Question

Assertion :Th equilibrium constant of the exothermic reaction at high temperature decreases. Reason: Since $\ln \frac{K_2}{K_1}=\frac{{\Delta H}^o}{R}[\frac{1}{T_1}-\frac{1}{T_2}]$ and for exothermic reaction, ${\Delta H}^o=-ve$ and thereby; $\frac{K_2}{K_1}<1$

• A. Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
• B. Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
• C. Assertion is correct but Reason is incorrect
• D. Assertion is incorrect but Reason is correct

Answer 1

The correct option is A Both Assertion and Reason are correct and Reason is the correct explanation for Assertion

From integrated form of van't Hoff equation, vz.

$log\frac{K_2}{K_1}=\frac{\Delta H^{\circ }}{2.303R}\left[\frac{T_2-T_1}{T_1{T}_2}\right]$

We may conclude that:

(i) If $\Delta H^{\circ }=0$, i.e. no heat is evolved or absorbed in the reaction.

$log\left(K_2/K_1\right)=0$ i.e., $K_2/K_1=1$ or $K_2=K_1$

So, equilibrium constant does not change with temperature.

(ii) If $\Delta H^{\circ }=+ve$ i.e. heat is abosrbed in the reaction, then

$log\left(K_2/K_1\right)=+ve$ or $log{K}_2>log{K}_{1}or{K}_2>K_1$

So, equilibrium constant increases with increase in temperature.

(iii) If $\Delta H^{\circ }=-ve$, i.e.,heat is evolved in the reaction , then

$log\left(K_2/K_1\right)=-ve.i.e.K_2<log{K}_1$ or $K_2<K_1$

So, equilibrium constant decreases with increase in temperature.