The rate constant of a reaction increases from $0.02\, s^{-1}$ to $0.04\, s^{-1}$ by increasing the temperature from 500 K to 600 K. The activation energy ($E_a$) of the reaction is (given $\log_{10} 2 = 0.3010$)

$17.289\, kJ\, mol^{-1}$

The correct answer is Option (2) → $17.289\, kJ\, mol^{-1}$

Use the Arrhenius equation (log form):

$\log\left(\frac{k_2}{k_1}\right) = \frac{E_a}{2.303\,R}\left(\frac{T_2-T_1}{T_1T_2}\right)$

Given:

• $k_1 = 0.02\ \text{s}^{-1}$
• $k_2 = 0.04\ \text{s}^{-1}$
• $T_1 = 500\ \text{K}$
• $T_2 = 600\ \text{K}$
• $\log_{10} 2 = 0.3010$
• $R= 8.314\ \text{J mol}^{-1}\text{K}^{-1}$

Step 1: Substitute values

$0.3010 = \frac{E_a}{2.303 \times 8.314} \left(\frac{600-500}{500 \times 600}\right)$

$0.3010 = \frac{E_a}{19.15} \times \frac{100}{300000}$

$0.3010 = \frac{E_a}{19.15} \times \frac{1}{3000}$

Step 2: Solve for $E_a$​

$E_a = 0.3010 \times 19.15 \times 3000$

$E_a \approx 17\,289\ \text{J mol}^{-1} = 17.289\ \text{kJ mol}^{-1}$