1. Chapter 6 Thermochemistry Study Guide
1. A gas absorbs 0.0 J of heat and then performs 30.7 J of work.
The change in internal energy of the gas is
A)
61.4 J
B)
30.7 J
C)
–61.4 J
D)
–30.7 J
E)
none of these
ANS: D
2. What is the kinetic energy of a 1.56-kg object moving at
94.0 km/hr?
A)
B)
C)
–4 kJ
D)
E)
ANS: A
3. Which of the following statements correctly describes the
signs of q and w for the following exothermic process at P = 1
atm and T = 370 K?

2. A)
q and w are negative.
B)
q is positive, w is negative.
C)
q is negative, w is positive.
D)
q and w are both positive.
E)
q and w are both zero.
ANS: C
4. Which of the following statements is correct?
A)
The internal energy of a system increases when more work is
done by the system than heat was flowing into the system.
B)
The internal energy of a system decreases when work is done on
the system and heat is flowing into the system.
C)
The system does work on the surroundings when an ideal gas
expands against a constant external pressure.
D)
All statements are true.
E)
All statements are false.
ANS: C
5. For a particular process q = –17 kJ and w = 21 kJ. Which of
the following statements is false?
A)
Heat flows from the system to the surroundings.
B)

3. The system does work on the surroundings.
C)
D)
The process is exothermic.
E)
None of the above is false.
ANS: B
6. One mole of an ideal gas is expanded from a volume of 1.00
liter to a volume of 8.93 liters against a constant external
pressure of 1.00 atm. How much work (in joules) is performed
on the surroundings? Ignore significant figures for this problem.
(T = 300 K; 1 L·atm = 101.3 J)
A)
402 J
B)
803 J
C)
D)
905 J
E)
none of these
ANS: B
7. A fuel-air mixture is placed in a cylinder fitted with a piston.
The original volume is 0.310-L. When the mixture is ignited,
gases are produced and 935 J of energy is released. To what
volume will the gases expand against a constant pressure of 635
mmHg, if all the energy released is converted to work to push
the piston?
A)
10.7 L

4. B)
8.02 L
C)
11.4 L
D)
11.0 L
E)
1.78 L
ANS: C
8. Calculate the work associated with the compression of a gas
from 121.0 L to 80.0 L at a constant pressure of 13.1 atm.
A)
–537 L atm
B)
537 L atm
C)
3.13 L atm
D)
–3.13 L atm
E)
101 L atm
ANS: B
9. A gas absorbs 825 J of heat and then has 841 J of work done
upon it. The change in internal energy of the gas is
A)
1666 J
B)
16 J
C)
–16 J
D)
–1666 J

5. E)
none of these
ANS: A
10. Of energy, work, enthalpy, and heat, how many are state
functions?
A)
0
B)
1
C)
2
D)
3
E)
4
ANS: C
11. Which of the following properties is (are) intensive
properties?
I. mass
II. temperature
III. volume
IV. concentration
V. energy
A)
I, III, and V
B)
II only
C)
II and IV
D)
III and IV

6. E)
I and V
ANS: C
can be considered to be
A)
the heat flow required to maintain a constant temperature
B)
the work done in pushing back the atmosphere
C)
the difference in the H–O bond energy in H2O(l) compared to
H2O(g)
D)
E)
none of these
ANS: B
13. Consider the reaction:
C2H5OH(l) + –
Consider the following propositions:
I. The reaction is endothermic
II. The reaction is exothermic.
III. The enthalpy term would be different if the water formed
was gaseous.
Which of these propositions is (are) true?
A)
I
B)
II
C)

7. III
D)
I, II
E)
II, III
ANS: E
14. How much heat is required to raise the temperature of a
5.75-g sample of iron (specific heat = 0.450 J/g°C) from 25.0°C
to 79.8°C?
A)
2.54 J
B)
315 J
C)
700 J
D)
848 J
E)
142 J
ANS: E
15. A 32.5 g piece of aluminum (which has a molar heat
capacity of 24.03 J/°C·mol) is heated to 82.4°C and dropped
into a calorimeter containing water (specific heat capacity of
water is 4.18 J/g°C) initially at 22.3°C. The final temperature of
the water is 24.2°C. Ignoring significant figures, calculate the
mass of water in the calorimeter.
A)
212 g
B)
5.72 kg
C)
6.42 g

8. D)
1.68 kg
E)
none of these
ANS: A
16. A 45.9 g sample of a metal is heated to 95.2°C and then
placed in a calorimeter containing 120.0 g of water (c = 4.18
J/g°C) at 21.6°C. The final temperature of the water is 24.5°C.
Which metal was used?
A)
Aluminum (c = 0.89 J/g°C)
B)
Iron (c = 0.45 J/g°C)
C)
Copper (c = 0.20 J/g°C)
D)
Lead (c = 0.14 J/g°C)
E)
none of these
ANS: B
17. You take 295.5 g of a solid at 30.0°C and let it melt in 425
g of water. The water temperature decreases from 85.1°C to
30.0°C. Calculate the heat of fusion of this solid.
A)
160 J/g
B)
166 J/g
C)
331 J/g
D)
721 J/g
E)

9. cannot solve without the heat capacity of the solid
ANS: C
18. 30.0 mL of pure water at 282 K is mixed with 50.0 mL of
pure water at 306 K. What is the final temperature of the
mixture?
A)
294 K
B)
297 K
C)
342 K
D)
588 K
E)
24 K
ANS: B
19. Consider the reaction
= –286 kJ
Which of the following is true?
A)
The reaction is exothermic.
B)
The reaction is endothermic.
C)
The enthalpy of the products is less than that of the reactants.
D)
Heat is absorbed by the system.
E)
Both A and C are true.
ANS: E

10. 20. What is the specific heat capacity of a metal if it requires
178.1 J to change the temperature of 15.0 g of the metal from
25.00°C to 32.00°C?
A)
0.590 J/g°C
B)
11.9 J/g°C
C)
25.4 J/g°C
D)
1.70 J/g°C
E)
283 J/g°C
ANS: D
21. A 140.0-g sample of water at 25.0°C is mixed with 111.7 g
of a certain metal at 100.0°C. After thermal equilibrium is
established, the (final) temperature of the mixture is 29.6°C.
What is the specific heat capacity of the metal, assuming it is
constant over the temperature range concerned?
A)
0.34 J/g°C
B)
0.68 J/g°C
C)
0.22 J/g°C
D)
2.9 J/g°C
E)
none of these
ANS: A
22. If 5.0 kJ of energy is added to a 15.5-g sample of water at
10.°C, the water is

11. A)
boiling
B)
completely vaporized
C)
frozen solid
D)
decomposed
E)
still a liquid
ANS: E
23. A chunk of lead at 91.6°C was added to 200.0 g of water at
15.5°C. The specific heat of lead is 0.129 J/g°C, and the
specific heat of water is 4.18 J/g°C. When the temperature
stabilized, the temperature of the mixture was 17.9°C.
Assuming no heat was lost to the surroundings, what was the
mass of lead added?
A)
1.57 kg
B)
170 g
C)
204 g
D)
211 g
E)
none of these
ANS: D
24. What is the specific heat capacity of silver if it requires
86.3 J to raise the temperature of 15 grams of silver by 25°C?
A)
4.3 J/g°C

12. B)
0.23 J/g°C
C)
0.14 J/g°C
D)
0.60 J/g°C
E)
none of these
ANS: B
25. If a student performs an endothermic reaction in a
the actual value if the heat exchanged with the calorimeter is
not taken into account?
A)
absorbs heat from the reaction.
B)
absorb heat from the reaction.
C)
reaction absorbs
heat from the calorimeter.
D)
from the calorimeter.
E)
does not absorb heat.
ANS: D
26. Consider the reaction:
When a 21.1-g sample of ethyl alcohol (molar mass = 46.07

13. g/mol) is burned, how much energy is released as heat?
A)
0.458 kJ
B)
0.627 kJ
C)
D)
E)
2.18 kJ
ANS: C
27. The total volume of hydrogen gas needed to fill the
Hindenburg was 2.11 108 L at 1.00 atm and 24.7°C. How
much energy was evolved when it burned?
A)
B)
C)
D)
E)
ANS: D
28. What is the enthalpy change when 49.4 mL of 0.430 M
sulfuric acid reacts with 23.3 mL of 0.309 M potassium
hydroxide?
–111.6 kJ/mol

14. A)
–0.402 kJ
B)
–3.17 kJ
C)
–2.37 kJ
D)
–0.803 kJ
E)
–112 kJ
ANS: A
29. How much heat is liberated at constant pressure when 2.35 g
of potassium metal reacts with 5.68 mL of liquid iodine
monochloride (d = 3.24 g/mL)?
–740.71 kJ/mol
A)
B)
C)
D)
E)
ANS: D
30. Which of the following statements is/are true?

15. II. When 50.0 g of aluminum at 20.0°C is placed in 50.0 mL of
water at 30.0°C, the H2O will undergo a smaller temperature
change than the aluminum. (The density of H2O = 1.0 g/mL,
specific heat capacity of H2O = 4.18 J/g°C, specific heat
capacity of aluminum = 0.89 J/g°C)
III. When a gas is compressed, the work is negative since the
surroundings are doing work on the system and energy flows out
of the system.
IV. For the reaction (at constant pressure)
2N2(g) + 5O2(g) 2N2O5(g), the change in enthalpy is the
same whether the reaction takes place in one step or in a series
of steps.
A)
I, II, IV
B)
II, III
C)
II, III, IV
D)
II, IV
E)
All of the above statements are true.
ANS: D
31. Consider the following processes:
–15 kJ/mol
A)
0 kJ/mol

16. B)
10 kJ/mol
C)
–10 kJ/mol
D)
–20 kJ/mol
E)
20 kJ/mol
ANS: C
32. At 25°C, the following heats of reaction are known:
167.4
341.4
–43.4
A)
–217.5 kJ/mol
B)
–130.2 kJ/mol
C)
+217.5 kJ/mol
D)
–108.7 kJ/mol
E)
none of these
ANS: D
33. Given the heats of the following reactions:

17. I.
–1225.6
II.
–2967.3
III.
–84.2
IV.
PCl3(g)
–285.7
A)
–110.5 kJ
B)
–610.1 kJ
C)
–2682.2 kJ
D)
–7555.0 kJ
E)
None of these is within 5% of the correct answer.
ANS: B
34. One of the main advantages of hydrogen as a fuel is that:
A)
The only product of hydrogen combustion is water.
B)
It exists as a free gas.
C)

18. It can be economically supplied by the world's oceans.
D)
Plants can economically produce the hydrogen needed.
E)
It contains a large amount of energy per unit volume of
hydrogen gas.
ANS: A
35. Consider the following standard heats of formation:
P4O10(s) = –3110 kJ/mol
H2O(l) = –286 kJ/mol
H3PO4(s) = –1279 kJ/mol
Calculate the change in enthalpy for the following process:
ANS:
–290 kJ
36. For the complete combustion of 1.000 mole of ethane gas at
-1560 kJ/mol. What will be
the heat released when 4.42 g of ethane is combusted under
these conditions?
A)
–230 kJ
B)
230 kJ
C)
10588 kJ
D)
–10588 kJ
E)
none of these
ANS: B
37. For the complete combustion of 1.000 mole of propane gas

19. -2220 kJ/mol. What will be
the heat released when 4.13 g of propane is combusted under
these conditions?
A)
–208 kJ
B)
208 kJ
C)
23651 kJ
D)
–23651 kJ
E)
none of these
ANS: B
38. A 36.2 g piece of metal is heated to 81°C and dropped into a
calorimeter containing 50.0 g of water (specific heat capacity
of water is 4.18 J/g°C) initially at 21.7°C. The empty
calorimeter has a heat capacity of 125 J/K. The final
temperature of the water is 29.7°C. Ignoring significant figures,
calculate the specific heat of the metal..
A)
1.439 J/gK
B)
0.900 J/gK
C)
0.360 J/gK
D)
0.968 J/gK
E)
none of these
ANS: A

20. 1
Chapter 6
Thermochemistry
Section 6.1
The Nature of Energy
Copyright © Cengage Learning. All rights reserved
*Capacity to do work or to produce heat.Law of conservation of
energy – energy can be converted from one form to another but
can be neither created nor destroyed.The total energy content of
the universe is constant.
Energy
*
Section 6.1
The Nature of Energy
Copyright © Cengage Learning. All rights reserved
*Potential energy – energy due to position or
composition.Kinetic energy – energy due to motion of the object
and depends on the mass of the object and its velocity.
Energy

21. *
Section 6.1
The Nature of Energy
Copyright © Cengage Learning. All rights reserved
*In the initial position, ball A has a higher potential energy than
ball B.
Initial Position
*
Section 6.1
The Nature of Energy
Copyright © Cengage Learning. All rights reserved
*After A has rolled down the hill, the potential energy lost by A
has been converted to random motions of the components of the
hill (frictional heating) and to the increase in the potential
energy of B.
Final Position
*

22. Section 6.1
The Nature of Energy
Copyright © Cengage Learning. All rights reserved
*Heat involves the transfer of energy between two objects due
to a temperature difference.Work – force acting over a
distance.Energy is a state function; work and heat are notState
Function – property that does not depend in any way on the
system’s past or future (only depends on present state).
Energy
*
Section 6.1
The Nature of Energy
Copyright © Cengage Learning. All rights reserved
*System – part of the universe on which we wish to focus
attention.Surroundings – include everything else in the
universe.
Chemical Energy
*
Section 6.1
The Nature of Energy
Copyright © Cengage Learning. All rights reserved

23. *Endothermic Reaction:Heat flow is into a system.Absorb
energy from the surroundings.Exothermic Reaction:Energy
flows out of the system.Energy gained by the surroundings must
be equal to the energy lost by the system.
Chemical Energy
*
Section 6.1
The Nature of Energy
Copyright © Cengage Learning. All rights reserved
*
Is the freezing of water an endothermic or exothermic process?
Explain.
CONCEPT CHECK!
*
Exothermic process because you must remove energy in order to
slow the molecules down to form a solid.
Section 6.1
The Nature of Energy
Copyright © Cengage Learning. All rights reserved
*
Classify each process as exothermic or endothermic. Explain.
The system is underlined in each example.
Your hand gets cold when you touch ice.

24. The ice gets warmer when you touch it.
Water boils in a kettle being heated on a stove.
Water vapor condenses on a cold pipe.
Ice cream melts.
Exo
Endo
Endo
Exo
Endo
CONCEPT CHECK!
* Exothermic (heat energy leaves your hand and moves to the
ice) Endothermic (heat energy flows into the ice) Endothermic
(heat energy flows into the water to boil it) Exothermic (heat
energy leaves to condense the water from a gas to a liquid)
Endothermic (heat energy flows into the ice cream to melt it)
Section 6.1
The Nature of Energy
Copyright © Cengage Learning. All rights reserved
*
For each of the following, define a system and its surroundings
and give the direction of energy transfer.
Methane is burning in a Bunsen burner in a laboratory.
Water drops, sitting on your skin after swimming, evaporate.
CONCEPT CHECK!
* System – methane and oxygen to produce carbon dioxide and
water; Surroundings – everything else around it; Direction of
energy transfer – energy transfers from the system to the

25. surroundings (exothermic) System – water drops; Surroundings
– everything else around it; Direction of energy transfer –
energy transfers from the surroundings (your body) to the
system (water drops) (endothermic)
Section 6.1
The Nature of Energy
Copyright © Cengage Learning. All rights reserved
*
Hydrogen gas and oxygen gas react violently to form water.
Explain.
Which is lower in energy: a mixture of hydrogen and oxygen
gases, or water?
CONCEPT CHECK!
*
Water is lower in energy because a lot of energy was released in
the process when hydrogen and oxygen gases reacted.
Section 6.1
The Nature of Energy
ThermodynamicsThe study of energy and its interconversions is
called thermodynamics.Law of conservation of energy is often
called the first law of thermodynamics.
Copyright © Cengage Learning. All rights reserved
*
Section 6.1
The Nature of Energy

26. Copyright © Cengage Learning. All rights reserved
*
Internal energy E of a system is the sum of the kinetic and
potential energies of all the “particles” in the system.To change
the internal energy of a system: ΔE = q + w
q represents heat
w represents work
Internal Energy
*
Section 6.1
The Nature of Energy
Work vs Energy Flow
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*
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Section 6.1
The Nature of Energy
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*Thermodynamic quantities consist of two parts:Number gives
the magnitude of the change.Sign indicates the direction of the
flow.
Internal Energy

27. *
Section 6.1
The Nature of Energy
Internal EnergySign reflects the system’s point of
view.Endothermic Process:q is positiveExothermic Process:q is
negative
Copyright © Cengage Learning. All rights reserved
*
Section 6.1
The Nature of Energy
Copyright © Cengage Learning. All rights reserved
*Sign reflects the system’s point of view.System does work on
surroundings:w is negativeSurroundings do work on the
system:w is positive
Internal Energy
*
Section 6.1
The Nature of Energy
*Work = P × A × Δh = PΔVP is pressure.A is area.Δh is the
piston moving a distance.ΔV is the change in volume.

28. Work
*
Section 6.1
The Nature of Energy
Copyright © Cengage Learning. All rights reserved
*For an expanding gas, ΔV is a positive quantity because the
volume is increasing. Thus ΔV and w must have opposite signs:
w = –PΔV To convert between L·atm and Joules,
use 1 L·atm = 101.3 J.
Work
*
Section 6.1
The Nature of Energy
Copyright © Cengage Learning. All rights reserved
*
Which of the following performs more work?
a) A gas expanding against a pressure of 2 atm from 1.0
L to 4.0 L.
A gas expanding against a pressure of 3 atm from 1.0
L to 3.0 L.

29. They perform the same amount of work.
EXERCISE!
*
They both perform the same amount of work. w = -PΔV
a) w = -(2 atm)(4.0-1.0) = -6 L·atm
b) w = -(3 atm)(3.0-1.0) = -6 L·atm
Section 6.1
The Nature of Energy
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*
Determine the sign of ΔE for each of the following with the
listed conditions:
a) An endothermic process that performs work. |work| > |heat|
|work| < |heat|
b) Work is done on a gas and the process is exothermic.
|work| > |heat| |work| < |heat|
Δ E = negative
Δ E = positive
Δ E = positive
Δ E = negative
CONCEPT CHECK!
*
a) q is positive for endothermic processes and w is negative
when system does work on surroundings; first condition – ΔE is
negative; second condition – ΔE is positive
b) q is negative for exothermic processes and w is positive when
surroundings does work on system; first condition – ΔE is
positive; second condition – ΔE is negative

30. Section 6.2
Enthalpy and Calorimetry
Change in EnthalpyState functionΔH = q at constant pressureΔH
= Hproducts – Hreactants
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*
*
Section 6.2
Enthalpy and Calorimetry
Consider the combustion of propane:
C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(l)
ΔH = –2221 kJ
Assume that all of the heat comes from the combustion of
propane. Calculate ΔH in which 5.00 g of propane is burned in
excess oxygen at constant pressure.
–252 kJ
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*
EXERCISE!
*
(5.00 g C3H8)(1 mol / 44.094 g C3H8)(-2221 kJ / mol C3H8)
ΔH = -252 kJ

31. Section 6.2
Enthalpy and Calorimetry
CalorimetryScience of measuring heatSpecific heat
capacity:The energy required to raise the temperature of one
gram of a substance by one degree Celsius.Molar heat
capacity:The energy required to raise the temperature of one
mole of substance by one degree Celsius.
Copyright © Cengage Learning. All rights reserved
*
*
Section 6.2
Enthalpy and Calorimetry
CalorimetryIf two reactants at the same temperature are mixed
and the resulting solution gets warmer, this means the reaction
taking place is exothermic.An endothermic reaction cools the
solution.
Copyright © Cengage Learning. All rights reserved
*
*
Section 6.2
Enthalpy and Calorimetry

32. A Coffee–Cup Calorimeter Made of Two Styrofoam Cups
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*
*
Section 6.2
Enthalpy and Calorimetry
CalorimetryEnergy released (heat) = s × m × ΔT
s = specific heat capacity (J/°C·g)
m = mass of solution (g)
ΔT = change in temperature (°C)
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*
*
Section 6.2
Enthalpy and Calorimetry
A 100.0 g sample of water at 90°C is added to a 100.0 g sample
of water at 10°C.
The final temperature of the water is:
a) Between 50°C and 90°C

33. b) 50°C
c) Between 10°C and 50°C
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*
CONCEPT CHECK!
*
The correct answer is b).
Section 6.2
Enthalpy and Calorimetry
A 100.0 g sample of water at 90.°C is added to a 500.0 g sample
of water at 10.°C.
The final temperature of the water is:
a) Between 50°C and 90°C
b) 50°C
c) Between 10°C and 50°C
Calculate the final temperature of the water.
23°C
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*
CONCEPT CHECK!
*
The correct answer is c). The final temperature of the water is
23°C.
- (100.0 g)(4.18 J/°C·g)(Tf – 90.) = (500.0 g)(4.18 J/°C·g)(Tf –
10.)
Tf = 23°C

34. Section 6.2
Enthalpy and Calorimetry
You have a Styrofoam cup with 50.0 g of water at 10.°C. You
add a 50.0 g iron ball at 90. °C to the water. (sH2O = 4.18
J/°C·g and sFe = 0.45 J/°C·g)
The final temperature of the water is:
a) Between 50°C and 90°C
b) 50°C
c) Between 10°C and 50°C
Calculate the final temperature of the water.
18°C
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*
CONCEPT CHECK!
*
The correct answer is c). The final temperature of the water is
18°C.
- (50.0 g)(0.45 J/°C·g)(Tf – 90.) = (50.0 g)(4.18 J/°C·g)(Tf –
10.)
Tf = 18°C
Section 6.3
Hess’s Law
In going from a particular set of reactants to a particular set of
products, the change in enthalpy is the same whether the
reaction takes place in one step or in a series of steps.
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*

35. *
Section 6.3
Hess’s Law
This reaction also can be carried out in two distinct steps, with
enthalpy changes designated by ΔH2 and ΔH3.
N2(g) + O2(g) → 2NO(g) ΔH2 = 180 kJ
2NO(g) + O2(g) → 2NO2(g) ΔH3 = – 112 kJ
N2(g) + 2O2(g) → 2NO2(g) ΔH2 + ΔH3 = 68 kJ
ΔH1 = ΔH2 + ΔH3 = 68 kJ
N2(g) + 2O2(g) → 2NO2(g) ΔH1 = 68 kJ
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*
*
Section 6.3
Hess’s Law
The Principle of Hess’s Law
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*
*

36. Section 6.3
Hess’s Law
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*
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Section 6.3
Hess’s Law
Characteristics of Enthalpy ChangesIf a reaction is reversed, the
sign of ΔH is also reversed.The magnitude of ΔH is directly
proportional to the quantities of reactants and products in a
reaction. If the coefficients in a balanced reaction are multiplied
by an integer, the value of ΔH is multiplied by the same integer.
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*
*
Section 6.3
Hess’s Law
ExampleConsider the following data:

37. Calculate ΔH for the reaction
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Section 6.3
Hess’s Law
Problem-Solving StrategyWork backward from the required
reaction, using the reactants and products to decide how to
manipulate the other given reactions at your disposal.Reverse
any reactions as needed to give the required reactants and
products.Multiply reactions to give the correct numbers of
reactants and products.
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*
*

38. Section 6.3
Hess’s Law
ExampleReverse the two reactions:
Desired reaction:
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*
Section 6.3
Hess’s Law
ExampleMultiply reactions to give the correct numbers of
reactants and products:
4( ) 4( )
3( ) 3( )
Desired reaction:

39. *
Section 6.3
Hess’s Law
ExampleFinal reactions:
Desired reaction:
ΔH = +1268 kJ
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*
Section 6.4
Standard Enthalpies of Formation
Standard Enthalpy of Formation (ΔHf°)Change in enthalpy that

40. accompanies the formation of one mole of a compound from its
elements with all substances in their standard states.
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Section 6.4
Standard Enthalpies of Formation
Conventional Definitions of Standard StatesFor a CompoundFor
a gas, pressure is exactly 1 atm.For a solution, concentration is
exactly 1 M.Pure substance (liquid or solid)For an ElementThe
form [N2(g), K(s)] in which it exists at 1 atm and 25°C.
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Section 6.4
Standard Enthalpies of Formation
A Schematic Diagram of the Energy Changes for the Reaction
CH4(g) + 2O2(g) CO2(g) + 2H2O(l)
ΔH°reaction = –(–75 kJ) + 0 + (–394 kJ) + (–572 kJ) = –891 kJ
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41. *
Section 6.4
Standard Enthalpies of Formation
Problem-Solving Strategy: Enthalpy Calculations
When a reaction is reversed, the magnitude of ΔH remains the
same, but its sign changes.
When the balanced equation for a reaction is multiplied by an
integer, the value of ΔH for that reaction must be multiplied by
the same integer.
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*
*
Section 6.4
Standard Enthalpies of Formation
Problem-Solving Strategy: Enthalpy Calculations
The change in enthalpy for a given reaction can be calculated
from the enthalpies of formation of the reactants and products:
ΔH°rxn = ΣnpΔHf°(products) - ΣnrΔHf°(reactants)
4. Elements in their standard states are not included in the
ΔHreaction calculations because ΔHf° for an element in its

42. standard state is zero.
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*
Section 6.4
Standard Enthalpies of Formation
Calculate ΔH° for the following reaction:
2Na(s) + 2H2O(l) → 2NaOH(aq) + H2(g)
Given the following information:
ΔHf° (kJ/mol)
Na(s) 0
H2O(l) –286
NaOH(aq) –470
H2(g) 0
ΔH° = –368 kJ
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EXERCISE!
*
[2(–470) + 0] – [0 + 2(–286)] = –368 kJ
ΔH = –368 kJ
Section 6.5
Present Sources of Energy
Fossil FuelsPetroleum, Natural Gas, and
CoalWoodHydroNuclear

43. Copyright © Cengage Learning. All rights reserved
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*
Section 6.5
Present Sources of Energy
Energy Sources Used in the United States
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*
*
Section 6.5
Present Sources of Energy
The Earth’s AtmosphereTransparent to visible light from the
sun.Visible light strikes the Earth, and part of it is changed to
infrared radiation.Infrared radiation from Earth’s surface is
strongly absorbed by CO2, H2O, and other molecules present in
smaller amounts in atmosphere.Atmosphere traps some of the
energy and keeps the Earth warmer than it would otherwise be.
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*

44. *
Section 6.5
Present Sources of Energy
The Earth’s Atmosphere
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*
*
Section 6.6
New Energy Sources
Coal ConversionHydrogen as a FuelOther Energy
AlternativesOil shaleEthanolMethanolSeed oil
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45. *
*
*
*
Exothermic process because you must remove energy in order to
slow the molecules down to form a solid.
* Exothermic (heat energy leaves your hand and moves to the
ice) Endothermic (heat energy flows into the ice) Endothermic
(heat energy flows into the water to boil it) Exothermic (heat
energy leaves to condense the water from a gas to a liquid)
Endothermic (heat energy flows into the ice cream to melt it)
* System – methane and oxygen to produce carbon dioxide and
water; Surroundings – everything else around it; Direction of
energy transfer – energy transfers from the system to the
surroundings (exothermic) System – water drops; Surroundings
– everything else around it; Direction of energy transfer –
energy transfers from the surroundings (your body) to the
system (water drops) (endothermic)
*
Water is lower in energy because a lot of energy was released in
the process when hydrogen and oxygen gases reacted.
*
*
*
*
*
*

46. They both perform the same amount of work. w = -PΔV
a) w = -(2 atm)(4.0-1.0) = -6 L·atm
b) w = -(3 atm)(3.0-1.0) = -6 L·atm
*
a) q is positive for endothermic processes and w is negative
when system does work on surroundings; first condition – ΔE is
negative; second condition – ΔE is positive
b) q is negative for exothermic processes and w is positive when
surroundings does work on system; first condition – ΔE is
positive; second condition – ΔE is negative
*
*
(5.00 g C3H8)(1 mol / 44.094 g C3H8)(-2221 kJ / mol C3H8)
ΔH = -252 kJ
*
*
*
*
*
The correct answer is b).
*
The correct answer is c). The final temperature of the water is
23°C.
- (100.0 g)(4.18 J/°C·g)(Tf – 90.) = (500.0 g)(4.18 J/°C·g)(Tf –
10.)
Tf = 23°C
*
The correct answer is c). The final temperature of the water is
18°C.
- (50.0 g)(0.45 J/°C·g)(Tf – 90.) = (50.0 g)(4.18 J/°C·g)(Tf –
10.)

47. Tf = 18°C
*
*
*
*
*
*
*
*
*
*
*
*
*
*
*
[2(–470) + 0] – [0 + 2(–286)] = –368 kJ
ΔH = –368 kJ
*
*

48. *
*
*
322
222
13
22
NH()N()H() H = 46 kJ
2 H()O()2 HO() H = 484 kJ
¾¾®+D
+¾¾®D-
ggg
ggg
2223
2 N()6 HO()3 O()4 NH()
+¾¾®+
gggg
223
222
13
22
N()H()NH() H = 46 kJ
2 HO()2 H()O() H = +484 kJ
+¾¾®D-
¾¾®+D
ggg
ggg
223
222
2 N()6 H()4 NH() H = 184 kJ
6 HO()6 H()3 O() H = +1452 kJ
+¾¾®D-
¾¾®+D

49. ggg
ggg