Question

The correct relationship between free energy change in a reaction and the corresponding equilibrium constant, ${{\text{K}}_{\text{C}}}$is:
A. $\Delta {\text{G}}\, = \,{\text{RT}}\,{\text{ln}}{{\text{K}}_{\text{C}}}$
B. $ - \Delta {\text{G}}\, = \,{\text{RT}}\,{\text{ln}}{{\text{K}}_{\text{C}}}$
C. $\Delta {{\text{G}}^0}\, = \,{\text{RT}}\,{\text{ln}}{{\text{K}}_{\text{C}}}$
D. $ - \Delta {{\text{G}}^0}\, = \,{\text{RT}}\,{\text{ln}}{{\text{K}}_{\text{C}}}$

Answer 1

Hint: We have to determine the relationship between Gibbs free energy and equilibrium constant. Gibbs free energy is related to the emf of the cell and equilibrium constant is also related to the emf of the cell. So, by comparing both equations we will determine the relationship between Gibbs free energy and equilibrium constant.

Complete step-by-step answer:The relation between equilibrium constant and e.m.f of the cell is given by the Nernst equation which is as follows:
${{\text{E}}_{{\text{cell}}}}\,{\text{ = }}\,{\text{E}}_{{\text{cell}}}^0 - \dfrac{{{\text{RT}}}}{{{\text{nF}}}}{\text{ln}}\,\dfrac{{{\text{oxidised}}}}{{{\text{reduced}}}}$
Where,
${\text{n}}$is the number of electrons transferred.
F is the Faraday constant.
${{\text{E}}_{{\text{cell}}}}$is the e.m.f of the cell.
${\text{E}}_{{\text{cell}}}^0$is the standard reduction potential of the cell.
R is the gas constant.
T is the temperature.
The ratio of the concentration of oxidized and reduced species at equilibrium is written as ${{\text{K}}_{\text{C}}}$. So,
${{\text{E}}_{{\text{cell}}}}\,{\text{ = }}\,{\text{E}}_{{\text{cell}}}^0 - \dfrac{{{\text{RT}}}}{{{\text{nF}}}}{\text{ln}}\,\,{{\text{K}}_{\text{C}}}$…..$(1)$
We will multiply the equation $(1)$with -nF as follows:
$ - {\text{nF}}{{\text{E}}_{{\text{cell}}}}\,{\text{ = }}\, - {\text{nFE}}_{{\text{cell}}}^0 - \left( { - {\text{nF}}\dfrac{{{\text{RT}}}}{{{\text{nF}}}}{\text{ln}}\,{{\text{K}}_{\text{C}}}} \right)$
$ - {\text{nF}}{{\text{E}}_{{\text{cell}}}}\,{\text{ = }}\, - {\text{nFE}}_{{\text{cell}}}^0 + \,{\text{RT}}\,{\text{ln}}\,{{\text{K}}_{\text{C}}}$….$(2)$
The relation between Gibbs free energy and the e.m.f of the cell is as follows:
$\Delta {\text{G}}\,{\text{ = }}\,\, - \,{\text{nF}}{{\text{E}}_{{\text{cell}}}}$….$(3)$
Where,
$\Delta {\text{G}}$ is the Gibbs free energy
At equilibrium, $\Delta {\text{G}}$ is written as $\Delta {{\text{G}}^0}$ and ${{\text{E}}_{{\text{cell}}}}$ is written as ${\text{E}}_{{\text{cell}}}^0$. So,
$\Delta {{\text{G}}^0}{\text{ = }}\,\, - \,{\text{nFE}}_{{\text{cell}}}^0$….$(4)$
On substituting the value of $\Delta {\text{G}}$ and$\Delta {{\text{G}}^0}$ from equation $(3)$and $(4)$ in equation $(2)$,
$\Delta {\text{G}}\,{\text{ = }}\,\Delta {{\text{G}}^0} + \,{\text{RT}}\,{\text{ln}}\,{{\text{K}}_{\text{C}}}$
At equilibrium $\Delta {\text{G}}$becomes zero so,
$\Delta {{\text{G}}^0} = \, - \,{\text{RT}}\,{\text{ln}}\,{{\text{K}}_{\text{C}}}$
Or
$ - \Delta {{\text{G}}^0} = \,\,{\text{RT}}\,{\text{ln}}\,{{\text{K}}_{\text{C}}}$
So, the correct relationship between free energy change in a reaction and the corresponding equilibrium constant, ${{\text{K}}_{\text{C}}}$is $ - {\text{\Delta }}{{\text{G}}^0}\, = \,{\text{RT}}\,{\text{ln}}{{\text{K}}_{\text{C}}}$.

Therefore, option (D) $ - {{\Delta }}{{\text{G}}^0}\, = \,{\text{RT}}\,{\text{ln}}{{\text{K}}_{\text{C}}}$, is correct.

Note:The Gibbs free energy is the energy which is used in work. The Gibbs free energy is related to the enthalpy and entropy as follows:
$\Delta {\text{G}}\,{\text{ = }}\,\Delta {\text{H}}\, - {\text{T}}\Delta {\text{S}}$
At equilibrium, the rate of forward direction and backward direction reaction becomes equal, so the concentration on both sides of the reaction becomes equal. So, entropy and enthalpy change tends to zero so the free energy change also tends to zero. The negative value of Gibbs free energy tells the reaction is spontaneous. The positive value of Gibbs free energy tells the reaction is nonspontaneous.