2 Enthalpy, Entropy, and Spontaneous Processes: A Brief Review
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16.2 ENTHALPY, ENTROPY, AND SPONTANEOUS PROCESSES: A BRIEF REVIEW
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and O2 is partly converted to heat, which flows from the system (reactants plus products) to the surroundings:
CH 4(g) + 2 O2(g) ¡ CO2(g) + 2 H 2O(l)
¢H° = -890.3 kJ
Because heat is lost by the system, the reaction is exothermic and the standard
enthalpy of reaction is negative ( ¢H° = -890.3 kJ). The total energy is conserved, so
all the energy lost by the system shows up as heat gained by the surroundings.
Because spontaneous reactions so often give off heat, the nineteenth-century
French chemist Marcellin Berthelot proposed that spontaneous chemical or physical
changes are always exothermic. But Berthelot’s proposal can’t be correct. Ice, for
example, spontaneously absorbs heat from the surroundings and melts at temperatures above 0 °C. Similarly, liquid water absorbs heat and spontaneously boils at
temperatures above 100 °C. As further examples, gaseous N2O4 absorbs heat when it
decomposes to NO2 at 400 K, and table salt absorbs heat when it dissolves in water at
room temperature:
H 2O(s)
H 2O(l)
N2O4(g)
NaCl(s)
¡ H 2O(l)
¡ H 2O(g)
¡ 2 NO2(g)
¡ Na + (aq) + Cl -(aq)
H2O
CO2
¢Hfusion = +6.01 kJ
¢Hvap = +40.7 kJ
¢H° = +55.3 kJ
¢H° = +3.88 kJ
All these processes are endothermic, yet all are spontaneous. In all cases, the system
moves spontaneously to a state of higher enthalpy by absorbing heat from the
surroundings.
Because some spontaneous reactions are exothermic and others are endothermic,
enthalpy alone can’t account for the direction of spontaneous change; a second factor
must be involved. This second determinant of spontaneous change is nature’s tendency to move to a condition of maximum randomness (Section 8.12).
Molecular randomness is called entropy and is denoted by the symbol S. Entropy
is a state function (Section 8.2), and the entropy change ¢S for a process thus depends
only on the initial and final states of the system:
¢S = Sfinal - Sinitial
When the randomness of a system increases, ¢S has a positive value; when randomness decreases, ¢S is negative.
The randomness of a system comes about because the particles in the system
(atoms, ions, and molecules) are in incessant motion, moving about in the accessible volume, colliding with each other and continually exchanging energy.
Randomness—and thus entropy—is a probability concept, related to the number of
ways that a particular state of a system can be achieved. A particular state of a macroscopic system, characterized by its temperature, pressure, volume, and number of
particles, can be achieved in a vast number of ways in which the fluctuating positions and energies of the individual particles differ but the volume and total energy
are constant.
We’ll examine the relationship between entropy and probability in the next section, but first let’s take a qualitative look at the four spontaneous endothermic
processes mentioned previously (melting of ice, boiling of liquid water, decomposition of N2O4, and dissolving of NaCl in water). Each of these processes involves an
increase in the randomness of the system. When ice melts, for example, randomness
increases because the highly ordered crystalline arrangement of tightly held water
molecules collapses and the molecules become free to move about in the liquid.
When liquid water vaporizes, randomness further increases because the molecules
can now move independently in the much larger volume of the gas. In general,
processes that convert a solid to a liquid or a liquid to a gas involve an increase in
randomness and thus an increase in entropy (Figure 16.3).
O2
CH4
᭡ The combustion of natural
gas (mainly CH4) in air is a
spontaneous, exothermic
reaction.
Remember...
A state function is a function or property
whose value depends only on the present
state (condition) of the system, not on the
path used to arrive at that condition. Pressure, volume, temperature, enthalpy, and
entropy are state functions. (Section 8.2)
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Chapter 16 THERMODYNAMICS: ENTROPY, FREE ENERGY, AND EQUILIBRIUM
Figure 16.3
How molecular randomness—and thus
entropy—changes when solids, liquids,
and gases interconvert.
Less randomness
(less entropy)
More randomness
(more entropy)
𝚫S > 0
𝚫S > 0
𝚫S < 0
𝚫S < 0
Melting
Vaporization
Solid
Gas
Liquid
Freezing
Condensation
The decomposition of N2O4 (O2N ¬ NO2) is accompanied by an increase in randomness because breaking the N ¬ N bond allows the two gaseous NO2 fragments
to move independently. Whenever a molecule breaks into two or more pieces, the
amount of molecular randomness increases. More specifically, randomness—and
thus entropy—increases whenever a reaction results in an increase in the number of
gaseous particles (Figure 16.4).
Figure 16.4
How molecular randomness—and thus
entropy—changes when the number of
gaseous particles changes.
Less randomness
(less entropy)
More randomness
(more entropy)
𝚫S > 0
𝚫S < 0
Reaction increases number of gas particles.
N2O4
Remember...
Hydrated ions are surrounded and stabilized
by an ordered shell of solvent water molecules. The stabilization results from ion–dipole
attractions. (Section 11.2)
Reaction decreases number of gas particles.
2 NO2
The entropy change on dissolving sodium chloride in water occurs because the
crystal structure of solid NaCl is disrupted and the Na + and Cl - ions become
hydrated (Section 11.2). Disruption of the crystal increases randomness because the
Na + and Cl - ions are tightly held in the solid but are free to move about in the liquid.
The hydration process, however, decreases randomness because the polar, hydrating
water molecules adopt an orderly arrangement about the Na + and Cl - ions. It turns
out that the overall dissolution process for NaCl results in a net increase in randomness, and ¢S is thus positive (Figure 16.5). This is usually the case for the dissolution
of molecular solids, such as HgCl2, and salts that contain +1 cations and -1 anions.
For salts such as CaSO4, which contain more highly charged ions, the hydrating
water molecules are more strongly attached to the ions and the dissolution process
often results in a net decrease in entropy. The following dissolution reactions illustrate the point:
HgCl2(s) ¡ HgCl2(aq)
NaCl(s) ¡ Na + (aq) + Cl -(aq)
CaSO4(s) ¡ Ca2 + (aq) + SO 42-(aq)
¢S = +9 J/(K # mol)
¢S = +43 J/(K # mol)
¢S = -140 J/(K # mol)
16.2 ENTHALPY, ENTROPY, AND SPONTANEOUS PROCESSES: A BRIEF REVIEW
Less randomness
(less entropy)
More randomness
(more entropy)
H2O
H2O
+
−
− + − + −
+ − + − +
− + − + −
+ − + − +
NaCl − + − + −
NaCl(s) + H2O(l)
The polar H2O molecules are
oriented such that the partially
positive H atoms are near the
anions and the partially negative
O atoms are near the cations.
ΔS > 0
ΔS < 0
−
+
Na+(aq) + Cl–(aq)
Disruption of the crystal increases the entropy, but the hydration process decreases
the entropy. For the dissolution of NaCl, the net effect is an entropy increase.
Figure 16.5
Dissolution of sodium chloride. When NaCl dissolves in water, the crystal breaks up, and
the Na+ and Cl - ions are surrounded by hydrating water molecules.
WORKED EXAMPLE 16.1
PREDICTING THE SIGN OF ≤S
Predict the sign of ¢S in the system for each of the following processes:
(a) CO2(s) ¡ CO 2(g) (sublimation of dry ice)
(b) CaSO 4(s) ¡ CaO(s) + SO 3(g)
(c) N2(g) + 3 H 2(g) ¡ 2 NH 3(g)
(d) I 2(s) ¡ I 2(aq) (dissolution of iodine in water)
STRATEGY
To predict the sign of ¢S, look to see whether the process involves a phase change, a
change in the number of gaseous molecules, or the dissolution (or precipitation) of a
solid. Entropy generally increases for phase transitions that convert a solid to a liquid
or a liquid to a gas, for reactions that increase the number of gaseous molecules, and for
the dissolution of molecular solids or salts with +1 cations and -1 anions.
SOLUTION
(a) The molecules in a gas are free to move about randomly, whereas the molecules in
a solid are tightly held in a highly ordered arrangement. Therefore, randomness
increases when a solid sublimes and ¢S is positive.
(b) One mole of gaseous molecules appears on the product side of the equation and
none appears on the reactant side. Because the reaction increases the number of
gaseous molecules, the entropy change is positive.
(c) The entropy change is negative because the reaction decreases the number of
gaseous molecules from 4 mol to 2 mol. Fewer particles can move independently
after reaction than before.
(d) Iodine molecules are electrically neutral and form a molecular solid. The dissolution process destroys the order of the crystal and enables the iodine molecules to
move about randomly in the liquid. Therefore, ¢S is positive.
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Chapter 16 THERMODYNAMICS: ENTROPY, FREE ENERGY, AND EQUILIBRIUM
Ī PROBLEM 16.2
(a)
(b)
(c)
(d)
Predict the sign of ¢S in the system for each of the following processes:
H 2O(g) ¡ H 2O(l) (formation of rain droplets)
I 2(g) ¡ 2 I(g)
CaCO 3(s) ¡ CaO(s) + CO 2(g)
Ag +(aq) + Br -(aq) ¡ AgBr(s)
WORKED CONCEPTUAL EXAMPLE 16.2
PREDICTING THE SIGN OF ≤S FOR A GAS-PHASE REACTION
Consider the gas-phase reaction of A2 molecules (red) with B atoms (blue):
(a) Write a balanced equation for the reaction.
(b) Predict the sign of ¢S for the reaction.
STRATEGY
To determine the stoichiometry of the reaction, count the number of reactant A2 molecules and B atoms and the number of product AB molecules. To predict the sign of the
entropy change, see if the reaction increases or decreases the number of gaseous particles.
SOLUTION
(a) In this reaction, 3 A2 molecules and 6 B atoms are consumed and 6 AB molecules
are formed (3 A2 + 6 B : 6 AB). Dividing by 3 to reduce the coefficients to their
smallest whole number values gives the balanced equation A2(g) + 2 B(g) :
2 AB(g).
(b) Because the reaction decreases the number of gaseous particles from 3 mol to 2 mol,
the entropy change is negative.
CONCEPTUAL PROBLEM 16.3
Consider the gas-phase reaction of AB3 and A2 molecules:
(a) Write a balanced equation for the reaction.
(b) What is the sign of the entropy change for the reaction?
16.3
᭡ Shaking a box that contains 20 quarters
gives a random arrangement of heads
and tails.
ENTROPY AND PROBABILITY
Why do systems tend to move spontaneously to a state of maximum randomness?
The answer is that a random arrangement of particles is more probable than
an ordered arrangement because a random arrangement can be achieved in more
ways. To begin with a simple example, suppose that you shake a box containing
20 identical coins and then count the number of heads (H) and tails (T). It’s very
unlikely that all 20 coins will come up heads; that is, a perfectly ordered arrangement
of 20 heads (or 20 tails) is much less probable than a random mixture of heads
and tails.
The probabilities of the ordered and random arrangements are proportional to
the number of ways that the arrangements can be achieved. The perfectly ordered
arrangement of 20 heads can be achieved in only one way because it consists of a
16.3 ENTROPY AND PROBABILITY
single configuration. In how many ways, though, can a random arrangement be
achieved? If there were just two coins in the box, each of them could come up in one
of two ways (H or T), and the two together could come up in 2 * 2 = 22 = 4 ways
(HH, HT, TH, or TT). Three coins could come up in 2 * 2 * 2 = 23 = 8 ways (HHH,
THH, HTH, HHT, HTT, THT, TTH, or TTT), and so on. For the case of 20 coins, the
number of possible arrangements is 220 = 1,048,576.
Because the ordered arrangement of 20 heads (or 20 tails) can be achieved in
only one way and a random mixture of heads and tails can be achieved in
220 - 2 L 220 ways, a random arrangement is 220 times more probable than a perfectly ordered arrangement. If you begin with an ordered arrangement of 20 heads
and shake the box, the system will move to a state with a random mixture of heads
and tails because that state is more probable. (Note that the state with a random
arrangement of coins includes all possible arrangements except the two perfectly
ordered arrangements.)
An analogous chemical example is a crystal containing diatomic molecules such
as carbon monoxide in which the two distinct ends of the CO molecule correspond to
the heads and tails of a coin. Let’s suppose that the long dimensions of the molecules
are oriented vertically (Figure 16.6) and that the temperature is 0 K, so that the molecules are locked into a fixed arrangement. The state in which the molecules pack
together in a perfectly ordered “heads-up” arrangement (Figure 16.6a) can be
achieved in only one way, whereas the state in which the molecules are arranged randomly with respect to the vertical direction can be achieved in many ways—220 ways
for a hypothetical crystal containing 20 CO molecules (Figure 16.6b). Therefore, a
structure in which the molecules are arranged randomly is 220 times more probable
than the perfectly ordered heads-up structure.
The Austrian physicist Ludwig Boltzmann proposed that the entropy of a particular state is related to the number of ways that the state can be achieved, according to
the formula
S = k ln W
where S is the entropy of the state, ln W is the natural logarithm of the number of
ways that the state can be achieved, and k, now known as Boltzmann’s constant, is a
universal constant equal to the gas constant R divided by Avogadro’s number
(k = R/NA = 1.38 * 10-23 J/K). Because a logarithm is dimensionless, the Boltzmann equation implies that entropy has the same units as the constant k, joules per
kelvin.
Now let’s apply Boltzmann’s formula to our hypothetical crystal containing
20 CO molecules. Because a perfectly ordered state can be achieved in only one way
(W = 1 in the Boltzmann equation) and because ln 1 = 0, the entropy of the perfectly
ordered state is zero:
S = k ln W = k ln 1
= 0
The more probable state in which the molecules are arranged randomly can be
achieved in 220 ways and thus has a higher entropy:
S = k ln W = k ln 220
= (1.38 * 10-23 J/K) (20) (ln 2)
= 1.91 * 10-22 J/K
where we have made use of the relation ln x a = a ln x (Appendix A.2).
If our crystal contained 1 mol of CO molecules, the entropy of the perfectly
ordered state (6.02 * 1023 C atoms up) would still be zero, but the entropy of the
state with a random arrangement of CO molecules would be much higher because
Avogadro’s number of molecules
can be arranged randomly in a huge number of
23
ways (W = 2NA = 26.02 * 10 ).
(a) The perfectly ordered
“heads-up” structure.
20 “heads”
0 “tails”
(b) The molecules arranged
randomly in one of the 220
ways in which a disordered
structure can be obtained.
9 “heads”
11 “tails”
Figure 16.6
A hypothetical crystal containing
20 CO molecules.
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Chapter 16 THERMODYNAMICS: ENTROPY, FREE ENERGY, AND EQUILIBRIUM
According to Boltzmann’s formula, the entropy of the state with a random
arrangement of CO molecules is
S = k ln W = k ln 2NA = kNA ln 2
Because k = R/NA,
S = R ln 2 = (8.314 J/K) (0.693)
= 5.76 J/K
Remember...
The dipole moment (M) is a measure of
the net polarity of a molecule and is defined
as m = Q * r, where Q is the magnitude of
the charge at either end of the molecular
dipole and r is the distance between the
charges. (Section 10.1)
Dipole–dipole forces result from electrical
interactions among neighboring polar molecules. (Section 10.2)
Based on experimental measurements, the entropy of 1 mol of solid carbon monoxide near 0 K is about 5 J/K, indicating that the CO molecules adopt a nearly random
arrangement. Entropy associated with a random arrangement of molecules in space
is sometimes called positional, or configurational, entropy.
The nearly random arrangement of CO molecules in crystalline carbon monoxide
is unusual but can be understood in terms of molecular structure. Because CO molecules have a dipole moment of only 0.11 D, intermolecular dipole–dipole forces are
unusually weak (Sections 10.1 and 10.2), and the molecules therefore have little
preference for a slightly lower energy, completely ordered arrangement. By contrast,
HCl, with a larger dipole moment of 1.11 D, forms an ordered crystalline solid, and
so the entropy of 1 mol of solid HCl at 0 K is 0 J/K.
Boltzmann’s formula also explains why a gas expands into a vacuum. If the two
bulbs in Figure 16.1 have equal volumes, each molecule has one chance in two of
being in bulb A (heads, in our coin example) and one chance in two of being in bulb
B (tails) when the stopcock is opened. It’s exceedingly unlikely that all the molecules
in 1 mol of gas will be in bulb A because that state can be achieved in only one way.
The state in which Avogadro’s number of molecules
are randomly distributed
23
between bulbs A and B can be achieved in 26.02 * 10 ways, and the entropy of that
state is therefore higher than the entropy of the ordered state by the now familiar
amount, R ln 2 = 5.76 J/K. Thus, a gas expands spontaneously because the state of
greater volume is more probable.
We can derive a general equation for the entropy change that occurs on the
expansion of an ideal gas at constant temperature by considering the distribution of
N molecules among B hypothetical boxes, or cells, each having an equal volume v.
Volume per box = v
Number of accessible boxes = B
Total volume, V = Bv
Remember...
According to the kinetic–molecular theory,
the kinetic energy of 1 mol of an ideal
gas equals 3RT/2 and is independent of
pressure and volume. (Section 9.6)
Since the energy of an ideal gas depends only on the temperature (Section 9.6), ¢E for
the expansion of an ideal gas at constant temperature is zero. To calculate the entropy
change ¢S = Sfinal - Sinitial using the Boltzmann equation, we have only to find the
number of ways N molecules can be distributed among the B boxes.
A single molecule can go into any one of the boxes and can thus be assigned to
B boxes in B ways. Two molecules can occupy the boxes in B * B = B 2 ways, three
molecules can fill the boxes in B * B * B = B 3 ways, and so on. The number of ways
that N molecules can occupy B boxes is W = B N.
Now suppose that the initial volume comprises Binitial boxes and the final volume
consists of Bfinal boxes:
Vinitial = Binitialv
and
Vfinal = Bfinalv
Then the probabilities of the initial and final states—that is, the number of ways they
can be achieved—are
Winitial = (Binitial) N
and
Wfinal = (Bfinal) N
16.4 ENTROPY AND TEMPERATURE
According to the Boltzmann equation, the entropy change due to a change in volume is
¢S = Sfinal - Sinitial = k ln Wfinal - k ln Winitial = k ln
= k lna
Wfinal
Winitial
Bfinal N
Bfinal
b = kN ln
Binitial
Binitial
Because Vinitial = Binitial v and Vfinal = Bfinalv,
¢S = kN lna
Vfinal/v
Vfinal
b = kN ln
Vinitial/v
Vinitial
Finally, because k = R/NA and the number of particles equals the number of
moles of gas times Avogadro’s number (N = nNA), then
kN = a
R
b(nNA) = nR
NA
and so the entropy change for expansion (or compression) of n moles of an ideal gas
at constant temperature is
¢S = nR ln
Vfinal
Vinitial
For a twofold expansion of 1 mol of an ideal gas at constant temperature, ¢S = R ln 2,
the same result as we obtained previously.
Because the pressure and volume of an ideal gas are related inversely (P =
nRT/V), we can also write
¢S = nR ln
Pinitial
Pfinal
Thus, the entropy of a gas increases when its pressure decreases at constant temperature, and the entropy decreases when its pressure increases. Common sense tells us
that the more we squeeze the gas, the less space the gas molecules have and so
randomness decreases.
Ī PROBLEM 16.4
Which state has the higher entropy? Explain in terms of probability.
(a) A perfectly ordered crystal of solid nitrous oxide (N ‚ N ¬ O) or a disordered crystal in which the molecules are oriented randomly
(b) Quartz glass (Section 10.10) or a quartz crystal
(c) 1 mol of N2 gas at STP or 1 mol of N2 gas at 273 K in a volume of 11.2 L
(d) 1 mol of N2 gas at STP or 1 mol of N2 gas at 273 K and 0.25 atm
16.4
ENTROPY AND TEMPERATURE
Thus far we’ve seen that entropy is associated with the orientation and distribution
of molecules in space. Disordered crystals have higher entropy than ordered crystals,
and expanded gases have higher entropy than compressed gases.
Entropy is also associated with molecular motion. As the temperature of a substance increases, random molecular motion increases and there is a corresponding
increase in the average kinetic energy of the molecules. But not all the molecules
have the same energy. As we saw in Section 9.6, there is a distribution of molecular
speeds in a gas, a distribution that broadens and shifts to higher speeds with increasing temperature (Figure 9.12, page 328). In solids, liquids, and gases, the total energy
of a substance can be distributed among the individual molecules in a number of
ways that increases as the total energy increases. According to Boltzmann’s formula,
the more ways that the energy can be distributed, the greater the randomness of the
state and the higher its entropy. Therefore, the entropy of a substance increases with
increasing temperature (Figure 16.7).
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Chapter 16 THERMODYNAMICS: ENTROPY, FREE ENERGY, AND EQUILIBRIUM
Higher temperature:
•Greater molecular motion
•Broader distribution of individual molecular energies
•More randomness
•Higher entropy
Lower temperature:
•Less molecular motion
•Narrower distribution of individual molecular energies
•Less randomness
•Lower entropy
Figure 16.7
A substance at a higher temperature has greater entropy than the same substance at a lower temperature.
A typical plot of entropy versus temperature is shown in Figure 16.8. At absolute
zero, every substance is a solid whose particles are tightly held in a crystalline structure. If there is no residual orientational disorder, like that in carbon monoxide
(Figure 16.6b), the entropy of the substance at 0 K will be zero, a general result
summarized in the third law of thermodynamics:
Third Law of Thermodynamics The entropy of a perfectly ordered
crystalline substance at 0 K is zero.
(The first law of thermodynamics was discussed in Section 8.1. We’ll review the first
law and discuss the second law in Section 16.6.)
Entropy
Gas
Liquid
Figure 16.8
Entropy versus temperature. The
entropy of a pure substance, equal to
zero at 0 K, shows a steady increase with
rising temperature, punctuated by
discontinuous jumps in entropy at the
temperatures of the phase transitions.
Solid
mp
Temperature (K)
bp
16.5 STANDARD MOLAR ENTROPIES AND STANDARD ENTROPIES OF REACTION
651
As the temperature of a solid is raised, the added energy increases the vibrational
motion of the molecules about their equilibrium positions in the crystal. The number
of ways in which the vibrational energy can be distributed increases with rising temperature, and the entropy of the solid thus increases steadily as the temperature
increases.
At the melting point, there is a discontinuous jump in entropy because there are
many more ways of arranging the molecules in the liquid than in the solid. Furthermore, the molecules in the liquid can undergo translational and rotational as well as
vibrational motion, and so there are many more ways of distributing the total energy
in the liquid. (Translational motion is motion of the center of mass.) An even greater
jump in entropy is observed at the boiling point because molecules in the gas are free
to occupy a much larger volume. Between the melting point and the boiling point, the
entropy of a liquid increases steadily as molecular motion increases and the number
of ways of distributing the total energy among the individual molecules increases. For
the same reason, the entropy of a gas rises steadily as its temperature increases.
16.5
STANDARD MOLAR ENTROPIES AND
STANDARD ENTROPIES OF REACTION
We won’t describe how the entropy of a substance is determined, except to note that
two approaches are available: (1) calculations based on Boltzmann’s formula and (2)
experimental measurements of heat capacities (Section 8.7) down to very low temperatures. Suffice it to say that standard molar entropies, denoted by S°, are known
for many substances.
Remember...
The molar heat capacity is the amount of
heat needed to raise the temperature of
1 mol of a substance by 1 °C. (Section 8.7)
Standard Molar Entropy, S° The entropy of 1 mol of the pure substance
at 1 atm pressure and a specified temperature, usually 25 °C.
Values of S° for some common substances at 25 °C are listed in Table 16.1, and
additional values are given in Appendix B. Note that the units of S° are joules (not
kilojoules) per kelvin mole [J/(K # mol)]. Standard molar entropies are often called
absolute entropies because they are measured with respect to an absolute reference
point—the entropy of the perfectly ordered crystalline substance at 0 K
[S° = 0 J/(K # mol) at T = 0 K].
Standard molar entropies make it possible to compare the entropies of different
substances under the same conditions of temperature and pressure. It’s apparent
from Table 16.1, for example, that the entropies of gaseous substances tend to be
TABLE 16.1
Standard Molar Entropies for Some Common Substances at 25 °C
Substance
Formula
Gases
Acetylene
Ammonia
Carbon dioxide
Carbon monoxide
Ethylene
Hydrogen
Methane
Nitrogen
Nitrogen dioxide
Dinitrogen tetroxide
Oxygen
C2H2
NH3
CO2
CO
C2H4
H2
CH4
N2
NO2
N2O4
O2
#
S° [J/(K mol)]
200.8
192.3
213.6
197.6
219.5
130.6
186.2
191.5
240.0
304.3
205.0
Substance
Liquids
Acetic acid
Ethanol
Methanol
Water
Solids
Calcium carbonate
Calcium oxide
Diamond
Graphite
Iron
Iron(III) oxide
Formula
CH3CO2H
CH3CH2OH
CH3OH
H2O
CaCO3
CaO
C
C
Fe
Fe2O3
#
S° [J/(K mol)]
160
161
127
69.9
91.7
38.1
2.4
5.7
27.3
87.4
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Chapter 16 THERMODYNAMICS: ENTROPY, FREE ENERGY, AND EQUILIBRIUM
larger than those of liquids, which, in turn, tend to be larger than those of solids.
Table 16.1 also shows that S° values increase with increasing molecular complexity.
Compare, for example, CH3OH, which has S° = 127 J/(K # mol), and CH3CH2OH,
which has S° = 161 J/(K # mol).
Once we have values for standard molar entropies, it’s easy to calculate the
entropy change for a chemical reaction. The standard entropy of reaction ( ≤S°) can
be obtained simply by subtracting the standard molar entropies of all the reactants
from the standard molar entropies of all the products:
¢S° = S°(products) - S°(reactants)
Because S° values are quoted on a per-mole basis, the S° value for each substance
must be multiplied by the stoichiometric coefficient of that substance in the balanced
chemical equation. Thus, for the general reaction
aA + bB ¡ cC + dD
the standard entropy of reaction is
¢S° = [c S°(C) + d S°(D)] - [a S°(A) + b S°(B)]
where the units of the coefficients are moles, the units of S° are J/(K # mol), and the
units of ¢S° are J/K.
As an example, let’s calculate the standard entropy change for the reaction
N2O4(g) ¡ 2 NO2(g)
Using the appropriate S° values from Table 16.1, we find that ¢S° = 175.7 J/K:
¢S° = 2 S°(NO2) - S°(N2O4)
J
J
= (2 mol)a240.0 #
b - (1 mol)a304.3 #
b
K mol
K mol
= 175.7 J/K
Although the standard molar entropy of N2O4 is larger than that of NO2, as expected
for a more complex molecule, ¢S° for the reaction is positive because 1 mol of N2O4
is converted to 2 mol of NO2. As noted earlier, we expect an increase in entropy
whenever a molecule breaks into two or more pieces.
WORKED EXAMPLE 16.3
CALCULATING THE STANDARD ENTROPY OF REACTION
Calculate the standard entropy of reaction at 25 °C for the Haber synthesis of ammonia:
N2(g) + 3 H 2(g) ¡ 2 NH 3(g)
STRATEGY
To calculate ¢S° for the reaction, subtract the standard molar entropies of all the reactants from the standard molar entropies of all the products. Look up the S° values in
Table 16.1 or Appendix B, and remember to multiply the S° value for each substance by
its coefficient in the balanced chemical equation.
SOLUTION
¢S° = 2 S°(NH 3) - [S°(N2) + 3 S°(H 2)]
J
J
J
b - c (1 mol) a191.5 #
b + (3 mol) a130.6 #
bd
= (2 mol)a192.3 #
K mol
K mol
K mol
= -198.7 J/K
BALLPARK CHECK
As predicted in Worked Example 16.1c, ¢S° should be negative because the reaction
decreases the number of gaseous molecules from 4 mol to 2 mol.
16.6 ENTROPY AND THE SECOND LAW OF THERMODYNAMICS
Ī PROBLEM 16.5 Calculate the standard entropy of reaction at 25 °C for the decomposition of calcium carbonate:
CaCO3(s) ¡ CaO(s) + CO 2(g)
16.6
ENTROPY AND THE SECOND LAW
OF THERMODYNAMICS
We’ve seen thus far that molecular systems tend to move spontaneously toward a
state of minimum enthalpy and maximum entropy. In any particular reaction,
though, the enthalpy of the system can either increase or decrease. Similarly, the
entropy of the system can either increase or decrease. How, then, can we decide
whether a reaction will occur spontaneously? In Section 8.13, we said that it is the
value of the free-energy change, ¢G, that is the criterion for spontaneity, where
¢G = ¢H - T¢S. If ¢G 6 0, the reaction is spontaneous; if ¢G 7 0, the reaction is
nonspontaneous; and if ¢G = 0, the reaction is at equilibrium. In this section and the
next, we’ll see how that conclusion was reached. Let’s begin by looking at the first
and second laws of thermodynamics:
First Law of Thermodynamics
In any process, spontaneous or nonspontaneous, the total energy of a
system and its surroundings is
constant.
Second Law of Thermodynamics
In any spontaneous process, the total
entropy of a system and its surroundings always increases.
The first law is simply a statement of the conservation of energy (Section 8.1). It
says that energy (or enthalpy) can flow between a system and its surroundings but
the total energy of the system plus the surroundings always remains constant. In an
exothermic reaction, the system loses enthalpy to the surroundings; in an endothermic reaction, the system gains enthalpy from the surroundings. Because energy is
conserved in all chemical processes, spontaneous and nonspontaneous, the first law
helps us keep track of energy flow between the system and the surroundings but it
doesn’t tell us whether a particular reaction will be spontaneous or nonspontaneous.
The second law, however, provides a clear-cut criterion of spontaneity. It says
that the direction of spontaneous change is always determined by the sign of the
total entropy change:
¢Stotal = ¢Ssystem + ¢Ssurroundings
Specifically,
If ¢Stotal 7 0, the reaction is spontaneous.
If ¢Stotal 6 0, the reaction is nonspontaneous.
If ¢Stotal = 0, the reaction mixture is at equilibrium.
All reactions proceed spontaneously in the direction that increases the entropy of the
system plus surroundings. A reaction that is nonspontaneous in the forward direction
is spontaneous in the reverse direction because ¢Stotal for the reverse reaction equals
- ¢Stotal for the forward reaction. If ¢Stotal is zero, the reaction doesn’t go spontaneously in either direction, and so the reaction mixture is at equilibrium.
To determine the value of ¢Stotal, we need values for the entropy changes in the
system and the surroundings. The entropy change in the system, ¢Ssys, is just the
entropy of reaction, which can be calculated from standard molar entropies (Table
16.1), as described in Section 16.5. For a reaction that occurs at constant pressure, the
entropy change in the surroundings is directly proportional to the enthalpy change
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