Burdge/Overby, Chemistry: Atoms First, 2e Ch14 | Page 23

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CHAPTE R 14? Entropy and Free Energy

We can make a reliable estimate of that temperature as follows. First we calculate ?H° and ?S° for
the reaction at 25°C, using the data in Appendix 2. To determine ?H°, we apply Equation 10.18:

?H° = [?H °(CaO) + ?H °(CO2)] ? [?H °(CaCO3)]
f
f
f

= [(–635.6 kJ/mol) + (–393.5 kJ/mol)] ? (–1206.9 kJ/mol)
= 177.8 kJ/mol
Next we apply Equation 14.6 to find ?S°:

?S° = [S°(CaO) + S°(CO2)] ? S°(CaCO3)

= [(39.8 J/K ? mol) + (213.6 J/K ? mol)] ? (92.9 J/K ? mol)
= 160.5 J/K ? mol
From Equation 14.10, we can write
?G° = ?H° ? T?S°
and we obtain
Student Annotation: ?Be careful with
units in problems of this type. S° values are
tabulated using joules, whereas ?H° values
f
are tabulated using kilojoules.

?G° = (177.8 kJ/mol) ? (298 K)(0.1605 kJ/K ? mol)

= 130.0 kJ/mol
Because ?G° is a large positive number, the reaction does not favor product formation at 25°C
(298 K). And, because ?H° and ?S° are both positive, we know that ?G° will be negative (product
formation will be favored) at high temperatures. We can determine what constitutes a high temperature for this reaction by calculating the temperature at which ?G° is zero.
0 = ?H° ? T?S°
or
?H°
T = ??____??
?
?S°
(177.8 kJ|??mol)
_______________??
??
= ????
0.1605 kJ/K ? mol
= 1108 K (835°C)
At temperatures higher than 835°C, ?G° becomes negative, indicating that the reaction would then
favor the formation of CaO and CO2. At 840°C (1113 K), for example,
?G° = ?H° ? T?S°

(?

)

1 kJ
= 177.8 kJ/mol ? (1113 K)(0.1605 kJ/K ? mol) ? ______????
??
??
1000 J
= –0.8 kJ/mol
At still higher temperatures, ?G° becomes increasingly negative, thus favoring product formation
even more. Note that in this example we used the ?H° and ?S° values at 25°C to calculate changes
to ?G° at much higher temperatures. Because both ?H° and ?S° actually change with temperature,
this approach does not give us a truly accurate value for ?G°, but it does give us a reasonably good
estimate.
Equation 14.10 can also be used to calculate the change in entropy that accompanies a phase
change. Recall that at the temperature at which a phase change occurs both phases of a substance
are present. For example, at the freezing point of water, both liquid water and solid ice coexist in
a state of equilibrium [9 Section 12.5], where ?G is zero. Therefore, Equation 14.10 becomes
0 = ?H ? T?S
or
?H
?S = ___?
?? ??
T
Consider the ice-water equilibrium. For the ice-to-water transition, ?H is the molar heat of fusion
(see Table 12.7) and T is the melting point. The entropy change is therefore
?Sice

water

6010 J/mol
??
???
?=
= __________?22.0 J/K ? mol
273 K

Thus, when 1 mole of ice melts at 0°C, there is an increase in entropy of 22.0 J/K ? mol. The
increase in entropy is consistent with the increase in the number of possible arrangements of

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