Chem 122
Chapter 11: Sheet 60 Multi-Step Energy Calculations (Answers)
Page 1 of 2
1. For the following combustion, what mass of carbon dioxide is produced when 1500 kJ of
energy is released?
2 C2H6(g) + 7 O2(g) —> 4 CO2(g) + 6 H2O(g) + 2502 kJ
n (CO2(g)) = ∆HC = -1500 kJ 4 mol 44.01 g = 105.5 g
HC
‰-2502 kJ‰ mol 
2. How much energy is released when 1.00 t of sulfur trioxide is produced by the following
reaction?
2 SO2(g) + O2(g) —> 2 SO3(g)
∆HC = -192.8 kJ
3
3
∆HC (SO3(g)) = nHC = 1.00 t 10 kg 10 g mol -192.8 kJ = -1.20 x 106 kJ
‰ t ‰ kg ‰80.06 g‰ 2 mol  (-1.2040969 x 106 kJ)
3. In respiration, glucose is oxidized by oxygen gas to produce carbon dioxide, liquid water, and
energy. What is the energy released when 18.0 g of glucose is consumed?
Step 1: Find HC from the balanced reaction
C6H12O6(s) + 6 O2(g) —> 6 CO2(g) + 6 H2O(l)
∆HC° = ΣnH°fp - ΣnH°fr
= (6 mol CO2(g)) -393.5 kJ + (6 mol H2O(l)) -285.8 kJ
—
‰mol CO2(g)
‰mol H2O(l)  ž
_ (1 mol C6H12O6(s)) -1273.1 kJ
+ (6 mol O2(g)) 0 kJ
—
‰mol C6H4O6(s)
‰mol O2(g) ž
= [ -2361 kJ + (-1714.8 kJ) ] - [ -1273.1 kJ ]
= -2802.8 kJ (this is the enthalpy of the balanced reaction ∆HC)
To find molar enthalpy:
HC° = ∆HC° = -2802.8 kJ = -2802.7 kJ/mol
n
1 mol
Step 2: Use the HC from step 1 to find ∆HC for 18.0 g C6H12O6(s)
∆HC (C6H12O6(s)) = nHC = 18.0 g mol
-2802.7 kJ = -280 kJ = -2.80 x 102 kJ
‰180.18 g‰
mol 
4. Methanol is burned in a bomb calorimeter. Liquid water is formed as a product. If 3.40 g of
methanol reacts, what is the expected temperature change in a calorimeter with a heat
capacity of 6.75 kJ/°C?
Step 1: Find HC from the balanced reaction
2 CH3OH(l) + 3 O2(g) —> 2 CO2(g) + 4 H2O(l)
∆HC° = ΣnH°fp - ΣnH°fr
= (2 mol CO2(g)) -393.5 kJ + (4 mol H2O(l)) -285.8 kJ
—
‰mol CO2(g)
‰mol H2O(l)  ž
_ (2 mol CH3OH(l)) -239.1 kJ
+ (3 mol O2(g)) 0 kJ
—
‰mol CH3OH(l)
‰mol O2(g) ž
= [ -787.0 kJ + (-1143.2 kJ) ] - [ -478.2 kJ ]
= [-1930.2 kJ] - [-478.2 kJ]
= -1452.0 kJ (this is the enthalpy of the balanced reaction ∆HC)
To find molar enthalpy:
HC° = ∆HC° = -1452.0 kJ = -726.0 kJ/mol
n
2 mol
(continued next page)
Chem 122
Chapter 11: Sheet 60 Multi-Step Energy Calculations (Answers)
Page 2 of 2
Step 2: Use HC° in calorimetry to find ∆t
energy gained by bomb calorimeter = energy lost by methanol burning
q (calorimeter) = ∆HC° (methanol)
C∆t = nHC°
6.75 kJ ∆t = 3.40 g mol 726.0 kJ
‰ °C 
‰32.05 g‰ mol 
6.75 kJ ∆t = 77.017
‰ °C 
∆t = 77.017 °C
6.75
∆t = 11.4 °C
5. A waste heat exchanger is used to absorb the energy from the complete combustion of
hydrogen sulfide gas. What volume of water undergoing a temperature change of 65°C is
required to absorb all of the energy from the burning of 15 kg of hydrogen sulfide?
Step 1: Find HC from the balanced reaction
2 H2S(g) + 3 O2(g) —> 2 SO2(g) + 2 H2O(g)
∆HC° = ΣnH°fp - ΣnH°fr
= (2 mol SO2(g)) -296.8 kJ + (2 mol H2O(g)) -241.8 kJ
—
‰mol SO2(g)
‰mol H2O(g)  ž
_ (2 mol H2S(g)) -20.6 kJ + (3 mol O2(g)) 0 kJ
—
‰mol H2S(g)
‰mol O2(g) ž
= [ -593.6 kJ + (-483.6 kJ) ] - [ -41.2 kJ ]
= [-1077.2 kJ] - [-41.2 kJ]
= -1036.0 kJ (this is the enthalpy of the balanced reaction ∆HC)
To find molar enthalpy:
HC° = ∆HC° = -1036.0 kJ = -518.0 kJ/mol
n
2 mol
Step 2: Use HC° in calorimetry to find v of water in heat exchanger
energy gained by water in heat exchanger = energy lost by H2S(g) burning
q (water) = ∆HC° (H2S(g))
vc∆t = nHC°
v 4.19 kJ 64°C = 15 kg mol 103 g 518.0 kJ
‰ L·°C 
‰34.08 g‰ kg ‰ mol 
268.16 v = 227,992.9577
L
v = 227,992.9577 L
268.16
v = 850 L