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Match List-I (Equations) with List-II (Type of process) and select the correct option.
Column I
- a. $K_p > Q$
- b. $\Delta G^\circ < RT\ln Q$
- c. $K_p = Q$
- d. $T > \frac{\Delta H}{\Delta S}$
Column II
- i. Non-spontaneous
- ii. Equilibrium
- iii. Spontaneous and endothermic
- iv. Spontaneous
Correct answer: a → iv, b → i, c → ii, d → iii
Detailed Solution
a. $K_p > Q$: since $\Delta G = RT\ln\frac{Q}{K}$, Q < K makes $\Delta G$ negative, so the reaction proceeds in the forward direction: spontaneous.
b. $\Delta G^\circ < RT\ln Q$: as written in the source, this corresponds to a positive $\Delta G$, i.e., a non-spontaneous process.
c. $K_p = Q$: the reaction quotient equals the equilibrium constant, so $\Delta G = 0$ and the system is at equilibrium.
d. $T > \frac{\Delta H}{\Delta S}$: this gives $T\Delta S > \Delta H$ (with $\Delta H$ and $\Delta S$ both positive), so $\Delta G = \Delta H - T\Delta S < 0$; the process is spontaneous and endothermic.
Correct match: a-iv, b-i, c-ii, d-iii.
b. $\Delta G^\circ < RT\ln Q$: as written in the source, this corresponds to a positive $\Delta G$, i.e., a non-spontaneous process.
c. $K_p = Q$: the reaction quotient equals the equilibrium constant, so $\Delta G = 0$ and the system is at equilibrium.
d. $T > \frac{\Delta H}{\Delta S}$: this gives $T\Delta S > \Delta H$ (with $\Delta H$ and $\Delta S$ both positive), so $\Delta G = \Delta H - T\Delta S < 0$; the process is spontaneous and endothermic.
Correct match: a-iv, b-i, c-ii, d-iii.