7.0 Relationship Between $\Delta {G^ \circ }\ and\ K$

$$\Delta G = \Delta {G^ \circ } + RT\ln Q$$

At equilibrium $$\Delta G = 0$$

$$\therefore \Delta {G^ \circ } = - RT\ln Q$$

Under equilibrium condition, $$Q = {K_p} = {K_c} = K$$ and $$\Delta G = 0$$ Therefore, we get $$\begin{equation} \begin{aligned} \Delta {G^ \circ } = - RT\ln K \\ \Delta {G^ \circ } = - 2.303RT\log K \\\end{aligned} \end{equation} $$

Antilog will give

$$K = {e^{ - \frac{{\Delta {G^ \circ }}}{{RT}}}}$$ and therefore,

1. For ${\Delta {G^ \circ } < 0} \ \ ,\ \ \ \ \ K > 1$

2. For ${\Delta {G^ \circ } > 0} \ \ ,\ \ \ \ \ K < 1$