Download The First Law of Thermodynamics Does Not Predict Spontaneous

20.1 The Second Law of
Thermodynamics: Entropy,
Free Energy, and the
Direction of Chemical
Reactions
Thermodynamics: Predicting Spontaneous Change
The First Law Does Not Predict Spontaneous Change
The Sign of H Does Not Predict Spontaneous Change
Freedom of Motion and Dispersal of Kinetic Energy
Entropy and the Number of Microstates
Entropy and the Second Law
Standard Molar Entropies and the Third Law
Predicting Relative S ° of a System
20.2 Calculating the Change in Entropy of a Reaction
Standard Entropy of Reaction
Entropy Changes in the Surroundings
Entropy Change and the Equilibrium State
Spontaneous Exothermic and Endothermic Reactions
20.3 Entropy, Free Energy, and Work
Free Energy Change and Reaction Spontaneity
Standard Free Energy Changes Free Energy and Work
Temperature and Reaction Spontaneity
Coupling of Reactions
20.4 Free Energy, Equilibrium, and Reaction Direction
Concepts and Skills to Review Before You Study This Chapter
 internal energy, heat, and work (Section 6.1)
 state functions (Section 6.1) and standard states (Section 6.6)
 enthalpy, H, and Hess's law (Sections 6.2 and 6.5)
 entropy and solution formation (Section 13.3)
 comparing Q and K to find reaction direction (Section 17.4)
In the last few chapters, we’ve examined some fundamental questions about change, whether
chemical or physical:
How fast does it occur? How far does it go toward completion? How is a change affected by concentration and temperature?
And we’ve explored these questions in systems ranging from the stratosphere to a limestone cave to the cells of your body.
But why does the change occur in the first place? Some changes seem to happen in one direction but not the other. A fuel
burns in air and forms carbon dioxide and water vapor as it runs a jet engine (photo), but those products will never react to reform
the fuel and oxygen. A new shovel left outside slowly rusts, but bring the rusty one inside and it won’t become shiny. A cube of sugar
dissolves in a cup of coffee after a little stirring, but wait for another millennium and the dissolved sugar won’t form the cube again.
Chemists call these spontaneous changes. Most release energy, but some absorb it.
The principles of thermodynamics, developed almost 200 years ago, explain why a spontaneous change occurs. And, as far as
we know, they apply to every change in every system in the universe—every working machine, every growing plant, every thinking
student, every moving continent, every exploding star.
IN THIS CHAPTER... We discuss why changes occur and how we can utilize them by focusing on
the concepts of entropy and free energy and their relation to the direction of a spontaneous
change.
 We discuss the need for a criterion to predict the direction of a spontaneous change.
 We review the first law of thermodynamics and see that it accounts for the energy of a change but not the direction.
 We see that the sign of the enthalpy change does not predict direction either.
 We find the criterion for predicting the direction of a spontaneous change in the second law of thermodynamics, which
focuses on entropy (S), a state function based on the natural tendency of a system's energy to become dispersed.
 We examine entropy changes for exothermic and endothermic processes.
 We develop the concept of free energy as a simplified criterion for spontaneous change and see how it relates to the
work a system can do.
 We explore the key relationship between the free energy change of a reaction and its equilibrium constant.
20.1 THE SECOND LAW OF THERMODYNAMICS: PREDICTING SPONTANEOUS CHANGE
A spontaneous change of a system is one that occurs under specified conditions without a continuous input of energy from outside
the system. The freezing of water, for example, is spontaneous at 1 atm and -5°C. A spontaneous process, such as burning or falling,
may need a little “push” to get started—a spark to ignite gasoline vapors in your car’s engine, a shove to knock a book off your
desk—but once the process begins, the process itself supplies the energy to continue. In contrast, a nonspontaneous change occurs
only if the surroundings continuously supply the system with an input of energy. Under given conditions, if a change is spontaneous
in one direction, it is not spontaneous in the other.
20.1 • The Second Law of Thermodynamics: Predicting Spontaneous Change
The term spontaneous does not mean instantaneous nor does it reveal anything about how long a process takes to occur;
it means that, given enough time, the process will happen by itself. Many processes are spontaneous but slow—ripening, rusting,
and aging.
Can we predict the direction of a spontaneous change in cases that are not as obvious as burning gasoline or falling books?
In this section, we first briefly consider two concepts that do not predict the direction of a spontaneous change and then focus on
one that does—the concept of entropy.
The First Law of Thermodynamics Does Not Predict Spontaneous Change
Let’s see whether energy changes can predict the spontaneous direction. Recall from Chapter 6 that the first law of
thermodynamics (law of conservation of energy) states that the internal energy (E) of a system, the sum of the kinetic and potential
energies of its particles, changes when heat (q) and/or work (w) are absorbed or released:
E = q + w
Whatever is not part of the system (sys) is part of the surroundings (surr); thus, the change in energy and, therefore, heat and/or
work absorbed by the system is released by the surroundings, and vice versa:
E sys = -E surr or (q + w) sys = -(q + w) surr
Since the system plus the surroundings is the universe (univ), it follows that the total energy of the universe is constant, so the
change in energy of the universe is zero:
 E sys + E surr = - E surr +  E surr = 0 =  E univ
The first law accounts for the energy, but not the direction, of a process. When gasoline burns in a car engine, the first law
states that the potential energy difference between the bonds in the fuel mixture and those in the exhaust gases is converted to
the kinetic energy of the moving car and its parts plus the heat released to the environment. But, why doesn’t the heat released in
the engine convert exhaust fumes back into gasoline and oxygen? When an ice cube melts in your hand, the first law tells that
energy from your hand is converted to kinetic energy as the solid changes to a liquid. But, why doesn’t the pool of water in your
cupped hand transfer the heat back to your hand and refreeze? Neither of these events violates the first law—if you measure the
work and heat in each case, you find that energy is conserved—but these reverse changes never happen. That is, the first law by
itself does not predict the direction of a spontaneous change.
The Sign of H Does Not Predict Spontaneous Change
Perhaps the sign of the enthalpy change (H), the heat gained or lost at constant pressure (qp), is the criterion for spontaneity; in
fact, leading scientists thought so through most of the 19th century. If so, we would expect exothermic processes ( H < 0) to be
spontaneous and endothermic processes (H > 0) nonspontaneous. Let’s examine some examples to see if this is true.
1. Spontaneous processes with H < 0. All freezing and condensing processes are exothermic and spontaneous at certain
conditions: H2O(l)  H2O(s) H°rxn = -H°fus = -6.02 kJ (1 atm; T = 0°C)
The burning of methane and all other combustion reactions are spontaneous and exothermic:
CH4(g) + 2O2(g)  CO2(g) + 2H2O(g) H°rxn = -802 kJ
Oxidation of iron and other metals occurs spontaneously and exothermically:
3
2Fe(s) + O2(g)  Fe2O3(s) H°rxn = -826 kJ
2
Ionic compounds form spontaneously and exothermically from their elements:
1
Na(s) + Cl2(g)  NaCl(s) H°rxn = -411 kJ
2
2. Spontaneous processes H>0. In many cases, an exothermic process occurs spontaneously
under one set of conditions, whereas the opposite, endothermic process occurs spontaneously
under another set. All melting and vaporizing processes are endothermic and spontaneous at
certain conditions:
H2O (s)  H2O (l)
H°rxn = H°fus = +6.02 kJ (1 atm; T = 0°C)
At ordinary pressure, water vaporizes spontaneously:
H2O (l)  H2O (g)
H°rxn = H°vap = +44.0 kJ (1 atm; T = 100°C)
Figure 20.1 A spontaneous endothermic reaction.
Most soluble salts dissolve endothermically and spontaneously:
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Chapter 20 • Thermodynamics: Entropy, Free Energy, and the Direction of Chemical Reactions
NaCI(s) + H20  Na+(aq) + Cl-(aq)
H°soln = +3.9 kJ
RbCl03 (s) + H20  Rb+(aq) +
H°soln = +47.7 kJ
NH4 N03 (s) + H20 
NH4+
ClO3-(aq)
(aq)
+NO3-(aq)
H°soln = +25.7 kJ
Even some endothermic reactions are spontaneous:
1
N2 O5(S)  2NO2(g) + O2(g)
H°rxn = +109.5 kJ
2
Ba(OH)2 .8H2O (s) + 2NH4NO3 (s)  Ba2 (aq) + 2NO3 - (aq) +2NH3 (aq) + 10H2 O(l)
+
H°rxn = +62.3 kJ
In the latter reaction, ionic solids are mixed (Figure 20.1A) and the waters of hydration released by one of them solvate the
ions. The reaction mixture absorbs heat from the surroundings so quickly that the beaker freezes to a wet board (Figure 20.1B).
Given just these few examples, we see that, as for the first law, the sign of ∆H by itself does not predict the direction of a
spontaneous change.
Freedom of Particle Motion and Dispersal of Kinetic Energy
When we look closely at the previous examples of spontaneous endothermic processes, they have one major feature in
common: in every case, the chemical entities— atoms, molecules, or ions—have more freedom of motion after the change. Put
another way, after the change, the particles have a wider range of energy of motion (kinetic energy); we say that the energy has
become more dispersed, distributed, or spread out:
• The phase changes convert a solid, in which motion is restricted, to a liquid, in which particles have more freedom to move
around each other, and then to a gas, in which the particles have much greater freedom of motion. Thus, the energy of motion is
more dispersed.
• Dissolving a salt changes a crystalline solid and a pure liquid into separate ions and solvent molecules moving and interacting, so
their freedom of motion is greater and their energy of motion more dispersed.
• In the chemical reactions, fewer moles of crystalline solids produce more moles of gases and/or solvated ions, so once again, the
freedom of motion of the particles increases and their energy of motion is more dispersed:
less freedom of particle motion  more freedom of particle motion
localized energy of motion  dispersed energy of motion
Phase change: solid  liquid  gas
Dissolving of salt: crystalline solid + liquid  ions in solution
Chemical change:
crystalline solids l gases + ions in solution
In thermodynamic terms, a change in the freedom of motion of particles in a system, that is, in the dispersal of their energy
of motion, is a key factor for predicting the direction of a spontaneous process.
Entropy and the Number of Microstates
Earlier, we discussed the quantized electronic energy levels of an atom (Chapter 7) and a molecule (Chapter 11), and mentioned
the quantized kinetic energy levels—vibrational, rotational, and translational—of a molecule (see Chapter 9, Tools of the
Laboratory, pp. 374-375). Now, we’ll see that the energy state of a whole system of atoms or molecules is quantized, too.
Energy Dispersal and the Meaning of Entropy Let’s see why freedom of motion and dispersal of energy relate to
spontaneous change:
• Quantization of energy. Picture a system of, say, 1 mol of N2 gas and focus on one molecule. At any instant, it is moving
through space (translating) at some speed and rotating at some frequency, and its atoms are vibrating at some frequency. In the
next instant, the molecule collides with another or with the container, and these motional (kinetic) energy states change to
different values. The complete quantum state of the molecule at any instant consists of its electronic states and these
translational, rotational, and vibrational states. In this discussion, we focus on the latter three, that is, on the kinetic energy
states.
• Number of microstates. The energy of all the molecules in a system is similarly quantized. Each quantized state of the system is
called a microstate, and at any instant, the total energy of the system is dispersed throughout one microstate. In the next instant,
it is dispersed throughout a different microstate. The number of microstates possible for a system of 1 mol of molecules is
23
staggering, on the order of1010 .
• Dispersal of energy. At a given set of conditions, each microstate has the same total energy as any other. Therefore, each
microstate is equally possible for the system, and the laws of probability say that, over time, all microstates are equally likely. The
number of microstates for a system is the number of ways it can disperse (distribute or spread) its kinetic energy among the
various motions of all its particles.
In 1877, the Austrian mathematician and physicist Ludwig Boltzmann related the number of microstates (W) to the entropy (S) of
a system:
20.1 • The Second Law of Thermodynamics: Predicting Spontaneous Change
S = k ln W
(20.1)
where k, the Boltzmann constant, is the universal gas constant (R) divided by Avogadro’s number (NA), or R/NA, and equals 1.38
X10 2 3 J/K. The term W is the number of microstates, so it has no units; therefore, S has units of joules/kelvin (J/K). Thus,

A system with fewer microstates (smaller W) has lower entropy (lower S).

A system with more microstates (larger W) has higher entropy (higher S).
For our earlier examples of endothermic processes,
lower entropy (fewer microstates)  higher entropy (more microstates)
Phase change:
solid liquid  gas
Dissolving of salt:
crystalline solid + liquid  ions in solution
Chemical change:
crystalline solids  gases + ions in solution
(Recall from Chapter 13 that entropy is a key factor in the formation of solutions.)
Entropy as a State Function If a change results in a greater number of microstates, there are more ways to disperse the energy
of the system and the entropy increases:
Smore microstates > Sfewer microstates
If a change results in a lower number of microstates, the entropy decreases. Like internal energy (E) and enthalpy (H), entropy is a
state function, so it depends only on the present state of the system, not on how it arrived at that state (see Chapter 6 , pp.
256-257). Therefore, the change in entropy of the system (Ssys) depends only on the difference between its final and initial values:
Ssys = Sfinal - Sinitial
As with any state function, Ssys > 0 when its value increases during a change. For example, for the phase change when dry ice
sublimes, the entropy increases:
CO2(s)  CO2(g)
Ssys = Sfinal - Sinitial = SgaseousCO2 - SsolidCO2 > 0
And, Ssys <0 when the Entropy decrese, as when water vapor condenses:
H2O(g)  H2O(l)
Ssys = Sfinal - Sinitial = Sliquid H2O - S gaseousH2O <0
As an example of a reaction during which entropy increases, consider the decomposition of dinitrogen tetroxide (N2O4 ,
written here as O2N—N02 ):
O2N—N02 (g)  2N02(g)
When the N—N bond in 1 mol of dinitrogentetroxide breaks, the 2 mol of N02 molecules have many more possible motions;
thus, at any instant, the energy of the system is dispersed into any one of a larger number of microstates. Thus, the change
in entropy of the system, which is the change in entropy of the reaction (Srxn), goes up:
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Chapter 20 • Thermodynamics: Entropy, Free Energy, and the Direction of Chemical Reactions
Ssys = Srxn = Sfinal - Sinitial = Sproducts - Sreactants = 2SNO2 – SO2N-NO2 > 0
Quantitative Meaning of an Entropy Change There are two approaches for quantifying an entropy change, but they give
the same result. The first is based on counting the number of microstates possible for a system, and the second on the heat
absorbed (or released) by the system. We’ll examine both with a system of 1 mol of a gas expanding from 1 L to 2 L and behaving
ideally, much as neon does at 298 K:
1
1.
mol neon (initial: 1 L and 298 K)  1 mol neon (final: 2 L and 298 K)
Quantifying Ssys from the number of microstates. Figure 20.2 shows two flasks connected by a stopcock—the right flask is
evacuated, and the left flask contains 1 mol of neon. When we open the stopcock, the gas expands until each flask contains 0.5mol
but why? Opening the stopcock increases the volume, which increases the number of translational energy levels the particles can
occupy as they move to more locations. Thus, the number of microstates-and the entropy-increases.
Figure 20.2 Spontaneous expansion
of a gas. A, With the
contains 1 mol of Ne. B,
expands and each flask
stopcock closed, the left flask
With the stopcock open, the gas
contains 0.5 mol of Ne.
Figure 20.3 The entropy increase due to expansion of a gas. Energy levels are shown as lines in a box of narrow width (left). Each
distribution of energies for the 21 particles is one microstate. When the stopcock is opened, the box is wider (volume increases, right),
and the particles have more energy levels available.
Figure 20.3 presents this idea with particles on energy levels in a box of changeable volume. When the stopcock opens, there are
more energy levels, and they are closer together on average, so more distributions of particles are possible.
In Figure 20.4 , the number of microstates is represented by the placement of particles in the left and/or right flasks:
 One Ne atom. At a given instant, an Ne atom in the left flask has its energy in one of some number (W ) of microstates. Opening the
stopcock increases the volume, which increases the number of possible locations and the number of translational energy level.
Thus, the system has 21, or 2, times as many microstates available when the atom moves through both flasks (final state, Wfinal) as
when it is confined to the left flask (initial state, Winitial).
 Two Ne atoms. For atoms A and B moving through both flasks, there are 22, or 4, times as many microstates as when the atoms were
initially in the left flask—some number of microstates with both A and B in the left, that many with A in the left and B in the right or
with B in the left and A in the right, and that many with both in the right.
 Three Ne atoms. Add another atom, and there are 23, or 8, times as many microstates when the stopcock is open.
 Ten Ne atoms. With 10 Ne atoms, there are 210, or 1024, times as many microstates for the atoms in both flasks as there were for the
10 atoms in the left flask.
 One mole of Ne atoms. With 1 mol (NA) of Ne, there are 2𝑁𝐴 times as many microstates for the atoms in the larger volume (Wfinal)
than in the smaller (Winitia)
𝑊𝐹𝑖𝑛𝑎𝑙
𝑊𝑖𝑛𝑖𝑡𝑖𝑎𝑙
= 2𝑁𝐴
20.3 • Entropy, Free Energy, and Work
Figure 20.4 Expansion of a gas and the increase
in number of microstates. Each set of particle
locations represents a different microstate.
When the volume increases (stopcock opens),
the relative number of microstates is 2n, where n
is the number of particles.
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Chapter 20 • Thermodynamics: Entropy, Free Energy, and the Direction of Chemical Reactions
Now let’s find Ssys through the Boltzmann equation, S = k ln W. From the properties of logarithms (Appendix A), we know that
ln A — ln B = ln A/B. Thus,
Ssys = Sfjnal - Sjnitial = k ln Wfinal - k ln Winitial = k ln
𝑊𝐹𝑖𝑛𝑎𝑙
𝑊𝑖𝑛𝑖𝑡𝑖𝑎𝑙
= k ln 26.02𝑥10
23
23
Also, from Appendix A, ln Ay = y ln A; so we have Ssys = k ln 26.02𝑥10 = k (6.02X 1023) (ln 2)
= (1.38X 10-23 J/K)(6.02X1023)(0.693) = 5.76 J/K
2. Quantifying Ssys from the changes in heat. Now let’s compare the Ssys we just found for a gas expanding into an evacuated
flask with the Ssys for a gas that is heated and does work on the surroundings. This approach uses the relationship
Ssys =
𝑞𝑟𝑒𝑣
𝑇
(20.2)
where T is the temperature at which the heat change occurs and q is the heat absorbed. The subscript
“rev” refers to a reversible process, one that occurs in such tiny increments that the system remains at
equilibrium and the direction of the change can be reversed by an infinitesimal reversal of conditions.
We’ll approximate a reversible expansion by placing 1 mol of Ne gas in a piston- cylinder assembly
within a heat reservoir (to maintain a constant T of 298 K) and start by confining the gas to a volume of
1 L by the “pressure” of a beaker of sand on the piston (Figure 20.5). We remove one grain of sand (an
“infinitesimal” decrease in pressure) with a pair of tweezers, and the gas (system) expands a tiny
amount, raising the piston and doing work, —w. Assuming Ne behaves ideally, it absorbs an equivalent
tiny increment of heat, q, from the heat reservoir. With each grain of sand removed, the expanding gas
absorbs another tiny increment of heat. This process simulates a reversible expansion because we can
reverse it by putting back a grain of sand, which causes the surroundings to do a tiny
quantity of work compressing the gas, thus releasing a tiny quantity of heat to the reservoir.
Figure 20.5 Simulating a reversible process.
If we continue this nearly reversible expansion to 2 L and use calculus to integrate the tiny increments of heat together, qrev is
1718 J. From Equation 20.2,
Ssys = 𝑞𝑟𝑒𝑣
𝑇
=
1718 𝐽
298 𝐾
= 5.76 J/K
This is the same value we obtained from counting the number of microstates.
Entropy and the Second Law of Thermodynamics
The change in entropy determines the direction of a spontaneous process, but we must consider more than just the
entropy change of the system. After all, at 20°C, solid water melts spontaneously and Ssys goes up, while at — 10°C,
liquid water freezes spontaneously and Ssys goes down. But when we consider both the system and its surroundings, we
find that all real processes occur spontaneously in the direction that increases the entropy of the universe (system plus
surroundings). This is one way to state the second law of thermodynamics.
The second law says that either the entropy change of the system or of the surroundings may be negative. But, for a
process to be spontaneous, the sum of the two entropy changes must be positive. If the entropy of the system decreases,
the entropy of the surroundings must increase even more to offset that decrease, so that the entropy of the universe
(system plus surroundings) increases. A quantitative statement of the second law is, for any real spontaneous process,
Suniv = Ssys + Ssurr > 0
(20.3)
(We return to this idea in Section 20.2.)
Standard Molar Entropies and the Third Law
Entropy and enthalpy are state functions, but their values differ in a fundamental way:
• For enthalpy, there is no zero point, so we can measure only changes.
• For entropy, there is a zero point, and we can determine absolute values by applying the third law of thermodynamics:
A perfect crystal has zero entropy at absolute zero: Ssys = 0 at 0 K.
A “perfect” crystal means all the particles are aligned flawlessly. At absolute zero, the particles have minimum energy, so there is
only one microstate. Thus, in Equation 20.1,
W = 1 so S = k ln 1 = 0
20.3 • Entropy, Free Energy, and Work
When we warm the crystal to any temperature above 0 K, the total energy increases, so it can be dispersed into more than one
microstate. Thus,
W > 1 and ln W > 0 so S > 0
In principle, to find S of a substance at a given temperature, we cool it to as close to 0 K as possible. Then we heat it in small
increments, dividing q by T to get the increase in S for each increment, and add up all the entropy increases to the temperature
of interest, usually 298 K. Therefore, S of a substance at a given temperature is an absolute value. As with other thermodynamic
variables,
• We compare entropy values for substances at the temperature of interest in their standard states: 1 atm for gases,
1 M for solutions and the pure substance in its most stable form for solids or liquids.
• Because entropy is an extensive property—one that depends on the amount of substance—we specify the standard molar
entropy (S°), in units of J/mol-K (or J mol1 K—1 ). (S ° values at 298 K for many elements, compounds, and ions appear, with other
thermodynamic variables, in Appendix B.)
Predicting Relative S° of a System
Let’s see how the standard molar entropy of a substance is affected by several parameters: temperature, physical state,
dissolution, and atomic size or molecular complexity. (Unless stated otherwise, the S° values refer to the system at 298 K.)
A
B
C
Figure 20.6 Visualizing the effect of temperature on entropy.
A, Computer simulations show each particle in a crystal moving about its lattice position. Adding heat increases T and the total energy,
so the particles have greater freedom of motion, and their energy is more dispersed. Thus, S increases.
B, At any T, there is a range of occupied energy levels and, thus, a certain number of microstates. Adding heat increases the total energy
(area under curve), so the range of occupied energy levels becomes greater, as does the number of microstates (higher S).
C, A system of 21 particles occupy energy levels (lines) in a box whose height represents the total energy. When heat is added, the total
energy increases (box is higher) and becomes more dispersed (more lines), so S increases.
Temperature Changes Temperature has a direct effect on entropy—for any substance, S° increases as T rises, as these
values for copper metal show:
T (K):
S °:
273
295
298
31.0 32.9
33.2
As heat is absorbed (q > 0), temperature, which is a measure of the average kinetic energy of the particles, increases. Recall that
the kinetic energies of gas particles are distributed over a range, which becomes wider as T rises (Figure 5.14, p. 225); liquids and
solids behave the same. Thus, at any instant, there are more microstates available in which the energy can be dispersed, so the
entropy of the substance goes up. Figure 20.6 presents three ways to view the effect of temperature on entropy.
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Chapter 20 • Thermodynamics: Entropy, Free Energy, and the Direction of Chemical Reactions
Physical States and Phase Changes In melting or vaporizing, heat is absorbed (q > 0). The particles have more
freedom of motion, and their energy is more dispersed. Thus, S ° increases as the physical state of a substance
changes from solid to liquid to gas:
Na
51.4(s)
153.6
S ° (s or l):
S ° (g):
H2O
69.9(l)
188.7
C (graphite)
5.7(s)
158.0
Figure 20.7 plots entropy versus T as solid O2 is heated and changes to liquid and then to gas, with S ° values at
various points; this behavior is typical of many substances. At the molecular scale, several stages occur:



Particles in the solid vibrate about their positions but, on average, remain fixed. The energy of the solid is
dispersed least, that is, has the fewest microstates, so the solid has the lowest entropy.
As T rises, the entropy increases gradually as the particles’ kinetic energy increases.
When the solid melts, the particles move much more freely between and around each other, so there is an
abrupt increase in entropy (Sfus).
Figure 20.7 The increase in entropy during phase changes from solid to liquid to gas.
• Further heating of the liquid increases the speed of the particles, and the entropy increases gradually.
• When the liquid vaporizes and becomes a gas, the particles undergo a much larger, abrupt entropy increase
(Sovap); the increase in entropy from liquid to gas is much larger than from solid to liquid: Sovap >> Sofus
• Finally, with further heating of the gas, the entropy increases gradually.
Dissolving a Solid or a Liquid in Water Recall from Chapter 13 that, in general, entropy increases when a solid
or liquid solute dissolves in a solvent: Ssoln > (Ssolute + Ssolvent). But, when water is the solvent, the entropy may also
depend on the nature of solute and solvent interactions and includes two opposing events:
1. For ionic solutes, when the crystal dissolves in water, the ions have much more freedom of motion and their
energy is dispersed into more microstates. That is, the entropy of the ions themselves is greater in the
solution. However, water molecules become arranged around the ions (Figure 20.8), which limits the
molecules’ freedom of motion (see also Figure 13.2, p. 519). In fact, around small, multiply charged ions, H2O
molecules become so organized that their energy of motion becomes less dispersed. This negative portion of
the total entropy change can lead to negative S° values for the ions in solution. In the case of AlCl3, the Al3+
(aq) ion has such a negative S° value (-313 J/mol-K) that when AlCl3 dissolves in water, even though S° of
Cl-(aq) is positive, the entropy of aqueous AlCl3 is lower than that of solid AlCl3.*
S ° (s or l):
S ° (aq):
NaCl
72.1(s)
115.1
AlCl3
167(s)
-148
CH3OH
127(l)
132
20.3 • Entropy, Free Energy, and Work
*An S° value for a hydrated ion can be negative because it is relative to the S° value for the hydrated proton, H+
(aq), which is assigned a value of 0. In other words, Al3+ (aq) has a lower entropy than H+ (aq).
MIX
Figure 20.8 The entropy change accompanying the dissolution of a salt. The entropy of a salt solution is
usually greater than that of the solid and of water, but it is affected by water molecules becoming
organized around each ion.
A Ethanol
B Water
C Solution of water and ethanol
Figure 20.9 The small increase in entropy when ethanol dissolves in water. In pure ethanol (A) and water (B),
molecules form many H bonds to other like molecules. C, In solution, these two kinds of molecules form H bonds to
each other, so their freedom of motion does not change significantly.
2. For molecular solutes, the increase in entropy upon dissolving is typically much smaller than for
ionic solutes. After all, for a solid such as glucose, there is no separation into ions, and for a
liquid such as methanol or ethanol (Figure 20.9), there is no breakdown of a crystal structure.
Furthermore, in these small alcohols, as in pure water, the molecules form many H bonds, so
there is relatively little change in their freedom of motion either before or after they are mixed.
Dissolving a Gas in a Liquid The particles in a gas already have so much freedom of motion—and, thus,
such highly dispersed energy—that they lose some when they dissolve in a liquid or solid. Therefore, the
entropy of a solution of a gas in a liquid or a solid is always less than the entropy of the gas itself. For
instance, when gaseous O2 [S°(g) = 205.0 J/mol.K] dissolves in water, its entropy decreases considerably [S °(aq)
= 110.9 J/mol.K] (Figure 20.10). When a gas dissolves in another gas, however, the entropy increases as a result
of the separation and mixing of the molecules.
Figure 20.10 The entropy of a gas dissolved in a liquid.
11
12
Chapter 20 • Thermodynamics: Entropy, Free Energy, and the Direction of Chemical Reactions
Atomic Size and Molecular Complexity Differences in S° values for substances in the same phase are usually based on
atomic size and molecular complexity.
1. Within a periodic group, energy levels become closer together for heavier atoms, so the number of
microstates, and thus the molar entropy, increase:
Li
Na
K
Rb
Cs
Molar mass (g/mol):
6.941
22.99
39.10
85.47
132.9
S°(s):
29.1
51.4
64.7
69.5
85.2
The same trend of increasing entropy holds for similar compounds down a group:
HF
20.01
173.7
Molar mass (g/mol):
S°(g):
HCl
36.46
186.8
HBr
80.91
198.6
HI
127.9
206.3
2. For different forms of an element (allotropes), the entropy is higher in the form that allows the atoms more
freedom of motion. For example, S° of graphite is 5.69 J/mol.K, whereas S° of diamond is 2.44 J/mol.K. In
graphite, covalent bonds extend within a two-dimensional sheet, and the sheets move past each other
relatively easily; in diamond, covalent bonds extend in three dimensions, allowing the atoms little movement
(Table 12.6, p. 490).
3. For compounds, entropy increases with chemical complexity (that is, with number of atoms in the formula),
and this trend holds for ionic and covalent substances:
NaCl
AlCl3
P4O10
S °(s):
72.1
167
229
S °(g):
NO
NO2
N2O4
211
240
304
The trend is also based on the different motions that are available. Thus, for example, among the nitrogen oxides
listed, the number of different vibrational motions increases with the number of atoms in the molecule (Figure
20.11; see also Chapter 9, p. 374).
NO
NO2
N2O4
Figure 20.11 Entropy, vibrational motion, and molecular complexity.
For larger compounds, we also consider the motion of different parts of a molecule. A long hydrocarbon chain
can rotate and vibrate in more ways than a short one, so entropy increases with chain length. A ring compound, such
as cyclopentane (C5 H10), has lower entropy than the chain compound with the samNoe molar mass, pentene
(C5 H10), because the ring structure restricts freedom of motion:
S °:
CH4(g)
C2H6(g)
C3H8(g)
C4H1()(g)
C5H1()(g)
C5H1()(cyclo, g)
C2H5OH(/ )
186
230
270
310
348
293
161
Remember, these trends hold only for substances in the same physical state. Gaseous methane (CH4) has higher
entropy than liquid ethanol (C2 H5 OH), even though ethanol molecules are more complex. When gases are
compared with liquids, the effect of physical state dominates the effect of molecular complexity.
20.3 • Entropy, Free Energy, and Work
SAMPLE PROBLEM 20.1 Predicting Relative Entropy Values
Problem Select the substance with the higher entropy in each pair, and explain your choice
[Assume constant temperature, except in part (e)]:
(a)
(b)
(c)
(d)
(e)
(f)
1 mol of SO2 (g) or 1 mol of SO3 (g)
1 mol of CO2 (s) or 1 mol of CO2 (g)
3 mol of O2 (g) or 2 mol of O3 (g)
1 mol of KBr(s) or 1 mol of KBr(aq)
Seawater at 2°C or at 23 °C
1 mol of CF4 (g) or 1 mol of CCl4 (g)
Plan In general, particles with more freedom of motion have more microstates in which to disperse their kinetic energy, so
they have higher entropy. We know that either raising temperature or having more particles increases entropy. We apply
the general categories described in the text to choose the member with the higher entropy.
Solution
(a) 1 mol of SO3 (g). For equal numbers of moles of substances with the same types of atoms in the same physical state, the
more atoms in the molecule, the more types of motion available, and thus the higher the entropy.
(b) 1 mol of CO2 (g). For a given substance, entropy increases in the sequence S < l < g
(c) 3 mol of O2 (g). The two samples contain the same number of oxygen atoms but different numbers of molecules.
Despite the greater complexity of O3, the greater number of molecules dominates because there are many more
microstates possible for 3 mol of particles than for 2 mol.
(d) 1 mol of KBr(aq). The two samples have the same number of ions, but their motion is more limited and their energy
less dispersed in the solid than in the solution.
(e) Seawater at 23°C. Entropy increases with rising temperature.
(f) 1 mol of CCl4 (g). For similar compounds, entropy increases with molar mass.
FOLLOW-UP PROBLEMS
Brief Solutions to all Follow-up Problems appear at the end of the chapter.
20.1A Select the substance with the higher entropy in each pair, and explain your choice
(assume 1 mol of each at the same T ):
(a) PCl3 (g) or PCl5 (g)
(b) CaF2 (s) or BaCl2 (s)
(c) Br2 (g) or Br2 (l)
20.1B Select the substance with the lower entropy in each pair, and explain your choice
(assume 1 mol of each at the same T ):
( a ) LiBr(aq) or NaBr(aq)
( b ) quartz or glass
(c) ethylcyclobutane or cyclohexane
SOME SIMILAR PROBLEMS 20.22-20.29
 Summary of Section 20.1
 A change is spontaneous under specified conditions if it occurs without a continuous input of energy.
 Neither the first law of thermodynamics nor the sign of DH predicts the direction of a spontaneous change.
 Many spontaneous processes involve an increase in the freedom of motion of the system's particles and, thus, in the
dispersal of the system's energy of motion.
 Entropy is a state function that measures the extent of energy dispersed into the number of microstates possible for
a system. Each microstate consists of the quantized energy levels of the system at a given instant.
 The second law of thermodynamics states that, in a spontaneous process, the entropy of the universe (system plus
surroundings) increases.
 Absolute entropy values can be determined because perfect crystals have zero entropy at 0 K (third law of
thermodynamics).
 Standard molar entropy, S° (J/mol.K), is affected by temperature, phase changes, dissolution, and atomic size or
molecular complexity.
13
14
Chapter 20 • Thermodynamics: Entropy, Free Energy, and the Direction of Chemical Reactions
20.2 CALCULATING THE CHANGE IN ENTROPY OF A REACTION
Chemists are especially interested in learning how to predict the sign and calculate the value of the entropy change that
occurs during a reaction.
Entropy Changes in the System: Standard Entropy of Reaction (Sorxn)
The standard entropy of reaction, Sorxn is the entropy change that occurs when all reactants and products are in their
standard states.
Predicting the Sign of Sorxn Changes in structure and especially in amount (mol) of gas help us predict the sign of
Sorxn. Because gases have such great freedom of motion and, thus, high molar entropies, if the number of moles of gas
increases, Sorxn is positive; if the number decreases, Sorxn is negative. Here are a few examples:
• Increase in amount of gas. When gaseous H2 reacts with solid I2 to form gaseous HI, the total number of moles of substance
stays the same. Nevertheless, the sign of Sorxn is positive (entropy increases) because the number of moles of gas increases:
H2 (g) + I2 (s)  2HI(g) Sorxn = Soproduct - Soreactant > 0
• Decrease in amount of gas. When ammonia forms from its elements, 4 mol of gas produce 2 mol of gas, so Sorxn is negative
(entropy decreases):
N(g) + 3H2(g) = 2NH3(g) (reversible Reaction) Sorxn = Soproduct - Soreactant < 0
• No change in amount of gas, but change in structure. When the amount (mol) of gas doesn’t change, we cannot predict
the sign of Sorxn . But, a change in one of the structures can make it easier to predict the sign of Sorxn. For example, when
cyclopropane is heated to 500oC, the ring opens and propene forms. The chain has more freedom of motion than the ring, so
Sorxn is positive:
𝐻𝑒𝑎𝑡
1 mole of cyclopropane (g)
CH3-CH=CH2 (g) Sorxn = Soproduct - Soreactant > 0
−−>
Keep in mind, however, that in general we cannot predict the sign of the entropy change unless the reaction involves a
change in number of moles of gas.
Calculating Sorxn from S° Values By applying Hess’s law (Chapter 6), we combined Hof values to find Horxn.
Similarly, we combine S° values to find the standard entropy of reaction, Sorxn :
Sorxn = mSoroducts - nSoreactants (20.4)
where m and n are the amounts (mol) of products and reactants, respectively, given by the coefficients in the
balanced equation. For the formation of ammonia, we have
Sorxn = [(2 mol NH3 )(S° of NH3 ) ] - [(1 mol N2 )(S° of N2) + (3 mol H2 )(5° of H2 ) ]
From Appendix B, we find the S° values:
Sorxn = [(2 mol)(193 J/mol-K)] - [(1 mol)(191.5 J/mol-K) + (3 mol)(130.6 J/mol-K)] = -197 J/K
As we predicted from the decrease in number of moles of gas, Sorxn < 0.
SAMPLE PROBLEM 20.2 Calculating the Standard Entropy of Reaction Sorxn
Problem Predict the sign of Sorxn, if possible, and calculate its value for the combustion of 1 mol of propane at 25°C:
C3H8(g) + 5O2(g)  3CO2(g)+4H2O(l)
Plan We use the change in the number of moles of gas to predict the sign of S°rxn. To find S °xn, we apply Equation 20.4.
Solution The amount (mol) of gas decreases ( 6 mol yields 3 mol), so the entropy should decrease (S°xn < 0). Calculating S°rxn.
Using Appendix B values,
S°xn = [(3 mol CO2 )(S° of CO2 ) + (4 mol H2 O)(S° of H2 O)]- [(1 mol C3 H8 )(S° of C3 H8 )+ (5 mol O2 )(S° of O2 )]
= [(3 mol)(213.7 J/mol-K) + (4 mol)(69.940 J/mol-K)]- [(1 mol)(269.9 J/mol-K) 1 (5 mol)(205.0 J/mol-K)]
= -374.0 J/K
Check S°xn < 0, so our prediction is correct. Rounding gives [3(200) + 4(70)] - [270 + 5(200)] = 880 - 1270 = -390, close to the
calculated value.
20.3 • Entropy, Free Energy, and Work
FOLLOW-UP PROBLEMS
20.2A Balance each equation, predict the sign of Sr°xn if possible, and calculate its value at 298 K:
(a) NO(g)  N2O(g) + N2O3 (g)
(b) CH3OH(g)  CO(g) + H2(g)
20.2B Balance each equation, predict the sign of S°rxn if possible, and calculate its value at 25°C:
(a) NaOH(s) + CO2 (g) Na2 CO3 (s) + H2 O(l)
(b) Fe(s) + H2O(g) Fe2O3 (s) + H2(g)
SOME SIMILAR PROBLEMS 20.33-20.38
Entropy Changes in the Surroundings: The Other Part of the Total
In the synthesis of ammonia, the combustion of propane, and many other spontaneous reactions, the entropy of the
system decreases (S°xn < 0). Remember that the second law dictates that, for a spontaneous process, a decrease in the
entropy of the system is outweighed by an increase in the entropy of the surroundings. In this section, we examine the
influence of the surroundings—in particular, the addition (or removal) of heat and the temperature at which this heat
flow occurs—on the total entropy change.
The Role of the Surroundings In essence, the surroundings add heat to or remove heat from the system. That is, the
surroundings function as an enormous heat source or heat sink, one so large that its temperature remains constant, even though
its entropy changes. The surroundings participate in the two types of enthalpy changes as follows:
1. In an exothermic change, heat released by the system is absorbed by the surroundings. More heat increases the freedom
of motion of the particles and makes the energy more dispersed, so the entropy of the surroundings increases:
For an exothermic change:
qsys < 0,
qsurr > 0, and
Ssurr > 0
2. In an endothermic change, heat absorbed by the system is released by the surroundings. Less heat reduces the freedom of
motion of the particles and makes the energy less dispersed, so the entropy of the surroundings decreases:
For an endothermic change:
qsys > 0,
qsurr < 0, and
ASsurr < 0
Temperature at Which Heat Is Transferred The temperature of the surroundings when the heat is transferred also
affects Ssurr. Consider the effect of an exothermic reaction at a low or at a high temperature:
• At a low T, such as 20 K, there is little motion in the surroundings due to little energy. This means there are few energy levels in
each microstate and few microstates in which to disperse the energy. Transferring a given quantity of heat to these
surroundings causes a relatively large change in how much energy is dispersed.
• At a high T, such as 298 K, the surroundings have a large quantity of energy dispersed. There are more energy levels in each
microstate and a greater number of microstates. Transferring the same given quantity of heat to these surroundings causes a
relatively small change in how much energy is dispersed.
In other words, Ssurr is greater when heat is added at a lower T. Putting these ideas together, Ssurr is directly related to
an opposite change in the heat of the system (q sys) and inversely related to the temperature at which the heat is
transferred:
𝒒𝒔𝒚𝒔
Ssurr = 𝑻
For a process at constant pressure, the heat (qP) is AH (Section 6.2), so
Ssurr = -
𝑯𝒔𝒚𝒔
𝑻
(20.5)
Thus, we find Ssurr by measuring Hsys and T at which the change takes place.
The main point: If a spontaneous reaction has a negative Ssys (fewer microstates into which energy is dispersed),
Ssurr must be positive enough (even more microstates into which energy is dispersed) for Suniv to be positive
(net increase in number of microstates for dispersing the energy).
SAMPLE PROBLEM 20.3 Determining Reaction Spontaneity
Problem At 298 K, the formation of ammonia has a negative Sosys:
N2 (g) + 3H2 (g) 2NH3(g)
Sosys = -197 J/K
Calculate ASuniv, and state whether the reaction occurs spontaneously at this temperature.
15
16
Chapter 20 • Thermodynamics: Entropy, Free Energy, and the Direction of Chemical Reactions
Plan For the reaction to occur spontaneously, Suniv > 0, so Ssurr must be greater than +197 J/K. To find Ssurr, we need H osys,
which is the same as H orxn. We use Hof values from Appendix B to find H orxn. Then, we divide H orxn by the given T (298 K) to
find Ssurr. To find Suniv, we add the calculated Ssurr to the given Sosys (-197 J/K).
Solution Calculating H osys:
H osys = H orxn
= [(2 mol NH3 )(-45.9 kJ/mol)] - [(3 mol H2 )(0 kJ/mol)+ (1 mol N2 )(0 kJ/mol)]
= -91.8 kJ
Calculating Ssurr :
Ssurr = -
𝑜
𝐻𝑠𝑦𝑠
𝑇
−91.8 𝑘𝐽 ×
=-
−1000 𝐽
1 𝑘𝐽
298 𝐾
= 308 J/K
Determining Suniv:
Suniv = Sosys + Ssurr = -197 J/K + 308 J/K = 111 J/K
Suniv > 0 , so the reaction occurs spontaneously at 298 K (see margin).
Check Rounding to check the math, we have
H orxn  2(-45 kJ) = -90 kJ
Ssurr  -(-90,000 J)/300 K = 300 J/K
Suniv  -200 J/K + 300 J/K = 100 J/K
Given the negative H orxn, Le Chatelier’s principle says that low T favors NH3 formation, so the answer is reasonable (see
Section 17.6).
Comment
1. Because H° has units of kJ, and S has units of J/K, don’t forget to convert kJ to J, or you’ll introduce a large error.
2. This example highlights the distinction between thermodynamics and kinetics. NH3 forms spontaneously, but so slowly
that catalysts are required to achieve a practical rate.
FOLLOW-UP PROBLEMS
20.3A Gaseous phosphorus trichloride forms from the elements through this reaction:
P4 (s) + 6Cl2(g)  4PCl3(g)
Calculate H orxn, Ssurr , and Sorxn to determine if the reaction is spontaneous at 298 K.
20.3B Does the oxidation of FeO(s) to Fe2O3(s) occur spontaneously at 298 K? (Show the calculation for 1 mol Fe2O3.)
SOME SIMILAR PROBLEMS 20.39-20.42
Do Organisms Obey the Laws of Thermodynamics? Taking the surroundings into account is crucial, not only for
determining reaction spontaneity, as in Sample Problem 20.3, but also for understanding the relevance of thermodynamics to
biology. Let's examine the first and second laws to see if they apply to living systems.
Figure 20.12 A whole-body calorimeter. In this room-sized apparatus, a person exercises while respiratory gases, energy
input and output, and other physiological variables are monitored.
1. Do organisms comply with the first law? The chemical bond energy in food and oxygen is converted into the
mechanical energy of jumping, flying, crawling, swimming, and countless other movements; the electrical energy of nerve
conduction; the thermal energy of warning the body; and so forth. Many experiments have demonstrated that the total
energy is conserved in these situations. Some of the earliest were performed by Lavoisier, who showed that “animal heat”
was produced by slow, continual combustion. In experiments with guinea pigs, he invented a calorimeter to measure the
heat released from intake of food and 02 and output of C02 and water, and he included respiration in his new theory of
20.3 • Entropy, Free Energy, and Work
combustion. Modem, room-sized calorimeters measure these and other variables to confirm the conservation of energy for
an exercising human (Figure 20.12).
2. Do organisms comply with the second law? Mature humans are far more complex than the egg and sperm cells from
which they develop, and modem organisms are far more complex than the one-celled ancestral specks from which they
evolved. Are the growth of an organism and the evolution of life exceptions to the spontaneous tendency of natural
processes to increase freedom of motion and disperse energy? For an organism to grow or a species to evolve, many moles
of oxygen and nutrients— carbohydrates, proteins, and fats—undergo exothermic reactions to form many more moles of
gaseous CO2 and H2O. Formation of these waste gases and the accompanying release of heat result in an enormous
increase in the entropy of the surroundings. Thus, the localization of energy and synthesis of macromolecular structures
required for the growth and evolution of organisms (system) cause a far greater dispersal of energy and freedom of motion
in the environment (surroundings). When system and surroundings are considered together, the entropy of the universe,
as always, increases.
The Entropy Change and the Equilibrium State
For a process approaching equilibrium, Suniv > 0. When the process reaches equilibrium, there is no further net change, Suniv =0,
because any entropy change in the system is balanced by an opposite entropy change in the surroundings:
At equilibrium: Suniv = Ssys + Ssurr = 0 so Ssys = -Ssurr
As an example, let’s calculate Suniv for the vaporization-condensation of 1 mol of water at 100°C (373 K),
H2O(l; 373 K) = H2O(g; 373 K)
First, we find S°ys for the forward change (vaporization) of 1 mol of water:
S°sys = mS°products - mS°reactants
= S° of H2 O (g; 373 K) - S° of H2 O (l; 373 K)
= 195.9 J/K - 86.8 J/K
= 109.1 J/K
As we expect, the entropy of the system increases (Sosys> 0) as the liquid absorbs heat and changes to a gas.
For Ssurr of the vaporization step, we have
𝑜
∆𝐻𝑠𝑦𝑠
Ssurr = 𝑇
𝑜
𝑜
Where ∆𝐻𝑠𝑦𝑠
= ∆𝐻𝑣𝑎𝑝
at 373 K = 40.7 kJ/mol =40.7 x 103 J/mol. For 1 mol of water, we have
Ssurr = -
𝑜
∆𝐻𝑠𝑦𝑠
𝑇
=-
40.7𝑥103 𝐽
373𝐾
= -109 J/K
The surroundings lose heat, and the negative sign means that the entropy of the surroundings decreases. The two entropy
changes have the same magnitude but opposite signs, so they cancel:
Suniv = 109 J/K + (-109 J/K) = 0
For the reverse change (condensation), Suniv also equals zero, but Ssys and Ssurr have signs opposite those for vaporization.
A similar treatment of a chemical change shows the same result: the entropy change of the forward reaction is equal in
magnitude but opposite in sign to the entropy change of the reverse reaction. Thus, when a system reaches equilibrium,
neither the forward nor the reverse reaction is spontaneous, and so there is no net reaction in either direction.
Spontaneous Exothermic and Endothermic Changes
No matter what its enthalpy change, a reaction occurs because the total entropy of the reacting system and its surroundings
increases. There are two ways this can happen:
1. In an exothermic reaction (Hsys < 0), the heat released by the system increases the freedom of motion and dispersal of
energy in the surroundings; thus, Ssurr > 0.
• If the entropy of the products is more than that of the reactants (Ssys > 0), the total entropy change (Ssys + Ssurr) will
be positive (Figure 20.13A). For example, in the oxidation of glucose, an essential reaction for all higher organisms,
C6H12O6(s) + 6O2(g)  6CO2(g) + 6H2O(g) + heat
6 mol of gas yields 12 mol of gas; thus, Ssys > 0, Ssurr > 0, and Suniv > 0.
• If the entropy of the products is less than that of the reactants (Ssys < 0), the entropy of the surroundings must increase
even more (Ssurr >> 0) to make the total S positive (Figure 20.13B). For example, when calcium oxide and carbon dioxide
form calcium carbonate, the amount (mol) of gas decreases from 1 to 0 :
CaO(s) + C02 (g)  CaC03 (s) + heat
However, even though the system’s entropy goes down, the heat released increases the entropy of the surroundings even
more; thus, Ssys < 0, but Ssurr >> 0, so Suniv > 0
17
18
Chapter 20 • Thermodynamics: Entropy, Free Energy, and the Direction of Chemical Reactions
A
B
C
Figure 20.13 Components of ASuniv for spontaneous reactions. For a reaction to occur spontaneously, Suniu must be positive.
A, An exothermic reaction in which Ssys increases; the size of Ssurr is not important.
B, An exothermic reaction in which Ssys decreases; Ssurr must be larger than Ssys.
C, An endothermic reaction in which Ssys increases; Ssurr must be smaller than Ssys.
2. In an endothermic reaction (Hsys > 0), the heat absorbed by the system decreases molecular freedom of motion
and dispersal of energy in the surroundings; so Ssurr < 0. Thus, the only way an endothermic reaction can occur
spontaneously is if Ssys is positive and large enough to outweigh the negative Ssurr (Figure 20.13C).
• In the solution process for many ionic compounds, heat is absorbed to form the solution, so the entropy of the
surroundings decreases (Ssurr < 0). However, when the crystalline solid becomes freely moving ions, the entropy
increase is so large (Ssys » 0) that it outweighs the negative Ssurr. Thus, Suniv is positive.
• Spontaneous endothermic reactions are similar. Recall the reaction between barium hydroxide octahydrate and
ammonium nitrate (Figure 20.1, p. 879),
heat + Ba(OH)2.8H2O(s) + 2NH4NO3(s)  Ba2+ (aq) + 2𝑁𝑂3− (aq) + 2NH3(aq) + 10H2O (l)
3 mol of crystalline solids absorb heat from the surroundings (Ssurr < 0) and yields 15 mol of dissolved ions and
molecules, which have much more freedom of motion and, therefore, much greater entropy (Ssys >> 0).
 Summary of Section 20.2







𝑜 is calculated from S° values.
The standard entropy of reaction, 𝑆𝑟𝑥𝑛
𝑜 > 0.
When the amount (mol) of gas increases in a reaction, usually 𝑆𝑟𝑥𝑛
𝑜
Ssurr is related directly to -𝐻𝑠𝑦𝑠 and inversely to the T at which the change occurs.
In a spontaneous change, the entropy of the system can decrease only if the entropy of the surroundings increases even
more, so that Suniv > 0 .
The second law is obeyed in living systems when we consider system plus surroundings.
𝑜 = -S .
For a system at equilibrium, Suniv = 0, so 𝑆𝑠𝑦𝑠
surr
𝑜
𝑜 < 0) is spontaneous (S
Even if 𝑆𝑠𝑦𝑠 < 0, an exothermic reaction (𝐻𝑟𝑥𝑛
univ > 0) if Ssurr >> 0 ; an endothermic reaction
𝑜 > 0 ) is spontaneous only if 𝑆 𝑜 > S .
(𝐻𝑟𝑥𝑛
surr
𝑠𝑦𝑠
20.3 ENTROPY, FREE ENERGY, AND WORK
By measuring both Ssys and Ssurr, we can predict whether a reaction will be spontaneous at a particular temperature. It would
be useful to have one criterion for spontaneity that we can determine by measuring the system only. The Gibbs free energy,
or simply free energy (G), combines the system’s enthalpy and entropy:
G = H - TS
One of the greatest but least recognized American scientists, Josiah Willard Gibbs (1839-1903), established chemical
thermodynamics as well as major principles of equilibrium and electrochemistry. Although the great European scientists of his time,
James Clerk Maxwell and Henri Le Chatelier, realized Gibbs’s achievements, he was not recognized by his American colleagues until
nearly 50 years after his death!
Free Energy Change and Reaction Spontaneity
The free energy change (G) is a measure of the spontaneity of a process and of the useful energy available from it.
Deriving the Gibbs Equation Let’s examine the meaning of G by deriving it from the second law. By definition, the entropy
change of the universe is the sum of the entropy changes of the system and the surroundings:
20.3 • Entropy, Free Energy, and Work
Suniv = Ssys + Ssurr
At constant pressure,
𝐻
Ssurr = - 𝑇𝑠𝑦𝑠
Substituting for Ssurr gives a relationship that relies solely on the system:
𝐻
Suniv = Ssys - 𝑇𝑠𝑦𝑠
Multiplying both sides by -T and rearranging gives
-TSumv = Hsys — TSsys
Using the Gibbs free energy relationship, G = H — TS, we obtain the Gibbs equation for the change in the free energy of the
system (DGsys) at constant T and P:
Gsys = Hsys — TSsys
(20.6)
Combining Equation 20.6 with the one preceding it shows that
-TSuniv = Gsys = Hsys - TSsys
Spontaneity and the Sign of G Let’s see how the sign of G tells if a reaction is spontaneous. According to the second law,
• Suniv > 0 for a spontaneous process
• Suniv < 0 for a nonspontaneous process
• Suniv = 0 for a process at equilibrium
Since the absolute temperature is always positive, for a spontaneous process,
TSumv > 0 so -TSuniv < 0
From our derivation above, G = - TSuniv, so we have
• G < 0 for a spontaneous process
• G > 0 for a nonspontaneous process
• G =0 for a process at equilibrium
Without incorporating any new ideas, we can now predict spontaneity with one variable (Gsys) rather than two (Ssys and
Ssurr).
Calculating Standard Free Energy Changes
The sign of G reveals whether a reaction is spontaneous, but the magnitude of G tells how spontaneous it is. Because free
energy (G) combines three state functions, H, S, and T, it is also a state function. As we do with enthalpy, we focus on the free
energy change (G). As we do with other thermodynamic variables, to compare the free energy changes of different reactions,
we calculate the standard free energy change (G°), which occurs when all components of the system are in their standard
states.
Using the Gibbs Equation to Find AGo One way to calculate G0 is by writing the Gibbs equation (20.6) at standard-state
0 and ∆𝑆 0 . Adapting the Gibbs equation, we have
conditions and using Appendix B to find ∆𝐻𝑠𝑦𝑠
𝑠𝑦𝑠
0
0
0
∆𝐺𝑠𝑦𝑠 = ∆𝐻𝑠𝑦𝑠 -T∆𝑆𝑠𝑦𝑠
(20.7)
This important relationship is used to find any one of these three variables, given the other two, as in Sample Problem 20.4.
SAMPLE PROBLEM 20.4 Calculating ∆𝑮𝟎𝒔𝒚𝒔 from Enthalpy and Entropy Values
Problem Potassium chlorate, a common oxidizing agent in fireworks and matchheads, undergoes a solid-state
disproportionation reaction when heated:
+5
+7
-1
4KClO3 (s)  3KClO4 (s) + KCl(s)
Use ∆𝐻𝑓0 and S0 values to calculate 𝑮𝟎𝒔𝒚𝒔 (which is 𝑮𝟎𝒓𝒙𝒏 ) at 25oC for this reaction.
Plan To solve for G0, we need values from Appendix B. We use ∆𝐻𝑓0 values to calculate 𝑯𝟎𝒓𝒙𝒏 (𝑯𝟎𝒔𝒚𝒔 ), use S0 values to
calculate 𝑮𝑺𝟎𝒓𝒙𝒏 (𝑺𝟎𝒔𝒚𝒔 ), and then apply Equation 20.7.
Solution Calculating 𝑯𝟎𝒔𝒚𝒔 , from 𝑯𝟎𝒇 values (with Equation 6.9, p. 271):
𝑯𝟎𝒔𝒚𝒔 = 𝑯𝟎𝒓𝒙𝒏 = m𝑯𝟎𝒇(𝒑𝒓𝒐𝒅𝒖𝒄𝒕𝒔) - n𝑯𝟎𝒇(𝒓𝒆𝒂𝒄𝒕𝒂𝒏𝒕𝒔)
= [(3 mol KClO4 )( ∆𝐻𝑓0 of KClO4 ) + (1 mol KCl)( ∆𝐻𝑓0of KCl)] - [(4 mol KClO3 )( ∆𝐻𝑓0 of KClO3 )]
= [(3 mol)(-432.75 kJ/mol) + (1 mol)(-436.7 kJ/mol)] - [(4 mol)(-397.7 kJ/mol)]
= -144.2 kJ
Calculating 𝑺𝟎𝒔𝒚𝒔 from S° values (with Equation 20.4):
𝑺𝟎𝒔𝒚𝒔 = 𝑯𝟎𝒓𝒙𝒏 = m𝑺𝟎𝒑𝒓𝒐𝒅𝒖𝒄𝒕𝒔 - m𝑺𝟎𝒓𝒆𝒂𝒄𝒕𝒂𝒏𝒕𝒔
= [(3 mol KClO4 )(S° of KClO4 ) + (1 mol KCl)(S° of KCl)] - [(4 mol KClO3 )(S° of KClO3 )]
= [(3 mol)(151.0 J/mol-K) + (1 mol)(82.6 J/mol-K)] - [(4 mol)(143.1 J/mol-K)]
19
20
Chapter 20 • Thermodynamics: Entropy, Free Energy, and the Direction of Chemical Reactions
= 36.8 J/K
Calculating G°sys at 298 K:
0
0
0
𝐺𝑠𝑦𝑠
= 𝐻𝑠𝑦𝑠
- T𝑆𝑠𝑦𝑠
= -144.2 kJ = [(298 K)(236.8 J/K)(
1 𝑘𝐽
1000𝐽
)] = −140 𝑘𝐽
Check Rounding to check the math:
H0  [3(-433 kJ) +(-440 kJ)] - [4(-400 kJ)] = -1740 kJ + 1600 kJ = -140 kJ
S°  [3(150 J/K) + 85 J/K] - [4(145 J/K)] = 535 J/K - 580 J/K = -45 J/K
G°  -140 kJ - 300 K(-0.04 kJ/K)
= -140 kJ + 12 kJ
= -128 kJ
Comment
1. Recall from Section 20.1 that reaction spontaneity tells nothing about rate. Even though this reaction is spontaneous, the rate
is very low in the solid. When KClO3 is heated slightly above its melting point, the ions can move and the reaction occurs readily.
2.Under any conditions, a spontaneous reaction has a negative change in free energy: G < 0. Under standard-state conditions, a
spontaneous reaction has a negative standard free energy change: G ° < 0.
FOLLOW-UP PROBLEMS
0
0
0
20.4A Use ∆𝐻𝑟𝑥𝑛
and ∆𝑆𝑟𝑥𝑛
to calculate ∆𝐺𝑟𝑥𝑛
at 298 K for this reaction:
2NO(g) + Cl2 (g)  2NOCl(g)
20.4B Determine the standard free energy change at 298 K for this reaction:
2NO(g) + O2 (g)  2NO2 (g)
SOME SIMILAR PROBLEMS 20.53 and 20.54
0 Another way to calculate ∆𝐺 0
Using Standard Free Energies of Formation to Find ∆𝐺𝑟𝑥𝑛
𝑟𝑥𝑛 is with values for the
standard free energy of formation (∆𝐺𝑓0 ) of the components. Analogously to the standard enthalpy of formation, ∆𝐻𝑓0 (Section
6 .6 ), ∆𝐺𝑓0 is the free energy change that occurs when 1 mol of compound is made from its elements, with all components in
their standard states. Because free energy is a state function, we can apply Hess’s law and combine ∆𝐺𝑓0 values of reactants
0 , no matter how the reaction takes place:
and products to calculate ∆𝐺𝑟𝑥𝑛
0
0 = 𝑚∆𝐺 0
∆𝐺𝑟𝑥𝑛
𝑓(𝑝𝑟𝑜𝑑𝑢𝑐𝑡𝑠) - 𝑛∆𝐺𝑓(𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑡𝑠)
(20.8)
∆𝐺𝑓0 Values have properties similar to ∆𝐻𝑓0 values:
• ∆𝐺𝑓0 of an element in its standard state is zero.
• An equation coefficient (m or n above) multiplies ∆𝐺𝑓0 by that number.
• Reversing a reaction changes the sign of ∆𝐺𝑓0 .
Many ∆𝐺𝑓0 values appear along with those for ∆𝐻𝑓0 and S° in Appendix B.
SAMPLE PROBLEM 20.5 Calculating ∆𝑮𝟎𝒓𝒙𝒏 from ∆𝑮𝟎𝒇 Values
0 for the reaction in Sample Problem 20.4:
Problem Use ∆𝐺𝑓0 values to calculate ∆𝐺𝑟𝑥𝑛
4KClO3 (s)  3KClO4 (s) + KCl(s)
0
Plan We apply Equation 20.8 to calculate ∆𝐺𝑟𝑥𝑛
Solution Applying Equation 20.8 with values from Appendix B:
0
0 = 𝑚∆𝐺 0
∆𝐺𝑟𝑥𝑛
𝑓(𝑝𝑟𝑜𝑑𝑢𝑐𝑡𝑠) - 𝑛∆𝐺𝑓(𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑡𝑠)
= [(3 mol KClO4 )( ∆𝐺𝑓0 of KClO4) + (1 mol KCl)( ∆𝐺𝑓0 of KCl)]- [(4 mol KClO3 )( ∆𝐺𝑓0 of KClO3)]
= [(3 mol)(-303.2 kJ/mol) + (1 mol)(-409.2 kJ/mol)] - [(4 mol)(-296.3 kJ/mol)]
= -134 kJ
Check Rounding to check the math:
0
∆𝐺𝑟𝑥𝑛
 [3(-300 kJ) + 1(-400 kJ)] - 4(-300 kJ)
= -1300 kJ + 1200 kJ = -100 kJ
Comment The slight discrepancy between this answer and the one obtained in Sample Problem 20.4 is due to rounding. As you
can see, when ∆𝐺𝑓0 values are available, this method is simpler arithmetically than the approach in Sample Problem 20.4.
FOLLOW-UP PROBLEMS
0
20.5A Use ∆𝐺𝑓0 values to calculate ∆𝐺𝑓𝑟𝑥𝑛
at 298 K:
(a) 2NO(g) + Cl2 (g)  2NOCl(g) (from Follow-up Problem 20.4A)
(b) 3H2 (g) + Fe2O3(s)  2Fe(s) + 3H2 O(g)
0 at 25fC:
20.5B Use ∆𝐺𝑓0 values to calculate ∆𝐺𝑟𝑥𝑛
(a) 2NO(g) + O2 (g)  2NO2 (g) (from Follow-up Problem 20.4B)
(b) 2C(graphite) + O2 (g)  2CO(g)
SOME SIMILAR PROBLEMS 20.51, 20.52, 20.55(b), and 20.56(b)
20.3 • Entropy, Free Energy, and Work
The Free Energy Change and the Work a System Can Do
Thermodynamics developed after the invention of the steam engine, a major advance that spawned a new generation of
machines. Thus, some of the field’s key ideas applied the relationships between the free energy change and the work a system
can do:
• G is the maximum useful work that can be done by a system during a spontaneous process at constant T and P:
G = Wmax
(20.9)
• G is the minimum work that must be done to a system to make a nonspontaneous process occur at constant T and P.
The free energy change is the maximum work the system can possibly do. But the work it actually does is always less and
depends on how the free energy is released. Let’s consider the work done by two common systems—a car engine and a battery.
1. “Useful” work done by a car engine. When gasoline (represented by octane, C8 H18) is burned in a car engine,
C8H18(l) +
25
2
O2(g)  8CO2(g) + 9H2O(g)
a large amount of energy is released as heat (Hsys < 0), and because the number of moles of gas increases, the entropy of the
system increases (Ssys > 0). Therefore, the reaction is spontaneous (Gsys < 0). The free energy released turns the wheels, moves
the belts, plays the radio, and so on—all examples of “useful” work. However, the maximum work is done by any spontaneous
process only if the free energy is released reversibly, that is, in an infinite number of steps. Of course, in any real process,
work is performed in a finite number of steps, that is, irreversibly, so the maximum work is never done. Any free energy not
used for work is lost to the surroundings as heat. For a car engine, much of the free energy released just warms the engine and
the outside air, which increases the freedom of motion of the particles in the universe, in accord with the second law.
2. “Useful” work done by a battery. As you’ll see in Chapter 21, a battery is essentially a packaged spontaneous redox
reaction that releases free energy in the form of an electric current to the surroundings (flashlight, computer, motor, etc.). If we
connect the battery terminals to each other through a short piece of wire, the free energy change is released all at once but does
no work—it just heats the wire and battery. If we connect the terminals to a motor, a significant portion of the free energy runs
the motor, but some is still converted to heat. If we connect the battery to a more “efficient” device, one that discharges the free
energy still more slowly, more of the energy does work and less is converted to heat. However, as with all systems, only when
the battery discharges infinitely slowly can it do the maximum work.
Efficiency can be defined as the percentage of work output relative to the energy input. The range of efficiencies among
common devices is very large: an incandescent bulb converts <7% of incoming electricity to light, the rest being given off as heat.
At the other extreme, an electrical generator converts 95% of the incoming mechanical energy to electricity. Here are the
efficiencies of some other devices: home oil furnace, 65%; hand-tool motor, 63%; liquid fuel rocket, 50%; car engine, <30%;
compact fluorescent bulb, 18%; solar cell, ~15%. Therefore, all engineers must face the fact that no real process uses all the
available free energy to do work because some is always “wasted” as heat.
Let’s summarize the relation between the free energy change of a reaction and the work it can do:
• A spontaneous reaction (Gsys < 0) will do work on the surroundings (-w). For any real machine, the actual work done is
Always less than the maximum because some of the G is released as heat.
• A nonspontaneous reaction (Gsys > 0) will occur only if the surroundings do work on the system (+w). For any real machine,
the actual work done on the system is always more than the minimum because some of the added free energy is wasted as
heat.
• A reaction at equilibrium (Gsys = 0) can no longer do any work.
The Effect of Temperature on Reaction Spontaneity
In most cases, the enthalpy contribution (H) to the free energy change (G) is much larger than the entropy contribution (TS).
In fact, the reason most exothermic reactions are spontaneous is that the large negative H makes G negative. However, the
temperature of a reaction influences the magnitude of the T S term, so, for many reactions, the overall spontaneity depends
on the temperature. From the signs of H and S, we can predict how the temperature affects the sign of G. (The values we’ll
use below for the thermodynamic variables are standard-state values from Appendix B, but we show them without the degree
sign to emphasize that the relationships among G, H, and S are valid at any conditions. Also, we assume that H and S
change little with temperature, which is true as long as no phase change occurs.)
Let’s examine the four combinations of positive and negative H and S—two that are independent of temperature and
two that are dependent on temperature:
• Temperature-independent cases. When H and S have opposite signs, the reaction occurs spontaneously either at all
temperatures or at none (nonspontaneous).
1. Reaction is spontaneous at all temperatures: H < 0, S > 0. Since S is positive, -TS is negative; thus, both contributions
favor a negative G. Most combustion reactions are in this category. The decomposition of hydrogen peroxide, a common
disinfectant, is also spontaneous at all temperatures:
2H2O2(l)  2H2O(l) + O2(g)
H = -196 kJ and S = 125 J/K
21
22
Chapter 20 • Thermodynamics: Entropy, Free Energy, and the Direction of Chemical Reactions
2.Reaction is nonspontaneous at all temperatures: H > 0, S < 0. Both contributions oppose spontaneity: H is positive and
S is negative, so —TS is positive; thus, G is always positive. The formation of ozone from oxygen requires a continual
energy input, so it is not spontaneous at any temperature:
3O2(g)  2O3(g)
 H = 286 kJ and S = -137 J/K
• Temperature-dependent cases. When H and S have the same sign, the relative magnitudes of -TS and H determine
the sign of G. In these cases, the direction of the change in T is crucial.
3. Reaction becomes spontaneous as temperature increases: H > 0 and S > 0. With a positive H, the reaction will occur
spontaneously only when -TS becomes large enough to make DG negative, which will happen as the temperature rises. For
example,
2N2O(g) + O2 (g)  4NO(g)
H = 197.1 kJ and S = 198.2 J/K
The oxidation of N2O occurs spontaneously at any T > 994 K.
4. Reaction becomes spontaneous as temperature decreases: H < 0 and S < 0. Here, H favors spontaneity, but S does
not (-TS > 0). The reaction will occur spontaneously only when -TS becomes smaller than H, and this happens as the
temperature drops. For example,
4Fe(s) + 3O2(g)  2Fe2 O3 (s)
H = -1651 kJ and S = -549.4 J/K
The production of iron (III) oxide occurs spontaneously at any T < 3005 K.
Table 20.1 summarizes these four possible combinations of H and S, and Sample
Problem 20.6 applies them.
Table 20 Reaction Spontaneity and the Signs of H, S, and G
Description
H
S
-TS
G
+
Spontaneous at all T
+
+
+
Nonspontaneous at all T
+
+
+ or Spontaneous at higher T; nonspontaneous at lower T
+
+ or Spontaneous at lower T; nonspontaneous at higher T
SAMPLE PROBLEM 20.6 Using Molecular Scenes to Determine the Signs of H, S and G
Problem The scenes below represent a familiar phase change for water (blue spheres):
(a) What are the signs of H and S for this process? Explain.
(b) Is the process spontaneous at all T, no T, low T, or high T? Explain.
Plan (a) From the scenes, we determine any change in amount of gas and/or any change in the freedom of motion of the
particles, which will indicate the sign of S.
Also, since the scenes represent a physical change, freedom of particle motion indicates whether heat is absorbed or released,
and thus tells us the sign of H. (b) The question refers to the sign of G ( + or -) at the different temperature possibilities, so we
apply Equation 20.6 and refer to the previous text discussion and Table 20.1.
Solution
(a) The scenes represent the condensation of water vapor, so the amount of gas decreases dramatically, and the separated
molecules give up energy as they come closer together. Therefore,S< 0 and H < 0.
(b) With S negative, the -TS term is positive. In order for G < 0, the magnitude of T must be small. Therefore, the process is
spontaneous at low T.
23
Chapter 20 • Thermodynamics: Entropy, Free Energy, and the Direction of Chemical Reactions
Check The answer in part (b) seems reasonable based on our analysis in part (a). The answer makes sense because we know from
everyday experience that water condenses spontaneously, and it does so at low temperatures.
FOLLOW-UP PROBLEMS
20.6AThe scenes below represent the reaction of X2 Y2 to yield X2 (red) and Y2 (blue):
(a)
(b)
What is the sign of S for the reaction?
If the reaction is spontaneous only above 325°C, what is the sign of DH? Explain.
20.6B The scenes below represent the reaction of AB to yield A (gray) and B2 (red):
What are the signs of S and H for the reaction?
Is G < 0 at higher T, lower T, all T, or no T? Explain.
SOME SIMILAR PROBLEMS 20.47 and 20.48
(a)
(b)
As you saw in Sample Problem 20.4, one way to calculate G is from enthalpy and entropy changes. As long as phase changes
don’t occur, H and S usually change little with temperature, so we use their values at 298 K in Sample Problem 20.7 to examine
the effect of T on G and, thus, on reaction spontaneity.
SAMPLE PROBLEM 20.7 Determining the Effect of Temperature on G
Problem A key step in the production of sulfuric acid is the oxidation of SO2 (g) to SO3(g):
2 SO2 (g) + O2 (g) 2 SO3 (g)
At 298 K, G = -141.6 kJ; H = -198.4 kJ; and S = -187.9 J/K.
(a) Use the data to decide if this reaction is spontaneous at 25°C, and predict how G will change with increasing T.
(b) Assuming that H and S are constant with T (no phase change occurs), is the reaction spontaneous at 900°C?
Plan (a) We note the sign of G to see if the reaction is spontaneous and the signs of H and S to see the effect of T.
(b) We use Equation 20.7 to calculate G from the given H and S at the higher T (in K).
Solution
(a) G < 0, so the reaction is spontaneous at 298 K: SO2 and O2 will form SO3 spontaneously. With S < 0, the term -TS > 0, and
this term will become more positive at higher T. Therefore, G will become less negative, and the reaction less spontaneous, with
increasing T.
(b) Calculating G at 900°C (T = 273 + 900 = 1173 K):
G = H -TS = -198.4 kJ - [(1173 K)(-187.9 J/K)(1 kJ/1000 J)] = 22.0 kJ G > 0, so the reaction is nonspontaneous at the higher T.
Check The answer in part (b) seems reasonable based on our prediction in part (a). The arithmetic seems correct, given
considerable rounding:
G  -200 kJ -[(1200 K)(-200 J/K)(1 kJ/1000 J)] = +40 kJ
FOLLOW-UP PROBLEMS
20.7A For the following reaction at 298 K, S = -308.2 J/K and H = -192.7 kJ:
4NO(g) N2 O(g) + N2O3A(g)
(a) Is the reaction spontaneous at 298 K?
(b) Would the reaction become more or less spontaneous at higher T?
(c) Assuming that S and H don’t change with T, find G at 500°C.
20.7B A reaction is nonspontaneous at room temperature but is spontaneous at -40°C. What can you say about the signs and
relative magnitudes of H, S, and -TS?
SOME SIMILAR PROBLEMS 20.57(b) and 20.58(b)
The Temperature at Which a Reaction Becomes Spontaneous As you’ve just seen, when the signs are the same for H and S
of a reaction, it can be nonspontaneous at one temperature and spontaneous at another. The “crossover” temperature occurs
when a positive G switches to a negative G because of the magnitude of the -TS term. We find this temperature by setting G
equal to zero and solving for T:
Therefore,
G = H- TS = 0
H = TS and T =
∆𝐻
𝑆
(20.10)
Consider the reaction of Copper (I) oxide with carbon. It does not occur at very low temperatures but does at higher temperatures
and is used to extract copper from one of its ores:
Cu2 O(s) + C(s) + Heat  2Cu(s) + CO(g)
We predict that this reaction has a positive S because the number of moles of gas increases; in fact, S = 165 J/K. Furthermore,
because the reaction is nonspontaneous at lower temperatures, it must have a positive H; the actual value is 58.1 kJ. As the - TS
term becomes more negative with higher T, it eventually outweighs the positive H term, so G becomes negative and the
reaction occurs spontaneously.
SAMPLE PROBLEM 20.8 Finding the Temperature at Which a Reaction Becomes Spontaneous
Problem At 25°C (298 K), the reduction of copper(I) oxide to copper is nonspontaneous (G = 8.9 kJ). Calculate the temperature
at which the reaction becomes spontaneous.
Plan As just discussed, we want the temperature at which G crosses over from a positive to a negative value. We set G equal
to zero, and use Equation 20.10 to solve for T, using the H (58.1 kJ) and S (165 J/K) values from the text.
Solution From G = H — TS = 0, we have
1000𝐽
∆𝐻 58.1 𝑘𝐽 × 1 𝑘𝐽
T=
=
𝑆
165 𝐽/𝐾
= 352 K
Thus, at any temperature above 352 K (79°C), which is a moderate temperature for extracting a metal from its ore, G < 0, so the
reaction becomes spontaneous.
T=
60,000𝐽
= 400 K
150 𝐽/𝐾
which is close to the answer.
FOLLOW-UP PROBLEMS
20.8A Find the temperature (in °C) above which the reaction in Follow-up Problem 20.7A is no longer spontaneous.
20.8B Use Appendix B values to find the temperature at which the following reaction becomes spontaneous (assume H and S
are constant with T):
CaO(s) + CO2 (g)  CaCO3 (s)
SOME SIMILAR PROBLEMS 20.59(d) and 20.60(d)
24
20.3 • Entropy, Free Energy, and Work
Figure 20.14 shows that the line for TS rises steadily (and thus the -TS term becomes more negative) with increasing
temperature. This line crosses the relatively constant H line at 352 K, as we found in Sample Problem 20.8. At any higher
temperature, the -TDS term is greater than the H term, so G is negative.
Figure 20.14 The effect of temperature on reaction spontaneity. At low T,G > 0 because H dominates. At 352 K, H = TS, so G = 0.
At any higher T, G < 0 because -TS dominates.
Coupling of Reactions to Drive a Nonspontaneous Change
In a complex, multistep reaction, we often see a nonspontaneous step driven by a spontaneous step. In such a coupling of
reactions, one step supplies enough free energy for the other to occur, just as burning gasoline supplies enough free energy to
move a car.
Look again at the reduction of copper(I) oxide by carbon. In Sample Problem 20.8, we found that the overall reaction becomes
spontaneous at any temperature above 352 K. Dividing the reaction into two steps, however, we find that even at a higher
temperature, say 375 K, copper(I) oxide does not spontaneously decompose to its elements:
1
Cu2 O(s)  2Cu(s) + 2O2 (g)
G375= 140.0 kJ
However, the oxidation of carbon to CO at 375 K is quite spontaneous:
1
C(s) + O2(g)  CO(g)
G375= -143.8 kJ
2
Coupling these reactions means having the carbon in contact with the Cu2 O, which allows the reaction with the larger negative G
1
to “drive” the one with the smaller positive G. Adding the reactions together and canceling the common substance ( O2) gives an
overall reaction with a negative G:
2
Cu2 O(s) + C(s)  2Cu(s) + CO(g) G375= -3.8 kJ
Many biochemical reactions are also nonspontaneous, including key steps in the syntheses of proteins and nucleic acids,
the formation of fatty acids, the maintenance of ion balance, and the breakdown of nutrients. Driving a nonspontaneous step by
coupling it to a spontaneous one is a life-sustaining strategy common to all organisms, as you’ll see in the upcoming Chemical
Connections essay.

Summary of Section 20.3
 The sign of the free energy change, G = H — TS, is directly related to reaction spontaneity: a negative G
corresponds to a positive Suniv.
 We use the standard free energy of formation (DG°) to calculate DGrxn at 298 K.
 The maximum work a system can do is never obtained from a real (irreversible) process because some free energy is
always converted to heat.
 The magnitude of T influences the spontaneity of a temperature-dependent reaction (same signs of H and S) by
affecting the size of TS. For such a reaction, the T at which the reaction becomes spontaneous can be found by setting
G = 0.
 A nonspontaneous reaction (G > 0) can be coupled to a more spontaneous one (G << 0) to make it occur. For example,
in organisms, the hydrolysis of ATP drives many reactions that have a positive G.
25
CHEMICAL CONNECTIONS TO BIOLOGICAL ENERGET
The Universal Role of ATP
Despite their incredible diversity, virtually all organisms use the same amino acids to make their proteins, the same
nucleotides to make their nucleic acids, and the same carbohydrate (glucose) to provide energy.
In addition, all organisms use the same spontaneous reaction to drive a variety of nonspontaneous ones.
This reaction is the hydrolysis of adenosine triphosphate (ATP) to adenosine diphosphate (ADP):*
ATP4- + H2O = ADP3- + HPO42- + H+ (Reversible Reaction) G°' = -30.5 kJ
In the metabolic breakdown of glucose, for example, the first step, addition of HPO 42- to glucose, is nonspontaneous:
Glucose + HPO42- + H+ = [glucose phosphate]- + H2O
AG°' = 13.8 kJ
Coupling cannot occur if reactions are physically separated, so these reactions take place on an enzyme (Section 16.7)
that simultaneously binds glucose and ATP, and the phosphate group of ATP that will be transferred lies next to the –OH
group of glucose that will bind it (Figure B20.1).
Figure B20.1 The coupling of a nonspontaneous reaction to the hydrolysis of ATP. Glucose lies next to ATP
(shown as ADP-O-PO3H) in the enzyme’s active site. ADP (shown as ADP-OH) and glucose phosphate are released.
The ADP produced in energy-releasing reactions combines with phosphate to regenerate ATP in energy-absorbing
reactions catalyzed by other enzymes. Thus, there is a continuous cycling of ATP to ADP and back to ATP again to
supply energy to the cells (Figure B20.2).
Figure B20.2 The cycling of metabolic free energy.
Why is ATP a “high-energy” molecule? By examining the phosphate portions of ATP, ADP, and HPO 42-, we can see two basic
chemical reasons why ATP hydrolysis supplies so much free energy (Figure B20.3):
26
20.3 • Entropy, Free Energy, and Work
1. Charge repulsion. At physiological pH (~7), the triphosphate group of ATP has four negative charges close together.
This high charge repulsion is reduced in ADP (Figure B20.3, part A).
2. Electron delocalization. Once HPO42- is free, there is extensive delocalization and resonance stabilization of the 𝜋 electrons
(Figure B20.3, part B).
Figure B20.3 ATP is a high-energy molecule.
Thus, greater charge repulsion and less electron delocalization make ATP higher in energy (less stable) than the sum of the energies of
ADP and HPO42-. When ATP is hydrolyzed, some of this additional energy is released and harnessed by the organism to drive metabolic
reactions that could not otherwise take place.
Problems
B20.1 The oxidation of 1 mol of glucose supplies enough metabolic energy to form 36 mol of ATP. Oxidation of 1 mol of a typical dietary
fat like tristearin (C57H116O6) yields enough energy to form 458 mol of ATP. How many molecules of ATP can form per gram of (a) glucose;
(b) tristearin?
B20.2 Nonspontaneous processes like muscle contraction, protein synthesis, and nerve conduction are coupled to the spontaneous
hydrolysis of ATP to ADP. ATP is then regenerated by coupling its synthesis to energy-yielding reactions such as
Creatine phosphate  creatine + phosphate
G °' = -43.1 kJ/mol
Find G°' for the overall reaction that regenerates ATP from ADP and HPO42- by coupling with the breakdown of creatine phosphate.
27
28
Chapter 20 • Thermodynamics: Entropy, Free Energy, and the Direction of Chemical Reactions
20.4 FREE ENERGY, EQUILIBRIUM, AND REACTION DIRECTION
A chemical reaction proceeding toward equilibrium is a spontaneous change. But why is there a drive to attain equilibrium?
And what determines the value of the equilibrium constant?
As you know from earlier discussions, the sign of G is not the only way to predict reaction direction. In Chapter 17, we did so
by comparing the values of the reaction quotient (Q) and the equilibrium constant (K). Recall that
• If Q < K (Q/K < 1): reaction proceeds spontaneously to the right.
• If Q > K (Q/K > 1): reaction proceeds spontaneously to the left.
• If Q = K (Q/K = 1): reaction has attained equilibrium and no longer proceeds spontaneously in either direction.
It is easier to see the relation between these two ways to predict reaction spontaneity the sign of G and the magnitude of Q/K
when we compare the sign of the natural logarithm of Q/K (ln Q/K) with the sign of G (refer to Appendix A if necessary):
• If Q/K < 1, then ln Q/K < 0: reaction proceeds spontaneously to the right (G < 0).
• If Q/K > 1, then ln Q/K > 0: reaction proceeds spontaneously to the left (G > 0).
• If Q/K = 1, then ln Q/K = 0: reaction is at equilibrium (G = 0).
Note that the signs of G and of ln Q/K are the same for a given direction; in fact, G equals ln Q/K multiplied by the proportionality
constant RT:
𝑄
G = RT ln 𝐾 = RT ln Q - RT ln Q
(20.11)
Q represents the concentrations (or pressures) of a system’s components at any time during the reaction, and K represents these
quantities at equilibrium. Therefore,
Equation 20.11 says that G is a measure of how different the concentrations at any time, Q, are from the concentrations at equilibrium, K:
• If Q and K are very different, the reaction releases (or absorbs) a lot of free energy.
• If Q and K are nearly the same, the reaction releases (or absorbs) relatively little.
The Standard Free Energy Change and the Equilibrium Constant When we choose standard-state values for Q (1 atm for
gases, 1 M for solutions, etc.) in Equation 20.11, G becomes, by definition, G° and Q equals 1:
G° = RT ln 1 - RT ln K
0 ) from its
Since ln 1 = 0, the term “RT ln Q” drops out, which allows us to find the standard free energy change of a reaction (𝐺𝑟𝑥𝑛
equilibrium constant, or vice versa:
G° = -RT ln K
(20.12)
The following sample problem applies this central relationship.
SAMPLE PROBLEM 20.9 Exploring the Relationship between AG° and K
Problem
(a) Use Appendix B to find K at 298 K for the following reaction:
1
NO(g) + O2 (g) = NO2(g)
2
(b) Use the equilibrium constant to calculate G° at 298 K for the following reaction:
2HCl(g) = H2(g) + Cl2(g)
K = 3.89X10-34 at 298 K
Plan
(a) We use 𝐺𝑓0 values from Appendix B with Equation 20.8 to find G°, and then we use Equation 20.12 to solve for K.
(b) In this calculation, we do the reverse of the second calculation in part (a): we are given that K = 3.89X10-34, so we use Equation 20.12 to
solve for G °.
Solution
(a) Solving for G° using 𝐺𝑓0 values from Appendix B and Equation 20.8 (see Sample Problem 20.5):
1
G ° = [(1 mol NO2)( 𝐺𝑓0 of NO2)] - [(1 mol NO)( 𝐺𝑓0 of NO) +( mol of O2)( 𝐺𝑓0 of O2)]
1
2
= [(1 mol)(51 kJ/mol)] - [(1 mol)(86.60 kJ/mol) + ( mol)(0 kJ/mol)]
2
= -36 kJ
Solving for K using Equation 20.12:
G° = - RT ln K
20.4 • Free Energy, Equilibrium, and Reaction Direction
So
ln K = -
𝐺 0
𝑅𝑇
=-
(−36 𝑘𝐽)(
(8.314
1000𝐽
)
1𝑘𝐽
𝐽
)(298𝐾)
𝑚𝑜𝑙.𝐾
= 14.53
K = e14.53 = 2.0 X106
(b) Solving for G° using Equation 20.12:
K = 3.89X10-34 so ln K = -76.9
G° = -RT ln K = -(8.314 J/mol.K)(298 K)(-76.9)
= 1.91X105 J = 191 kJ
Check
(a) A negative free energy change is consistent with a positive K. Rounding to confirm the value of G° gives 50 kJ - 90 kJ = -40 kJ,
close to the calculated value.
(b) A very negative K is consistent with a highly positive G°. Looking up 𝐺𝑓0 values to check the value of G° gives
(0 kJ + 0 kJ) - [2(-95.30 kJ)] = -190.6 kJ.
FOLLOW-UP PROBLEMS
20.9A Use Appendix B to find K at 298 K for the following reaction:
2C(graphite) + 02(g) = 2CO(g)
20.9B Use the given value of K to calculate AG° at 298 K for the following reaction:
2HCl(g) + Br2(l) = 2HBr(g) + Cl2(g)
SOME SIMILAR PROBLEMS 20.67-20.70
K = 2.22X10-15 at 298 K
29
30
Chapter 20 • Thermodynamics: Entropy, Free Energy, and the Direction of Chemical Reactions
Table 20.2 The Relationship Between AG° and K at 298 K
G° (kJ)
200
K
9x10-36
Essentially no forward reaction;
100
3x10-18
reverse reaction goes to completion.
50
2x10-9
10
2x10-2
1
7x10-1
0
1
-1
1.5
-10
5x101
-50
6x108
-100
3x1017
Forward reaction goes to completion
-200
1x1035
Essentially no reverse reaction.
Significance
REWARD REACTION
to sextent.
FORWARD REACTION
Forward and reverse reaction proceed
Table 20.2 shows that, due to their logarithmic relationship, a small change in G° causes a large change in K. Note just these two examples:
1
• As G° becomes more positive, K becomes smaller: if G° = +10 kJ, K is 0.02, so the product terms are the size of the reactant terms
50
(see Figure 17.2A).
• As G becomes more negative, K becomes larger: if G° = -10 kJ, the product terms are 50 times larger than the reactant terms
(see Figure 17.2B).
Finding the Free Energy Change Under Any Conditions In reality, reactions rarely begin with all components in their standard
states. By substituting the relationship between G° and K (Equation 20.12) into the expression for G (Equation 20.11), we obtain a
relationship that applies to any starting concentrations:
G = G ° + RT ln Q
(20.13)
Sample Problem 20.10 uses molecular scenes to explore these ideas, and Sample Problem 20.11 applies them to an important industrial
reaction.
SAMPLE PROBLEM 20.10 Using Molecular Scenes to Find AG for a Reaction at Nonstandard Conditions
Problem These molecular scenes represent three mixtures in which A2 (black) and B2 (green) are forming AB. Each molecule represents 0.10
atm. The equation is
A2(g) + B2(g) = 2AB(g)
G° = -3.4 kJ/mol
(a) If mixture 1 is at equilibrium, calculate K.
(b) Which mixture has the most negative DG, and which has the most positive?
Plan (a) Mixture 1 is at equilibrium, so we first write the expression for Q and then find the partial pressure of each substance from the
numbers of molecules and calculate K.
(b) To find G, we apply Equation 20.13. We are given G° (-3.4 kJ/mol) and know R (8.314 J/mol-K), but we still need to find T. We calculate
T from Equation 20.12 using K from part (a), and substitute the partial pressure of each substance (by counting particles) to get Q.
Solution (a) Writing the expression for Q and calculating K:
A2(g) + B2(g) = 2AB(g)
Q=
2
𝑃𝐴𝐵
𝑃𝐴2 ×𝑃𝐵2
K=
0.402
0.2 ×0.2
= 4.0
20.4 • Free Energy, Equilibrium, and Reaction Direction
(b) Calculating T from Equation 20.12 for use in Equation 20.13:
G ° = - RT ln K = -
T
∆𝐺 0
= −𝑅 ln 𝐾
3.4 𝑘𝐽
𝑚𝑜𝑙
=
= -(
8.314 𝐽
𝑚𝑜𝑙.𝐾
) T ln 4.0
−3.4𝑘𝐽 1000𝐽
(
)
𝑚𝑜𝑙
1 𝑘𝐽
8.314𝐽
−(
) 𝑙𝑛 4.0
𝑚𝑜𝑙.𝐾
= 295 K
Calculating G from Equation 20.13 for each reaction mixture:
Mixture 1:
G = G° + RT ln Q = -3.4 kJ + RT ln 4.0
= -3.4 kJ (
1000𝐽
1𝑘𝐽
)+ (
8.314𝐽
𝑚𝑜𝑙.𝐾
) (295𝐾)ln 4.0
= -3400 J + 3400 J = 0.0 J
Mixture 2:
(0.2)2
G = -3.4 kJ + RT ln (0.3)(0.3)
= -3.4 kJ (
1000𝐽
1𝑘𝐽
)+ (
8.314𝐽
𝑚𝑜𝑙.𝐾
) (295𝐾)ln 0.44
= -5.4 x 10 3 J
Mixture 3:
(0.6)2
G = -3.4 kJ + RT ln (0.1)(0.1)
= -3.4 kJ (
1000𝐽
1𝑘𝐽
)+ (
8.314𝐽
𝑚𝑜𝑙.𝐾
) (295𝐾)ln 36
= 5.4 x 10 3 J
Mixture 2 has the most negative G, and mixture 3 has the most positive G.
Check In (b), round to check the arithmetic; for example, for mixture 3,
DG  -3000 J + (8 J/mol.K)(300 K)4  7000 J, which is in the correct ballpark.
Comment
1. By using the properties of logarithms, we did not have to calculate T and G in (b). For mixture 2, Q < 1, so ln Q is negative,
which makes G more negative. Also, note that Q(0.44) < K(4.0), so G < 0. For mixture 3, Q > 1 (and is greater than it is for
mixture 1), so ln Q is positive, which makes G positive. Also, Q(36) > K(4.0), so G > 0.
2.In (b), the value of zero for DG of the equilibrium mixture (mixture 1) makes sense, because a system at equilibrium has
released all of its free energy.
3. Note especially that, by definition, when the components are in their standard states, G = G°:
(1.0)2
G = G° + RT ln Q = -3.4 kJ/mol + RT ln (1.0)(1.0)
= -3.4 kJ/mol + RT ln 1.0
= -3.4 kJ/mol
FOLLOW-UP PROBLEMS
20.10A The scenes below depict three mixtures in which A (orange) and B (green) are forming AB3 in a reaction for which
G° = - 4.6 kJ/mol. Assume that each molecule represents 0.10 mol and the volume is 1.0 L
31
32
Chapter 20 • Thermodynamics: Entropy, Free Energy, and the Direction of Chemical Reactions
(a) Write a balanced equation for the reaction.
(b) If K = 8.0, which mixture is at equilibrium?
(c) Rank the three mixtures from the highest (most positive) G to the lowest (most negative) G.
20.10B The scenes below depict mixtures in which X2 (tan) and Y2 (blue) are forming XY2. Each molecule represents 0.10 mol, and
the volume is 0.10 L. The equation is X2(g) + 2Y2(g) = 2XY2(g); G° = -1.3 kJ/mol.
(a) If K = 2.0, which mixture is at equilibrium?
(b) Rank the three mixtures from the lowest (most negative) G to highest (most positive) G.
(c) What is the sign of G for the change that occurs as each nonequilibrium mixture approaches equilibrium?
SOME SIMILAR PROBLEMS 20.81 and 20.82
SAMPLE PROBLEM 20.11 Calculating G at Nonstandard Conditions
Problem The oxidation of SO2(g), 2SO2(g) + O2(g)  2SO3(g), is the key reaction in the manufacture of sulfuric acid.
0
(a) Calculate K at 298 K. (𝐺298
= -141.6 kJ/mol of reaction as written.)
(b) A container is filled with 0.500 atm of SO2, 0.0100 atm of O2, and 0.100 atm of SO3 and kept at 25°C. In which direction, if any,
will the reaction proceed to reach equilibrium?
(c)
Calculate G for the system in part (b).
Plan
(a) We know G°, T, and R, so we can calculate K from Equation 20.12 [see Sample Problem 20.9(a)].
(b) To determine if a net reaction will occur, we find Q from the given partial pressures and compare it with K from part (a).
(c) These are not standard-state pressures, so we find G at 298 K with Equation 20.13 from the values of G° (given) and Q
[from part (b)].
Solution (a) Calculating K at 298 K:
G ° = -RT ln K so K = 𝑒
−∆𝐺0
𝑅𝑇
At 298 K, the exponent is
𝑘𝐽
1000𝐽
×
𝑚𝑜𝑙
1 𝑘𝐽
𝐽
8.314
×298 𝐾
𝑚𝑜𝑙.𝐾
−141.6
-(G°/RT) = -(
so
K=e
−∆G0
RT
) = 57.2
= e57.2 = 7 x 1024
(b) Calculating the value of Q:
Q=
2
𝑃𝑆𝑂
3
2
𝑃𝑆𝑂2 𝑥 𝑃𝑂2
=
(0.100)2
0.5002 𝑥 0.0100
= 4.00
Because Q < K, the denominator will decrease and the numerator will increase more SO3 will form—until Q equals K. To reach
equilibrium, the reaction will proceed to the right.
(c)
Calculating G, the nonstandard free energy change, at 298 K:
G298 = G° + RT In Q
= -141.6 kJ/mol + (8.314 J/mol.K x
1 𝑘𝐽
1000𝐽
x 298 K x ln 4.00)
= -138.2 kJ/mol
Check Note that in parts (a) and (c), we made the free energy units (kJ) consistent with the units in R (J). For significant figures in
addition and subtraction, we retain one digit to the right of the decimal place in part (c).
Comment As in the synthesis of NH3 (Section 17.6), where the yield is high but the rate is low at a lower temperature, this process
is carried out at higher temperature with a catalyst to attain a higher rate. We discuss the details of the industrial production of
sulfuric acid in Chapter 22.
Chapter 20 • Problems
FOLLOW-UP PROBLEMS
20.11A At 298 K, G° = -33.5 kJ/mol for the formation of chloroethane from ethylene and hydrogen chloride:
C2H4C?) + HCl(g) = C2H5Cl(g)
(a) Calculate K at 298 K.
(b) Calculate G at 298 K if [C2H5Cl] = 1.5 M, [C2H4] = 0.50 M, and [HCl] = 1.0 M.
20.11B At 298 K, hypobromous acid (HBrO) dissociates in water with Ka = 2.3X10—9.
(a) Calculate G° for the dissociation of HBrO.
(b) Calculate G if [H3O+] = 6.0X10—4 M, [BrO-] = 0.10 M, and [HBrO] = 0.20 M.
SOME SIMILAR PROBLEMS 20.79 and 20.80
Another Look at the Meaning of Spontaneity At this point, we introduce two terms related to spontaneous and nonspontaneous:
1. Product-favored reaction. For the general reaction
A=B
K = [B]/[A] > 1
and therefore, the reaction proceeds largely from left to right (Figure 20.15A). From pure A to equilibrium, Q < K and the
curved green arrow in the figure indicates that the reaction is spontaneous (G < 0). From there on, the curved red arrow
indicates that the reaction is nonspontaneous (G > 0). Similarly, from pure B to equilibrium, Q > K and the reaction is also
spontaneous (G < 0), but not thereafter. In either case, free energy decreases until the reaction reaches a minimum at the
equilibrium mixture: Q = K and G = 0. For the overall reaction A = B (starting with all components in their standard states),
∆𝐺𝐵0 is smaller than ∆𝐺𝐴0 , so AG0 is negative, which corresponds to K > 1. We call this a product-favored reaction because in
its final state the system contains mostly product.
A
B
Figure 20.15 Figure 20.15 Free energy and the extent of reaction. Each reaction proceeds spontaneously (curved green arrows)
from reactants (A or C) or products (B or D) to the equilibrium mixture, at which point G = 0. After that, the reaction is
nonspontaneous (curved red arrows). A, For the product-favored reaction A = B, 𝐺𝐴0 > 𝐺𝐵0 so G ° <0 and K > 1. B, For the
reactant-favored reaction C=D, 𝐺𝐷0 > 𝐺𝐶0 so G °>0 and K < 1.
2. Reactant-favored reaction. For the opposite reaction,
C=D
K = [D]/[C] < 1
And the reaction proceeds slightly from left to right (Figure 20.15B). Here, too, whether we start with pure C or pure D, the
reaction is spontaneous (G < 0) until equilibrium. In this case, however, the equilibrium mixture contains mostly C (the
reactant), so we say the reaction is reactant favored. Here, 𝐺𝐷0 is larger than 𝐺𝐶0 , so G° is positive, which corresponds to K < 1.
Thus, “spontaneous” refers to that portion of a reaction in which the free energy decreases—from the starting mixture to
the equilibrium mixture. A product-favored reaction goes predominantly, but not completely, toward product, and a reactantfavored reaction goes relatively little toward product (see Table 20.2).
 Summary of Section 20.4




Two ways of predicting reaction spontaneity are from the sign of AG or from the value of Q/K. These variables are related to
each other by G = RT ln Q/K. When Q = K, Q/K = 1 and ln Q/K = 0. Thus, the system is at equilibrium and can release no more
free energy.
Beginning with Q at the standard state, the free energy change is G ° and is related to the equilibrium constant: G ° = -RT ln K.
Any nonequilibrium mixture of reactants and products moves spontaneously (AG < 0) toward the equilibrium mixture.
A product-favored reaction goes predominantly toward product and, thus, has K > 1 and G ° < 0; a reactant-favored reaction has K
< 1 and G ° > 0.
33
34
Chapter 20 • Thermodynamics: Entropy, Free Energy, and the Direction of Chemical Reactions
CHAPTER REVIEW GUIDE
 Learning Objectives: Relevant section (§) and/or sample problem (SP) numbers appear in parentheses.
Understand These Concepts
How the tendency of a process to occur by itself is distinct from how long it takes to occur (Introduction)
The distinction between a spontaneous and a nonspontaneous change (§20.1)
Why the first law of thermodynamics and the sign of H° cannot predict the direction of a spontaneous process (§20.1)
How the entropy ( S) of a system is defined by the number of microstates over which its energy is dispersed (§20.1)
How entropy is alternatively defined by the heat absorbed (or released) at constant T in a reversible process (§20.1)
The criterion for spontaneity according to the second law of thermodynamics: that a change increases Suniv (§20.1)
How absolute values of standard molar entropies (S°) can be obtained because the third law of thermodynamics
provides a “zero point” (§20.1)
8. How temperature, physical state, dissolution, atomic size, and molecular complexity influence S° values (§20.1)
0
9. How 𝑆𝑟𝑥𝑛
KJJ, is based on the difference between the summed S° values for the reactants and those for products (§20.2)
0
10. How the surroundings add heat to or remove heat from the system and how Ssurr influences overall 𝑆𝑟𝑥𝑛
(§20.2)
11. The relationship between Ssurr and Hsys (§20.2)
12. How reactions proceed spontaneously toward equilibrium (Suniv > 0) but proceed no further at equilibrium (Sumv = 0) (§20.2)
13. How the free energy change (G) combines a system’s entropy and enthalpy changes (§20.3)
14. How the expression for the free energy change is derived from the second law (§20.3)
1.
2.
3.
4.
5.
6.
7.
15. The relationship between G and the maximum work a system can perform and why this quantity of work is never performed in a real
process (§20.3)
16. How temperature determines spontaneity for reactions in which S and H have the same sign (§20.3)
17. Why the temperature at which a reaction becomes spontaneous occurs when G = 0 (§20.3)
18. How a spontaneous change can be coupled to a nonspontaneous change to make it occur (§20.3)
19. How G is related to the ratio of Q to K (§20.4)
20. The meaning of G° and its relation to K (§20.4)
21. The relation of G to G° and Q (§20.4)
22. Why G decreases, no matter what the starting concentrations, as the reacting system moves toward equilibrium (§20.4)
23. The distinction between product-favored and reactant- favored reactions (§20.4)
Master These Skills
1. Predicting relative S ° values of systems (§20.1 and SP 20.1)
0 for a chemical change (SP 20.2)
2. Calculating ∆𝑆𝑟𝑥𝑛
3. Finding reaction spontaneity from Ssurr and H °sys (SP 20.3)
0 from ∆𝐻 0 and S ° values (SP 20.4)
4. Calculating ∆𝐺𝑟𝑥𝑛
𝑓
0 from ∆𝐺 0 values (SP 20.5)
5. Calculating ∆𝐺𝑟𝑥𝑛
𝑓
6. Predicting the signs of H, S, and G (SP 20.6)
7. Determining the effect of temperature on G (SP 20.7)
8. Calculating the temperature at which a reaction becomes spontaneous (§20.3 and SP 20.8)
9. Calculating K from AG° and vice versa (§20.4 and SP 20.9)
10. Using AG° and Q to calculate AG at any conditions (SPs 20.10, 20.11)
Key Terms: Page numbers appear in parentheses.
Section 20.1
Section 20.2
spontaneous change (877)
microstate (880)
entropy (S) (880)
second law of thermodynamics
(883)
third law of thermodynamics (883)
standard molar entropy (S0) (883)
0
standard entropy of reaction (∆𝑆𝑟𝑥𝑛
)
(888)
Section 20.3
free energy (G) (893)
standard free energy change (∆𝐺 0 ) (894)
standard free energy of formation (∆𝐺𝑓𝑜 )
(895)
adenosine triphosphate (ATP) (902)
Key Equation and Relationships: Page numbers appear in parentheses.
20.1 Quantifying entropy in terms of the number of microstates (W) over which the energy of a system can be dispersed (880) :
S = k ln W
20.2 Quantifying the entropy change in terms of heat absorbed (or released) in a reversible process (882):
∆𝑆𝑠𝑦𝑠 =
𝑄𝑟𝑒𝑣
𝑇
20.3 Stating the second law of thermodynamics, for a spontaneous process (883):
∆𝑆𝑢𝑛𝑖𝑣 = ∆𝑆𝑠𝑦𝑠 +∆𝑆𝑠𝑢𝑟𝑟 > 0
20.4 Calculating the standard entropy of reaction from the standard molar entropies of reactants and products (888):
Chapter 20 • Problems
Sorxn = mSoroducts - nSoreactants
20.5 Relating the entropy change in the surroundings to the enthalpy change of the system and the temperature (890):
Ssurr = -
Hsys
T
20.6 Expressing the free energy change of the system in terms of its component enthalpy and entropy changes (Gibbs equation) (894):
Gsys = Hsys — TSsys
20.7 Calculating the standard free energy change from the standard free energies of formation (894):
0 = ∆𝐻 0 -T∆𝑆 0
∆𝐺𝑠𝑦𝑠
𝑠𝑦𝑠
𝑠𝑦𝑠
20.8 Calculating the standard free energy change from the standard free energies of formation (895):
0
0
0
∆𝐺𝑟𝑥𝑛
= 𝑚∆𝐺𝑓(𝑝𝑟𝑜𝑑𝑢𝑐𝑡𝑠)
- 𝑛∆𝐺𝑓(𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑡𝑠)
20.9 Relating the free energy change to the maximum work a process can do (896):
G = Wmax
20.10 Finding the temperature at which a reaction becomes spontaneous (900):
T=
∆𝐻
𝑆
20.11 Expressing the free energy change in terms of Q and K (903):
G = RT ln
Q
K
= RT ln Q - RT ln Q
20.12 Expressing the free energy change when Q is at standard state conditions (904):
G° = -RT ln K
20.13 Expressing the free energy change for nonstandard initial conditions (905):
G = G ° + RT ln Q
BRIEF SOLUTION TO FOLLOW-UP PROBLEM
20.1A (a) PCl5(g): higher molar mass and more complex molecule; (b) BaCl2Cr): higher molar mass; (c) Br2(g): gases have
more freedom of motion and dispersal of energy than liquids.
20.1B (a) LiBr(aq): same number of ions, but lower molar mass of ions; (b) quartz: the particles have less freedom of
motion in the crystalline structure; (c) cyclohexane: molecule with no side chain has less freedom of motion.
20.2A (a) 4NO(g)  N2O(g) + N2O3(g)
0 <0
ngas = -2, so 𝑆𝑟𝑥𝑛
0
𝑆𝑟𝑥𝑛
= [(1 mol N2O)(219.7 J/mol.K)+ (1 mol N2O3)(314.7 J/mol.K)] – [(4 mol NO)(210.65 J/mol.K)]
= -308.2 J/K
(b) CH3OH(g)  CO(g) + 2H2(g)
0
ngas = 2 so 𝑠𝑟𝑥𝑛
>0
0
𝑠𝑟𝑥𝑛
= [(1 mol CO)(197.5 J/mol-K)+(2 mol H2)(130.6 J/mol-K)]- [(1 mol CH3OH)(238 J/mol-K)]
= 221 J/K
20.2B (a) 2NaOH(s) + CO2(g)  Na2CO3(s) + H2O(l)
0
ngas = -1 so 𝑠𝑟𝑥𝑛
<0
0
𝑠𝑟𝑥𝑛
= [(1 mol H2O)(69.940 J/mol.K)+ (1 mol Na2CO3)(139 J/mol.K)]- [(1 mol CO2)(213.7 J/mol.K) + (2 mol NaOH)(64.454 J/mol.K)]
= -134 J/K
(b) 2Fe(s) + 3H2O(g) Fe2O3(s)+ 3H2(g)
0
ngas = 0, so cannot predict sign of 𝑠𝑟𝑥𝑛
0
𝑠𝑟𝑥𝑛 = [(1 mol Fe2O3)(87.400 J/mol.K) + (3 mol H2)(130.6 J/mo.K) ] -[(2 mol Fe)(27.3 J/mol.K) + (3 mol H2O)(188.72 J/mol-K)]
= -141.6 J/K
0
20.3A To find Ssurr, calculate 𝐻𝑟𝑥𝑛
and divide it by T:
0
𝐻𝑟𝑥𝑛
= [(4 mol PCl3)(-287 kJ/mol)] - [(1 mol P4)(0 kJ/mol) + (6 mol Cl2)(0 kJ/mol)]
= -1148 kJ
Ssurr = = -
Hsys
T
1000𝐽
(1148 𝑘𝐽)(
)
1𝑘𝐽
298 𝐾
= 3850 J/K
0
To find Suniv, calculate 𝑆𝑟𝑥𝑛
and add it to Ssurr:
0
𝑆𝑟𝑥𝑛
= [(4 mol PCl3)(312 J/mol.K)] - [(1 mol P4)(41.1 J/mo.K)+ (6 mol Cl2)(223.0 J/mol.K)]
35
36
Chapter 20 • Thermodynamics: Entropy, Free Energy, and the Direction of Chemical Reactions
= -131 J/K
Suniv = Ssurr + S °xn = 3850 J/K + (-131 J/K) = 3719 J/K Reaction is spontaneous at 298 K.
1
20.3B 2FeO(s) + 2 O2(g)  Fe2O3(s)
1
S°sys = (1 mol Fe2O3)(87.400 J/mol.K) - [(2 mol FeO)(60.75 J/mol-K) + (2 mol O2)(205.0 J/mol-K)]
= -136.6 J/K
1
H°sys = (1 mol Fe2O3)(-825.5 kJ/mol) - [(2 mol FeO)(-272.0 kJ/mol)+ (2 mol O2) (0 kJ/mol)
= -281.5 kJ
Ssurr = -
Hsys
T
1000 𝐽
)
1 𝑘𝐽
(−281.5 𝑘𝐽)(
=
298 𝐾
= +945 J/K
Suniv = S °sys +Ssurr = -136.6 J/K + 945 J/K = 808 J/K Reaction is spontaneous at 298 K.
20.4A H°rxn = [(2 mol NOCl)(51.71 kJ/mol)] - [(2 mol NO)(90.29 kJ/mol) + (1 mol Cl2)(0 kJ/mol)]
= -77.16 kJ
S°rxn = [(2 mol NOCl)( 261.6 J/mol.K)])] - [(2 mol NO)( 210.65 J/mol.K) + (1 mol Cl2)( 223.0 J/mol.K)]
= -121.16 J/K
G°rxn = H°rxn - TS°rxn
= 277.16 kJ - [(298 K)(-121.1 J/K)(1 kJ/1000 J)]
= -41.1 kJ
20.4B Using 𝐻𝑓0 and S° values from Appendix B,
H°rxn = -114.2 kJ and S°rxn =-146.5 J/K
G°rxn = H°rxn - TS°rxn
= -114.2 kJ - [(298 K)(-146.5 J/K)(1 kJ/1000 J)]
= -70.5 kJ
20.5A (a) G°rxn = [(2 mol NOCl)(66.07 kJ/mol)] - [(2 mol NO)(86.60 kJ/mol) + (1 mol Cl2)(0 kJ/mol)]
= -41.06 kJ
(b) G°rxn = [(2 mol Fe)(0 kJ/mol) + (3 mol H2O)(-228.60 kJ/mol)] - [(3 mol H2)(0 kJ/mol) + (1 mol Fe2O3)(-743.6 kJ/mol)]
= 57.8 kJ
20.5B (a) G°rxn = (2 mol NO2)(51 kJ/mol) - [(2 mol NO)(86.60 kJ/mol) + (1 mol O2)(0 kJ/mol)]
= -71 kJ
(b) G°rxn = (2 mol CO)(-137.2 kJ/mol) - [(2 mol C)(0 kJ/mol) + (1 mol O2)(0 kJ/mol)]
= -274.4 kJ
20.6A (a) More moles of gas are present after the reaction, so S > 0.
(b) The problem says the reaction is spontaneous (G < 0) only above 325°C, which implies high T. If S > 0, - TS < 0, so
G will become negative at higher T only if H > 0.
20.6B (a) A solid forms a gas and a liquid, so S >0, and a crystalline array breaks down, so H > 0.
(b) For the reaction to occur spontaneously (G < 0), - TS must be greater than H, which would occur only at higher T.
20.7A (a) G = H - TS
= -192.7 kJ - [(298 K)(-308.2 J/K)(1 kJ/1000 J)] = -100.9 kJ
= -100.9 kJ, The reaction is spontaneous at 298 K.
(b) As T increases, - TS becomes more positive, so the reaction becomes less spontaneous.
(c) G = H - TS
= -192.7 kJ- [(500 + 273 K)(-308.2 J/K)(1 kJ/1000 J)]
= 45.5 kJ
20.7B G becomes negative at lower T, so H < 0, S < 0, and -TS > 0. At lower T, the negative H value becomes larger than
the positive - TS value.
20.8A G = 0 when T =
∆𝐻
∆𝑆
=
(−192.7 kJ)(1000 J/1 kJ)
−308.2 𝐽/𝐾
= 625.2 K
T (°C) = 625.2 K - 273.15 = 352.0°C
20.8B H = H°f CaCO3 - (H°f CaO + H°f CO2)
= - 1206.9 kJ - (- 635.1 kJ- 393.5 kJ)
= - 178.3 kJ
Chapter 20 • Problems
S = S° CaCO3 - (S° CaO + S° CO2)
= 92.9 J/K - (38.2 J/K + 213.7 J/K) = -159.0 J/K
T=
∆𝐻
∆𝑆
=
(−178.3 kJ)(1000 J/1 kJ)
= 1121 K
−159.0 𝐽/𝐾
Reaction becomes spontaneous (G < 0) at any T < 1121 K.
20.9A G° = [(2 mol CO)(-137.2 kJ/mol)] - [(2 mol C)(0 kJ/mol) + (1 mol O2)(0 kJ/mol)]
= -274.4 kJ
∆𝐺 0
𝑅𝑇
ln K = -
=
𝐽
)
1𝑘𝐽
(−274.4 𝑘𝐽)(1000
𝐽
)(298 𝐾)
𝑚𝑜𝑙.𝐾
(8.314
= 111
K = 1.61x1048
20.9B
K = 2.22 x 10-15, so ln K = -33.7
G° = -RT ln K = -(8.314 J/mol.K)(298 K)(-33.7)
= 8.35 x104 J = 83.5 kJ
20.10A (a) A(g) + 3B(g) = AB3(g); G° = -4.6 kJ/mol
(b) For mixture 1, Q1 =
[𝐴𝐵3 ]
[𝐴]1 [𝐵]3
0.20
= (0.4)(0.80)3 = 0.98
For mixture 2, Q2 =
0.30
(0.3)(0.50)3
= 8.0
For mixture 3, Q3 =
0.40
(0.2)(0.20)3
= 250
Mixture 2 is at equilibrium.
(c) Based on Comments 1 and 2 in Sample Problem 20.10, G1 < 0, G2 = 0, and G3 > 0, so G3 > G2 > GV
20.10B (a) Mixture 2 is at equilibrium.
(b) 3 (most negative) < 2 < 1 (most positive)
( C ) Any reaction mixture moves spontaneously toward equilibrium, so both changes have a negative G.
20.11A (a) G° = -RT ln K, so
ln K =
∆𝐺 0
𝑅𝑇
=
(−33.5 𝑘𝐽)(1000
𝐽
)
1𝑘𝐽
𝐽
)(298 𝐾)
𝑚𝑜𝑙.𝐾
(8.314
= 13.521
K = 7.45x105
[𝐶2𝐻5𝐶𝑙]
1.50
(b) Q = [𝐶2𝐻4][HCl] = (0,5)(1,0) = 3.0
G = G° + RT ln Q
= -33.5 kJ + [(8.314 J/mol.K)(1 kJ/1000 J)(298 K)(ln 3.0)]
= -30.8 kJ
20.11B (a) AG° = - RT ln K
= -8.314 J/mol.K x
= 49 kJ/mol
[𝐻3 𝑂 + ][𝐵𝑟𝑂 − ]
(b) Q =
[𝐻𝐵𝑟𝑂]
=
1000 𝐽
1 𝑘𝐽
x298 𝐾 x ln (2.3x10-9)
(6.0𝑥10−4 )(0.10)
0.20
= 3.0x10-4
G = G° + RT ln Q
= 49 kJ + [(8.314 J/mol.K)(1 kJ/1000 J)(298 K)(ln 3.0 x10-4)]
= 29 kJ/mol
PROBLEM
Problems with colored numbers are answered in Appendix E and worked in detail in the Student Solutions Manual. Problem sections
match those in the text and give the numbers of relevant sample problems. Most offer Concept Review Questions, Skill -Building
Exercises (grouped in pairs covering the same concept), and Problems in Context. The Comprehensive Problems are based on material
from any section or previous chapter.
Note: Unless stated otherwise, problems refer to systems at 298 K (25°C). Solving these problems may require values from Appendix B.
37
38
Chapter 20 • Thermodynamics: Entropy, Free Energy, and the Direction of Chemical Reactions
The Second Law of Thermodynamics:
Predicting Spontaneous Change
(Sample Problem 20.1)
Concept Review Questions
20.1 Distinguish between the terms spontaneous and instantaneous. Give an example of a process that is spontaneous
but very slow, and one that is very fast but not spontaneous.
20.2 Distinguish between the terms spontaneous and nonspontaneous. Can a nonspontaneous process occur? Explain.
20.3 State the first law of thermodynamics in terms of (a) the energy of the universe; (b) the creation or destruction of energy;
(c) the energy change of system and surroundings. Does the first law reveal the direction of spontaneous change? Explain.
20.4 State qualitatively the relationship between entropy and freedom of particle motion. Use this idea to explain why you will
probably never (a) be suffocated because all the air near you has moved to the other side of the room; (b) see half the water in
your cup of tea freeze while the other half boils.
20.5 Why is Svap of a substance always larger than Sfus?
20.6 How does the entropy of the surroundings change during an exothermic reaction? An endothermic reaction? Other than the
examples in text, describe a spontaneous endothermic process.
20.7 (a) What is the entropy of a perfect crystal at 0 K?
(b) Does entropy increase or decrease as the temperature rises?
(c) Why is 𝐻𝑓0 = 0 but S ° > 0 for an element?
(d) Why does Appendix B list 𝐻𝑓0 values but not 𝑆𝑓0 values?
Skill-Building Exercises (grouped in similar pairs)
20.8 Which of these processes are spontaneous? (a) Water evaporates from a puddle. (b) A lion chases an antelope. (c) An isotope
undergoes radioactive disintegration.
20.9 Which of these processes are spontaneous? (a) Earth moves around the Sun. (b) A boulder rolls up a hill. (c) Sodium metal and
chlorine gas form solid sodium chloride.
20.10 Which of these processes are spontaneous? (a) Methane burns in air. (b) A teaspoonful of sugar dissolves in a cup of hot coffee.
(c) A soft-boiled egg becomes raw.
20.11 Which of these processes are spontaneous? (a) A satellite falls to Earth. (b) Water decomposes to H 2 and O2 at 298 K and 1 atm.
(c) Average car prices increase.
20.12 Predict the sign of Ssys for each process: (a) A piece of wax melts. (b) Silver chloride precipitates from solution. (c) Dew forms on
a lawn in the morning.
20.13 Predict the sign of Ssys for each process: (a) Gasoline vapors mix with air in a car engine. (b) Hot air expands.
(c)Humidity condenses in cold air.
20.14 Predict the sign of Ssys for each process: (a) Alcohol evaporates. (b) A solid explosive converts to a gas. (c) Perfume vapors diffuse
through a room.
20.15 Predict the sign of Ssys for each process: (a) A pond freezes in winter. (b) Atmospheric CO2 dissolves in the ocean. (c) An apple
tree bears fruit.
20.16 Without using Appendix B, predict the sign of S ° for
(a) 2K(s) + F2(g) 2KF(s)
(b) NH3(g) + HBr(g)  NH4Br(s)
(c) NaClO3(s)  Na+(aq) + ClO3—(aq)
20.17 Without using Appendix B, predict the sign of S° for
1
1
(a) H2S(g) + O2(g)  S8 (s) + H2O(g)
2
8
(b) HCl(aq) + NaOH(aq)  NaCl(aq) + H2O(l)
(c) 2NO2(g) — N2O4(g)
20.18 Without using Appendix B, predict the sign of S° for
(a) CaCO30) + 2HCl(aq)  CaCl2(aq) + H2O(l) + CO2(g)
(b) 2NO(g) + O2(g)  2NO2(g)
(c) 2KClO3(s)  2KCl(s) + 3O2(g)
20.19 Without using Appendix B, predict the sign of S ° for
(a) Ag+(aq) + Cl-(aq)  AgCl(s)
(b) KBr(s) KBr(aq)
(c) CH3CH=CH2(g)  Ciclopropane (g)
20.20 Predict the sign of S for each process:
(a) C2H5OH(g) (350 K and 500 torr)  C2H5OH(g) (350 K and 250 torr)
(b) N2(g) (298 K and 1 atm)  N2(aq) (298 K and 1 atm)
(c) O2(aq) (303 K and 1 atm)  O2(g) (303 K and 1 atm)
20.21 Predict the sign of S for each process:
(a) O2(g) (1.0 L at 1 atm) O2(g) (0.10 L at 10 atm)
(b) Cu(s) (350°C and 2.5 atm)  Cu(s) (450°C and 2.5 atm)
(c) Cl2(g) (100°C and 1 atm)  Cl2(g) (10°C and 1 atm)
Chapter 20 • Problems
20.22 Predict which substance has greater molar entropy. Explain.
(a) Butane CH3CH2CH2CH3(g) or 2-butene CH3CH=CHCH3(g)
(b) Ne(g) or Xe(g)
(c) CH4(g) or CCl4(l)
20.23 Predict which substance has greater molar entropy. Explain.
(a) NO2(g) or N2O4(g)
(b) CH3OCH3(l) or CH3CH2OH(l)
(c) HCl(g) or HBr(g)
20.24 Predict which substance has greater molar entropy. Explain.
(a) CH3OH(l) or C2H5OH(l)
(b) KClO3(s) or KClO3(aq)
(c) Na(s) or K(s)
20.25 Predict which substance has greater molar entropy. Explain.
(a) P4(g) or P2(g)
(b) (HNO3(aq) or HNO3(l)
(c) CuSO4(s) or CuSO4.5H2O(s)
20.26 Without consulting Appendix B, arrange each group in order of increasing standard molar entropy (S°). Explain.
(a) Graphite, diamond, charcoal
(b) Ice, water vapor, liquid water
(c) O2, O3, O atoms
20.27 Without consulting Appendix B, arrange each group in order of increasing standard molar entropy (S°). Explain.
(a) Glucose (C6H12O6), sucrose (C12H22O11), ribose (C5H10O5)
3
(b) CaCO3, Ca +C + O2, CaO +CO2
2
(c) SF6(g), SF4(g), S2F10(g)
20.28 Without consulting Appendix B, arrange each group in order of decreasing standard molar entropy (S°). Explain.
(a) ClO4-(aq), ClO2- (aq), ClO3- (aq)
(b) NO2(g), NO(g), N2(g)
(c) Fe2O3(s), Al2O3(s), Fe3O4(s)
20.29 Without consulting Appendix B, arrange each group in order of decreasing standard molar entropy (S°). Explain.
(a) Mg metal, Ca metal, Ba metal
(b) Hexane (C6H14), benzene (C6H6), cyclohexane (C6H12)
(c) PF2Cl3(g), PF5(g), PF3(g)
Calculating the Change in Entropy of a Reaction
(Sample Problems 20.2 and 20.3)
Concept Review Questions
20.30 For the reaction depicted in the molecular scenes, X is red and Y is green.
(a) Write a balanced equation.
(b) Determine the sign of Srxn.
(c) Which species has the highest molar entropy?
20.31 Describe the equilibrium condition in terms of the entropy changes of a system and its surroundings. What does this
description mean about the entropy change of the universe?
20.32 For the reaction H2O(g) + Cl2O(g)  2HClO(g), you know S °rxn and S ° of HClO(g) and of H2O(g). Write an expression that can
be used to determine S ° of Cl2O(g).
Skill-Building Exercises (grouped in similar pairs)
20.33 For each reaction, predict the sign and find the value of S°rxn:
(a) 3NO(g)  N2O(g) + NO2(g)
(b) 3H2(g) + Fe2O3(s)  2Fe(s) + 3H2O(g)
(c) P4(s)+ 5O2(g)  P4O10(S)
20.34 For each reaction, predict the sign and find the value of S°rxn:
(a)
3NO2g) + H2O(l) 2HNO3(l) + NO(g)
(b)
N2(g) + 3F2(g)  2NF3(g)
(c) C6H12O6(s) +6O2(g)  6CO2(g) + 6H2O(g)
39
40
Chapter 20 • Thermodynamics: Entropy, Free Energy, and the Direction of Chemical Reactions
20.35 Find S °rxn for the combustion of ethane (C2H6) to carbon dioxide and gaseous water. Is the sign of S °rxn as expected?
20.36 Find S °rxn for the combustion of methane to carbon dioxide and liquid water. Is the sign of S °rxn as expected?
20.37 Find S° rxn for the reaction of nitrogen monoxide with hydrogen to form ammonia and water vapor. Is the sign of S °rxn as
expected?
20.38 Find S° rxn for the combustion of ammonia to nitrogen dioxide and water vapor. Is the sign of S °rxn as expected?
20.39 (a) Find S° rxn for the formation of Cu2O(s) from its elements.
(b) Calculate S univ, and state whether the reaction is spontaneous at 298 K.
20.40 (a) Find S°rxn for the formation of HI(g) from its elements.
(b) Calculate S univ, and state whether the reaction is spontaneous at 298 K.
20.41 (a) Find S°rxn for the formation of CH3OH(l) from its elements.
(b) Calculate S univ, and state whether the reaction is spontaneous at 298 K.
20.42 (a) Find S°rxn for the formation of N2O(g) from its elements.
(b) Calculate S univ, and state whether the reaction is spontaneous at 298 K.
Problems in Context
20.43 Sulfur dioxide is released in the combustion of coal. Scrubbers use aqueous slurries of calcium hydroxide to remove the SO 2 from
flue gases. Write a balanced equation for this reaction and calculate S °rxn at 298 K [S ° of CaSO3(s) = 101.4 J/mol-K].
20.44 Oxyacetylene welding is used to repair metal structures, including bridges, buildings, and even the Statue of Liberty. Calculate S
°rxn for the combustion of 1 mol of acetylene (C2H2).
Entropy, Free Energy, and Work
(Sample Problems 20.4 to 20.8)
Concept Review Questions
20.45 What is the advantage of calculating free energy changes rather than entropy changes to determine reaction spontaneity?
20.46 Given that Gsys = -TSuniv, explain how the sign of Gsys correlates with reaction spontaneity.
20.47 (a) Is an endothermic reaction more likely to be spontaneous at higher temperatures or lower temperatures? Explain.
(b) The change depicted below occurs at constant pressure. Explain your answers to each of the following:
(1) What is the sign of Hsys?
(2) What is the sign of Ssys?
(3) What is the sign of Ssurr?
(4) How does the sign of G sys vary with temperature?
20.48 Explain your answers to each of the following for the change depicted below.
(a) What is the sign of Hsys?
(b) What is the sign of S sys?
(c) What is the sign of S surr?
(d) How does the sign of G sys vary with temperature?
20.49 With its components in their standard states, a certain reaction is spontaneous only at high T. What do you know about
the signs of H° and S°? Describe a process for which this is true.
20.50 How can S° be relatively independent of T if S° of each reactant and product increases with T?
Skill-Building Exercises (grouped in similar pairs)
20.51 Calculate G° for each reaction using G°f values:
(a) 2Mg(s) +O2(g)  2MgO(s)
(b) 2CH3OH(g) +3O2(g)  2CO2(g) +CO2(g)
(c) BaO(s) + CO2(g)  BaCO3(s)
20.52 Calculate G° for each reaction using G°f values:
(a) H2(g) + I2(s)  2HI(g)
(b) MnO2(s) + 2CO(g)  Mn(s) + 2CO2(g)
(b) NH4Cl(s)  NH3(g) + HCl(g)
20.53 Find G ° for the reactions in Problem 20.51 using H°f and S° values.
Chapter 20 • Problems
20.54 Find AG° for the reactions in Problem 20.52 using H°f and S ° values.
1
20.55 Consider the oxidation of carbon monoxide: CO(g) + O2(g) CO2(g)
2
(a) Predict the signs of AS ° and AH°. Explain.
(b) Calculate AG ° by two different methods.
20.56 Consider the combustion of butane gas:
13
C4H10(g) + O2(g)  4CO2(g) + 5H2O(g)
2
(a) Predict the signs of S ° and H°. Explain.
(b) Calculate G ° by two different methods.
20.57 For the gaseous reaction of xenon and fluorine to form xenon hexafluoride:
(a) Calculate S ° at 298 K (H ° = -402 kJ/mol and G ° = -280. kJ/mol).
(b) Assuming that S ° and (H ° change little with temperature, calculate G° at 500. K.
20.58 For the gaseous reaction of carbon monoxide and chlorine to form phosgene (COCl2):
(a) Calculate S ° at 298 K (H ° = -220. kJ/mol and G ° = -206 kJ/mol).
(b) Assuming that S ° and (H ° change little with temperature, calculate G° at 450. K.
20.59 One reaction used to produce small quantities of pure H2 is
CH3OH (g) = CO(g) + 2H2(g)
(a) Determine H° and S ° for the reaction at 298 K.
(b) Assuming that these values are relatively independent of temperature, calculate G° at 28°C, 128°C, and 228°C.
(c) What is the significance of the different values of G°?
(d) At what temperature (in K) does the reaction become spontaneous?
20.60 A reaction that occurs in the internal combustion engine is
N2(g ) + O2(g ) =2NO(g)
(a) Determine H° and S ° for the reaction at 298 K.
(b) Assuming that these values are relatively independent of temperature, calculate G ° at 100.°C, 2560.°C, and 3540.°C.
(c) What is the significance of the different values of G °?
(d) At what temperature (in K) does the reaction become spontaneous?
Problems in Context
20.61 As a fuel,H2(g) produces only nonpolluting H2O(g) when it burns. Moreover, it combines with O2(g) in a fuel cell
(Chapter 21) to provide electrical energy.
(a) Calculate H °, S °, and G ° per mole of H2 at 298 K.
(b) Is the spontaneity of this reaction dependent on T ? Explain.
(c) At what temperature does the reaction become spontaneous?
20.62 The U.S. government requires automobile fuels to contain a renewable component. Fermentation of glucose from
corn yields ethanol, which is added to gasoline to fulfill this requirement:
C6H12O6(s)  2C2H5OH(l) + 2CO2(g)
Calculate H°, S °, and G ° for the reaction at 25°C. Is the spontaneity of this reaction dependent on T ? Explain.
Free Energy, Equilibrium, and Reaction Direction
(Sample Problems 20.9 to 20.11)
Concept Review Questions
20.63 (a) If K << 1 for a reaction, what do you know about the sign and magnitude of G °? (b) If G ° << 0 for a reaction,
what do you know about the magnitude of K? Of Q ?
20.64 How is the free energy change of a process related to the work that can be obtained from the process? Is this
quantity of work obtainable in practice? Explain.
20.65 The scenes and the graph relate to the reaction of X2(g) (black) with Y2(g) (orange) to form XY(g).
(a) If reactants and products are in their standard states, what quantity is represented on the graph by x?
(b) Which scene represents point 1? Explain.
(c) Which scene represents point 2? Explain.
41
42
Chapter 20 • Thermodynamics: Entropy, Free Energy, and the Direction of Chemical Reactions
20.66 What is the difference between G ° and G? Under what circumstances does G = G °?
Skill-Building Exercises (grouped in similar pairs)
20.67 Calculate K at 298 K for each reaction:
(a) MgCO3(s) = Mg2+(aq) 1 CO32-(aq)
(b) H2(g) + O2(g) = H2O2 (l)
20.68 Calculate G° at 298 K for each reaction:
(a) 2H2S(g) + 3O2(g) = 2H2O(g) + 2SO2(g); K = 6.57X10173
(b) H2SO4(l) = H2O(l) +SO3(g); K = 4.46X10-15
20.69 Calculate K at 298 K for each reaction:
(a) HCN(aq) + NaOH(aq) = NaCN(aq) + H2O(l)
(b) SrSO4(s) = Sr2+(aq) + SO42-(aq)
20.70 Calculate G° at 298 K for each reaction:
(a) 2NO(g) + Cl2(g) = 2NOCl(g); K = 1.58X107
(b) Cu2S(s) + O2(g) = 2Cu(s) + SO2(g); K = 3.25 X1037
20.71 Use H0 and S0 values for the following process at 1 atm to find the normal boiling point of Br2: Br2(l) = Br2(g)
20.72 Use H0 and S0 values to find the temperature at which these sulfur allotropes reach equilibrium at 1 atm:
S(rhombic) = S(monoclinic)
20.73 Use Appendix B to determine the Ksp of Ag2S.
20.74 Use Appendix B to determine the Ksp of CaF2.
20.75 For the reaction I2(g) + Cl2(g) = 2ICl(g), calculate Kp at 25oC [𝐺𝑓0 of ICl(g) = -6.075 kJ/mol].
20.76 For the reaction CaCO3(s) = CaO(s) + CO2(g), calculate the equilibrium PCO2 at 25oC.
20.77 The Ksp of PbCl2 is 1.7x10-2 at 25oC. What is G0? Is it possible to prepare a solution that contains Pb2+(aq) and Cl-(aq),
at their standard-state concentrations?
20.78 The Ksp of ZnF2 is 3.0x10-2 at 25oC. What is G0? Is it possible to prepare a solution that contains Zn2+(aq) and F-(aq)
at their standard-state concentrations?
20.79 The equilibrium constant for the reaction 2Fe3+(aq) + Hg2+ (aq) = 2Fe2+(aq) + 2Hg2+(aq) is Kc = 9.1x10-6 at 298 K.
(a) What is G0 at this temperature?
(b) If standard-state concentrations of the reactants and products are mixed, in which direction does the reaction proceed?
(c) Calculate G when [Fe3+] = 0.20 M, [𝐻𝑔22+ ] = 0.010 M, [Fe2+] = 0.010 M, and [Hg2+] = 0.025 M. In which direction will
the reaction proceed to achieve equilibrium?
20.80 The formation constant for the reaction Ni2+(aq) + 6NH3(aq) = Ni(NH3)62+(aq) is Kf = 5.6x108 at 25oC.
(a) What is G0 at this temperature?
(b) If standard-state concentrations of the reactants and products are mixed, in which direction does the reaction proceed?
(c) Determine G when [Ni(NH3)62+] = 0.01 M, [Ni2+] = 0.001M, and [NH3] = 0.0050 M. In which direction will the reaction
proceed to achieve equilibrium?
20.81 The scenes below depict three gaseous mixtures in which A is reacting with itself to form A2. Assume that each particle
represents 0.10 mol and the volume is 0.10 L.
(a) If K = 0.33, which mixture is at equilibrium? (b) Rank the mixtures from the most positive G to the most negative G.
20.82 The scenes below depict three gaseous mixtures in which X (orange) and Y2 (black) are reacting to form XY and Y.
Assume that each gas has a partial pressure of 0.10 atm.
(a) If K = 4.5, which mixture is at equilibrium?
(b) Rank the mixtures from the most positive G to the most negative G.
Problems in Context
20.83 High levels of ozone (O3) cause rubber to deteriorate, green plants to turn brown, and many people to have difficulty breathing.
(a) Is the formation of O3 from O2 favored at all T , no T , high T , or low T ?
Chapter 20 • Problems
(b) Calculate G° for this reaction at 298 K.
(c) Calculate G at 298 K for this reaction in urban smog where [O2] = 0.21 atm and [O3] = 5X10-7 atm.
20.84 A BaSO4 slurry is ingested before the gastrointestinal tract is x-rayed because it is opaque to x-rays and defines the
contours of the tract. Ba2+ ion is toxic, but the compound is nearly insoluble. If G° at 37°C (body temperature) is 59.1
kJ/mol for the process; BaSO4(s) = Ba2+(aq) + SO42-(aq) what is [Ba21] in the intestinal tract? (Assume that the only source
of SO42- is the ingested slurry.)
Comprehensive Problems
20.85 According to advertisements, “a diamond is forever.”
(a) Calculate H°, S°, and G° at 298 K for the phase change ; Diamond  graphite
(b) Given the conditions under which diamond jewelry is normally kept, argue for and against the statement in the ad.
(c) Given the answers in part (a), what would need to be done to make synthetic diamonds from graphite?
(d) Assuming H° and S° do not change with temperature, can graphite be converted to diamond spontaneously at 1 atm?
20.86 Replace each question mark with the correct information:
Srxn
Hrxn
(a)
+
(b)
?
0
(c)
+
(d)
0
?
(e)
?
0
(f)
+
+
Grxn
?
+
?
Comment
?
Spontaneous
Not Spontaneous
Spontaneous
?
TS > H
20.87 Among the many complex ions of cobalt are the following: Co(NH3)63+(aq) + 3en(aq) = Co(en)331(aq) + 6NH3(aq)
where “en” stands for ethylenediamine, H2NCH2CH2NH2. Six Co—N bonds are broken and six Co—N bonds are formed
in this reaction, so H°rxn < 0; yet K > 1. What are the signs of S° and G °? What drives the reaction?
20.88 What is the change in entropy when 0.200 mol of potassium freezes at 63.7°C (Hfus = 2.39 kJ/mol)?
20.89 Is each statement true or false? If false, correct it.
(a) All spontaneous reactions occur quickly.
(b) The reverse of a spontaneous reaction is nonspontaneous.
(c) All spontaneous processes release heat.
(d) The boiling of water at 100°C and 1 atm is spontaneous.
(e) If a process increases the freedom of motion of the particles of a system, the entropy of the system decreases.
(f) The energy of the universe is constant; the entropy of the universe decreases toward a minimum.
(g) All systems disperse their energy spontaneously.
(h) Both Ssys and Ssurr equal zero at equilibrium.
20.90 Hemoglobin carries O2 from the lungs to tissue cells, where the O2 is released. The protein is represented as Hb in its
unoxygenated form and as Hb.O2 in its oxygenated form. One reason CO is toxic is that it competes with O2 in binding to Hb:
Hb • O2(aq) + CO(g) = Hb.CO(aq) +O2(g)
(a) If G °  -14 kJ at 37°C (body temperature), what is the ratio of [Hb.CO] to [Hb.O2] at 37°C with [O2] = [CO]?
(b) How is Le Chatelier’s principle used to treat CO poisoning?
20.91 Magnesia (MgO) is used for fire brick, crucibles, and furnace linings because of its high melting point. It is produced by
decomposing magnesite (MgCO3) at around 1200°C.
(a) Write a balanced equation for magnesite decomposition.
(b) Use H° and S° values to find G° at 298 K.
(c) Assuming that H° and S° do not change with temperature, find the minimum temperature at which the reaction is
spontaneous.
(d) Calculate the equilibrium PCo2 above MgCO3 at 298 K.
(e) Calculate the equilibrium PCo2 above MgCO3 at 1200 K.
20.92 To prepare nuclear fuel, U3O8 (“yellow cake”) is converted to UO2(NO3)2, which is then converted to UO3 and finally
UO2. The fuel is enriched (the proportion of the 235U is increased) by a two-step conversion of UO2 into UF6, a volatile solid,
followed by a gaseous-diffusion separation of the 235U and 238U isotopes:
UO2CO + 4HF(g)  UF4(S) + 2H2O(g)
UF4(S) + F2(g)  UF6(S)
Calculate G ° for the overall process at 85°C:
∆𝐻𝑓0 (kJ/mol)
S0 (J/mol.K)
∆𝐺𝑓0 (kJ/mol)
UO2(s)
-1085
77.0
-1032
UF4(S)
-1921
152
-1830
UF6(s)
-2197
225
-2068
20.93 Methanol, a major industrial feedstock, is made by several catalyzed reactions, such as CO(g) + 2H2(g)  CH3OH(l).
(a) Show that this reaction is thermodynamically feasible.
(b) Is it favored at low or at high temperatures?
(c) One concern about using CH3OH as an auto fuel is its oxidation in air to yield formaldehyde, CH2O(g), which poses a
health hazard. Calculate G° at 100.°C for this oxidation.
43
44
Chapter 20 • Thermodynamics: Entropy, Free Energy, and the Direction of Chemical Reactions
20.94 (a) Write a balanced equation for the gaseous reaction between N2O5 and F2 to form NF3 and O2.
0
(b) Determine ∆𝐺𝑟𝑥𝑛
.
(c) Find Grxn at 298 K if PN2O5 = PF2 = 0.20 atm, PNF3 = 0.25 atm, and PO2 = 0.50 atm.
20.95 Consider the following reaction:
2NOBr(g) = 2NO(g) + Br2(g)
K = 0.42 at 373 K
0 and ∆𝐻 0 are constant with temperature, find
Given that S ° of NOBr(g) = 272.6 J/mol.K and that ∆𝑆𝑟𝑥𝑛
𝑟𝑥𝑛
(a) S°rxn at 298 K
(b) G°rxn at 373 K
(c) H°rxn at 373 K (d) H°f of NOBr at 298K
(e) G °rxn at 298 K (f) G°f of NOBr at 298K
20.96 Hydrogenation is the addition of H2 to double (or triple) carbon-carbon bonds. Peanut butter and most commercial baked
goods include hydrogenated oils. Find H°, S°, and G° for the hydrogenation of ethene (C2H4) to ethane (C2H6) at 25°C.
20.97 Styrene is produced by catalytic dehydrogenation of ethylbenzene at high temperature in the presence of superheated
steam.
(a) Find H°rxn, G°rxn, and S°rxn given these data at 298 K:
Compound
∆𝐻𝑓0 (kJ/mol)
∆𝐺𝑓0 (kJ/mol)
S0 (J/mol.K)
Ethylbenzene, C6H5—CH2CH3
-12.5
119.7
255
Styrene, C6H5—CH=CH2
103.8
202.5
238
(b) At what temperature is the reaction spontaneous?
(c) What are G°rxn and K at 600.°C?
(d) With 5.0 parts steam to 1.0 part ethylbenzene in the reactant mixture and the total pressure kept constant at 1.3 atm, what is
G at 50.% conversion, that is, when 50.% of the ethylbenzene has reacted?
20.98 Propylene (propene; CH3CH=CH2) is used to produce polypropylene and many other chemicals. Although most is obtained
from the cracking of petroleum, about 2% is produced by catalytic dehydrogenation of propane (CH3CH2CH3):
Pt/Al2O3
CH3CH2CH3
 CH3CH=CH2 + H2
Because this reaction is endothermic, heaters are placed between the reactor vessels to maintain the required temperature.
(a) If the molar entropy, S°, of propylene is 267.1 J/mol? K, find its entropy of formation, 𝑆𝑓0 .
(b) Find ∆𝐺𝑓0 of propylene (∆𝐻𝑓0 for propylene = 20.4 kJ/mol).
0 and ∆G°
(c) Calculate ∆𝐻𝑟𝑥𝑛
rxn for the dehydrogenation.
(d) What is the theoretical yield of propylene at 580°C if the initial pressure of propane is 1.00 atm?
(e) Would the yield change if the reactor walls were permeable to H2? Explain.
(f) At what temperature is the dehydrogenation spontaneous, with all substances in the standard state?
Note: Problems 20.99 and 20.100 relate to the thermodynamics of adenosine triphosphate (ATP).
Refer to the Chemical Connections essay on pp. 902-903.
20.99 Find K for (a) the hydrolysis of ATP,
(b) the dehydration- condensation to form glucose phosphate, and (c) the coupled reaction between ATP and glucose.
(d) How does each K change when T changes from 25°C to 37°C?
20.100 Energy from ATP hydrolysis drives many nonspontaneous cell reactions:
ATP4-(aq) + H2O(l) = ADP3—(aq) + HPO42-(aq) + H+(aq)
∆𝐺 𝑜′ = -30.5 kJ
(a) Find K for the hydrolysis of ATP at 37°C.
0′ for metabolism of 1 mol of glucose.
(b) Find 𝐺𝑟𝑥𝑛
(c) How many moles of ATP can be produced by metabolism of 1 mol of glucose?
(d) If 36 mol of ATP is formed, what is the actual yield?
20.101 From the following reaction and data, find (a) S° of SOCl2 and (b) T at which the reaction becomes nonspontaneous:
SO3(g) + SCl2(l)  SOCl2(l) + SO2(g)
G°rxn = -75.2 kJ
SO3(g)
SCl2(l )
SOCl2(l )
SO2(g)
(kJ/mol)
-396
-50.0
-245.6
-296.8
S ° (J/mol-K)
256.7
184
-
248.1
𝐻𝑓0
20.102 Write equations for the oxidation of Fe and of Al. Use ∆𝐺𝑓0 to determine whether either process is spontaneous at 25°C.
20.103 The molecular scene depicts a gaseous equilibrium mixture at 460°C for the reaction of H2 (blue) and I2 (purple) to form HI.
Each molecule represents 0.010 mol and the container volume is 1.0 L.
(a) Is Kc >, =, or < 1?
(b) Is Kp >, =, or < Kc?
(c) Calculate G°rxn.
(d) How would the value of G°rxn change if the purple molecules represented H2 and the blue I2? Explain.
20.104 A key step in the metabolism of glucose for energy is the isomerization of glucose-6-phosphate (G6P) to fructose6-phosphate (F6P): G6P = F6P; K = 0.510 at 298 K.
(a) Calculate AG ° at 298 K.
(b) Calculate AG when Q, the [F6P]/[G6P] ratio, equals 10.0.
Chapter 20 • Problems
(c) Calculate G when Q = 0.100.
(d) Calculate Q if G = -2.50 kJ/mol.
20.105 A chemical reaction, such as HI forming from its elements, can reach equilibrium at many temperatures. In contrast,
a phase change, such as ice melting, is in equilibrium at a given pressure and temperature. Each of the graphs below depicts
Gsys vs. extent of change.
(a) Which graph depicts how Gsys changes for the formation of HI? Explain.
(b) Which graph depicts how Gsys changes as ice melts at 1 °C and 1 atm? Explain.
20.106 When heated, the DNA double helix separates into two random coil single strands. When cooled, the random coils
reform the double helix: double helix = 2 random coils.
(a) What is the sign of S for the forward process? Why?
(b) Energy must be added to break H bonds and overcome dispersion forces between the strands. What is the sign of G for
the forward process when TS is smaller than H?
(c) Write an expression for T in terms of H and S when the reaction is at equilibrium. (This temperature is called the
melting temperature of the nucleic acid.)
20.107 In the process of respiration, glucose is oxidized completely. In fermentation, O 2 is absent and glucose is broken
down to ethanol and CO2. Ethanol is oxidized to CO2 and H2O.
(a) Balance the following equations for these processes :
Respiration: C6H12O6(s) + O2(g)  CO2(g) + H2O(l)
Fermentation: C6H12O6(s)  C2H5OH(l) + CO2(g)
Ethanol oxidation: C2H5OH(l) 1 O2(g)  CO2(g) + H2O(l)
(b) Calculate G°rxn for respiration of 1.00 g of glucose.
(c) Calculate G°rxn for fermentation of 1.00 g of glucose.
(d) Calculate G°rxn for oxidation of the ethanol from part (c).
20.108 Consider the formation of ammonia:
N2(g) + 3H2(g) = 2NH3(g)
(a) Assuming that H° and S° are constant with temperature, find the temperature at which Kp = 1.00.
(b) Find Kp at 400.°C, a typical temperature for NH3 production.
(c) Given the lower Kp at the higher temperature, why are these conditions used industrially?
20.109 Kyanite, sillimanite, and andalusite all have the formula Al2SiO5. Each is stable under different conditions
(see the graph at right). At the point where the three phases intersect:
(a) Which mineral, if any, has the lowest free energy?
(b) Which mineral, if any, has the lowest enthalpy?
(c) Which mineral, if any, has the highest entropy?
(d) Which mineral, if any, has the lowest density?
20.110 Acetylene is produced commercially by the partial oxidation of methane. At 1500°C and pressures of 1-10 bar, the
yield of acetylene is about 20%. The major side product is carbon monoxide, and some soot and carbon dioxide also form.
(a) At what temperature is the desired reaction spontaneous:
1
2CH4 + O2  C2H2 + 2H2 + H2O
2
(b) Acetylene can also be made by the reaction of its elements, carbon (graphite) and hydrogen. At what temperature is this
formation reaction spontaneous?
(c) Why must this reaction mixture be immediately cooled?
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Chapter 20 • Thermodynamics: Entropy, Free Energy, and the Direction of Chemical Reactions
20.111 Synthesis gas, a mixture that includes the fuels CO and H2, is used to produce liquid hydrocarbons and methanol. It
is made at pressures up to 100 atm by oxidation of methane followed by the steam reforming and water gas shift reactions.
Because the process is exothermic, temperatures reach 950-1100°C, and the conditions are such that the amounts of H2,
CO, CO2, CH4, and H2O leaving the reactor are close to the equilibrium amounts for the steam reforming and water gas shift
reactions:
CH4(g) + H2O(g) = CO(g) + 3H2(g) (steam re-forming)
CO(g) + H2O(g) = CO2(g) + H2(g) (water-gas shift)
(a) At 1000.°C, what are G ° and H° for the steam re-forming reaction and for the water-gas shift reaction?
(b) By doubling the steam re-forming step and adding it to the water-gas shift step, we obtain the following combined
reaction:
2CH4(g) + 3H2O(g) = CO2(g) + CO(g) + 7H2(g)
Is this reaction spontaneous at 1000.°C in the standard state?
(c) Is it spontaneous at 98 atm and 50.% conversion (when 50.% of the starting materials have reacted)?
(d) Is it spontaneous at 98 atm and 90.% conversion?