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Q.
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
Equilibrium
Solution:
$log \frac{K_{2}}{K_{1}}=\frac{\Delta H}{2.303\times R} \left[\frac{T_{2}-T_{1}}{T_{1}T_{2}}\right]$
$\Delta H=\frac{2.303\times RT_{1}T_{2}}{\left(T_{2}-T_{1}\right)}log\frac{K_{2}}{K_{1}}$
$=\frac{2.303\times8.314\times300\times400}{400-300} log \frac{10}{25}$
$=\frac{2.303\times8.314\times300\times400}{100} log \frac{10}{25} <\,0$
As $\Delta H$ is -ve, the reaction is exothermic
Hence, $E_{a}$ (backward) is more than $E_{a}$ (forward)
