Question

For the reaction $B{r}_2\rightleftharpoons 2Br$, the equilibrium constant at ${327}^{\circ }C$ is lesser than the equilibrium constant at ${527}^{\circ }C$. What is the nature of reaction ?

• A. Endothermic
• B. Exothermic
• C. Heat is evolved
• D. Can't product

Answer 1

The correct option is A Endothermic

equilibrium constants at ${327}^{\circ }C$. or (600 K) is $K_{P_1}$ and at ${527}^{\circ }C\left(800K\right)$ is $K_{P_2}$
$\to Wehavelog\left(\frac{K_{P_2}}{K_{P_1}}\right)=\frac{\Delta H^{\circ }}{2.303R}\times \frac{T_2-T_1}{T_2.T_1}$
$\Rightarrow log\left(\frac{K_{P_2}}{K_{P_1}}\right)=\frac{\Delta H^{\circ }}{2.303R}\times \frac{200}{800\times 600}$
As $K_{P_2}>K_{P_1}$, then L.H.S. is positive
So, $\Delta H^{\circ }$ becomes positive.
$\therefore$ It is an endothermic reaction.