SPONTANEOUSCHANGE: ENTROPYAND FREE ENERGY

In Chapter 15, we noted that the reaction of nitrogen and oxygengases, which does not occur appreciably in the forward directionat room temperature, produces significant equilibrium amounts

of NO(g) at high temperatures.

Another reaction involving oxides of nitrogen is the conversion ofNO(g) to

This reaction, unlike the first, yields its greatest equilibrium amountsof at low temperatures.

What is there about these two reactions that causes the forward re-action of one to be favored at high temperatures but the forwardreaction of the other to be favored at low temperatures? Our primaryobjective in this chapter is to develop concepts to help us answerquestions like this. This chapter, taken together with ideas fromChapter 7, shows the great power of thermodynamics to provideexplanations of many chemical phenomena.

NO2(g)

2 NO(g) + O2(g) ∆ 2 NO2(g)

NO2(g):

N2(g) + O2(g) ∆ 2 NO(g)

19-1 Spontaneity: TheMeaning of SpontaneousChange

19-2 The Concept of Entropy19-3 Evaluating Entropy and

Entropy Changes19-4 Criteria for Spontaneous

Change: The Second Lawof Thermodynamics

19-5 Standard Free EnergyChange,

19-6 Free Energy Change andEquilibrium

19-7 and K as Functionsof Temperature

19-8 Coupled Reactions➣ FOCUS ON Coupled

Reactions in BiologicalSystems

¢G°

¢G°

780

C O N T E N T S

Thermodynamics originated in the early nineteenth century in attempts toimprove the efficiency of steam engines. However, the laws of thermodynamicsare widely useful throughout the field of chemistry and in biology as well, as wediscover in this chapter.

PETRMC19_780-821-hr 12/30/05 2:01 PM Page 780

SPONTANEITY: THE MEANING

OF SPONTANEOUS CHANGE

Most of us have played with spring-wound toys, whether a toy automobile,top, or music box. In every case, once the wound-up toy is released, it keepsrunning until the stored energy in the spring has been released; then the toystops. The toy never rewinds itself. Human intervention is necessary (windingby hand). The running down of a wound-up spring is an example of aspontaneous process. The rewinding of the spring is a nonspontaneous process.Let’s explore the scientific meaning of these two terms.

A spontaneous process is a process that occurs in a system left to itself;once started, no action from outside the system (external action) is necessaryto make the process continue. Conversely, a nonspontaneous process will notoccur unless some external action is continuously applied. Consider the rust-ing of an iron pipe exposed to the atmosphere. Although the process occursonly slowly, it does so continuously. As a result, the amount of iron decreasesand the amount of rust increases until a final state of equilibrium is reached inwhich essentially all the iron has been converted to iron(III) oxide. We say thatthe reaction

is spontaneous. Now consider the reverse situation: the extraction of pure ironfrom iron(III) oxide. We should not say that the process is impossible, but it iscertainly nonspontaneous. In fact, this nonspontaneous reverse process is in-volved in the manufacture of iron from iron ore.

We will consider specific quantitative criteria for spontaneous change laterin the chapter, but even now we can identify some spontaneous processes in-tuitively. For example, in the neutralization of NaOH(aq) with HCl(aq), thenet change that occurs is

There is very little tendency for the reverse reaction (self-ionization) to occur,so the neutralization reaction is a spontaneous reaction. The melting of ice,however, is spontaneous at temperatures above 0 °C but nonspontaneousbelow 0 °C.

From our discussion of spontaneity to this point, we can reach theseconclusions.

• If a process is spontaneous, the reverse process is nonspontaneous.

• Both spontaneous and nonspontaneous processes are possible, but onlyspontaneous processes will occur without intervention. Nonspontaneousprocesses require the system to be acted on by an external agent.

We would like to do more, however. We want to be able to predict whetherthe forward or the reverse direction is the direction of spontaneous change in aprocess, so we need a criterion for spontaneous change. To begin, let’s look tomechanical systems for a clue. A ball rolls downhill, and water flows to a lowerlevel. A common feature of these processes is that potential energy decreases.

For chemical systems, the property analogous to the potential energy of amechanical system is the internal energy or the closely related propertyenthalpy In the 1870s, P. Berthelot and J. Thomsen proposed that thedirection of spontaneous change is the direction in which the enthalpy of a sys-tem decreases. In a system in which enthalpy decreases, heat is given off by thesystem to the surroundings. Bertholet and Thomsen concluded that exothermicreactions should be spontaneous. In fact, many exothermic processes are spon-taneous, but some are not. Also, some endothermic reactions are spontaneous.

1H2. 1U2

H3O+(aq) + OH-(aq) ¡ 2 H2O(l)

4 Fe(s) + 3 O2(g) ¡ 2 Fe2O3(s)

19-1

19-1 Spontaneity: The Meaning of Spontaneous Change 781

Spontaneous: “proceedingfrom natural feeling ornative tendency withoutexternal constraintdeveloping without apparentexternal influence, force,cause, or treatment.”Merriam-Webster’s CollegiateDictionary, on-line, 2000.

Á ;

▲

The amount of steel destroyedeach year by rust isapproximately equal to 10%

of the world’s annual production.

The melting of an ice cubeoccurs spontaneously attemperatures above 0 °C.

▲

No change takes place fora system at equilibrium untilsome mechanism is provided

to couple it to the surroundings. Forexample, gasoline is stable until anengine converts its chemical energy toheat and work. The engine providesthe coupling to the surroundings.

Spontaneous vsNonspontaneous activity

When the springpowering this mechanical toyunwinds, the toy stopsmoving. The spring cannotspontaneously rewind.

▲

AAAAAAAAAAAAAAAABBBBWWWWTTTT0000ccccrrrrooooppppppppeeeedddd ttttooooppppttttoooo 11111111pppp6666

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782 Chapter 19 Spontaneous Change: Entropy and Free Energy

Thus, we cannot predict whether a process is spontaneous from its enthalpychange alone. Here are three examples of spontaneous, endothermic processes.

• the melting of ice at room temperature• the evaporation of liquid diethyl ether from an open beaker• the dissolving of ammonium nitrate in water

We will have to look to thermodynamic functions other than enthalpy changeas criteria for spontaneous change.

CONCEPT ASSESSMENT ✓Is it correct to say that a spontaneous process is a fast process and a nonspontaneousprocess is a very slow one? Explain.

THE CONCEPT OF ENTROPY

To continue our search for criteria for spontaneous change, consider Figure 19-1,which depicts two identical glass bulbs joined by a stopcock. Initially, the bulbon the left contains an ideal gas at 1.00 atm pressure, and the bulb on the right isevacuated. When the valve is opened, the gas immediately expands into theevacuated bulb. After this expansion, the molecules are dispersed throughoutthe apparatus, with essentially equal numbers of molecules in both bulbs and apressure of 0.50 atm. What causes this spontaneous expansion of the gas ata constant temperature?

One of the characteristics of an ideal gas is that its internal energy doesnot depend on the gas pressure, but only on the temperature. Therefore, in thisexpansion Also, the enthalpy change is zero: This meansthat the expansion is not caused by the system dropping to a lower energystate. A convenient mental image to “explain” the expansion is that the gasmolecules tend to spread out into the larger volume available to them at the re-duced pressure. A more fundamental description of the underlying cause isthat, for the same total energy, in the expanded volume there are more avail-able translational energy levels among which the gas molecules can be distrib-uted. The tendency is for the energy of the system to spread out over a largernumber of energy levels.

A similar situation—the mixing of ideal gases—is depicted in Figure 19-2. Inthis case, the two bulbs initially are filled with different ideal gases at 1.00 atm.When the stopcock is opened, the gases mix. The resulting change is essentiallythat of the expansion of the ideal gas pictured in Figure 19-1, but twice over.That is, each gas expands into the new volume available to it, without regardfor the other gas (recall Dalton’s law of partial pressures, page 198). Again, eachexpanded gas has more translational energy levels available to its molecules—the energy of the system has spread out. And again, the internal energy andenthalpy of the system are not changed by the expansion.

ENTROPY

The thermodynamic property related to the way in which the energy of asystem is distributed among the available microscopic energy levels iscalled entropy.

The greater the number of configurations of the microscopic particles (atoms, ions,molecules) among the energy levels in a particular state of a system, the greater theentropy of the system.

¢H = 0.¢U = 0.

1U2

19-2

1¢H2

(a) Initial condition

(b) After expansion into vacuum

FIGURE 19-1Expansion of an ideal gasinto a vacuum(a) Initially, an ideal gas isconfined to the bulb onthe left at 1.00 atm pressure.(b) When the stopcock isopened, the gas expands intothe identical bulb on theright. The final condition isone in which the gas isequally distributed betweenthe two bulbs at a pressureof 0.50 atm.

▲

KEEP IN MIND that U and H are related asfollows: and

For afixed amount of an ideal gasat a constant temperature,

and ¢H = ¢U.PV = constant, ¢1PV2 = 0,

¢H = ¢U + ¢1PV2.H = U + PV,

▲

Jansen, Michael P. “The Costof Converting a Gasoline-

Powered Vehicle to Propane.A Practical Review Problem forSenior High School or IntroductoryChemistry.” J. Chem. Educ. 2000: 77,1578 (December 2000).

Take two dice and show thatrolling a 7 is most common

( chance) and a 2 or 12 leastcommon ( ). For 10 dice, thechance of all ten showing a 1 is 1 in60 million! The random roll is mostlikely and dominates the process.Use this to convince students thatthe random states dominate andthe ordered states can be ignored.

1>361>6

Mixing of Gases activity

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19-2 The Concept of Entropy 783

(a) Before mixing

(b) After mixing

Gas A Gas B

FIGURE 19-2The mixing of ideal gasesThe total volume of thesystem and the total gaspressure remain fixed. The netchange is that (a) beforemixing, each gas is confinedto half of the total volume(a single bulb) at a pressure of1.00 atm, and (b) after mixing,each gas has expanded intothe total volume (both bulbs)and exerts a partial pressureof 0.50 atm.

▲

A bust marking LudwigBoltzmann’s tomb in ViennaBoltzmann’s famous equationis inscribed on the tomb.At the time of Boltzmann’sdeath, the term “log” wasused for both naturallogarithms and logarithms tothe base ten; the symbol “ln”had not yet been adopted.

▲

Entropy is denoted by the symbol S. Like internal energy and enthalpy,entropy is a function of state (see page 239). It has a unique value for a systemwhose temperature, pressure, and composition are specified. The entropychange, is the difference in entropy between two states of a system, and italso has a unique value.

In the gas expansion of Figure 19-1, the entropy of the gas increases andIn the mixing of gases as carried out in Figure 19-2, entropy also

increases, a fact that we can represent symbolically.

Because both of these expansions occur spontaneously and neither is accom-panied by a change in internal energy or enthalpy, it seems possible thatincreases in entropy underlie spontaneous processes. This is a proposition that wewill have to examine more closely later, but let’s accept it tentatively for now.

THE BOLTZMANN EQUATION FOR ENTROPY

The connection between macroscopic changes, such as the mixing of gases,and the microscopic nature of matter was enunciated by Ludwig Boltzmann.The conceptual breakthrough that Boltzmann made was to associate the num-ber of energy levels in the system with the number of ways of arranging theparticles (atoms, ions, or molecules) in these energy levels. The microscopicenergy levels are also called states, and the particular way a number of parti-cles is distributed among these states is called a microstate. The more states agiven number of particles can occupy, the more microstates the system has.The more microstates that exist, the greater the entropy. Boltzmann derivedthe relationship

(19.1)

where S is the entropy, k is the Boltzmann constant, and W is the number ofmicrostates. We can think of the Boltzmann constant as the gas constant permolecule; that is, (Although we didn’t specifically introduce k inthe discussion of kinetic–molecular theory, appears in equation 6.21.)The number of microstates, W, is the number of ways that the atoms or mole-cules can be positioned in the states available and still give rise to the sametotal energy. Each permissible arrangement of the particles constitutes one ofthe microstates, so W is the total number of microstates that correspond to thesame energy.

How can we use Boltzmann’s equation to think about the distribution of mi-croscopic particles among the energy levels of a system? Let’s consider againthe particle-in-a-box model for a matter wave (page 301). Specifically, we canuse the equation to calculate a few energy levels for a matterwave in a one-dimensional box. Representative energy levels are shown in theenergy-level diagrams in Figure 19-3a for a particle in boxes of lengths L, 2L,and 3L. We see the following relationship between the length of the box and thenumber of levels: L, three levels; 2L, six levels; and 3L, nine levels. As the bound-aries of the box are expanded, the number of available energy levels increasesand the separation between levels decreases. Extending this model to three-dimensional space and large numbers of gas molecules, we find less crowdingof molecules into a limited number of energy levels when the pressure ofthe gas drops and the molecules expand into a larger volume. Thus, there is agreater spreading of the energy, and the entropy increases.

We can use the same particle-in-a-box model to understand the effect of rais-ing the temperature of a substance on the entropy of the system. We will con-sider a gas, but our conclusions are equally valid for a liquid or solid. At lowtemperatures, the molecules have a low energy, and the gas molecules can oc-cupy only a few of the energy levels; the value of W is small, and the entropy is

n2 h2>8mL2=EW

R>NA

k = R>NA.

S = k ln W

¢S = Smix of gases - [SA(g) + SB(g)] 7 0

A(g) + B(g) ¡ mixture of A(g) and B(g)

¢S 7 0.

≤S,

PETRMC19_780-821-hr 12/30/05 2:02 PM Page 783

784 Chapter 19 Spontaneous Change: Entropy and Free Energy

low. As the temperature is raised, the energy of the molecules increases and themolecules have access to a larger number of energy levels. Thus the number ofaccessible microstates increases and the entropy rises (Fig. 19-3b).

In summary, the state of a thermodynamic system can be described in twoways: the macroscopic description, in terms of state functions P, V, and T; andthe microscopic description, requiring a knowledge of the position and veloc-ity of every particle (atom or molecule) in the system. Boltzmann’s equationprovides the connection between the two.

ENTROPY CHANGE

An entropy change is based on two measurable quantities: heat and tem-perature Both of these factors affect the availability of energy levels to themicroscopic particles of a system. The following equation relates these factorsto an entropy change,

(19.2)

where T is the Kelvin temperature. Notice that is directly proportional tothe quantity of heat because the more energy added to a system (as heat), thegreater the number of energy levels available to the microscopic particles.Raising the temperature also increases the availability of energy levels, but fora given quantity of heat the proportional increase in number of energy levelsis greatest at low temperatures. That is why is inversely proportional to theKelvin temperature.

¢S

¢S

¢S =qrev

T

1T2. 1q2

1W2

E9E3 E6

E8

E7E5

E6E4E2

E3E5

E4

E2E1

E1

E3

E2E1

3L2LL

(a)

E6

E5

E4

E3

E2

E1

2L

E6

E5

E4

E3

E2

E1

2L

(b)

Ene

rgy

Ene

rgy

High TLow T

FIGURE 19-3Energy levels for a particle in a one-dimensional box(a) The energy levels of a particle in a box become more numerous and closer togetheras the length of the box increases. The range of thermally accessible levels is indicatedby the tinted band. The solid circles signify a system consisting of 15 particles. Eachdrawing represents a single microstate of the system. Can you see that as the boxlength increases there are many more microstates available to the particles? As thenumber of possible microstates for a given total energy increases, so does the entropy.(b) More energy levels become accessible in a box of fixed length as the temperature israised. Because the average energy of the particles also increases, the internal energyand entropy both increase as the temperature is raised.

▲

Sattar, Simeen.“Thermodynamics of Mixing

Real Gases.” J. Chem. Educ. 2000:77, 1361 (October 2000).

Note that whenThis can only

happen at absolute zero.Noting this fact can preparestudents for the third law ofthermodynamics and theobservation of residual entropyat T = 0 K.

W = 1, S = 0.

The number of accessiblestates, W, is determined bythe properties of a system.

When a process takes place, thecoupling must make the systemlarger; hence W must increase asthe system spontaneously moves toa different equilibrium.

PETRMC19_780-821-hr 12/30/05 2:02 PM Page 784

Are You Wondering . . .If there is a natural way to incorporate the notion of infinitesimal changes inderiving equation (19.2)?

The infinitesimal change in entropy, dS, that accompanies an infinitesimal rever-sible heat flow, is Now imagine the change in a system fromstate 1 to state 2 is carried out in a series of such infinitesimal reversible steps.Summation of all these infinitesimal quantities through the calculus technique ofintegration yields

If the change of state is isothermal (carried out at constant temperature), we canwrite

and we have recovered the definition of entropy change in (19.2).Starting with equivalent expressions for can be related to other

system properties. For the isothermal, reversible expansion of an ideal gas, leading to equation (19.10) on page 798, which describes in terms of

gas volumes. For a reversible change in temperature, which leads tothe entropy change for a change in temperature (see Exercise 86).

CP dT,=dqrev

¢S-dwrev ,=dqrev

¢Sdqrev ,

¢S = L

dqrev

T=

1T

L dqrev =qrev

T

¢S = L

dqrev

T

¢S.

dqrev>T.=dSdqrev ,

19-2 The Concept of Entropy 785

Equation (19.2) appears simple, but it is not. If S is to be a function of state,for a system must be independent of the path by which heat is lost or

gained. Conversely, because the value of q ordinarily depends on the path cho-sen (recall page 243), equation (19.2) holds only for a carefully defined path.The path must be reversible, that is, As we learned in Chapter 7, a re-versible process can be made to reverse its direction when just an infinitesimalchange in the opposite direction is made in a system property (review Figure7-12). Because q has the unit J and has the unit the unit of entropychange, is or

In some instances, it is difficult to construct mental pictures to assess howthe entropy of a system changes during a process. However, in many cases, anincrease or decrease in the accessibility of energy levels for the microscopic par-ticles of a system parallels an increase or decrease in the number of microscopicparticles and the space available to them. As a consequence, we can often makequalitative predictions about entropy change by focusing on those two factors.Let’s test this idea by considering again the three spontaneous, endothermicprocesses listed at the conclusion of Section 19-1 and illustrated in Figure 19-4.

In the melting of ice, a crystalline solid is replaced by a less structured liq-uid. Molecules that were relatively fixed in position in the solid, being limitedto vibrational motion, are now free to move about a bit. The molecules havegained some translational and rotational motion. The number of accessiblemicroscopic energy levels has increased, and so has the entropy.

In the vaporization process, a liquid is replaced by the even less structuredgas. Molecules in the gaseous state, because they can move within a large freevolume, have many more accessible energy levels than do those in the liquidstate. In the gas, energy can be spread over a much greater number of micro-scopic energy levels than in the liquid. The entropy of the gaseous state ismuch higher than that of the liquid state.

In the dissolving of ammonium nitrate in water, for example, a crystallinesolid and a pure liquid are replaced by a mixture of ions and water moleculesin the liquid (solution) state. This situation is somewhat more involved thanthe first two because some decrease in entropy is associated with the clusteringof water molecules around the ions due to ion–dipole forces. The increase in

JK-1.J>K,¢S,K-1,1>T

q = qrev .

¢S

Concerning reversibleprocesses, draw an isothermon a P–V plot. If the system

moves from one state to anothersuch that the system remains onthe isotherm, the process isreversible. If the ideal gas law isnot obeyed, then the system movesoff the isotherm as it starts itsirreversible process, and returns tothe isotherm when completed.This is quite easy to draw andmakes the distinction betweenreversible and irreversibleprocesses easier to grasp.

Lattice Vibrations activity

PETRMC19_780-821-hr 12/30/05 2:02 PM Page 785

786 Chapter 19 Spontaneous Change: Entropy and Free Energy

FIGURE 19-4Three processes in which entropy increasesEach of the processes pictured—(a) the melting of asolid, (b) the evaporation of a liquid, and (c) the dissolving of a solute—results in an increase inentropy. For part (c), the generalization works bestfor nonelectrolyte solutions, in which ion–dipoleforces do not exist.

▲

entropy that accompanies the destruction of the solid’s crystalline lattice pre-dominates, however, and for the overall dissolution process, In each ofthe three spontaneous, endothermic processes discussed here, the increase inentropy outweighs the fact that heat must be absorbed and each process is spontaneous.

In summary, there are four situations that generally produce an increasein entropy.

• Pure liquids or liquid solutions are formed from solids.

• Gases are formed from either solids or liquids.

• The number of molecules of gas increases as a result of a chemical reaction.

• The temperature of a substance increases. (Increased temperature meansan increased number of accessible energy levels for the increased molec-ular motion, whether it be vibrational motion of atoms or ions in a solid,or translational and rotational motion of molecules in a liquid or gas.)

We apply these generalizations in Example 19-1.

E X A M P L E 1 9 - 1Making Qualitative Predictions of Entropy Changes in Chemical Processes.Predict whether each of the following processes involves an increase or decrease inentropy or whether the outcome is uncertain.

(a) The decomposition of ammonium nitrate (a fertilizer and a highlyexplosive compound):

(b) The conversion of to (a key step in the manufacture of sulfuricacid):

(c) The extraction of sucrose from cane sugar juice:

(d) The “water gas shift” reaction (involved in the gasification of coal):

Solution(a) Here, a solid yields a large quantity of gas. Entropy increases.(b) Three moles of gaseous reactants produce two moles of gaseous prod-

ucts. The loss of one mole of gas indicates a loss of volume available to asmaller number of gas molecules. This loss reduces the number of pos-sible configurations for the molecules in the system and the number ofaccessible microscopic energy levels. Entropy decreases.

H2(g).+CO2(g)¡H2O(g)+CO(g)

C12H22O11(s)¡C12H22O11(aq)

2 SO3(g).¡O2(g)+2 SO2(g)SO3SO2

O2(g).+4 H2O(g)+2 N2(g)¡2 NH4NO3(s)

1¢H 7 02,1¢S 7 02¢S 7 0.

Solid Liquid

(a) Melting: Sliquid � Ssolid

Liquid Vapor

(b) Vaporization: Svapor � Sliquid

�

SolutionSolventSolute

(c) Dissolving: Ssoln � (Ssolvent � Ssolute)

Enthalpy of SolutionFormation activity

Craig, N. C. “Campbell’sRule for Estimating Entropy

Changes in Gas-Producing andGas-Consuming Reactions andRelated Generalizations aboutEntropies and Enthalpies” J. Chem.Educ. 2003: 80, 1432.

Lambert, F. L. “EntropyIs Simple, Qualitatively”

J. Chem. Educ. 2002: 79, 1241.

Jensen, W. B. “Entropyand Constraint of Motion”

J. Chem. Educ. 2004: 81, 639.

Estimation of EntropyChanges activity

PETRMC19_780-821-hr 12/30/05 2:02 PM Page 786

19-3 Evaluating Entropy and Entropy Changes 787

(c) The sucrose molecules are reduced in mobility and in the number offorms in which their energy can be stored when they leave the solutionand arrange themselves into a crystalline state. Entropy decreases.

(d) The entropies of the four gases are likely to be different because theirmolecular structures are different. The number of moles of gases is thesame on both sides of the equation, however, so the entropy change islikely to be small if the temperature is constant. On the basis of just thegeneralizations listed on page 786, we cannot determine whether en-tropy increases or decreases.

Practice Example A: Predict whether entropy increases or decreases in each ofthe following reactions. (a) The Claus process for removing from natural gas:

(b) the decomposition of mercury(II)oxide:

Practice Example B: Predict whether entropy increases or decreases orwhether the outcome is uncertain in each of the following reactions. (a)

(b) the chlor-alkali process,

CONCEPT ASSESSMENT ✓Figure 19-1 illustrates a spontaneous process through the expansion of an ideal gasinto an evacuated bulb. Use the one-dimensional particle-in-a-box model torepresent the initial condition of Figure 19-1. Use a second particle-in-a-box modelto represent the system after expansion into the vacuum. Use these models to explainon a microcopic basis why this expansion is spontaneous.[Hint: Assume that the volume of the bulbs is analogous to the length of the box.]

EVALUATING ENTROPY AND ENTROPY CHANGES

The difficulty of calculating an entropy change with equation (19.2) was men-tioned on page 785. The emphasis then shifted to making qualitative predictionsabout entropy changes. In this section we will see that in a few instances a simpledirect calculation of is possible. Also, we will find that, unlike the case withinternal energy and enthalpy, it is possible to determine absolute entropy values.

PHASE TRANSITIONS

In the equilibrium between two phases, the exchange of heat can be carried outreversibly, and the quantity of heat proves to be equal to the enthalpy change forthe transition, In this case, equation (19.2) can be written as

(19.3)

Rather than use the general symbol “tr” to represent a transition, we can bemore specific about just which phases are involved, such as “fus” for the melt-ing of a solid and “vap” for the vaporization of a liquid. If the transitionsinvolve standard-state conditions ( pressure), we also use thedegree sign (°). Thus for the melting (fusion) of ice at its normal melting point,

the standard entropy change is

Entropy changes depend on the quantities of substances involved and are usu-ally expressed on a per-mole basis.

= 22.0 J mol-1 K-1

¢Sfus° =¢Hfus°Tmp

=6.02 kJ mol-1

273.15 K= 2.20 * 10-2 kJ mol-1 K-1

H2O(s, 1 atm) ∆ H2O(l, 1 atm) ¢Hfus° = 6.02 kJ at 273.15 K

1 bar � 1 atm

¢Str =¢Htr

Ttr

¢Htr .

¢S

19-3

Cl2(g).+H2(g)+2 H2O(l) electrolysis

" 2 OH-(aq)

+2 Cl-(aq)2 Ag(s);+ZnO(s)¡Ag2O(s)+Zn(s)

O2(g).+2 Hg(l)¡2 HgO(s)2 H2O(g);+3 S(s)¡SO2(g)+2 H2S(g)

H2S

Brown, R. J. C., Brown, R. F. C. “Melting Point and

Molecular Symmetry.” J. Chem.Educ. 2000: 77, 724 (June 2000).

KEEP IN MINDthat the normal melting pointand normal boiling point aredetermined at 1 atm pressure.The difference between 1 atmand the standard state pres-sure of 1 bar is so small thatwe can usually ignore it.

▲

Remind students thatenthalpy change varieswith temperature. However,

in most cases these changes areignored as discussed later on inthis chapter.

PETRMC19_780-821-hr 12/30/05 2:02 PM Page 787

Are You Wondering . . .If the thermodynamic entropy (equation 19.2), can be derivedfrom the statistical entropy (equation 19.1),

It can, but a rigorous derivation is very complex. However, by using a system ofevenly spaced energy levels for the surroundings, a simplified derivation can beshown. Consider just two energy levels from the numerous energy levels repre-senting the surroundings. Now, as suggested in Figure 19-5, construct two states,A and B, which are the most probable initial state (state A) and final state (state B)after the addition of an infinitesimal amount of heat

The temperature of B is slightly higher than the temperature of A. The amountof heat used is just the energy difference between the two levelsand therefore a single particle jumps from level i to level j. The number of Amicrostates, is almost exactly the same as the number of B microstates, The entropy change for the surroundings is

However the ratio of probabilities is related to the population of each state asshown in the following relationship

Boltzmann derived the following distribution law

which relates the population of level i to that of level j. We use this to determinethe entropy change of the surroundings along with the substitution of for

We have shown the relationship of the statistical entropy (equation 19.1) to thethermodynamic entropy (equation 19.2) to be true for the entropy change of the sur-roundings; however, this can be shown to hold for the system as well.

¢Ssur = k ln ¢WB

WA≤ = k ln 1eqrev>kT2 = k

qrev

kT=

qrev

T

¢e.qrev

nj

ni= e-¢e>kT

WB

WA=

ni

nj + 1L

ni

nj

¢Ssur = k ln ¢WB

WA≤

WB.WA,

1¢e21q21qrev2.

S � k ln W ?≤S � qrev/T,

788 Chapter 19 Spontaneous Change: Entropy and Free Energy

FIGURE 19-5Energy levels and populationsrepresenting the initial state of thesurroundings (State A) and thestate of the surroundings (State B)upon the addition of a quantityof heat (q). The spacing betweenthe energy levels is . Thepopulations of the energy levelsare indicated by and .njni

¢e

▲ �� � q

ni

nj nj � 1

ni � 1

State A State B

q

PETRMC19_780-821-hr 12/30/05 2:02 PM Page 788

19-3 Evaluating Entropy and Entropy Changes 789

Phase transitions are goodexamples of almostreversible processes at the

melting or boiling points. Forexample, at the melting point ofwater heat slowly moves betweenthe system and surroundings andeither freezes or melts the water. Ifthe temperature is above or belowthe melting point, then one phaseis favored over the other.

E X A M P L E 1 9 - 2Determining the Entropy Change for a Phase Transition. What is the standardmolar entropy for the vaporization of water at 373 K given that the standard molarenthalpy of vaporization is

SolutionAlthough a chemical equation is not necessary, writing one can help us see theprocess for which we seek the value of

Practice Example A: What is the standard molar entropy of vaporization,for a chlorofluorocarbon that once was heavily used in refrigeration

systems? Its normal boiling point is and

Practice Example B: The entropy change for the transition from solid rhombicsulfur to solid monoclinic sulfur at 95.5 °C is What is thestandard molar enthalpy change, for this transition?

A useful generalization known as Trouton’s rule states that for many liq-uids at their normal boiling points, the standard molar entropy of vaporizationhas a value of about

(19.4)

For instance, the values of for benzene and octane are87.1 and respectively. If the increased accessibility of micro-scopic energy levels produced in transferring one mole of molecules from liq-uid to vapor at 1 atm pressure is roughly comparable for different liquids,then we should expect similar values of

Instances in which Trouton’s rule fails are also understandable. In waterand in ethanol, for example, hydrogen bonding among molecules produces alower entropy than would otherwise be expected in the liquid state. Conse-quently, the entropy increase in the vaporization process is greater than nor-mal, and so

The entropy concept helps explain Raoult’s law (Section 13-6). Recall thatfor an ideal solution, and intermolecular forces of attraction arethe same as in the pure liquid solvent (page 533). Thus, we expect the molar

to be the same whether vaporization of solvent occurs from an ideal so-lution or from the pure solvent at the same temperature. So, too, should be the same because When one mole of solvent is trans-ferred from liquid to vapor at the equilibrium vapor pressure P°, entropy in-creases by the amount As shown in Figure 19-6, because the entropy ofthe ideal solution is greater than that of the pure solvent, the entropy of thevapor produced by the vaporization of solvent from the ideal solution is alsogreater than the entropy of the vapor obtained from the pure solvent. For thevapor above the solution to have the higher entropy, its molecules must havea greater number of accessible microscopic energy levels. In turn, the vapormust be present in a larger volume and, hence, must be at a lower pressurethan the vapor coming from the pure solvent. This relationship corresponds toRaoult’s law: xA PA° .=PA

¢Svap.

¢Hvap>T.=¢Svap

¢Svap

¢Hvap

¢Hsoln = 0

K-1.¢Svap° 7 87 J mol-1

¢Svap° .

86.2 J mol-1 K-1,(C8H18)(C6H6)¢Svap°

¢Svap° =¢Hvap°

TbpL 87 J mol-1 K-1

K-1.87 J mol-1

¢Htr° ,1.09 J mol-1 K-1.=¢Str°

¢Hvap° = 20.2 kJ mol-1.-29.79 °C,CCl2F2,¢Svap° ,

= 109 J mol-1 K

¢Svap° =¢Hvap°

Tbp=

40.7 kJ mol-1

373 K= 0.109 kJ mol-1K-1

¢Svap° = ?

¢Hvap° = 40.7 kJ>mol H2OH2O(l, 1 atm) ∆ H2O(g, 1 atm)

¢Svap° .

40.7 kJ mol-1?

Vapor at PP � P �

�Svap�Svap

�Ssoln

Puresolvent

Idealsolution

Vapor at P�

Ent

ropy

FIGURE 19-6An entropy-basedrationale of Raoult’s lawIf has the same valuefor vaporization from thepure solvent and from anideal solution, the equilibriumvapor pressure is lowerabove the solution: P 6 P°.

¢Svap

▲

We found the value offor water at 373 K

to be 109 J inExample 19-2.

mol-1 K-1¢Svap°

▲

PETRMC19_780-821-hr 12/30/05 2:02 PM Page 789

790 Chapter 19 Spontaneous Change: Entropy and Free Energy

ABSOLUTE ENTROPIES

To establish an absolute value of the entropy of a substance, we look for a con-dition in which the substance is in its lowest possible energy state, called thezero-point energy. The entropy of this state is taken to be zero. Then we evaluateentropy changes as the substance is brought to other conditions of temperatureand pressure. We add together these entropy changes and obtain a numericalvalue of the absolute entropy. The principle that permits this procedure is thethird law of thermodynamics, which can be stated as follows:

The entropy of a pure perfect crystal at 0 K is zero.

Figure 19-7 illustrates the method outlined in the preceding paragraph fordetermining absolute entropy as a function of temperature. Where phase tran-sitions occur, equation (19.3) is used to evaluate the corresponding entropychanges. Over temperature ranges in which there are no transitions, valuesare obtained from measurements of specific heats as a function of temperature.

The absolute entropy of one mole of a substance in its standard state iscalled the standard molar entropy, Standard molar entropies of a numberof substances at 25 °C are tabulated in Appendix D. To use these values to cal-culate the entropy change of a reaction, we use an equation with a familiarform (recall equation 7.21).

(19.5)

The symbol means “the sum of,” and the terms added are the products ofthe standard molar entropies and the corresponding stoichiometric coeffi-cients, Example 19-3 shows how to use this equation.

E X A M P L E 1 9 - 3Calculating Entropy Changes from Standard Molar Entropies. Use data fromAppendix D to calculate the standard molar entropy change for the conversion ofnitrogen monoxide to nitrogen dioxide (a step in the manufacture of nitric acid).

2 NO(g) + O2(g) ¡ 2 NO2(g) ¢S298 K° = ?

n.

a¢S° = ca np S°(products) - a nr S°(reactants) d

S°.

¢S°

Temperature Dependenceof Entropy simulation

Stan

dard

mol

ar e

ntro

py, J

mol

�1 K

�1

200

50 100 150 200 250 300Temperature, K

100

150

250

50 Solid

(s) � (1)

(1) � (g)

Liquid

Gas

FIGURE 19-7Molar entropy as a function of temperatureThe standard molar entropy of methylchloride, is plotted at varioustemperatures from 0 to 300 K, withthe phases noted. The vertical segmentbetween the solid and liquid phasescorresponds to the other verticalsegment, to By the third lawof thermodynamics, an entropy of zero isexpected at 0 K. Experimental methodscannot be carried to that temperature,however, so an extrapolation is required.

¢Svap.¢Sfus ;

CHCl3,

▲

PETRMC19_780-821-hr 12/30/05 2:02 PM Page 790

19-4 Criteria for Spontaneous Change: The Second Law of Thermodynamics 791

SolutionEquation (19.5) takes the form

Some qualitative reasoning can be applied as a useful check on this calculation.Because three moles of gaseous reactants produce only two moles of gaseous prod-ucts, the entropy should decrease; that is, should be negative.

Practice Example A: Use data from Appendix D to calculate the standardmolar entropy change for the synthesis of ammonia from its elements.

Practice Example B: is an unstable oxide that readily decomposes. Thedecomposition of 1.00 mol of to nitrogen monoxide and nitrogen dioxide at25 °C is accompanied by the entropy change What is the stan-dard molar entropy of at 25 °C?

Example 19-3 used the standard molar entropies of and NO(g).Why is the value for greater than that of NO(g),210.8 J Entropy increases when a substance absorbs heat (recallthat ), and some of this heat goes into raising the average transla-tional kinetic energies of molecules. But there are other ways for heat energyto be used. One possibility, pictured in Figure 19-8, is that the vibrational ener-gies of molecules can be increased. In the diatomic molecule NO(g), only onetype of vibration is possible; in the triatomic molecule three types arepossible. Because there are more possible ways of distributing energy among

molecules than among NO molecules, has a higher molar en-tropy than does NO(g) at the same temperature. Thus the following statementshould be added to the generalizations about entropy on page 786.

• In general, the more complex their molecules (that is, the greater thenumber of atoms present), the greater the molar entropies of substances.

NO2(g)NO2

NO2(g),

¢S = qrev>Tmol-1 K-1?

NO2(g), 240.1 J mol-1 K-1,NO2(g)

N2O3(g)¢S° = 138.5 J K-1.

N2O3

N2O3

N2(g) + 3 H2(g) ¡ 2 NH3(g) ¢S298 K° = ?

¢S°

= 12 * 240.12 - 12 * 210.82 - 205.1 = -146.5 J K-1

¢S° = 2SNO2(g)° - 2SNO(g)° - SO2(g)°

In general, at lowtemperatures, because thequanta of energy are so small,translational energies are mostimportant in establishing theentropy of gaseous molecules.As the temperature increasesand the quanta of energybecome larger, first rotationalenergies become important,and finally, at still highertemperatures, vibrationalmodes of motion start to contribute to the entropy.

▲

(a) (b)

N N N NO

O O O OO O

Methane, CH4S � � 186.3 J mol�1 K�1

Ethane, C2H6S � � 229.6 J mol�1 K�1

Propane, C3H8S � � 270.3 J mol�1 K�1

CRITERIA FOR SPONTANEOUS CHANGE:THE SECOND LAW OF THERMODYNAMICS

In Section 19-2, we came to the tentative conclusion that processes in which theentropy of a system increases should be spontaneous and that processes inwhich entropy decreases should be nonspontaneous. But this statement canpresent difficulties, for example, how to explain the spontaneous freezing of

19-4

FIGURE 19-8Vibrational energy and entropyThe movement of atoms is suggested by the arrows. (a) The NO molecule has only onetype of vibrational motion, whereas (b) the molecule has three. This differencehelps account for the fact that the molar entropy of is greater than that of NO(g).NO2(g)

NO2

▲

Formation of AluminumBromide movie

Molecular Motions activity

PETRMC19_780-821-hr 12/30/05 5:01 PM Page 791

792 Chapter 19 Spontaneous Change: Entropy and Free Energy

Calculate a residual entropyfor, say, one mole of CO.Carbon monoxide has a

small dipole and it freezes withsome residual order. That is, ratherthan a lattice at K of COCOCOCO somemolecules are flipped COOCOCOCCO Each CO canbe either CO or OC in the lattice,so there is a residual order of

giving a residualentropy of

R ln 2, where R is the gasconstant. This agrees quite wellwith experimental results.

=ln 2ln122N = kNS = k

W = 122NÁ .

ÁÁ ,

ÁT = 0

water at Because crystalline ice has a lower molar entropy than doesliquid water, the freezing of water is a process for which entropy decreases. Theway out of this dilemma is to recognize that two entropy changes must alwaysbe considered simultaneously—the entropy change of the system itself and theentropy change of the surroundings. The criteria for spontaneous change mustbe based on the sum of the two, called the entropy change of the “universe”:

(19.6)

Although it is beyond the scope of this discussion to verify the followingexpression, the expression provides the basic criterion for spontaneous change.In a spontaneous change,

(19.7)

Equation (19.7) is one way of stating the second law of thermodynamics.Another way is through the following statement.

All spontaneous processes produce an increase in the entropy of the universe.

According to expression (19.7), if a process produces positive entropychanges in both the system and its surroundings, the process is surely sponta-neous. And if both these entropy changes are negative, the process is just assurely nonspontaneous. If one of the entropy changes is positive, and theother negative, whether the sum of the two is positive or negative depends onthe relative magnitudes of the two changes. The freezing of water produces anegative entropy change in the system, but in the surroundings, which absorbheat, the entropy change is positive. As long as the temperature is below 0 °C,the entropy of the surroundings increases more than the entropy of the systemdecreases. Because the total entropy change is positive, the freezing of waterbelow 0 °C is indeed spontaneous.

FREE ENERGY AND FREE ENERGY CHANGE

We could use expression (19.7) as the basic criterion for spontaneity (sponta-neous change), but it would be very difficult to apply. To evaluate a total en-tropy change we always have to evaluate for the surroundings. Atbest, this process is tedious, and in many cases it is not even possible, becausewe cannot figure out all the interactions between a system and its surround-ings. Surely it is preferable to have a criterion that can be applied to the systemitself, without having to worry about changes in the surroundings.

To develop this new criterion, let us explore a hypothetical process conductedat constant temperature and pressure and with work limited to pressure–volume work. This process is accompanied by a heat effect, which is equal to

for the system as seen in Section 7-6. The heat effect experienced bythe surroundings is the negative of that for the system: Furthermore, if the hypothetical surroundings are large enough, the path bywhich heat enters or leaves the surroundings can be made reversible. That is,the quantity of heat can be made to produce only an infinitesimal change in thetemperature of the surroundings. In this case, according to equation (19.2), theentropy change in the surroundings is * Now substitute thisvalue of into equation (19.6), then multiply by T to obtain

Finally, multiply by (change signs).

(19.8)-T ¢Suniv = ¢Hsys - T ¢Ssys

-1

T ¢Suniv = T ¢Ssys - ¢Hsys = -1¢Hsys - T ¢Ssys2¢Ssurr

- ¢Hsys>T.=¢Ssurr

- ¢Hsys .=qsurr = -qp

(¢Hsys)¢Hqp,

¢S(¢Suniv),

¢Suniv = ¢Ssys + ¢Ssurr 7 0

¢Stotal = ¢Suniverse = ¢Ssystem + ¢Ssurroundings

-10 °C.

Many aspects of free energycan be illustrated by water

falling and pooling at variouslevels. Construct an apparatus withglass bowls that you can cascadewater into and over. This analogyis often helpful for students tovisualize spontaneous processes:Variations in the height of thefalling water, equilibrium,maintenance of production (flow),and local pooling can all be easilyfollowed and related to free energy.

*We cannot similarly substitute for A process that occurs spontaneously isgenerally far removed from an equilibrium condition and is therefore irreversible. We cannotsubstitute q for an irreversible process into equation (19.2).

¢Ssys .¢Hsys>T

PETRMC19_780-821-hr 12/30/05 2:02 PM Page 792

19-4 Criteria for Spontaneous Change: The Second Law of Thermodynamics 793

This is the significance of equation (19.8). The right side of this equation hasterms involving only the system. On the left side appears the term which embodies the criterion for spontaneous change, that for a spontaneousprocess,

Equation (19.8) is generally cast in a somewhat different form which re-quires a new thermodynamic function, variously called the Gibbs energy orthe free energy, G. The free energy for a system is defined by the equation

The free energy change, for a process at constant T is

(19.9)

In equation (19.9), all the terms apply to measurements on the system. All refer-ence to the surroundings has been eliminated. Also, when we compare equa-tions (19.8) and (19.9), we get

Now, by noting that is negative when is positive, we have our finalcriterion for spontaneous change based on properties of only the system itself.

For a process occurring at constant T and P, these statements hold true.

• If (negative), the process is spontaneous.

• If (positive), the process is nonspontaneous.• If (zero), the process is at equilibrium.

Evaluation of the units in equation (19.9) shows that free energy is indeedan energy term. has the unit joules (J), and the product has the units

is the difference in two quantities with units of energy.

APPLYING THE FREE ENERGY CRITERIA FOR SPONTANEOUS CHANGE

Later, we will look at quantitative applications of equation (19.9), but for nowwe can use the equation to make some qualitative predictions. Altogether thereare four possibilities for on the basis of the signs of and These pos-sibilities are outlined in Table 19.1 and demonstrated in Example 19-4.

If is negative and is positive, the expression is neg-ative at all temperatures. The process is spontaneous at all temperatures. Thiscorresponds to the situation noted previously in which both and are positive and is also positive.

Unquestionably, if a process is accompanied by an increase in enthalpy (heatis absorbed) and a decrease in entropy, is positive at all temperatures andthe process is nonspontaneous. This corresponds to a situation in which both

and are negative and is also negative.¢Suniv¢Ssurr¢Ssys

¢G

¢Suniv

¢Ssurr¢Ssys

¢H - T ¢S=¢G¢S¢H

¢S.¢H¢G

¢GK * J>K = J.T ¢S¢H

¢G = 0¢G 7 0

¢G 6 0

¢Suniv¢G

¢G = -T ¢Suniv

¢G = ¢H - T ¢S

≤G,

G = H - TS

¢Suniv 7 0.

¢Suniv ,

J. Willard Gibbs(1839–1903)—a great“unknown” scientist of theUnited StatesGibbs, a Yale Universityprofessor of mathematicalphysics, spent most of hiscareer without recognition,partly because his work wasabstract and partly becausehis important publicationswere in little-read journals.Yet today, Gibbs’s ideasserve as the basis of most ofchemical thermodynamics.

▲

Sometimes the termis referred to as

“organizational energy,”because is related to theway the energy of a systemis distributed among theavailable energy levels.

¢S

“T ¢S”▲

TABLE 19.1 Criteria for Spontaneous Change:

Case Result Example

1. spontaneous at all temp.

2.

3.

4. nonspontaneous at all temp. 3 O2(g) ¡ 2 O3(g)+-+

2 NH3(g) ¡ N2(g) + 3 H2(g)nonspontaneous at low temp.spontaneous at high temp.

rb +-

++

H2O(l) ¡ H2O(s)spontaneous at low temp.nonspontaneous at high temp.

rb -+

--

2 N2O(g) ¡ 2 N2(g) + O2(g)-+-

≤G≤S≤H

≤G � ≤H � T≤S

The Balance BetweenEnthalpy and Entropy

PETRMC19_780-821-hr 12/30/05 2:02 PM Page 793

794 Chapter 19 Spontaneous Change: Entropy and Free Energy

Only because of the highactivation energy for the

decompositionreaction does exist at all.

NH4NO3(s)NH4NO3(s)

▲

A related observation isthat only a small fraction ofthe mass of the universe is inmolecular form.

▲

For cases 2 and 3, there isa particular temperature atwhich a process switches frombeing spontaneous to beingnonspontaneous. Section 19-6explains how to determinesuch a temperature.

▲

The questionable cases are those in which the entropy and enthalpy factorswork in opposition—that is, with and both negative or both positive.In these cases, whether a reaction is spontaneous or not (that is, whether is negative or positive) depends on temperature. In general, if a reactionhas negative values for both and it is spontaneous at lower tempera-tures, whereas if and are both positive, the reaction is spontaneous athigher temperatures.

E X A M P L E 1 9 - 4Using Enthalpy and Entropy Changes to Predict the Direction of SpontaneousChange. Under what temperature conditions would the following reactions occurspontaneously?

(a)(b)

Solution

(a) The reaction is exothermic, and in Example 19-1(a) we concluded thatbecause large quantities of gases are produced. With

and this reaction should be spontaneous at all temperatures(case 1 of Table 19.1).

(b) Because one mole of gaseous reactant produces two moles of gaseousproduct, we expect entropy to increase. But what is the sign of Wecould calculate from enthalpy of formation data, but there is noneed to. In the reaction, covalent bonds in are broken and no newbonds are formed. Because energy is absorbed to break bonds, mustbe positive. With and case 3 in Table 19.1 applies. is larger than at low temperatures, and the reaction is nonsponta-neous. At high temperatures, the term becomes larger than

becomes negative, and the reaction is spontaneous.

Practice Example A: Which of the four cases in Table 19.1 would apply to eachof the following reactions:

(a)(b)

Practice Example B: Under what temperature conditions would the followingreactions occur spontaneously? (a) The decomposition of calcium carbonate intocalcium oxide and carbon dioxide. (b) The “roasting” of zinc sulfide in oxygen toform zinc oxide and sulfur dioxide. This exothermic reaction releases 439.1 kJ forevery mole of zinc sulfide that reacts.

Example 19-4(b) illustrates why there is an upper temperature limit for thestabilities of chemical compounds. No matter how positive the value of for dissociation of a molecule into its atoms, the term will eventually ex-ceed in magnitude as the temperature increases. Known temperaturesrange from near absolute zero to the interior temperatures of stars (about

). Molecules exist only at limited temperatures (up to aboutor about 0.03% of this total temperature range).

The method of Example 19-4 is adequate for making predictions about thesign of but we will also want to use equation (19.9) to calculate numericalvalues. We will do that in Section 19-5.

¢G,

1 * 104 K3 * 107 K

¢HT ¢S

¢H

2 C(graphite) + 2 H2(g) ¡ C2H4(g), ¢H° = 52.26 kJ?N2(g) + 3 H2(g) ¡ 2 NH3(g), ¢H° = -92.22 kJ;

¢G¢H,T ¢S

T ¢S¢H¢S 7 0,¢H 7 0

¢HI2(g)

¢H¢H?

¢S 7 0,¢H 6 0¢S 7 0

I2(g) ¡ 2 I(g)2 NH4NO3(s) ¡ 2 N2(g) + 4 H2O(g) + O2(g) ¢H° = -236.0 kJ

¢S¢H¢S,¢H

¢G¢S¢H

Changes in enthalpy andentropy vary withtemperature, but the

differences can be ignored at smalltemperature intervals. Thetemperature dependence of freeenergy change is overwhelmedby the T factor in equation (19.9).

PETRMC19_780-821-hr 12/30/05 2:02 PM Page 794

Are You Wondering . . .What the term free energy signifies?

We might think that the quantity of energy available to do work in the surround-ings as a result of a chemical process is This would be the same as the quan-tity of heat that an exothermic reaction releases to the surroundings. (In thinkingalong those lines, we would say that an endothermic reaction is incapable of doingwork.) However, that quantity of heat must be adjusted for the heat requirementin producing the entropy change in the system If an exothermicreaction is accompanied by an increase in entropy, the amount of energy availableto do work in the surroundings is greater than If entropy decreases in theexothermic reaction, the amount of energy available to do work is less than But notice that in either case, this amount of energy is equal to Thus,the amount of work that can be extracted from a chemical process is so theGibb’s function G is called the free energy function. Notice also that this interpreta-tion of free energy allows for the possibility of work being done in an endothermicprocess if exceeds In Chapter 20, we will see how the free energy changeof a reaction can be converted to electrical work. In any case, do not think of freeenergy as being “free.” Costs are always involved in tapping an energy source.

¢H.T ¢S

- ¢G,- ¢G.

- ¢H.- ¢H.

1qrev = T ¢S2.

- ¢H.

19-5 Standard Free Energy Change, 795≤G°

STANDARD FREE ENERGY CHANGE, Because free energy is related to enthalpy we cannot establishabsolute values of G any more than we can for H. We must work with free en-ergy changes, We will find a special use for the free energy change corre-sponding to reactants and products in their standard states—the standard freeenergy change, The standard state conventions were introduced and ap-plied to enthalpy change in Chapter 7.

The standard free energy of formation, is the free energy change fora reaction in which a substance in its standard state is formed from its ele-ments in their reference forms in their standard states. And, as was the casewhen we established enthalpies of formation in Section 7-8, this definitionleads to values of zero for the free energies of formation of the elements in theirreference forms at a pressure of 1 bar. Other free energies of formation are re-lated to this condition of zero and are generally tabulated per mole of sub-stance (see Appendix D).

Some additional relationships involving free energy changes are similar tothose presented for enthalpy in Section 7-7: (1) changes sign when aprocess is reversed; and (2) for an overall process can be obtained by sum-ming the values for the individual steps. The two expressions that followare useful in calculating values, depending on the data available. We canuse the first expression at any temperature for which and values areknown. We can use the second expression only at temperatures at which values are known. The only temperature at which tabulated data are com-monly given is 298.15 K. The first expression is applied in Example 19-5 andPractice Example 19-5A, and the second expression in Practice Example 19-5B.

¢G° = ca np ¢Gf°(products) - a nr ¢Gf°(reactants) d¢G° = ¢H° - T¢S°

¢Gf°¢S°¢H°

¢G°¢G

¢G¢G

¢Gf° ,

≤G°.

¢G.

1G = H - TS2,≤G°19-5

Ask the students why isit called free energy—the explanation here gives

the reason.

KEEP IN MINDthat a pressure of 1 bar isvery nearly the same as1 atm. The difference in thesetwo pressures on the valuesof properties is generally sosmall that we can use thetwo pressure units almostinterchangeably.

▲

Many substances do notexist at the standardconditions that define their

free energy. This does not mattersince we can calculate the standardfree energy from nonstandardconditions. Dealing withnonstandard conditions isdiscussed on page 797. Freeenergies are listed under standardconditions for ease andconciseness.

PETRMC19_780-821-hr 12/30/05 2:02 PM Page 795

796 Chapter 19 Spontaneous Change: Entropy and Free Energy

This example says that allthe reactants and productsare maintained at 25 °C and

1 atm pressure. Under theseconditions, the free energy changeis for oxidizing two molesof NO to two moles of To dothis, it is necessary to replenish theNO so as to maintain the standardconditions.

NO2.-70.4 kJ

�H

�G � 0

�G � 0

T�S

Ene

rgy,

kJ

�G � 0

Temperature, K

FIGURE 19-9Free energy change as a function oftemperatureThe value of is given by the distancebetween the two lines; that is,

In this illustration, both and have positive values at all temperatures.If is greater than the value, and the reaction is nonspontaneous. If isless than and the reaction isspontaneous. At equilibrium, thisoccurs at the temperature at which the twolines intersect. An assumption made here isthat and are essentially independent of temperature.

¢S¢H

¢G = 0;T ¢S, ¢G 6 0

¢H¢G 7 0T ¢S¢H

¢S¢H¢G = ¢H - T ¢S.

¢G

▲

E X A M P L E 1 9 - 5Calculating for a Reaction. Determine at 298.15 K for the reaction

SolutionBecause we have values of and the most direct method of calculating is to use the expression In doing so, we must first convert allthe data to a common energy unit (for instance, kJ).

Practice Example A: Determine at 298.15 K for the reaction and

Practice Example B: Determine for the reaction in Example 19-5 by usingdata from Appendix D. Compare the two results.

FREE ENERGY CHANGE AND EQUILIBRIUM

We have seen that for spontaneous processes and that fornonspontaneous processes. If the forward and reverse processesshow an equal tendency to occur, and the system is at equilibrium. At thispoint, even an infinitesimal change in one of the system variables (such astemperature or pressure) will cause a net change to occur. But if a system atequilibrium is left undisturbed, no net change occurs with time.

Let’s consider the hypothetical process outlined in Figure 19-9. If we start atthe left-hand side of the figure, we see that exceeds and that ispositive; the process is nonspontaneous. The magnitude of decreases withincreasing temperature. At the right-hand side of the figure, exceeds and is negative; the process is spontaneous. At the temperature at which thetwo lines intersect, and the system is at equilibrium.

For the vaporization of water, with both liquid and vapor in their standard states(which means that ), the intersection of the two lines in Figure 19-9is at (100.00 °C). That is, for the vaporization of water at 1 atm,

H2O(l, 1 atm) ∆ H2O(g, 1 atm) ¢G° = 0 at 373.15 K

T = 373.15 K¢G°=¢G

¢G = 0¢G

¢HT ¢S¢G

¢GT ¢S¢H

¢G = 0,¢G 7 0¢G 6 0

19-6

¢G°

-549.3 J K-1.=¢S°-1648 kJ=¢H°2 Fe2O3(s).¡3 O2(g)+4 Fe(s)¢G°

= -70.4 kJ = -114.1 kJ + 43.68 kJ

¢G° = -114.1 kJ - (298.15 K * -0.1465 kJ K-1)

¢H° - T ¢S°.=¢G°¢G°¢S°,¢H°

¢S° = -146.5 J K-1

2 NO(g) + O2(g) ¡ 2 NO2(g) (at 298.15 K) ¢H° = -114.1 kJ

¢G°¢G°

PETRMC19_780-821-hr 12/30/05 2:02 PM Page 796

19-6 Free Energy Change and Equilibrium 797

At 25 °C, the line lies above the line in Figure 19-9. This means that

The positive value of does not mean that vaporization of water will notoccur. From common experience, we know that water spontaneously evapo-rates at room temperature. What the positive value means is that liquid waterwill not spontaneously produce at 1 atm pressure at 25 °C. Instead,

is produced with a vapor pressure that is less than 1 atm pressure. Theequilibrium vapor pressure of water at 25 °C is that is,

Figure 19-10 offers a schematic summary of ideas concerning the transitionbetween liquid and gaseous water at 25 °C.

CONCEPT ASSESSMENT ✓Redraw Figure 19-9 for case 2 in Table 19.1. How does your drawing compare withFigure 19-9? State the similarities and differences.

RELATIONSHIP OF TO FOR NONSTANDARD CONDITIONS

If you think about the situation just described for the vaporization of water,there is not much value in describing equilibrium in a process in terms of its

value. At only one temperature are the reactants in their standard statesin equilibrium with products in their standard states; that is, at only one tem-perature does We want to be able to describe equilibrium for a vari-ety of conditions, typically nonstandard conditions. Many reactions, such asprocesses occurring under physiological conditions, take place under non-standard conditions. How, under such circumstances, can a biochemist decidewhich processes are spontaneous? For this we need to work with not

To obtain the relationship between and we will consider a reactionbetween ideal gas molecules. We assume this is the case in the reaction of ni-trogen with hydrogen to produce ammonia.

N2(g) + 3 H2(g) ∆ 2 NH3(g)

¢G°,¢G¢G°.¢G,

¢G° = 0.

¢G°

≤G≤G°

H2O(l, 0.03126 atm) ∆ H2O(g, 0.03126 atm) ¢G = 0

23.76 mmHg = 0.03126 atm;H2O(g)

H2O(g)

¢G°

H2O(l, 1 atm) ¡ H2O(g, 1 atm) ¢G° = +8.590 kJ at 298.15 K

¢G° 7 0.T ¢S¢H°

H2O(g, 1 atm)

(a)

H2O(1)

H2O(g, 23.76 mmHg)

(b)

H2O(1)

H2O(g, 10 mmHg)

(c)

H2O(1)

FIGURE 19-10Liquid–vapor equilibrium and the direction of spontaneous change(a) For the vaporization of water at 298.15 K and 1 atm,

The direction of spontaneous change is thecondensation of (b) At 298.15 K and 23.76 mmHg, the liquid and vapor are inequilibrium and (c) At 298.15 K and 10 mmHg, the vaporization of occurs spontaneously: and ¢G 6 0.H2O(g, 10 mmHg),¡H2O(l, 10 mmHg)

H2O(l)¢G° = 0.H2O(g).¢G° = 8.590 kJ.H2O(g, 1 atm),

¡H2O(l, 1 atm)

▲

Here, we have used astandard-state pressure of1 atm. See Feature Problem 88for an appraisal of this matterusing a standard-statepressure of 1 bar.

▲

The liquid–vaporequilibrium representedhere is out of contact withthe atmosphere. In thepresence of the atmosphere,the pressure on the liquidwould be barometricpressure, whereas that ofthe vapor would remainessentially unchanged at0.03126 atm.

▲

Water droplets in theatmosphere frequentlybecome supercooled

because of the absence of a dustparticle upon which to nucleate.As soon as the supercooleddroplet finds a bit of dust it isimmediately and spontaneouslyconverted to ice.

Erné, B. H.“Thermodynamics of

Water Superheated in theMicrowave Oven.” J. Chem. Educ.2000: 77, 1309 (October 2000).

PETRMC19_780-821-hr 12/30/05 2:02 PM Page 797

798 Chapter 19 Spontaneous Change: Entropy and Free Energy

The expressions for and are and respectively. First consider how the enthalpy terms and

are related for an ideal gas. The enthalpy of an ideal gas is a function oftemperature only; it is independent of pressure. Thus, under any mixing con-ditions for an ideal gas, We can write

(ideal gas)

We now need to obtain a relationship between and To do so, weconsider the isothermal expansion of an ideal gas for which and

If the expansion occurs reversibly (recall Figure 7-12), the work ofexpansion for one mole of an ideal gas is given by an equation derived inFeature Problem 108 in Chapter 7.

(reversible, isothermal)

The reversible, isothermal heat of expansion is

From equation (19.2), we obtain the entropy change for the isothermal expan-sion of one mole of an ideal gas.

(19.10)

Using equation (19.10), we can now evaluate the entropy of an ideal gas underany conditions of pressure. From the ideal gas equation, we know that the vol-ume of an ideal gas is inversely proportional to the pressure, so we can recastequation (19.10) as

where and are the initial and final pressures, respectively. If we set bar and designate as P° and as S°, we obtain for the entropy at anypressure P

(19.11)

Now, let’s return to the ammonia synthesis reaction and calculate the en-tropy change for that reaction. We begin by applying equation (19.11) to eachof the three gases.

Then we substitute the above values into the equation to obtain

By rearranging the terms, we get

and, since the first three terms on the right-hand side of the above equationrepresent we have

¢S = ¢S° + R ln

PN2 P3 H2

P2 NH3

¢S = ¢S° - R ln P2 NH3 + R ln PN2 + R ln P3

H2

¢S = ¢S° - 2R ln PNH3 + R ln PN2 + 3R ln PH2

¢S°,

¢S = 2SNH3° - SN2

° - 3SH2° - 2R ln PNH3 + R ln PN2 + 3R ln PH2

¢S = 2SNH3° - 2R ln PNH3 - SN2

° + R ln PN2 - 3SH2° + 3R ln PH2

3SH2

-SN2-2SNH3=¢S

SNH3 = SNH3° - R ln PNH3 SN2 = SN2

° - R ln PN2 SH2 = SH2° - R ln PH2

S = S° - R ln PP°

= S° - R ln P1

= S° - R ln P

SiPi

Pi = 1PfPi

¢S = Sf - Si = R ln

Vf

Vi= R ln

Pi

Pf= -R ln

Pf

Pi

¢S =qrev

T= R ln

Vf

Vi

qrev = -w = RT ln

Vf

Vi

w = -RT ln

Vf

Vi

¢U = 0.q = -w

¢S°.¢S

¢G = ¢H° - T ¢S

¢H°.=¢H

¢H°¢H¢H° - T ¢S°,

=¢G°¢H - T ¢S=¢G¢G°¢G

PETRMC19_780-821-hr 12/30/05 2:02 PM Page 798

Are You Wondering . . .If there is a microscopic approach to obtaining equation (19.10)?

To do this, we use the ideas of Ludwig Boltzmann. Consider an ideal gas at an ini-tial volume and allow the gas to expand isothermally to a final volume Using the Boltzmann equation, we find that for the change in entropy,

where k is the Boltzmann constant, and are the initial and final entropies, re-spectively, and and are the number of microstates for the initial and finalmacroscopic states of the gas, respectively. We must now obtain a value for theratio To do that, suppose that there is only a single gas molecule in a con-tainer. The number of microstates available to this single molecule should be pro-portional to the number of positions where the molecule can be and, hence, to thevolume of the container. That is also true for each molecule in a system of particles—Avogadro’s number of particles. The number of microstates availableto the whole system is

Because the number of microstates for each particle is proportional to the vol-ume V of the container, the number of microstates for (Avogadro’s number)ideal gas molecules is

Thus, the ratio of the microstates for isothermal expansion is

We can now calculate as follows:

where R is the ideal gas constant. This equation, which gives the entropy changefor the expansion of one mole of ideal gas, is simply equation (19.10).

¢S = k ln

Wf

Wi= k ln¢Vf

Vi≤NA

= NA k ln¢Vf

Vi≤ = R ln¢Vf

Vi≤

¢S

Wf

Wi= ¢Vf

Vi≤NA

W r VNA

NA

Wtotal = Wparticle 1 * Wparticle 2 * Wparticle 3 * Á

NA

Wf>Wi .

WfWi

SfSi

¢S = k ln

Wf

Wi

¢S = Sf - Si = k ln Wf - k ln Wi

Vf .Vi

19-6 Free Energy Change and Equilibrium 799

Finally, we can write the equation for by substituting the expression for into the equation

(ideal gas)

This leads to

To simplify, we designate the quotient in the logarithmic term as the reactionquotient Q (recall page 636).

(19.12)¢G = ¢G° + RT ln Q

¢G = ¢G° + RT ln

P2 NH3

PN2 P3 H2

¢G = ¢H° - T ¢S°5

+ RT ln

P2 NH3

PN2 P3 H2

¢G = ¢H° - T ¢S° - RT ln

PN2 P3 H2

P2 NH3

¢G = ¢H° - T ¢S

¢S¢G

It is commonly stated thatwe are headed for a state ofmaximum entropy.

Although this is true, it gives theincorrect notion that entropy, asdefined here, and time are related.They are not. Entropy is a propertyof an equilibrium state. Here wecalculate the difference betweenthe entropies of two differentequilibrium states.

Free Energy and ReactionMixture activity

PETRMC19_780-821-hr 12/30/05 2:02 PM Page 799

800 Chapter 19 Spontaneous Change: Entropy and Free Energy

Equation (19.12) is the relationship between and that we have beenseeking in this section, and we see that the key term in the equation is the re-action quotient formulated for the actual, nonstandard conditions. We can useequation (19.12) to decide on the spontaneity of a reaction under any condi-tions of composition, provided that the temperature and pressure at which weobserve the reaction are constant. We turn now to describing how the standardGibbs free energy change is related to the equilibrium constant.

RELATIONSHIP OF TO THE EQUILIBRIUM CONSTANT KWe encounter an interesting situation when we apply equation (19.12) to areaction at equilibrium. We have learned that at equilibrium and inChapter 15 we saw that if a system is at equilibrium, or or,more generally, So, we can write that, at equilibrium,

which means that

(19.13)

If we have a value of at a given temperature, we can use equation(19.13) to calculate an equilibrium constant K. This means that the tabulationof thermodynamic data in Appendix D can serve as a direct source of count-less equilibrium constant values at 298.15 K.

We need to say a few words about the units required in equation (19.13). Be-cause logarithms can be taken of dimensionless numbers only, K has no units;neither does ln K. The right-hand side of equation (19.13) has the unit of “RT”:

on the left-hand side of the equation, musthave the same unit: The part of this unit means “per mole ofreaction.” One mole of reaction is simply the reaction based on the stoichio-metric coefficients chosen for the balanced equation. When a value is ac-companied by a chemical equation, the portion of the unit is oftendropped, but there are a few times when we need it, as for the proper cancel-lation of units in equation (19.13).

CRITERIA FOR SPONTANEOUS CHANGE: OUR SEARCH CONCLUDED

The graphs plotted in Figure 19-11 represent the culmination of our quest forcriteria for spontaneous change. Unfortunately, to construct these plots in alltheir detail is beyond the scope of this text. However, we can rationalize theirgeneral shape on the basis of two ideas: (1) Every chemical reaction consists ofboth a forward and a reverse reaction, even if one of these occurs only to avery slight extent. (2) The direction of spontaneous change in both the forwardand reverse reactions is the direction in which free energy decreases

As a consequence, free energy reaches a minimum at some pointbetween the left-hand and right-hand sides of the graph. This minimum is theequilibrium point in the reaction.

Now consider the vertical distance between the two end points of thegraph; this distance represents of the reaction. If, as in Figure 19-11(a),

of a reaction is small, either positive or negative, the equilibrium condi-tion is one in which significant amounts of both reactants and products willbe found. If is a large, positive quantity, as in Figure 19-11(b), the equilib-rium point lies far to the left—that is, very close to the reactants side. We cansay that the reaction occurs hardly at all. If is a large, negative quantity, asin Figure 19-11(c), the equilibrium point lies far to the right—that is, veryclose to the products side. We can say that the reaction goes essentially tocompletion. Table 19.2 summarizes the conclusions of this discussion, givesapproximate magnitudes to the terms small and large, and relates valuesto values of K.

¢G°

¢G°

¢G°

¢G°¢G°

1¢G 6 02.

“mol-1”¢G°

“mol-1”J mol-1.¢G°,J mol-1.=K-1 * KJ mol-1

¢G°

¢G° = -RT ln K

¢G = ¢G° + RT ln K = 0

Q = K.Q = Kp,Q = Kc

¢G = 0,

≤G°

¢G°¢GEquation 19.12 shows thatthe value of the reactionquotient affects whether

the forward or reverse reaction isfavored under a particular set ofconditions. Remind students thatall reactions proceed towardequilibrium where free energychange is equal to zero.

From a theoreticalstandpoint, as can besee from Figure 19-11, all

chemical reactions reachequilibrium and no chemicalreaction goes totally to completion.

PETRMC19_780-821-hr 12/30/05 2:02 PM Page 800

19-6 Free Energy Change and Equilibrium 801

See the note on page 796.Here the system starts atstandard conditions and

runs down to equilibrium becausethe reactants are not replenished.At equilibrium there is no drivingforce for the reaction, so there is nofurther reaction and the free energymust be at a minimum.

EquilibriumanimationNO2–N2O4

�G�

Free

ene

rgy,

G

Extent of reaction

Equilibrium

Reactants(std states)

Products(std states)

(a)

�G�

Free

ene

rgy,

G

Extent of reactionReactants(std states)

Products(std states)

(c)

�G�

Free

ene

rgy,

G

Extent of reactionReactants(std states)

Products(std states)

(b)

FIGURE 19-11Free energy change, equilibrium, and the direction of spontaneous changeFree energy is plotted against the extent of reaction for a hypothetical reaction. isthe difference between the standard molar free energies of formation of products andreactants. The equilibrium point lies somewhere between pure reactants and pureproducts. (a) is small, so the equilibrium mixture lies about midway between thetwo extremes of pure products or reactants in their standard states. The effect ofnonstandard conditions can be deduced from the slope of the curve. Mixtures with

are to the right of the equilibrium point, and undergo spontaneous changein the direction of lower free energy, eventually coming to equilibrium. Similarly,mixtures with are to the left of the equilibrium point and spontaneouslyyield more products before reaching equilibrium. (b) is large and positive, so theequilibrium point lies close to the extreme of pure reactants in their standard states.Consequently, very little reaction takes place before equilibrium is reached. (c)is large and negative, so the equilibrium point lies close to the extreme of pureproducts in their standard states; the reaction goes essentially to completion.

¢G°

¢G°Q 6 K

Q 7 K

¢G°

¢G°

▲

TABLE 19.2 Significance of the Magnitudeof (at 298 K)

K Significance

�200 kJ/mol 9.1 10�36

�100 3.0 10�18

�50 1.7 10�9

�10 1.8 10�2

�1.0 6.7 10�1

No reaction

Equilibriumcalculation

isnecessary

Reaction goesto completion

0 1.0

�1.0 1.5�10 5.6 101

�50 5.8 108

�100 3.3 1017

�200 1.1 1035

≤G°

≤G°

AND PREDICTING THE DIRECTION OF CHEMICAL CHANGE

We have considered both and in relation to the spontaneity of chemi-cal reactions, and this is a good time to summarize some ideas about them.

signifies that a reaction or process is spontaneous in the forwarddirection (to the right) for the stated conditions.¢G 6 0

¢G¢G°≤G:≤G°

The huge range of values ofthe equilibrium constant areseen to arise from the

natural logarithm term. Choose afew reasonable values of the freeenergy, both negative and positive,to illustrate this fact.

Free Energy andChemical Equilibriumactivity

PETRMC19_780-821-hr 12/30/05 2:02 PM Page 801

802 Chapter 19 Spontaneous Change: Entropy and Free Energy

KEEP IN MINDthat the ideal gas equation

can be rewrittenas

so that pressure is aneffective concentration.1P>RT2 =[concentration]=1n>V2PV = nRT

▲

KEEP IN MINDthat the precise definition ofpH is

By using this definition ofactivity, we remove the unitsand can employ the log termin a correct manner.

= - log[H+]

= - loga [H+]M1 M

b pH = - log aH+

▲

signifies that the forward reaction is spontaneous when reac-tants and products are in their standard states. It further signifies that

whatever the initial concentrations or pressures of reactants andproducts.

signifies that the reaction is at equilibrium under the statedconditions.

signifies that the reaction is at equilibrium when reactants andproducts are in their standard states. It further signifies that whichcan occur only at a particular temperature.

signifies that the reaction or process is nonspontaneous in the for-ward direction under the stated conditions.

signifies that the forward reaction is nonspontaneous when reac-tants and products are in their standard states. It further signifies that

whatever the initial concentrations or pressures of reactants andproducts.

only when all reactants and products are in their standardstates. Otherwise,

THE THERMODYNAMIC EQUILIBRIUM CONSTANT: ACTIVITIES

When we derived the equation we used the relation-ship given in equation (19.11),

where for a gas, we defined a standard state of 1 bar, the reference state for val-ues of entropy. The ratio is dimensionless, which is essential for a termappearing in a logarithm. At the time, we were considering a gas-phase reac-tion, but we must be able to discuss reactions in solutions also, so we need amore general approach. For this, we need to return to the concept of activity in-troduced in Section 15-2, which requires us to write

where a is the activity, defined as

In a gas-phase reaction, we express pressure in bars and take the reference stateto be 1 bar. In this way and as expected, activity is a dimensionless quantity.

In extending the activity concept to solutions, we define the reference stateas a 1 M solution, so the activity of a substance is the numerical value of itsmolarity. Thus, the activity of protons in a 0.1 M solution of HCl in water is

Another situation that we have encountered is that of a heterogeneous equi-librium, such as

Recall that we choose the pure solids as the reference states, but the effectiveconcentrations of the and CaO(s) in the system are also those of thepure solids. Consequently, the activity of a solid is unity. This conclusionagrees with our observation in Figure 15-3 that the addition of either CaO(s) or

to an equilibrium mixture of and CaO(s) has noeffect on the pressure of The activities of CaO(s) and areconstant (unity).

CaCO3(s)CO2(g).CO2(g),CaCO3(s),CaCO3(s)

CaCO3(s)

CaCO3(s) ∆ CaO(s) + CO2(g)

aH+ =0.1 M1 M

= 0.1

a =the effective concentration of a substance in the system

the effective concentration of that substance in a standard reference state

S = S° - R ln a

P>P°

S = S° - R ln PP°

= S° - R ln P1

¢G = ¢G° + RT ln Q,

RT ln Q.+¢G°=¢G¢G = ¢G°

K 6 1,

¢G° 7 0

¢G 7 0

K = 1,¢G° = 0

¢G = 0

K 7 1,

¢G° 6 0

PETRMC19_780-821-hr 12/30/05 2:02 PM Page 802

Are You Wondering . . .How to deal with activities under nonideal conditions?

For nonideal conditions, we sweep all deviations from ideality into an experimen-tally determined correction factor called the activity coefficient, Thus, for real(nonideal) systems, we write

For a gas:For a solute:For a pure solid or liquid:

More advanced treatments of this topic show how is related to the compositionof the system. For dilute solutions or near ideal gases, we will assume that However, keep in mind that the results we obtain are limited by the validity ofthis assumption.

g = 1.g

aL = aS = 1aX = g[X]aG = gPG

g.

19-6 Free Energy Change and Equilibrium 803

In summary, we can make these statements.

• For pure solids and liquids: The activity . The reference state is thepure solid or liquid.

• For gases: With ideal gas behavior assumed, the activity is replaced bythe numerical value of the gas pressure in bars. The reference state is thegas at 1 bar at the temperature of interest. Thus, the activity of a gas at0.50 bar pressure is 0.50. (Recall also, that 1 bar ofpressure is almost identical to 1 atm.)

• For solutes in aqueous solution: With ideal solution behavior assumed (forexample, no interionic attractions), the activity is replaced by the numer-ical value of the molarity. The reference state is a 1 M solution. Thus, theactivity of the solute in a 0.25 M solution is 0.25.

When an equilibrium expression is written in terms of activities, the equi-librium constant is called the thermodynamic equilibrium constant. Thethermodynamic equilibrium constant is dimensionless and thus appropriatefor use in equation (19.13).

Thermodynamic equilibrium constants, K, are sometimes identical to and values, as in parts (a) and (b) of Example 19-6. In other instances, as inpart (c), this is not the case. In working through Example 19-6, keep in mindthat the sole reason for writing thermodynamic equilibrium constants in thistext is to get the proper value to use in equation (19.13). Note that the reactionquotient Q must also be written in the same manner as K when used in equa-tion (19.12), as demonstrated in Example 19-7.

E X A M P L E 1 9 - 6Writing Thermodynamic Equilibrium Constant Expressions. For the followingreversible reactions, write thermodynamic equilibrium constant expressions, mak-ing appropriate substitutions for activities. Then equate K to or where thiscan be done.

(a) The water gas reaction

(b) Formation of a saturated aqueous solution of lead(II) iodide, a veryslightly soluble solute

(c) Oxidation of sulfide ion by oxygen gas (used in removing sulfides fromwastewater, as in pulp and paper mills)

O2(g) + 2 S2-(aq) + 2 H2O(l) ∆ 4 OH-(aq) + 2 S8(s)

PbI2(s) ∆ Pb2+(aq) + 2 I-(aq)

C(s) + H2O(g) ∆ CO(g) + H2(g)

Kp,Kc

Kp

Kc

=(0.25 M)>(1 M)=a

=(0.50 bar)>(1 bar)=a

a = 1

PETRMC19_780-821-hr 12/30/05 2:02 PM Page 803

804 Chapter 19 Spontaneous Change: Entropy and Free Energy

SolutionIn each case, once we have made the appropriate substitutions for activities, if allterms are molarities, the thermodynamic equilibrium constant is the same as Ifall terms are partial pressures, If both molarities and partial pressures ap-pear in the expression, however, the equilibrium constant expression can be desig-nated only as K.

(a) The activity of solid carbon is 1. Partial pressures are substituted for theactivities of the gases.

(b) The activity of solid lead(II) iodide is 1. Molarities are substituted for ac-tivities of the ions in aqueous solution.

(c) The activity of both the solid sulfur and the liquid water is 1. Molaritiesare substituted for the activities of and The partialpressure of is substituted for its activity. Thus, the resulting K isneither a nor a

Practice Example A: Write thermodynamic equilibrium constant expressionsfor each of the following reactions. Relate these to or where appropriate.

(a)

(b)

Practice Example B: Write a thermodynamic equilibrium constant expressionto represent the reaction of solid lead(II) sulfide with aqueous nitric acid to producesolid sulfur, a solution of lead(II) nitrate, and nitrogen monoxide gas. Base the ex-pression on the balanced net ionic equation for the reaction.

Cl2(g) + H2O(l) ∆ HOCl(aq) + H+(aq) + Cl-(aq)

Si(s) + 2 Cl2(g) ∆ SiCl4(g)

KpKc

K =a4

OH-(aq) a2 S(s)

aO2(g) a2 S2-(aq) a2

H2O(l)=

[OH-]4 # (1)2

PO2# [S2-]2 # (1)2 =

[OH-]4

PO2# [S2-]2

Kp.Kc

O2(g)S2-(aq).OH-(aq)

K =aPb2+(aq) a2

I-(aq)

aPbI2(s)= [Pb2+][I-]2 = Kc = Ksp

K =aCO(g)

aC(s) aH2O(g)=

(PCO)(PH2)

(PH2O)=

(PCO)(PH2)

(PH2O)= Kp

K = Kp.Kc .

E X A M P L E 1 9 - 7Assessing Spontaneity for Nonstandard Conditions. For the decomposition of 2-propanol to form propanone(acetone) and hydrogen,

the equilibrium constant is 0.444 at 452 K. Is this reaction spontaneous under standard conditions? Will the reaction bespontaneous when the partial pressures of 2-propanol, propanone, and hydrogen are 0.1 bar, respectively?

Solution

(CH3)2CHOH(g) ∆ (CH3)2CO(g) + H2(g)

In each case, we first must obtain thevalue of which we can get fromequation (19.13).

¢G°, = 3.05 * 103 J mol-1

¢G° = -RT ln K = -8.3145 J mol-1 K-1 * 452 K * ln10.4442

This result enables us to state categorical-ly that the reaction will not proceed spon-taneously if all reactants and products arein their standard states—that is, with thepartial pressures of reactants and prod-ucts at 1 bar.

To determine whether the reaction isspontaneous under the nonstandard-state conditions given, we must calculate

We first write Q in terms of activitiesand then substitute the partial pressuresof the gases for the activities of the gases.

¢G.Q =

a(CH3)2CO aH2

a(CH4)2CHOH=

P(CH3)2CO PH2

P(CH3)2CHOH

PETRMC19_780-821-hr 12/30/05 2:02 PM Page 804

We have now acquired all the tools with which to perform one of the mostpractical calculations of chemical thermodynamics: determining the equilibriumconstant for a reaction from tabulated data. Example 19-8, which demonstratesthis application, uses thermodynamic properties of ions in aqueous solutionas well as of compounds. An important idea to note about the thermodynam-ic properties of ions is that they are relative to which, by convention, isassigned values of zero for and S°. This means that entropies listedfor ions are not absolute entropies, as they are for compounds. Negativevalues of S° simply denote an entropy less than that of H+(aq).

¢Gf°,¢Hf°,H+(aq),

19-6 Free Energy Change and Equilibrium 805

Then use this expression in equation(19.12).

= -1.43 * 104 J mol-1

= 3.05 * 103 J mol-1 + a8.3145 J mol-1 K-1 * 452 K * ln 0.1 * 0.1

1b

¢G = ¢G° + RT ln

P(CH3)2CO PH2

P(CH3)2CHOH

The value of is negative, so we canconclude that under this second set ofconditions the reaction should proceedspontaneously. Remember, however, thatthermodynamics says nothing about therate of the reaction, only that the reactionwill proceed in its own good time!

¢G

Practice Example A: Use the data in Appendix D to decide whether the following reaction is spontaneous understandard conditions at 298.15 K.

Practice Example B: If a gaseous mixture of and both at a pressure of 0.5 bar, is introduced into a pre-viously evacuated vessel, which of the two gases will spontaneously convert into the other at 298.15 K?

NO2,N2O4

N2O4(g) ¡ 2 NO2(g)

E X A M P L E 1 9 - 8Calculating the Equilibrium Constant of a Reaction from the Standard Free Energy Change. Determine the equi-librium constant at 298.15 K for the dissolution of magnesium hydroxide in an acidic solution.

Solution

Mg(OH)2(s) + 2 H+(aq) ∆ Mg2+(aq) + 2 H2O(l)

The key to solving this problem is tofind a value of and then to use theexpression We canobtain from standard free energiesof formation listed in Appendix D. Notethat because its value is zero, the term

is not included.¢Gf°[H+(aq)]

¢G°¢G° = -RT ln K.

¢G°

= 2(-237.1 kJ mol-1) + (-454.8 kJ mol-1) - (-833.5 kJ mol-1)

¢G° = 2 ¢Gf°[H2O(l)] + ¢Gf°[Mg2+(aq)] - ¢Gf°[Mg(OH)2(s)]

Now solve for ln K and K.

K = e38.5 = 5 * 1016

ln K =- ¢G°

RT=

-(-95.5 * 103 J mol-1)

8.3145 J mol-1 K-1 * 298.15 K= 38.5

¢G° = -RT ln K = -95.5 kJ mol-1 = -95.5 * 103 J mol-1

PETRMC19_780-821-hr 12/30/05 2:02 PM Page 805

806 Chapter 19 Spontaneous Change: Entropy and Free Energy

When a question requires the use of thermodynamic properties, it is a goodidea to think qualitatively about the problem before diving into calculations.The dissolution of in acidic solution considered in Example 19-8is an acid–base reaction that was used as an example in Chapter 18 to illustratethe effect of pH on solubility. It was also mentioned in Chapter 5 as the basisfor using milk of magnesia as an antacid. We should certainly expect the reac-tion to be spontaneous. This means that the value of K should be large, whichwe found to be the case. If we had made an error in sign in our calculation(an easy thing to do using the expression ), we would haveobtained But we would have seen immediately that this is thewrong answer. This erroneous value suggests a reaction in which the concen-tration of products is extremely low at equilibrium.

The data listed in Appendix D are for 25 °C. Thus, values of and Kobtained with these data are also at 25 °C. Most chemical reactions are carriedout at temperatures other than 25 °C, however. In Section 19-7, we will learnhow to calculate values of equilibrium constants at various temperatures.

AND K AS FUNCTIONS OF TEMPERATURE

In Chapter 15, we used Le Châtelier’s principle to make qualitative predic-tions of the effect of temperature on an equilibrium condition. We can nowdescribe a quantitative relationship between the equilibrium constant and tem-perature. In the method illustrated in Example 19-9, we assume that ispractically independent of temperature. Although absolute entropies dependon temperature, we assume that the entropy change for a reaction is alsoindependent of temperature. Yet, the term is strongly temperature-dependent because of the temperature factor T. As a result, which isequal to is also dependent on temperature.

E X A M P L E 1 9 - 9Determining the Relationship Between an Equilibrium Constant and Tempera-ture by Using Equations for Free Energy Change. At what temperature will theequilibrium constant for the formation of NOCl(g) be Datafor this reaction at 25 °C are

SolutionTo determine an unknown temperature from a known equilibrium constant, weneed an equation in which both of these terms appear. The required equation is

However, to solve for the unknown temperature, we need the value ofat that temperature. We know the value of at 25 °C but(-40.9 kJ mol-1),¢G°¢G°

-RT ln K.=¢G°

¢S° = -121.3 J mol-1 K-1

¢G° = -40.9 kJ mol-1 ¢H = -77.1 kJ mol-1

2 NO(g) + Cl2(g) ∆ 2 NOCl(g)

1.00 * 103?=K = Kp

T ¢S°,-¢H°¢G°,

“T¢S”¢S°

¢H°

≤G°19-7

¢G°

K = 2 * 10-17.-RT ln K=¢G°

Mg(OH)2(s)

The value of K obtained here is thethermodynamic equilibrium constant.According to the conventions wehave established, the activities of both

and are 1, and mo-larities can be substituted for the activi-ties of the ions.

H2O(l)Mg(OH)2(s)K =

aMg2+(aq) a2 H2O(l)

aMg(OH)2(s) a2 H+(aq)

=[Mg2+]

[H+]2 = Kc = 5 * 1016

Practice Example A: Determine the equilibrium constant at 298.15 K for Compareyour answer to the for AgI in Appendix D.

Practice Example B: At 298.15 K, should manganese dioxide react to an appreciable extent with 1 M HCl(aq),producing manganese(II) ion in solution and chlorine gas?

Ksp

I-(aq).+Ag+(aq)∆AgI(s)

In general, equilibriumconstants for exothermicreactions will be lower at

higher temperatures and higherat lower temperatures.Alternatively, equilibriumconstants for endothermicreactions are lower at lowertemperatures and higher athigher temperatures. This is seenby evaluating equation 19.15.

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19-7 and K as Functions of Temperature 807≤G°

we also know that this value will be different at other temperatures. We can assume,however, that the values of and will not change much with temperature.This means that we can obtain a value of from the equation

where T is the unknown temperature and the values of and are thoseat 25 °C. Now we have two equations that we can set equal to each other. That is,

We can gather the terms with T on the right,

and solve for T.

Now substitute values for R, and ln K.

Although the answer shows three significant figures, the final result shouldprobably be rounded to just two significant figures. The assumption we made aboutthe constancy of and is probably no more valid than that.

Practice Example A: At what temperature will the formation of fromNO(g) and have For the reaction

at 25 °C, and

Practice Example B: For the reaction whatis the value of K at (a) 25 °C; (b) 75 °C? Use data from Example 19-9.[Hint: The solution to part (a) can be done somewhat more simply than that for (b).]

An alternative to the method outlined in Example 19-9 is to relate the equi-librium constant and temperature directly, without specific reference to a freeenergy change. We start with the same two expressions as in Example 19-9,

and divide by

(19.14)

If we assume that and are constant, equation (19.14) describes astraight line with a slope of and a y-intercept of Table 19.3 listsequilibrium constants as a function of the reciprocal of Kelvin temperature for

¢S°>R.- ¢H°>R¢S°¢H°

ln K =- ¢H°

RT+

¢S°R

-RT.-RT ln K = ¢G° = ¢H° - T ¢S°

2 NOCl(g),∆Cl2(g)+2 NO(g)

K-1.-146.5 J mol-1=¢S°-114.1 kJ mol-1=¢H°2 NO2(g)∆O2(g)+2 NO(g)Kp = 1.50 * 102?O2(g)

NO2(g)

¢S°¢H°

=-77.1 * 103 J mol-1

-178.7 J mol-1 K-1 = 431K

=-77.1 * 103 J mol-1

-121.3 J mol-1 K-1 - 18.3145 * 6.9082 J mol-1 K-1

T =-77.1 * 103 J mol-1

-121.3 J mol-1 K-1 - 38.3145 J mol-1 K-1 * ln 11.00 * 10324

¢S°,¢H°,

T =¢H°

¢S° - R ln K

¢H° = T ¢S° - RT ln K = T(¢S° - R ln K)

¢G° = ¢H° - T ¢S° = -RT ln K

¢S°¢H°T ¢S°,-¢H°=¢G°¢G°

¢S°¢H°

TABLE 19.3 Equilibrium Constants, for the Reaction

at Several Temperatures

T, K 1 T,

800 6.81850 5.14900 3.74950 2.30

1000 1.161050 0.0011001170 -2.121.2 * 10-18.5 * 10-4

-0.943.9 * 10-19.09 * 10-41.0 * 10-49.52 * 10-43.2 * 10010.0 * 10-41.0 * 10110.5 * 10-44.2 * 10111.1 * 10-41.7 * 10211.8 * 10-49.1 * 10212.5 * 10-4

ln KpKpK�1/

2 SO2(g) � O2(g) ∆ 2 SO3(g)Kp ,

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808 Chapter 19 Spontaneous Change: Entropy and Free Energy

the reaction of and that forms The ln and datafrom Table 19.3 are plotted in Figure 19-12 and yield the expected straight line.

Now we can follow the procedure used in Appendix A-4 to derive the Clausius–Clapeyron equation. We can write equation (19.14) twice, for twodifferent temperatures and with the corresponding equilibrium constants. Then,if we subtract one equation from the other, we obtain the result shown here,

(19.15)

where and are two Kelvin temperatures; and are the equilibriumconstants at those temperatures; is the enthalpy of reaction, expressed in

and R is the gas constant, expressed as Jacobusvan’t Hoff (1852–1911) derived equation (19.15), which is often referred to asthe van’t Hoff equation.

E X A M P L E 1 9 - 1 0Relating Equilibrium Constants and Temperature Through the van’t Hoff Equa-tion. Use data from Table 19.3 and Figure 19-12 to estimate the temperature atwhich for the reaction

SolutionSelect one known temperature and equilibrium constant from Table 19.3 and the en-thalpy change of the reaction, from Figure 19-12. The data to be substitutedinto equation (19.15) are and

J mol-1.¢H° = -1.8 * 105K2 = 9.1 * 102;T2 = 800 K,K1 = 1.0 * 106;T1 = ?,

¢H°,

2 SO2(g) + O2(g) ∆ 2 SO3(g)

1.0 * 106=Kp

8.3145 J mol-1 K-1.J mol-1;¢H°

K1K2T1T2

ln

K2

K1=

¢H°R

¢ 1T1

-1T2≤

1>TKpSO3(g).O2(g)SO2(g)

For simplicity, we havedropped the units in thissetup, but you should be ableto show that units cancelproperly.

▲

1T, K�1

8.0

6.0

4.0

2.0

0.0

7.6

7.6

8.0

Slope �

9.0 10.0 11.0 12.0

3.4 10�4 K�1

� 2.2 104 K3.4 10�4 K�1

13.0 10�4

�2.0

�4.0

ln K

p

KEEP IN MINDthat the Clausius–Clapeyronequation (12.2) is just aspecial case of equation(19.15) in which theequilibrium constantsare equilibrium vaporpressures and ¢H° = ¢Hvap.

▲

FIGURE 19-12Temperature dependence of the equilibriumconstant for the reaction

This graph can be used to establish theenthalpy of reaction, (see equation 19.14).

= -1.8 * 102 kJ mol-1 = -1.8 * 105 J mol-1

¢H° = -8.3145 J mol-1 K-1 * 2.2 * 104 K slope = ¢H°>R = 2.2 * 104 K

¢H°

2 SO2(g) � O2(g) ∆ 2 SO3(g)

Kp

▲

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19-8 Coupled Reactions 809

Practice Example A: Estimate the temperature at which forthe reaction in Example 19-10. Use data from Table 19.3 and Figure 19-12.

Practice Example B: What is the value of for the reaction at 235 °C? Use data from Table 19.3, Figure 19-12, and the

van’t Hoff equation (19.15).

CONCEPT ASSESSMENT ✓The normal boiling point of water is 100 °C. At 120 °C and 1 atm, is or greater for the vaporization of water?

COUPLED REACTIONS

We have seen two ways to obtain product from a nonspontaneous reaction: (1)Change the reaction conditions to ones that make the reaction spontaneous(mostly by changing the temperature), and (2) carry out the reaction by elec-trolysis. But there is a third way also. Combine a pair of reactions, one with apositive and one with a negative to obtain a spontaneous overall reac-tion. Such paired reactions are called coupled reactions. Consider the extrac-tion of a metal from its oxide.

When copper(I) oxide is heated to 673 K, no copper metal is obtained. Thedecomposition of to form products in their standard states (for instance,

) is nonspontaneous at 673 K.

(19.16)

Suppose this nonspontaneous decomposition reaction is coupled with the par-tial oxidation of carbon to carbon monoxide—a spontaneous reaction. Theoverall reaction (19.17), because it has a negative value of is spontaneouswhen reactants and products are in their standard states.

(19.17) Cu2O(s) + C(s) ¡ 2 Cu(s) + CO(g) ¢G673 K° = -50 kJ

C(s) + 12

O2(g) ¡ CO(g) ¢G673 K° = -175 kJ

Cu2O(s) ¡ 2 Cu(s) + 12

O2(g) ¢G673 K° = +125 kJ

¢G°,

Cu2O(s) ¢

" 2 Cu(s) +12

O2(g) ¢G673K° = +125 kJ

PO2 = 1.00 barCu2O

¢G,¢G

19-8

T ¢S¢H

2 SO3(g)∆O2(g)+2 SO2(g)Kp

5.8 * 10-2=Kp

T1 =1

1.57 * 10-3 = 6.37 * 102 K

1T1

= 13.2 * 10-42 + 11.25 * 10-32 = 1.57 * 10-3

-7.00

-2.2 * 104 +1

800=

1T1

-7.00 = -2.2 * 104¢ 1T1

-1

800≤

ln 9.1 * 102

1.0 * 106 =-1.8 * 105

8.3145¢ 1

T1-

1800≤

ln

K2

K1=

¢H°R

¢ 1T1

-1T2≤

In general chemistry, simpleexamples of reactions aregenerally used. In fact, in

almost all interesting cases, onereaction is coupled to another, andso forth. No better example existsthan the complex cycles of coupledchemical reactions in biologicalprocesses, such as illustrated in theFocus On feature.

Reduction of CuO movie

PETRMC19_780-821-hr 12/30/05 2:02 PM Page 809

FOCUS ON COUPLED REACTIONS IN BIOLOGICAL SYSTEMS

An important example of coupled reactions in living or-ganisms involves the breakdown of glucose accompanied by the conversion of adenosine diphos-phate (ADP) to adenosine triphosphate (ATP)* in themitochondria of cells (Fig. 19-13). ATP serves as anenergy-storage compound that is broken down to ADP,which releases energy that can be used in cell processes.For example, it is used in the ribosomes to produce pro-teins. The ATP-forming reaction

is spontaneous under standard conditions at 37 °C, sowhy does the cell need to use glucose to make ATP? Theanswer is that cells do not operate with as re-quired by standard conditions; in fact, the pH in a cell isabout 7. When we estimate for the reaction at that pHand assume all other species are at 1.0 M (still far from ac-tuality), we get

¢G = ¢G° + RT ln ¢ aATP aH2O

aADP aPi aH+≤

¢G

[H+] = 1 M

PO O�

O�

Adenosine PO

O�

OO

ADP3�

PO O

O�

Adenosine PO

O�

OO

ATP4�

P O�

O�

O

¢G = -9.2 kJ mol-1

ADP3- + HPO4

2- + H+ ¡ ATP4- + H2O

(C6H12O6)

Mitochondria and endoplasmic reticulumA colorized scanning electron micrograph of mitochondria (blue)and rough endoplasmic reticulum (yellow) in a pancreatic cell.Mitochondria are the powerhouses of the cell. They oxidize sugarsand fats, producing energy for the conversion of ADP to ATP.Rough endoplasmic reticulum is a network of folded membranescovered with protein-synthesizing ribosomes (small dots).

▲

*The structural formulas for and are in accord with the discussion of expandedvalence shells on page 394. Often, these structures are written with one P-to-O double bond perphosphate unit.

ATP4-ADP3-

Note that reactions (19.16) and (19.17) are not the same, even though eachhas Cu(s) as a product. The purpose of coupled reactions, then, is to produce aspontaneous overall reaction by combining two other processes: one nonspon-taneous and one spontaneous. Many metallurgical processes employ coupledreactions, especially those that use carbon or hydrogen as reducing agents.

To sustain life, organisms must synthesize complex molecules from simplerones. If carried out as single-step reactions, these syntheses would generallybe accompanied by increases in enthalpy, decreases in entropy, and increasesin free energy—in short, they would be nonspontaneous and would not occur.In living organisms, changes in temperature and electrolysis are not viable op-tions for dealing with nonspontaneous processes. Here, coupled reactions arecrucial, as described in the Focus On feature.

810

PETRMC19_780-821-hr 12/30/05 2:02 PM Page 810

Gibbs energy change of In the cell, thecoupling of this conversion involves ten chemical reac-tions that are enzyme-catalyzed and form 2 moles of ATP.The overall process can be represented by two steps.

The overall reaction is therefore spontaneous. The couplingof the partial oxidation of glucose by way of glycolysis isspontaneous and provides a source of ATP. The ATP is thenused in the formation of proteins. That is, the reaction

is coupled with a nonspontaneous reaction to produce aprotein in a ribosome as suggested by Figure 19-13.

¢G° = -32.4 kJ mol-1

ATP4- + H2O ¡ ADP3- + HPO4

2- + H+

¢G°¿ = -153 kJ mol-12 lactate- + 2 ATP4- + 2 H2O

glucose + 2 ADP3- + 2 HPO4

2- ¡

¢G°¿ = 2 * 32.4 = 64.8 kJ mol-1

2 ADP3- + 2 HPO4

2- + 2 H+ ¡ 2 ATP4- + 2 H2O

¢G°¿ = -218 kJ mol-1 glucose ¡ 2 lactate- + 2 H+

-218 kJ mol-1.where is a common abbreviation for We nowuse and to obtain

In biochemistry, this value of is called the biologicalstandard state, designated Thus, we can write

Note that the proton concentration is divided by togive the activity because we have revised our definition ofstandard state.

Under standard biochemical conditions, the conversionof ADP to ATP is not spontaneous. Consequently, in cellsthis conversion must be coupled with another reaction toprovide the necessary energy. The reaction, in the absenceof oxygen (the anaerobic conditions inside a cell), isglycolysis—a metabolic process that converts glucose tothe lactate anion. This reaction has a biochemical standard

10-7

¢G = ¢G°¿ + RT ln ¢ aATP aH2O

aADP aPi([H+]>10-7)

≤

¢G°¿.¢G

¢G = -9.2 kJ mol-1 + 41.6 kJ mol-1 = +32.4 kJ mol-1

10-7=aH+aH2O = 1=aPi=aADP=aATP

HPO4

2-.Pi

Mitochondria RibosomesC6H12O6 Proteins

2 CH3CH(OH)CO2� 2 H� Amino acids�

ADP

ATP

ADP

ATP

FIGURE 19-13Schematic of coupled reactionswithin a cell

▲

Summary19-1 Spontaneity: The Meaning of SpontaneousChange—A process that proceeds without need of exter-nal intervention is said to be a spontaneous process. Anonspontaneous process cannot occur without externalintervention. A process that is spontaneous in one direc-tion is nonspontaneous in the reverse direction. Somespontaneous processes are exothermic, others are en-dothermic so that criteria other than enthalpy change areneeded to define spontaneity.

19-2 The Concept of Entropy—Entropy is a thermody-namic property related to the distribution of a system’senergy among the available microscopic energy levels.Boltzmann’s formula (equation 19.1) illustrates the rela-tionship between entropy and the number of microstatesof a system. The thermodynamic definition of an entropy

change is of a quantity of heat divided by a Kelvintemperature (equation 19.2), and having the units The quantity of heat released (or absorbed) during theprocess must be from a reversible process in order forthe entropy change to be path-independent.

19-3 Evaluating Entropy and Entropy Changes—Thethird law of thermodynamics states that the entropy of apure, perfect crystal at 0 K is zero. Thus there are absoluteentropies, unlike internal energy and enthalpy. The en-tropy of a substance in its standard state is called standardmolar entropy, S°. Standard molar entropies of reactantsand products can be used to calculate standard entropychanges in chemical reactions (equation 19.5). Anotherimportant entropy-related relationship is Trouton’s rule,which states that the standard entropy of vaporization at

J K-1.(qrev)

811

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812 Chapter 19 Spontaneous Change: Entropy and Free Energy

Integrative ExampleThe synthesis of methanol is of great importance because methanol can be used directly as a motor fuel, mixed with gaso-line for fuel use, or converted to other organic compounds. The synthesis reaction, carried out at about 500 K, is

What is the value of at 500 K?

StrategyOur approach to this problem begins with determining from free energy of formation data and using to find at 298 K. The next step is to calculate from enthalpy of formation data, and use this value together with at 298 Kin expression (19.15) to find at 500 K.

Solution

Kp

Kp¢H°Kp¢G°¢G°

Kp

CO(g) + 2 H2(g) ∆ CH3OH(g)

To calculate at 298 K, use at 298 K, written asin the expression

-RT ln Kp.=¢G°-24.8 * 103 J mol-1,

¢G°Kp

the normal boiling point is approximately constant at(equation 19.4).

19-4 Criteria for Spontaneous Change: The Second Lawof Thermodynamics—The basic criterion for spontaneouschange is that the entropy change of the universe, which isthe sum of the entropy change of the system plus that ofthe surroundings (equation 19.6), must be greater thanzero (equation 19.7). This statement is known as thesecond law of thermodynamics. An equivalent criterionapplied to the system alone is based on a thermodynamicfunction known as the Gibbs energy, G, or the free energy.The free energy change, , is the enthalpy change forthe system minus the product of the temperatureand entropy change for the system (equation 19.9).Table 19.1 summarizes the criteria for spontaneous changebased on free energy change.

19-5 Standard Free Energy Change, —The standardfree energy change, is based on the conversion ofreactants in their standard states to products in their stan-dard states. Tabulated free energy data are usually standardmolar free energies of formation, and usually at298.15 K.

¢Gf° ,

¢G°,≤G°

1T ¢S21¢H2 ≤G

K-187 J mol-119-6 Free Energy Change and Equilibrium—The relation-ship between the standard free energy change and theequilibrium constant for a reaction is The constant, K, is called a thermodynamic equilibriumconstant. It is based on the activities of reactants andproducts, but these activities can be related to solutionmolarities and gas partial pressures by means of a fewsimple conventions.

19-7 and K as Functions of Temperature—Bystarting with the relationship between standard freeenergy change and the equilibrium constant, the van’tHoff equation—relating the equilibrium constant andtemperature—can be written. With this equation, tabulat-ed data at 25 °C can be used to determine equilibrium con-stants not just at 25 °C but at other temperatures as well.

19-8 Coupled Reactions—Nonspontaneous processes canbe made spontaneous by coupling them with spontaneousreactions and by taking advantage of the state functionproperty of G. Coupled reactions, that is, paired reactionsthat yield a spontaneous overall reaction, occur in metal-lurgical processes and in biochemical transformations.

≤G°

-RT ln K.=¢G°

Write the equation for methanol synthesis; place freeenergy of formation data from Appendix D under for-mulas in the equation, and use these data to calculate

at 298 K.¢G° -1 mol CO * (-137.2 kJ>mol CO) = -24.8 kJ

¢G° = 1 mol CH3OH * (-162.0 kJ>mol CH3OH)

-162.00-137.2¢Gf° , kJ mol-1CO(g) + 2 H2(g) ∆ CH3OH(g)

Kp = e10.0 = 2.2 * 104

ln Kp = - ¢G°>RT =-(-24.8 * 103 J mol-1)

8.3145 J mol-1 K-1 * 298 K= 10.0

To determine at 298 K, use standard enthalpy offormation data from Appendix D, applied in the samemanner as was previously used for ¢G°.

¢H°

-1 mol CO * (-110.5 kJ>mol CO) = -90.2 kJ ¢H° = 1 mol CH3OH * (-200.7 kJ>mol CH3OH)

-200.70-110.5¢Hf°, kJ mol-1CO(g) + 2 H2(g) ∆ CH3OH(g)

Use the van’t Hoff equation with at 298 K and Solve for

at 500 K.Kp

-90.2 * 103 J mol-1.=¢H°Kp = 2.2 * 104

Kp

2.2 * 104 = e-14.7 = 4 * 10-7 Kp = 9 * 10-3

= -14.7

ln

Kp

2.2 * 104 =-90.2 * 103 J mol-1

8.3145 J mol-1 K-1 a 1298 K

-1

500 Kb

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Exercises 813

ExercisesSpontaneous Change and Entropy

1. Indicate whether each of the following changes repre-sents an increase or decrease in entropy in a system,and explain your reasoning: (a) the freezing ofethanol; (b) the sublimation of Dry Ice; (c) the burningof a rocket fuel.

2. Arrange the entropy changes of the followingprocesses, all at 25 °C, in the expected order ofincreasing and explain your reasoning: (a)

(b)(c)

3. Use ideas from this chapter to explain this famousremark attributed to Rudolf Clausius (1865): “Die En-ergie der Welt ist konstant; die Entropie der Weltstrebt einem Maximum zu.” (“The energy of theworld is constant; the entropy of the world increasestoward a maximum.”)

4. Comment on the difficulties of solving environmentalpollution problems from the standpoint of entropychanges associated with the formation of pollutantsand with their removal from the environment.

5. Indicate whether the entropy of the system would in-crease or decrease in each of the following reactions. Ifyou cannot be certain simply by inspecting the equa-tion, explain why.(a)(b)

CuSO4 # 5 H2O(s)CuSO4 # 3 H2O(s) + 2 H2O(g) ¡CCl4(l) ¡ CCl4(g)

H2O(g, 10 mmHg).¡H2O(l, 1 atm)CO2(g, 10 mmHg);¡

CO2(s, 1 atm)H2O(g, 1 atm);¡H2O(l, 1 atm)¢S,

(c)(d)

(not balanced)6. Which substance in each of the following pairs would

have the greater entropy? Explain.(a) at 75 °C and 1 atm: 1 mol or 1 mol (b) at 5 °C and 1 atm: 50.0 g Fe(s) or 0.80 mol Fe(s)(c) 1 mol (l, 1 atm, 8 °C) or 1 mol (s, 1 atm,

)(d) 0.312 mol (g, 0.110 atm, 32.5 °C) or 0.284 mol

(g, 15.0 atm, 22.3 °C)7. For each of the following reactions, indicate whether

for the reaction should be positive or negative. If itis not possible to determine the sign of from theinformation given, indicate why.(a)(b)(c)(d)(e)

8. By analogy to and how would you defineentropy of formation? Which would have the largestentropy of formation: or

First make a qualitative prediction; then testyour prediction with data from Appendix D.CS2(l)?

CH3CH2OH(l),CH4(g),

¢Gf°¢Hf°Si(s) + 2 Cl2(g) ¡ SiCl4(g)Fe2O3(s) + 3 CO(g) ¡ 2 Fe(s) + 3 CO2(g)2 NaCl(l) ¡ 2 Na(l) + Cl2(g)2 HgO(s) ¡ 2 Hg(l) + O2(g)CaO(s) + H2O(l) ¡ Ca(OH)2(s)

¢S¢S

O2

SO2

-8 °CBr2Br2

H2O(g)H2O(l)

H2S(g) + O2(g) ¡ H2O(g) + SO2(g)SO3(g) + H2(g) ¡ SO2(g) + H2O(g)

AssessmentWe have been successful in determining the equilibrium constant for the synthesis of methanol at 500 K using tabulatedthermodynamic data and the van’t Hoff equation. Note that increasing the temperature favors the reverse reaction, aswe should expect from Le Châtelier’s principle. That is, an increase in temperature in a reversible reaction favors the heatabsorbing (endothermic) reaction. Here, the forward reaction is exothermic and the reverse reaction is endothermic.

Phase Transitions

9. In Example 19-2, we dealt with and forwater at 100 °C.(a) Use data from Appendix D to determine valuesfor these two quantities at 25 °C.(b) From your knowledge of the structure of liquidwater, explain the differences in values and in

values between 25 °C and 100 °C.10. Pentane is one of the most volatile of the hydrocar-

bons in gasoline. At 298.15 K, the following enthalpiesof formation are given for pentane:

(a) Estimate the normal boiling point of pentane.(b) Estimate for the vaporization of pentane at298 K.(c) Comment on the significance of the sign of at298 K.

¢G°

¢G°

-146.9 kJ mol-1.=¢Hf°[C5H12(g)]-173.5 kJ mol-1;=¢Hf°[C5H12(l)]

¢Svap°¢Hvap°

¢Svap°¢Hvap° 11. Which of the following substances would obey Trou-ton’s rule most closely: HF, (toluene), or

(methanol)? Explain your reasoning.12. Estimate the normal boiling point of bromine, in

the following way: Determine for fromdata in Appendix D. Assume that remainsconstant and that Trouton’s rule is obeyed.

13. In what temperature range can the following equilib-rium be established? Explain.

14. Refer to Figures 12-19 and 19-9. Which has the lowestfree energy at 1 atm and solid, liquid, orgaseous carbon dioxide? Explain.

-60 °C:

H2O(l, 0.50 atm) ∆ H2O(g, 0.50 atm)

¢Hvap°Br2¢Hvap°

Br2,CH3OH

C6H5CH3

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814 Chapter 19 Spontaneous Change: Entropy and Free Energy

Free Energy and Spontaneous Change

15. Which of the following changes in a thermodynamicproperty would you expect to find for the reaction

at all temperatures: (a)(b) (c) (d) Explain.

16. If a reaction can be carried out only by electrolysis,which of the following changes in a thermodynamicproperty must apply: (a) (b)(c) (d) Explain.

17. Indicate which of the four cases in Table 19.1 appliesto each of the following reactions. If you are unable todecide from only the information given, state why.(a)(b)

(c)

18. Indicate which of the four cases in Table 19.1 appliesto each of the following reactions. If you are unable todecide from only the information given, state why.

(a)

¢H° = +105.5 kJ

H2O(g) +12

O2(g) ¡ H2O2(g)

¢H° = +159.2 kJNH4CO2NH2(s) ¡ 2 NH3(g) + CO2(g)

¢H° = +41.2 kJCO2(g) + H2(g) ¡ CO(g) + H2O(g)

¢H° = -87.9 kJPCl3(g) + Cl2(g) ¡ PCl5(g)

¢G 7 0?¢G = ¢H;¢S 7 0;¢H 7 0;

¢S 6 0?¢G 6 0;¢S 7 0;¢H 6 0;2 Br(g)¡Br2(g)

(b)

(c)

19. For the mixing of ideal gases (see Figure 19-2), explainwhether a positive, negative, or zero value is expectedfor and

20. What values of and would you expect forthe formation of an ideal solution of liquid compo-nents? (Is each value positive, negative, or zero?)

21. Explain why (a) some exothermic reactions donot occur spontaneously, and (b) some reactions inwhich the entropy of the system increases do notoccur spontaneously.

22. Explain why you would expect a reaction of the typealways to be spontaneous at

high rather than at low temperatures.B(g)+A(g)¡AB(g)

¢G¢S,¢H,¢G.¢S,¢H,

¢H° = -38.54 kJ

NO(g) +12

Cl2(g) ¡ NOCl(g)

¢H° = -3135 kJ

C6H6(l) +152

O2(g) ¡ 6 CO2(g) + 3 H2O(g)

Standard Free Energy Change

23. From the data given in the following table, determinefor the reaction

All data are at 298 K.

HCl(g)

24. Use data from Appendix D to determine values offor the following reactions at 25 °C.

(a)(b)(c)(d)

25. At 298 K, for the reaction and the standard molar

entropies are and Determine (a) at 298 Kand (b) whether the reaction proceeds spontaneouslyin the forward or the reverse direction when reactantsand products are in their standard states.

26. At 298 K, for the reaction

and the standard molar entropies are

Determine (a)at 298 K and (b) whether the reaction proceeds sponta-neously in the forward or the reverse direction whenreactants and products are in their standard states.

27. The following standard free energy changes are givenfor 25 °C.(1)

¢G° = -33.0 kJN2(g) + 3 H2(g) ¡ 2 NH3(g)

¢G°135.6 J K-1.HNO2(aq),152.2 J K-1;Br2(l),240.1 J K-1;NO2(g),82.4 J K-1;Br-(aq),0 J K-1;

H+(aq),-61.6 kJ=¢H°2 HNO2(aq),+Br2(l)¡2 NO2(g)+

2 Br-(aq)+2 H+(aq)

¢G°222.4 J K-1.POCl3(l),205.1 J K-1;O2(g),311.8 J K-1;PCl3(g),

-620.2 kJ=¢H°2 POCl3(l),¡O2(g)+2 PCl3(g)

2 Al(s) + 6 H+(aq) ¡ 2 Al3+(aq) + 3 H2(g)Fe3O4(s) + 4 H2(g) ¡ 3 Fe(s) + 4 H2O(g)2 SO3(g) ¡ 2 SO2(g) + O2(g)C2H2(g) + 2 H2(g) ¡ C2H6(g)

¢G°

-202.9-314.4NH4Cl(s)-95.30-92.31-16.48-46.11NH3(g)

¢Gf°, kJ mol�1¢Hf°, kJ mol�1

NH4Cl(s).¡HCl(g)+NH3(g)¢S°

(2)

(3)

(4)

(5)

Combine the preceding equations, as necessary, to ob-tain values for each of the following reactions.

(a)

(b)(c)

Of reactions (a), (b), and (c), which would tend to go tocompletion at 25 °C, and which would reach an equi-librium condition with significant amounts of all reac-tants and products present?

28. The following standard free energy changes are givenfor 25 °C.(1)

(2)

(3)

(4)

Combine the preceding equations, as necessary, to ob-tain values for the following reactions.(a)

¢G° = ?SO2(g) + CO(g) + 2 H2(g)COS(g) + 2 H2O(g) ¡¢G°

¢G° = -28.6 kJCO(g) + H2O(g) ¡ CO2(g) + H2(g)

¢G° = +1.4 kJCO(g) + H2S(g) ¡ COS(g) + H2(g)

¢G° = -41.5 kJCS2(g) + H2O(g) ¡ COS(g) + H2S(g)

¢G° = -246.4 kJSO2(g) + 3 CO(g) ¡ COS(g) + 2 CO2(g)

¢G° = ?2 NH3(g) + 2 O2(g) ¡ N2O(g) + 3 H2O(l)

¢G° = ?2 H2(g) + O2(g) ¡ 2 H2O(l)

¢G° = ?N2O(g) +32

O2(g) ¡ 2 NO2(g)

¢G°

¢G° = +208.4 kJ2 N2(g) + O2(g) ¡ 2 N2O(g)

¢G° = +102.6 kJN2(g) + 2 O2(g) ¡ 2 NO2(g)

¢G° = +173.1 kJN2(g) + O2(g) ¡ 2 NO(g)

¢G° = -1010.5 kJ4 NH3(g) + 5 O2(g) ¡ 4 NO(g) + 6 H2O(l)

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Exercises 815

(b)

(c)

Of reactions (a), (b), and (c), which is spontaneous inthe forward direction when reactants and productsare present in their standard states?

29. Write an equation for the combustion of one mole ofbenzene, and use data from Appendix D todetermine at 298 K if the products of the combus-tion are (a) and and (b) and

Describe how you might determine thedifference between the values obtained in (a) and(b) without having either to write the combustionequation or to determine values for the combus-tion reactions.

30. Use molar entropies from Appendix D, together withthe following data, to estimate the bond-dissociationenergy of the molecule.

Compare your result with the value listed in Table 10.3.F2(g) ¡ 2 F(g) ¢G° = 123.9 kJ

F2

¢G°

H2O(g).CO2(g)H2O(l),CO2(g)

¢G°C6H6(l),

¢G° = ?COS(g) + H2O(g) ¡ CO2(g) + H2S(g)

¢G° = ?SO2(g) + CO2(g) + 3 H2(g)COS(g) + 3 H2O(g) ¡ 31. Assess the feasibility of the reaction

by determining each of the following quantities forthis reaction at 25 °C.(a) (The standard molar entropy of is

)(b) (Use data from Table 10.3 and and

bond energies of 222 and respectively.)(c)Is the reaction feasible? If so, is it favored at high orlow temperatures?

32. Solid ammonium nitrate can decompose to dinitrogenoxide gas and liquid water. What is at 298 K? Isthe decomposition reaction favored at temperaturesabove or below 298 K?

¢G°

¢G°

301 kJ mol-1,N ¬ FF ¬ O¢H°

301.2 J K-1.N2F4(g)¢S°

N2H4(g) + 2 OF2(g) ¡ N2F4(g) + 2 H2O(g)

The Thermodynamic Equilibrium Constant

33. For one of the following reactions, Iden-tify that reaction. For the other two reactions, what isthe relationship between and K? Explain.(a)

(b)

(c)34. can be prepared by passing steam over hot iron:

4 H2(g).+Fe3O4(s)∆4 H2O(g)+3 Fe(s)H2(g)

NH4HCO3(s) ∆ NH3(g) + CO2(g) + H2O(l)

HI(g) ∆12

H2(g) +12

I2(g)

2 SO2(g) + O2(g) ∆ 2 SO3(g)Kp,Kc ,

Kp = K.=Kc (a) Write an expression for the thermodynamic equi-librium constant for this reaction.(b) Explain why the partial pressure of is inde-pendent of the amounts of Fe(s) and present.(c) Can we conclude that the production of from could be accomplished regardless of theproportions of Fe(s) and present? Explain.Fe3O4(s)

H2O(g)H2(g)

Fe3O4(s)H2(g)

Relationships Involving Q, and K¢G, ¢G°,

35. Use data from Appendix D to determine at 298 K

for the reaction

36. Use data from Appendix D to establish for the reac-tion (a) at 298 K for the reaction as written;(b) at 298 K.

37. Use data from Appendix D to determine values at298 K of and K for the following reactions. (Note:The equations are not balanced.)(a)(b)(c)

38. In Example 19-1, we were unable to conclude byinspection whether for the reaction

should be positive ornegative. Use data from Appendix D to obtain at298 K.

39. Use thermodynamic data at 298 K to decide in whichdirection the reaction

is spontaneous when the partial pressures of and are 0.20, and 0.10 atm, respectively.1.0 * 10-4,SO3

O2,SO2,

2 SO2(g) + O2(g) ∆ 2 SO3(g)

¢S°H2(g)+CO2(g)¡H2O(g)

+CO(g)¢S°

Ag+(aq) + SO4

2-(aq) ∆ Ag2SO4(s)Fe2O3(s) + H2(g) ∆ Fe3O4(s) + H2O(g)HCl(g) + O2(g) ∆ H2O(g) + Cl2(g)

¢G°

Kp

¢G°2 N2O5(g):∆O2(g)+2 N2O4(g)

2 NO(g).∆12

O2(g)+N2O(g)

Kp 40. Use thermodynamic data at 298 K to decide in whichdirection the reaction

is spontaneous when the partial pressures of and HCl are all 0.5 atm.

41. The standard free energy change for the reaction

is at 298 K. Use this thermodynamicquantity to decide in which direction the reaction is spon-taneous when the concentrations of

and are 0.10 M, M,and M, respectively.

42. The standard free energy change for the reaction

is at 298 K. Use this thermodynamicquantity to decide in which direction the reaction isspontaneous when the concentrations of

and are 0.10 M, and respectively.1.0 * 10-3 M,

1.0 * 10-3 M,OH-(aq)NH4

+(aq),NH3(aq),

29.05 kJ mol-1

NH3(aq) + H2O(l) ∆ NH4

+(aq) + OH-(aq)

1.0 * 10-31.0 * 10-3H3O+(aq)CH3CO2

-(aq),CH3CO2H(aq),

27.07 kJ mol-1

CH3CO2

-(aq) + H3O+(aq)CH3CO2H(aq) + H2O(l) ∆

Cl2,H2,

H2(g) + Cl2(g) ∆ 2 HCl(g)

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816 Chapter 19 Spontaneous Change: Entropy and Free Energy

43. For the reaction all but one of the following equations is correct.Which is incorrect, and why? (a) (b)

(c) (d)

44. Why is such an important property of a chemicalreaction, even though the reaction is generally carriedout under nonstandard conditions?

45. At 1000 K, an equilibrium mixture in the reactioncontains

0.276 mol 0.276 mol 0.224 mol CO, and0.224 mol (a) What is at 1000 K?(b) Calculate at 1000 K.(c) In which direction would a spontaneous reactionoccur if the following were brought together at1000 K: 0.0750 mol 0.095 mol 0.0340 mol CO,and 0.0650 mol

46. For the reaction at 1000 K.

(a) What is at 1000 K?[Hint: What is ](b) If 0.40 mol 0.18 mol and 0.72 mol aremixed in a 2.50-L flask at 1000 K, in what directionwill a net reaction occur?

47. For the following equilibrium reactions, calculate at the indicated temperature.[Hint: How is each equilibrium constant related to athermodynamic equilibrium constant, K?](a) at 445 °C

(b)

at 25 °C(c)

at 25 °C

(d)

at 25 °C48. Two equations can be written for the dissolution of

in acidic solution.

¢G° = -95.5 kJ mol-1

Mg(OH)2(s) + 2 H+(aq) ∆ Mg2+(aq) + 2 H2O(l)

Mg(OH)2(s)

Kc = 9.14 * 10-6

2 Fe2+(aq) + 2 Hg2+(aq)

2 Fe3+(aq) + Hg2

2+(aq) ∆

Kc = 4.61 * 10-3

N2O4(g) ∆ 2 NO2(g)Kc = 1.7 * 10-13

N2O(g) +12

O2(g) ∆ 2 NO(g)

Kc = 50.2H2(g) + I2(g) ∆ 2 HI(g)

¢G°

SO3O2,SO2,Kp ?

¢G°Kc = 2.8 * 102

2 SO3(g),∆O2(g)+2 SO2(g)H2O?

H2,CO2,

¢G°Kp

H2O.CO2,H2,

H2O(g)+CO(g)∆H2(g)+CO2(g)

¢G°RT ln Q.

+¢G°=¢Ge-¢G°>RT;=Kp1¢G° - ¢H°2>T;=¢S°Kp ;=K

2 NO2(g)¡O2(g)+2 NO(g)

(a) Explain why these two equations have differentvalues.

(b) Will K for these two equations be the same ordifferent? Explain.(c) Will the solubilities of in a buffer so-lution at depend on which of the two equa-tions is used as the basis of the calculation? Explain.

49. At 298 K, and for the reaction

Use these data to determine and compare your result with the value in Appendix D.

50. Use thermodynamic data from Appendix D to calcu-late values of for the following sparingly solublesolutes: (a) AgBr; (b) (c)[Hint: Begin by writing solubility equilibriumexpressions.]

51. To establish the law of conservation of mass, Lavoisiercarefully studied the decomposition of mercury(II)oxide:

At and (a) Show that the partial pressure of in equilib-rium with HgO(s) and Hg(l) at 25 °C is extremely low.(b) What conditions do you suppose Lavoisier usedto obtain significant quantities of oxygen?

52. Currently, is being studied as a source of carbonatoms for synthesizing organic compounds. One pos-sible reaction involves the conversion of tomethanol,

With the aid of data from Appendix D, determine(a) if this reaction proceeds to any significant extentat 25 °C;(b) if the production of is favored by rais-ing or lowering the temperature from 25 °C;(c) for this reaction at 500 K;(d) the partial pressure of at equilibrium if

and each initially at a partial pressureof 1 atm, react at 500 K.

H2(g),CO2(g)CH3OH(g)

Kp

CH3OH(g)

CO2(g) + 3 H2(g) ¡ CH3OH(g) + H2O(g)

CH3OH.CO2

CO2

O2(g)¢G° = +58.54 kJ.¢H° = +90.83 kJ25 °C,

HgO(s) ¡ Hg(l) +12

O2(g)

Fe(OH)3.CaSO4 ;Ksp

¢Gf°[COCl2(g)],COCl2(g).∆Cl2(g)+CO(g)6.5 * 1011

=Kp-137.2 kJ>mol=¢Gf°[CO(g)]

pH = 8.5Mg(OH)2(s)

¢G°

¢G° = -47.8 kJ mol-1

12

Mg(OH)2(s) + H+(aq) ∆12

Mg2+(aq) + H2O(l)

and K as Functions of Temperature≤G°

53. Use data from Appendix D to establish at 298 Kfor the reaction:

(a) (b) (c) (d) K.54. A possible reaction for converting methanol to

ethanol is

(a) Use data from Appendix D to calculate and for this reaction at 25 °C.¢G°

¢S°,¢H°,

C2H5OH(g) + H2O(g)CO(g) + 2 H2(g) + CH3OH(g) ¡

¢G°;¢H°;¢S°;

2 NaHCO3(s) ¡ Na2CO3(s) + H2O(l) + CO2(g)

(b) Is this reaction thermodynamically favored athigh or low temperatures? At high or low pressures?Explain.(c) Estimate for the reaction at 750 K.

55. What must be the temperature if the followingreaction has and

56. Estimate at 100 °C for the reaction Use data from Table 19.3 and

Figure 19-12.2 SO3(g).∆O2(g)

+2 SO2(g)Kp

Fe2O3(s) + 3 CO(g) ¡ 2 Fe(s) + 3 CO2(g)

¢S° = 15.2 J K-1?¢H° = -24.8 kJ,¢G° = -45.5 kJ,

Kp

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Integrative and Advanced Exercises 817

57. The synthesis of ammonia by the Haber process oc-curs by the reaction at 400 °C. Using data from Appendix D and assumingthat and are essentially unchanged in thetemperature interval from 25 to 400 °C, estimate at 400 °C.

58. Use data from Appendix D to determine (a)and at 298 K and (b) at 875 K for the

water gas shift reaction, used commercially to pro-duce [Hint: Assume that and are essentiallyunchanged in this temperature interval.]

59. In Example 19-10, we used the van’t Hoff equation todetermine the temperature at which for the reaction Obtain another estimate of this temperature withdata from Appendix D and equations (19.9) and(19.13). Compare your result with that obtained inExample 19-10.

60. The following equilibrium constants have been deter-mined for the reaction

at 448 °C and 66.9 at 350 °C. Use these datato estimate for the reaction.

61. For the reaction and at 298 K.

(a) What is at 0 °C?(b) At what temperature will Kp = 1.00?

Kp

Kp = 0.113¢H° = +57.2 kJ mol-12 NO2(g),∆N2O4(g)

¢H°Kp = 50.0

2 HI(g):∆I2(g)+H2(g)

2 SO3(g).∆O2(g)+2 SO2(g)Kp = 1.0 * 106

¢S°¢H°H2(g).+CO2(g)∆H2O(g)+CO(g)H2(g):

Kp¢G°¢S°,¢H°,

Kp

¢S°¢H°

2 NH3(g)∆3 H2(g)+N2(g)62. Use data from Appendix D and the van’t Hoff equa-

tion (19.15) to estimate a value of at 100 °C for thereaction [Hint: First determine at 25 °C. What is for thereaction?]

63. For the reaction

determine by using the van’t Hoff equation(19.15) and by using tabulated data in Appendix D.Compare the two results, and comment on how goodthe assumption is that is essentially independentof temperature in this case.

64. Sodium carbonate, an important chemical used in theproduction of glass, is made from sodium hydrogencarbonate by the reaction

Data for the temperature variation of for this reac-tion are at 30 °C; at50 °C; at 70 °C; and at 100 °C.(a) Plot a graph similar to Figure 19-12, and determine

for the reaction.(b) Calculate the temperature at which the total gaspressure above a mixture of and

is 2.00 atm.Na2CO3(s)NaHCO3(s)

¢H°

2.31 * 10-16.27 * 10-33.90 * 10-4Kp = 1.66 * 10-5

Kp

2 NaHCO3(s) ∆ Na2CO3(s) + CO2(g) + H2O(g)

¢H°

¢H°

Kp = 4.56 * 108 at 260 °C

Kp = 2.15 * 1011 at 200 °C

CO(g) + 3 H2(g) ∆ CH4(g) + H2O(g),

¢H°Kp

2 NO2(g).∆O2(g)+2 NO(g)Kp

Coupled Reactions

65. Titanium is obtained by the reduction of whichin turn is produced from the mineral rutile (a) With data from Appendix D, determine at298 K for this reaction.

(b) Show that the conversion of to with reactants and products in their standard states, isspontaneous at 298 K if the reaction in (a) is coupledwith the reaction

2 CO(g) + O2(g) ¡ 2 CO2(g)

TiCl4(l),TiO2(s)

TiO2(s) + 2 Cl2(g) ¡ TiCl4(l) + O2(g)

¢G°(TiO2).

TiCl4(l), 66. Following are some standard free energies of forma-tion, per mole of metal oxide at 1000 K: NiO,

MnO, The standardfree energy of formation of CO at 1000 K is per mol CO. Use the method of coupled reactions(page 809) to determine which of these metal oxidescan be reduced to the metal by a spontaneous reactionwith carbon at 1000 K and with all reactants and prod-ucts in their standard states.

-250 kJ-630 kJ.TiO2,-280 kJ;-115 kJ;

¢Gf°,

Integrative and Advanced Exercises67. Use data from Appendix D to estimate (a) the normal

boiling point of mercury and (b) the vapor pressure ofmercury at 25 °C.

68. Dinitrogen pentoxide, is a solid with ahigh vapor pressure. Its vapor pressure at 7.5 °C is100 mmHg, and the solid sublimes at a pressure of1.00 atm at 32.4 °C. What is the standard free energychange for the process at 25 °C?

69. Consider the vaporization of water: at 100 °C, with in its standard state,

but with the partial pressure of at 2.0 atm.Which of the following statements about this vapor-ization at 100 °C are true? (a) (b)(c) (d) Explain.¢G 7 0?¢G° 7 0,

¢G = 0,¢G° = 0,

H2O(g)H2O(l)H2O(g)

H2O(l) ¡N2O5(g)¡N2O5(s)

N2O5,

70. At 298 K, 1.00 mol BrCl(g) is introduced into a 10.0-Lvessel, and equilibrium is established in the reaction

Calculate the

amounts of each of the three gases present when equi-librium is established.[Hint: Use data from Appendix D as necessary.]

71. Use data from Appendix D and other informationfrom this chapter to estimate the temperature atwhich the dissociation of becomes appreciable[for example, with the 50% dissociated into I(g)at 1 atm total pressure].

I2(g)I2(g)

12

Cl2(g).+12

Br2(g)∆BrCl(g)

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818 Chapter 19 Spontaneous Change: Entropy and Free Energy

72. The following table shows the enthalpies and freeenergies of formation of three metal oxides at 25 °C.

PbO(red)

ZnO

(a) Which of these oxides can be most readily decom-posed to the free metal and (b) For the oxide that is most easily decomposed, towhat temperature must it be heated to produce at 1.00 atm pressure?

73. The following data are given for the two solid formsof at 298 K.

180.(?) (?)

Estimate values for the two missing entries. To dothis, assume that for the transition

the values of and at 25 °Chave the same values that they do at the equilibriumtemperature of 127 °C.

74. Oxides of nitrogen are produced in high-temperaturecombustion processes. The essential reaction is

At what approximate temper-ature will an equimolar mixture of and be1.0% converted to NO(g)?[Hint: Use data from Appendix D as necessary.]

75. Use the following data, as appropriate, to estimate themolarity of a saturated aqueous solution of

234

118.4

76. A plausible reaction for the production of ethyleneglycol (used as antifreeze) is

The following thermodynamic properties ofat 25 °C are given:

and Usethese data, together with values from Appendix D, toobtain a value of S°, the standard molar entropy of

at 25 °C.77. Use the following data

221.3146.0

together with other data from the text to determinethe temperature at which the equilibrium pressure of

-918.1-1085.8CuSO4 # H2O(s)-1400.0-1684.3CuSO4 # 3 H2O(s)

J mol�1 K�1kJ mol�1kJ mol�1S°,¢G°f ,¢H°f ,

CH2OHCH2OH(l),

-323.1 kJ mol-1.=¢Gf°-454.8 kJ mol-1=¢Hf°CH2OHCH2OH

2 CO(g) + 3 H2(g) ¡ CH2OHCH2OH(l)

-128.0-221.3IO3

-(aq)-32.6-599.5-545.8Sr2+(aq)

-855.1-1019.2Sr(IO3)2(s)

J mol�1 K�1kJ mol�1kJ mol�1S°,¢G°f ,¢H°f ,

Sr(IO3)2 .

O2(g)N2(g)2 NO(g).∆O2(g)+

N2(g)

¢S°¢H°HgI2(yellow),¡HgI2(red)

-102.9HgI2(yellow)-101.7-105.4HgI2(red)

J mol�1 K�1kJ mol�1kJ mol�1S°,¢G°f ,¢H°f ,

HgI2

O2(g)

O2(g)?

-318.3-348.3-11.20-31.05Ag2O

-188.9-219.0

¢Gf°, kJ mol�1¢Hf°, kJ mol�1

water vapor above the two solids in the followingreaction is 75 Torr.

78. For the dissociation of at 25 °C, A sam-

ple of pure is placed in a flask and connect-ed to an ultrahigh vacuum system capable ofreducing the pressure to mmHg.(a) Would produced by the decomposition of

at 25 °C be detectable in the vacuum systemat 25 °C?(b) What additional information do you need to de-termine as a function of temperature?(c) With necessary data from Appendix D, determinethe minimum temperature to which wouldhave to be heated for to become detectable inthe vacuum system.

79. Introduced into a 1.50-L flask is 0.100 mol of the flask is held at a temperature of 227 °C until equi-librium is established. What is the total pressure of thegases in the flask at this point?

[Hint: Use data from Appendix D and appropriate re-lationships from this chapter.]

80. From the data given in Exercise 64, estimate a value ofat 298 K for the reaction

81. The normal boiling point of cyclohexane, is80.7 °C. Estimate the temperature at which the vaporpressure of cyclohexane is 100.0 mmHg.

82. The term thermodynamic stability refers to the sign ofIf is negative, the compound is stable with

respect to decomposition into its elements. Use thedata in Appendix D to determine whether isthermodynamically stable at (a) 25 °C and (b) 200 °C.

83. At 0 °C, ice has a density of and an ab-solute entropy of At this tem-perature, liquid water has a density of and an absolute entropy of The pressure corresponding to these values is 1 bar.Calculate and for the melting oftwo moles of ice at its normal melting point.

84. The decomposition of the poisonous gas phosgene isrepresented by the equation

Values of for this reaction are at 99.8 °C and at 395 °C.

At what temperature is 15% dissociated whenthe total gas pressure is maintained at 1.00 atm?

85. Use data from Appendix D to estimate the aqueoussolubility, in milligrams per liter, of AgBr(s) at 100 °C.

86. The standard molar entropy of solid hydrazine at itsmelting point of 1.53 °C is The en-thalpy of fusion is For in theinterval from 1.53 °C to 298.15 K, the molar heat ca-pacity at constant pressure is given by the expression

Determine the stan-dard molar entropy of at 298.15 K.[Hint: The heat absorbed to produce an infinitesimalchange in the temperature of a substance is

]Cp dT.=dqrev

N2H4(l)97.78 + 0.05861T - 2802.=Cp

N2H4(l)12.66 kJ mol-1.K-1.67.15 J mol-1

COCl2

4.44 * 10-2=Kp6.7 * 10-9=KpKpCl2(g).+CO(g)∆COCl2(g)

¢H°¢S°,¢G°,¢G,

59.94 J mol-1 K-1.1.000 g mL-1

37.95 J mol-1 K-1.0.917 g mL-1

Ag2O(s)

¢Gf°¢Gf° .

C6H12,

2 NaHCO3(s) ¡ Na2CO3(s) + H2O(g) + CO2(g)

¢S°

PCl5(g) ∆ PCl3(g) + Cl2(g)

PCl5(g);

CO2(g)CaCO3(s)

PCO2

CaCO3(s)CO2(g)

10-9

CaCO3(s)+131 kJ mol-1.=¢G°CO2(g)+CaO(s)∆

CaCO3(s)CaCO3(s)

CuSO4 # 3 H2O(s) ∆ CuSO4 # H2O(s) + 2 H2O(g)

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Feature Problems 819

87. Use the following data to estimate the standard molarentropy of gaseous benzene at 298.15 K; that is,

For (s, 1 atm) at its meltingpoint of 5.53 °C, S° is The enthalpyof fusion is From the melting pointto 298.15 K, the average heat capacity of liquid benzeneis The enthalpy of vaporizationof at 298.15 K is and in the33.85 kJ mol-1,C6H6(l)

134.0 J K-1 mol-1.

9.866 kJ mol-1.128.82 J mol-1 K-1.

C6H6S°[C6H6(g, 1 atm)].

vaporization, is produced at a pressure of95.13 Torr. Imagine that this vapor could be compressedto 1 atm pressure without condensing and while behav-ing as an ideal gas. Calculate [Hint: Refer to Exercise 86, and note the following:For infinitesimal quantities, for the com-pression of an ideal gas, and for pressure–volume work, ]dw = -P dV.

-dw;=dqdq>dT;=dS

S°[C6H6(g, 1 atm)].

C6H6(g)

Feature Problems88. A tabulation of more precise thermodynamic data

than are presented in Appendix D lists the followingvalues for and at 298.15 K, at a stan-dard-state pressure of 1 bar.

69.91188.825

(a) Use these data to determine, in two differentways, at 298.15 K for the vaporization:

The value you obtainwill differ slightly from that on page 797, becausehere, the standard state pressure is 1 bar, and there, itis 1 atm.(b) Use the result of part (a) to obtain the value of Kfor this vaporization and, hence, the vapor pressure ofwater at 298.15 K.(c) The vapor pressure in part (b) is in the unit bar.Convert the pressure to millimeters of mercury.(d) Start with the value given onpage 797, and calculate the vapor pressure of water at298.15 K in a fashion similar to that in parts (b) and(c). In this way, demonstrate that the results obtainedin a thermodynamic calculation do not depend on theconvention we choose for the standard-state pressure,as long as we use standard-state thermodynamic dataconsistent with that choice.

89. The graphs show how varies with temperaturefor three different oxidation reactions: the oxidationsof C(graphite), Zn, and Mg to CO, ZnO, and MgO,respectively. Graphs such as these can be used toshow the temperatures at which carbon is an effectivereducing agent to reduce metal oxides to the free met-als. As a result, such graphs are important to metallur-gists. Use these graphs to answer the followingquestions.(a) Why can Mg be used to reduce ZnO to Zn at alltemperatures, but Zn cannot be used to reduce MgOto Mg at any temperature?(b) Why can C be used to reduce ZnO to Zn at sometemperatures but not at others? At what temperaturescan carbon be used to reduce zinc oxide?(c) Is it possible to produce Zn from ZnO by its directdecomposition without requiring a coupled reaction?If so, at what approximate temperatures might thisoccur?(d) Is it possible to decompose CO to C and in aspontaneous reaction? Explain.

O2

¢G°

¢G° = 8.590 kJ,

1 bar).H2O(g,∆(l, 1 bar)H2O¢G°

-228.572-241.818H2O(g)-237.129-285.830H2O(l)

J mol�1 K�1kJ mol�1kJ mol�1S°,¢G°f ,¢H°f ,

H2O(g)H2O(l)

(e) To the set of graphs, add straight lines represent-ing the reactions

given that the three lines representing the formationof oxides of carbon intersect at about 800°C.[Hint: At what other temperature can you relate and temperature?]The slopes of the three lines described above differsharply. Explain why this so—that is, explain theslope of each line in terms of principles governing freeenergy change.(f) The graphs for the formation of oxides of othermetals are similar to the ones shown for Zn and Mg;that is, they all have positive slopes. Explain why car-bon is such a good reducing agent for the reduction ofmetal oxides.

90. In a heat engine, heat is absorbed by a workingsubstance (such as water) at a high temperature Part of this heat is converted to work and the rest

is released to the surroundings at the lower tem-perature The efficiency of a heat engine is the ratio

The second law of thermodynamics establishesthe following equation for the maximum efficiency ofa heat engine, expressed on a percentage basis.

efficiency =wqh

* 100% =Th - T1

Th* 100%

w>qh.1Tl2.

1ql21w2, 1Th2.

1qh2

¢G°

2 CO(g) + O2(g) ¡ 2 CO2(g)

C(graphite) + O2(g) ¡ CO2(g)

�1000

500 1000 1500 2000

�800

�600

�400

�200

0 2 Zn � O2 2 ZnO

2 C � O2 2 CO

2 Mg � O2 2 MgO

mp

bp

bp

mp

Temperature, �C

�G

�, k

J

for three reactions as a function of temperature.The reactions are indicated by the equations writtenabove the graphs. The points noted by arrows are themelting points (mp) and boiling points (bp) of zinc andmagnesium.

¢G°▲

PETRMC19_780-821-hr 12/30/05 2:02 PM Page 819

820 Chapter 19 Spontaneous Change: Entropy and Free Energy

In a particular electric power plant, the steam leaving asteam turbine is condensed to liquid water at 41 °C and the water is returned to the boiler to be regeneratedas steam. If the system operates at 36% efficiency,(a) What is the minimum temperature of the steam

used in the plant?(b) Why is the actual steam temperature probablyhigher than that calculated in part (a)?(c) Assume that at the is in equilibriumwith Estimate the steam pressure at the tem-perature calculated in part (a).(d) Is it possible to devise a heat engine with greaterthan 100 percent efficiency? With 100 percent efficien-cy? Explain.

91. The free energy available from the complete combus-tion of 1 mol of glucose to carbon dioxide and water is

(a) Under biological standard conditions, computethe maximum number of moles of ATP that could

¢G° = -2870 kJ mol-1C6H12O6(aq) + 6 O2(g) ¡ 6 CO2(g) + 6 H2O(l)

H2O(l).H2O(g)Th

[H2O(g)]

1T12

Self-Assessment Exercises92. In your own words, define the following symbols:

(a) (b) (c) K.93. Briefly describe each of the following ideas, methods,

or phenomena: (a) absolute molar entropy; (b) coupledreactions; (c) Trouton’s rule; (d) evaluation of an equi-librium constant from tabulated thermodynamic data.

94. Explain the important distinctions between each ofthe following pairs: (a) spontaneous and nonsponta-neous processes; (b) the second and third laws of ther-modynamics; (c) and

95. For a process to occur spontaneously, (a) the entropyof the system must increase; (b) the entropy of thesurroundings must increase; (c) both the entropy ofthe system and the entropy of the surroundings mustincrease; (d) the net change in entropy of the systemand surroundings considered together must be a pos-itive quantity; (e) the entropy of the universe must re-main constant.

96. The free energy change of a reaction can be used to as-sess (a) how much heat is absorbed from the sur-roundings; (b) how much work the system does onthe surroundings; (c) the net direction in which the re-action occurs to reach equilibrium; (d) the proportionof the heat evolved in an exothermic reaction that canbe converted to various forms of work.

97. The reaction, , is expected to be (a) spontaneous at

all temperatures; (b) spontaneous at low tempera-tures, but nonspontaneous at high temperatures; (c)nonspontaneous at all temperatures; (d) spontaneousat high temperatures only.

98. If for a reaction, it must also be true that(a) (b) (c) (d) (e) theequilibrium activities of the reactants and products donot depend on the initial conditions.

¢S° = 0;¢H° = 0;K = 1;K = 0;¢G° = 0

¢H = -161 kJO2(g)+2 Cl2(g)¡2 Cl2O(g)

¢G°.¢G

¢Gf°;¢Suniv ;99. Two correct statements about the reversible reaction

are (a) (b) theequilibrium amount of NO increases with an in-creased total gas pressure; (c) the equilibriumamount of NO increases if an equilibrium mixture istransferred from a 10.0-L container to a 20.0-L con-tainer; (d) (e) the composition of an equilib-rium mixture of the gases is independent of thetemperature.

100. Suppose a graph similar to Figure 19-9 were drawnfor the process at 1 atm.(a) Refer to Figure 12-18 and determine the temper-ature at which the two lines would intersect.(b) What would be the value of at this tempera-ture? Explain.

101. Without performing detailed calculations, indicatewhether any of the following reactions would occurto a measurable extent at 298 K.(a) Conversion of dioxygen to ozone:

(b) Dissociation of to

(c) Formation of BrCl:

102. Explain briefly why(a) the change in entropy in a system is not always asuitable criterion for spontaneous change;(b) is so important in dealing with the questionof spontaneous change, even though the conditionsemployed in a reaction are very often nonstandard.

103. A handbook lists the following standard en-thalpies of formation at 298 K for cyclopentane,

and-77.2 kJ>mol.¢Hf°[C5H10(g)] =

-105.9 kJ>mol=¢Hf°[C5H10(l)]C5H10 :

¢G°

Br2(l) + Cl2(g) ¡ 2 BrCl(g)

N2O4(g) ¡ 2 NO2(g)NO2 :N2O4

3 O2(g) ¡ 2 O3(g)

¢G°

I2(s) ¡ I2(l)

K = Kc ;

K = Kp ;2 NO(g)∆O2(g)+N2(g)

form from ADP and phosphate if all the energy ofcombustion of 1 mol of glucose could be utilized.(b) The actual number of moles of ATP formed by acell under aerobic conditions (that is, in the presenceof oxygen) is about 38. Calculate the efficiency of en-ergy conversion of the cell.(c) Consider these typical physiological conditions.

Calculate for the conversion of 1 mol ADP to ATPand for the oxidation of 1 mol glucose under theseconditions.(d) Calculate the efficiency of energy conversion forthe cell under the conditions given in part (c). Com-pare this efficiency with that of a diesel engine thatattains 78% of the theoretical efficiency operatingwith and Suggest a reasonfor your result.[Hint: See Problem 90.]

T1 = 873 K.Th = 1923 K

¢G¢G

[ATP] = [ADP] = [Pi] = 0.00010 M.[glucose] = 1.0 mg>mL; pH = 7.0;PCO2 = 0.050 bar; PO2 = 0.132 bar;

PETRMC19_780-821-hr 12/30/05 2:02 PM Page 820

eLaboratory Exercise 821

(a) Estimate the normal boiling point of cyclopentane.(b) Estimate for the vaporization of cyclopen-tane at 298 K.(c) Comment on the significance of the sign of at 298 K.

¢G°

¢G°104. Consider the reaction

at 298 K.(a) Is the forward reaction endothermic or exothermic?(b) What is the value of at 298 K?(c) What is the value of K at 298 K?(d) Does the reaction tend to occur spontaneouslyat temperatures above 298 K, below 298 K, both,or neither?

¢G°

2 H2O(l)+N2O(g)¡NH4NO3(s)

eMedia Exercises

105. Observe the motion of atoms in the Mixing of Gasesactivity (Activebook 19-2). Describe why entropyincreases in this process. If the choice of each atom’slocation in either of the two halves of the system(boxes) is considered a microstate, W, calculate thechange in entropy between the initial (separated)and final (mixed) states.

106. (a) Describe how temperature causes a change inentropy on the atomic scale in the Lattice Vibrationsactivity (Activebook 19-2). (b) What is the property ofthe crystal that represents the degree of its order(or disorder)?

107. (a) In the Temperature Dependence of Entropy sim-ulation (Activebook 19-3), what point of the curve cor-responds to the greatest relative increase in entropy?(b) How does this relate to the values of enthalpychanges observed at this point, compared to else-where on the curve?

108. (a) Write out a balanced chemical equation for thereaction illustrated in the Formation of AluminumBromide movie (Activebook 19-3). (b) What informa-tion is needed to calculate the entropy change associ-ated with this reaction? (c) Using the tabulatedthermodynamic values in Appendix D, calculate theentropy change that would occur if 25.0 grams ofaluminum reacted with an excess of liquid bromineat 298 K.

109. (a) In the – Equilibrium animation (Active-book 19-6), under which condition(s) shown will thefree energy of the system be greater than the stan-dard free energy? (b) Which set of conditions maxi-mizes the free energy of the system?

N2O4NO2

eLaboratory ExerciseThese exercises allow students to apply the concepts andskills for this chapter in the simulated laboratory of VirtualChemLab. Worksheets for each of these exercises are foundin the Virtual ChemLab Workbook in your MediaPak.110. (VCL 19-1) The Balance Between Enthalpy and

EntropyFor chemical reactions, we say a reaction proceeds tothe right when is negative and the reaction pro-ceeds to the left when is positive. At equilibrium,¢G

¢G

is zero. The Gibbs–Helmholtz equation specifiesthat at constant temperature or, inother words, the sign and size of are governedby the balance between enthalpic and entropic

considerations. In this assignment you will dis-solve several different salts in water, measure the re-sulting temperature changes, and then make somedeductions about the thermodynamic driving forcesbehind the dissolving process.

1¢S2 1¢H2¢G¢G = ¢H - T¢S

¢G

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