What drives chemical reactions? Reactivity 1
You should know:
✔ chemical reactions involve a transfer of energy between the system and the surroundings, while total energy is conserved
✔ reactions are described as endothermic or exothermic, depending on the direction of energy transfer between the system and the surroundings
✔ the relative stability of reactants and products determines whether reactions are endothermic or exothermic
✔ the standard enthalpy change for a chemical reaction, ΔH ○ , refers to the heat transferred at constant pressure under standard conditions and states. It can be determined from the change in temperature of a pure substance.
You should be able to:
✔ understand the difference between heat and temperature
✔ understand the temperature change (decrease or increase) that accompanies endothermic and exothermic reactions, respectively
✔ sketch and interpret energy profiles for endothermic and exothermic reactions
✔ apply the equations Q = mcΔT and ΔH = Q n in the calculation of the enthalpy change of a reaction.
Thermochemistry studies temperature changes and heat transfer during chemical or physical processes. At constant pressure, the amount of heat released or absorbed in a chemical reaction or change of state is equal to the enthalpy change, ΔH, of that reaction or change of state.
The first law of thermodynamics states that energy cannot be created or destroyed; it can only be converted from one form to another. A specific physical or chemical change will always absorb or release the same amount of energy when it is carried out under the same conditions. This energy can be determined by measuring the amount of heat transferred between the system and the surroundings
Heat, temperature and specific heat capacity
When the same amount of heat is transferred to a sample of water and to a block of copper of the same mass, the temperature rise of the copper is approximately ten times greater than that of the water. In other words, the specific heat capacity (c) of water is approximately ten times higher than that of copper (table 1). In the general case, the amount of heat (Q) transferred to the sample and the temperature change (ΔT) of that sample are related as follows:
Q = mcΔT
• Enthalpy change, ΔH, in a system refers to the heat absorbed or released by that system at constant pressure. The enthalpy change of a chemical reaction is associated with the breaking and forming of chemical bonds.
• A system is all the reactants, products and solvents involved in the change of interest.
• The surroundings are everything else in the universe outside the system.
• An open system c an exchange matter and energy with the surroundings.
1.1 Measuring enthalpy changes Samplecontent
• A closed system c an exchange energy but not matter with the surroundings.
• An isolated system c annot exchange matter or energy with the surroundings.
▾ Table 1 Specific heat capacities of selected substances at 25 °C
• Heat, Q, is a form of energy (SI unit: joule, J).
• Temperature, T, is a measure of the average kinetic energy of particles (SI unit: kelvin, K).
• Specific heat capacity, c, of a substance is the amount of heat (in J) required to raise the temperature of 1 g of that substance by 1 K. It is usually expressed in J g 1 K 1 or kJ kg 1 K 1
The relationship between temperature and kinetic energy of particles is discussed in Structure 1.1
Example 1
A block of aluminium with the mass 34.5 g and initial temperature 23.5 °C absorbs 650 J of heat. C alculate the final temperature, in °C, of the block. Refer to table 1.
Solution
First, solve the equation Q = mcΔT for ΔT using the specific heat capacity of aluminium from table 1:
ΔT = Q mc = 650 J 34.5 g × 0.903 g 1 K 1 ≈ 20.9 K
The change in temperature has the same numerical value when expressed in K or °C (see Tool 3.2). Therefore:
ΔT = T2 T1 = 20.9 °C
T2 = 20.9 °C + 23.5 °C = 44.4 °C
The equation Q = mcΔT and the specific heat capacity of water, c w = 4.18 kJ kg 1 K 1 , are given in sections 1 and 2, respectively, of the data booklet.
Exothermic and endothermic reactions
Exothermic reactions produce heat while endothermic reactions absorb heat. When thermal energy is transferred between the system and surroundings, it can be measured and used to calculate the enthalpy change (ΔH) of the reaction or any other process, such as a change of state. The standard enthalpy change (ΔH ○ ) of a process is the amount of heat transferred to a closed system during that process under standard conditions. Both ΔH and ΔH ○ are usually expressed in kJ mol 1
• In an exothermic reaction, heat is transferred from the system to the surroundings (ΔH < 0).
• In an endothermic reaction, heat is transferred from the surroundings to the system (ΔH > 0).
• The standard enthalpy change, ΔH ○ , of a chemical reaction is the amount of heat transferred to a closed system during that reaction under standard conditions. It is usually expressed in kJ mol 1 .
• Standard state refers to the most stable state of a substance at a given temperature, usually 298.15 K (25 °C), and a pressure of 100 kPa. A standard enthalpy change requires all participating species to be in their standard states, unless specified otherwise.
• Standard conditions for gases, also known as standard temperature and pressure (STP), are a temperature of 273.15 K (0 °C) and a pressure of 100 kPa.
• Standard ambient temperature and pressure (SATP) are a temperature of 298.15 K (25 °C) and a pressure of 100 kPa.
At constant pressure, the enthalpy change and the amount of heat produced or consumed by a given process have equal values but opposite signs (table 2). The reason for this is the way we define these quantities: the heat flow is measured with respect to the surroundings while the enthalpy change is measured with respect to the system. For an exothermic reaction, the surroundings receive thermal energy from the system (Q > 0) while the system loses thermal energy to the surroundings (ΔH < 0). Conversely, an endothermic reaction draws thermal energy from the surroundings (Q < 0), so the enthalpy of the system increases (ΔH > 0).
Samplecontent
Process Heat Heat flowEnthalpy change exothermicreleasedfrom system to surroundings ΔH < 0 endothermicconsumedfrom surroundings to system ΔH > 0
▴ Table 2 Heat flow and enthalpy change
The transfer of heat between the system and surroundings is not an instantaneous process. Therefore, an exothermic reaction typically raises the temperature of the reaction mixture while an endothermic reaction lowers that temperature. However, external cooling or heating can be used to keep the temperature of the system constant or even reverse the usual temperature changes associated with exothermic and endothermic processes.
Changes of state often occur at a constant temperature because the heat flow between the system and surroundings exactly matches the energy consumed or released by the change. For example, a mixture of ice and liquid water will have a temperature of 0 °C until all the ice melts or all the water freezes (figure 1). Similarly, the temperature of liquid water under standard pressure cannot be raised above 100 °C because any external energy supplied to the water will be consumed by the phase change H2O(l) → H2O(g)
▴ Figure 1 Heating curve of water
The thermal effect of a chemical reaction depends on the relative stability of reactants and products: energy is released when the potential energy of the products is lower than that of the reactants, and it is consumed when the products have higher potential energy than the reactants. The change in potential energy along the reaction pathway can be represented as a reaction energy profile (figure 2). For an exothermic reaction, the products are more stable than the reactants, so the overall change in energy of the system is negative (ΔH < 0). Conversely, the products of an endothermic reaction are less stable than the reactants, so the overall change in energy of the system is positive (ΔH > 0).
Energy profiles of chemical reactions, transition states and activation energy will be discussed further in Reactivity 2.2
Assessment tip
Calorimetry experiments typically show a smaller change in temperature than that expected from theoretical calculations. The main reason for this is the loss of heat to the surroundings, which can be minimized by improving the insulation of the reaction chamber (Tool 1.3) or compensated for by extrapolating the thermal curve (Tool 3.4, figure 32).
Assessment tip
A sketched energy profile for a chemical reaction or change of state should include the following features: relative positions of the reactants, products and transition state; a curve representing the reaction pathway; an arrow representing the enthalpy change (pointing downward for ΔH < 0 or upward for ΔH > 0); and an upward arrow representing the activation energy (E a). The reactants must be positioned higher than the products for an exothermic process, and lower than the products for an endothermic process. The axes should be labelled as reaction coordinate (x) and potential energy (y).
Calorimetry
The standard enthalpy change for a chemical reaction can be determined from the change in temperature of a pure substance, such as water, using a calorimeter (see Tool 1.3). A typical calorimetry experiment is described in example 2.
Example 2
To measure the enthalpy change of combustion for ethanol, ΔH c, a student heated a copper calorimeter containing 0.100 dm3 of water. The following data were recorded:
• initial temperature of water, T1 = 18.0 °C
• final temperature of water, T2 = 81.4 °C
• mass of ethanol combusted, m e = 1.19 g
• density of water, d = 1.00 g cm 3
a) Calculate the enthalpy change of combustion of ethanol, ΔH c, in kJ mol 1
b) Compare your answer with the ΔH c value for ethanol from section 14 of the data booklet, giving two reasons for the difference.
c) Sketch and annotate a diagram for the experimental set-up.
Solution
a) First, use the equation Q = mcΔT to calculate the amount of heat produced by the combustion of ethanol: ΔT = T2 T1 = 81.4 18.0 = 63.4 K
(Remember that the change in temperature has the same numerical value on the Celsius and Kelvin scales.)
V(H2O) = 0.100 dm3 = 100 cm3
m(H2O) = d × V = 1.00 g cm 3 × 100 cm3 = 100 g
Q = mcΔT = 100 g × 4.18 J g 1 K 1 × 63.4 K ≈ 2.65 × 104 J = 26.5 kJ
Next, calculate the amount of ethanol, in mol, using the expression n = m M :
M(CH3CH2OH) = 2 × 12.01 + 6 × 1.01 + 16.00 = 46.08 g mol 1
n = 1.19 g 46.08 g mol 1 ≈ 0.0258 mol
Finally, calculate the enthalpy change of combustion:
ΔH c = − Q n = − 26.5 kJ 0.0258 mol ≈ −1030 kJ mol 1
Notice that the combustion of ethanol is exothermic, so ΔH c < 0
b) ΔH c for ethanol given in the data booklet ( 1367kJmol 1) is larger in magnitude (more negative) than the result of our calculations. There are several potential reasons for this difference, including:
• heat loss to the surroundings
• heat absorbed by the copper calorimeter (rather than water)
• incomplete combustion of ethanol
• the presence of impurities, such as water, in the ethanol sample.
All these factors would reduce the experimental ΔH c value.
c) See figure 3.
burner containing ethanol copper calorimeter water thermometer screen to reduce draughts insulating card ▴ Figure 3 Experimental set-up for determining the heat of combustion for ethanol
Practice problems for R1.1
Problem 1
When equal masses of samples X and Y absorb the same amount of heat, their temperatures rise by 5 °C and 10 °C, respectively. Which statement is correct?
A. The specific heat capacity of X is twice that of Y
B. The specific heat capacity of X is half that of Y.
C. The specific heat capacity of X is five times as high as that of Y
D. The specific heat capacity of X is one-fifth of that of Y.
Problem 2
Which of the following statements are correct?
I An endothermic reaction reduces the potential energy of the system.
II An exothermic reaction has a negative enthalpy change.
III The enthalpy change is positive when the products are less stable than reactants.
A. I and II only
B. I and III only
C. II and III only
D. I, II and III
Problem 3
Sketch and label the energy profile for an endothermic reaction.
Problem 4
50.0 cm3 of 1.00 mol dm 3 copper(II) sulfate solution is placed in a polystyrene cup, and excess zinc powder is added at t = 100 s.
Cu2+(aq) + Zn(s) → Cu(s) + Zn2+(aq)
The temperature of the reaction mixture is plotted against time, as shown below.
The amount of heat (Q) produced by the reaction can be calculated from the temperature change (ΔT) and mass (m) of the reaction mixture using the expression Q = mc w ΔT, where c w is the specific heat capacity of water.
a) State two assumptions that you make by using this expression.
b) Determine the temperature change, ΔT, that would be observed if the reaction took place instantaneously and with no heat loss to the surroundings.
c) Calculate the amount of heat, in kJ, produced during the reaction.
d) Calculate the enthalpy change, in kJ mol 1, for this reaction.
1.2 Energy cycles in reactions
You should know:
✔ bond breaking absorbs energy, and bond forming releases energy
✔ Hess’s law states that the enthalpy change for a reaction is independent of the pathway between the initial and final states
✔ standard enthalpy changes of formation, ΔH ○ , and combustion, ΔH c ○ , data are used in thermodynamic calculations
✔ an application of Hess’s law uses enthalpy of formation data or enthalpy of combustion data to calculate the enthalpy change of a reaction
✔ a Born–Haber cycle is an application of Hess’s law, used to show energy changes in the formation of an ionic compound.
Bond enthalpies
You should be able to:
✔ calculate the enthalpy change of a reaction from given average bond enthalpy data
✔ apply Hess’s law to calculate enthalpy changes in multistep reactions
Samplecontent
✔ deduce equations and solutions to problems involving ΔHf
and
✔ calculate enthalpy changes of a reaction using ΔH
or
data
✔ interpret and determine values from a Born–Haber cycle for compounds composed of univalent and divalent ions.
Energy is absorbed when chemical bonds are broken, and it is released when bonds are formed. The difference between the total energy of chemical bonds in the reactants and products determines whether a chemical reaction is exothermic (ΔH r < 0) or endothermic (ΔH r > 0).
The average bond enthalpy, or simply bond enthalpy (BE) is the energy required to break one mole of a particular covalent bond in a gaseous molecule. The BE value for each type of bond is obtained by averaging the slightly different values for the experimentally determined bond enthalpies in various molecules. For example, the actual enthalpies of C H bonds in organic compounds vary from 370 to 440 kJ mol 1, giving the average value of 414 kJ mol 1
The standard enthalpy change for a chemical reaction (ΔH r ○ ) can be estimated using the BE values for all participating species:
ΔH r ○ = Σ (BE bonds broken) Σ (BE bonds formed)
Since bond enthalpies are average values, the standard enthalpy change for a particular reaction calculated in this way may differ slightly from the experimentally determined ΔH r ○ value. If the reaction involves solid and/or liquid substances, the difference will be greater because all BE values are calculated for gaseous species.
Example 3
a) Calculate the standard enthalpy change, in kJ mol 1, for the complete combustion of ethane, C2H6(g).
b) Compare and contrast the value obtained in part (a) with that given in section 14 of the data booklet.
Bond enthalpy (BE) values are listed in section 12 of the data booklet.
Covalent bonds and their characteristics are discussed in Structure 2.2.
Solution
a) First, formulate a balanced chemical equation for the complete combustion of 1 mol of ethane:
C2H6(g) + 3.5O2(g) → 2CO2(g) + 3H2O(l)
List the bonds that are broken and formed in this reaction, together with the corresponding BE values from section 12 of the data booklet:
Bonds broken:
1 × C C, 6 × C H and 3.5 × O O
ΔH = 1 × 346 + 6 × 414 + 3.5 × 498 = 4573 kJ mol 1
Bonds formed:
4 × C O and 6 × O H
ΔH = 4 × 804 + 6 × 463 = 5994 kJ mol 1
ΔH c ○ (C2H6, g) = Σ (BE bonds broken) Σ (BE bonds formed) = 4573 − 5994 = −1421 kJ mol 1
Notice that the enthalpy change is negative because the combustion of ethane is exothermic.
b) According to section 14 of the data booklet, the actual ΔH c ○ value for C2H6(g) is 1561 kJ mol 1. The two ΔH c ○ values are close but not identical because bond enthalpies are averages, so they might differ slightly in specific molecules. In addition, all BE values are given for gaseous species while one of the products in our reaction, H2O(l), is liquid under standard conditions.
Bond enthalpy is a quantitative measure of bond strength: the higher the BE value, the stronger the bond. In turn, bond strength is inversely related to bond length: the closer the two atoms together, the stronger the bond. Multiple bonds are shorter than single bonds, so bond enthalpies generally increase along with the bond order: single < double < triple (table 3).
Nucleophilic substitution reactions of halogenoalkanes are discussed in Reactivity 3.4.
▴ Table 3 Characteristics of carbon–carbon covalent bonds
The BE values of carbon–halogen bonds decrease down the group, from 492 kJ mol 1 for C F to 228 kJ mol 1 for C I. As a result, the reactivity of halogenoalkanes in nucleophilic substitution reactions increases in the same order: RF < RCl < RBr < RI
Hess’s law is particularly useful for calculating enthalpy changes that cannot be determined experimentally. The unknown enthalpy change for a chemical reaction can be found by combining other reactions with known enthalpy changes, as shown in the example below.
Example 4
Consider these two combustion reactions with known enthalpy changes:
2NO(g) + O2(g) → 2NO2(g) ΔH1 ○ = −116 kJ mol 1
2CO(g) + O2(g) → 2CO2(g) ΔH2 ○ = −566 kJ mol 1
Using these data, calculate the standard enthalpy change, in kJ mol 1, for the following reaction:
CO(g) + NO2(g) → CO2(g) + NO(g) ΔH ○ 3 = ?
Solution
Chemical equations can be combined in the same way as mathematical equations. To obtain the third equation, we need to subtract the first equation from the second, cancel out O2(g) and halve the stoichiometric coefficients of the remaining species:
2CO(g) + O2(g) 2NO(g) O2(g) → 2CO2(g) 2NO2(g)
2CO(g) + 2NO2(g) → 2CO2(g) + 2NO(g)
CO(g) + NO2(g) → CO2(g) + NO(g)
The enthalpy changes of the reactions must be combined in the same way as the equations. Therefore:
ΔH ○ 3 = 0.5 × (ΔH ○ 2 −ΔH ○ 1 ) = 0.5 × [−566 ( 116)] = 225 kJ mol 1
Standard enthalpy changes of formation and combustion
The standard enthalpy change of formation (ΔH ○ f ) for a given compound is the enthalpy change of a real or hypothetical chemical reaction in which one mole of that compound in its standard state is formed from elementary substances, also in their standard states. For example, the ΔH ○ values for carbon dioxide, CO2(g), and methanol, CH3OH(l), correspond to the standard enthalpy changes of the following processes:
C(s, graphite) + O2(g) → CO2(g) ΔH ○ r ≡ ΔH ○ f (CO2, g)
C(s, graphite) + 2H2(g) + 0.5O2(g) → CH3OH(l) ΔH ○ r ≡ ΔH ○ f (CH3OH, l)
Elementary substances in their standard states, such as C(s, graphite) or O2(g), cannot be formed from simpler species, so their ΔH ○ f values are assumed to be 0. However, only one allotrope of each element is assigned a zero ΔH ○ value while other allotropes have different structures and thus different (non-zero) standard enthalpies of formation. For example, the ΔH ○ f values for graphite and diamond are 0 and 1.8 kJ mol 1, respectively.
• Hess’s law states that, regardless of the route of a chemical or physical transformation, the enthalpy change will always be the same provided that the initial and final states of the system are identical.
In many cases, the chemical transformation of reactants into products can be carried out in more than one way. In figure 4, product B can be synthesized from reactant A either directly or through intermediate C. Both routes lead from the same initial state of the system to the same final state, so the thermal effects of these routes must be identical: ΔH1 = ΔH2 + ΔH3. This is the basis of Hess’s law, which states that the enthalpy change for a chemical or physical process is independent of the pathway between the initial and final states.
The standard enthalpy change of combustion (ΔH ○ c ) for a given compound is the enthalpy change of a chemical reaction in which one mole of that compound in its standard state reacts with oxygen gas to produce the most stable combustion products, such as carbon dioxide and water, also in their
• The standard enthalpy change of formation (ΔH ○ f ) of a substance is the enthalpy change that occurs when one mole of that substance is formed from its constituent elements, with all participating species in their standard states.
Assessment tip
The equations representing ΔH ○ f values must be balanced for one mole of the product. All participating species must be in their standard states at SATP (p = 100 kPa and T = 25 °C or 298.15 K), and these states must be shown in the equations.
Nature of science
Strictly speaking, standard enthalpy changes can be defined for any temperature, and that temperature must be stated explicitly, e.g. ΔH ○ f, 298. In our course, all ΔH ○ f values refer to SATP, so the temperatures are omitted.
Hess’s law
• The standard enthalpy change of combustion (ΔH ○ c ) of a substance is the enthalpy change that occurs when one mole of that substance is burned completely in oxygen, with all participating species in their standard states.
Assessment tip
The equations representing ΔH ○ c values must be balanced for one mole of the substance undergoing combustion, and the states of all participating species must be shown.
standard states. For example, the ΔH ○ c values for graphite and methanol correspond to the standard enthalpy changes of the following processes:
C(s, graphite) + O2(g) → CO2(g) ΔH ○ r ≡ ΔH ○ c (C, graphite)
CH3OH(l) + 1.5O2(g) → CO2(g) + 2H2O(l) ΔH ○ r ≡ ΔH ○ c (CH3OH, l)
Notice that the combustion of graphite and the formation of carbon dioxide under standard conditions are represented by the same equation, so ΔH ○ c (C, graphite) ≡ ΔH ○ (CO2, g)
Complete combustion of organic compounds containing nitrogen usually produces nitrogen gas rather than nitrogen oxides, for example:
CH3NH2(g) + 2.25O2(g) → CO2(g) + 2.5H2O(l) + 0.5N2(g) ΔH
H
c (CH3NH2, g)
The standard enthalpy change for any reaction can be calculated using enthalpy of formation or enthalpy of combustion data for all participating species. The following expressions can be derived using Hess’s law:
ΔH ○ r = Σ (ΔH ○ f products) Σ (ΔH ○ f reactants)
ΔH ○ r =
c reactants) Σ (
H
c products)
Notice that the order of subtraction for enthalpies of combustion is opposite to that for enthalpies of formation.
The expressions for determining the enthalpy change of a reaction using ΔH ○ f or ΔH ○ c data are given in section 1 of the data booklet.
Standard enthalpies of formation and combustion are listed in sections 13 and 14, respectively, of the data booklet.
Calculate the enthalpy of combustion for propanoic acid, CH3CH2COOH(l), using section 13 of the data booklet and the following information:
ΔH ○ (CH3CH2COOH,l)=−511 kJ mol 1
Solution
Assessment tip
Most combustion reactions are spontaneous and exothermic, so their enthalpy change values are negative. One of the few exceptions is the combustion of nitrogen, N2(g), which is unfavourable because of the particularly high enthalpy of the triple N N bond
(BE = 945 kJ mol 1). As a result, the ΔH ○ f values for all oxides of nitrogen are positive.
First, formulate a balanced equation for the complete combustion of one mole of propanoic acid:
CH3CH2COOH(l) + 3.5O2(g) → 3CO2(g) + 3H2O(l) ΔH
Next, list the ΔH ○ r values for all participating species:
c (CH3CH2COOH, l)
CH3CH2COOH(l) + 3.5O2(g) → 3CO2(g) + 3H2O(l)
ΔH ○ / kJmol 1 5110 394 286
Notice that O2(g) is an elementary substance in its standard state, so
ΔH ○ (O2, g)=0
Finally, use the expression
H
products)
ΔH ○ c (CH3CH2COOH, l) = 3 × ( 394) + 3 × ( 286) ( 511) =−1529 kJ mol 1
f reactants):
Sample student answer
a) Determine the enthalpy change of reaction, ΔH r, in kJmol 1, when 1.00mol of gaseous hydrazine decomposes to its elements. Use bond enthalpy values in section 12 of the data booklet. [3]
N2H4(g) → N2(g) + 2H2(g)
b) The standard enthalpy of formation of N2H4(l) is +50.6kJmol 1
Calculate the enthalpy of vaporization, ΔH vap, of hydrazine. Give your answer in kJmol 1. [2]
This answer could have achieved 3/5 marks:
a) Bonds broken:
(1 × N N) + (4 × N H) = 158 + 4 × 391 = 1722 kJ mol–1
Bonds formed:
(1 × N N) + (2 × H H) = 945 + 2 × 436 = 1817 kJ mol–1
ΔH r = 1817 1722 = 95 kJ mol–1
vap
b) N2H4(l) N2H4(g) N2(g) + 2H2(g)
H vap =
Born–Haber cycles (AHL only)
Correct Σ(BE bonds broken).
Correct Σ(BE bonds formed).
The subtraction is reversed; correct ΔH r value is 95 kJ mol–1
Correct diagram.
H gives
H for the process N2H4(g) → N2H4(l), whereas the vaporization is the reverse process; also, an incorrect value of ΔH (+95 kJ mol–1 instead of 95 kJ mol–1) is used. Correct answer:
Born–Haber cycles are an application of Hess’s law to ionic compounds. Each Born–Haber cycle involves several enthalpy changes associated with stepwise formation of a solid ionic lattice from its constituent elements, typically a metal and a non-metal (figure 5). The individual steps in the cycle include ionization energy (ΔH ○ IE ), enthalpy of atomization (ΔH ○ at ), electron affinity (ΔH ○ EA), lattice enthalpy (ΔH ○ lat) and enthalpy of formation (ΔH ○ f ).
• A Born–Haber cycle is a hypothetical series of steps representing the formation of an ionic compound.
• Ionization energy (IE, or ΔH ○ IE ) is the standard enthalpy change that occurs when one mole of electrons is removed from one mole of gaseous atoms or positively charged ions:
M(g) → M+(g) + e (IE1) M+(g) → M2+(g) + e (IE2)
• Enthalpy of atomization (ΔH ○ at ) is the standard enthalpy change that occurs when one mole of gaseous atoms is formed from an elementary substance. For a solid monatomic species, ΔH ○ at is identical to the enthalpy of sublimation, ΔH ○ sub:
M(s) → M(g)
Ionization energy, electron affinity and lattice enthalpy were introduced in Structure 1.3, Structure 3.1 and Structure 2.1, respectively. The values of ionization energy and electron affinity for most elements are listed in section 9 of the data booklet, and the values of lattice enthalpy in section 16.
Example 5
Assessment
tip
Construction of complete Born–Haber cycles is a time-consuming process, so it will not be required in examination papers. However, you should be able to annotate Born–Haber cycles involving singly and doubly charged ions, interpret the enthalpy changes associated with individual steps and apply Hess’s law to calculate unknown
values.
• Electron affinity (EA, or ΔH ○ EA) is the standard enthalpy change that occurs when one mole of electrons is added to one mole of gaseous atoms:
X(g) + e → X (g) (EA1)
X (g) + e → X2 (g) (EA2)
• Lattice enthalpy (ΔH ○ lat) is the standard enthalpy change that occurs when one mole of structural units of a solid lattice is separated into gaseous ions:
M a Xb(s) → a Mb+(g) + bXa (g)
Assessment tip
Lattice enthalpies are sometimes quoted as negative values that represent the exothermic formation of the solid lattice from gaseous ions. However, in DP Chemistry, the lattice enthalpy always refers to the endothermic formation of gaseous ions from the solid lattice.
▴ Figure 5 A generalized Born–Haber cycle
Any unknown ΔH ○ value in a Born–Haber cycle can be calculated if all other ΔH ○ values are known. This is particularly important for lattice enthalpies, which cannot be determined experimentally. According to Hess’s law, the direct route
MX(s) → M+(g) + X (g)
must have the same enthalpy change as the alternative route involving all other steps in the Born–Haber cycle, i.e.
Therefore:
Notice that the sign of ΔH ○ f must be changed because the reaction
MX(s) → M(s) + 1 2 X2(g) is opposite to the formation of the ionic compound MX(s) from its constituent elements, M(s) and X2(g)
a) Formulate an equation that represents the standard enthalpy change of formation for sodium chloride.
b) Construct a Born–Haber cycle for sodium chloride.
c) Calculate the lattice enthalpy for sodium chloride, in kJmol 1, using sections 9 and 12 of the data booklet and the following information (“sub” refers to sublimation):
ΔH ○ f (NaCl, s) = −411 kJ mol 1
ΔH ○ sub(Na) = +108 kJ mol 1
Solution
a) Na(s) + 1 2 Cl2(g) → NaCl(s)
ΔHat (Cl)
ΔHIE (Na) +121 kJ mol‒1 +496 kJ mol‒1
Na+(g) + Cl(g) + e‒
Na+(g) + Cl2(g) + e‒
Na(g) + Cl2(g)
ΔHsub(Na) +108 kJ mol‒1
Na(s) + Cl2(g)
Na+(g) + Cl‒ (g)
b) x kJ mol‒1
ΔHEA(Cl) ‒349 kJ mol‒1 1 2 1 2 1 2
ΔHlat(NaCl)
ΔHf (NaCl)
‒411 kJ mol‒1
NaCl(s)
▴ Figure 6 Born–Haber cycle for sodium chloride
Notice that the atomization of chlorine produces one mole of Cl(g) atoms by breaking Cl Cl bonds in half a mole of Cl2(g) molecules, so ΔH ○ at (Cl) ≡ 1 2 BE(Cl Cl). For sodium, ΔH ○ at (Na) ≡ ΔH ○ sub(Na)
c) To determine the lattice enthalpy for sodium chloride, the pathway in figure 6 from NaCl(s) to Na+(g) + Cl (g) should be followed anticlockwise, and the sign of the ΔH ○ (NaCl, s) should be changed:
ΔH ○ lat(NaCl) = −ΔH ○ f (NaCl, s) + ΔH ○ sub(Na) + ΔH ○ IE (Na) + 1 2 BE(Cl Cl) + ΔH ○ EA(Cl)
ΔH ○ lat(NaCl) = − ( 411) + 108 + 496 + 1 2 (242) + ( 349) =+787 kJ mol 1
Notice that the final value is very close to that given in section 16 of the data booklet (+790 kJmol 1).
Practice problems for R1.2
Problem 1
Which equation corresponds to the bond enthalpy of the H Br bond?
A. HBr(g) → 1 2 H2(g) + 1 2 Br2(g)
B. HBr(g) → 1 2 H2(g) + 1 2 Br2(l)
C. HBr(g) → H+(g) + Br (g)
D. HBr(g) → H(g) + Br(g)
Example 6
Problem 2
Consider the following equations:
2Al(s) + 1.5O2(g) → Al2O3(s) ΔH ○ r = 1675 kJ mol 1
2Fe(s) + 1.5O2(g) → Fe2O3(s) ΔH ○ r = 825 kJ mol 1
What is the standard enthalpy change, in kJ mol 1, of the following reaction?
Fe2O3(s) + 2Al(s) → 2Fe(s) + Al2O3(s)
A. 850
B. +850
C. 2500
D. +2500
Problem 3
According to experimental data, each of the two oxygen–oxygen bonds in ozone, O3, has a bond order of 1.5. Using sections 11 and 12 of the data booklet, predict the lengths and BE values for these bonds.
Problem 4
a) Calculate the enthalpy change, in kJ mol 1, for the evaporation of water. Use section 13 of the data booklet.
b) Explain how the ΔH ○ c value obtained in part (a) of example 3 could be corrected using your answer to part (a) of this problem.
Problem 5
a) Formulate the equations that represent ΔH ○ c for Ch3OH(l) and ΔH ○ f for H2O(l).
b) The equations representing ΔH ○ values for CH3OH(l) and CO2(g) are given in the section “Standard enthalpy changes of formation and combustion” of this chapter. Demonstrate, by combining these equations with your answer to part (a), the validity of the expression ΔH ○ r = Σ (ΔH ○ f products) Σ (ΔH ○ f reactants) for the combustion of methanol.
Problem 6
Outline the physical and/or chemical processes associated with the standard enthalpies of atomization, ΔH ○ at , for the following elements at 298 K:
a) aluminium
b) nitrogen
c) bromine
Problem 7
a) Formulate an equation that represents the standard enthalpy change of formation for calcium oxide.
b) Construct a Born–Haber cycle for calcium oxide.
c) Calculate the lattice enthalpy for calcium oxide, in kJ mol 1, using sections 9 and 12 of the data booklet and the following information (“sub” refers to sublimation):
ΔH ○ (CaO, s) = 635 kJ mol 1
ΔH ○ sub(Ca) = +155 kJ mol 1
ΔH ○ IE (Ca) = +1146 kJ mol 1
1.3 Energy from fuels
You should know:
✔ reactive metals, non-metals and organic compounds undergo combustion reactions when heated in oxygen
✔ incomplete combustion of organic compounds, especially hydrocarbons, produces carbon monoxide and carbon
✔ fossil fuels include coal, crude oil and natural gas, which have different advantages and disadvantages
✔ biofuels are produced from the biological fixation of carbon over a short period of time through photosynthesis
✔ a fuel cell can be used to convert chemical energy from a fuel directly to electrical energy.
Combustion
reactions
You should be able to:
✔ deduce equations for reactions of combustion, including hydrocarbons and alcohols
✔ deduce equations for the incomplete combustion of hydrocarbons and alcohols
✔ evaluate the amount of carbon dioxide added to the atmosphere when different fuels burn
✔ understand the link between c arbon dioxide levels and the greenhouse effect
✔ understand the difference between renewable and non-renewable energy sources
✔ consider the advantages and disadvantages of biofuels
✔ deduce half-equations for the electrode reactions in a fuel cell.
Reactive metals and most non-metals react with excess oxygen gas to produce their most stable oxides; for example:
2Mg(s) + O2(g) → 2MgO(s)
C(s) + O2(g) → CO2(g) (complete combustion)
If oxygen is the limiting reactant, incomplete combustion might occur:
C(s) + 1 2 O2(g) → CO(g) (incomplete combustion)
Combustion of hydrocarbons and many other organic compounds in excess oxygen produces carbon dioxide and water:
CH4(g) + 2O2(g) → CO2(g) + 2H2O(l)
CH3CH2OH(l) + 3O2(g) → 2CO2(g) + 3H2O(l)
Depending on the reaction conditions, water can be produced as a liquid or gas (steam).
Organic fuels must have a high activation energy of combustion in order to prevent accidental fires and to increase the efficiency of internal combustion engines (Reactivity 2.2).
Incomplete combustion of organic compounds produces elemental carbon (soot) and/or carbon monoxide, depending on the reactant ratio:
CH4(g) + 1.5O2(g) → CO(g) + 2H2O(l)
CH4(g) + O2(g) → C(s) + 2H2O(l)
The soot produced by the incomplete combustion of methane in a laboratory gas burner can form a dark powdery deposit on a crucible or other laboratory glassware, which increases the mass of the container and potentially introduces a systematic error (Tool 1.2).
Nearly all combustion reactions are exothermic, so their enthalpy changes are negative (Reactivity 1.1). If you obtain a positive ΔH ○ c value, doublecheck your calculations for possible errors.
In combustion reactions, oxygen is the oxidizing agent while the combusted substance is the reducing agent (Reactivity 3.2).
Nature of science
Carbon monoxide is a poisonous gas that irreversibly binds to hemoglobin in the red blood cells, reducing the oxygencarrying capacity of the blood.
Soot particles can cause irritation of skin, eyes and lungs, aggravate chronic respiratory conditions and increase the risk of cancer.