Unlike the mass action expression, the values for the concentrations are not necessarily those at
equilibrium. Because the concentrations of the species change as the reaction progresses towards
equilibrium, the value of Q, and hence also the value of ΔG, will change as the reaction progresses.
As pointed out above, when the system is at equilibrium ΔG is zero, and in that case we would
obtain from the equation
0 = ΔGº + RT ln Qeqwhere Qeq is the value of the reaction quotient at equilibrium. But that is just K, the equilibrium
constant! Hence we can rewrite and rearrange to get
ΔGº = −RT ln Kor equivalently,
K = e−∆G ̊/RTThus, we have obtained the final missing link to put it all together: the more negative ΔG° is, the
larger the equilibrium constant, and hence the more the products are favored at equilibrium.
Examples
H 2 O (l) + heat → H 2 O (g)When water boils, hydrogen bonds (H-bonds) are broken. Energy is absorbed (the reaction is
endothermic), and thus ΔH is positive. Entropy increases as the closely packed molecules of the
liquid become the more randomly moving molecules of a gas; thus, TΔS is also positive. Since ΔH
and TΔS are both positive, the reaction will proceed spontaneously only if TΔS > ΔH. For the
particular values of ΔS and ΔH applicable for this reaction, this condition is true only at
temperatures above 100°C. Below 100°C, ΔG is positive and the water remains a liquid. At 100°C, ΔH
= TΔS and ΔG = 0: an equilibrium is established between water and water vapor. The opposite is true
a. Vaporization of water at one atmosphere pressure