A first order reaction is 50% completed in 20 minutes at 27 °C and in 5 minutes at 47 °C. The energy of activation of the reaction is:

(Given: ln 4=1.386)

For a first-order reaction:

$$t_{1/2}=\frac{0.693}{k}$$

Since the reaction is 50% complete:

At $$27^\circ\mathrm{C}$$:

$$t_{1/2}=20\,\mathrm{min}$$

At $$47^\circ\mathrm{C}$$:

$$t_{1/2}=5\,\mathrm{min}$$

Therefore,

$$\frac{k_2}{k_1}=\frac{20}{5}=4$$

Using the Arrhenius equation:

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

$$\ln 4=\frac{E_a}{8.314}\left(\frac{1}{300}-\frac{1}{320}\right)$$

$$E_a=\frac{8.314\times1.3863}{\frac{1}{300}-\frac{1}{320}}$$

$$E_a\approx55311.38\,\mathrm{J\,mol^{-1}}$$

$${E_a=55.31\,\mathrm{kJ\,mol^{-1}}}$$

Correct option: (B)