AP Chemistry Lab Investigation #12: _Determining the Enthalpy of Chemical Substances Dissolved in Wa

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Determining the Enthalpy of Chemical Substances Dissolved in Water Abstract: The purpose of this lab is to learn how to determine the enthalpy change of different chemical substances dissolved in water stirred by the magnetic stirrer and ultimately to find the most appropriate chemical substance for hand warmer. This was accomplished by measuring the temperature change of dilutions, substituting the given data (mass, ΔT, specific heat capacity) into the mathematical equation: q=mcΔT. The enthalpy change for each chemical was then calculated by comparing the flow of heat (q) for each chemical.

BY: AIDEN MOON, GLORIA YOO, JAMES LEE

AP Chemistry Lab Investigation #12: Heat enthalpy 2017-09-29

By Aiden Moon, Gloria Yoo, James Lee

TABLE OF CONTENTS

Materials

2

Procedure

3

Pre-Lab Questions

3

Data Collection and Computation

5

Argumentation and Documentation

7

Post-Lab Questions

8

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AP Chemistry Lab Investigation #12: Heat enthalpy 2017-09-29

By Aiden Moon, Gloria Yoo, James Lee

MATERIALS: -

Thermometer

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Balance

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Scoops

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100mL graduated cylinder

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150mL beaker

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150mL Styrofoam calorimeter with Styrofoam cover (see Figure 1)

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Weighing boat

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Magnetic stirrer (see Figure 1) and stir bar

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Anhydrous calcium chloride (CaCl2)*

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Anhydrous sodium carbonate (N2CO3)

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Anhydrous sodium acetate (NaC2H3O2)

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Lithium chloride (LiCl)*

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Magnesium sulfate (MgSO4)*

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Sodium chloride (NaCl)

(*Among the seven chemicals, we’ve used three of them for the experiment, which are boldly highlighted: anhydrous calcium chloride, lithium chloride, and magnesium sulfate.)

Figure 1: Styrofoam calorimeter on the magnetic stirrer.

PROCEDURES: 2

AP Chemistry Lab Investigation #12: Heat enthalpy 2017-09-29

By Aiden Moon, Gloria Yoo, James Lee

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Part 1 (calorimetry practice):

We made a series of solutions by putting 5 grams of each chemical substance (in our case: MgSO4, LiCl, and CaCl2) into 100mL of distilled water (about 24-25 ℃) within a Styrofoam calorimeter. The Styrofoam calorimeter contained one magnetic bar inside and it sat on the magnetic stirrer (see Figure 1). Then, we turned the magnetic stirrer on, covered the lid, and fixed the thermometer with the container, observing the temperature change until the temperature stopped changing. We repeated this experiment twice for each chemical substance. Finally, given the temperature change, we calculated the flow of heat (q=mc ΔT) and then calculated the enthalpy change per mole (ΔH kJ/mol) by dividing the calculated value of q by molar mass of each chemical substance.

Part 2 (calorimeter calibration procedure): We first measured 50mL of distilled water at room temperature (about 24-25℃) in a graduated cylinder. Then, we prepared for hot water that was approximately 60-70℃ and poured into the same graduated cylinder that contained cold water little by little. We did this until the temperature reached 50℃. When the desired temperature was approached, we measured the mixed water to be 100mL and then poured it into the Styrofoam calorimeter, observing the temperature change until it stopped changing. Given the temperature change, we calculated the q of cold water and hot water so that we could calculate the q of calorimeter by using the equation: qhot=-(qcold+qcal).

PRE-LAB QUESTIONS: 1. Describe the changes you observe in the animation, including changes in the bonds and particulate attractions and changes in the amount of disorder in the system.

Through the animation, we could observe that the salt, NaCl, was separated by many water molecules. Since Na+ itself is a cation, or positively charged, the partially negative charged oxygens of the water molecules gets attached to Na +. Since the oxygen molecules are just partially negative, not completely, it required about five water molecules to separate Na+ from NaCl salt. Within the same context, since Cl - is an anion, or negatively charged, the partially positive charged hydrogen side was attached to Cl -.

As the particles were separated by the water molecules, the attractions between Na + and Cl- called ‘ionic bond’ was changed to another attraction called ‘ion-dipole interaction’.

At the beginning, the water was bonded by the attraction called ‘hydrogen bond’, but we could see that some of the O-H bonds were broken in order to separate NaCl throughout the process. As the result of this process, the disorder, or entropy, increased because of the separation of highly ordered molecule.

2. When sodium chloride is dissolved in water, the temperature of the resulting solution is lower than the temperature of the water before the salt dissolves. How can this result be explained based on bond breaking and bond making that is occurring?

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AP Chemistry Lab Investigation #12: Heat enthalpy 2017-09-29

By Aiden Moon, Gloria Yoo, James Lee

Breaking of bonds absorbs heat, an endothermic process, and the formation of bonds releases heat, an exothermic process. In this case, the final temperature was lower possibly because the heat absorbed by the surroundings to break the ionic bond of NaCl crystal was stronger than the heat released to the surrounding when the ions and water molecules formed new attractions between them.

3. Why do some salts, such as sodium chloride, dissolve spontaneously even though the process is endothermic overall? ΔG = ΔH - T ΔS When ΔG<0, the reaction is observed to be spontaneous. When ΔH >0, the reaction is observed to be endothermic. According to the above formula, T ΔS should be bigger than ΔH in order to make the reaction spontaneous when ΔH >0. We know that the change in temperature is very small when the ionic salt dissolves in water, thus we can predict the change in disorder to be greater, ΔS>0. In conclusion, if a reaction is spontaneous even though the change in temperature is small, this is because the change in disorder is high, that it makes TΔS bigger than ΔH. 4. When some ionic salts are dissolved in water the temperature of the resulting solution is higher than the temperature of the water before the salt dissolves. What do you think determines whether the resulting solution is cooler or warmer than the starting water?

The type of ionic salt determines the final temperature of the resulting solution.

According to the mathematical formula of the change in heat, which is Q= mc*ΔT, change in heat is directly proportional to ‘c’, the specific heat capacity. Since specific heat capacity is different for each type of ionic salt and molecules, we can conclude that the type of ionic salt would determine the final temperature of the resulting solution.

DATA COLLECTION AND COMPUTATION:

Substance LiCl CaCl2 MgSO4

2012 Cost per 500g ($) 32.75 6.55 70

*All solutions had 100.0mL of water and 5.00gram of certain solution

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AP Chemistry Lab Investigation #12: Heat enthalpy 2017-09-29

By Aiden Moon, Gloria Yoo, James Lee Table 1: Cost of chemical substances chosen.

Chemical LiCl CaCl2 MgSO4

Before (1st trial) 24℃ 24℃ 24℃

After (1st trial) 32℃ 28℃ 27℃

Before (2nd trial) 24℃ 24℃ 24℃

After (2nd trial) 31℃ 28℃ 26℃

*All solutions had 100.0mL of water and 5.00gram of certain solution

Table 2: Temperature change observed for each trial.

1.

a) Calculate the enthalpy change of the cold water using the equation qhot=mhotc∆thot. Assume that the density of water is exactly 1g/mL. Is this an endothermic or exothermic process? Explain. q=mcΔT qcold=100mL⋅1g/mL ⋅4.18J/℃º*g ⋅(36℃-24℃)  qcold= 5016J The reaction is endothermic as the cold water absorbs heat from the hot water.

b) Calculate the enthalpy change of the hot water using the equation qcold=mcoldc∆tcold. Assume that the density of water is exactly 1g/mL. Is this an endothermic or exothermic process? Explain. q=mcΔT qhot=100mL ⋅1g/mL ⋅ (4.18J/℃º*g) ⋅ (36-49)  qhot=-5434J The reaction is exothermic as the hot water loses its heat to the cold water.

c) These amounts are not equal because the calorimeter (the coffee cups) absorbs some of the thermal energy transferred by the hot water. Thus under the real conditions observed in the laboratory the law of conservation of energy equation becomes qhot= -(qcold+qcal), where qcal is the enthalpy change of the calorimeter. Use this equation to calculate the enthalpy change of the calorimeter. qhot=-(qcold+qcal) -5434J=-(5016J+qcal)  -5434J=-5016J-qcal  qcal=418J The calorimeter absorbs about 418J of heat from the water solution.

d) The calorimeter constant, C, is the heat absorbed by the calorimeter per degree of temperature change, C=qcal/∆tcal. Assuming the starting of the temperature of the calorimeter is the same as that of the cold water, calculate the calorimeter constant in units of joules per degree Celsius.

C=qcal/ΔTcal ΔTcal=36℃-24℃=12℃ C=418J/12℃  C=34.8J/℃ (3s.f.)

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AP Chemistry Lab Investigation #12: Heat enthalpy 2017-09-29

By Aiden Moon, Gloria Yoo, James Lee 2.

MgSO4 qrxn=105g(4.18J/℃º*g) ⋅ (26-24)  qrxn=877.8J qsoln=-(877.8J+34.8J/℃*12℃)  qsoln=-1295.4J CaCl2 qrxn=105g ⋅(4.18 J/℃º*g) ⋅ (27-24)  qrxn=1316.7J qsoln=-(1316.7J+34.8J/℃*12)  qsoln=-1734.3J LiCl qrxn=105g((4.18 J/℃º*g)(32-24)  qrxn=3511.2J qsoln=-(3511.2J+34.8J/℃*12)  qsoln=-3928.8J

3.

By convention, scientists report enthalpy changes for dissolution (and many other processes) in units of kilojoules per mole of solute dissolved. Using your values of qsoln, calculate the enthalpy in units of kilojoules per mole. This quantity has the symbol ∆Hsoln. Calculate ∆Hsoln for each of the three solids you tested.

MgSO4 qsoln=-1295.4J Number of moles in MgSO4: 5g⋅1mol/120.366g= 0.0415mol ΔHsoln=(-1295.4J*1KJ/1000J)/0.0415mol=-31.20KJ/mol ΔHsoln=31.73KJ/mol of MgSO4 CaCl2 5g⋅1mol/110.98g=0.04505mol ΔHsoln=(-1734.3J *1KJ/1000J)/0.04505mol=-38.49KJ/mol ΔHsoln=38.49KJ/mol of CaCl2 LiCl 5g ⋅1mol/42.394g=0.1179mol ΔHsoln=(-3928.8J*1KJ/1000J)/0.1179mol=-33.32KJ/mol ΔHsoln=33.32KJ /mol of LiCl

4.

Based on the cost information provided, and your experimental work and calculations, select which chemical you believe will make the most cost-effective hand warmer. The hand warmer you are designing needs to increase in temperature by 20°C. Calculate the amount of the compound you selected that would be required for a hand warmer that meets this requirement.

Among the three chemicals that we used, CaCl2 was thought to be the most effective and economic chemical to use as a source of hand warmer. CaCl2 is the cheapest chemical (see Table 1) among the three and releases about -39 kJ/mol of heat, which is the highest among the three chemicals.

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AP Chemistry Lab Investigation #12: Heat enthalpy 2017-09-29

By Aiden Moon, Gloria Yoo, James Lee qrxn = mc∆T qrxn=50(4.18J/℃º*g)(20℃)=4180J qrxn=4180J qsoln= -(qrxn + C∆T) C∆T= 34.8J/°C⋅(4°C) = 139.2J qsoln= -(4180J + 139.2J) qsoln= -4.32 kJ ΔHsoln=-38.49 kJ/mol of CaCl2 qsoln/∆Hsoln= number of mol of CaCl2 -4.32 kJ/-38.49 kJ/mol of CaCl2= number of mol of CaCl2 0.112 mol of CaCl2⋅110.984g/mol of CaCl2= 12.46 g of CaCl2

ARGUMENTATION AND DOCUMENTATION: Out of all the substances that we have tested in the experiment, we came up with a conclusion that CaCl2 is the most appropriate chemical that qualifies all aspects -safety, cost, and heatto make a hand warmer. This is because CaCl 2 released the most amount of energy, at same conditions, compared with other substances. In other words, it had the highest Enthalpy (kJ/mol) of an average of -38.49; while other chemicals such as LiCl had -33.32 and also MgSO 4 with -31.73. CaCl2 is not only suitable to make an effective hand warmer for its amount of heat released per mole, but also for its cheap price. CaCl2 costs $6.55 per 500 grams which is relatively cheaper than other substances like LiCl –the second highest enthalpy- which costs $32.75 per 500 grams (see Table 2).

According to the Material Safety Data Sheets (MSDS), CaCl 2 is non-flammable, but it has potential effects to harm human health in case of skin contact, eye contact, ingestion, and inhalation in extreme rare cases. Therefore, the precautions might be to keep locked up, do not ingest, breathe dust, avoid contact with eye and skin, and keep away from incompatibles such as moisture. Overall, the data values of enthalpies were all measured based on the experimented results, and the safety and environmental impact were based on MSDS. All dilutions, including those of other classmates, were made with same volume of Styrofoam calorimeters. The class-average value of heat enthalpy of CaCl 2 turns out to be about -40.23 kJ/mol, which is not numerically distant from our data: -38.49 kJ/mol. However, this value seems far from the published value, which is -81.3 kJ/mol, a percentage error of about 50.5% error.

The equipment and methods utilized for this experiment were generally same for whole students.

Therefore, neither the amounts of CaCl 2 nor the indicated heat enthalpy of CaCl 2 was not appropriate for this experiment. This would not, however, change our conclusion since we know that the actual heat enthalpy for CaCl2 is over -80 kJ/mol, which is found to be more effective than we expected.

POST-LAB ASSESSMENT:

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AP Chemistry Lab Investigation #12: Heat enthalpy 2017-09-29

By Aiden Moon, Gloria Yoo, James Lee

1. Are the dissolving processes you carried out endothermic or exothermic or neither? Explain your thinking. i. Lithium chloride (LiCl) Exothermic. The change in enthalpy is negative, which means the reaction releases the heat. ii. Magnesium sulfate (MgSO4) Exothermic. The change in enthalpy is negative, which means the reaction releases the heat. iii. Calcium chloride (CaCl2) Exothermic. The change in enthalpy is negative, which means the reaction releases the heat.

2. Dissolving ionic compounds involves the separation of the solid ionic compound into cations and anions in water. This process can be represented by an equation showing the solid as a reactant and the aqueous ions as products. The heat of reaction ∆H soln is written after the products, typically in units of kJ/mol.

Example: sodium hydroxide dissolves exothermically, releasing 44.2 kilojoules per mole dissolved. This process is represented as NaOH(s) + Na + (aq) + OH- (aq), ∆Hsoln =-44.2 kJ/mol.

Write an equation to represent the dissolving process for each salt you studied. Include your calculated heat of reaction as in the example. i. LiClLi++Clii. MgSO4Mg2++SO42iii. CaCl2Ca2++2Cl-

∆Hsoln = -27.9 kJ/mol ∆Hsoln = -31.73 kJ/mol ∆Hsoln = -39 kJ/mol

3. Changes in matter are generally classified as physical or chemical based on whether new substances are formed through the process. Does dissolving represent a physical change, a chemical change, or an intermediate change? Explain your reasoning, including evidence from the animation you viewed.

For instance, when you dissolve a lithium chloride crystal in water, the lithium chloride dissociates in Li+ ions and Cl- ions, which may be written as: LiCl (s)→ Li+(aq)+Cl-(aq) Since such chemical equation can be derived from the reaction, dissolution should be represented as a chemical change, changing the chemical identity of solution. We can barely see the ionic compounds broken into separate ions from the animation, which indicates a chemical change.

4. Share your calculated values of ∆Hsoln with your classmates and obtain their values. a) Determine the class average value and standard deviation* for each solid.

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AP Chemistry Lab Investigation #12: Heat enthalpy 2017-09-29

By Aiden Moon, Gloria Yoo, James Lee 2

SE (σ) * = √Σ(xi-xmv) /n n= the number of data plot xi= each of the values of data xmv= the mean of the xi values Σ= sum of the values i. Lithium chloride (LiCl): Millie & Min Su Edward, Paul, Cindy Dana, Anthony, Stan Evan, Julia Her, Julia Kim Aiden, James, Gloria

LiCl LiCl LiCl LiCl LiCl

-22.3 kJ/mol -27 kJ/mol -29.8 kJ/mol -27.92 kJ/mol -27.9 kJ/mol

MgSO4 MgSO4

-26.44 kJ/mol -31.73 kJ/mol

CaCl2 CaCl2 CaCl2 CaCl2

-41.4 kJ/mol -41.51 kJ/mol -39.01 kJ/mol -39 kJ/mol

Average: -26.98 kJ/mol Standard deviation: 2.514 ii. Magnesium sulfate (MgSO4): Evan, Julia Huh, Julia Kim Aiden, James, Gloria Average: -29.09 kJ/mol Standard deviation: 2.645 iii. Calcium chloride (CaCl2): Edward, Paul, Cindy Dana, Anthony, Stan Evan, Julia Huh, Julia Kim Aiden, James, Gloria Average: -40.23 kJ/mol Standard deviation: 1.226 b) Find the published value of ∆Hsoln for each solid and determine the percent error in the class average.

Lithium chloride (LiCl) ((-37 kJ/mol) –(-26.98 kJ/mol) /(37 kJ/mol))⋅100=27.1% error Na2CO3 (Sodium Carbonate) ((-26.7 kJ/mol)–(-29.08 kJ/mol) /(-26.7 kJ/mol))⋅100=8.9% error CaCl2 (Calcium Chloride) ((-81.3 kJ/mol )–(-40.23 kJ/mol) /(-81.3 kJ/mol ))⋅100=50.5% error

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AP Chemistry Lab Investigation #12: Heat enthalpy 2017-09-29

By Aiden Moon, Gloria Yoo, James Lee

5. What possible sources of error could affect the accuracy of your calculated value of the amount of solid in your hand warmer? List at least two and what effect they would have on the temperature change. i. There could have been temperature change after measuring the temperature of the hot water due to the lower room temperature. Moreover, the temperature change of hot water could have been resulted from the speed of pouring it into the Styrofoam calorimeter. As a result, the experimental temperature change might have been slightly higher than the theoretical temperature change. ii. The magnetic stirrer could have absorbed the heat released from the chemical reactions. If the magnetic stirrer did absorb heat, the experimental temperature change would be smaller than the theoretical temperature change, which would have made the calculated enthalpy of exothermic reactions be higher. iii. Perhaps we might not have put all the chemicals into the water from the weighing boat. In addition, some of the amounts of chemicals were remained attached to the paper towel and not put into the water appropriately. This would have reduced the temperature change, which would have made the calculated enthalpy of exothermic reactions higher. iv. We didn’t measure the water temperature to its specific decimal points, which might have made the calculation inaccurate. v. The water might not have been exactly 100.0mL due to parallax error. If there was more water, it would have made the calculated enthalpy of exothermic reactions higher, and vice versa.

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