11.4 HESS' LAW

11.4 HESS’ LAW
Practice
(Pages 504-505)
3
1. 2 Al(s) + O
2
(g) Al 2
O 3
(s) f H = 1675.7 kJ
2
3
Fe2O 3(s) 2 Fe(s) + O
2
(g) d H
= +824.2 kJ
2
Fe2O 3(s) + 2 Al(s) Al 2
O 3
(s) + 2 Fe(s) r H = 851.5 kJ
1
2. C(s) + O
2(g) CO(g) f H = 110.5 kJ
2
1
HO(g)
2
H(g) 2
+ O(g) 2
d H
= +241.8 kJ
2
H2O(g) + C(s) CO(g) + H
2
(g) r H
= +131.3 kJ
1
3. CO(g) C(s) + O
2(g) d H
= +110.5 kJ
C(s) + O
2(g) 2
1
H
2(g) + O
2(g) 2
2
2
CO (g) f H = 393.5 kJ
HO(g) f H
= -241.8 kJ
CO(g) + H
2(g) + O
2(g) CO 2
(g) + H 2
O(g) r H
= -524.8 kJ
1
4. CO(g) C(s) + O
2(g) d H
= +110.5 kJ
2
CO
2(g) + 2 H2O(g) CH 4
(g) + 2 O 2
(g) r H
= +802.7 kJ
C(s) + O
2(g) CO 2
( g)
f H
= -393.5 kJ
3
3 H
2(g) + O
2(g)
3 H 2
O(g) f H
= -725.4 kJ
2
3 H
2(g) + CO(g) CH 4
(g) + H 2
O(g) r H
= -205.7 kJ
Web Activity: SimulationHess’ Law
(Page 506)
[No written response is required.]
Lab Exercise 11.B: Testing Hess’ Law
(Page 506)
Purpose
The purpose of this investigation is to test Hess’ law.
Problem
What is the standard molar enthalpy of combustion of pentane?
Prediction
According to Hess’ law and the data given, the standard molar enthalpy of combustion of pentane
is –3 488.7 kJ/mol.
430 Unit 6 Solutions Manual Copyright © 2007 Thomson Nelson

C 5 H 12 (l) 5 C(s) + 6 H 2 (g) d H
= +173.5 kJ
5 C(s) + 5 O 2 (g) 5 CO 2 (g) f H
= -1 967.5 kJ
6 H 2 (g) + 3 O 2 (g) 6 H 2 O(g) f H
= -1 450.8 kJ
6 H 2 O(g) 6 H 2 O(l) con H
= -243.9 kJ
C 5 H 12 (l) + 8 O 2 (g) 5 CO 2 (g) + 6 H 2 O(l) c H = 3 488.7 kJ
3 488.7 kJ
H = = 3 488.7 kJ/mol
1 mol
c m
CH 5 12
Analysis
Q
mct
1.24 kg
4.19 J
g•°C
(37.6 18.4) °C
= 99.8 kJ
99.8 kJ
cH
m
CH
1mol
5 12
2.15g
72.17g
3.35MJ/mol
According to the evidence collected in this experiment and the law of conservation of energy, the
molar enthalpy of combustion of pentane is reported as –3.35 MJ/mol.
Evaluation
The experimental design is adequate. Calorimetry is an obvious way to test a prediction from
Hess’ law. The major uncertainties related to the design are due to measurement uncertainties and
the difference in conditions for obtaining the predicted and experimental answers. The predicted
value corresponds to standard ambient temperature and pressure (SATP) conditions, whereas the
experimental value does not. However, I feel confident enough in the evidence to use it to test the
prediction.
The percent difference of this experiment is 4.0%, as shown by the following calculation:
3.35 MJ/mol 3.4887 MJ/mol
% difference = 100 = 4.0%
3.4887 MJ/mol
The prediction is judged to be verified. The percent difference is reasonably low and is
likely due to the uncertainties mentioned. Hess’ law appears to be acceptable since the prediction
was verified.
The purpose of this investigation was accomplished for only one example, which is not
sufficient to test Hess’ law. Additional investigations, including a variety of chemical reactions,
are required to better achieve the purpose.
Lab Exercise 11.C: Analysis Using Hess’ Law
(Page 506)
Purpose
The purpose of this exercise is to use Hess’ law to determine an enthalpy change.
Problem
What is the standard enthalpy change for the production of hydrogen from methane and steam?
CH 4 (g) + H 2 O(g) CO(g) + 3 H 2 (g) r H = ?
Copyright © 2007 Thomson Nelson Unit 6 Solutions Manual 431

Analysis
[Many solutions are possible. Two are given below. Encourage students to find other solutions to
emphasize this point when using Hess’ law: the net enthalpy change is independent of the route.]
(from 2) CH 4 (g) + 2 O 2 (g) CO 2 (g) + 2 H 2 O(g) c H = -802.5 kJ
(from 3) CO 2 (g) + H 2 (g) CO(g) + H 2 O(g) r H = +41.2 kJ
(from 4) 4 H 2 O(g) 4 H 2 (g) + 2 O 2 (g) d H
= +967.2 kJ
CH 4 (g) + H 2 O(g) CO(g) + 3 H 2 (g) r H
= +205.9 kJ
According to the evidence from equations 2, 3, and 4 and Hess’ law, the standard enthalpy
change for the production of hydrogen from methane and steam is +205.9 kJ.
1
(from 1) C(s) + O
2(g) CO(g) c H = -110.5 kJ
2
1
(from 4) H 2 O(g) H 2
(g) + O 2
(g) d H
= +241.8 kJ
2
(from 5) CH 4 (g) C(s) + 2 H 2 (g) d H
= +74.6 kJ
CH 4 (g) + H 2 O(g) CO(g) + 3 H 2 (g) r H
= +205.9 kJ
According to the evidence from equations 1, 4, and 5 and Hess’ law, the standard enthalpy
change for the production of hydrogen from methane and steam is +205.9 kJ.
[Other solutions include reactions using equations 5 and 6, equations 2, 7, and 8, equations 1, 5,
7, and 8, equations 2, 4, and 8, and equations 2, 4, and 7.]
Investigation 11.3: Applying Hess’ Law
(Pages 507, 517)
Purpose
The purpose of this investigation is to use Hess’ law to determine a molar enthalpy of
combustion.
Problem
What is the standard molar enthalpy of combustion for magnesium?
Procedure
1. Measure precisely about 1.00 g of magnesium oxide in a clean, dry calorimeter.
2. Obtain 50.0 mL of 1.00 mol/L HCl(aq) in a 50 mL graduated cylinder.
3. Measure the initial temperature of the HCl(aq).
4. Add the acid to the magnesium oxide and quickly cover the calorimeter.
5. Stir and record the maximum temperature of the mixture.
6. Dispose of the contents in the sink, rinse and dry the calorimeter cup.
7. Obtain about 15 cm of magnesium ribbon and clean it with steel wool.
8. Measure the mass of the clean magnesium ribbon.
9. Place 50.0 mL of HCl(aq) into a 50 mL graduated cylinder.
10. Place the acid into the calorimeter and record the initial temperature.
11. Add the magnesium ribbon to the calorimeter, making sure all of the metal is in the acid.
12. Cover and stir; record the maximum temperature.
13. Dispose of the contents in the sink.
432 Unit 6 Solutions Manual Copyright © 2007 Thomson Nelson

Evidence
Reacting MgO(s) and Mg(s) with HCl(aq)
Reactant Mass (g) Initial temperature of
HCl(aq) (°C)
Final temperature of
mixture (°C)
magnesium oxide 0.91 23.6 36.0
magnesium metal 0.16 23.6 37.7
50.0 mL of 1.00 mol/L HCl(aq) were used for each reaction.
Analysis
Q
(calorimeter)
mct
50.0g
4.19J
g•°C
(36.0 23.6) °C
2.60 kJ
2.60 kJ
rH
m
MgO
1mol
0.91g
40.31g
0.12MJ/mol
According to the evidence, the molar enthalpy of reaction for magnesium oxide is reported as
–0.12 MJ/mol.
Q mct
(calorimeter)
4.19J
50.0g (37.7 23.6) °C
g•°C
2.95 kJ
2.95 kJ
rH
m
Mg
1mol
0.16g
24.31g
0.45MJ/mol
According to the evidence, the molar enthalpy of reaction for magnesium is reported as
–0.45 MJ/mol.
According to Hess’ law, the three equations can be combined as follows.
MgCl 2 (aq) + H 2 O(l) MgO(s) + 2 HCl(aq)
Mg(s) + 2 HCl(aq) MgCl 2 (aq) + H 2 (g)
1
H
2(g) + O
2(g) H2O(l)
2
r H = 1 mo × +0.12 MJ/mol = +0.12 MJ
r H = 1 mol × (0.45 MJ/mol) = 0.45 MJ
r H = 0.2858 MJ
1
Mg(s) + O
2
(g) MgO(s)
c H = 0.62 MJ
2
0.62 kJ
Therefore, cH
m
= 0.62 MJ/mol
Mg 1mol
According to the evidence collected and Hess’ law, the molar enthalpy of combustion for
magnesium is 0.62 MJ/mol.
Evaluation
[Parts 1, 2 and 3 of the Evaluation should have been checked in the Report Checklist.]
The design using a simple polystyrene calorimeter is judged to be adequate because the
problem was answered with no obvious flaws. A more sophisticated calorimeter could be used as
Copyright © 2007 Thomson Nelson Unit 6 Solutions Manual 433

an alternative. However, the design used in this experiment is simpler and less expensive. The
materials were adequate for the chosen design, although a more accurate balance (e.g., milligram
balance) would improve the overall certainty to three significant digits. The procedure was
adequate, although additional trials would improve the certainty of the answer. The technological
skills were simple and adequate.
Overall, I am moderately certain of the results. Sources of uncertainty include limitations
of the measurements made, heat losses by the calorimeter, and the non-SATP conditions under
which the results were obtained. I would estimate a 5% total for these uncertainties.
The accuracy of the results obtained is reflected by the percent difference shown.
0.62 MJ/mol - 0.6016 MJ/mol
% difference = 100 = 3.0%
0.6016 MJ/mol
The prediction is judged to be verified as supported by the low (3.0%) percent difference.
Hess’ law is judged to be acceptable because the prediction was verified.
The purpose of this investigation has been met. Hess’ law passed a severe test—being
able to predict. Further tests of Hess’ law in other contexts need to be done if the law is to gain
even greater acceptance.
Explore an Issue: Alternative Energy Sources and Technologies
(Pages 507–508)
Issue
What is the most effective use of government funding for research in energy technology?
Resolution
The provincial and federal governments should direct all their research funding to bio-based
energy technology.
Design
Within small groups, research the pros and cons of using public money to fund bio-based versus
fossil-fuel-based energy technology.
Evidence
Pros
Bio-based energy technology:
is cleaner
is renewable
is sustainable
may create more jobs
may improve our quality of life
will likely diversify our economy
will benefit farmers growing grain
will benefit owners of wood lots
may improve our environment
preserves fossil fuels for future generations
Cons
Bio-based energy technology:
would be costly to research and develop
may not make use of readily available energy
resources, including other renewable energy
sources
will likely not fully meet all energy demands
lacks the infrastructure for distribution
will likely produce an expensive bio-fuel
may increase the cost and availability of food
will bring oil and gas companies less money
may mean that oil and gas investors lose
money
Analysis
On the basis of the evidence collected and discussion within the group, there is no clear
consensus for or against the resolution. There was a general agreement that the resolution should
not say “all” and would be better changed to “significant” or “increased”.
Evaluation
The design is adequate, but some improvements could include a voting system within the group,
assigning ratings to distinguish more important from less important perspectives, and collating
the results of several groups.
434 Unit 6 Solutions Manual Copyright © 2007 Thomson Nelson

Section 11.4 Questions
(Pages 508–509)
1. C 6 H 12 O 6 (s) 2 C 2 H 5 OH(l) + 2 CO 2 (g) r H = ?
C 6 H 12 O 6 (s) + 6 O 2 (g) 6 CO 2 (g) + 6 H 2 O(l)
c H = –2 803.1 kJ
4 CO 2 (g) + 6 H 2 O(l) 2 C 2 H 5 OH(l) + 6 O 2 (g) r H = +2 733.6 kJ
net: C 6 H 12 O 6 (s) 2 CO 2 (g) + 2 C 2 H 5 OH(l)
r H = 69.5 kJ
2. (a) C 2 H 6 (g) C 2 H 4 (g) + H 2 (g) r H = ?
C 2 H 6 (g) + 7 2 O 2(g) 2 CO 2 (g) + 3 H 2 O(l)
r H = 1 560.4 kJ
2 CO 2 (g) + 2 H 2 O(l) C 2 H 4 (g) + 3 O 2 (g) r H = +1 411.0 kJ
H 2 O(l) H 2 (g) + 1 2 O 2(g)
r H = +285.8 kJ
net: C 2 H 6 (g) C 2 H 4 (g) + H 2 (g)
r H = +136.4 kJ
(b) C 2 H 4 (g) + H 2 O(l) C 2 H 5 OH(l) r H = ?
C 2 H 4 (g) + 3 O 2 (g) 2 CO 2 (g) + 2 H 2 O(l)
r H = 1 411.0 kJ
2 CO 2 (g) + 3 H 2 O(l) C 2 H 5 OH(l) + 3 O 2 (g) r H = +1366.8 kJ
net: C 2 H 4 (g) + H 2 O(l) C 2 H 5 OH(l)
r H = 44.2 kJ
(c) This technique works because the addition of any chemical equations that yield the
intended net chemical reaction will provide the net enthalpy charge. All of the reaction
paths start and end with the same chemicals and the same chemical potential energy (and
thus the same change in enthalpy).
3. CH 4 (g) + H 2 O(l) CH 3 OH(l) + H 2 (g) r H = ?
CH 4 (g) + H 2 O(l) CO(g) + 3 H 2 (g)
r H = +249.9 kJ
CO(g) + 2 H 2 (g) CH 3 OH(l)
r H = 128.7 kJ
net: CH 4 (g) + H 2 O(l) CH 3 OH(l) + H 2 (g)
r H = +121.2 kJ
121.2 kJ
Therefore, rH
m
= +121.2 kJ/mol
CH 1mol
3OH
4.
Question Fuel (a)
Net synthesis
enthalpy
change (kJ)
(b)
Molar enthalpy
of combustion
(kJ/mol)
(c)
Molar enthalpy
of synthesis
(kJ/mol)
(d)
Net molar
enthalpy
(kJ/mol)
1 ethanol from glucose –1 366.8 69.5 1 436.3
69.5
(exothermic)
2 ethanol from ethane –1 366.8 +92.2 1 274.6
+92.2
(endothermic)
(from 2(a) & (b))
3 methanol from methane
+121.2
(endothermic)
–725.9 +121.2 604.7
(c) In most cases, the production of these fuels is slightly endothermic, whereas the
combustion is always very exothermic. Therefore, on the basis of this information the net
process is very exothermic.
(e) On the basis of this information, producing both bio-fuels makes sense from an energy
perspective. Ethanol from glucose makes the most sense, as it releases the most net
energy per mole.
5. The supply of trees for methanol and grain for ethanol also affects the choice. Other costs
incurred by this process include the harvesting and processing of the original starting
Copyright © 2007 Thomson Nelson Unit 6 Solutions Manual 435

materials, and the production and maintenance of transportation and manufacturing
equipment. Also, the calculations above do not take into account the expense of increasing
the rates of reaction by increasing the temperature of the reaction system.
6. Some technological solutions would include: more fuel-efficient vehicles, better public
transportation systems, more and better recycling technologies, more energy-efficient
housing, and technologies to improve the energy efficiency of industries (especially those
industries that are producing energy-related products).
7. C 2 H 2 (g) + 2 H 2 (g) C 2 H 6 (g) r H = ?
C 2 H 2 (g) + 5 2 O 2(g) 2 CO 2 (g) + H 2 O(l)
r H° = 1299 kJ
2 H 2 (g) + O 2 (g) 2 H 2 O(l) r H° = 572 kJ
2 CO 2 (g) + 3 H 2 O(l) C 2 H 6 (g) + 7 2 O 2(g) r H° = +1560 kJ
net: C 2 H 2 (g) + 2 H 2 (g) C 2 H 6 (g)
r H° = 311 kJ
1mol 311kJ
rH
200g
CH
26.04g 1 mol
2 2
2.39 MJ
1 mol 205.7 kJ
rH
300g
8. CO
28.01g 1 mol
2.20 MJ
9. 8 C(s) + 9 H 2 (g) C 8 H 18 f H =?
8 C(s) + 8 O 2 (g) 8 CO 2 (g) r H = –3148 kJ
9 H 2 (g) + 9 2 O 2(g) 9 H 2 O(g) r H = –2572.2 kJ
8 CO 2 (g) + 9 H 2 O(g) C 8 H 18 (l) + 25 O 2(g) c H = +5470.1 kJ
2
net: 8 C(s) + 9 H 2 (g) C 8 H 18 (l)
f H = 250.1 kJ
10. The reference equations are:
(1) C 2 H 5 OH(l) + 3 O 2 (g) 2 CO 2 (g) + 3 H 2 O(l) 1 H = 1367 kJ
(2) CH 3 COOH(l) + 2 O 2 (g) 2 CO 2 (g) + 2 H 2 O(l) 2 H = 875 kJ
Applying Hess’ Law:
C 2 H 5 OH(l) + 3 O 2 (g) 2 CO 2 (g) + 3 H 2 O(l)
r H = 1367 kJ
2 CO 2 (g) + 2 H 2 O(l) CH 3 COOH(l) + 2 O 2 (g) r H = +875 kJ
net: C 2 H 5 OH(l) + O 2 (g) CH 3 COOH(l) + H 2 O(l)
r H = 492 kJ
11. Purpose
The purpose of this investigation is to use Hess’ law to determine a molar enthalpy of
combustion.
Problem
What is the enthalpy change for the reaction of aqueous potassium hydroxide with aqueous
hydrobromic acid?
Prediction
According to Hess’ law, the enthalpy change of the direct reaction of aqueous potassium
hydroxide and aqueous hydrobromic acid should equal the sum of the enthalpy changes of the
two other reactions that produce the net reaction equation for aqueous potassium hydroxide
and aqueous hydrobromic acid.
The enthalpy change required to answer the problem is the first given equation:
436 Unit 6 Solutions Manual Copyright © 2007 Thomson Nelson

HBr(aq) + KOH(aq) H 2 O(l) + KBr(aq) 1 H = ?
This chemical equation can be obtained by adding the third given equation to the reverse of
the second given equation:
KOH(s) + HBr(aq) H 2 O(l) + KBr(aq) 3 H = ?
KOH(aq) KOH(s) 2 H = ?
Therefore, on the basis of Hess’ law, 1 H = 3 H + ( 2 H).
Analysis
Experiment 1:
HBr(aq) + KOH(aq) H 2 O(l) + KBr(aq) 1 H = ?
Q mct
calorimeter
H
n m
HBr
200.0 g
4.19 J
g• C
(22.5 20.0) C
2.10 kJ
2.10 kJ
1.00 mol
0.100L
1 L
21.0 kJ/mol
According to the evidence and the reaction of one mole of HBr(aq):
HBr(aq) + KOH(aq) H 2 O(l) + KBr(aq)
1 H = 21.0 kJ
Experiment 2: [in 200 mL of solution]
KOH(s) KOH(aq) 2 H = ?
Q mct
calorimeter
200.0g
4.19J
g•°C
(24.120.0) °C
3.44 kJ
3.44 kJ
rH
m
KOH
1mol
5.61g
56.11g
34.4 kJ/mol
According to the evidence and the dissolving of one mole of KOH(aq):
KOH(s) KOH(aq)
2 H = 34.4 kJ
Experiment 3:
KOH(s) + HBr(aq) H 2 O(l) + KBr(aq) 3 H = ?
Copyright © 2007 Thomson Nelson Unit 6 Solutions Manual 437

Q
calorimeter
mct
4.19J
200.0g (26.7 20.0) °C
g•°C
5.61 kJ
5.61 kJ
rH
m
KOH
1mol
5.61g
56.11g
56.2 kJ/mol
According to the evidence and the reaction of one mole of KOH(s):
KOH(s) + HBr(aq) H 2 O(l) + KBr(aq)
On the basis of the results of the three experiments and Hess’ law,
KOH(s) + HBr(aq) H 2 O(l) + KBr(aq)
KOH(aq) KOH(s)
net: HBr(aq) + KOH(aq) H 2 O(l) + KBr(aq)
3 H = 56.2 kJ
3 H = 56.2 kJ
2 H = (34.4 kJ)
net H = 21.8 kJ
Evaluation
The experimental design is based on Hess’ law, and provides the evidence necessary to
provide an analytical answer against which to test the prediction. The Materials, Procedure
and skills are not provided and/or not witnessed, so they cannot be judged here. On the basis
of the experimental design and the evidence provided, it appears that one can be confident
enough to use this evidence and the Analysis above to judge Hess’ law.
The percent difference between the experimental result of measuring the
neutralization of aqueous potassium hydroxide and aqueous hydrobromic acid directly, and
the experimental results for the same net reaction obtained from calorimetry for two different
reactions and Hess’ law is as follows:
21.8 kJ ( 21.0 kJ)
% difference 100 4%
21.0 kJ
This percent difference is relatively low and clearly within expected experimental
uncertainties. Therefore, the prediction is verified and Hess’ law has passed the test in this
investigation. I am reasonably confident in this judgment.
The purpose of the investigation has been achieved—Hess’ law was tested.
11.5 MOLAR ENTHALPIES OF FORMATION
Lab Exercise 11.D: Testing r H° from Formation Data
(Page 513)
[Only Evaluation Parts 2 and 3 can be completed by students.]
Purpose
The purpose of this problem is to test the use of molar enthalpies of formation as a method of
predicting the enthalpy change of a reaction.
Problem
What is the standard molar enthalpy of combustion of methanol?
438 Unit 6 Solutions Manual Copyright © 2007 Thomson Nelson