Chemistry Student Edition - Basic
Answer Key
Chapter 17: Thermochemistry
Heat Flow
Questions
1.
2.
3.
4.
5.
6.
7.
8.
9.
What is true of all chemical processes?
How can you distinguish a system and its surroundings?
What are the required characteristics of an isolated system?
Explain the difference between exothermic and endothermic processes.
Indicate which of the following processes are exothermic and which are endothermic:
a. ice melting
b. burning a candle
c. cooking an egg
d. the evaporation of sweat
Calculate the amount of heat needed to increase the temperature of 125 g of water from
22°C to 59°C.
Calculate the specific heat of copper, given that 204.75 J of energy raises the
temperature of 15 g of copper from 35°C to 70°C.
432 J of energy is required to raise the temperature of a block of aluminum from 20°C to
60°C. Calculate the mass of aluminum present.
300 g of liquid ethanol at 22°C is supplied with 6480 J of heat. What is the final
temperature of the ethanol? (The specific heat of ethanol is 2.44 J/g·°C).
Answers
1. All chemical processes involve the transfer of heat to or from the system.
2. The system includes the components involved in a reaction, while the surroundings
include everything else.
3. An isolated system can exchange neither heat or matter with its surroundings
4. Exothermic processes release heat to the surroundings, while endothermic processes
require heat from the surroundings.
5.
a. Endothermic
b. Exothermic
c. Endothermic
d. Endothermic
6. mass = 125 g
ΔT = 37°C
specific heat of water = 4.18 J/g °C
4.18J
x 370 C
g 0C
= 19332.5 J or 19.3 kJ
ΔH = 125g x
C = ms
C
m
204.75J
=
(15g)(350 C)
s=
7.
= 0.39J / g0C
C = ms
m=
8.
=
C
s
432 J
0
(0.9J / g0C)(40 C)
= 12 grams Al
heat = (mass)(specific heat)(Δtemp)
heat
Δtemp =
(spec heat)(mass)
6480K
=
(2.44J / g 0C)(300g)
= 8.80
9.
final temperature = 220 + 8.80
final temperature = 30.80 C
Enthalpy
Questions
1. Use standard enthalpies of formation from the supplemental link below to calculate ΔHrxn
for the reaction: 2CO(g) + O2(g) → 2CO2(g).
2. Use standard enthalpies of formation from the supplemental link below to calculate ΔHrxn
for the reaction: 4NH3(g) + 5O2(g) → 6H2O(g) + 4NO(g).
3. Complete combustion of 1.00 mol of acetone (C3H6O) liberates 1790 kJ of heat: C3H6O(l)
+ 4 O2(g) → 3 CO2(g) + 3 H2O(l); ΔH = -1790 kJ. Using this information and ΔHf data for
CO2 and H2O, calculate the standard enthalpy of formation for acetone.
4. Determine ΔHf for IF, given the following information:
IF7 + I2(s) → IF5 + 2IF; ΔHf = -89 kJ
ΔHf for IF7 = -941 kJ
ΔHf for IF5 = -840 kJ
5. Write the balanced equation for the complete combustion of C2H6, and calculate the heat
of combustion per mole of gaseous water formed using standard enthalpies of formation.
Answers
1. Use standard enthalpies of formation from the supplemental link below to calculate ΔHrxn
for the reaction: 2CO(g) + O2(g) → 2CO2(g).
ΔHreaction = ΣnΔHproducts – ΣmΔHreactants
= 2(-393.5 kJ/mol) –[2(-110.5) + 0] kJ/mol oxygen is an element, so has a ΔH of zero
= -787 + 221) kJ/mol = -566 kJ/mol
2. 4NH3(g) + 5O2(g) → 6H2O(g) + 4NO(g).
ΔHreaction = ∑n ΔHproducts – ∑m ΔHreactants
= 6(–241.8) kJ/mol + 4(90.3) kJ/mol–[ 4(–46.2 kJ/mol) – 0 ]
= –1450.8 kJ/mol + 361.2 kJ/mol + 184.8 kJ/ mol
= –905.8 kJ/mol
Note that we used the enthalpy value for water as a gas since that was included in the
reaction conditions.
3. ΔHreaction = ΣnΔHproducts – ΣmΔHreactants
-1790 kJ/mol = [3(-393.5) + 3(-285.8)] – ΔHacetone
-1790 kJ/mol = (-1180.5 – 857.4)kJ/mol - ΔHacetone
-1790kJ/mol = -2037.9 kJ/mol - ΔHacetone
ΔHacetone = - 247.5 kJ/mol
4. ΔHreaction = ΣnΔHproducts – ΣmΔHreactants
let x = ΔH for IF
then -89 kJ= [(-840)+ (2x)]kJ - (-941)kJ
-89kJ = 101kJ + 2x
2x = -190
x = -95 kJ
5. The enthalpy of the reaction is:
ΔH = [(2 CO2)(-393.5 kJ/mole) + (6 H2O)(-241.8 kJ/mole)] - [(2 C2H6)(-84.68 kJ/mole) +
(7 O2)(0 kJ/mole)]
= -2238 kJ - (-169 kJ) = -2069 kJ for 6 moles water
-2069 kJ / 6 moles H2O = -345 kJ / mole H2O
Enthalpy and Phase Transitions
Questions
1. Mercury melts at -39°C. How much energy is required to melt 150. grams of mercury if
its molar heat of fusion is 2.29 kJ/mol?
2. Sulfur has a boiling point of 444.6°C. If the heat of vaporization for sulfur is 45 kJ/mol,
how much heat is released when 42 grams of sulfur vapor is converted to liquid sulfur?
3. Calculate the total enthalpy change when 200. grams of water vapor at 120°C is
converted to liquid water at 42°C.
4. How much energy is involved in dissolving 76 grams of NaCl in water? Is this an
exothermic or endothermic process?
5. Calculate the total energy involved in converting 50. grams of ice at -10°C to liquid water
at 95°C.
6. If the heat of solution for LiBr is -49 kJ/mol, how many grams of LiBr must be dissolved
in order to release 3,500 joules of energy?
Answers
1. total energy = (ΔHfusion) )(mass)
= (11.8 J/g)(150 g) = 1770 joules or 1.7 kJ
2. total energy released = (ΔHvap)(mass)
= (0.33 J/g)(42 g) = 13.9 joules
3. enthalpy change = (cooling for 120°C to 100°C) + (transition from vapor to liquid) +
cooling from 100°C to 42°C
cooling for 120°C to 100°C = (200 g)(20°)(2.01 J/g°C) = -8360 joules or 8.36 kJ. The
number is negative since this is an exothermic process as are the others in this problem
for the transition for vapor to liquid, the heat of vaporization units are in kJ/mol, so we
need to convert grams of water to moles:
200 grams water
= 11.1 moles
18 grams / mole
then the heat of vaporization = (40.7 kJ/mol)(11.1 moles) = - 451.8 kJ (note the negative
value)
cooling from 100°C to 42°C = (58°C)(200 g)(4.18 J/g°C) = 48488 joules or 48.5 kJ
total energy change = -8.36kJ + (-451.8 kJ) + (-48.5 kJ) = 508.7 kJ
4. Heats of solution are in kJ/mol, so we first need to convert 76 grams NaCl to moles:
76 grams
= 1.3 moles NaCl
58.5 grams / mole
Then (1.3 moles)(3.9 kJ/mole) = 5.07 kJ. The number is positive, so this is an
endothermic reaction
5. step a: ice at -10°C to ice at 0°C = (2.09 j/g°C)(50 g)(10°C) = 1045 joules or 1.05 kJ
to calculate the heat needed to melt the ice, we need to get moles of water:
moles of water =
50 grams
= 2.8 moles
18 grams / mole
step b: ice at 0°C to water at 0°C = (6.0 kJ/mol)(2.8 mol) = 16.8 kJ
step c: water at 0°C to water at 95°C = (4.18 J/g°C)(50 g)(95°C) = 19855 joules or 19.9
kJ
total energy = 1.05 kJ + 16.8 kJ + 19.9 kJ = 37.8 kJ energy input needed.
6. moles LiBr =
3500 joules 1kJ
x
= 0.071 moles LiBr
49kJ / mol 1000j
Hess’s Law
Questions
1. Calculate the heat released by the burning of sulfur in oxygen: 2S(s)+3O2(g)→2SO3(g). ΔH
values are known for the following reactions:
S(s)+O2(g)→SO2(g)
ΔH=−296 kJ
2SO2(g)+O2(g)→2SO3(g)
ΔH=−198 kJ
2. Calculate ΔH for the following reaction, which describes the production of syn-gas from
carbon:
H2O(g)+C(s)→CO(g)+H2(g).
The following enthalpy changes are known:
H2(g)+12O2(g)→H2O(g)
ΔH=−242.0 kJ
2CO(g)→2C(s)+O2(g)
ΔH=+221.0 kJ
3. Calculate the heat of reaction for the following equation:
C3H8(g)+5O2(g)→3CO2(g)+4H2O(g).
The following formation reactions have known ΔH values:
3C(s)+4H2(g)→C3H8(g)
ΔH=−103.8 kJ
2H2(g)+O2(g)→2H2O(g)
ΔH=−484 kJ
C(s)+O2(g)→CO2(g)
ΔH=−393 kJ
Answers
1. S(s) + O2 (g) → SO2 (g) ΔH= -297 kJ is for one mole S. We need two moles so we
multiply the equation times 2:
2 S(s) + 2O2 (g) → 2SO2 (g)
ΔH = - 594 kJ Now add in the second equation
2SO2 (g) + O2 (g) → 2SO3 (g)
Δ H= - 198 kJ
We see that the 2 SO2 molecules on both the reactant side and the product side cancel,
so we have
2S(s) + 3O2 (g) → 2SO3(g)
ΔH = - 792 kJ
2. H2O is a reactant in the original equation, but a product in the first given reaction, so we
need to reverse that reaction:
H2O (g) → H2 (g) + 1/2 O2 (g)
Δ H = 242.0 kJ
in given equation 2 CO is a reactant, but is a product in the original reaction. Note that
there are also two moles CO in the given equation, but only one in the original reaction.
For this equation, we reverse the direction for the reaction and we halve the materials.
C(s) + 1/2 O2 (g) → CO (g)
ΔH = + 110.5 kJ
Now we have
H2O (g) → H2 (g) + 1/2 O2 (g)
Δ H = 242.0 kJ
C(s) + 1/2 O2 (g) → CO (g)
ΔH = - 110.5 kJ
The ΔH is then + 131.6 kJ
3. Given the following steps in the reaction mechanism.
Step #1: 3C (s)+ 4 H2 (g) → C3H8 (g)
ΔH = - 103.8 kJ
Step #2: 2 H2(g) + O2 (g) → 2H2O (g) ΔH = -484 kJ
Step #3: C (s) + O2 (g) → CO2 (g)
ΔH = - 393 kJ
Since propane (C3H8) is in the reactant side in the original equation, we reverse equation
1 and changing the sign of ΔH:
C3H8 (g) → 4 H2 (g) + 3C (s)
ΔH = +103.8 kJ
We need 4 H2O in the product side, so we multiply equation two by 2:
4 H2 (g)+ 2 O2 (g) → 4 H2O (g) ΔH = -968 kJ
And equation three is multiplied by 3 to obtain 3 CO2 molecules:
3 C (s) + 3 O 2 (g)→ 3 CO 2 (g) ΔH = -1180.5 kJ
So we now have:
C3H8 (g) → 4 H2 (g) + 3C (s)
ΔH = +103.8 kJ
4 H2 (g) + 2 O2 (g) → 4 H2O (g) ΔH = -968 kJ
3 C (s) + 3 O 2 (g)→ 3 CO 2 (g) ΔH = -1180.5 kJ
giving us: C3H8 (g) + 5 O2 (g) -----> 3 CO2 (g) + 4 H2O(g) with a ΔH of -2044.7 kJ