Match List-I (Equations/Conditions) with List-II (Type of processes) and select the correct option.

**List-I

Question

Match List-I (Equations/Conditions) with List-II (Type of processes) and select the correct option.

**List-I (Equations)**
(a) $K_p > Q$
(b) $T < \frac{\Delta H}{\Delta S}$ (for $\Delta H > 0$)
(c) $K_p = Q$
(d) $T > \frac{\Delta H}{\Delta S}$ (for $\Delta H > 0$)

**List-II (Type of processes)**
(i) Non-spontaneous
(ii) Equilibrium
(iii) Spontaneous and endothermic
(iv) Spontaneous

Accepted Answer

1. **(a) $K_p > Q$**: When the reaction quotient ($Q$) is less than the equilibrium constant ($K_p$), the reaction proceeds in the forward direction to reach equilibrium. Therefore, the forward process is **Spontaneous (iv)**. [NCERT Chem 11, Equilibrium, Sec 6.3] 2. **(c) $K_p = Q$**: When $Q$ equals $K_p$, the system is at **Equilibrium (ii)**. [NCERT Chem 11, Equilibrium, Sec 6.3] 3. **(d) $T > \frac{\Delta H}{\Delta S}$**: For an endothermic reaction ($\Delta H > 0$), spontaneity requires $\Delta G = \Delta H - T\Delta S < 0$. This occurs when $T\Delta S > \Delta H$, or $T > \frac{\Delta H}{\Delta S}$. Thus, the process is **Spontaneous and endothermic (iii)**. [NCERT Chem 11, Thermodynamics, Sec 5.6] 4. **(b) $T < \frac{\Delta H}{\Delta S}$**: Conversely, if the temperature is low ($T\Delta S < \Delta H$) for an endothermic reaction, $\Delta G$ will be positive. Thus, the process is **Non-spontaneous (i)**.

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