For a reaction, $A \rightleftharpoons P$, the plots of $[A]$ and $[P]$ with time at temperatures $T_

Question

For a reaction, $A ightleftharpoons P$, 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 $\Delta H^{\circ}$ and $\Delta S^{\circ}$ are independent of temperature and ratio of $\ln K$ at $T_{1}$ to $\ln K$ 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.)\

Professor PC Thomas Classes · Answer

$(\mathrm{a}, \mathrm{c}): \frac{\ln K_{1}}{\ln K_{2}}>\frac{T_{2}}{T_{1}}$ On increasing temperature, concentration of product decreases and hence, $K$ decreases.\ Since, reaction is exothermic, therefore, $\Delta H^{\circ}<0$\ From the graph, $[P]_{ ext {eq }}>5,[A]_{ ext {eq }}<5$\ $K_{ ext {eq }}=\frac{[P]}{[A]}>1$\ $\Delta G^{\circ}=-R T \ln K_{ ext {eq }} \Rightarrow \Delta G^{\circ}<0$\ $\frac{\ln K_{1}}{\ln K_{2}}=\frac{\frac{-\Delta H^{\circ}}{T_{1} R}+\frac{\Delta S^{\circ}}{R}}{\frac{-\Delta H^{\circ}}{T_{2} R}+\frac{\Delta S^{\circ}}{R}}>\frac{T_{2}}{T_{1}}$ $$ \begin{aligned} & \frac{\left(-\Delta H^{\circ}+T_{1} \Delta S^{\circ} ight)}{\left(-\Delta H^{\circ}+T_{2} \Delta S^{\circ} ight)} \frac{T_{2}}{T_{1}}>\frac{T_{2}}{T_{1}} \ & \quad \quad-\Delta H^{\circ}+T_{1} \Delta S^{\circ}>-\Delta H^{\circ}+T_{2} \Delta S^{\circ} \ & \Rightarrow \Delta S^{\circ}<0 \end{aligned} $$