Chemical and Ionic Equilibrium - Result Question 28
28. 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 $\ln K$ 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.)
(2018 Adv.)
(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) $\Delta G^{\ominus} < 0, \Delta S^{\ominus} > 0$
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Answer:
Correct Answer: 28. (a, c)
Solution:
- For the reaction, $A \rightleftharpoons P$
Given,
$T_1 < T_2 $
$\dfrac{\ln K_1}{\ln K_2} > \dfrac{T_2}{T_1} \hspace{15mm} …(i)$
It shows, On increasing the temperature, $K$ decreases so reaction is exothermic i.e., $\Delta H^{\circ} < 0$
Besides, graph shows $K > 1$
So $ \hspace{5mm} \Delta G^{\circ} < 0$
Now from equation (i)
$T_1 \ln K_1 > \mathrm{T}_2 \ln K_2 $
$ -\Delta G^{\circ}{ }_1 > -\Delta G^{\mathrm{o}}{ }_2 $
$ \text { Likewise }\left(-\Delta H^{\mathrm{o}}+T_1 \Delta S^{\mathrm{o}}\right) > \left(-\Delta H^{\mathrm{o}}+T_2 \Delta S^{\mathrm{o}}\right) $
$ \text { or simply } T_1 \Delta S^{\circ} > T_2 \Delta S^{\mathrm{o}} $
$ \text { So, } \left(T_2-T_1\right) \Delta S^{\circ} < 0 $
$ \therefore \Delta S^{\circ} < 0$
In other words, increase of $\Delta G$ with increase in temperature is possible only when $\Delta S^{\circ} < 0 $. Hence, options (a) and (c) are correct.
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