Chemistry: Thermochemistry: Enthalpy and Hess's Law Worksheet

Question 1

The combustion of methane is represented by CH4(g) + 2O2(g) -> CO2(g) + 2H2O(l), delta H = -890 kJ per mole of CH4. How much heat is released when 2.50 mol of CH4 burns completely?

Accepted Answer

When 2.50 mol of CH4 burns, the enthalpy change is 2.50 x -890 kJ = -2225 kJ. This means 2225 kJ of heat is released.

Question 2

A reaction warms 150.0 g of water from 22.0°C to 31.5°C. Use q = mc delta T and c = 4.184 J/g°C to find the heat gained by the water. Then state the heat change for the reaction.

Accepted Answer

The temperature change is 9.5°C, so q = 150.0 g x 4.184 J/g°C x 9.5°C = 5962 J, or 5.96 kJ. The water gains 5.96 kJ, so the reaction loses 5.96 kJ and has a heat change of -5.96 kJ.

Question 3

When ammonium nitrate dissolves in water, the temperature of the solution decreases. State whether the process is endothermic or exothermic and explain the sign of delta H.

Accepted Answer

The process is endothermic because the dissolving ammonium nitrate absorbs heat from the water and surroundings. The delta H value is positive.

Question 4

Use the data to find delta H for CO(g) + 1/2O2(g) -> CO2(g). Given: C(s) + O2(g) -> CO2(g), delta H = -393.5 kJ. C(s) + 1/2O2(g) -> CO(g), delta H = -110.5 kJ.

Accepted Answer

Reverse the CO formation reaction and add it to the CO2 formation reaction. The enthalpy change is -393.5 kJ + 110.5 kJ = -283.0 kJ, so delta H for the reaction is -283.0 kJ.

Question 5

Use standard enthalpies of formation to calculate delta H for C2H5OH(l) + 3O2(g) -> 2CO2(g) + 3H2O(l). Values: C2H5OH(l) = -277.7 kJ/mol, CO2(g) = -393.5 kJ/mol, H2O(l) = -285.8 kJ/mol, O2(g) = 0 kJ/mol.

Accepted Answer

The products have enthalpy 2(-393.5) + 3(-285.8) = -1644.4 kJ. The reactants have enthalpy -277.7 kJ. Delta H = -1644.4 - (-277.7) = -1366.7 kJ, so the reaction is exothermic.

Question 6

For N2(g) + 3H2(g) -> 2NH3(g), delta H = -92.4 kJ. What is delta H for 4NH3(g) -> 2N2(g) + 6H2(g)?

Accepted Answer

The target reaction is the reverse of the original reaction multiplied by 2. Reversing makes delta H positive, and multiplying by 2 gives delta H = +184.8 kJ.

Question 7

Use Hess's Law to find delta H for 2SO2(g) + O2(g) -> 2SO3(g). Given: 2S(s) + 3O2(g) -> 2SO3(g), delta H = -790.4 kJ. S(s) + O2(g) -> SO2(g), delta H = -296.8 kJ.

Accepted Answer

Double the second equation to get 2S(s) + 2O2(g) -> 2SO2(g), delta H = -593.6 kJ. Reverse it and add it to the first equation. Delta H = -790.4 kJ + 593.6 kJ = -196.8 kJ.

Question 8

Estimate delta H for H2(g) + Cl2(g) -> 2HCl(g) using bond enthalpies. H-H = 436 kJ/mol, Cl-Cl = 243 kJ/mol, H-Cl = 431 kJ/mol.

Accepted Answer

Bonds broken require 436 + 243 = 679 kJ. Bonds formed release 2(431) = 862 kJ. Delta H = bonds broken - bonds formed = 679 - 862 = -183 kJ, so the reaction is exothermic.

Question 9

In a coffee-cup calorimeter, 50.0 mL of 1.00 M HCl reacts with 50.0 mL of 1.00 M NaOH. The temperature rises from 22.0°C to 28.7°C. Assume the solution has a mass of 100.0 g and c = 4.184 J/g°C. Calculate delta H in kJ per mole of water formed.

Accepted Answer

The solution gains q = 100.0 g x 4.184 J/g°C x 6.7°C = 2803 J, or 2.80 kJ. The reaction releases 2.80 kJ. Since 0.0500 mol of water forms, delta H = -2.80 kJ / 0.0500 mol = -56.1 kJ/mol of water.

Question 10

Explain why the state symbols in a thermochemical equation matter. Use H2O(l) and H2O(g) as part of your explanation.

Accepted Answer

State symbols matter because different physical states have different enthalpies. H2O(l) and H2O(g) do not have the same energy, so a reaction that forms liquid water has a different delta H than one that forms water vapor.

Question 11

Use standard enthalpies of formation to calculate delta H for CaCO3(s) -> CaO(s) + CO2(g). Values: CaCO3(s) = -1207.0 kJ/mol, CaO(s) = -635.1 kJ/mol, CO2(g) = -393.5 kJ/mol.

Accepted Answer

The products have enthalpy -635.1 + -393.5 = -1028.6 kJ. The reactant has enthalpy -1207.0 kJ. Delta H = -1028.6 - (-1207.0) = +178.4 kJ, so the decomposition is endothermic.

Question 12

Use Hess's Law to calculate delta H for N2(g) + 2O2(g) -> 2NO2(g). Given: N2(g) + O2(g) -> 2NO(g), delta H = +180.5 kJ. 2NO(g) + O2(g) -> 2NO2(g), delta H = -114.1 kJ.

Accepted Answer

Add the two given equations. The 2NO cancels because it is produced in the first equation and used in the second equation. Delta H = +180.5 kJ + -114.1 kJ = +66.4 kJ.

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Chemistry: Thermochemistry: Enthalpy and Hess's Law

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