355
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 kJ
a. 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 
4
? kJ = 1 lb fat 


 = 1.72 x 10 kJ
1
lb
885.458
g
fat
2
mol
fat




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.)
 −2000 kJ   2 mol fat   885.458 g fat   1 lb 
? lb fat = 20 laps 
 = 0.0233 lb



4
 100 laps   −6.70 x 10 kJ   1 mol fat   453.6 g 
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.


 0.0725 kJ 0.00418 kJ

0.00418 kJ
q v = − C cal +
m w  ∆T = − 
+
1200 g  2.16 °C = − 11.0 kJ
g °C
g °C


 °C

∆E° =
? kJ
−11.0 kJ  2.01594 g H2 
kJ
=

 = − 222
mol H2
0.100 g H2  mol H2
mol H2

DH° = DE° + (Dn)RT
∆H° = − 222 kJ + (0 − 1.5) mol
H2(g) + ½O2(g) → H2O(l )
0.008314 kJ
298.15 K = − 226 kJ/mol H 2
K mol
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 )
∆E° = ∆H° − ( ∆n ) RT = − 3924 kJ − (6 − 9) mol
0.008314 kJ
kJ
298.15 K = − 3917
K mol
mol C 6 H12
 1 mol C 6 H12   −3917 kJ 
q v = ? kJ = 8.25 g C 6 H12 

 = − 384 kJ
 84.162 g C 6 H12   1 mol C 6 H12 


0.00418 kJ
q v = − C cal +
m w  ∆T
g °C


C cal = −
qv
0.00418 kJ
−384 kJ
0.00418 kJ
−
mw = −
−
825.0 g = 48 kJ/ C
∆T
g °C
g °C
(25.6 − 18.2 ) °C
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) +
H
2O(l )
→
HNO (l )
3
+
NO(g)
DH° = -23.84 kJ/mole
∆E° = ∆H° − (∆n) RT= − 23.84
kJ
2
kJ
−  −  0.008314
(298.15 K ) = − 22.19 kJ/mol
mol
K mol
 3
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) +
∆E° =
O
2(g)
→ 4CO2(g) + 5H2O(l )
−85.99 kJ  58.124 g C 4 H10 

 = − 1428 kJ/mol
3.500 g C 4 H10  1 mol C 4 H10 
∆H° = ∆E° + (∆n) RT= − 1428
kJ
15 
kJ
+  4 −
(298.15 K ) = − 1437 kJ/mol
 0.0083144
mol 
2
K 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 O
2(g)
→ 7CO2(g)
∆E° = ∆H° − ( ∆n ) RT = − 322.7 kJ − (7 − 7.5) mol
+
3H2O(l )
0.008314 kJ
kJ
298.15 K = − 321.5
K mol
mol
 1 mol C 6 H5 COOH 

−321.5 kJ
q v = ? kJ = 8.174 g C 6 H5 COOH 

 = − 21.52 kJ
122.124
g
C
H
COOH
1
mol
C
H
COOH
6
5
6
5





0.00418 kJ
q v = − C cal +
m w  ∆T
g °C


C cal = −
qv
0.00418 kJ
−21.52 kJ
0.00418 kJ
−
mw = −
−
200.0 g = 0.315 kJ/°C
∆T
g °C
g °C
(35.72 − 17.02 ) °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)


 0..315 kJ 0.00418 kJ

0.00418 kJ
q v = − C cal +
m w  ∆T = − 
+
210.0 g  (27.69 − 22.33 ) °C
g °C
g °C


 °C

= − 6.39 kJ
∆E° =
? kJ
−6.39 kJ  65.37 g Zn 
kJ
=

 = − 782
mol Zn 0.534 g Zn  1 mol Zn 
mol Zn
∆H° = ∆E° + ( ∆n ) RT = − 782 kJ + (0 − 0) mol
0.008314 kJ
298.15 K = − 782 kJ/mol
K 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)
Cu2O(s) → CuO(s) + Cu(s)
DH° = -238 kJ
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 = -11.3 kJ + (-119 kJ) = -130 kJ
DH°f = DH°1 + DH°2
2. Given the following;
Fe2O3(s) + 3CO(g) → 2Fe(s) + 3CO2(g)
3Fe2O3(s) + CO(g) → 2Fe3O4(s) + CO2(g)
Fe3O4(s) + CO(g) → 3FeO(s) + CO2(g)
DH° = -28 kJ
DH° = -59 kJ
DH° = 39 kJ
Calculate the DH° for;
FeO(s) + CO(g) → Fe(s) + CO2(g)
FeO(s) +
CO
½Fe2O3(s) +
2(g)
→
3 2 CO(g)
Fe O (s) + CO(g)
3 4
→ Fe(s) + 3 2 CO2(g)
DH°1 = - (39 kJ)
DH°2 = ½(-28 kJ)
Fe O (s) + CO (g) → ½Fe O (s) + CO(g)
DH°3 = - (-59 kJ)
3 4
2
2 3
_____________________________________________________________________
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(g raphite) + O2(g) → CO2(g) 2H2O(l ) → 2H2(g) + O2(g)
2C2H6(g) + 7O2(g) → 4CO2(g) + 6H2O(l ) DH° = -393.5 kJ
DH° = 571.6 kJ
DH° = -3119.6 kJ
calculate the enthalpy change for the following reaction:
2C(graphite)
+
3H2(g)
→
C2H6(g)
2C(g raphite) + 2O2(g) → 2CO2(g)
3H2(g) +
32O
2(g)
DH°1 = 2(-393.5 kJ)
→ 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(g raphite)
+
→
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)
2N2(g) + 5O2(g) + 2H2O(l ) → 4HNO3(aq) N2(g) + O2(g) → 2NO(g)
3NO2(g)
→
3NO(g) +
32O
2(g)
N2(g) + 5 2 O2(g) + H2O(l ) → 2HNO3(aq) 2NO(g) → N2(g) + O2(g)
________________________________________________
3NO2(g) + H2O(l ) → 2HNO3(aq) +
DH°rxn = DH°1 + DH°2 + DH°3 = -49 kJ
DH° = -173kJ
DH° = -255 kJ
DH° = 181 kJ
DH°1 = - 3 2 (-173kJ)
DH°2 = ½(-255 kJ)
DH°3 = -(181 kJ)
NO(g)
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) S8(s) + 8O2(g) → 8SO2(g)
2SO2(g) → S8(s) + 2O2(g)
S8(s) + 3O2(g) → 2SO3(g)
________________________________
2SO2(g) + O2(g) → 2SO3(g)
DH°rxn = DH°1 + DH°2 = -198.0 kJ
DH° = -3166 kJ
DH° = -2374 kJ
DH°1 = - (-2374 kJ)
DH°2 = (-3166 kJ)
360
Chapter 6 Worksheet Keys
CHEMISTRY 151 - HEAT OF FORMATION KEY
1. Calculate the molar heat of combustion of acetylene gas, C2H2(g).
52O
C2H2(g) +
2(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
 1 mol C 4 H10   −2876 kJ 
2
? kJ = 20.0 g C 4 H10 

 = − 9.90 x 10 kJ
 58.124 g C 4 H10   1 mol C 4 H10 
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) +
32O
2(g)
→ H2S(g)
→ H2O(l ) + SO2(g) DH°comb = -562 kJ
S8(rhombic) + 8O2(g) → 8SO2(g)
H2(g) + ½O2(g) → H2O(l )
S
8(rhombic)
DH°1 = DH°f H2O(l ) = -285.8 kJ
+ O2(g) → SO2(g)
H2O(l ) + SO2(g) → H2S(g) +
DH°comb = -2375 kJ
32O
2(g)
DH°2 =
(-2375
kJ)
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)