Q41 / 150 · MHT CET 2026 (19 April Shift 1) · Chemistry
Identify from following, the relation between standard free energy change in a reaction and equilibrium constant $K_c$.
- A$\triangle G^\circ = RT \ln K_c$
- B$-\triangle G^\circ = \dfrac{RT}{\ln K_c}$
- C$\triangle G^\circ = \dfrac{\ln K_c}{RT}$
- Correct answerD$-\triangle G^\circ = RT \ln K_c$
Solution
The relation between the free energy change and the reaction quotient is given by $\Delta G = \Delta G^\circ + RT \ln Q$.
At equilibrium, the free energy change $\Delta G = 0$ and the reaction quotient $Q = K_c$.
Substituting these values into the equation, we get $0 = \Delta G^\circ + RT \ln K_c$.
Rearranging the terms yields $-\Delta G^\circ = RT \ln K_c$.
Answer: $-\triangle G^\circ = RT \ln K_c$
At equilibrium, the free energy change $\Delta G = 0$ and the reaction quotient $Q = K_c$.
Substituting these values into the equation, we get $0 = \Delta G^\circ + RT \ln K_c$.
Rearranging the terms yields $-\Delta G^\circ = RT \ln K_c$.
Answer: $-\triangle G^\circ = RT \ln K_c$
Hence, the correct answer is option: 'D'.