The rate constant is doubled when temperature increases from $27^{\circ} \mathrm{C}$ to $37^{\circ} \mathrm{C}$. What is activation energy in kJ ?

The activation energy $E_a$ can be determined using the integrated Arrhenius equation for two temperature conditions:

$\ln \left(\frac{k_2}{k_1}\right) = \frac{E_a}{R} \left(\frac{1}{T_1} - \frac{1}{T_2}\right)$

Given $T_1 = 27^\circ\mathrm{C} = 300\,\mathrm{K}$, $T_2 = 37^\circ\mathrm{C} = 310\,\mathrm{K}$, and $\frac{k_2}{k_1} = 2$, with $R = 8.314\,\mathrm{J}\,\mathrm{mol}^{-1}\,\mathrm{K}^{-1}$.

Substituting values yields:

$\ln(2) = \frac{E_a}{8.314} \left(\frac{1}{300} - \frac{1}{310}\right)$

Evaluating the temperature term:

$\frac{1}{300} - \frac{1}{310} = \frac{310 - 300}{300 \times 310} = \frac{10}{93000}$

Using $\ln(2) \approx 0.693$, the equation becomes:

$0.693 = \frac{E_a}{8.314} \cdot \frac{10}{93000}$

Solving for $E_a$:

$E_a = \frac{0.693 \times 8.314 \times 93000}{10} = 53600.6\,\mathrm{J}\,\mathrm{mol}^{-1}$

Converting to kilojoules:

$E_a = \frac{53600.6}{1000} = 53.60\,\mathrm{kJ}\,\mathrm{mol}^{-1}$

The value corresponds to option C.