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)

greater than that of forward reaction

Answer 2

$log \frac{K_{2}}{K_{1}}=\frac{\Delta H}{2.303\times R} \left[\frac{T_{2}-T_{1}}{T_{1}T_{2}}\right]$ $\Delta H=\frac{2.303\times RT_{1}T_{2}}{\left(T_{2}-T_{1}\right)}log\frac{K_{2}}{K_{1}}$ $=\frac{2.303\times8.314\times300\times400}{400-300} log \frac{10}{25}$ $=\frac{2.303\times8.314\times300\times400}{100} log \frac{10}{25} Hence,$E_{a}$(backward) is more than$E_{a}$ (forward)