Question

Equilibrium constant, K changes with temperature. At 300 K, equilibrium constant is 25 and at 400 K it is 10. Hence, backward reaction will have energy of activation

• A. equal to that of forward reaction
• B. less than that of forward reaction
• C. greater than that of forward reaction
• D. given values are not sufficient to explain given statement

Answer 1

Answer: (C)

$log\frac{K_2}{K_1}=\frac{\Delta H}{2.303\times R}\left[\frac{T_2-T_1}{T_1{T}_2}\right]$
$\Delta H=\frac{2.303\times R{T}_1{T}_2}{\left(T_2-T_1\right)}log\frac{K_2}{K_1}$
$=\frac{2.303\times 8.314\times 300\times 400}{400-300}log\frac{10}{25}$
$=\frac{2.303\times 8.314\times 300\times 400}{100}log\frac{10}{25}<0$
As $\Delta H$ is -ve, the reaction is exothermic

Hence, $E_a$ (backward) is more than $E_a$ (forward)