The rate constant of a reaction increases by 5% when the temperature of the reaction is increased from 300 to 301 K whereas the equilibrium constant increases only by 2%. What will be the activation energy of forward as well as backward reaction?

36.636  kJ mol⁻¹, 21.767 kJ mol⁻¹

The correct answer is option 1. 36.636  kJ mol⁻¹, 21.767 kJ mol⁻¹

According to Arrhenius's equation,

\(log\frac{k_2}{k_1} = \frac{E_{a,f}}{2.303R}\left(\frac{T_2 − T_1}{T_1T_2}\right)\)

If \(k_1 = k\) at 300K, then at 301 K, \(k_2 = k + \frac{5}{100}k = 1.05 k\)

∴ \(log\frac{1.05k}{k} = \frac{E_{a,f}}{2.303 × 8.314 \text{ J K}^{−1}\text{mol}^{−1}}\left(\frac{301 K − 300 K}{300 K ×  301 K}\right)\)

∴\(E_{a,f} = log (1.05) × 2.303 ×(8.314 \text{ J K}^{−1}\text{mol}^{−1}) × 300K × 301K\)

⇒ \(E_{a,f} \approx 36636J mol^{−1}\)

⇒\(E_{a,f} = 36.636 kJ mol^{−1}\)

According to van't Hoff equation (giving the effect of temperature on equilibrium constant)

\(log\frac{k_2}{k_1} = \frac{\Delta H^o}{2.303R}\left(\frac{T_2 − T_1}{T_1T_2}\right)\)

If \(K_1 = K\) at 300K, then at 301 K, \(K_2 = K + \frac{2}{100}K = 1.02 K\)

\(log\frac{1.2K}{K} = \frac{\Delta H^o}{2.303 × 8.314 \text{ J K}^{−1}\text{mol}^{−1}}\left(\frac{T_2 − T_1}{T_1T_2}\right)\)

⇒ \(\Delta H^o = log (1.02) × 2.303 ×(8.314 \text{ J K}^{−1}\text{mol}^{−1}) × 300K × 301K\)

⇒ \(\Delta H^o \approx 14869 J mol^{−1}\)

∴ \(\Delta H^o = 14.869 kJ mol^{−1}\)

Thus, the reaction is endothermic. For such a reaction,

\(\Delta H^o = \text{(Activation energy of the forward reaction) − (Activation energy of the backward reaction  reaction)}\), i.e.,

\(\Delta H^o = E_{a,f} − E_{a,b}\)

∴ \(E_{a,b} = E_{a,f} − \Delta H^o\)

⇒ \(E_{a,b} = 36.636 − 14.869 \text{kJ mol}^{−1}\)

∴ \(E_{a,b} = 21.767 \text{kJ mol}^{−1}\)