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CHAPTER 5 | Thermochemistry
5.23. Collect and Organize
The work being done is due to an expansion of the gas from 250.0 mL to 750.0 mL.
Analyze
Work is expressed as –P∆V. In this case, P is constant at 1.00 atm and the volume change is 500.0 mL
or 0.5000 L. We are to express work in both L  atm and in joules. We can convert from one unit to
another using 101.32 J/L  atm.
Solve
w = –P∆V = (1.00 atm)(0.5000 L) = –0.500 L  atm
In joules, this work is
–0.500 L  atm 
101.32 J
 –50.7 J
L  atm
Think about It
Because work was done by the system on the surroundings, the sign of work is negative.
5.27. Collect and Organize
The change in internal energy for a system is
∆E = q + w
Analyze
The system releases energy to its surroundings so q is negative. Because the system does work on the
surroundings, w is negative.
Solve
∆E = q + w = –210 kJ + (–65.5 kJ) = –276 kJ
Think about It
Be careful to always define the signs for q and w from the point of view of the system.
5.29. Collect and Organize
Work will be done by a system on the surroundings when the volume of the system shown in
Figure P5.29 increases.
Analyze
The volume of gas is proportional to the number of moles of gas (n) at constant temperature and
pressure. Because both temperature and pressure are constant for each reaction, if n increases in going
from reactants to products, the volume of the system increases and work is done by the system on the
surroundings.
Solve
(a) In this reaction, 3 moles of gaseous reactants form 3 moles of gaseous products. The ∆n for the
reaction is 0, so this reaction does not do work on the surroundings.
(b) In the reaction, 6 moles of gaseous reactants form 7 moles of gaseous products. The ∆n for the
reaction is +1, so this reaction does work on the surroundings.
(c) In this reaction, 3 moles of gaseous reactants form 2 moles of gaseous products. The ∆n for the
reaction is –1, so this reaction does not do work on the surroundings.
215
216 | Chapter 5
Think about It
Reaction c has +w (from –P∆V where ∆V = Vf – Vi and Vi > Vf so ∆V is negative). This means that the
surroundings did work on the system. The system was compressed.
5.38. Collect and Organize
The formation of plaster of Paris from gypsum requires heating to 150˚C to drive off the water.
Analyze
Because energy input is required for this reaction, q is positive.
Solve
Because ∆H = qP, when qP is positive, ∆H is also positive.
Think about It
Reactions that require energy input are endothermic reactions.
5.40. Collect and Organize
The reaction of vinegar with baking soda inflates a balloon through the evolution of CO 2 gas.
Analyze
The volume of the balloon increases as CO2 is produced in the reaction. The work done by the system
is defined by the equation
w = –P∆V = –P(Vf – Vi)
Solve
In the case of the expanding balloon, Vf > Vi so ∆V is positive. Therefore –P∆V is a negative quantity
and work is negative. This means that work was done by the system on the surroundings.
Think about It
Compressing a gas gives a positive value for work, which means that the surroundings have done
work on the system to increase its internal energy.
5.50. Collect and Organize
A heating curve plots temperature as a function of energy added to a substance.
Analyze
Octane at –57˚C is a solid just about to melt. As energy is added the solid octane melts and its
temperature does not change until all the solid is melted. Only when octane is entirely liquid does
added energy increase the temperature of the liquid until the boiling point of octane is reached.
During boiling, the temperature of the octane does not change. Once all the octane is converted to
gaseous form, added energy increases the temperature of the gaseous octane.
Relevant equations for finding the energy required for each step are as follows:
q = ncp ∆T for energy added to the solid, liquid, and gas phases where n = 1 mol, cp = heat
capacity for that phase, and ∆T = temperature change for that phase
q = n∆Hfus energy for melting
q = n∆Hvap energy for vaporization
Solve
Step in
Heating Curve
Melting octane
ice
Warming liquid
octane
Boiling octane
Ti, ˚C
Tf, ˚C
q, kJ
–57
–56.8
qfus = 1 mol  20.7 kJ/mol = 20.7 kJ
–56.8
125.7
q = 1 mol  254.6 J/mol ˚C  182.5˚C = 46.5 kJ
67.2
125.7
125.7
qvap = 1 mol  41.5 kJ/mol = 41.5 kJ
108.7
Total q,
kJ
20.7
Thermochemistry | 217
Heating octane
gas
125.7
q = 1 mol  316.9 J/mol ˚C  24.3˚C = 7.70 kJ
150
116.4
Think about It
Notice that the temperature does not change throughout a phase change. All of the added energy is
being used by the system to melt the solid or boil the liquid.
5.54. Collect and Organize
Upon dropping the hot metals into water, the metals cool down and the water heats up until they are
both at thermal equilibrium, that is, they will have the same final temperature.
Analyze
The energy lost by the two metals equals the energy gained by the water. The energy equations
necessary are
qmetal = ncP∆T
since we are given the molar heat capacities of the metals, and
qwater = ncP∆T
given that 1.00 L = 1000 g and cP = 75.3 J/mol ˚C and the molar mass of water is 18.02 g/mol.
Solve
–qlost by metals = qgained by water
–qlost by iron + –qlost by gold = qgained by water
We are given molar heat capacities for the metals, and that the specific heat capacity of water is
75.3 J/mol ˚C
–(nFe cP,Fe TFe  nAu cP,Au TAu )  ncP TH2O
– nFe cP,Fe Tf – Ti  Fe  nAu cP,Au Tf – Ti  Au   ncP Tf – Ti  H O
2
The metals (initially at 100.0˚C) reach the same temperature along with the water (initially at 20.0˚C),
so this equation becomes
– nFe cP,Fe Tf –100.0 C  nAu cP,Au Tf –100.0 C   ncP Tf – 20.0 C
Fe
Au 

H2 O
Rearranging this equation and solving,
– nFe cP,Fe  nAu cP,Au Tf –100.0 C  ncP,H2O Tf – 20.0 C










H2 O

218 | Chapter 5
T – 100.0 C 
T – 20.0 C –  n
ncP,H2 O
f
c
Fe P,Fe
f
 nAu cP,Au

1 mol
75.3 J

18.02 g mol  C


T – 20.0 C –  20.0 g Fe  1 mol Fe  25.19 J    20.0 g Au  1 mol Au  25.41 J  
1000 g 
Tf – 100.0 C
f


55.845 g
mol  C 

196.967 g
T – 100.0 C  4.18  10 J/ C  –360.3
T – 20.0 C –11.6 J/ C
T – 100.0 C  –360.3  T – 20.0 C 
mol  C  
3
f
f
f
f
Tf – 100.0 C  –360.3Tf  7206
361.3Tf  7306
Tf  20.2 C
Think about It
This small increase in the temperature of the water goes along with everyday experience. The high
molar heat capacity of water, coupled with the fact that there is a large quantity of water to cool down
the metals, means that the metals cool down to nearly the initial temperature of the water.
5.62. Collect and Organize
In a bomb calorimeter qsystem = ∆Ecomb, but since the PV work is usually small, ∆Ecomb ≈ ∆Hcomb. So we
may assume that ∆Hcomb = –qcalorimeter since qrxn = –qcalorimeter.
Analyze
We can find ∆Hcomb through
Hcomb  –qcalorimeter  –Ccalorimeter T
The ∆Hcomb we find is for the combustion of 1.608 g of cymene. To find ∆Hcomb in terms of kilojoules
per mole we need to divide the calculated ∆Hcomb by the moles of cymene (C10H14).
Solve
H comb  3.640 kJ/ C  19.35 C  –70.43 kJ
molar H comb 
–70.43 kJ
kJ
 –5878
mol

1 mol 
 1.608 g  134.2 g 
Think about It
Expressing the enthalpies of reactions in terms of molar enthalpies allows us to compare the enthalpy
for reactions on a per mole basis.
5.67. Collect and Organize
The enthalpy of formation is reflected in a reaction when (1) one mole of the substance is produced,
(2) the substance is produced under standard state conditions, and (3) it is produced from the
substance’s constituent elements in their standard state.
Analyze
°
Each reaction must meet all the criteria for H rxn
to be classified as an enthalpy of formation.
Thermochemistry | 219
Solve
°
(a) One mole of CO2 is produced from elemental carbon and oxygen so H rxn
for this reaction
°
represents H f .
(b) Because two moles of CO are produced from CO 2 (which is not an element) and C, this reaction
does not represent H f° .
(c) Because two substances are produced and one of the reactants (CO 2) is not an element, this
reaction does not represent H f° .
(d) One mole of CH4 is produced from elemental carbon and hydrogen; therefore this reaction
represents H f° .
Think about It
Enthalpy of formation must involve the reaction of the elements to form compounds. Remember,
though, that some elements such as O2, H2, and N2 are diatomic in their elemental state.
5.70. Collect and Organize
The enthalpy of a reaction can be computed by finding the difference between the sum of the
enthalpies of formation of the products and the sum of the enthalpies of formation of the reactants.
Analyze
We have to take into account the moles of products formed and the moles of reactants used as well,
since enthalpy is a stoichiometric quantity.
Hrxn   n Hf, products –  m Hf, reactants
Values for H f° for the reactants and products are found in Appendix 4; the H f° for methylamine is
given in the problem as –22.97 kJ/mol.
Solve
H rxn   3 mol CH 4   74.8 kJ/mol   1 mol CO2   393.5 kJ/mol    4 mol NH 3    46.1 kJ/mol  
–  4 mol CH 3 NH 2   22.97 kJ/mol    2 mol H 2 O   285.8 kJ/mol  
H rxn  –138.8 kJ
Think about It
Be careful to note and find the appropriate H f° for a compound that may exist in different phases.
For example, H f° of H2O(g) = –241.8 kJ/mol but H f° of H2O ( ) = –285.8 kJ/mol.
5.73. Collect and Organize
We are given the balanced chemical equation for the explosive reaction of fuel oil with ammonium
nitrate in the presence of oxygen.
Analyze
°
To calculate H rxn
we use
Hrxn   n Hf, products –  m Hf, reactants
Solve
H rxn   3 mol N 2   0.0 kJ/mol   17 mol H 2 O   241.8 kJ/mol   10 mol CO2   393.5 kJ/mol  
H rxn
–  3 mol NH 4 NO3   365.6 kJ/mol   1 mol C10 H 22   249.7 kJ/mol   14 mol O2   0.0 kJ/mol  
 –7198 kJ
220 | Chapter 5
Think about It
This is a very exothermic reaction that occurs very fast and is therefore explosive.
5.74. Collect and Organize
We are given the balanced chemical equation for the decomposition of TNT. The enthalpy change
from Problem 5.73 for the explosion of ammonium nitrate with fuel oil is –7198 kJ for 3 moles of
NH4NO3. In this problem we are asked to determine the amount of TNT needed to equal the enthalpy
change for the explosion of 1 mole of NH4NO3.
Analyze
First, we need to find how much energy (in kJ) is released for 1 mol of NH4NO3. We need only divide
the enthalpy obtained in Problem 5.73 by 3. Next, we know from the given information that the
decomposition of 1 mol TNT releases 10,153 kJ. To calculate how much TNT is required to equal the
enthalpy released by the NH4NO3 reaction we need to divide the enthalpy of the NH4NO3 reaction by
the enthalpy of the TNT reaction. Then, we can convert moles of TNT to mass TNT using the molar
mass of TNT (C7H5N3O6, 227.13 g/mol).
Solve
–7198 kJ
 –2399 kJ
3 mol
Equivalent mole of TNT to equal enthalpy release of the NH4NO3 reaction
1 mol TNT
–2399 kJ 
 0.2363 mol
–10,153 kJ
227.13 g
The mass of TNT  0.2363 mol 
 53.67 g TNT
1 mol
Enthalpy for explosion of 1 mol NH4 NO3 
Think about It
One mole of NH4NO3 would weigh 80.04 g. Therefore, a lesser mass of TNT is needed to produce the
same enthalpy change as the explosion of NH4NO3.
5.89. Collect and Organize
To calculate H f° for SO2 from the equations given we use Hess’s law.
Analyze
The equation for H f° of SO2 has S and O2 as the reactants and SO2 as the product. The sum of the
other two reactions must add up to the overall H f° . In both reactions, SO3 is produced. If we reverse
the first equation and add the second equation, the overall reaction will consist of S and O 2 as the
°
reactants and SO2 as the product. When the first reaction is reversed, the H rxn
will change from
exothermic to endothermic.
Solve
2 SO3 (g)  2 SO2 (g) + O2 (g)
°
H rxn
 196 kJ
1
4
S8 (s)  3 O2 (g)  2 SO3 (g)
°
H rxn
 790 kJ
1
4
S8 (s)  2 O2 (g)  2 SO2 (g)
°
H rxn
  594 kJ
This will be twice that of the H f° for the formation reaction (for 1 mol of SO2 formed):
1
8
S8 (s)  O2 (g)  SO2 (g)
Therefore, H f° = –297 kJ/mol.
Think about It
Remember that enthalpy is stoichiometric. If 5 mol of SO2 were formed then ∆H˚ = –1485 kJ.
Thermochemistry | 221
5.92. Collect and Organize
The overall process of converting diamond into graphite is
Cdiamond(s)  Cgraphite(s)
Analyze
Because diamond is a reactant and graphite the product, the first and third reactions do not need to be
reversed. We want to cancel CO2 on the products side and CO on the reactants side, which can be
achieved using the second reaction.
Solve
C diamond (s) + O2 (g)  CO 2 (g)
H °  –395.4 kJ
2 CO2 (g)  2 CO(g) + O2 (g)
H °  566.0 kJ
2 CO(g)  C graphite (s) + CO2 (g)
H °  –172.5 kJ
Cdiamond (s)  C graphite (s)
H °  –1.90 kJ
Think about It
The fact that the enthalpy of conversion of graphite to diamond is exothermic and, therefore, a
favorable process means that diamond is unstable with respect to graphite.
5.93. Collect and Organize
The two chemical reactions must add up to the overall reaction in which NOCl decomposes to nitrogen,
°
.
oxygen, and chlorine. We can use Hess’s law to find H rxn
Analyze
NOCl must be on the reactants side so the second reaction must be reversed. This reaction will then
°
be endothermic ( H rxn
= 38.6 kJ). The first reaction also has to be reversed because we need to have
°
N2 and O2 on the product side of the overall equation. This will be an exothermic reaction ( H rxn
=
–90.3 kJ).
Solve
(a) The first reaction, which produces NO from N2 and O2, represents an enthalpy of formation
reaction.
(b)
°
NO(g)  12 N 2 (g) + 12 O 2 (g)
H rxn
 –90.3 kJ
NOCl(g)  NO(g) +
NOCl(g) 
1
2
1
2
N 2 (g) +
°
H rxn
 38.6 kJ
Cl 2 (g)
1
2
O 2 (g) +
1
2
Cl 2 (g)
Multiplying this chemical reaction by 2, we obtain the H
2 NOCl(g)  N 2 (g)  O2 (g)  Cl 2 (g)
°
rxn
H rxn  –51.7 kJ
for the desired overall equation:
°
H rxn
 –103 kJ
Think about It
This reaction, because it is exothermic, releases energy in the decomposition of NOCl into its
constituent elements.