THERMOCHEMISTRY WORKSHEET KEY

1. The following describes the reaction that takes place when a typical fat, glyceryl trioleate, is metabolized by the body:

2C57H104O6(s) + 160O2(g ) → 114CO2(g ) + 104H2O(l ) DH° = -6.70 x 104 kJa. How much heat energy must be expelled by the body to rid the body of one pound of fat?

4 453.6 g 1 mol fat 6.70 x 10 kJ? kJ = 1 lb fat = 1 lb 885.458 g fat 2 mol fat

−

41.72 x 10 kJ

b. If a swimmer burns off 2000 kJ for each 100 laps she swims, how much fat will she burn off (in pounds) if she swims 20 laps. (Assume all of the energy comes from the above reaction.)

4

885.458 g fat2000 kJ 2 mol fat 1 lb? lb fat = 20 laps = 100 laps 6.70 x 10 kJ 1 mol fat 453.6 g

− −

0.0233 lb

2. 0.100 g of H2 and an excess of O2 are compressed into a bomb calorimeter containing 1200 g of water. The temperature before the reaction is 25.00 °C, and after the reaction it goes to 27.16 °C. The heat capacity of the calorimeter is 72.5 J/K. Calculate the heat of combustion of hydrogen gas.

− ∆ − ° − ° ° °

v cal w0.00418 kJ 0.0725 kJ 0.00418 kJq = C + m T = + 1200 g 2.16 C = 11.0 kJ

g C C g C

2

2 2 2 2

2.01594 g H? kJ 11.0 kJ kJE = = = 222 mol H 0.100 g H mol H mol H

−∆ ° −

DH° = DE° + (Dn)RT H2(g ) + ½O2(g ) → H2O(l )

0.008314 kJH = 222 kJ + (0 1.5) mol 298.15 K = K mol

∆ ° − − 2226 kJ/mol H−

3. The heat of combustion of liquid cyclohexane, C6H12(l ), is -3924 kJ/mole. 8.25 g of cyclohexane is placed in the bomb of a bomb calorimeter with excess oxygen. The calorimeter contains 825.0 g of water. When the mixture is ignited, the temperature increases from 18.2 °C to 25.6 °C. Calculate the heat capacity of the calorimeter.

C6H12(l ) + 9O2(g ) → 6CO2(g ) + 6H2O(l )

( )6 12

0.008314 kJ kJE = H n RT = 3924 kJ (6 9) mol 298.15 K = 3917 K mol mol C H

∆ ° ∆ ° − ∆ − − − −

6 12v 6 12

6 12 6 12

1 mol C H 3917 kJq = ? kJ = 8.25 g C H = 384 kJ84.162 g C H 1 mol C H

−−

− ∆ °

v cal w0.00418 kJq = C + m T

g C

( )−

− − − −∆ ° − ° °

vcal w

q 0.00418 kJ 384 kJ 0.00418 kJC = m = 825.0 g = T g C 25.6 18.2 C g C

48 kJ/ C

355

356 Chapter 6 Worksheet Keys

CHEMISTRY 151 - DH AND DE WORKSHEET KEY

1. Nitrogen dioxide gas reacts with liquid water to yield liquid nitric acid and nitrogen monoxide gas. When one mole of nitrogen dioxide reacts at constant pressure, 23.84 kJ of heat are evolved. What are the molar DH° and DE° for the reaction?

NO2(g ) + H2O(l ) → HNO3(l ) + NO(g )

DH° = -23.84 kJ/mole

( )kJ 2 kJE° = H° ( n) RT= 23.84 0.008314 298.15 K = mol 3 K mol

∆ ∆ − ∆ − − − −

22.19 kJ/mol

2. When 3.500 g of the gas butane, C4H10, is burned in a bomb calorimeter at 25 °C, 85.99 kJ of heat are evolved. Calculate the molar DH° and DE° for this reaction.

C4H10(g ) + O2(g ) → 4CO2(g ) + 5H2O(l )

−∆ −

4 10

4 10 4 10

58.124 g C H85.99 kJE° = = 3.500 g C H 1 mol C H

1428 kJ/mol

( )kJ 15 kJH° = E° + ( n) RT= 1428 + 4 0.0083144 298.15 K = mol 2 K mol

∆ ∆ ∆ − − −

1437 kJ/mol

357

CHEMISTRY 151 CALORIMETER WORKSHEET

1. In the last step of the copper experiment you reacted metallic zinc with a water solution of copper sulfate to form copper and water-soluble zinc sulfate. The following two step calculation will allow you to calculate the reaction’s heat of reaction, DH°

a. The heat capacity of the bomb calorimeter in which the reaction is to be run is determined by running a reaction with a known heat of reaction. The enthalpy of combustion of benzoic acid, C6H5COOH(s), is commonly used as the standard. The heat of combustion of benzoic acid is -322.7 kJ/mol. When 8.174 g of benzoic acid is burned, the temperature in a bomb calorimeter containing 200.0 g of water rises from 62.64 °F to 96.30 °F. What is the heat capacity of the bomb calorimeter in kJ/°C?

C6H5COOH(s) + 15 2 O2(g ) → 7CO2(g ) + 3H2O(l )

( ) 0.008314 kJ kJE = H n RT = 322.7 kJ (7 7.5) mol 298.15 K = 321.5 K mol mol

∆ ° ∆ ° − ∆ − − − −

6 5v 6 5

6 5 6 5

1 mol C H COOH 321.5 kJq = ? kJ = 8.174 g C H COOH = 21.52 kJ122.124 g C H COOH 1 mol C H COOH

−−

v cal w0.00418 kJq = C + m T

g C

− ∆ °

( )v

cal wq 0.00418 kJ 21.52 kJ 0.00418 kJC = m = 200.0 g = T g C 35.72 17.02 C g C

−− − − − °

∆ ° − ° °0.315 kJ/ C

b. The reaction between the zinc and copper sulfate is then run in the same calorimeter with 210.0 g of water. 0.534 g of Zn(s) is reacted with excess CuSO4(aq). The temperature rises from 22.33 °C to 27.69 °C. What is the DH° for this reaction?

Zn(s) + CuSO4(aq) → Cu(s) + ZnSO4(aq)

( )v cal w0.00418 kJ 0..315 kJ 0.00418 kJq = C + m T = + 210.0 g 27.69 22.33 C

g C C g C = 6.39 kJ

− ∆ − − ° ° ° ° −

65.37 g Zn? kJ 6.39 kJ kJE = = = 782 mol Zn 0.534 g Zn 1 mol Zn mol Zn

− ∆ ° −

( ) 0.008314 kJH = E + n RT = 782 kJ + (0 0) mol 298.15 K = K mol

∆ ° ∆ ° ∆ − − − 782 kJ/mol

358 Chapter 6 Worksheet Keys

LAW OF HESS KEY

1. Given the following, calculate the heat of formation of CuO(s).

2Cu2O(s) + O2(g ) → 4CuO(s) DH° = -238 kJCu2O(s) → CuO(s) + Cu(s) DH° = 11.3 kJ

CuO(s) + Cu(s) → Cu2O(s) DH°1 = -(11.3 kJ)Cu2O(s) + ½O2(g) → 2CuO(s) DH°2 = ½(-238 kJ)____________________________________________________________

Cu(s) + ½O2(g ) → CuO(s) DH°f = DH°1 + DH°2

DH°f = -11.3 kJ + (-119 kJ) = -130 kJ

2. Given the following; Fe2O3(s) + 3CO(g ) → 2Fe(s) + 3CO2(g ) DH° = -28 kJ3Fe2O3(s) + CO(g ) → 2Fe3O4(s) + CO2(g ) DH° = -59 kJFe3O4(s) + CO(g) → 3FeO(s) + CO2(g) DH° = 39 kJ

Calculate the DH° for;

FeO(s) + CO(g ) → Fe(s) + CO2(g )

FeO(s) + CO2(g ) → Fe3O4(s) + CO(g ) DH°1 = - (39 kJ)½Fe2O3(s) + 3 2 CO(g ) → Fe(s) + 3 2 CO2(g ) DH°2 = ½(-28 kJ)

Fe3O4(s) + CO2(g ) → ½Fe2O3(s) + CO(g ) DH°3 = - (-59 kJ)_____________________________________________________________________

FeO(s) + CO(g ) → Fe(s) + CO2(g ) DH°rxn = DH°1 + DH°2 + DH°3 = -17 kJ

359

CHEMISTRY 151 - LAW OF HESS KEY

1. From the following,

C(graphite) + O2(g ) → CO2(g ) DH° = -393.5 kJ2H2O(l ) → 2H2(g ) + O2(g ) DH° = 571.6 kJ2C2H6(g ) + 7O2(g ) → 4CO2(g ) + 6H2O(l ) DH° = -3119.6 kJcalculate the enthalpy change for the following reaction:

2C(graphite) + 3H2(g ) → C2H6(g )2C(graphite) + 2O2(g ) → 2CO2(g ) DH°1 = 2(-393.5 kJ)3H2(g ) + 3 2 O2(g) → 3H2O(l ) DH°2 = - 3 2 (571.6 kJ)2CO2(g ) + 3H2O(l ) → C2H6(g ) + 7 2 O2(g ) DH°3 = -½(-3119.6 kJ)________________________________________________

2C(graphite) + 3H2(g ) → C2H6(g ) DH°rxn = DH°1 + DH°2 + DH°3 = -84.6 kJ

2. Calculate the standard enthalpy change for the reaction of nitrogen dioxide and water, 3NO2(g ) + H2O(l ) → 2HNO3(aq) + NO(g ) given that

2NO(g ) + O2(g ) → 2NO2(g ) DH° = -173kJ2N2(g ) + 5O2(g ) + 2H2O(l ) → 4HNO3(aq) DH° = -255 kJN2(g ) + O2(g ) → 2NO(g ) DH° = 181 kJ

3NO2(g ) → 3NO(g ) + 3 2 O2(g ) DH°1 = - 3 2 (-173kJ)N2(g ) + 5 2 O2(g ) + H2O(l ) → 2HNO3(aq) DH°2 = ½(-255 kJ)2NO(g ) → N2(g ) + O2(g ) DH°3 = -(181 kJ)________________________________________________

3NO2(g ) + H2O(l ) → 2HNO3(aq) + NO(g )DH°rxn = DH°1 + DH°2 + DH°3 = -49 kJ

3. The standard heat of combustion of S8(s) to SO3(g ) is -3166 kJ/mol S8. The standard enthalpy or heat of combustion of S8(s) to SO2(g ) is -2374 kJ/mol S8. Use this information and the Law of Hess to calculate the standard heat of reaction for the following reaction.

2SO2(g ) + O2(g ) → 2SO3(g )Given:

S8(s) + 12O2(g ) → 8SO3(g ) DH° = -3166 kJS8(s) + 8O2(g ) → 8SO2(g ) DH° = -2374 kJ

2SO2(g ) → S8(s) + 2O2(g ) DH°1 = - (-2374 kJ)S8(s) + 3O2(g ) → 2SO3(g ) DH°2 = (-3166 kJ)

________________________________

2SO2(g ) + O2(g ) → 2SO3(g )DH°rxn = DH°1 + DH°2 = -198.0 kJ

360 Chapter 6 Worksheet Keys

CHEMISTRY 151 - HEAT OF FORMATION KEY

1. Calculate the molar heat of combustion of acetylene gas, C2H2(g ).

C2H2(g ) + 5 2 O2(g ) → 2CO2(g ) + H2O(l )

DH°comb = 2 × DH°f CO2(g ) + DH°f H2O(l ) - DH°f C2H2(g )

= 2(-393.5 kJ) + (-285.8 kJ) - 226.6 kJ = -1299.4 kJ

2. Many cigarette lighters contain butane, C4H10(l ), for which the heat of formation is 127 kJ/mol. Calculate the heat evolved when 20.0 g of liquid butane are completely burned.

C4H10(g ) + O2(g ) → 4CO2(g ) + 5H2O(l )

DH°comb = 4 × DH°f CO2(g ) + 5 × DH°f H2O(l ) - DH°f C4H2(g )

= 4(-393.5 kJ) + 5(-285.8 kJ) - (-127 kJ) = -2876 kJ

4 104 10

4 10 4 10

1 mol C H 2876 kJ? kJ = 20.0 g C H = 58.124 g C H 1 mol C H

−

29.90 x 10 kJ−

3. The heat of combustion of hydrogen sulfide gas is -562 kJ, the heat of combustion of sulfur in the rhombic form, S8(rhombic), is -2375 kJ, and the heat of formation of liquid water is -285.8 kJ. Using only data from this problem, determine the DH° for the following reaction.

H2(g ) + 1/8S8(rhombic) → H2S(g )

H2S(g ) + 3 2 O2(g ) → H2O(l ) + SO2(g ) DH°comb = -562 kJ

S8(rhombic) + 8O2(g ) → 8SO2(g ) DH°comb = -2375 kJ

H2(g ) + ½O2(g ) → H2O(l ) DH°1 = DH°f H2O(l ) = -285.8 kJ

S8(rhombic) + O2(g ) → SO2(g ) DH°2 = (-2375 kJ)

H2O(l ) + SO2(g ) → H2S(g ) + 3 2 O2(g ) DH°3 = -(-562 kJ)

DH° = DH°1 + DH°2 + DH°3 = -21 kJ

4. When one mole of CaNCN(s) reacts with liquid water to form calcium carbonate and ammonia gas, 261.75 kJ of heat are evolved. Calculate the heat of formation of CaNCN(s).

CaNCN(s) + 3H2O(l ) → CaCO3(s) + 2NH3(g )

DH°rxn = DH°f CaCO3(s) + 2 × DH°f NH3(g ) - DH°f CaNCN(s) - 3 × DH°f H2O(l )

DH°f CaNCN(s) = DH°f CaCO3(s) + 2 × DH°f NH3(g ) - 3 × DH°f H2O(l ) - DH°rxn

= -1206.88 kJ + 2(-46.25 kJ) - 3(-285.8 kJ) - (-261.75 kJ)

= -180.2 kJ/mole CaNCN(s)