Study Guide: Thermochemistry
Thermochemistry (enthalpy, entropy, free energy)
• exothermic vs. endothermic reactions
• measuring heat flow
• calorimetry (coffee cup calorimeters; bomb
calorimeters)
• specific heat; heat capacity
• enthalpy, entropy, free energy
• heat in balanced chemical equations
• calculating H (Hess’ Law)
• enthalpies of formation
• spontaneous reactions
Equations:
q = mct
q = (Ccal) (t)
H = Σ(Hproducts) – Σ(Hreactants)
this equation also applies to S and G
G = H – TS
Phases of Matter
• heat transfer in solids and liquids
• melting, vaporization, sublimation
• phase diagrams (triple point, critical point, melting
curve, sublimation curve, vapor pressure curve)
• boiling point
• intermolecular forces
Equations:
q = mct
q = (mass) (heat of fusion)
q = (mass) (heat of vaporization)
Constants:
Specific heat of ice = 2.09 J/g•°C
Specific heat of water = 4.18 J/g•°C
Specific heat of steam = 2.03 J/g•°C
heat of fusion of water = 334 J/g
heat of vaporization of water = 2260 J/g
1. A copper pot with a mass of 772 grams absorbs 22.7 kJ of heat. It’s final temperature is 137.0°C. What was its
initial temperature? (the specific heat of copper is 0.385 J/g•°C)
q = mct
22,700 J = (772g)(0.385 J/g•°C) T
T = 76.4 °C
T = Tf - Ti
76.4 °C = 137.0 °C - Ti
Ti = 60.6°C
2. How much heat is absorbed by a 15.5 g piece of gold as it is heated from 4.5°C to 177.4°C? (the specific heat of
gold is 0.129 J/g•°C)
q = mct
q = (15.5g)(0.129 J/g•°C)(172.9 °C)
q = 346 J
3. A container full of water absorbs 64.4 kJ of heat and its temperature rises from 22.0°C to 73.4°C. What is the
volume of water in mL? (the density of water = 1 g/mL)
q = mct
64,400 J = (m)(4.184 J/g•°C)(51.4°C)
m = 299.45g
v = m/d = (299.45g)/(1.00 g/mL)
v = 299 mL
4. A sample of fructose, C6H12O6, weighing 7.55 g is burned in a bomb calorimeter. The heat capacity of the
calorimeter is 2.155 x 104 J/°C. The temperature in the calorimeter rises from 22.54°C to 29.56°C. (a) What is q
when the 7.55 g of fructose is burned? (b) What is q for the combustion of 1 mole of fructose?
a)
qcal = (Ccal)(t)
qcal = (2.155 x 104 J/°C)(7.02°C)
qcal = 150930 J = 151 kJ
qrxn = -151 kJ
b)
180.0 g
151 kJ

 -3600 kJ/mol = -3.60 x 103 kJ/mol
7.55 g C6H12O6
1 mol
5. Naphthalene, C10H8, is the compound present in moth balls. When one mole of naphthalene is burned, 5.15 x 10 3 kJ
of heat is evolved. A sample of naphthalene burned in a bomb calorimeter (heat capacity = 9832 J/°C) increases the
temperature in the calorimeter from 25.1°C to 28.4°C. How many milligrams of naphthalene were burned?
a)
qcal = (Ccal)(t)
qcal = (9832 J/°C)(3.3°C)
qcal = 32445.6 J = 32.4 kJ
qrxn = -32.4 kJ
1 mol C10H8  128.0 g  128,000 mg
b)
 32.4456 kJ  5.15  10 3 kJ

x mg
128,000mg
x  806 mg
6. Nitrogen monoxide (NO) has been found to react with oxygen gas (O 2) to produce the brown gas nitrogen dioxide
(NO2). When one mole of NO reacts with oxygen, 57.0 kJ of heat is evolved. (a) Write the thermochemical equation
for the reaction between one mole of nitrogen monoxide and oxygen to produce nitrogen dioxide. (b) Is the reaction
exothermic or endothermic? (c) What is H when 5.00 g of nitrogen monoxide reacts? (d) How many grams of
nitrogen monoxide must react with an excess of oxygen to produce 10.0 kJ of heat?
a) NO + ½ O2  NO2
H = -57.0 kJ
b) exothermic
c)
57.0 kJ 1 mol NO

 5.00 g  - 9.50 kJ
1 mol NO
30.0 g
9.50 kJ 10.0 kJ

x
d) 5.00 g
x  5.26 g
7. Strontium metal (Sr) combines with graphite (C) and oxygen gas (O2) to produce strontium carbonate (SrCO3).
The formation of one mole of SrCO3 releases 1.220 x 103 kJ of heat. (a) Write a balanced thermochemical equation
for the reaction resulting in the formation of one mole of SrCO3. (b) What is H when 10.00 g of strontium reacts
with excess graphite and oxygen? (c) What mass of SrCO3 forms when 2355 kJ of heat are also formed?
a) Sr + C +
b)
3
2
O2  SrCO3
H = -1. 220 x 103 kJ
 1.220  10 3 kJ 1 mol Sr

 10.00 g  - 139.3 kJ
1 mol Sr
87.6 g
1220 kJ
2355 kJ

x
c) 147.6 g SrCO3
x  284.9 g
8. Given: 2CuO(s)  2Cu(s) + O2(g) H = 314.6 kJ
(a) Determine the heat of formation of CuO.
2Cu(s) + O2(g)  2CuO(s) H = -314.6 kJ
Cu(s) + ½ O2(g)  CuO(s) H = -157.3 kJ
(Since H = Σ(Hproducts) – Σ(Hreactants) and the heats of formation of reactants are both
zero, heat of formation of one mole of CuO = -157.3 kJ)
(b) Calculate H for the formation of 13.58 g of CuO.
157.3 kJ 1 mol CuO

 13.58 g  - 26.87 kJ
1 mol CuO
79.5 g
9. Limestone, CaCO3, when subjected to a temperature of 900°C in a kiln, decomposes to calcium oxide and carbon
dioxide. (a) Write a balanced chemical equation for this reaction. (b) Determine H for the reaction using table 8.3.
(c) How much heat is evolved or absorbed when one gram of limestone decomposes?
a) CaCO3  CaO + CO2
b) H = [(-635.1) + (-393.5)] – (-1206.9) = (-1028.6) – (-1206.9) = 178.3 kJ/mol
c)
1 mol CaCO3
178.3 kJ

 1.000 g  1.781 kJ (absorbed)
1 mol CaCO3
100.1 g
10. How much energy is needed to heat a 164.5 g sample of ice at –27.6°C to liquid water at 72.3°C?
q-27.60 = (164.5g)(2.09 J/g•°C)(27.6°C) = 9,489 J
qfus
= (164.5g)( 334 J/g)
= 54,943 J
q072.3 = (164.5g)(4.184 J/g•°C)(72.3°C) = 49,762 J
qtotal =114,194 J = 114 kJ
11. How much energy is released when 52.3 g of steam at 136.5°C cools and condenses to form water at 93.2°C?
q136.5100= (52.3g)(-2.03 J/g•°C)(36.5°C) = -3,875 J
qvap
= (52.3g)(-2260 J/g)
= -118,198 J
q10093.2 = (52.3g)(-4.184 J/g•°C)(6.8°C)
= -1,488 J
qtotal = -123,561 J = -124 kJ
12. How much energy is needed to heat a 42.3 g sample of ice at –35.7°C to steam at 112.0°C?
q-35.70 = (42.3g)(2.09 J/g•°C)(35.7°C)
= 3,156 J
qfus
= (42.3g)( 334 J/g)
= 14,128 J
q0100 = (42.3g)(4.184 J/g•°C)(100.0°C) = 17,698 J
qvap
= (42.3g)(2260 J/g)
= 95,598 J
q100112 = (42.3g)(2.03 J/g•°C)(12.0°C)
= 1,030 J
qtotal = 131,610 J = 132 kJ
13. You are given the following data for ethanol, C2H5OH:
normal melting point = -114 °C
normal boiling point = 78.5°C
critical temperature = 243°C
critical pressure = 63.0 atm
Assume that the triple point is slightly lower in temperature than the melting point and that the vapor pressure at
the triple point is about 10-5 mm Hg. (a) Sketch a phase diagram for ethanol. (b) Ethanol at 1 atm and 140°C is
compressed to 70 atm. Are two phases present at any time during this process? Explain. (c) Ethanol at 1 atm and
270°C is compressed to 70 atm. Are two phases present at any time during this process? Explain.
a)
Phase Diagram for Ethanol
63
P (atm)
S
L
G
1
1.3x10
-8
-114
78.5
T (ºC)
b) yes, gas condenses into liquid (below critical temperature)
c) no – gas cannot be liquefied above critical temperature
243
14. Hess’s Law and ΔH:
a) Explain why Hess’s law is used in the chemistry laboratory?
This law can be used to allow us to calculate heats of reaction for reactions that cannot be carried
out in a simple calorimeter. Instead, we can add together the enthalpies of reaction for reactions
we can measure, then add them in such a way that we find the enthalpy of another reaction.
b)
How can ΔH be calculated for an equation in which the coefficients have been multiplied by a factor
of two?
Multiply ΔH by 2
c)
What happens to the sign of ΔH if a reaction is run in the reverse direction from the way it is
written?
Reverse the sign of ΔH
15. What is meant by the terms of heat of fusion and heat of vaporization?
The amount of energy released or absorbed to cause one gram of a substance to change state. Heat of fusion
relates to melting and freezing, while heat of vaporization relates to evaporating and condensing.
16. From the following enthalpy changes,
2PbS(s) + 3O2(g)  2PbO(s) + 2SO2(g)
ΔH° = -827.0 kJ
PbO(s) + C(s)  Pb(s) + CO(g)
ΔH° = +106.8 kJ
Calculate the value of ΔH° in the following reaction:
2PbS(s) + 3O2(g) + 2C(s)  2Pb(s) + 2CO(g) + 2SO2(g).
Is the reaction endothermic or exothermic?
Multiply second equation by two
Add the two equations together
ΔH = (-827.0 kJ) + (213.6) = -613.4 kJ
17. Determine the change in enthalpy for the following reaction: C (graphite) + 2H2(g)  CH4(g)
Use these reaction equations:
C (graphite) + O2(g)  CO2(g)
ΔH° = -394 kJ
H2(g) +
1
2
O2(g)  H2O(l)
ΔH° = -286 kJ
CO2(g) + 2H2O(l)  CH4(g) + 2O2(g)
ΔH° = +890.3 kJ
Multiply middle equation by 2
Add the three equations together
ΔH = (-394 kJ) + (2)(-286) + (+890.3) = -75.7 kJ
18. Draw a temperature vs. time graph of water being heated at a constant rate at 1 atm from -10°C to 100°C.
100
Temp (ºC)
0
Time