Kelter P., Mosher M., Scott A. Chemistry. The Practical Science

Подождите немного. Документ загружается.

608 Chapter 14 Thermodynamics: A Look at Why Reactions Happen

ADP ATP

ADP AMP

The change in the free energy of the reaction is zero. Is this reaction sponta-

neous? Is it nonspontaneous? We can look at the phase transition of water from

solid to liquid to help sort all this out.

O(s) n H

O(l)

At temperatures below the melting point of water, the reaction is spontaneous in

the direction of H

O(s); that is, water freezes. At temperatures above the melting

point of water, the reaction is spontaneous in the opposite direction; ice melts. At

0°C and 1 atm, the process is not spontaneous in either direction. At this temper-

ature, the change in free energy for the process is zero (G = 0) and the reaction

proceeds neither toward the products nor toward the reactants. We say that the

reaction has reached

equilibrium. We will focus on this important concept in

Chapter 16, but for now, we need to know that a reaction at equilibrium hasn’t

stopped; it is just at a point where moving toward either the reactants or toward

the products is not thermodynamically favored. The reaction continues to make

both products and reactants at equal rates.

Remember that G is temperature-dependent (G = H − TS), so the

value of G can be changed by adjusting the temperature. That is, we can melt ice

or freeze water, depending on the temperature we pick. Active muscles do not

change their temperature in order to drive the production of ATP from ADP, but

this information can be used to calculate the temperature at which a process be-

comes spontaneous. Back to Mt. Everest....The base camp experiences temper-

atures ranging from a high of about –3

C in July to a low of about –16

C in

January. High up on the slopes of the mountain, the temperature drops even

lower, to around –36

C. For trekkers huddled in a tent, it may take a lot of effort

to get a propane stove to light (especially if the temperature drops below the boil-

ing point of propane; see Figure 14.22). How cold must it get before the propane

gas liqueﬁes? To answer this question, we need to consider the phase transition

shown below.

(l) n CH

(g) H = 15.1 kJ/mol

S = 65.4 J/mol

A phosphate group can be transferred from an ADP molecule to make ATP.

FIGURE 14.22

Trying to boil water at a camp on the

Lhotse Face of Mt. Everest (23,500 ft).

Application

When this reaction is at equilibrium, neither the forward nor the backward

reaction will be spontaneous (∆G = 0). The temperature to which this corre-

sponds is the boiling point. Because we know the equation that relates ∆G to the

temperature, we can calculate the boiling point of this reaction.

G = H − T(S)

0 = 15,100 J/mol − T(65.4 J/mol·K)

T(65.4 J/mol·K) = 15,100 J/mol

T = 231 K

T =−42

◦

Our calculations suggest that if the temperature falls to −42°C (−44°F), the

change in the free energy of the process will be zero. The reaction will be at equi-

librium, and propane will be at its boiling point. If the temperature gets any lower

than −42°C, the propane will be liquid, and the portable stove will become very

difﬁcult to light.

EXERCISE 14.9 Temperature Dependence of G

In one of the ﬁnal reactions to produce copper metal from its ore (chalcocite), cop-

per(I) oxide reacts with carbon to produce gaseous carbon monoxide and copper

metal. Above what temperature does the reaction become spontaneous?

O(s) + C(s) n 2Cu(s) + CO(g) H = 59 kJ/mol

S = 165 J/mol

Solution

G = H − T(S)

0 = 59,000 J/mol − T(165 J/mol·K)

T(171 J/mol·K) = 59,000 J/mol

T = 358 K

T = 85

◦

Above 85°C, this reaction becomes spontaneous (G < 0). So, at room tempera-

ture, the reaction doesn’t proceed without outside assistance. In practice, this reac-

tion is often run at very high temperatures to produce copper metal. The copper can

be further puriﬁed by electrolytic deposition (see Chapter 19).

PRACTICE 14.9

What is the boiling point of methanol? ...ofwater?

OH(l) n CH

OH(g)H

O(l) n H

O(g)



vap

H = 35.27 kJ/mol 

vap

H = 40.66 kJ/mol



vap

S = 104.6 J/mol

K 

vap

S = 109.0 J/mol

See Problems 67 and 68.

Changes in Pressure Affect Spontaneity

As the temperature changes, the value of G for a reaction changes. In fact, a

nonspontaneous reaction can often be made spontaneous if the temperature is

adjusted enough, such that H – TS becomes negative (see Table 14.5). For a

muscle cell in need of a quick energy ﬁx, changing the temperature in order to

make the conversion of ADP into ATP and AMP (adenosine monophosphate)

spontaneous is not an option. Thankfully, temperature isn’t the only variable that

14.6 When ∆G = 0; A Taste of Equilibrium 609

can change the spontaneity of a reaction. Pressure and concentration also play a

major role in determining reaction spontaneity.

The entropy change of a reaction depends on its pressure. This occurs because

there are fewer microstates in a compressed gas than there are if the pressure of

the same sample is reduced by expanding the volume of the container. Mathe-

matically, it has been shown that the pressure causes a change in the standard free

energy of a reaction in accordance with the following equation:

∆G = ∆G° + RT lnQ

where G is the change in the nonstandard state free energy

G° is change in the standard state free energy

R is the universal gas constant, 8.3145 J/mol

T is the temperature in kelvins

is a term called the pressure reaction quotient

We’ll discuss the pressure reaction quotient in much greater detail in Chapter 16.

However, for purposes of our current discussion, you may consider the pressure

reaction quotient, Q

,to bethe ratio of the pressures of all of the gaseous products

to the pressures of the gaseous reactants, raised to their respective stoichiometric

coefﬁcients. For example, here is the equation for the Haber process for the

formation of ammonia (part 2 of Exercise 14.5), along with the expression for

its Q

(g) + N

(g) n 2NH

(g)

For a reaction such as the fermentation of glucose, Q

is equal to the pressure of

carbon dioxide squared, because neither glucose nor ethanol is represented as a

gas in the equation.

(s) n 2C

O(l) + 2CO

(g)

Glucose Ethanol

The brewmaster at a local brewpub knows about the effect of pressure on

the fermentation of glucose. If the vat is left open to the atmosphere (1 atm), the

yeast carries out the fermentation at 25

C. The reaction, as we saw earlier in this

chapter, is spontaneous (–229 kJ/mol). If the brewmeister shuts the hatch on

the vat and seals it, the formation of carbon dioxide begins to build up pressure

as the glucose is catabolized. At some point, the pressure could reach 52.0 atm

(assuming the vat is strong enough to hold that pressure). Is the reaction sponta-

neous at that pressure?

G = G° + RT lnQ

G =−229,000 J/mol + (8.3145 J/mol

K)(298 K) ln



(52.0)



G =−229,000 J/mol + 19,580 J/mol

G =−209,420 J/mol

G =−209 kJ/mol

The reaction is still spontaneous, as you might expect for a reaction that produces

energy for a living organism. However, the reaction has a lower free energy than

it had at lower pressure.

610 Chapter 14 Thermodynamics: A Look at Why Reactions Happen

Application

Changes in Concentrations Affect Spontaneity

In a manner similar to that for gaseous reactions, the free energy of a reaction

changes as the concentrations of the reactants and products change.

Does this

make sense?

The equation relating the concentrations of reactants and products

to the observed free energy change (G) is

G = G° + RT lnQ

Note that this equation is mathematically similar to the equation for determining

the effect of changing pressure.

The greatest difference between this equation and the one presented in the

previous section is the value of the

reaction quotient (Q). For reactions in solution,

this reaction quotient is calculated by multiplying the initial molar concentra-

tions of all of the products, raised to the power of their respective stoichiometric

coefﬁcients, divided by the initial molar concentrations of all of the reactants,

raised to the power of their stoichiometric coefﬁcients. Mathematically, for the

reaction

rA + sB n tC + uD

the reaction quotient is

Q =

[C]

[D]

[A]

[B]

where [ ]

= the initial molar concentration of the substance.

What does this mean? Reactions with large concentrations of reactants have

a Q value less than 1. This means that RTlnQ is negative and that the observed

free energy (∆G) is less than the standard free energy (G°). In other words, large

concentrations of reactants and small concentrations of products increase

the spontaneity of the forward reaction (this situation decreases the value of the

observed ∆G). When the concentration of reactants is small and the concentra-

tion of products is large (Q > 1.0), the value of RT lnQ is greater than zero and the

observed ∆G is more positive than G°. In this case, the reaction is less sponta-

neous. Again, we will examine a similar relationship in much greater detail in

Chapter 16 and will explore more deeply what happens when G

= G°, the

equilibrium condition.

Coupled Reactions

Many reactions, such as the formation of glucose-6-phosphate from glucose, are

nonspontaneous. Metabolic sequences have developed to account for this. As

we’ve seen in this chapter, biological reactions are often coupled with other reac-

tions to produce a spontaneous process. Organisms typically couple a reaction

that has a very large negative ∆G with reactions that have positive free energy

changes, as shown in Figure 14.23. The result is a spontaneous reaction.

In glycolysis, there are four coupled reactions. Two of these reactions use the

negative free energy change of ATP

n ADP (∆G = –30.5 kJ/mol) to place phos-

phate groups on a glucose molecule before it is broken down. Then, as the resulting

carbohydrate is catabolized, the organism uses the stored energy of the carbohy-

drate to drive the reverse of this reaction (ADP

n ATP; ∆G = +30.5 kJ/mol). In

doing so, the glycolytic pathway consumes two molecules of ATP but produces

four molecules of ATP using the stored energy of the carbohydrate. The net result

is the accumulation of two molecules of ATP for each glucose molecule.

In this chapter we have examined why chemical reactions proceed and have

shown that it is directly related to an increase in the entropy of the universe. We

14.6 When ∆G = 0; A Taste of Equilibrium 611

ADP

Nonspontaneous

reaction

ATP

Spontaneous

reaction

FIGURE 14.23

Biochemical reactions typically involve

the coupling of a spontaneous reaction

with a nonspontaneous reaction. Coup-

ling the reactions makes the nonsponta-

neous reaction spontaneous.

Application

Motor

Elevator

Counterweight

612 Chapter 14 Thermodynamics: A Look at Why Reactions Happen

The Bottom Line

■

The multiplicity of a system can be used to deter-

mine the behavior of the system. The most probable

macrostate is the one that has the most contributing

microstates. (Section 14.1)

■

Spontaneous processes occur without outside assis-

tance. The reverse of the spontaneous process is non-

spontaneous. (Section 14.2)

■

Entropy is a measure of how energy and matter can

be distributed in a chemical system. Entropy is not

disorder. (Section 14.2)

■

The second law of thermodynamics says that the en-

tropy of the universe (as we deﬁned the term uni-

verse in Chapter 5) continues to increase. Any process

that is spontaneous must correspond to an increase

in the entropy of the universe (∆S

universe

> 0).

(Section 14.2)

■

The third law of thermodynamics says that the en-

tropy of a pure perfect crystalline material is zero.

This law enables us to calculate the entropy of any

compound in any state. (Section 14.4)

■

The free energy, ∆G, can be calculated via the Gibbs

equation (∆G

= ∆H – T∆S) to determine the spon-

taneity of a process. (Section 14.5)

■

When ∆G = 0, the system is at equilibrium. The for-

ward and reverse reactions still proceed, but their

rates are equal. (Section 14.6)

■

Coupled reactions are used by the body to make non-

spontaneous processes become spontaneous. The

use of ATP in this manner assists in the overall pro-

duction of energy from glucose. (Section 14.6)

routinely consider the spontaneity of a reaction as a function of the values of ∆H,

∆S, and reaction conditions such as temperature, pressure, and concentration.

We’ve noted that the free energy change (∆G) depends on all of these things.

Although we now know whether a reaction will proceed, we haven’t discussed

how fast reactions proceed. We know that the production of energy from glucose

is a spontaneous process, as is the rusting of a nail. But the rates of these two re-

actions are quite different, with important effects on us and our surroundings. In

the next chapter, we’ll discover that the degree of spontaneity implies nothing

about the speed of a reaction.

Elevators are raised by a motor. To

facilitate the motion of the elevator, a

counterweight helps pull the elevator up.

The motor and the counterweight couple

their action to produce a working

elevator.

Focus Your Learning 613

Key Words

bioenergetics The study of the energy changes that

occur within a living cell. (p. 580)

catabolism The biological degradation of molecules to

provide an organism with smaller molecules and

energy. (p. 587)

entropy A measure of how the energy and matter of a

system are distributed throughout the system.

(p. 587)

equilibrium The state of a reaction when G = 0. The

reaction hasn’t stopped; rather, the rates of the for-

ward and reverse reactions are equal. (p. 608)

free energy (G) The state function that is equal to the

enthalpy minus the temperature multiplied by the

entropy. Used to determine the spontaneity of a

process. (p. 601)

Gibbs equation G = H − TS (p. 601)

glycolysis A series of biochemical reactions that convert

glucose into two molecules of pyruvate. The process

results in the formation of two molecules of ATP.

(p. 585)

macrostate The macroscopic state of a system that indi-

cates the properties of the entire system. (p. 580)

metabolism The biochemical reactions of an organism.

(p. 595)

microstate The state of the individual components within

a system. (p. 581)

multiplicity The number of microstates within a system.

(p. 587)

nonspontaneous process A process that occurs only with

continuous outside intervention. (p. 585)

pressure reaction quotient (Q

) The ratio of the pressures

of all of the gaseous products raised to their respec-

tive stoichiometric coefﬁcients, divided by the pres-

sures of all of the gaseous reactants raised to their

stoichiometric coefﬁcients. (p. 610)

reaction quotient (Q) The ratio of the initial molar con-

centrations of all of the products raised to the power

of their respective stoichiometric coefﬁcients, divided

by the initial molar concentrations of all of the reac-

tants raised to the power of their stoichiometric

coefﬁcients. (p. 611)

reversible conditions The conditions that occur when a

process proceeds in a series of inﬁnitesimally small

steps. (p. 593)

second law of thermodynamics A spontaneous process is

accompanied by an increase in the entropy of the

universe; ∆S

universe

> 0. (p. 588)

spontaneous process A process that occurs without con-

tinuous outside intervention. (p. 585)

standard molar free energy of formation (

G°) The free

energy of a compound formed from its elements in

their standard states. (p. 605)

thermodynamics The study of the changes in energy in a

reaction. (p. 586)

third law of thermodynamics The entropy of a pure per-

fect crystal at 0 K is zero. (p. 598)

The answers to the odd-numbered problems and some selected

problems appear at the back of the book, as represented by the

blue numbering.

Section 14.1

Probability as a Predictor of Chemical Behavior

Skill Review

1. Suppose that four pennies fall from your pocket to the ﬂoor.

How many microstates for the four pennies would exist?

(Consider only heads up versus tails up in your explanation.)

5. Contrast the meanings of the terms macrostate and

microstate.

6. Deﬁne the term multiplicity.

7. One hundred chemistry students are about to take an exam.

There are two equal 200-seat classrooms for the students.

Knowing that students must have vacant seats next to them

during the exam, describe at least two macrostate arrange-

ments you could predict for the two rooms.

8. In the situation deﬁned in Problem 7, how could the posi-

tions of the students allow you to use microstates to explain

your answer?

Chemical Applications and Practice

9. Even a sample of only 1,000 atoms is too small to weigh ac-

curately, and yet we can calculate a probable distribution of

their numbers between two equal containers. How many mi-

crostates are possible if 1,000 atoms of He become distrib-

uted between two containers of equal size? Explain why an

equal distribution becomes more probable as the number of

He atoms increases.

Focus Your Learning

2. What is the probability that all four coins from Problem 1

would land heads up?

3. What is the probability that the four coins from Problem 1

would land two heads up and two tails up?

4. What is the most probable outcome for the four coins in

Problem 1?

10. A certain lecture hall is 25 m long, 12 m wide, and 15 m tall.

The oxygen molecules necessary to sustain life in the room

are, of course, free to circulate throughout the room. Explain,

using probability and microstates, why a student sitting in

the back should not be concerned that all the freely moving

oxygen molecules might concentrate at the front row of the

classroom.

11. Some powdered creamer is added to a cup of hot coffee.

Compare the initial macrostate to the ﬁnal macrostate as

the creamer dissolves. Has the number of microstates for the

creamer and coffee increased from the initial state to the ﬁnal

state? Explain the evidence you would use to explain your

answer.

12. A glass of tea is cooled by placing ice cubes in it. However,

after a few minutes, the ice has melted. Compare the initial

macrostate to the ﬁnal macrostate. Has the number of mi-

crostates for the ice and tea increased from the initial state to

the ﬁnal state? Explain the evidence you would use to explain

your answer.

13. Which of these would you classify as increasing the number

of microstates?

a. Water in a glass evaporates.

b. A precipitate of AgCl forms from the addition of a drop of

a 0.10 M solution of AgNO

into a glass of chlorinated tap

water.

c. An icicle forms from a downspout.

14. Which of these would you classify as increasing the number

of microstates?

a. A chunk of ice melts when placed in a glass of tap water at

room temperature.

b. A glass of tap water freezes when placed in a refrigerator

freezer compartment.

c. The combustion of gasoline produces carbon dioxide and

water vapor.

Section 14.2 Why Do Chemical Reactions Happen?

Entropy and the Second Law of Thermodynamics

Skill Review

15. Which of these, if any, need outside intervention to take

place? Describe the necessary intervention.

a. Combustion (combination with oxygen) of a piece of

paper in the air

b. Making a hard-boiled egg

c. Increasing muscle mass

d. Formation of calcium carbonate (bathtub ring) from

evaporating hard water

16. Which of these, if any, need outside intervention to take

place? Describe the necessary intervention.

a. The melting of butter on a warm pancake

b. Opening this book to this page

c. Paddling a canoe upstream

d. Cooking eggs for breakfast

17. What is the relationship between entropy and a nonsponta-

neous process?

18. What is the relationship between entropy and rate of change?

19. What is the relationship between entropy and number of

microstates?

20. What is the relationship between a nonspontaneous process

and the number of microstates?

21. Suppose you were studying in a residence hall. Down the hall

someone is preparing popcorn. The irresistible aroma even-

tually reaches your room. To take your mind off this, explain

how the second law of thermodynamics enables you to detect

this tasty temptation from so far away.

22. A scientist on the planet Zoltan expresses the third law of

thermodynamics by saying that the entropy of any com-

pound under standard state conditions is zero. Would this

change the ∆S values for the reaction of pyruvate to lactate

mentioned in the chapter? Explain.

Chemical Application and Practice

23. While pumping gasoline into your car, you might spill some

gasoline on your hands. However, the gasoline quickly evap-

orates with a cooling sensation on your skin. How can this

process be spontaneous if it requires the input of the energy

from your skin?

24. In a previous problem you were asked about the freezing of

water in a refrigerator. Water poured in a tray will sponta-

neously freeze in a freezer. Explain why this process arranges

the water molecules in a more orderly crystal form and yet is

still spontaneous.

25. Solid carbon dioxide is called dry ice. At room temperature,

this material spontaneously changes from an orderly solid to

a gas, bypassing the liquid phase entirely. When undergoing

this change, the molecules absorb heat from the nearby

surroundings. This, in turn, slows down the molecules

of the surroundings.

What can be said about

the change in entropy

for the carbon dioxide

molecules relative to

nearby air molecules?

What can be said about

the change in the en-

tropy of the universe

for this process?

614 Chapter 14 Thermodynamics: A Look at Why Reactions Happen

Dry ice.

26. The entropy of molecules can also depend on their relative

atomic motions. The atoms in molecules can twist, rotate,

and vibrate around a chemical bond. For which molecule

would you be able to predict more microstates: a molecule of

nitrogen or a molecule of ammonia (NH

)? Explain the basis

of your choice.

27. The following equation illustrates a gaseous reaction that is

possible in the atmosphere. Which direction in the reaction

shows the greater entropy for the system? Explain your

choice.

28. Another reaction common in the lower atmosphere is the

formation of ozone from NO

. Which direction in the reac-

tion shows the greater entropy for the system? Explain your

choice.

(g) + NO

(g)





(g) + NO(g)

Section 14.3 Temperature and Spontaneous Processes

Skill Review

29. Classify each of these as representing either a positive change

in entropy for the universe or a negative change in entropy

for the universe, or indicate that you would need more infor-

mation to determine this.

a. S

sys

=+; S

surr

b. S

sys

=+; S

surr

=−

c. S

sys

=−; S

surr

30. Classify each of these as representing either a positive change

in entropy for the universe or a negative change in entropy

for the universe, or indicate that you would need more infor-

mation to determine this.

a. S

sys

=−; S

surr

=−

b. S

sys

=+; S

surr

= 0

c. S

sys

=−; S

surr

= 0

31. How does the multiplicity of a system change as the molecu-

lar motions increase in the system?

32. How does the molecular motion of the surroundings change

as energy from the surroundings ﬂows into a system?

33. How does the entropy of the surroundings change as energy

ﬂows out of a system?

34. Can we predict the change in entropy of the universe as en-

ergy ﬂows into the system? Explain.

Chemical Applications and Practice

35. Mercury remains a liquid over a fairly wide temperature

range, which is one reason why it has been used in ther-

mometers. Classify each of these changes as a positive or a

negative change in the entropy of a sample of mercury.

a. Liquid mercury changes to a solid.

b. Liquid mercury expands as it is heated.

c. A small amount of mercury evaporates from the surface

of a column of mercury in a sealed thermometer.

36. Chlorine gas (Cl

) is a very toxic compound that can form as

a result of the improper use of bleach and ammonia cleaning

solutions. Classify each of these changes as a positive or a

negative change in the entropy of a sample of this green gas.

a. Chlorine gas condenses to a liquid.

b. Chlorine gas is heated in a container.

c. Chlorine gas deposits on a cold surface.

37. If you are enjoying an iced carbonated soft drink while study-

ing, you may have noticed that the carbon dioxide used to

carbonate the beverage bubbles slowly out of the beverage as

it warms. Is this a positive or a negative entropy change for

the CO

? Explain.

38. In the bubbling of carbon dioxide mentioned in Problem 37,

did the surroundings gain or lose heat? Is the sign for

S

universe

, at this temperature, positive or negative?

Section 14.4

Calculating Entropy Changes in Chemical Reactions

Skill Review

39. Propane stoves make use of the following combustion

reaction:

___ C

(g) + ___ O

(g) → ___ CO

(g) + ___ H

O(g)

Balance the equation, report the number of moles of gas

molecules on each side of the equation, and predict the

change in entropy of the system.

40. Methane can be combusted by the following reaction:

___ CH

(g) + ___ O

(g) → ___ CO

(g) + ___ H

O(g)

Balance the equation, report the number of moles of gas

molecules on each side of the equation, and predict the

change in entropy of the system.

41. Repeat the prediction for Problem 39, but assume that the

reaction is performed under temperatures that cause

the propane and water to be liquid, yet the oxygen and car-

bon dioxide remain as gases.

42. Repeat the prediction for Problem 40, but assume that

the reaction is performed under temperatures that cause the

methane and water to be liquid, yet the oxygen and carbon

dioxide remain as gases.

43. Explain why the change in the number of moles of gases,

from reactant to product, is typically more inﬂuential in de-

termining the sign on entropy changes than the change in

moles of liquids and solids.

44. The third law of thermodynamics speciﬁes a particular con-

dition for a perfect crystal. Why must the temperature be 0 K

instead of 0°C?

45. Predict the sign on the entropy change for each of these

reactions.

a. NH

(g) + HCl(g) → NH

Cl(s)

b. 2HgO(s) → 2Hg(l) + O

(g)

c. Cd(s) +

⁄2O

(g) →CdO(s)

46. Using the values for S

found in the appendix, determine

the actual values for the standard change in entropy for the

reactions in Problem 45.

47. Predict the sign on the entropy change for each of these

reactions.

a. 2SO

(g) + O

(g) → 2SO

(g)

b. 2NH

(g) → N

(g) + 3H

(g)

c. CO(g) + 2H

(g) → CH

OH(l)

48. Using the values for S° found in the appendix, determine

the actual values for the standard change in entropy for the

reactions in Problem 47.

Chemical Applications and Practices

49. Several states have implemented regulations for fuel alterna-

tives in automobiles. One such gasoline alternative is ethanol

OH). Balance the following combustion reaction and

determine the change in entropy for the reaction.

___ C

OH(l) + ___ O

(g) →____CO

(g) + ____ H

O(g)

Focus Your Learning

615

50. One important source of salt (NaCl) is the evaporation of

ocean water. Calculate the change in the standard molar en-

tropy as the dissolved ions precipitate the solid.

(aq) + Cl

−

(aq) →NaCl(s)

51. Graphite and diamond are both made only of carbon. How-

ever, pencils (incorporating the graphite form) sell for a few

cents, whereas engagement rings (displaying diamonds)

often sell for thousands of dollars. Using the appendix, deter-

mine the change in entropy for the conversion of diamond to

graphite.

52. Using the appendix to note the standard S° value for a

metal, explain why the value for the metal alone is lower than

the value for any of the listed compounds of that metal.

Section 14.5 Free Energy

Skill Review

53. Predict the relative magnitude of the temperature that would

result in a spontaneous process in each of these combinations.

a. H =+; S =+

b. H =+; S =−

54. Predict the relative magnitude of the temperature that would

result in a spontaneous process in each of these combinations.

a. H =−; S =+

b. H =−; S =−

55. Determine the value of H in each of these.

a. G =−24.5 kJ/mol; S = 287 J/mol·K; T = 298 K

b. G = 1.38 kJ/mol; S = 24 J/mol·K; T = 298 K

c. G = 500.0 kJ/mol; S =−6439 J/mol·K; T = 325 K

d. G =−24.5 kJ/mol; S = 187 J/mol·K; T = 39.9°C

56. Determine the value of G in each of these cases.

a. H =−20.5 kJ/mol; S = 259 J/mol·K; T = 298 K

b. H = 350 kJ/mol; S = 73 J/mol·K; T = 157 K

c. H = 299 kJ/mol; S

=−639 J/mol·K; T = 325 K

d. H =−4505 kJ/mol; S = 107 J/mol·K; T = 19.3°C

57. Determine the value of S in each of these cases.

a. G =−20.5 kJ/mol; H = 259 kJ/mol; T = 298 K

b. G = 150 kJ/mol; H =−73 kJ/mol; T = 157 K

c. G = 209 kJ/mol; H =−639 kJ/mol; T = 35.0 K

d. G =−4505 kJ/mol; H = 107 kJ/mol; T = 135.8°C

58. Determine the value of T in each of these cases.

a. G =−20.5 kJ/mol; H = 259 kJ/mol;

S = 260 J/mol·K

b. G = 150 kJ/mol; H =−73 kJ/mol;

S =−290 J/mol·K

c. G = 209 kJ/mol; H = 639 kJ/mol; S = 560 J/mol·

d. G =−4505 kJ/mol; H = 107 kJ/mol;

S = 1160 J/mol·K

59. Determine the spontaneity of each of these reactions at

298 K.

a. NH

(g) + HCl(g) → NH

Cl(s)

b. 2HgO(s) → 2Hg(l) + O

(g)

c. Cd(s) +

⁄2O

(g) →CdO(s)

60. Determine the spontaneity of each of these reactions at

298 K.

a. 2SO

(g) + O

(g) → 2SO

(g)

b. 2NH

(g) → N

(g) + 3H

(g)

c. CO(g) + 2H

(g) → CH

OH(l)

Chemical Application and Practices

61. One of the authors of this text formerly drove an automobile

that could best be described as blue and rust. Consider that the

rust was formed, at least partially, by the reaction illustrated

by this equation:

(g) + 4Fe(s) → 2Fe

(s)

Assuming standard conditions, what would you calculate as

the G° for this reaction?

62. The impressive white cliffs of Dover in England have stood

without signiﬁcant change from their basic structure of cal-

cium carbonate for eons. One possible reaction for the de-

composition of calcium carbonate is

CaCO

(s) → CaO(s) + CO

(g)

Use the 

G° values of these compounds to determine the

value of G° for the reaction. Comment on how thermody-

namics helps explain the relative stability of this beautiful

landmark.

63. Sulfur dioxide and sulfur trioxide are important anhydrides

for making sulfurous and sulfuric acid, respectively. When

sulfur-containing coal is burned, sulfur gases pose environ-

mental threats to air quality. Given the following two com-

bustion reactions, determine the G° value for the change

from SO

to SO

S(s) +

⁄2O

(g) → SO

(g) G° =−371 kJ/mol

S(s) + O

(g) → SO

(g) G° =−300 kJ/mol

(g) +

⁄

(g) → SO

(g) G° = ?

64. As we note earlier, graphite and diamond are both composed

entirely of carbon atoms. Using the following reactions at

298 K, determine the value of G° for the conversion of

graphite into diamond.

diamond

(s) + O

(g) → CO

(g) G° =−397 kJ/mol

graphite

(s) + O

(g) → CO

(g) G° =−394 kJ/mol

graphite

(s) → C

diamond

(s) G° = ??

65. As we saw in earlier in this chapter, the production of glucose

requires the input of energy. This is typically shown as green

plants using sunlight to help combine carbon dioxide with

water to form glucose through photosynthesis. However, in

some deep areas of the ocean, sunlight does not reach bacte-

ria that may grow around thermal vents. These organisms

have been shown to produce energy through oxidation of a

sulfur compound, H

S. Determine H°, S°, and G° for

this biochemically important reaction. Assume sulfur is in

the standard form—rhombic allotrope.

S(aq) + O

(g) → 2H

O(l) + 2S(s)

66. Smog produced over a city is formed from trace amounts of

automobile exhaust in a reaction with atmospheric oxygen.

One of the steps in the formation of ozone is the oxidation of

oxygen gas by nitrogen dioxide (a by-product of the combus-

tion of gasoline). Determine H°, S°, and G° for this

reaction.

(g) + O

(g) → NO(g) + O

(g)

616 Chapter 14 Thermodynamics: A Look at Why Reactions Happen

Section 14.6

When G = 0; A Taste of Equilibrium

Skill Review

67. Determine the temperature at which the indicated system

reaches equilibrium.

a. H° = 46.4 kJ/mol; S° = 27.6 J/mol·K

b. H° = 10.6 kJ/mol; S° = 77 J/mol·K

c. H° = 124 kJ/mol; S° = 295.5 J/mol·K

68. Determine the temperature at which the indicated system

reaches equilibrium.

a. H° = 57.6 kJ/mol; S° = 17.4 J/mol·K

b. H° = 94 kJ/mol; S° = 306 J/mol·K

c. H° = 32.1 kJ/mol; S° = 552 J/mol·K

69. At equilibrium, what is the pressure for the following process

at 298 K?

CaCO

(s)





CaO(s) + CO

(g) G° = 131 kJ/mol

70. What is the free energy change for the following reaction at

298 K, given that the pressures of NH

and N

are 1.0 atm

each and the pressure of H

is 2.0 atm? (Hint: Use the appen-

dix to determine the free energy change under standard

conditions.)

2NH

(g)





(g) + 3H

(g)

Chemical Applications and Practices

71. At 1 atm of pressure the boiling point of pure water is

100.0

C. If 1000.0 g of water were brought to this point and

then completely boiled away, what would you calculate as the

entropy change for the water? (Note: The heat of vaporiza-

tion for water is 40.7 kJ/mol.)

72. There is a historical approximation known as Trouton’s rule.

This approximation states that at the normal boiling point

for nonpolar compounds, the standard molar entropy of va-

porization is 87 J/mol·K. Using Trouton’s rule, determine the

approximate heat of vaporization of water. The actual molar

entropy of vaporization for water is close to 110 J/mol·K.

Explain why the Trouton value is so much different for polar

substances such as water but is often found to be much closer

to the actual value for nonpolar substances.

73. Under proper conditions, phase changes can be classiﬁed

as reversible processes. The energy involved with melting

or freezing of a sample is called the enthalpy of fusion. The

enthalpy of fusion (

fus

H) for the rare radioactive element

actinium, Z = 89, is 10.50 kJ/mol. If the entropy of fusion is

9.6 J/mol·K, what would you calculate as the melting point

for this silvery white metal?

74. One of the reasons why you are able to read this question is

that the cellular functions necessary are supplied with energy

from the molecule ATP (adenosine triphosphate), which

couples with many biochemical reactions. The formation of

ATP can be accomplished as follows:

ADP(aq) + H

−

(aq)





ATP(aq) + H

O(l)

G° for the reaction is 31 kJ/mol. Using that information,

what would you estimate for the reaction quotient, Q,at

25°C, for the reaction at equilibrium?

75. The gas commonly used in chemical laboratory burners is

methane (CH

). Shown here is the combustion reaction that

takes place when you light your lab burner.

(g) + 2O

(g)





(g) + 2H

O(g)

a. Use 

data to calculate the G° value for the reaction.

b. What would the value of G be if the pressure of O

were

reduced to 0.20 atm from 1.00 atm, at room conditions,

with the remaining gases at standard conditions?

c. Explain why, if G has a negative value indicating a

spontaneous change, we still have to bring a ﬂame or spark

to the system to get it to react in the lab.

76. Nitrogen gas and oxygen gas make up essentially 100% of our

air. The following reaction has G° = 174 kJ. Fortunately,

the two do not easily combine through this reaction under

the Earth’s atmospheric conditions.

(g) + O

(g)





2NO(g)

a. What is the value of Q

at 298 K for the reaction at

equilibrium?

b. What temperature would be needed for the reaction to be

spontaneous if the pressures were returned to 1.0 atm?

(Assume that enthalpy and entropy values do not change

signiﬁcantly over this range.)

c. What would be the value of G at 25°C if the pressure of

both N

and O

were increased to 5.0 atm each, while the

pressure of NO

was decreased to 0.50 atm?

Comprehensive Problems

77. Butane, burned in small portable lighters, combusts accord-

ing to the following unbalanced equation:

___ C

(g) + ___ O

(g) → ___ CO

(g) + ___ H

O(g)

First balance the equation, and then explain your answer to

this question: “Does the combustion show an increase or a

decrease in entropy for the system?”

78. Suppose a particularly noxious compound was entering a

water supply at only one point. If it would stay at the entry

point, it might easily be removed. However, this does not

happen. Explain why environmental scientists must be aware

of the second law of thermodynamics.

79. Because it was already known that a positive change in the

entropy of the universe indicated a spontaneous reaction,

why was it important to develop the Gibbs equation, which

is also used to predict the spontaneity of a reaction? What

advantage does the Gibbs equation offer?

80. In the ongoing research to ﬁnd a cure or treatment for

Alzheimer’s disease, some investigators have focused on the

appearance of plaque-like formations in brain cells. The solid

masses form from protein ﬁbers called beta-amyloids. Use

Internet resources and journal sources to ﬁnd out if the

process is based on a positive or a negative value for the S°

of the plaque-forming reaction.

81. In calculations involving Gibbs energy and equilibrium, you

must always be aware of the differences between G and

G

.WhenG = 0, at 25°C, what is true about Q?

82. A certain reaction is nonspontaneous at one temperature. If

raising the temperature caused the reaction to become spon-

taneous, what could you conclude about the entropy change

of the reaction? Explain your answer.

Focus Your Learning

617