Page 164 - Modern physical chemistry

156 Relationships among Reactants

Z 72 Energetic Conditions on Equilibria

A given reaction may proceed spontaneously at temperature T and pressure P as long
as it releases net work. But net work is available for dissipation only as long as the Gibbs
free energy G can decrease. When G can no longer decrease, the point of equilibrium has
been reached.
At such a point, the derivative of G with respect to reaction coordinate A., must vanish.
Since this equals the Gibbs energy of reaction, we have

I'lG=(OGJ =0 [7.54]
OA TP
and equation (7.52) reduces to
O=I'lGo +RTlnK. [7.55]
Here K is the value quotient Q assumes at equilibrium.
Subtracting equation (7.55) from (7.52) yields

flG= RTIn!{. [7.56]
K
When Q > K, flG is positive and the reaction goes spontaneously to the left at the given
T and P. Thus, Q is reduced until it equals K. When Q < K, flG is negative and the reac-
tion proceeds spontaneously to the right at the given T and P. Now Q increases until it
equals K.
At equilibrium, Q equals K and
I m
aLaM =K. [7.57]
a%a~

From (7.55), we have
inK = _ I'lGo [7.58]
RT
or
I'lGo
10gK= [7.59]
2.303RT

ExampleZ8
Determine the equilibrium constant at 25° C for the reaction

1
H2 (g)+-02 (g)~ H20 (g)
2
In table 7.4, log Kr is the logarithm for formation of 1 mole of the compound from its
elements in their standard states. For the given reaction, we have

10gK = (logKf) -(logKf) -1.(logKf) = 40.053 - 0 -1.0 = 40.053.
H20 H2 2 02 2
So
K = 1.130 x 1040 bar- I12 .