Question

For a reaction, ${A \rightleftharpoons P}$, the plots of [A] and [P] with time at temperatures $T_1$ and $T_2$ are given below. If $T_2 > T_1$, the correct statement(s) is (are) (Assume $\Delta H^{\ominus}$ and $\Delta S^{\ominus}$ are independent of temperature and ratio of lnK at $T_1$ to lnK at $T_2$ is greater than $\frac{T_2}{T_1}$ . Here $H, S, G$ and $K$ are enthalpy, entropy, Gibbs energy and equilibrium constant, respectively.)

• A. $\Delta H^{\ominus} < 0 , \Delta S^{\ominus} < 0$
• B. $\Delta G^{\ominus} < 0 , \Delta H^{\ominus} > 0$
• C. $\Delta G^{\ominus} < 0 , \Delta S^{\ominus} < 0$
• D. $\Delta G^{\ominus} < 0 , \Delta S^{\ominus} > 0$

Answer 1

Answer: (C)

$\Delta G^{\ominus} < 0 , \Delta S^{\ominus} < 0$

Answer 2

The correct answer is option (C) : $\Delta G^{\ominus} < 0 , \Delta S^{\ominus} < 0$
${A \rightleftharpoons P}$
given $T_2 > T_1$
$\frac{In K_{1}}{In K_{2}} > \frac{T_{2}}{T_{1} }$
$\Rightarrow T_{1} In k_{1} > T_{2} In k_{2}$
$\Rightarrow - \Delta G^{\circ}_{1} > \- \Delta G^{\circ}_{2}$
$\Rightarrow \left(-\Delta H^{\circ} + T_{1}\Delta S^{\circ}\right) > \left( - \Delta H^{\circ} + T_{2} \Delta S^{\circ} \right)$
$\Rightarrow T_{1} \Delta S^{\circ} > T_{2} \Delta S^{\circ}$
$\Rightarrow \Delta S^{\circ} < 0$