of electricity when passed through molten AlCl3 is
(1) 4.5 g (2) 9 g
(3) 13.5 g (4) 2.7 g
Electrochemical Cells
38. The cell reaction of the galvanic cell, ()()()() 22 sl |||| ++ aqaq CuCuHgHg is
(1) Hg+Cu2+→Hg2++Cu
(2) Hg+Cu2+→Hg2++Cu+
(3) Hg+Cu2+→Cu+Hg
(4) Cu+Hg2+→Cu2++Hg
39. The cell for which the cell reaction is H2+Cu2+→2H++Cu is represented as
(1) Cu||Cu2+||H+|H2(g)
(2) H2(g)||H+||Cu2+|Cu
(3) Pt,H2(1 atm)|H+||Cu2+(aq)|Cu
(4) Pt,H2|H2+(aq) (1atm)|Cu2+|Cu
40. The chemical reaction
2AgCl ( s ) +H2 ( g )→2HCl(aq)+2Ag(s) taking place in a galvanic cell is represented by the notation
41. Which is correct for a spontaneous cell reaction?
(1) Zn+2Ag+→ Zn2++2Ag
(2) 2Ag+ Zn2+ → 2Ag++ Zn
(3) Cu + Zn2+→ Cu2++ Zn
(4) Fe + Zn2+→ Fe2++ Zn
42. Saturated solution of KNO 3 is used to construct salt bridge because
(1) velocity of K+ is greater than that of NO–3
(2) velocity of NO–3 is greater than that of K+
(3) velocities of both K+ and NO–3 are nearly the same
(4) KNO3 is sparingly soluble in water
43. The difference of potential of two electrodes in a galvanic cell is known as (1) EMF
(2) Potential difference
(3) Electrode difference
(4) Ionic difference
44. The potential of single electrode depends upon
A) the nature of the electrode
B) temperature
C) concentration of the ion with respect to which it is reversible
(1) A only (2) B only
(3) C only (4) A, B and C
45. Standard electrode potential for equation 1 is 2.8 V, determine the standard electrode potential for equation (2). () 2 22...1FeF +→ () 2 1 FeF..2 2 +→
(1) 2.8 V (2) 1.4 V
(3) – 2.8 V (4) – 1.4 V
46. E0 for Cl2(g)+2e–1→2Cl–1(aq) is +1.36 V; then E0 for Cl–1 (aq)→½ Cl2(g)+e–1 is:
(1) +1.36 V (2) – 1.36V
(3) – 0.68 V (4) +0.68V
47. The following statements is correct with respect to both electrolytic cell and Galvanic cell
(1) In both cells, the anode is a negative electrode.
(2) In both cells, the cathode is a negative electrode.
(3) in both cells, reduction reaction takes place at the cathode
(4) in both cells, oxidation reaction takes place at the cathode
48. The cell reaction of the galvanic cell, ()()()() 2+ 2+ saq aql ||| CuCuH | gHg is
(1) Hg+Cu2+→Hg2+ +Cu
(2) Hg+Cu2+→Hg++Cu+
(3) Hg+Cu2+→+Cu+Hg2+
(4) Cu+Hg2+→Cu2++Hg
49 Incorrect statement about Daniel cell among the following is
(1) When it acts like galvanic cell, copper electrode is the negative electrode
(2) When it acts like galvanic cell, zinc electrode is the anode
(3) When connected to external emf of lesser than 1.1 V, Zn(s) reduces () 2 Cuaq + ion
(4) When connected to external emf of greater than 1.1 V, it acts like electrolytic cell
Electrochemical Series
50. Zn gives H2 gas with H2SO4 and HCl but not with HNO3 because (1) Zn acts as oxidant when reacts with HNO3
(2) HNO3 is weaker acid than H2SO4 and HCl
(3) In electrochemical series Zinc is above hydrogen (4) NO –3 is reduced in preference to H3O+
52. At 298 K the standard reduction potentials for the following half reactions are given as
Zn2+(aq)+2e–→Zn(s);–0.762 V
Cr3+(aq)+3e–→Cr(s);–0.740 V 2H+(aq)+2e–→H2(s);–0.00 V Fe3+(aq)+e–→Fe2+(aq);+0.770 V
The strongest reducing agent is (1) Zn(S) (2) H2(g) (3) Cr(s) (4) Fe2+(aq)
53. A gas X at 1 atm is bubbled through a solution containing a mixture of 1M Y– and 1M Z – at 25°C. if the reduction potential of Z > Y > X, then (1) Y will oxidize X but not Z (2) Y will oxidize both X and Z (3) Y will oxidize Z but not X (4) Y will reduce both X and Z
54. E° of Fe|Fe2+ is 0.44 V, E° of Cu|Cu2+ is –0.32 V. Then in the cell
(1) Cu oxidises Fe2+ ion
(2) Cu2+ oxidises iron
(3) Cu reduces Fe2+ ion
(4) Cu2+ ion reduces Fe
55. The standard potentials at 25°C for the half reactions are given against them below:
Zn2++2e–→Zn; E°=–0.762V
Mg2++2e–→Mg; E°=–2.37V
When Zn dust is added to a solution of MgCl2
(1) Magnesium is precipitated
(2) Zinc dissolves in the solution
(3) Zinc chloride is formed
(4) No reaction takes place
56. When
2 00 || E0.80V and E0.76V++ ==− AgAgZnZn , which of the following is correct?
(1) Ag+ can be reduced by H2
(2) Ag can oxidise H2 into H+ ion
(3) Zn+2 can be reduced by H2 (4) Ag can reduce Zn+2 ion
57. Given 32 4 00 /| E0.74V;E1.51V +−+ =−= CrCrMnOMn
23 27 2 00 O/ / E1.33V;E1.36V −+ == CrCrClCl
Based on the data given above, strongest oxidizing agent will be
(1) Cl– (2) Cr3+ (3) Mn2+ (4) MnO–4
58. Adding powdered lead and iron to a solution that is 1M in both Pb2+ and Fe2+ ions would results to a reaction in which
60. Given:
Co3++e→Co2+ ; E°=+1.81 V
Pb4+2e–→Pb2+ ; E°=+1.67 V
Ce4++e–→Ce3+ ; E°=+1.61 V
Bi3++3e–→Bi ; E°=+0.20 V
Oxidizing power of the species will increase in the order:
(1) Ce4+<Pb4+<Bi3+<Co3+
(2) Co3+<Pb4+<Co4+<Bi3+
(3) Bi3+<Ce4+<Pb4+<Co3+ (4) Co3+<Ce4+<Bi3+<Pb4+
Nernst Equation
61. For 2 27O14H6eCr −+−++→
30 2 27HO E1.33V.++=+ Cr
[Cr3+]=15 millimole, E = 1.067V. The pH of the solution is nearly equal to (1) 2 (2) 3 (3) 5 (4) 4
62. The reduction potential at pH=14 for the Cu2+/Cu couple is
2+ 2 0 -19 s ) p Cu ( /C O u CuH ] [givenE=0.34V and K1×1 = 0
59. If a spoon of copper metal is placed in a solution of ferrous sulphate
(1) Cu will precipitate
(2) iron will precipitate
(3) Cu and Fe will precipitate
(4) no reaction will take place
63. A hydrogen electrode placed in a buffer solution of sodium cyanide and hydrogen cyanide in the ratio x:y and y:x has the electrode potential values “ a ” and “ b ” volts respectively at 25°C. If a – b = 35.52 mV. Then y:x is (1) 1 (2) 2 (3) 3 (4) 6
64. The EMF of the cell Ni/Ni 2+(0.01 M)// Cl–(0.01 M) Cl2 Pt is___V. If the SRP of nickel and chlorine electrodes are – 0.25 V and +1.36 V respectively (1) + 1.61 (2) – 1.61 (3) + 1.79 (4) – 1.79
65. The solution of nickel sulphate in which nickel rod is dipped is diluted 10 times.
The potential of nickel
(1) decreases by 60 mV
(2) increases by 30 V
(3) decreases by 30 m V
(4) decreases by 60 V
66. Ag + I– →AgI + e–; E° = 0.152 V
Ag → Ag+e–; E° = 0.800V
What is the approximate value of log Ksp for AgI? (2.303 RT/F = 0.059 V)
(1) – 37.83 (2) – 16.13
(3) – 8.12 (4) + 8.612
67. K sp of AgCl is 10–10 M2 Electrode potential of Ag/AgCl electrode in contact with 1M KCl at 25°C is ( 0 / E0.8V AgAg+=+ )
(1) 1.08 V (2) 0.592 V
(3) 0.8 V (4) 0.20 V
68. For a cell reaction involving a two electron change, the standard emf of the cell is found to be 0.295 V at 25°C. The equilibrium constant of the reaction at 25°C will be
(1) 1 ×10–10
(2) 29.5 ×10–2
(3) 10
(4) 1 ×1010
69. For the following cell Zn 2+ ||Cd 2+ |Cd E cell=0.30 V and E ° cell=0.36 V. Then the valueof [Cd2+]/[Zn2+] is (1) 10 (2) 1 (3) 0.1 (4) 0.01
70. The standard emf of a galvanic cell involving cell reaction with n = 2 is found to be 0.295 V at 25°C. The equilibrium constant of the reaction would be (F=96500 C mol–1; R=8.314 JK–1 mol–1) (1) 1.0 × 1010 (2) 2.0 × 1011 (3) 4.0 × 1012 (4) 1.0 × 102
71. The Gibbs energy for the decomposition of Al2O3 at 500° C is as follows 23 2 24 AlOAl+O 33 → ; D r G =+960 kJ. The voltage needed for the electrolytic reduction of aluminium oxide Al2O3 at 500° C at least.
(1) – 4.5 V (2) – 3.0 V (3) – 2.5 V (4) – 5.0 V
72. What is reaction quotient,Q, for the cell Ni(s) / Ni2+(0.190 M) //Cl–(0.40 M) /Cl2(g), Pt(s) (1atm) (1) 3.1×10-1 (2) 1.3×10-1
(3) 8.0×10-2 (4) 3.0×10-2
73. During the discharge of lead storage battery, weight of PbSO 4 obtained at cathode involving of 1F of electricity is(At.Wt. of Pb = 208)
(1) 304 g (2) 152 g (3) 456 g (4) 608 g
Batteries
74. In a fuel cell, methanol is used as fuel and oxygen gas is used as an oxidiser. The reaction is
At 298K, standard Gibbs energies of formation for CH 3 OH (I) ,H 2 O (l) and CO 2(g) are –166.2,–237.2 and –394.4 kJ respectively. If standard enthalpy of combustion of methanol is –726 kJ, efficiency of the fuel cell will be (1) 80% (2) 87% (3) 90% (4) 97%
75. For lead storage battery pick the correct statements
A) During charging of battery, PbSO4 on anode is converted into PbO2
B) During charging of battery, PbSO4on cathode is converte d into PbO2
C) Lead storage battery, consists of grid of lead packed with PbO2 as anode D) Lead storage battery has ~38% solution of sulphuric acid as an electrolyte
Choose the correct answer from the options given below:
(1) B, D only (2) B, C, D only (3) A, B, D only (4) B, C only
76. Equivalent mass of H2SO4 in lead storage battery is (M = Molar mass) (1) 2M (2) M (3) M/2 (4) M/3
FURTHER EXPLORATION
1. A layer of chromium metal 0.25 mm thick is to be plated on a auto bumper with a total area of 0.32 M2 from a solution containing CrO4–2? What current flow is required for this electro plating if the bumper is to be plated in 60 sec the density of chromium metal is 7.20 g/cc.
(1) 4.9 × 103 A (2) 1.78 × 103 A
(3) 5.3 × 104 A (4) 10.69 × 104 A
2. In the electrochemical conversion (Kolbe’s eletrolysis) of R–COONa to R–R, 1 A current was passed for 965 seconds. Calculate the amount of R–R formed in this process (Faraday constant = 96,500 C mol –1) (1) 10 m mol (2) 5 m mol
MATCHING TYPE QUESTIONS
1. Match the terms given in Column I with the items given in Column II.
Column - I Column - II
(a) Λm(Molar conductance) (I) intensive property
(b) E°cell (II) depends on number of ions per unit volume
77. The anodic half – cell of lead battery is recharged using electricity 0.05 Faraday. The amount of PbSO 4 electrolysed in grams during the process is (molar mass of PbSO4=303 g mol–1)
(1) 22.8 (2) 15.2 (3) 7.6 (4) 11.4
78. A lead storage battery has been used for one month (30 days) at the rate of one hour per day by drawing a constant current of 2 amperes. H2SO4 consumed by the battery is (1) 1.12 mol (2) 2.24 mol (3) 3.36 mol (4) 4.48 mol
(3) 100 m mol (4) 50 m mol
3. In H2–O2 fuel cell the reaction occurring at cathode is
(1) 2H2 + O2 → 2H2O(l)
(2) H+ + OH– → H2O
(3) O2 + 2H2O + 4e– → 4OH–
(4) H+ + e– → 0.5 H2
4. The time period to coat a metal surface of 80 cm2 with 5 × 10–3 cm thick layer of silver (density =1.05 g cm–3) with the passage of 3A current through a silver nitrate solution is
(1) 125 seconds (2) 250 seconds
(3) 175 seconds (4) 200 seconds
(c) k (Conductivity) (III) extensive property
(d) Δ r G cell (IV) increases with dilution
Choose the correct answer from the options given below:
(A) (B) (C) (D)
(1) II I III IV
(2) I II IV III
(3) IV I II III
(4) II IV I III
2. Match the terms given in Column I(conductance) with the items given in Column II(units).
Column-I
Column–II
(A) Conductance (p) cm–1
(B) Specific conductance (q) ohm–1cm2mol–1
(C) Cell constant (r) ohm–1
(D) Molar conductance (s) ohm–1cm–1
Choose the correct answer from the options given below:
(A) (B) (C) (D)
(1) r s p q
(2) s r p q
(3) r s q p
(4) p s q r
3. Match material in column I with corresponding conductivity (S m –1) in column II
Column-I
Column-II
(A) Silver (I) 6.2×103
(b) Teflon (II) 1×10–18
(c) Germanium (III) 3.5×10–5
(d) Pure water (IV) 2
Choose the correct answer from the options given below:
(A) (B) (C) (D)
(1) I II IV III
(2) II I IV III
(3) IV II I III
(4) I III II IV
4. Match the terms given in Column I with the items given in Column II.
Column- I
Column - II
(A) Dilute solution of HCl (P) O2 evolved at anode
(B) Dilute solution of HCl (Q) H2 evolved at cathode
(C) Concentrate solution of NaCl (R) Cl2 evolved at anode
(D) Fairly concentrate solution of AgNO3 (S) Ag deposition at cathode
Choose the correct answer from the options given below:
(A) (B) (C) (D)
(1) P,Q P,Q R,Q P,S
(2) S P R Q
(3) P,R P S Q,S
(4) Q P R S
5. Match the terms given in Column I with the items given in Column II.
Column–I Column- II
(A) Electrolyte is a paste of KOH and ZnO (p) H2–O2 fuel cell
(B) Electrolyte is 38% H2SO4 solution (q) Mercury cell
(C) Water is the final product of cell reaction (r) Lead storage battery
(D) Cell in which metal changes from +3 state to +2 state in reaction at cathode (s) Nickel–Cadmium cell
Choose the correct answer from the options given below: (A) (B) (C) (D)
(1) r q p s
(2) q p r s
(3) q r p s
(4) q r s p
6. Match column I(batteries) with column II(change in oxidation state at cathode)
Column - I Column - II
(A) Fuel cell (I) +4 to +2
(B) Dry cell (II) +3 to +4
(C) Lead storage battery (III) 0 to - 2
(D) Ni-Cd cell (IV) +4 to +3 (V) +3 to +2
Choose the correct answer from the options given below:
(A) (B) (C) (D)
(1) III IV I V
(2) III I IV V
(3) III IV V I
(4) V IV I III
7. Match column I (electrolyte) with column II (product of electrolysis using inert electrodes)
Column- I
Column- II
(A) aq. CuCl2 (I) Ag and O2
(B) aq. Na2SO4 (II) H2 and Cl2
(C) aq. AgNO3 (III) H2 and O2
(D) aq. NaCl (IV) Cu and Cl2
Choose the correct answer from the options given below:
STATEMENT TYPE QUESTIONS
Each question has two statements. Statement I and statement II. Mark the correct answer as
(1) if both statement I and statement II are correct,
(2) if both statement I and statement II are incorrect,
(3) if statement I correct but statement II is incorrect,
(4) if statement I incorrect but statement II is correct.
(A) (B) (C) (D)
(1) III IV I II
(2) IV III II I
(3) III IV II I
(4) IV III I II
8. Match the terms given in Column I with the items given in Column II.
Choose the correct answer from the options given below:
(A) (B) (C) (D)
(1) V I II III
(2) V II I IV
(3) II I V III
(4) II V I IV
1. S-I : In electrolysis the quantity of electricity needed for depositing 1 mole of silver (from AgNO3) is different from that required for 1 mole of copper (from CuSO4).
S-II : The atomic weights of silver and copper are different.
2. S-I : Conductivity always increases with decrease in the concentration of electrolyte
S-II : Molar conductivity always increases with decrease in the concentration of electrolyte.
3. S-I : The increase in molar conductance with dilution is more in HCl than in HCOOH
S-II : On dilution the degree of dissociation of strong electrolyte increases while that of weak electrolyte decreases
4. S-I : aq.CuSO4 on electrolysis using Pt electrodes gives Cu at cathode and O2 at anode
S-II : aq.CuSO4 on electrolysis using Cu
ASSERTION AND REASON QUESTIONS
In each of the following questions, a statement of Assertion (A) is given, followed by a corresponding statement of Reason (R). Mark the correct answer as
(1) if both (A) and (R) are true and (R) is the correct explanation of (A),
(2) if both (A) and (R) are true but (R) is not the correct explanation of (A),
(3) if (A) is true but (R) is false,
(4) if both (A) and (R) are false.
1. (A) : molar conductivity of weak electrolyte at infinite dilution cannot be determined experimentally
(R) : Kohlrausch law helps to find the molar conductivity of a weak electrolyte at infinite dilution
2. (A) : If an aqueous solution of NaCl is electrolysed, the product obtained at the cathode is H2 gas and not Na.
(R) : Gases are liberated faster than the metals during the electrolysis of an electrolyte
electrodes gives Cu at cathode and Cu+2 ions at anode
5. S-I : Copper cannot reduce dilute HNO3
S-II : Copper cannot reduce dilute HCl.
6. S-I : ° cellE should have a positive value for galvanic cell to function.
S-II : cathodeanodeEE < for a feasible reaction, with reference to SRP
7. S-I : E lectrolysis of Brine solution using inert electrodes liberates H 2 at cathode
S-II : SRP of value of NHE is greater than SRP value of Na+/Na
3. (A) : The cell potential of mercury cell remains constant during its life.
(R) : The overall reaction does not involve any ion in solution whose concentration can change during its life time.
4. (A) : In a galvanic cell, the cathode is a positive electrode.
(R) : The tendency of metal ions from solution to deposit on a metal electrode makes the cathode positively charged.
5. (A) : In equation D r G=–nFEcell value of D r G Depends on n
(R) : Ecell is intensive property and D r G is an extensive property.
6. (A) : Galvanised iron does not rust
(R) : Zinc has a more negative electrode potential than iron
7. (A) : Copper is dissolved at anode and deposited at cathode when Cu
electrodes are used and electrolyte is 1 M CuSO4(aq) solution.
(R) : SOP of Cu is less than SOP of water and SRP of Cu is greater than SRP of water.
8. (A) : Molar conductivity of CH3COOH increases with dilution
(R) : The graph between λmv/s (C)1/2 is a straight line
9. (A) : Molar conductivity of 0.1 NH4OH solution is less than that of 0.001 M NH4OH solution
(R) : Dilution increases the degree of
BRAIN TEASERS
1. The time period to coat a metal surface of 80 cm2 with 5×10–3 cm thick layer of silver (density 1.05 g cm –3) with the passage of 3 A current through a silver nitrate solution is
(1) 115 seconds (2) 125 seconds
(3) 135 seconds (4) 145 seconds
2. A factory produces 40 kg of calcium in two hours by electrolysis. How much aluminum can be produced by same current in 2 hours if current efficiency is 50%?
(1) 22 kg (2) 18 kg
(3) 9 kg (4) 27 kg
3. Consider the galvanic cell,Pt(s)|H 2 (1 bar)|HCl (aq) (1M)|Cl 2(g) (1 bar)|Pt(s) After running cell for some time, the concentration of the electrlolyte is automatically raised to 3 M HCl. Molar conductivity of the 3 M HCl is about 240 S cm2mol–1 and limiting molar conductivity of HCl is about 420 S cm2mol–1. If Kb of water is 0.52 K kg/mol, what is the boiling point of the electrolyte at the end of the experiment?
(1) 375.6 K (2) 377.8 K
ionisation of NH4OH
10. (A) : A blue colour is obtained when a copper wire is immersed in AgNO3 solution
(R) : Silver reduces Cu2+ to copper
11. (A) : The molar conductance of weak electrolytes is low as compared to that of strong electrolytes at moderate concentrations
(R) : Weak electrolytes at moderate concentrations dissociate to a much greater extent when compared to strong electrolytes
4. Conductivity of a saturated solution of a sparingly soluble salt AB at 298 K is 1.85×10–6 S–1. Solubility product of the salt AB at 298 K is [given λ°M(AB)=1.4×10–4 S–2 mol–1]
5. In a Cu voltameter, mass deposited in 30 seconds is 100 g. Carefully analyse the current-time graph shown below and identify the incorrect statement.
(1) Electrochemical equivalent for Cu is 50
(2) A constant current of 66.66 mA would also discharge the same amount in the same time
(3) 33.33 g got discharged in 10 seconds
(4) 50 g got discharged in 15 seconds
6. Cell notation of a galvanic cell is represented as Ni(s)|Ni +2 ( M 1 ) || Cl –
( M 2)|Cl 2(1atm),Pt(s).Emf of the cell is maximum when the value of M1 and M2 are
(1) M1 = M2= 0.01 M
(2) M1 = 1M; M2 = 1M
(3) M1 = 0.1 M; M2 = 0.01 M
(4) M1 = 0.01 M; M2 = 0.01 M
7. The potential of the cell containing two hydrogen electrodes as represented below Pt,H2(g)|H+(10–6 M)||H+(10–4 M)|H2(g)Pt at 298 K is
(1) –0.118 V (2) –0.0591 V
FLASH BACK (Previous NEET Questions)
1. Find the emf of the cell in which the following reaction takes place at 298 K
Ni(s)+2Ag+(0.001M)→Ni2+(0.001M)+2
Ag(s)
(Given
0 cell 2.303RT
E=10.5V,=0.059 at 298K F )
(1) 1.385 V (2) 0.9615 V
(3) 1.05 V (4) 1.0385 V
2. Given below are half cell reactions: 2 42 854 MnOHeMnHO −+−+ ++→+
4 2 2 / 1.510 + =− MnMnO EV 22 1 22 2 OHeHO +− ++→ 22 / 1.223 OHO EV =+ Will the permanganate ion, 4MnO liberate O 2 from water in the presence of an acid?
(1) No, because 0 0.287 cell EV =−
(2) Yes, because 0 cell E2.733V =−
(3) No, because 0 cell E2.733V =+
(3) 0.118 V (4) 0.0591 V
8. Iron powder is added to 1.0 M solution of CdCl2 at 298 K. The reaction occurring is Cd 2+ (aq) +Fe (s) → Cd (s) +Fe 2+ (aq) . If the standard potential of a cell producing this reaction is 0.037 V, the concentrations of Cd2+ and Fe2+ ions in the above reaction at equilibrium respectively will be (Hint : 101.3=20)
(1) 0.05 M, 0.95 M
(2) 0.95 M, 0.05 M
(3) 0.40 M, 0.60 M
(4) 0.60 M, 0.40 M
(4) Yes, because 0 cell E0.287V =+
3. At 298K, the standard electrode potentials of Cu2+/Cu,Zn2+/Zn,Fe2+/Fe and Ag+/Ag are 0.34 V,–0.76V,–0.44V and 0.80V respectively
On the basis of standard electrode potential predict Which of the following reaction can not occur?
(1) CuSO4(aq)Fe (s) →FeSO4(aq)+Cu (s)
(2) FeSO4(aq)Zn (s) →ZnSO4(aq)+Fe (s)
(3) 2CuSO4 (aq) +2Ag (s) →2Cu (s) +Ag2SO4 (aq)
(4) CuSO4 (aq) Zn (s) →ZnSO4 (aq) +Cu(s)
4. The molar conductance of NaCl, HCl and CH 3 COONa at infinite dilution are 126.45, 426.16 and 91.0 S cm 2 mol–1 respectively. The molar conductance of CH3COOH at infinite dilution is. Choose the right option for your answer
(1) 540.48 S cm2 mol-1
(2) 201.28 S cm2 mol-1
(3) 390.71 S cm2 mol-1
(4) 698.28 S cm2 mol-1
5. The molar conductivity of 0.007 M acetic acid is 20 S cm 2 mol -1 . What is
the dissociation constant of acetic acid? Choose the correct option.
(1) 2.50 x 10-5 mol L-1
(2) 1.75 x 10-4 mol L-1
(3) 2.50 x 10-4 mol L-1
(4) 1.75 x 10-5 mol L-1
6. On electrolysis of dil. sulphuric acid using Platinum (Pt) electrode, the product obtained at anode will be :
(1) Oxygen gas
(2) H2S gas
(3) SO2 gas
(4) Hydrogen gas
7. The number of Faradays (F) required to produce 20 g of calcium from molten CaCl2 (Atomic mass of Ca =40 g mol–1) is
(1) 2 (2) 3
(3) 4 (4) 1
8. For a cell involving one electron, E°cell=0.59V at 298 K, the equilibrium constant for the cell reaction is [Given that 2.303 0.059 RT F = V at T=298 K]
CHAPTER TEST
1. Correct statement(s) is/are:
A. Electrical conductance through metals is called metallic or electronic conductance.
B. The electronic conductance do not depend on temperature.
C. The conductance of electricity by ions present in the solutions is called electrolytic or ionic conductance.
D. The conductivity of electrolytic (ionic) solutions depends on the nature of the electrolyte added, the nature of the solvent and its viscosity, temperature etc.
(1) 1.0 ×1030
(2) 1.0 ×105
(3) 1.0 ×102
(4) 1.0 ×1010
9. In the electrochemical cell :
Zn|ZnSO 4 (0.1 M) ||CuSO 4 (1.0 M)|Cu the emf of this Daniel cell is E 1. When the concentration of ZnSO4 is changed to 10 M and that of CuSO 4 changed to 0.01 M,the emf changes to E2.From the following, which one is the relationship between E1 and E2 (Given, RT/F=0.059)
(1) E1<E2
(2) E1>E2
(3) E2=0≠E1
(4) E1 =E2
10. The pressure of H2 required to make the potential of H 2 electrode zero in pure water at 298 K
(1) 10–10 atm
(2) 10–4 atm
(3) 10–14 atm
(4) 10–12 atm
(1) A and B
(2) A, C and D
(3) Only A and D
(4) Only B
2. Given below are two statements. one is labelled as Assertion A and the other is labelled as Reason R.
(A) : Mercury cell does not give steady potential.
(R) : In the cell reaction, ions are not involved in solution
In the lig ht of the above statements, choose the most appropriate answer from the options given below.
(1) if both (A) and (R) are true and (R) is the correct explanation of (A),
(2) if both (A) and (R) are true but (R) is not the correct explanation of (A),
(3) if (A) is true but (R) is false,
(4) if both (A) and (R) are false.
3. Which of the following pair(s) is/are incorrectly matched?
(i) R(resistance) – Ohm
(ii) ρ (resistivity) – Ohm metre
(iii) C (conductance) - seimen or mho
(iv) k (conductivity) - Ohm metre-1
(1) i, ii and iii (2) ii and iii
(3) i, ii and iv (4) iv only
4. Given below are two statements.
S–I : Conductivity of weak electrolytes decreases with dilution whereas molar conductance increases.
S–II : On dilution, number of ions per milliliter decreases, but total number of ions increases considerably.
In the light of the above statements, choose the most appropriate answer from the options given below.
(1) if both statement I and statement II are correct,
(2) if both statement I and statement II are incorrect,
(3) if statement I correct but statement II is incorrect,
(4) if statement I incorrect but statement II is correct.
5. The correct match is List I List II
(A) Faraday’s first law (I) e × 96500
(b) Chemical equivalent (II) 12 12 mm EE =
(c) Electrolytic dissociation (III) S.H.E
(d) Pt,H2(atm)/ H+(1M) (IV) m=eQ
(V) Salt bridge
Choose the correct answer from the options given below: (A) (B) (C) (D)
(1) IV I III II
(2) IV V II III
(3) I I II III
(4) IV I II III
6. If the resistance and specific resistance of a solution have same magnitude, then its cell constant is equal to
(1) 0 (2) 10
(3) 100 (4) 1
7. Calculate the EMF of the cell, Na/Na+=(1 M)/Fe2+(1M)/Fe
Given, / + o NaNa E =–2.71V and E° Fe+2/Fe =–
0.44V
(1) +1.98 V
(2) –2.47 V
(3) –3.15 V
(4) +2.27 V
8. The pH of an acid solution in contact with hydrogen electrode is one. Reduction potential of Hydrogen electrode is
(1) –3.9 mV
(2) –118 mV
(3) –59 mV
(4) –11.8 mV
9. Which among the following is incorrect statement?
(1) When Lead storage battery is discharged ,H2SO4 is consumed
(2) When Lead storage battery is discharged H2SO4 is formed
(3) Zinc is used to protect corrosion of iron
(4) Efficiency of fuel cell is about 70%
10. The efficiency of a cell is 60%. Its cell reaction is A (s)+B 2+ (aq)→A 2+ (aq)+B; D H = –240 kJ The standard electrode potential of cell is
(1) 0.746 V (2) 0.37 V
(3) 1.48 V (4) 2.96 V
11. A dilute aqueous solution of CuSO 4 is electrolysed using platinum electrodes. The products at the anode and cathode are:
(1) O2, H2 (2) H2, O2
(3) O2, Cu (4) S2O82-, H2
12. Given,
Which of the following reactions under standard condition will not take place in the specified direction?
(1) Ni2+(aq)+Cu (s) →Ni (s) +Cu2+ (aq)
(2) Cu(s)+2Ag+ (aq) → Cu2+ (aq) +2Ag(s)
(3) Cu+2 (aq) +H2 (g) → Cu (s) +2H+(aq)
(4) Zn (s) +2H+ (aq) → Zn2+ (aq) +H2(g)
13. Molar conductance of 0.1 M weak mono basic acid(HA) is 36 mho cm 2 mol–1. At
the same temperature, molar conductance of HA at infinite dilution is 360 mho cm2 mol–1 pH of HA will be
(1) 2 (2) 3
(3) 2.7 (4) 3.3
14. For a cell involving two electron transfer,E°=0.59V at 298 K. Equilibrium constant for the cell reaction will be
(1) 105 (2) 1010
(3) 1020 (4) 108
15. Resistance of 0.2 M solution of an electrolyte is 50Ω. The specific conductance of the solution is 1.4 Sm–1 . The resistance of 0.5 M solution of the same electrolyte is 280 Ω The molar conductivity of 0.5 M solution of the electrolyte in Sm2 mol–1 is
(1) 5 ×10–4
(2) 5 ×10–3
(3) 5 ×103
(4) 5 ×102
16. Given below are two statements. one is labelled as Assertion A and the other is labelled as Reason R.
(A) : Cu is less reactive than hydrogen (R) : 2 0 / CuCu E + is negative.
In light of the above statements, choose the most appropriate answer from the options given below.
(1) Both A and R are true and R is the correct explanation of A
(2) Both A and R are true and R is not correct explanation of A
(3) A is true but R is false
(4) A is false but R is true
17. Match the laws given in Column I with expressions given in Column II
Column I
Column II
(A) Lead storage battery (p) During discharging density of H2SO4 decreases
(b) Mercury cell (q) Reaction at cathode 4e–+O2+2H2O → 4OH–
(c) Fuel cell (r) Reaction at cathode MnO2+NH4++e–→MnO(OH)+NH3
(d) Leclanche cell (s) Suitable for low current devices like hearing aids
(A) (B) (C) (D)
(1) s p q r
(2) p r q s
(3) q r p s
(4) p s q r
18. The weight of silver (at. wt=108) displaced by a quantity of electricity which displaces 5600 mL of O2 at STP will be
(1) 5.4 g (2) 10.8 g
(3) 54.0 g (4) 108.0 g
19. A graph is plotted between Λ c (y-axis) versus C (x-axis) for KCl solution. The nature of the graph will be ( C = concentration of KCl)
(1) A straight line with positive slope and negative y–intercept
(2) A straight line with negative slope and positive y–intercept
(3) A straight line passing through origin
(4) A straight line with positive slope and positive y–intercept
20. Given below are two statements.
S–I : Increase in molar conductance with dilution is more in HBr than in HCOOH.
S–II : HBr (aq) is a stronger electrolyte whereas HCOOH (aq) is a weaker electrolyte.
In the light of the above statements, choose the most appropriate answer from the options given below.
(1) Statements I and II are correct
(2) Statement I is correct and statement II is incorrect
(3) Statement I is incorrect and statement II is correct
(4) Statements I and II are incorrect
21. Given below are two statements.
S–I : In a galvanic cell cathode has a positive potential with respect to solution.
S–II : Potential of a individual half cell cannot be measured.
In the light of the above statements, choose the most appropriate answer from the options given below.
(1) Both statement I and statement II are correct
(2) Both statement I and statement II are incorrect
(3) Statement I is correct and statement II is incorrect
(4) Statement I is incorrect and statement II is correct
22. Given below are two statements. one is labelled as Assertion A and the other is labelled as Reason R.
(A) : The molar conductance of weak electrolytes is less when compared
to that of strong electrolytes at moderate concentration
(R) : The degree of dissociation of weak electrolytes is less than that of strong electrolytes at moderate concentration
In the light of the above statements, choose the most appropriate answer from the options given below.
(1) Both ‘A’ & ‘R’ are correct and ‘R’ is the correct explanation of ‘A’
(2) Both ‘A’ & ‘R’ are correct and ‘R’ is not the correct explanation of ‘A’
(3) ‘A’ is true & ‘R’ is false
(4) Both ‘A’ & ‘R’ are false
23. E 1 , E 2 and E 3 are the emf values of the three galvanic cells respectively
(i) Zn| Zn2+(1M)||Cu2+(0.1M)|Cu
(ii) Zn| Zn2+(1M)||Cu2+(M)|Cu
(iii) Zn| Zn2+(0.1M)||Cu2+(1M)|Cu
Which one of the following is true?
(1) E2>E 3 >E1
(2) E 3 >E2>E1
(3) E1>E2>E 3
(4) E1>E 3 >E2
24. The S.R.Ps of Cu2+/Cu,Hg2+/Hg and Zn2+/ Zn are respectively 0.34 V, 0.85 V and -0.76 V. The wrong statement is
(1) Cu reduces Hg2+
(2) Zn reduces Cu2+
(3) Hg reduces Zn2+
(4) Zn reduces both Cu2+ and Hg2+
25. Which of the following is a good conductor of electricity?
(1) Diamond
(2) Graphite
(3) Solid NaCl
(4) Wood
26. Given below are two statements. one is labelled as Assertion A and the other is labelled as Reason R.
(A) : Λ m = kV , on dilution k decreases but Λ m increases.
(R) : On dilution, decrease in k is more than compensated by increase in its V
In the light of the above statements, choose the most appropriate answer from the options given below .
(1) (A) is true but (R) is false
(2) Both (A) and (R) are true but (R) is not the correct explanation of (A)
(3) Both (A) and (R) are true and (R) is the correct explanation of (A)
(4) Both (A) and (R) are false
27. The products of electrolysis of aqueous NaCl solution using inert electrodes are
(1) Na at cathode and Cl2 at anode
(2) H2 at cathode and Cl2 at anode
(3) H2 at cathode and O2 at anode
(4) Na at cathode and O2 at anode
28. Match the items of Column I and Column II
Column I
Column II
(A) k (specific conductance) (I) current × time
(b) Λ m (molar conductance) (II) Λ c / Λ o
(c) α(degree of dissociation) (III) 1000 M × k
(d) Q (quantity of electricity) (IV) G R ∗
(V) Salt bridge
Choose the correct answer from the options given below:
(A) (B) (C) (D)
(1) IV III I IV
(2) I II III IV
(3) IV III II I
(4) III I II IV
29. Cost of Electricity to liberate 24 g of Oxygen is Rs 3000. Cost of Electricity to deposit 24 g of magnesium will be
(1) Rs 2000
(2) Rs 1500
(3) Rs 4300
(4) Rs 2500
30. S.R.P values of some electrodes are given below.
2 0 / E2.37V + =− MgMg
E1.36V; =+ ClCl
2 1 / 2
0 / E0.8V; +=+AgAg
2 0 / E0.76V + =− ZnZn
From the above data, identify the best reducing agent
(1) Zn (2) Mg
(3) Cl2 (4) Ag
31. Molar conductance at infinite dilution of Ammonium chloride, Sodium hydroxide and Sodium chloride are X , Y and Z respectively. Molar conductance at infinite dilution of Ammonium hydroxide at the same temperature would be (all measurements expressed in SI units)
(1) X–Y + Z
(2) X+ Y–Z
(3) Y + Z –X
(4) X + Z –Y
32. Brine solution on electrolysis using inert electrodes gives ----- and ----- at anode and cathode respectively.
(1) Cl2 , H2 (2) Cl2,Na (3) O2, H2 (4) O2 ,Na
33. The following statements is correct with respect to both electrolytic cell and Galvanic cell
(1) in both cells, anode is shown by +ve sign
(2) in both cells, cathode is shown by –ve sign
(3) in both cells, reduction reaction takes place at the cathode
(4) in both cells, oxidation reaction takes place at the cathode
34. Aqueous CuSO4 is subjected to electrolysis using copper electrodes by passing 0.2 F of electricity. Correct statement about the process is
(1) 0.2 mol of oxygen is liberated at anode
(2) 2.24 litres of hydrogen (at STP) is liberated at cathode
(3) 0.2 moles of copper decomposes at anode
(4) 0.1 mol of copper is deposited at cathode
35. 2 0 -12-1 224 ohmcmg eq Λ= ClCHCOONa
Λ= HCl
0 -12-1 203 ohmcmgeq
0 -12-1 203 ohmcmgeq Λ= HCl
What is the value of 2 0 ClCHCOOH Λ ?
(1) 288.5 Ohm–1 cm2 g eq–1
(2) 289.5 Ohm–1 cm2 g eq–1
(3) 388.8 Ohm–1 cm2 g eq–1
(4) 59.5 Ohm–1 cm2 g eq–1
36. During the electrolysis of acidulated water, the mass of hydrogen obtained is ‘ x’ times that of O2 and the volume of is ‘ y’ times that of O2. The ratio of ‘y’ and ‘ x’ is
(1) 16 (2) 8
(3) 0.125 (4) 0.25
37. S.R.P values of some electrodes are given below.
2 2 1 / / 2 2.37 V;1.36 V ; + =−=+oo MgMgClCl EE
2 // 0.8 V ;0.76 V ; ++=+=oo AgAgZnZn EE
From the above data, identify the best reducing agent
(1) Zn (2) Mg
(3) Cl2 (4) Ag
38. For the given cell representation ()()()() 23 |||| ++ saqaqs MgMgAlAl total number of electrons transferred is (1) 2 (2) 3 (3) 6 (4) 5
39. For a particular galvanic cell, the cell notation is Cu | Cu2+||Ag+|Ag. The metal which dissolves is (1) Cu (2) Ag (3) Both Cu and Ag (4) Neither Cu nor Ag
40. Reason for increase in electrical conduction of electrolyte with increase in temperature is A) increase in the number of ions B) increase in the speed of ions C) increase in the degree of dissociation of electrolyte
(1) A and B only (2) B and C only (3) A and C only (4) A, B and C
41. When an electric current is passed through acidified water, 112mL of hydrogen gas at N.T.P was collected at
the cathode in 965 seconds. The current passed in ampere is (1) 1.0 (2) 0.5 (3) 0.1 (4) 2.0
42. Given 32 00//0.72,0.42 ++ =−= CrCrFeFe EVEV
The potential for the cell ()() 32 /0.1||0.01| CrCrMFeMFe ++ is (1) 0.339 V (2) – 0.339 V (3) – 0.26 V (4) 0.26 V
43. How long (approximate) should water be electrolyzed by passing through 100 amperes current so that the oxygen released can completely burn 27.66 g of diborane (Atomic weight of B = 10.8 u)
(1) 6.4 hours
(2) 0.8 hours
(3) 3.2 hours
(4) 1.6 hours
44. Eight grams of aq.CuSO4 is subjected to electrolysis by passing 1.5×1022 electrons using inert electrodes. Weight of cupric ions left over after electrolysis is (Atomic wt. of copper = 64)
(1) 6.4 g (2) 7.2 g
(3) 1.6 g (4) 2.4 g
45. Electrolysis using inert electrodes neither changes the composition of the electrolyte nor changes the pH of the solution in the case of (1) aq NaCl
The feasibility of a reaction can be predicted by thermodynamics. Along with feasibility, the extent of reaction can be determined from chemical equilibrium. It is equally important to know the speed at which a reaction is occurring. This is called kinetic feasibility of a reaction.
The branch of chemistry which deals with the study of reaction rates and their mechanisms is called chemical kinetics or reaction kinetics. The word kinetics is derived from the Greek word ‘kinesis’, meaning movement.
3.1 RATE OF REACTION
We observe that certain chemical reactions are fast, some are slow, and a few other reactions occur at moderate pace. Fast reactions are those in which the time required for the completion of the reaction is very short. Reactions between ionic substances are fast, and they are called instantaneous reactions. Slow reactions are those in which the time required for the completion of the reaction is very long. Generally, reactions between covalent substances are slow and time consuming.
How fast or how slow the reactions occur is dependent on the nature of reactants, the nature of the reaction, and experimental conditions of the reaction. The relative speed of some chemical reactions are listed in Table 3.1.
The speed of an automobile is expressed in terms of change in the position or distance by it in a certain period of time. Similarly, the rate of a reaction, also called the speed of the reaction, is expressed in terms of change in the concentration in a certain period of time.
Table 3.1 Some chemical reactions and their speeds Equation
1. HCl(aq) + KOH(aq) → KCl(aq) + H2O(l)
The rate of a reaction is defined as the change in molar concentration of a reactant or a product per unit time. To be more specific, rate can be expressed as the rate of decrease in concentration of a reactant or the rate of increase in concentration of a product.
Let us consider a hypothetical reaction, where one mole of reactant ‘ R ’ is converted to one mole of product ‘P’, assuming that the volume of the system remains constant. If the concentrations of reactant and product at time t1, are [R]1 and [P]1, respectively, and at time t2, the concentrations are [R]2 and [P]2, then i. change in time, ∆ t = t2– t1 = time taken
ii. change in the concentration of reactant,
∆ [R] = [R]2–[R]1
iii. change in the concentration of product,
∆ [P] = [P]2–[P]1
(Here, the square brackets are used to express molar concentrations.)
Rate of disappearance of R =
Decreaseinconcentrationof [] timetaken -∆ = ∆ RR t
For the reaction of forming nitric oxide from its elements, N2 + O2 2NO, the rate measured with respect to N 2 is ∆∆ 2 [N] t ,
rate with respect to O 2 is ∆∆ 2 [O] t rate with respect to NO is ∆ + ∆ [NO] t .
The rate of the reaction, expressed in terms of each substance, can be different. For example, in the above reaction, the rate of formation of NO is twice the rate of removal of either N2 or O2. To express a common rate equation for any reaction, stoichiometric coefficients of different participant substances are written as denominators, to relate the rates measured.
∆∆∆ -=-=+ ∆∆∆ 22 [N][O][NO] 2 ttt
For the reaction
5Br–(aq) + BrO3–(aq)→ 3Br2(aq) + 3H2O(l)
Rate is given as -
-=-
Rate of appearance of P = +∆ = ∆
Timetaken P P t
Increaseinconcentrationof[]
Rate of a reaction is mathematically given as
Rate = ∆-∆+∆ == ∆∆∆ [][][] CRP ttt
Negative sign denotes that the reactant concentration decreases and positive sign denotes that the product concentration increases in the reaction with time.
The rates can be determined with respect to any reactant or any product present in a given chemical reaction. However, in arriving at the rate of reaction, the stoichiometric coefficients of the reactants or the products are to be taken into account.
∆∆ 3 1Br[BrO] 5 tt
∆∆ == ∆∆ 22 [Br][HO]11 33tt
3.1.1 Instant Rate
The rate of a reaction is not constant throughout the reaction. The rates obtained over an interval of time are called average rates of the reaction. Average rate depends upon the change in concentration of a substance and time taken for that change to occur.
The rate of a reaction at a specified time is called instant rate. It is also called instantaneous rate. The rate of a chemical reaction at any instant is the decrease in the
concentration of reactant or increase in the concentration of product at that specified time during the occurrence of the reaction.
Average and instant rates of a reaction with respective to reactants and products are shown graphically in Fig.3.1.
Instant rate can be determined graphically by drawing a tangent at time ‘t’ (t is the time at which instantaneous rate is to be determined) on either of the curves for concentration of reactant (or concentration of product) versus time and calculating its slope.
The units of rate are concentration time–1. If concentration is in mol L–1 and the time is given in seconds, the unit of rate is mol L–1 s–1 or mol dm –3 s –1. For gaseous reactions, partial pressures of the substances are considered. In that case, the unit of rate will be atm s –1 or Pa s–1. The unit of rate is also given as g L–1s–1 or in SI system, as kg m –3 s–1
1. In the reaction of formation of sulphur trioxide by contact process 2SO2 +O2 2SO3, the rate of reaction was measured as 2 []dO dt =–2.5 × 10–4 mol L–1 s–1. What is rate of reaction in terms of [SO2] in mol L–1s–1?
Sol. From rate law
3 22 [] [][] 11 22 dSOdOdSO dtdtdt-==
22 [][] 2 dSOdO dtdt \-=-× = –2 × 2.5 × 10–4
= –5 × 10–4 mol L–1s–1
Try yourself:
1. In the reaction, A+2B → 6C+2D, if the initial rate []dA dt - at t = 0 is 2.06 × 10–2 M s–1, what will be the value of []dB dt at t = 0?
Ans: 25. –2×10
TEST YOURSELF
1. The instantaneous rate of the reaction is expressed in the following way for a reaction 11 23 dCdD dtdt
The reaction is
(1) 4A+ B → 2C+3D
(2) B+3D → 4A + 3D
(3) A+ B → C+ D
(4) B+D → A+C
Concentration of reactions Concentration of reactions
Fig.3.1 Average rate and instantaneous rate of a chemical reaction
2. For the chemical reaction N2(g)+3H2(g)→ 2NH3(g), the correct option is:
(1) 1123 32 dHdNH dtdt +=
(2) 23 2 dNdNH dtdt - =
(3) 23 1 2
dNdNH dtdt - =
(4) 3223
dHdNH dtdt =
3. For 2SO2 + O2 →2SO3 rate of disappearance of SO2 is 4 × 10–3Ms–1 at t = 10 s. Then the amount of SO3 formed and amount of O2 consumed at t = 10 s respectively, are
(1) 0.32 g and 0.064 g
(2) 0.1 g and 0.2 g
(3) 0.016 g and 0.064 g
(4) 0.01 g and 0.1 g
4. For the reaction N 2 + 3H 2 → 2NH 3 , if
3 NH t ∆ ∆ = 2 × 10 –4 mole L –1s –1, the value
2 H t -∆
∆ would be
(1) 1 × 10–4 mole L–1s–1
(2) 3 × 10–4 mole L–1s–1
(3) 4 × 10–4 mole L–1s–1
(4) 6 × 10–4 mole L–1s–1
5. For the following reaction xA → yB loglog0.3, - =+ dAdB dtdt the ratio of x : y is
(1) 3 : 1 (2) 1 : 3 (3) 2 : 1 (4) 1 : 2
Answer Key
(1) 2 (2) 3 (3) 1 (4) 2 (5) 3
3.2 FACTORS INFLUENCING REACTION RATES
Na t ure of reactants: Reactions between ionic substances in their solution state are instantaneous, as oppositely charged ions combine easily. A white precipitate of silver chloride is obtained immediately by mixing solutions of sodium chloride and silver nitrate.
l(aq) + AgNO3(aq) fast → AgCl(s) + NaNO3(aq)
Reactants between covalent substances involve bond breaking and bond making. The more the number of bonds involved, the more is the time required for the reaction. Formation of ammonia between the elements nitrogen and hydrogen is very slow.
It is observed that smaller the particle size of the solid reactants, greater is the surface area and higher is the rate of reaction.
Concentration of reactants: At a given temperature, the rate of reaction depends on the concentration of the reactants. The effect of concentration on the rate of the reaction is known from the law of mass action. The rate of a reaction at any instant of time is directly proportional to the product of concentration of reactants taking part in the reaction at that instant.
In case of gaseous reactions, partial pressures are measured for the substances. The rate of the reaction is proportional to the partial pressure of each of the reactants involved in the reaction.
Temperature: The rate of reaction increases with increase in temperature of the reaction. 50% of nitrogen pentoxide decomposes in 10 days at 0ºC, in 5 hours at 25ºC and in 12 minutes at 50ºC.
Catalyst: Catalyst is a substance that can increase the rate of reaction, without itself undergoing any chemical change. When a suitable catalyst is added to reactants, rat e of reaction increases
NaC
TEST YOURSELF
1. Which of the following does not affect the rate of reaction?
(1) Amount of the reactant taken
(2) Physical state of the reactant
(3) ΔH of reaction
(4) Size of vessel
2. Burning of coal is represented as C(s)+O2(g)→ CO2(g). The rate of this reaction is increased by
(1) decrease in the concentration of oxygen (2) powdering the lumps of coal
(3) decreasing the temperature of coal
(4) providing inert atmosphere
3. Which of the following explains the increase of reaction rate by a catalyst ?
(1) Catalyst provides the necessary energy to the colliding molecules to cross the barrier.
(2) Catalyst decreases the rate of backward reaction so that the rate of forward reaction increases.
(3) Catalyst decreases the enthalpy change of the reaction.
(4) Catalyst provides an alternative path of lower activation energy.
(1) 3 (2) 2 (3) 4
3.3 THEORY OF REACTION RATES
It is necessary to understand the mechanism of reaction occurrence and the various factors capable of influencing reaction rates. The following theories can explain them.
3.3.1 Arrhenius T heory
The rate constant of a reaction is given in terms of frequency factor (A), energy of activation (Ea) and absolute temperature (T) as,
Arrhenius provided the physical justification for the rate of reaction. He interpreted that the main factor on which the rate of a chemical reaction depends is energy of activation (Ea).
Consider the formation of hydrogen iodide on heating gaseous hydrogen with iodine. According to Arrhenius, this reaction takes place only when a molecule of hydrogen and a molecule of iodine collide to from an unstable intermediate, as shown in Fig.3.2. The intermediate is called transition state, as it exists for a very short period of time. It breaks up to finally give product molecules.
Fig.3.2 Formation of hydrogen iodide from elements through transition state
The transition state is a hypothetical state of the substance at which bond breakage and bond formation simultaneously occur. The energy required to form this activated complex intermediate is called activation energy. Activated complex has more energy compared to reactants as well as products.
All the molecules of reacting substances do not have the same kinetic energy. In the Maxwell’s curve of the distribution of molecular velocities of gases, the peak position corresponds to most probable kinetic energy, as shown in Fig.3.3.
Fig.3.3 Distribution of energy among gaseous molecules
In the distribution curve, the fraction of molecules (NE/NT ) with a given kinetic energy (E) is plotted on y-axis and kinetic energy is placed on x-axis. Here, NE is the number of molecules with energy E and NT is the total number of molecules. With an increase in the temperature, fraction of molecules with lower kinetic energy decreases and that with higher kinetic energy increases as shown in Fig.3.4.
reaction is postulated to occur when molecules collide with one another.
The M ain Points of Collision Theory
The reactant molecules must collide together, if they are to react with each other
The number of collisions per second per unit volume of the reaction mixture is known as collision frequency.
Fig.3.4 Maxwell distribution curve showing dependence of kinetic energy on temperature
It is clear that at the temperature, (t +10), the molecule having energy greater than activation energy gets doubled in comparison to those at temperature, t, leading to doubling the rate of the reaction.
3.3.2 Collision T heory
Molecular collision theory of reaction rates was developed by Trautz and Lewis. It is based on kinetic theory of gases. It provides a better insight into the energetic and mechanistic aspects of chemical reaction. According to simpler molecular collision theory, the reactant molecules are assumed to be hard spheres. The
All collisions do not lead to chemical reactions. The collisions that bring about a reaction are called effective collisions or activated collisions or fruitful collisions. The molecules that involve in such collisions are called activated molecules. The collisions that do not give products are normal collisions. The molecules involve in them are called normal molecules.
To bring about a chemical reaction or activated collisions, molecules of reactant must cross two barriers. They are
1. threshold energy
2. orientation of molecules
To bring about the activated collisions the molecules must posses a minimum amount of energy, called threshold energy of the reaction.
The energy barrier diagram is given in Fig.3.5.
For a bimolecular elementary reaction,
A + B products, rate of reaction can be represented as,
Fig.3.5 Energy profile diagrams
Rate = ZAB.e–Ea/RT
where, Z AB represents the collision frequency of reactants A and B and e –Ea/RT represents the fraction of molecules with energies equal to or greater than E a .
The fraction of effective collision is given by= a kE/RT e.
The additional energy required by reacting molecules to attain threshold energy is called activation energy.
Activation energy = Threshold energy –Energy of colliding molecules.
In addition to the above energy requirement, the reactant molecules must be properly oriented to bring about the reaction, The proper orientation means that the molecules approach in proper direction in collision, such that the bond breakages occur and, thus, reaction occurs. Properly oriented reactant molecules lead to bond formation, as shown in Fig.3.6. Improper orientation makes them simply bounce back and no products are formed.
To account for effective (fruitful) collisions, another factor P, called the probability or steric factor, is introduced. It takes into account the fact that in a collision, molecules are properly oriented, i.e.,
Rate = PZAB e–Ea/RT
Thus, in collision theory, activation energy and proper orientation of molecules together determine the criteria for an effective collision and hence, the rate of a chemical reaction.
Energy of activation is a characteristic of reactions. It is independent of concentration of reactants. In a reversible reaction, if activation energy of forward reaction Ea(f) is less than that of the backward reaction E a (b), the reaction is an exothermic reaction.
The heat of the reaction, ∆H = Ea(f)-Ea(b).
Reactions with lower activation energy are fast and those with higher activation energy are slow.
Increasing the concentration of reactants increases the rate. This is because of the increase in the collision frequency and increase in the number of reactant molecules crossing the energy barrier.
2. A catalyst lowers the activation energy of a certain reaction from 83.314 to 75 kJ mol –1 at 500 K. What will be the rate of reaction as compared to uncatalysed reaction? Assume that other things are equal.
Hence, rate is increased by 7.38 times of the initial rate, when catalyst is used.
Try yourself:
2. The activation energy and heat of the reaction, A+B → C+D +38 kcal is 20 kcal. What would be the activation energy of the following reaction, C+D → A+B.
Ans: 2Ea =58.
TEST YOURSELF
1. For the reaction A+B ⇔ C+D , the forward reaction is exothermic. The activation energy of C + D is .......... that for the formation of A+ B
(1) equal to (2) less than (3) greater than (4) double
2. Which of the following statement is true?
(1) the existence of certain intermediates in a reaction mechanism can sometimes be proven because intermediates can sometimes be trapped and identified.
(2) intermediates in a reaction mechanism cannot be isolated because they do not have finite lifetimes.
(3) reaction mechanisms cannot have any more than one intermediate.
(4) intermediates in a reaction mechanism appear in the overall, balanced equation for the reaction.
3. Rate of general reaction A+B → Products can be expressed as follows, on the basis of collision theory: Rate=ZAB e− Ea/RT. Which of the following statements is not correct for the above expression?
(1) Z is collision frequency and it is equal to number of collisions per second per
unit volume of the reaction mixture.
(2) e−Ea/RT is the fraction of molecules with kinetic energy equal to or greater than Ea.
(3) Ea is activation energy of the reaction.
(4) All the molecules that collide with one other are effective collisions.
4. Which of the following statements is incorrect about the collision theory of chemical reaction?
(1) it considers reacting molecules or atoms to be hard spheres and ignores their structural features.
(2) number of effective collisions determines the rate of reaction.
(3) collision of atoms or molecules possessing only proper orientation results in the product formation.
(4) molecules with sufficient thresh old energy and proper orientation should collide for the collision to be effective.
5. Consider an endothermic reaction, X → Y, with the activation energies E b and E f for the backward and forward reactions, respectively. In general,
(1) Eb < Ef (2) Eb > Ef
(3) Eb = Ef
(4) there is no definite relation between Eb and Ef
Answer Key
(1) 3 (2) 1 (3) 4 (4) 3
(5) 1
3.4 TEMPERATURE DEPENDENCE OF RATE OF REACTION
The temperature dependence of the rate of a chemical reaction is explained by Arrhenius equation, which was originally called the van’t Hoff’s proposal.
= a kAeE/RT ,
Here, A is the Arrhenius factor, also called frequency factor or pre-exponential factor. It
is a constant for a given reaction. k is the rate constant, R is the molar gas constant and E a is the energy of activation.
Taking logarithms on bot h sides of the Arrhenius equation, we get
pre-exponential factors are same at different temperatures,
3.4.1 Effect of Catalyst on Rate
A plot of ln k versus T–1 is a straight line, (as shown in Fig.3.7, with a negative slope. The slope is given as –Ea /R.
ln k
Slope = –E a R 1 T
Fig. 3.7 An Arrhenius plot is a graph of ln k against 1 T
I t has been found that for a chemical reaction, the specific rate is approximately doubled for every 10 º C rise of temperature for many reactions. The ratio of two specific rates measured at temperatures that differ by 10°C is called the temperature coefficient of the reaction. The temperature coefficient is normally 2. It approaches a value of 3 in some reactions.
Temperature coefficient, µ = + == oo oo (t10)C35C tC25C kk 2 kk
If k 1 and k 2 are the rate constants of a given chemical reaction at two different temperatures, T1 and T2, and assuming that the
A catalyst is a substance that alters the rate of a reaction without itself undergoing any permanent change. Catalyst increases the rate of reaction by changing the path of the reaction. For example MnO 2 catalyses the following reaction so as to increase its rate considerably.
2KClO3 2KCl + 3O2
In the presence of a catalyst, the reaction takes place through a different path of lower activation energy, as shown in Fig.3.8. Hence rate increases.
Fig.3.8 Path of uncatalysed and catalysed reaction
3. T he rate constant (k) for a particular reactions is 1.3 × 10 –4 M –1s –1 at 100°C, and 1.3 × 10–3M–1s–1 at 150°C. What is the energy of activation (Ea) (in kJ) for this reaction?( R= molar gas constant = 8.314 JK –1mol–1)
Sol. According to Arrhenius equation,
E a = 60 kJ/mol
Try yourself:
3. The activation energy of one of the reaction in a biochemical process is 532611 J mol –1 When the temperature falls from 310 K to 300 K, the change in rate constant observed is k300 = x×10–3k310. The value of x is ________. [Given: In 10 = 2.3, R = 8.3 JK –1 mol–1]
Ans: 1
TEST YOURSELF
1. The graph obtained between ln k (k= Rate constant) on y-axis and 1/T on x-axis is a straight line. The slope of it is -4 ×10 4 k. The activation energy of the reaction (in kJ mol−1 ) is R=8.3JK−1mol−1 (1) 166 (2)332 (3) 765 (4) 382
2. The temperature coefficient of a reaction is two. If the temperature is raised from 35°C to 75°C, rate increases by (1) 8 times (2) 32 times (3) 16 times (4) 64 times
3. For a first order reaction A → P, the temperature (T) dependent rate constant (k) was found to follow the equation log () 1 20006.0 k T =-+ The pre -exponential factor A is
4. For Arrhenius equation K=A.e−Ea/RT, under what conditions K will be equal to A? (1) When Ea = 0 (2) When T = ∞ (3) Both A and B (4) When R = 0
5. Rate constant at 600 K is 10 times more than the rate constant at 300 K. What is the activation energy of the reaction? (R=gas constant)
(1) 600 ln10 (2) 600ln10 R
(3) 600 R (4) 600 R ln10
6. The decomposition of N 2 O into N 2 and O2 in presence of argon follows first order kinetics K=5.0×1011 e−2000/T. The activation energy is
(1) 16.628 kJ mol–1 (2) 166.28 kJ mol–1
(3) 166.28 J mol–1 (4) 16.628 J mol–1
Answer Key
(1) 2 (2) 3 (3) 1 (4) 3
(5) 4 (6) 1
3. 5 ORDER OF REACTION
The speed of a chemical reaction, in general, depends on the concentration of reacting species of the reaction. An early generalisation in this regard is due to Gulberg and wage. This generalisation is known as law of mass action and is stated as follows.
The rate of a chemical reaction is proportional to the product of effective concentrations (active masses) of the reacting species, each raised to a power that is equal to the corresponding stoichiometric numbers of the substances appearing in the chemical reaction.
Thus, for a general reaction aA+bB → cC+dD, we have
rate ∝ [A]a[B]b or r = k[A]a[B]b.....(1) where k is the constant of proportionality. If the rate of a reaction is determined experimentally, it is found that equation (2) is not always applicable. However, the experimental results can be fitted to satisfy a relation of th e type of equation(2) where
the expo nents may or may not be equal to the respective stoichiometric coefficients. In general, we may write the rate as
r = k[A]n[B]m. This equation is called rate equation or rate law.......(2), where the dimensionless exponents n and m may or may not be equal to a and b respectively. The values, n and m, may have positive, negative, integral or fractional values, or zero value. The value n is known as the order of the reaction with respect to A, m as the order of the reaction with respect to B and so on. The sum (n+m) is known as the overall order of the reaction. If (n+m) = 1, the reaction said to be first order, if (n+m) = 2, the reaction is said to be second order and so on.
Therefore, the order of reaction is the observed number of concentration terms which govern the rate of reaction.
3.5.1 Rate Constant
For an elementary reaction,
xA + yB Products, the rate equation is given as,
Rate [A]x [B]y (or)
Rate = k [A]x [B]y ,
Here, k is the proportionality constant called rate constant.
Rate of Reaction
1. Rate is the speed with which reactants are converted into products.
2. It is the change in concentration of a reactant or a product per unit time.
3. It is dependent on the concentration of reactants.
4. It can be given as average rate or instant rate for a given reaction.
5. Unit of reaction rate is mol L –1s–1
If the rate equation is given as,
Rate =k[C]n then the rate constant is k = Rate / [C]n
Here, ‘ n ’ depends upon the nature of reaction. It may be any integer or zero or a fraction.
If the concentration of each reactant is taken as 1.0 mol L–1, the rate of the reaction is equal to rate constant. The rate constant is also called specific reaction rate. Rate constant is independent of the concentration of reactant. It depends, for a given reaction, on temperature and catalyst. The main differences between rate and rate constant of a reaction are given in Table 3.2.
Rate constant or specific reaction rate is defined as the rate of reaction corresponding to unit concentration of all the reactants in a reaction.
Units of rate constant is = n unitsofrate
[unitsofconcentration]
Units of rate constant for nth order reaction
Unit of rate
Unit of concentration
In general, unit of rate constant is: L(n–1) mol(1–n) s–1.
Rate Constant
1. Rate constant is a proportionality constant
2. It is equal to the reaction rate at unit concentration of reactants.
3. It is independent on concentration of reactants.
4. It can be given only in a single fashion for a given reaction.
5. Unit of rate constant is: mole(1-n).L (n-1). sec –1 when n is the order of reaction
Table 3.2 Difference between rate of reaction and rate constant
Some examples of rate law and unit of rate constant are given in Table 3.3.
4. A+ 2B → C, the rate equation for this reaction is given as Rate = K[A][B]. If the concentration of A is kept the same but that of B is doubled, what will happen to the rate?
Sol. Rate = k [A][B]= R
R’= k[A][2B]
[][] [][] [][2]2[] ' [] kABkAB kABk R RAB ==
⇒2 R= R’, i.e., rate is doubled.
3.5.2 Molecularity
Majority of chemical reactions known are compl ex reactions. Only a few reactions are elementary reactions. An elementary reaction is that in which the reactants are directly converted into final products in a single step. In a complex reaction, reactants are converted into final products through different elementary steps.
A sequential representation of elementary steps in an overall chemical reaction is known as reaction mechanism. For example, the reaction A Z may take place in the elementary steps as A 2 B ; 2B I and I Z, where B and I are intermediates.
The number of atoms or molecules or ions participating in an elementary step which must collide simultaneously to bring about a chemical reaction is called molecularity. In a chemical reaction, every elementary step has a molecularity of its own. Elementary reactions take place one after another with different rates.
The slowest among different elementary steps of a reaction limits the rate of the reaction and it is called rate determining step. The rate of overall reaction is controlled by the rate limiting step. The number of atoms or molecules or ions participating in the rate
determining step of the reaction is called overall order of the reaction. Molecularity has no meaning for a complex reaction.
The reaction can be unimolecular when one reacting species is involved. Decomposition of ammonium nitrite to give nitrogen is an example of unimolecular reaction.
Bimolecular reactions involve simultaneous collision between two species, for example, dissociation of hydrogen iodide into elements. Trimolecular reactions involve simultaneous collision between three reacting species.
Trimolecular reactions are also called termolecular reactions. Molecularity of the oxidation of colourless nitric oxide to coloured nitrogen dioxide is three.
The probability that more than three molecules can collide and react simultaneously is very small. Hence, reactions with the molecularity three are very rare and slow to proceed.
5. The hypothetical reaction A2+B2→ 2AB; follows the following mechanism A2 Fast → ← A+A,
A +B2 slow → AB + B, A+B Fast → AB. The order of the overall reaction is Sol. A 2+B 2→2AB ; A2→ A+A(fast); A+B2→ AB+B(slow)
Rate law = k[A][B2]put value of [A] from first reaction since A is intermediate 2 [] c KAA =
\ Rate law equation = 22 [][] c kKAB
\ Order = 13 1 22 =+=
Try yourself:
4. Suppose, the following mechanism has been proposed for a reaction 2A +B →D+E, A+B → C+D(slow), A+C →E( fast). What is the rate law expression for the reaction?
Answer: Rate = k[A][B]
TEST YOURSELF
1. The reaction 2O3 → 3O 2 proceeds through the mechanism shown below.
O3 O2 + O (fast)
O3 + O → 2O2(slow)
The rate law is given as
(1) r = k[O3]2 [O2]−1
(2) r = k[O3]2 [O2]
(3) r = k[O3][O2]
(4) r = k[O3]2
2. Which of the following statement is incorrect
(1) A second order reaction must be a biomoleculer elementary reaction.
(2) A biomolecular elementary reaction must be a second order reaction.
(3) Zero order reaction must be a complex reaction.
(4) First order reaction may be complex or elementary reaction.
3. Overall order of the reaction is
A 3 3A (fast)
33 () slow ABABslow +→
A3+3B3→ 3AB3
(1) 1.5 (2) 1.33
(3) 2 (4) 4
4. 2A→ Products is a complex reaction. Choose the correct statement about the reaction.
(1) Neither molecularity nor order can be predicted.
(2) Molecularity is two but order cannot be predicted.
(3) Molecularity is two and order is also two
(4) Molecularity cannot be predicted but order is two.
Answer Key
(1) 1 (2) 2 (3) 2 (4) 1
3.5.3 Integrated Rate Equations
Differential rate equation upon integration gives a relation between directly measured concentrations at different times and rate constant. The integrated rate equations are different for the reactions of different orders.
■ Zero Order Reaction
Reaction in which rate of the reaction is independent of the concentration of the reactants is called zero order reaction.
Starting from an initial concentration of reactant is [R]0= a, if x mol L–1consumed in the reaction, in a time interval of t, the concentration of the reactant after time t is
[R]t = ‘(a – x)’.
Rate=-=d(ax)k. dt
∫--=∫d(ax)kdt or x = kt or specific rate, k = x/t or= 0t k[R][R] t
The decomposition of gaseous ammonia on a hot platinum surface is an example of zero order reaction.
2NH3(g) → Pt,1130K N2(g) + 3H2(g)
rate = k[NH3]o = k
In this reaction Pt acts as a catalyst. At high pressure, the metal surface gets saturated with gas molecules. So, a further change in reaction conditions is unable to change the amount of ammonia on the surface of the catalyst, making the rate of the reaction independent of its concentration. Similarly, the thermal decomposition of HI on gold surface is also a zero order reaction.
Some enzyme catalysed reactions and some reacti ons that occur on metal surfaces are examples of zero order reactions.
■ First Order Reaction
Reaction in which rate of the reaction is directly proportional to the single concentration term of the reacting substance is called first order reaction. Rate=-=-=dCd(ax)dx;k(ax)dtdtdt
When t = 0, x = 0,
–ln (a – 0) = k × 0 + c (or) –ln a = c
Hence, –ln (a – x) = kt –ln a =a ktln (ax) (or) =1a kln t(ax) or
The above reaction can also be written as
ot to 1RR Kln;lnkt tRR
or [R]t = [R]0.e–kt
where [ R ] 0 = initial concentration of reactant [ R ] t = concentration of reactant at time ‘t’, and - kt e = fraction of concentration of reactant left unreacted
■ First Order Gas Phase Reactions
Decomposition of N2O and N2O5 are examples of first order gas phase reactions.
Let Pi be the initial pressure of reactant gas, X and P t be the total pressure at time t for the decomposition of X.
X(g)
Y(g) + Z(g)
at t = 0, pi atm 0 atm 0 atm at time t, (pi – x) atm x atm x atm
pt = pi – x + x + x =(pi + x) atm
x = pt – pi
Pressure of reactant at time = t is
pA = pi – x = pi – (pt – pi) = 2pi – pt
k = i A 2.303p log tp or k =i it 2.303p log t2pp
A plot of ln (a – x) against time ‘t’ for a first order reaction is a straight line with a slope of –k and intercept of ln a. A plot of [ln a/(a–x)] against time is also a straight line with a slope k, as shown is Fig.3.9.
For a zero order reaction, the graphs are different as shown in Fig.3.10 . The rate of a zero order reaction is constant throughout the course of the reaction and equal to the rate constant of the reaction.
All natural and artificial radioactive decay of unstable nuclei take place by first order kinetics.
Hydrogenation of ethylene is also an example of first order reaction.
C2H4(g) + H2(g) C2H6(g)
Fig. 3.9 First order reaction graphs
Fig. 3.10 Zero order reaction graphs
6. The reaction A→B,follows first order kinetics. The time taken for 0.8 mole of A to produce 0.6 mole of B is 1 hour. What is the time taken for conversion of 0.9 mole of A to produce 0.675 mole of B?
Sol. (b) A → B For a first order reaction
Given a = 0.8 mol, (a-x) = 0.8-0.6= 0.2
2.3030.8 log 10.2 k = or k = 2.303 log 4
again a = 0.9, a-x = 0.9–0.675= 0.225
2.3030.9 log t0.2225 = k
2.303 log 4 = 2.303 t log 4 Hence, t = 1 hour
Try yourself:
5. A reaction which is of first order w.r.t reactant A, has a rate constant 6 min –1. If we start with [A] = 0.5 molL–1, when would [A] reach the value of 0.05 mol L –1?
Ans: = 0.384 min.
3.5.4 Half L ife
Half-life of a reaction is the time in which the concentration of a reactant is reduced to one half of its initial concentration. It is represented by t0.5 or t1/2.
The time ‘t’ will be ‘t1/2’ when (a – x) = a/2. or a zero order reaction, the rate constant (k) is given as = k x t or = k x t or = 1/2 a 2k t
Half- life of a zero order reaction is directly proportional to the initial concentration of reactants and inversely proportional to the specific rate. For a first order reaction, the rate constant (k) is =2.303a klog(or) a tx
=2.303a log ka t x
== 1/2 2.303a2.303log2 log ka/2k t or = 1/2 0.693 t k
Half-life of a first order reaction is a constant. It is independent of the initial concentration of reactant. It is only dependent on nature of the substances, temperature and catalyst.
The relation between initial amount of the substance and substance left unreacted following first order kinetics after ‘n’ halflives is given as,
Amount left unreacted
= n
Initialamounttaken . 2
Integrated rate laws, units of rate constants and half-life expressions for some reactions are listed in Table 3.4.
Table 3.4 Differential and integrated rate laws of some reactions
7. At 30°C, the half life for the decomposition of AB2 is 200 s and is independent of the initial concentration of AB2. The time required for 80% of the AB2 to decompose is (Given: log 2= 0.30; log3 = 0.48) Sol. For first order reaction 0 0
2.303 log 0.2 ka ta = also t1/2 = 0.693 k
0.6930.6932.3031 log 200200 0.2 k t =⇒=
2.3031 200log 0.6930.2 k =× = 466.675≈ 467seconds
Checkpoint
Q. The rate of a first order reaction is 1.5 × 10–2 mol L–1 min–1 at 0.5 M concentration of the reactant. What is half life of the reaction?
Answer: =23.1 min
3.5.5.
Pseudo First Order Reaction
When a second-order reaction shows the kinetics of a first-order reaction if one of the reactants is present in large excess. The order with respect to the reactant present in excess is zero. A reaction of this type is called a pseudo first order reaction. Common examples of such reactions are those in which one of the reactant is the solvent.
In hydrolysis of sucrose, sucrose is isolated by taking large excess of water. The concentration of water does not get altered much during the course of reaction.
The rate of the reaction is given as, rate = k [C12H22O11] [H2O]
The term [H2O] can be taken as constant. The concentration of water is 55.5 mol L –1 , which is almost constant. The equation thus becomes,
Rate = k[C12H22O11], where rate constant is, k = k [H2O]
The reaction behaves as first order reaction. Such reactions are called pseudo first order reactions or pseudo unimolecular reactions. In these reactions, pseudo first order rate constant depends upon concentration of excess reagent.
For example: Base hydrolysis of ester is a second order reaction. Acid catalysed hydrolysis of ethyl acetate is a pseudo first order reaction, because water is taken in large excess.
CH3COOC2H5 + H2O
CH3COOH + C2H5OH
Rate of the reaction, rate = k [CH3COOC2H5]
In such a pseudo unimolecular reaction, the molecularity is 2, but the order of reaction is taken as 1.
3.6 DETERMINATION OF ORDER
The order of a reaction is determined using integrated equation method, graphical method, half-life method, van’t Hoff’s differential method and Ostwald’s isolation method.
Integrated Form of Rate Equation Method
It is also called trial and error method. Starting from initial concentration ‘a’, if the change in concentration in time ‘t’ is ‘x’, the rate equation in terms of order of the reaction ‘n’ is given as,
Rate of reaction = =dxn dtk(ax)
The kinetic data is substituted in one of the integrated rate laws and the specific rates are calculated. If the same value of ‘k’ is obtained from the data of a reaction, the trail of choosing the rate law is successful and the corresponding order is obtained.
Graphical Method
Alternately, the order can also be ascertained graphically interpreting the results in terms of integrated rate law.
As shown in Fig.3.11 , x versus time is a straight line parallel to time axis, which denotes zero order. log [ a /( a – x )] versus time is a straight line with a positive slope passing through origin, which denotes first order. [ x / a ( a – x )]versus time is a straight line with a positive slope passing through origin, which denotes second order. (Here, a = initial concentration of reactant, a – x = Concentration after time t of reaction)
determined. For the purpose of determining orders, two experiments are conducted. In one experiment, concentration of Y is taken in large excess compared to that of X. Order, n x is determined by varying the concentration of X in one of the three methods described above. In the second experiment, concentration of X is taken in large excess. Order, n y is determined. The total order of the reaction is ( n x + n y).
TEST YOURSELF
1. For a first order reaction t99.9% = x × t50% the value of x is
(1) 2 (2) 4 (3) 5 (4) 10
2. Match the following columns and choose the correct option from the codes given below.
Column-I (order of reaction)
Column-II (Units of rate constant )
(a) Zero order reaction (P) mol.L-1.s-1
Fig 3.11 Graphical methods to determine the order of reaction
Half life method
In general, half life of n th order reaction is inversely proportional to ( n - 1) th power of initial concentration(a) of reactant.
Ostwald’s Isolation Method
The isolation method is useful for the determination of order with respect to each of the reactants separately.
X + Y Products
By applying Ostwald’s method, the order n x with respect to the reactant X, and the order n y with respect to reactant Y , are separately
(b) First order reaction (Q) s–1
(c) Second order reaction (R) mol–1.L.s–1.
(d) Third order reaction (S) mol–2.L2.s–1.
(a) (b) (c) (d)
(1) P Q R S
(2) Q P R S
(3) R Q P S
(4) P Q S R
3. t ½ for a first order reaction is 10 s. The value of rate constant ‘k’ for the same reaction is
(1) 6.93 s−1
(2) 0.00693 s−1
(3) 0.0693 s−1
(4) 0.693 s−1
4. First order reaction among the following is (1) thermal decomposition of H 2O2
(2) 2NO + O2→ 2NO2
(3) decomposition of NH3 on metal surface at high temperature (4) CHCl3 +Cl2 → CCl4 + HCl
5. Half life of a first order reaction is 2 x . Time taken for the completion of 87.5% of the same reaction is (1) 1.5x (2) x (3) 1.66 x (4) 3x
6. If 3 + = dxn kHO dt and rate becomes 100 times when pH changes from 2 to 1, then the order of reaction is (1) 1 (2) 2 (3) 3 (4) 0
7. For a first order reaction, the rate constant is 6.909 min –1 . The time taken for 75% conversion in minutes is (1) 3 log2 2 (2) 2 log3 3 (3) 2
8. For a first order reaction A→Products, initial concentration of A is 0.1 M, which becomes
CHAPTER REVIEW
Rate of Reaction
■ When a chemical reaction is in progress, the concentration of the reactants decreases and the concentration of the products increases with the progress of time.
■ There is a change in the macroscopic detectable properties, like colour, concentration etc., of the system during a reaction.
■ The rate of a reaction is defined as the decrease in the concentration of the reactant
0.001 M after 5 minutes. Rate constant for the reaction in min–1 is
(1) 0.2303 (2) 1.3818 (3) 0.9212 (4) 0.4606
9. The half life period of a first order reaction is 60 minutes. What percentage will be left over after 240 minutes?
(1) 6.25% (2) 4.25% (3) 5% (4) 6%
10. For a chemical reaction A+B→ Product, the order is 1 with respect to A and B
Rate molL–1 s–1 [A] mol L–1 [B] mol L–1
0.10 20 0.5
0.40 x 0.5 0.80 40 y
What are the value of x and y?
(1) 160 and 4 (2) 80 and 4
(3) 40 and 4 (4) 80 and 2
Answer Key
(1) 4 (2) 1 (3) 3 (4) 1 (5) 1 (6) 2 (7) 3 (8) 3
(9) 1 (10) 4
or the increase in the concentration of the product in unit time.
■ The unit of rate of a reaction is: mole lit-1s–1.
■ In gaseous reaction the unit of rate of reaction is taken as atm s –1
■ Rate of reaction generally decreases nonuniformly as time proceeds.
■ For a reaction A B
The rate in terms of the concentration of
reactant d[A] A dt=
■ For a reaction, A B
The rate in terms of the concentration of product d[B] B dt + =
■ In the reaction N2 + 3H2 2NH3, the rate = 3 22 d[NH] d[N]d[H]11 dt3dt2dt=-=+
■ The slope of the tangent of the conc entration time curve at any point gives the rate of the reaction at that time.
Factors Influencing Reaction Rates
■ The rate of a reaction depends upon the nature of the reactants.
■ Ionic substances in aqueous solution contain free ions. Ions can directly participate in reaction as bonds need not be broken. Ionic reactions are very fast or instantaneous.
■ Reactions between covalent substances are slow because in such reactions, breaking of old bonds and formation of new covalent bonds occur.
■ With an increase in the concentration of the reactants, the rate of the reaction increases (except zero order reactions)
■ According to the law of mass action, rate = k[Reactants]n where k is rate constant or specific rate and n is order of the reaction
■ With an increase in temperature, the rate of a reaction always increases, whether the reaction is exothermic or endothermic.
■ A catalyst increases the rate of reaction by carrying the reaction in a new path involving lower activation ene rgy.
■ A catalyst alters the rate of reaction, path of reaction, activation energy, threshold energy and rate constant.
■ Catalyst does not alter H, S, G of reaction, energy of reactants and energy of products.
■ A catalyst favours both forward and backward reactions equally. Hence it does not affect equilibrium constant.
Theory of Reaction Rates
■ For a chemical reaction to take place collisions between the reacting molecules are essential.
■ Only a fr action of the total number of collisions lead to reaction. This fraction is known as effective or fruitful collisions.
■ The minimum energy that the molecules should possess so that their collisions lead to chemical reaction is called threshold energy.
■ The minimum extra energy that the molecules should possess, over and above the average energy, to enable them to react is called the energy of activation.
■ The greater the energy of activation of a reaction the lesser will be the rate constant and also the rate of the reaction.
■ The lesser the energy of activation of a reaction, the higher will be the rate constant and also the rate of the reaction.
■ Molecules possessing the threshold energy are called activated molecules. When activated molecules collide, an activated complex, also called transition state, is formed, which changes into the products.
■ The energy of the activated complex is greater than the energy of the reactants and also the energy of the products.
■ The fraction of molecules that possess the energy of activation is given by the Boltzmann factor, e–Ea /RT
■ With an increase in concentration, the number of activated molecules increases.
■ The number of effective collisions increases with temperature, due to which the rate of the reaction increases.
■ Heat of reaction E = (Ea)f – (Ea)b, where (Ea) f is the energy of activation of the forward reaction an d (Ea) b is the energy of activation of the backward reaction.
■ In an exothermic reaction, the energy of activation of the backward reaction is greater than the energy of activation of the forward reaction.
■ In an endothermic reaction, the energy of activation of the backward reaction is less than the energy of activation of the forward reaction.
Temperature Dependence of Rate of Reaction
■ The relationship between the rate constant of a reaction and temperature is given by Arrhenius equation.
■ k = Ae–Ea/RT, where ‘A’ is frequency factor.
■ The frequency factor and energy of activation are constant for a given reaction. They vary very slightly with temperature
■ When a graph is drawn between log k and 1/T, a straight line (linear) is obtained with a negative slope.
■ At two temperatures T 1 and T 2, the rate constants K1 and K2 are given by
Order of Reaction
■ The mathematical dependence of the rate of a reaction on the molar concentrations of the reactants is called the rate law or rate equation.
■ For a reaction, aA + bB cC + dD, rate = K[A] x[B] y, where K is rate constant or specific rate, x is order of reaction with respect to A and y is order of reaction with respect to B.
■ The rate constant of a reaction becomes equal to the rate of the reaction when the concentration of all the reactants are unity.
■ The rate constant is also known as the specific reaction rate, if concentration is unity.
■ Rate constant does not change with change of concentration of reactants, products, volume of vessel, and coefficients of reactants.
■ The rate constant of a reaction is always a characteristic value at a given temperature.
■ The unit of the rate constant of n th order reaction is: lit(n-1)mole(1-n)sec-1
■ The ratio between the rate constants of a reaction at two temperatures, which differ by 10°, is called the temperature coefficient of the reaction.
■ For most of the reactions, the value of the temperature coefficient is found to be 2 or sometimes 3.
■ The rate constant of a reaction generally increases by 100 to 200% with 10 oC raise in temperature.
Molecularity of Reaction
■ The slowest of different steps in the reaction mechanism is called rate limiting or rate determining step of a reaction.
■ Molecularity of a reaction is sum of number of atoms or ions or molecules in the rate determining step of the reaction.
■ Molecularity may be 1 or 2 or 3. It can be neither zero nor fractional.
■ The sum of the powers of the concentration terms of the reactants in the rate equation of the reaction is known as the order of the reaction.
■ The rate equation is determined experimentally, from which we can know the order of the reaction.
■ Rate is independent on the concentration of reactant in a zero order reaction.
■ For a zero order reaction, rate is equal to the rate constant.
■ Unit of rate constant of zero order reaction is: mol L–1 s–1
■ Rate is dependent on single concentration term in a first order reaction.
rate = k × concentration
■ Units of rate constant of first order reaction is: s–1
■ Rate is dependent on two concentration terms in a second order reaction, rate = k × (Concentration) 2
■ Units of rate constant of second order reaction are: L mol–1 s–1
■ If ‘a’ is initial concentration of the reactant, reduced to (a – x) in time t, the rate constant of a first order reaction, klog2.303a tax = -
■ Time taken for 50% completion of a reaction is called halflife (t 1/2).
■ Half life of a first order reaction is independent of the initial concentration, 1/2 0.693 t k =
■ Time required for 75% completion of a first order reaction is 2 half lives. For 87.5% completion it is 3 half lives, and for 99.9%, completion is 10 halflives.
■ Half life of a reaction is directly proportional to a1–n, where ‘a’ is initial concentration and ‘n’ is order or reaction.
■ For a zero order reaction, t1/2 is taken as a/2k. For a second order reaction t 1/2 is taken as 1/k a .
■ The order of reaction is experimentally determined.
■ Order of the reaction can be obtained by integrated equation method, half life method, Van’t Hoff differential method and Ostwald’s isolation method.
■ Molecularity is obtained for the reaction mechanism and not from stoichiometry.
Exercises
NEET DRILL
Level - I
Rates of chemical reactions
1. Which of the following reaction is spontaneous at room temperature
(1) I2 + H2O →
(2) Pt 222 2H+O2HO →
(3) N2 + O2 → 2NO
(4) 2HCl → H2 + Cl2
2. Among the following slowest reaction under identical conditions is
3. The rate of reaction for N2 + 3H2 → 2NH3 may be represented as
(1)
3 22 d[NH] d[N]d[H]11 r=-=-+ dt3dt2dt
(2) 3 22 d[NH] d[N]d[H]11 r=-=-=+ dt3dt2dt
(3) 3 22 d[NH] d[N]d[H] 1 r=-=3+ dtdt2dt
(4) 3 22 d[NH] d[N]d[H] 1 r=-=-=+2 dt3dtdt
4. The chemical reaction occurring between covalent molecules involve
(1) breaking of existing bonds (2) formation of new bonds
(3) evolution of heat energy
(4) both (1) and (2)
5. The rate of a chemical reaction decreases as the reaction proceeds. This is because (1) the reactant concentration remains constant as the reaction proceeds.
(2) the product concentration remains constant as the reaction proceeds.
(3) concentration of reactant decreases from time to time, as the reaction progresses (4) concentration of product decreases from time to time, as the reaction progresses
6. In a reaction 2A + B → A2B , the reactant ‘A’ will disappear at (1) half the rate at which B disappears (2) the same rate at which B disappears
(3) the same rate at which A 2B is formed (4) twice the rate at which B disappears
7. Burning of coal is represented as C (s) + O2(g) → CO2(g). The rate of this reaction is increased by (1) decrease in the concentration of oxygen.
(2) powdering the lumps of coal.
(3) decreasing the temperature of coal.
(4) providing inert atmosphere.
8. For a hypothetical reaction; A → L the rate expression is, rate A dC dt =-
(1) negative sign represents that rate is negative
(2) negative sign pertains to decrease in the concentrations of reactant
(3) negative sign indicates the attractive forces between reactants
(4) all of the above are correct
9. For the reaction; 2HI → H 2 + I 2 , the expression HI 1 2 d dt - represents
(1) the rate of formation of HI.
(2) the rate of disappearance of HI.
(3) the instantaneous rate of the reaction.
(4) the average rate of reaction.
Factors influencing reaction rates
10. The rate of reaction which does not involve gases, does not depend upon
(1) Temperature
(2) Concentration
(3) Pressure
(4) Catalyst
11. A catalyst accelerates the reaction, because (1) it brings the reactants closer.
(2) it lowers the activation energy.
(3) it changes the heat of reaction.
(4) it increases the activation energy.
12. Which of the following parameters of a chemical reaction are increased when a catalyst is used ?
(1) Rate and activation energy
(2) Rate constant and enthalpy
(3) Enthalpy and time duration
(4) Rate and Rate constant
13. Which of the following influence the rate of reaction
A) Nature of reactants
B) Concentration of reactants
C) Temperature
D) Molecularity
(1) A, B
(2) B, C, D
(3) C, D
(4) A, B, C
14. Which of the following does not affect the rate of reaction ?
(1) Amount of the reactants taken
(2) Physical state of the reactants
(3) ∆ H of reaction
(4) Size of the vessel
15. Increase in the concentration of the reactants leads to a change in
(1) heat of reaction
(2) aqctivation energy
(3) collision frequency
(4) threshold energy
16. The reactions 2NO + O 2 → 2NO 2 ; 2CO + O2 → 2CO2 look to be identical, yet the first is faster than the second. The reason is that
(1) The first reaction has lower enthalpy change than the second.
(2) The first reaction has lower internal energy change than the second.
(3) The first reaction has lower activation energy than the second.
(4) The first reaction has higher activation energy than the second.
17. The specific rate constant of a reaction depends on the (1) concentration of the reactant.
(2) time.
(3) temperature.
(4) concentration of the product.
18. The temperature coefficient of a reaction is (1) the rate constant at a fixed temperature.
(2) the ratio of rate constants at two temperatures.
(3) the ratio of rate constants at two different temperatures differing by 10°C.
(4) the ratio of rate constants at two pressures.
Theory of Reaction Rates
19. Increase of temperature will increase the reaction rate due to
(1) increase of number of effective collisions. (2) increase of mean free path.
(3) increase of number of molecules.
(4) increase of number of collisions .
20. The minimum energy possessed by reacting molecules to enter into chemical reaction is called
(1) kinetic energy
(2) potential energy
(3) threshold energy
(4) activation energy
21. In a reaction, threshold energy is equal to (1) activation energy
(2) normal energy of the reactants
(3) activation energy + energy of reactants
(4) activation energy – energy of reactants
22. The value of activation energy for a chemical reaction primarily depends on (1) temperature
(2) nature of the reacting species
(3) the collision frequency
(4) concentration of the reacting species
23. Collision theory satisfactorily explains (1) first order reaction.
(2) zero order reaction.
(3) bimolecular reaction.
(4) any order reaction.
24. According to collision theory of reaction rates, the activation energy is
(1) the energy gained by the molecule on colliding with other molecules.
(2) the energy that molecule should possess in order to undergo reaction.
(3) the energy it should possess so that it can enter into an effective collision.
(4) the energy it has to acquire so that it can enter into an effective collision.
25. Pick the appropriate choice about collision theory of reaction rates.
I) It explains the effect of temperature on rate of reaction
II) It assumes that the reactants must be in correct orientation to react.
III) It says rate depends upon the frequency at which reactants collide.
IV) The collisions having energy higher than the threshold value will give successful reactions.
The correct points are (1) I, II, III and IV
(2) II and IV
(3) I and IV
(4) I, III and IV
26. According to arrhenius plot, in the collision theory the intercept is equal to (1) log10K (2) ln A
(3) ln K (4) R a E -
Temperature Dependence of Rate of Reaction
27. The temperature coefficient of most of the reactions lies between (1) 1 and 3 (2) 2 and 3 (3) 1 and 4 (4) 2 and 4
28. For a reaction, K10°C K t t x + →=
when the temperature is increased from 10°C to 100°C, the rate constant (K) increased by a factor of 512. Then, value of x is
(1) 1.5 (2) 2.5 (3) 3 (4) 2
29. The rate constants of a reaction at 300 K and 280 K respectively are K 1 and K2.Then
(1) K1 = 20 K2
(2) K2 = 4 K1
(3) K1 = 4 K2
(4) K1 = 0.5 K2
30. The rate constant K1 of a reaction is found to be double that of rate constant K 2 of another reaction. The relationship between corresponding activation energies of the two reactions E1 and E2 can be represented as
(1) E1 > E2
(2) E1 < E2
(3) E1 = E2
(4) E1 = 2E2
31. The Arrhenius equation expressing the effect of temperature on the rate constant of reaction is
(1) K = E a /RT
(2) R KAe aE T=
(3) Kloge R a E T =
(4) R Ke aE T=
32. Activation energy depends on (1) pressure.
(2) concentration of reactants.
(3) concentration of products.
(4) nature of reactants.
33. In general the rate of a given reaction can be increased by all the factors except (1) increasing the temperature.
(2) increasing the concentration of reactants. (3) increasing the activation energy.
(4) using a positive catalyst.
34. The effect of temperature on a reaction rate for which E a is zero is given by
(1) with increase of temperature rate increases.
(2) with increase of temperature rate decreases.
(3) rate is independent of temperature.
(4) reaction never occurs.
35. In the graph drawn between log K and 1 T , intercept equals to
(1) 2.303 R a E -
(2) log A
(3) ln A
(4) 3 () 2 lgA .30 o
Order of Reaction
36. Wrong statement among the following is (1) effective collisions are more if activation energy is less.
(2) zero order reaction proceeds at a constant rate independent of concentration or time.
(3) reactions with highest rate constant values have lowest activation energies.
(4) if initial concentration increases half life decreases in zero order.
37. If both dc dt & specific rate have same units, then rate law is
(1) R = K[A]2
(2) 1 RK[A]2 =
(3) R = K[A]–2
(4) R = K
38. The half life for a given reaction was doubled as the initial concentration of the reactant was doubled. The order of the reaction is (1) Zero (2) 1st (3) 2nd (4) 3rd
39. The inversion of cane sugar into glucose and fructose is
(1) 1st order (2) 2nd order (3) 3rd order (4) zero order
40. The half–life of a first order reaction is (1) independent of the initial concentration of the reactant.
(2) directly proportional to the initial concentration of the reactant.
(3) inversely proportional to the initial concentration of the reactant.
(4) directly proportional to the square of the initial concentration of the reactant.
41. HCl 2 RCOOR + HORCOOH + ROH →
follows ...... reaction kinetics
(1) second order
(2) unimolecular
(3) pseudo unimolecular
(4) zero order
42. 2A → B + C would be a zero order reaction when the rate of reaction
(1) is directly proportional [A].
(2) is directly proportional [A] 2
(3) is independent of change of [A].
(4) is independent of [B] & [C].
43. Which of the following is a first order reaction?
(1) 2N2O5 → 4NO2 + O2
(2) 2H2O2 → 2H2O + O2
(3) H+ CHCOOCH+3252HO products →
(4) All the above
44. If a reaction obeys the following equation
2.303 log ka tax =, the order of the reaction will be
(1) zero (2) one (3) two (4) three
45. A reaction involves two reactants. The rate of the reaction is directly proportional to the concentration of one of them and inversely proportional to the concentration of the other. The overall order of reaction will be?
(1) One (2) Two
(3) Zero (4) fractional
46. Which of the following represents the expression for 3/4th life of 1st order reaction
(1) 4 K 2.3033 log
(2) 3 K 2.303 log
(3) 4 K 2.303 log
(4) log 4 2.303 K
47. Give relation between half reaction time ( t 1/2) and initial concentration of reactant for (n–1) order reaction.
(1) 2 1/20 [] tRn∝
(2) 2 1/20 [] tRn∝
(3) 1 1/20 [] tRn + ∝
(4) 1/20 []tR ∝
48. If ‘ a ’ is the initial concentration of the reactant, the time taken for completion of the reaction is a/K. The order of the reaction, will be
(1) 0
(2) 1
(3) 1.5
(4) 2
49. For the reaction A → B, the rate law expression is: rate = K[ A ]. Which of the following statements is incorrect?
(1) The reaction follows first order kinetics
(2) The t1/2 of reaction depends on initial concentration of reactants
(3) K is constant for the reaction at a constant temperature
(4) The rate law provides a simple way of predicting the concentration.
50. Which of the following is correct for a first order reaction? (K = rate constant; t1/2= half–life)
(1) t1/2 = 0.693 × K
(2) 1/2 1 K 0.693 t ×=
(3) K × t1/2 = 0.693
(4) 6.93 × K × t1/2 = 1
51. If a substance with half life 3 days is taken at other place in 12 days. What amount of substance is left now, if it follows first order kinetics
(1) 1 32 (2) 1 8
(3) 1 16 (4) 1 4
52. Which one is not correct?
(1) t1/2 of zero order reaction is dependent of initial concentration of reactant.
(2) t1/2 of first order reaction is independent of initial concentration of reactant.
(3) Rate of zero order reaction does not depend upon initial concentration of reactant.
(4) Rate of zero order reaction depends upon initial concentration of reactant.
53. In the case of a first order reaction, the ratio of the time required for 99.9% completion of the reaction to its half life is nearly (1) 1 (2) 10 (3) 20 (4) 8
54. If the rate of gaseous reaction is independent of pressure, the order of reaction is (1) 0 (2) 1 (3) 2 (4) 3
55. The half life of a 1st order reaction is 1 min 40 seconds. Calculate its rate constant (K)
(1) K = 6.9×10–3 min–1
(2) K = 6.9×10–3 s–1
(3) K = 6.9×10–3 s
(4) K = 100 s
56. The unit of rate constant depends on (1) number of reactants
(2) concentration terms (3) order of reaction
(4) molecularity of reaction
57. If concentration of reactants is made ‘ x ‘ times, the rate constant, K becomes (1) eK/x (2) K/x (3) unchanged (4) x/K
58. The rate equation for the reaction 2A + B → C is found to be : rate = K[A][B]. The correct statement in relation to this reaction is (1) unit of K must be sec–1
(2) value of K is independent of the initial concentrations of A and B
(3) rate of formation of C is twice the rate of disappearance of A
(4) t1/2 is a constant
Molecularity of Reaction
59. For the following elementary step, ()()()() +3(aq) 3 aqaq 3 CHCBrCH C+ Br → , the molecularity is
(1) Zero (2) 1
(3) 2 (4) fractional
60. .......... of a reaction cannot be determined experimentally.
(1) Order (2) Rate
(3) Rate constant (4) Molecularity
61. Which of the following is correct?
(1) molecularity of a reaction is always same as the order of a reaction.
(2) in some cases molecularity of the reaction is same as the order of a reaction.
(3) molecularity of the reaction is always more than order of a reaction.
(4) molecularity never be equal to order of a reaction.
62. (A) : Order of the reaction is related to molecularity of the reaction.
(R) : Molecularity of reaction can be fractional.
(1) Both (A) and (R) are correct and (R) is the correct explanation of (A).
(2) Both (A) and (R) are correct but (R) is not the correct explanation of (A).
(3) (A) is correct but (R) is wrong.
(4) Both (A) and (R) are wrong.
63. I 22 22 Alkaline medium 2HO 2HO + O → Rate for the above reaction is given as 22 22 HO Rate HOI d k dt -
Incorrect statement among the following is
(1) Order with respect to H2O2 is 1
(2) Slowest step is unimolecular
(3) Order with respect to iodide ion is 1
(4) Molecularity of slowest step is 2
64. For consider the reaction H 2(g) + Br 2(g) → 2HBr(g), with the reaction 1/2 22 rate KHBr = . Which of the following statement is true about this reaction?
(1) the reaction is of second order.
(2) molecularity of the reaction is 1.5.
(3) the unit of K is sec–1 .
(4) molecularity of the reaction is 2.
65. For a reaction 2H2O2 → 2H2O + O2, the order and molecularity respectively of the reaction can be
(1) 1, 0 (2) 2, 0.5
(3) 1, 1 (4) 0.5, 0.5
Level - II
Rate of Chemical Reactions
1. For N 2 + 3H 2 2NH 3 , the rate of disappearance of H 2 is 0.01 M.min –1 The amount of NH3 formed at that instant will be (1) – 0.02 mol (2) Zero (3) 0.1132 g (4) 0.17 g
2. Consider the following reaction N2(g) + 3H2(g) →2NH3(g). The rate of the reaction in terms of N2 at T (K) is - 2 11 N 0.02 molL s d dt
= What is the value of () 2 11 H m s olL d dt
at the same temperature?
(1) 0.02 (2) 50 (3) 0.06 (4) 0.04
3. For the reaction, 4NH 3 + 5O 2 → 4NO + 6H 2 O, the rate of reaction with respect to NH3 is 2 × 10–3 M s–1. Then, the rate of the reaction with respect to oxygen is __ M s–1
(1) 4 × 10–3 (2) 1.5 × 10–3
(3) 2.5 × 10–3 (4) 3 × 10–3
4. Concentration of a reactant ‘A’ is changed from 0.044 M to 0.032 M in 25 minutes, the average rate of the reaction during this interval is
(1) 0.0048 mol L–1 min–1
(2) 0.00048 mol L–1 s–1
(3) 4.8×10−4 mol L–1 min–1
(4) 0.0048 mol L–1 s–1
5. In the reaction, A → 2B, the concentration of A falls from 1.0 M to 0.9982 M in one minute. What is the rate in mol L –1 s–1
(1) 1.8 × 10–3 (2) 3.0 × 10–5
(3) 3.6 × 10–3 (4) 6.0 × 10–5
6. The rate of formation of SO3 in the reaction 2SO2 + O2 → 2SO3 is 100 g min–1. Hence, rate of disappearance of O 2 is
(1) 50 g min–1 (2) 100 g min–1
(3) 20 g min–1 (4) 40 g min–1
7. 1 dm 3 of 2 M CH 3COOH is mixed with 1 dm3 of 3 M ethanol to form ester. The initial rate if each solution is diluted with an equal volume of water, would become
(1) 2 times (2) 4 times
(3) 0.25 times (4) 0.5 times
8. Decomposition of NH 3 on gold surface follows zero order kinetics. If rate constant (K) is 5×10 −4 M s −1 , rate of formation of N2 will be
(1) 10−3 M s−1 (2) 2.5×10−4 M s−1 (3) 5×10−4 M s−1 (4) Zero
9. The rate of reaction A → products is 10 mol L–1 min–1 at time (t1) = 2min. What will be the rate in mol L–1 min–1 at time (t2) – 12 min
(1) more than 10 (2) 10 (3) less than 10 (4) 20
10. For A→B, [A] changed from 4.4 × 10–2 M to 3.2 × 10–2 M in 25 min. Now A t -∆ ∆ will be
(1) 4.8 × 10−4 M min–1
(2) 4.8 × 10−4 M s–1
(3) 9.6 × 10–3 M s–1 (4) 2.4× 10–4 M min–1
Factors Influencing Reaction Rates
11. The rate law of the reaction, RCl + NaOH → ROH + NaCl is given by Rate = k [RCl]. Which is correct with respect to the rate of this reaction
(1) is doubled by doubling the concentration of NaOH
(2) is halved by reducing the concentration of RCl to half
(3) is increased by increasing the temperature of the reaction
(4) is unaffected by change in temperature Which is correct ?
12. In a reaction A→B, when the concentration of reactant is made 8 times, the rate got doubled. The order of reaction is (1) 1/3 (2) 1 (3) 1/2 (4) 2
13. The initial rates for gaseous reaction A + 3B → AB3are given below [A] (M) [B] (M) Rate (M s–1)
16. If + 3 HO dxn k dt = and rate becomes 100 times when pH changes from 3 to 2, the order of reaction is (1) 1 (2) 2 (3) 3 (4) 0
17. For an elementary reaction, 2A + B → C + D, the active mass of B is kept constant but that of A is tripled. The rate of reaction will (1) decrease by 3 times.
(2) increase by 9 times.
(3) increase by 3 times. (4) decrease by 6 times.
18. Which of the following statement is incorrect?
(1) The rate law for any reaction cannot be determined experimentally.
(2) Complex reactions have fractional order.
(3) Bimolecular reactions involve simultaneous collision between two species.
(4) Molecularity is only applicable for elementary reaction.
19. I 22 22 OH 2HO2HOO→+ , the intermediate in the reaction is (1) IO (2) IO2–(3) IO3– (4) IO4–
Order of the reaction is (1) zero (2) three (3) one (4) two
14. A hypothetical reaction, A 2 + B 2 → 2AB, follows the mechanism as: A 2 A + A (fast), A + B2 → AB + B(slow), A + B → AB(fast).
The order of the overall reaction is (1) 2 (2) 1 (3) 1.5 (4) 0
15. Depletion of ozone occurs as; 2O 3→ 3O2
Step 1: O3 ck O2+(O) (fast)
Step 2: k 32 O+(O)2O (slow) →
What is the order of the reaction? (1) 1 (2) 2 (3) 3 (4) 0.5
20. An endothermic reaction A → B has an activation energy 15 kcal/mol and the heat of reaction is 5 kcal/mol. The activation energy of the reaction B → A is
(1) 20 kcal/mol
(2) 15 kcal/mol
(3) 10 kcal/mol
(4) zero
21. For producing the effective collisions, the colliding molecules must have
(1) a certain minimum amount of energy. (2) energy equal to or greater than threshold energy.
(3) proper orientation.
(4) both threshold energy and proper orientation.
22. In Arrhenius equation, Arrhenius fac tor is equal to
(1) (P+Z) (2) Z P (3) (P–Z) (4) PZ
23. Molecular collision theory is applicable to (1) Bimolecular reactions.
(2) Ionic reaction.
(3) Only organic reaction.
(4) Instantaneous reaction.
24. For a reaction, A → B+300 cal, the activation energy for the conversion of B → A in cal/ mol is
(1) 250 (2) 350
(3) 100 (4) 298
25. According to molecular collision theory, the collisions are effective if
a) molecules are having energy greater than threshold energy.
b) molecules should be properly oriented.
c) the reaction should be exothermic.
d) the reaction should be endothermic.
The correct answers
(1) Only a and b
(2) only a, b and c
(3) Only c
(4) Only a and d
26. Given below are two statements. One is labelled Assertion (A) and the other is labelled Reason (R).
(A) : Every collision between molecules with sufficient energy lead to formation of product
(R) : At the instant of impact colliding molecules should not be properly oriented.
In light of the given statements, choose the correct answer from the options given below.
(1) Both (A) and (R) are true and (R) is the correct explanation of (A)
(2) Both (A) and (R) are true and (R) is not the correct explanation of (A)
(3) (A) is true and (R) is false
(4) Both (A) and (R) are false
27. For a reaction, 2A +B → A 2B, the possible mechanism is Slow ABABHeat +→+ Fast 2 ABAABHeat +→-
The possible energy profile diagram will be (1) (2) (3) (4)
28. The correct reaction profile diagram for a positive catalyst reaction (1) (2) (3)
(4) Temperature Dependence of Rate of Reaction
29. The rate of reaction becomes 2 times for every 10ºC rise in temperature. How many times the rate of reaction will increase when the temperature is increased from 30ºC to 80ºC
(1) 16 (2) 32 (3) 64 (4) 28
30. The activation energy of a reaction is 58.3 kJ / mole. The ratio of the rate constants at 305 K and 300 K is about (R = 8.3 J K–1 mol–1)
(Antilog 0.1667 = 1.468)
(1) 1.25 (2) 1.75 (3) 1.5 (4) 2.0
31. For 10 ,3 + →= t t k XY k , if the rate constant at
300 k is ‘Q’ min–1, at what temperature the
rate constant becomes ‘9Q’ min-1?
(1) 47°C (2) 320°C
(3) 280 K (4) 9300 K ×
32. For a reaction the activation energy is zero. What is the value of rate constant at 300 K.[Given rate constant at 280 K is 1.6×10 6 s–1; R = 8.314 J mol–1 K–1]
(1) 2.08 × 10–3 (2) 2.08 × 10–2
(3) 1.6 × 106 (4) 6.93 × 10–2
33. The temperature coefficient of a reaction is 2.5. If its rate constant at T1 K is 2.5 × 10–3 s–1, the rate constant at T2 K in s–1 is (T2 = 10 + T1)
34. In the equilibrium reaction, A + B ⇔ C + D, the activation energy for the forward reaction is 25 kcal mol –1 and that of the
backward reaction is 15 kcal mol –1. Which one of the following statements is correct ?
(1) It is an exothermic process. (2) It is an endothermic process. (3) It is a reaction for which ∆H= 0. (4) It is a sublimation process.
35. For a first order process, a R E T - value is –23.03, then the value of K A is
(1) 102.303 (2) 10–10 (3) 10–23.03 (4) 1010
36. K, A and E a of a process at 25°C respectively are 5 × 10–4 s–1, 6 × 1014 s–1 and 108 kJ/mol. Then the value of rate constant as time →∞,will be (in s–1) (1) 1.2×1018 (2) Zero (3) 6×1014 (4) 5×10–4
37. The half life of a first order reaction is 100 seconds at 280 K. If the temperature coefficient is 3.0, its rate constant at 290 K (in s–1) is
38. The temperature coefficient of a reaction is 2. When the temperature is increased from 30°C to 90°C, the rate of reaction is increased by
(1) 150 times (2) 410 times (3) 72 times (4) 64 times
Order of Reaction
39. 3/4th of first order reaction was completed in 32 min, 15/16th part will be completed in (1) 24 min (2) 64 min (3) 16 min (4) 32 min
40. Initial concentration of the reactant is 1.0 M. The concentration becomes 0.9 M, 0.8 M and 0.7 M in 2 hours, 4 hours and 6 hours respectively. Then the order of reaction is (1) 2 (2) 1
(3) zero (4) 3
41. For a first order reaction, t0.75 is 138.6 s. Its specific rate constant is (in s –1) (1) 10–2 (2) 10–4 (3) 10–5 (4) 10–6
42. 20% first order reaction is completed in 50 minutes. Time required for the completion of 60% of the reaction is ....... min (1) 100 (2) 150 (3) 262 (4) 205
43. In a first order reaction, 20% reaction is completed in 24 minutes. The percentage of reactant remaining after 48 minutes is (1) 60 (2) 64 (3) 81 (4) 80
44. A first order reaction is half-completed in 45 minutes. How long does it need for 99.9% of the reaction to be completed?
45. For a first order reaction A → B, the reaction rate at reactant concentration of 0.01 M is found to be 2.0 × 10–5 mol L–1 s–1.The half life period of the reaction is (1) 220 s (2) 30 s (3) 374 s (4) 347 s
46. 99% of a first order reaction was completed in 32 min. When will 99.9% of the reaction complete?
(1) 50 min (2) 46 min (3) 49 min (4) 48 min
47. For a first order reaction with half-life of 150 seconds, the time taken for the concentration of the reactant to fall from M/10 to M/100 will be approximately
(1) 1500 s (2) 500 s
(3) 900 s (4) 600 s
48. A reaction which is of first order with respect to reactant A, has a rate constant of 6 min –1 . If we start with [A] = 0.5 mol L –1 , when would [A] reach the value of 0.05 mol L–1.
(1) 0.384 min (2) 15 min
(3) 20 min (4) 3.84 min
49. For a first order process A → B, rate constant K 1 = 0.693 min –1 and another first order process C → D, K2 =x min–1. If 99.99% of C → D requires time same as 50% of reaction A → B , then the value of x ?(in min–1)
(1) 0.0693 (2) 6.93
(3) 23.03 (4) 1.386
50. For () H 122211 aq CHO[ sucrose ] + → Products, the concentration of sucrose changes from 0.06 M to 0.03 M in 30 minutes. Then, the concentration of sucrose at the end of 60 minutes will be
(1) Zero (2) 0.015 M
(3) 0.09 M (4) 0.12 M
51. 99% of a 1st order reaction completed in 2.303 minutes. What is the rate constant and half-life of the reaction
(1) 2.303 and 0.3010 (2) 2 and 0.3465
(3) 2 and 0.693 (4) 0.3010 and 0.693
52. Out of 300 g substance [decomposes as per 1 st order], how much will remain after 18 hr? (t0.5=3 hr)
(1) 4.6 g (2) 5.6 g
(3) 9.2 g (4) 6.4 g
53. 75% of a first order process is completed in 30 min. The time required for 93.75% completion of same process (in hr)
(1) 1 (2) 120
(3) 2 (4) 0.25
54. For a first order reaction at 27°C, the ratio of time required for 75% completion to 25% completion of reaction is
(1) 3.0 (2) 2.303
(3) 4.8 (4) 0.477
55. The half life period of a first order chemical reaction is 6.93 minutes. The time required for the completion of 99 % of the chemical reaction will be (log 2 = 0.3010)
(1) 23.03 minutes (2) 46.06 minutes
(3) 460.6 minutes (4) 230.3 minutes
56. In a first order reaction, the concentration of the reactant, decreases from 0.8M to 0.4M in 15 minutes. Then, 0.1M becomes 0.025M in
(1) 7.5 minutes
(2) 15 minutes
(3) 30 minutes
(4) 60 minutes
57. In the following reaction A → B + C, the rate constant is 0.001 M/s. If we start with 1 M of A, then concentration of A and B after 10 minutes are respectively
(1) 0.5 M, 0.5 M
(2) 0.4 M, 0.6 M
(3) 0.6 M, 0.4 M
(4) 0.1 M, 0.9 M
58. 90% of first order process X→Y is completed in a time equals to 99% of another first order process Y→Q . If K value of Y→Q is 0.09 s–1, K value of X→Y will be (in s –1)
(1) 0.27
(2) 0.3
(3) 0.03
(4) 0.045
59. Based on the following data for a reaction what is its order (A→product)
Conc. A 2 M 0.2 M 0.02 M 0.00
Time in min 0 10 20 ∞
(1) 1st
(2) 2nd
(3) 3rd
(4) zero
60. If the initial concentration is reduced to1/4 th of the initial value of a zero order reaction, the half life of the reaction
(1) remain constant
(2) becomes 1/4th
(3) becomes double (4) becomes fourfold
61. The initial concentration of cane sugar in presence of an acid was reduced from 0.20 to 0.10 M in 5 hours and to 0.05 M in 10 hours, value of K ? (in hr –1)
(1) 0.693
(2) 1.386
(3) 0.1386
(4) 3.465
Molecularity of Reaction
62. A reaction involving two different reactants can never be:
(1) unimolecular reaction
(2) first order reaction
(3) second order reaction
(4) bimolecular reaction
63. Identify incorrect statements.
(1) Molecularity is not always an integral quantity.
(2) Molecularity can be defined for elementary reactions only.
(3) Molecularity is not always an integral quantity.
(4) Molecularity became zero.
64. Which of the following is correct?
(1) Molecularity of a reaction is always same as the order of reaction.
(2) In some cases molecularity of the reaction is same as the order of reaction.
(3) Molecularity of the reaction is always more than order of reaction.
(4) Molecularity never be equal to order.
65. _______ of a reaction cannot be determined experimentally.
(1) Order
(2) Rate
(3) Rate constant
(4) Molecularity
FURTHER EXPLORATION
1. A first order reaction is 50% completed in 20 minutes at 270C and in 5 minutes at 470C. The energy of activation of the reaction is :
(1) 43.85 kJ/mol
(2) 55.33 kJ/mol
(3) 11.97 kJ/mol
(4) 6.65 kJ/mol
2. Match the rate expressions in List-I for the decomposition of NH3 with the corresponding profiles provided in List-II.1, k and k 1 are constants having appropriate u nits.
List - I List - II
(I)
3 13 rate 1 kNH kNH = + under all possible initial concentrations of NH3 (P)
(II)
3 13 rate 1 kNH kNH = + under high initial concentrations of NH3 such that k1[NH3]>>1 (Q)
(III)
3 13 rate 1 kNH kNH = + NH3 under high initial concentrations of such that k1[NH3]<<1 (R)
(IV)
2 3 13 rate 1 kNH kNH = + under low initial concentrations of NH3 such that k1[NH3]<<1 (S) (T)
Choose the correct answer from the options given below:
(I) (II) (III) (IV)
(1) R P T Q (2) R T P Q
(3) T S P R
(4) S P Q T
3. Unit of frequency factor A in= aE KAeRT
(1) time-1
(2) mole.litre-1.time-1
(3) mole-1.litre.time-1
(4) Depend upon order of reaction
4. The energy of activation for a reaction is 100 kJ mol–1. Presence of a catalyst lowers the activation energy by 75 %. What will be effect on rate of reaction at 20°C, other things being equal?
(1) Increases by a factor of 2.34 x 10 13
(2) Increases by a factor of 2.34 x 10 10
(3) Increases by a factor of 10 15
(4) No effect
MATCHING TYPE QUESTIONS
1. Match the following columns and choose the correct option from the codes given below.
Column–I (Order of reaction )
Column–II (Units of rate constant)
(A) Zero order reaction (p) mol L–1 s–1
(B) First order reaction (q) s–1
(C) Second order reaction (r) mol–1 L s–1
(D) Third order reaction (s) mol–2 L2 s–1
Choose the correct answer from the options given below:
(A) (B) (C) (D)
(1) p q r s
(2) q p r s
STATEMENT TYPE QUESTIONS
Directions for following questions
Each question has two statements. Statement I (S-I) and statement II (S-II). Mark the correct answer as
(1) if both statement I and statement II are correct,
(2) if both statement I and statement II are incorrect,
(3) if statement I correct but statement II is incorrect,
(4) if statement I incorrect but statement II is correct.
1. S-I : A first order reaction can be bimolecular.
S-II : Radio active disintegration follows first order kinetics.
2. S–I : A unimolecular reaction is always a first order reaction.
(3) r q p s
(4) p q s r
2. Match the following columns and choose the correct option from the codes given below.
Column–I (Half–Life)
Column-II (Order)
(A) t1/2 =constant (p) First order
(B) 1/2 ta ∝ (q) Pseudo first order
(C) 1 1/2 tp∝ (r) Zero order
(D) 1/2 1/ ta ∝ (s) Second order
Choose the correct answer from the options given below:
(A) (B) (C) (D)
(1) p r s s
(2) p r s s
(3) q r s s, p
(4) p r s s, q
3.
S–II : A first order reaction is always a unimolecular reaction.
S–I : According Arrhenius equation, the lower the value of Ea, faster will be the rate of reaction.
S–II : 99% of a first order reaction was completed in 32 min, but 99.9% of a first order reaction will take 64 min to complete.
4. S–I : With increase in temperature rate of reaction increases
S–II : Increasing the temperature of the reaction decreases the fraction of reacting molecules which collide with energies greater than Ea
5. S–I : Reaction with molecularity ≥ 4 are common
S–II : Order of reaction is derived from chemical equation.
ASSERTION AND REASON QUESTIONS
In each of the following questions, a statement of Assertion (A) is given, followed by a corresponding statement of Reason (R). Mark the correct answer as
(1) if both (A) and (R) are true and (R) is the correct explanation of (A)
(2) if both (A) and (R) are true but (R) is not the correct explanation of (A)
(3) if (A) is true but (R) is false
(4) if both (A) and (R) are false.
1. (A) : A catalyst helps in attaining the equilibrium faster
BRAIN TEASERS
1. For a reaction 10 K K t t xc + = , when temperature is increased from 10°C to 100°C, rate
constant (K) increased by a factor of 512. Then, the value of x is
(1) 1.5 (2) 2.5
(3) 3 (4) 2
2. For a consecuitive first order reaction
12 ABKK C →→ , K1 = 2 × 10–3 s–1, K2 = 5 × 10–5 s–1 Determine the time at which [B] will be maximum.
(1) 1892 s (2) 946 s (3) 238 s (4) 5 min
3. Depletion of ozone occurs as; 2O 2→3O2
Step 1 : () c K
32 OOO + (fast)
Step 2 : () c K
32 OO2O + (slow)
What is the order of the reaction?
(1) 1 (2) 2
(3) 3 (4) 0.5
4. The following results have been obtained during the kinetic studies of the reaction
(R) : It catalyzes the forward as well as the back ward reactions to the same extent
2. (A) : Activation energy of every reaction can be zero.
(R) : The minimum extra amount of energy absorbed by reactant molecules so that their energy becomes equal to threshold value, is called heat of reaction
3 (A) : Decompostion of Ammonia on platinum surface at high pressure is a zero order reaction
(R) : At high pressure, the metal surface gets saturated with gas molecules
The correct option for the rate equation for above reaction is
(1) Rate = K[P]2[Q] (2) Rate = K[P]2[Q]°
(3) Rate = K[P][Q] (4) Rate = K[P]1[Q]2
5. Mechanism of a hypothetical reaction.
X2 → Y2 → 2XY is given below
(a) X2 X + X(fast)
(b) X + Y2 → XY + T (slow)
(c) X + Y → XY(fast)
The overall order of the reaction will be
(1) 2 (2) 0
(3) 1.5 (4) 1
6. In the reaction
()()()() 2 32 BrO5Br6H3Br3HO aqaqll --+→+++
2P + Q → R + S
The rate of appearance of Br2 is related to rate of disappearance of bromide ions as following
(1)
(2)
(3)
(4)
2 Br Br 3 5 dd dtdt -
=
2 Br Br 3 5 dd dtdt -
=-
2 Br Br 5 3 dd dtdt -
=-
2 Br Br 5 3 dd dtdt -
=
FLASH BACK (Previous NEET Questions)
1. For a reaction, activation energy Ea = 0 and the rate constant at 200 K is1.6 × 106 s–1. The rate constant at 400 K will be [Given that gas constant, R = 8.314 J K–1 mol–1] (2019-O)
(1) 3.2 × 106 s–1 (2) 3.2 × 104 s–1
(3) 1.6 × 106 s–1 (4) 1.6 × 103 s–1
2. The given graph is a representation of kinetics of a reaction
The y and x axes for zero and first order reactions, respectively are (2022)
(1) zero order (y = concentration and x = time), first order (y = rate constant and x = concentration)
(2) zero order (y = rate and x = concentration), first order (y = t1/2 and x = concentration)
(3) zero order (y = rate and x = concentration), first order (y = rate and x = t1/2)
(4) zero order (y = concentration and x
7. The correct rate law for the reaction ()()() 1130K,High pressure 3 22 Pt catalyst 2NH N3H ggg →+ is
(1) rate = k[NH3]2
(2) rate = kNH3]
(3) rate = k[NH3]0
(4) 1 2 3 rate NH k =
= ti me), first order (y = t 1/2 and x = concentration)
3. For a first order reaction A → Products, initial concentration of A is 0.1 M, which becomes 0.001 M after 5 minutes. Rate constant for the reaction in min–1 is (2022)
(1) 0.2303
(2) 1.3818
(3) 0.9212
(4) 0.4606
4. For a reaction A → B, enthalpy of reaction is -4.2 kJ mol–1 and enthalpy of activation is 9.6 kJ mol–1. The correct potential energy profile for the reaction is s hown in option. (2021) (1) (2) (3) (4)
5. The slope of Arrhenius Plot 1 ln / kvs T
of
first oder reaction is –5 x 103 K . The value of Ea of the reaction is. Choose the correct option for your answer [Given R = 8.314 J K-1 mol-1] (2021)
(1) –83 kJ mol-1
(2) 41.5 kJ mol-1
(3) 83.0 kJ mol-1
(4) 166 kJ mol-1
6. For the chemical reaction,N 2(g) + 3H2(g) 2NH3(g), the correct option is (2019)
(1) 23 HNH 32 dd dtdt
(2) 23 11HNH 32 dd dtdt
(3) 23 NNH 1 2 dd dtdt
(4) 23 NNH 2 dd dtdt
7. If the rate constant for a first order reaction is K, the time ( t ) required for the completion of 99% of the reaction is given by (2019)
(1) t = 2.303/K
(2) t = 0.693/K
(3) t = 6.909/K
(4) t = 4.606/K
8. When initial concentration of the reactant is doubled, the half-life period of a zero order reaction (2018)
(1) is halved
(2) is doubled
(3) is tripled
(4) remains unchaged
9. The correct difference between first and second order reactions is that (2018)
(1) the rate of a first-order reaction does
not depend on reactant concentrations; the rate of a second-order reaction does depend on reactant concentrations
(2) the half-life of a first-order reaction does not depend on [A]o the half-life of a second-order reaction does depend on [A]0
(3) a first-order reaction can be catalysed; a second order reaction cannot be catalysed
(4) the rate of a first-order reaction does depend on reactant concentrations; the rate of a second-order reaction does not depend on reactant concentrations.
10. Mechanism of a hypothetical reaction
X2 + Y2 → 2XY is given below : (2017)
(i) X2 → X + X (fast)
(ii) X → Y2 XY + Y(slow)
(iii) X + Y → XY(fast)
The overall order of the reaction will be
(1) 2 (2) 0
(3) 1.5 (4) 1
11. Consider the reaction between chlorine and nitric oxide: Cl2(g) + 2NO(g) → 2NOCl(g).
On doubling the concentration of both reactants, the rate of the reaction increases by a factor of 8. However, if only the concentration of Cl 2 is doubled, the rate increases by a factor of 2. The order of this reaction with respect to NO is: (2017-R)
(1) 0 (2) 1
(3) 2 (4) 3
12. For the reaction, XA + YB → ZC, If AB1.5Cddd dtdtdt
then the correct statement among the following is : (2017-R)
(1) The value of X = Y = Z = 3
(2) The value of X = Y = 3
(3) The value of X = 2
(4) The value of Y = 2
13. The rate of first-order reaction is 0.04 mol L–1 s–1 at 10 seconds and 0.03 mol L–1 s–1 at 20 seconds after initiation of the reaction. The half-life period of the reaction is (2016-I)
(1) 44.1 s
(2) 54.1 s
(3) 24.1 s
(4) 34.1 s
14. When initial concentration of a reactant is doubled in a reaction, its half-life period is not affected. The order of the reaction is (2015-C)
(1) second
(2) more than zero but less than first (3) zero
(4) first
15. The rate constant of the reaction A → B is 0.6 × 10–3 mol L-1 s–1. If the concentraction of A is 5 M, then concentration of B after 20 minutes is (2015-Re-exam)
(1) 3.60 M (2) 0.36 M
(3) 0.72 M (4) 1.08 M
16. The activation energy of a reaction can be determined from the slope of which of the following graphs? (2015-C)
(1) 1 ln vs k T
(2) 1 vs ln T kT
CHAPTER TEST
1. For the reaction 2A → B, what will be the rate of reaction w.r.t ‘A’ if the rate of appearance of ‘B’ is 0.1 mol L–1.s–1 ?
2. The rate law of a reaction, A+B → Product is: rate = K[A] n [B]m. On doubling the concentration of A and halving the concentration of B, the ratio of new rate to the earlier rate of reaction will be (1) n–m (2) 2 n–m
(3) In k vs T (4) ln vs k T T
17 For a reaction between A and B the order with respect to A is 2 and the order with respect to B is 3 . The concentrations of both A and B are doubled, the rate will increase by a factor of (2013-K)
(1) 12 (2) 16
(3) 32 (4) 10
18. A reaction is 50% complete in 2 hours and 75% complete in 4 hours. The order of reaction is (2013-K)
(1) 1 (2) 2 (3) 3 (4) 0
19. What is the activation energy for a reaction if its rate doubles when the temperature is raised from 20°C to 35°C? (2013)
(R = 8.314 J mol–1 K–1)
(1) 342 kJ mol–1
(2) 269 kJ mol–1
(3) 34.7 kJ mol–1
(4) 15.1 kJ mol–1
20. A reaction having equal energies of activation for forward and reverse reactions has: (2013)
(1) ∆ S=0 (2) ∆ G=0
(3) ∆ H=0 (4)
(3) 1 2mn + (4) m+n
3. Identify the false statement about catalyst (1) It increases the activation energy of the reaction
(2) It does not alter the magnitude of enthalpy change of the reaction
(3) It has no influence on energy of the reactants and products
(4) It does not alter the magnitude of Gibb’s free energy change of the reaction
4. A graph of ln K vs 1 T is given:
Where K is rate constant (in sec –1), T is temperature (in K K
Then fraction of molecules having energy equal to or greater than threshold energy is (1) e20/ T (2) e20/ RT (3) e –20/ RT (4) e –20/ T
5. For the reaction A → products, rate becomes 4 times the initial rate when concentration of ‘A’ is doubled. Order of the reaction is (1) –1 (2) 2
(3) 0.5 (4) 0.25
6. For the reaction H2(g)+ Br2(g) → 2HBr(g), the reaction rate = K[H 2] [Br 2] 1/2 . Which of the following statement is true about this reaction
(1) The reaction is of second order
(2) Molecularity of the reaction is 1.5
(3) The unit of K is sec–1
(4) Molecularity of the reaction is 2
7. Rate constant of a first order reaction is 0.03465 min –1 . Time taken for the completion of 87.5 % of the reaction is (1) 40 min (2) 60 min (3) 80 min (4) 50 min
8. 93.75% of a first order reaction is completed in 600 minutes. Time taken for 99.9% of the reaction to get completed is (1) 1200 min (2) 1500 min
(3) 1333.33 min (4) 900 min
9. In case of first order reactions, which of the following is NOT correct.
(1) t3/4 = 2t1/2 (2) t7/8 = 3t1/2
(3) t99.9 = 10t1/2 (4) t99 = 3t1/2
10. Identify correct statements from the following
A) Units of rate of second order reaction are L. mol–1 time–1
B) Rate constant of a reaction does not vary with concentration changes of reactants.
C) Catalyst increases the magnitude of rate constant of a reaction.
(1) A only (2) B and C only
(3) A and C only (4) C only
11. Hydrogenation of vegetable ghee at 25°C reduces pressure of H 2 from 2 atm to 1.2 atm in 50 minute. The rate of reaction in terms of molarity per second is
(1) 1.09 × 10–6 (2) 1.09 × 10–5
(3) 1.09 × 10–7 (4) 1.09 × 10–9
12. Which of the following reactions occurs at measurable rate?
(1) reaction between H+ and OH– ions in aqueous solution
(2) reaction between AgNO 3 and NaCl aqueous solutions
(3) hydrolysis of methyl acetate
(4) reaction between hydrogen and oxygen gases at room temperature
13. In a gas phase reaction ()()() 2 1 2 ABC ggg
the
increase in pressure from 100 mm of Hg to 120 mm of Hg is noticed in 5 minutes. If decomposition of A2 follows 1st order kinetics then what is rate of decomposition of A 2 in mm of Hg per minute, at 10 th minute
(1) 3.679 (2) 2.318
(3) 1.612 (4) 2.875
14. Consider the gaseous reaction, A(g) → 2B(g). The initial pressure of ‘ A ’ is ‘ p o ’ mm and total pressure after time ‘ t ’ is ‘p t ’ mm. If the reaction follows first order kinetics, the correct expression for rate constant (K) is
(1) 0 0 2.303 log 2 t p tpp -
(2) 0 0 2 2.303 log 2 t p tpp -
(3) 0 0 2 2.303 log 3 t p tpp -
(4) 0 0 2.303 log t p tpp -
15. Two gases ‘A’ and ‘B’ are in a container. The experimental rate law for the reaction between them has been found to be rate = k[A]2[B]. Predict the effect on the rate of the reaction when the partial pressure of each reactant is doubled :
(1) the rate is doubled (2) rate becomes four times (3) the rate becomes six times (4) the rate becomes eight times
16. Let the equilibrium constant of Step I be 2 × 10–3 mol–1 L and the rate constants for the formation of A2X and A2 in Steps II and III are 3.0 × 10–2) mol–1 L min–1 and 1 × 103 mol–1 L min–1( all data at 25°C), then what is the overall rate constant mol–2 L2 min–1 of the consumption of B2 ?
(1) 6 × 10–5
(2) 1.2 × 10–4
(3) 3 × 10–5
(4) 1.5 × 10–5
17. How many times the rate of formation of A 2 will increases if concentrations of AX is doubled and that of B2 is increased three fold?
(1) 36 (2) 12 (3) 6 (4) 8
18. The rate of a certain biochemical reaction at physiological temperature (T) occurs 106 times faster with enzyme than without. The change in the activation energy upon adding enzyme is: (2020)
(1) +6(2.303)RT (2) +6RT
(3) –6(2.303)RT (4) –6RT
19. Consider the following plots of rate constant(k) versus 1 T for four different r eactions( a,b,c and d ). Which of the following orders is correct for the activation energies of these reactions? ( 2020)
(1) E a > E c > Ed > Eb
(2) E c > E a > Ed > Eb
(3) Eb > E a > Ed > E c
(4) Eb > Ed > E c > E a
20. For following reactions
700 Product K A →
500 Product K catalyst A →
It was found that the activation energy is decreased by 30 kJ/mol in the presence of catalyst. If the rate remains unchanged, the activation energy for catalysed reaction is (Assume pre exponential factor is same): (2020)
(1) 198 kJ/mol
(2) 135 kJ/mol
(3) 75 kJ/mol
(4) 105 kJ/mol
21. At 518°C, the rate of decomposition of a sample of gaseous acetaldehyde, initially at a pressure of 363 Torr, was 1.00 Torr s −1 when 5% had reacted and 0.5 Torr s−1 when 33% had reacted. The order of the reaction is: (2018)
(1) 2 (2) 3 (3) 0 (4) 1
22. For the reaction A → B; following curves represent conc. vs time
The correct curves are (1) 1, 2 only (2) 2, 3 only (3) 1, 4 only (4) 3, 4 only
23. For 222 1 2 XYXY +→ , relative rates of species is given as
(1)
(2)
(3)
(4)
24. For N2 + 3H2 → 2NH3, rates of disappearance of N2 and N2 and rate of appearance of NH3 respectively are a,b and c, then
(1) a > b > c (2) a < c < b (3) a = b > c (4) a = b = c
25. From the graph.
Now, correct relationship is (1) 1 23 12 o CCCC CC XYQ== (2) 23 01 12 CCCC CC QYX>> (3) 01 23 12 CCCC CC
26. For a chemical reaction Y 2 + 2Z → Product, rate controlling step is Y + 1/2Z → Q. If the concentration of Z is doubled, the rate of reaction will
(1) Remain the same
(2) Become four times
(3) Become 1.414 times (4) Become double
27. The initial rates for gaseous reaction A + 3B→ AB 3 are given below
(1) zero (2) three (3) one (4) two
28. 50% completion of a first order reaction takes place in 16 minutes. Then fraction that would react in 32 minutes from the beginning
(1) 1 2 (2) 1 4 (3) 1 8 (4) 3 4
29. For a first order reaction, the half–life is 50 s. Identify the correct statement from the following.
(1) the reaction is almost complete in 500 s
(2) the same quantity of reactant is consumed for every 50 s of the reaction
(3) quantity of reactant remaining after 100 s is half of what remains after 50 s
(4) the same quantity of reactant is consumed for every 25 s of the reaction
30. For a first order reaction, if the time taken for completion of 50% of the reaction is ‘ t’ seconds, the time required for completion of 99.9% of the reaction is:
(1) 10t (2) 5t (3) 100t (4) 2t
31. Rate constant K and Half–life period respectively for the reaction, A + B
35. The activation energy of a reaction is 58.3 kJ/mol. The ratio of the rate constants at 305 K and 300 K is about (R = 8.3 J K–1 mol–1)
(Antilog 0.1667 = 1.468) (1) 1.25 (2) 1.75
(3) 1.5 (4) 2.0
36. For X → Y, 10 3 t t k k + = . If the rate constant at 300 K is ‘Q’ min–1, at what temperature rate constant becomes ‘9Q’ min–1?
(1) 47°C (2) 320°C
(3) 280 K (4) 9300K ×
37. For N2O5(in CCl4) 4-1
(1) 1 s–1, 0.5 s (2) 0.5 s–1, 1.386 s (3) 0.5 s–1, 0.05 s (4) 0.05 s–1, 1.386 s
32. A reaction 2A → 3B takes place according to first order kinetics the time in which the concentration of reactant and product become equal is (K = 4.6 × 10 –2 min –1, log 5 = 0.7 log 3 = 0.48, log 2 = 0.3)
(1) 11 min (2) 15 min
(3) 9 min (4) 25 min
33. For the decompostion of N2O5(g), it is given that: N2O5(g) → 4NO2(g) + O2(g). Activation energy ()()() 2 522 1 ,2 2 aggg ENONOO →+ ;
Activation energy E a’, (1) E a = E a ’ (2) E a > E a ’ (3) E a < E a ’ (4) E a = 2E a ’
34. For a first order reaction A → P, the temperature (T) dependent rate constant (K) was found to follow the equation
() 1 log20006.0 =-+ K T .
The pre–exponential factor, A and the activation energy E a, respectively, are
(1) 1 × 106 s–1 and 9.2 kJ mol–1
(2) 6 s–1 and 16.6 kJ mol–1
(3) 1 × 106 s–1 and 16.6 kJ mol–1
(4) 1 × 106 s–1 and 38.3 kJ mol–1
22 1 2,610s 2 NOOK→+=× at 35 0 K and K = 1.2 × 10–3 s–1 at 360 K. Then, when temperature is changed to 380 K, value of K (in s–1)
38. In the Arrhenius equation, the Boltzmann factor R e aE Trepresents the ____ of the molecules possessing energy in excess of activation energy (1) number (2) fraction (3) weight (4) percentage
39. The graph of log K versus 1/T is given below. Here ‘x’ is equal to [Where K = rate constant and T = absolute temperature] slope=x log K 1 T (1) 2.303 a E(2) 2.303 a E R(3) 2.303 . a ER(4) 2.303 a R E -
40. A first order reaction is 50% completed in20 minutes at 27°C and in 5 minutes at 47°C. The energy of activation of the reaction is :
41. The initial concentration of cane sugar in presence of an acid was reduced from 0.20 to 0.10 M in 5 hours and to 0.05 M in 10 hours, value of K ? (in hr –1) (1) 0.693 (2) 1.386 (3) 0.1386 (4) 3.465
42. From the graph the value c t ∆ ∆ and the value of rate of reaction at X respectively are called
(1) Average rate and instantaneous rate
(2) Instantaneous rate and average rate
(3) Average rate only
(4) Instantaneous rate only
ANSWER KEY
43. In the reversible reaction 1 2 24 2 k k NONO
The rate of disappearance of NO2 is equal to
(1) 2 1 2 2 2k NO k
(2) 2k1[NO2]2 – 2k2 [N2O4]
(3) 2k1[NO2]2 – k2 [N2O4]
(4) (2k1– k2) [NO4]
44. The variation of the concentration of the products with time is given by the curve :
(1) x (2) y (3) z (4) w
45. For the following elementary step; (CH 3) 3 CBr (aq) → (CH 3) C + (aq) + Br (aq) – , the molecularity is
