Page 143 - Essentials of physical chemistry

Gibbs’ Free Energy and Equilibria 105

We need to make a point here about equilibrium reactions, they are dynamic. A naive idea of an
equilibrium is that it oozes in one direction, sets up concentrations, and then coagulates into a sort of
semistable pudding. That is far from the truth; the double arrow ( ƒ ) is a very active process in a
ƒ!
never ceasing reaction in both directions. Modern spectroscopy has shown that even the H atoms in
organic compounds can exchange with other H atoms, so a lot of activity is going on in seemingly
stable compounds and if there is some sort of reaction it can often reverse itself. Equilibrium
reactions are constantly going in both directions even though one direction may be favored.
Of course, temperature can affect the extent of the equilibrium in either direction.

Example 1
Suppose we place 0.300 mol of H 2 and 0.100 mol of D 2 in a 2.00 L vessel at 258C.

2HD;
H 2(g) þ D 2(g) ƒ! DG ¼ 0:700 kcal ¼ 2:929 kJ ½2Š
ƒ
(0:300 x) (0:100 x) (2x)

þ2929 J=mol
DG 0 ¼ RT ln K P ,so ¼ ln K P ffi 1:1816. Note that the logarithm
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(8:314 J= K mol)(298:15 K)

comes out to be a pure number after all the units cancel as it must, which is a good way to check the
units. The 1 mol value for DG 0 is based on the balanced reaction for 1 mol H 2 . This reaction is
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chosen in our idea of essential physical chemistry to illustrate several points at once, but let us not
miss the points. This is an exchange reaction of H and D where only gas phase collisions occur and
this implies there is some sort of microscopic mechanism in operation, although as usual the
thermodynamics does not give any information on the mechanism. Second, the K P value is a
constant meaning that there is a fixed relationship between the concentrations. Third, the gases
are in the same container so we can convert pressures directly to concentrations. Fourth, note that, as
written, the DG 0 value is negative. The main idea of DG 0 is that it is negative for a reaction
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which ‘‘tends’’ to the right as written based on standard state values, but that does not mean it will go
to completion. A positive value for DG 0 would mean that the reaction ‘‘tends’’ to go to the left, but
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not that it would not occur at all. Generally, all reactions that have a negative DG 0 will occur to
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some extent, but here we are dealing with the case of DG 0 ¼ 0, the condition for an equilibrium.
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This example also introduces the idea of ‘‘x’’ as a ‘‘molar reaction coordinate, the extent of the
reaction’’; it is the molar amount of the reaction that occurs on the left side and shows up as moles of
product on the right side of the reaction. We will use this same idea when we treat reaction kinetics.
Keeping in mind our emphasis on the idea of an equilibrium as a dynamic process, we are actually
treating this in a similar way to a kinetics problem but without any mention of time. Let us solve
for ‘‘x.’’

2
RT
2 n HD 2
P HD V (2x)
K P ¼ e 1:1816 ¼ 3:2596 ¼ ¼ ¼
RT RT (0:3 x)(0:1 x)
P H 2 Š P D 2
½ ½ Š
n H 2 n D 2
V V

RT
Note that the pressures are low enough that the ideal gas law is accurate and the factors of
V
2
2
all cancel in the equilibrium expression. It is easy to see that (3:2596)(0:03 0:4x þ x ) ¼ 4x ,
2
which leads to 0:2271x þ 0:400x 0:030 ¼ 0 and can be solved using the quadratic