Question
The correct relationship between free energy and equilibrium constant $$K$$ of a reaction is
A.
$$\Delta {G^ \circ } = - RT\,\ln \,K$$
B.
$$\Delta G = RT\,\ln \,K$$
C.
$$\Delta {G^ \circ } = RT\,\ln \,K$$
D.
$$\Delta G = - RT\,\ln \,K$$
Answer :
$$\Delta {G^ \circ } = - RT\,\ln \,K$$
Solution :
The Gibbs free energy of a reaction, $${\Delta _r}G$$ is related to the composition of the reaction mixture and the standard reaction Gibbs free energy $${\Delta _r}{G^ \circ }$$ as
$${\Delta _r}G = {\Delta _r}{G^ \circ } + RT\,\ln \,Q$$
where, $$Q = $$ reaction quotient
At equilibrium $$Q = K$$ and $${\Delta _r}G = 0.$$
Therefore, the above reaction becomes
$$\eqalign{
& 0 = {\Delta _r}{G^ \circ } + RT\,\ln \,K \cr
& {\Delta _r}{G^ \circ } = - RT\,\ln \,K \cr
& {\text{or}}\,\,{\Delta _r}{G^ \circ } = - 2.303\,RT\,\log \,K \cr
& K = {\text{equilibrium constant}} \cr} $$