Question

The equilibrium constant of a reaction at $298K$ is $5\times {10}^{-3}$ and at $1000K$ is $2\times {10}^{-5}$. What is the sign of $\Delta H$ for the reaction?

• A. $\Delta H$ is +ve
• B. $\Delta H$ is -ve
• C. $\Delta H\propto 0$
• D. $\Delta H$ is $\pm$ve

Answer 1

The correct option is A $\Delta H$ is +ve

Relation between equilibrium constant and $\Delta H$ is as follows-

$\log \frac{K_2}{K_1}=\frac{{\Delta }_H}{2.303R}[\frac{1}{T_1}-\frac{1}{T_2}]$

$\log \frac{2\times {10}^{-5}}{5\times {10}^{-3}}=\frac{\Delta H}{2.303R}\frac{1}{T_1}-\frac{1}{T_2}$

$\log 4\times {10}^{-3}=\frac{\Delta H}{2.303R}\times 2.3\times {10}^{-3}$

So Sign of $\Delta H$ would be positive.