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 $T_2 / T_1$ . 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: (A), (C)

$A \rightleftharpoons P$
given $T_2 > T_1$
$\frac{\ln K_{1}}{\ln K_{2}} > \frac{T_{2}}{T_{1} }$
$ \Rightarrow T_{1} \ln k_{1} > T_{2} \ln 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 $