Explanation

(d)
$k_{27{^\circ}C} = \frac{0.693}{30}\ \min^{- 1},$
$k_{47{^\circ}C} = \frac{0.693}{10}\ \min^{- 1}$
$\therefore k_{47{^\circ}C}/\ k_{27{^\circ}C} = 3$
Or $\log\frac{k_{47{^\circ}C}}{k_{27{^\circ}C}} = \log 3$
We know that
$\log\frac{k_{2}}{k_{1}} = \frac{E_{a}}{2.303\ R}\left( \frac{1}{T_{1}} - \frac{1}{T_{2}} \right)$
So, $E_{a} = \frac{\log{3\ \times 2.303\ R\ \times T_{1}T_{2}}}{T_{2} - T_{1}}$
$= \frac{0.4771 \times 2.303 \times 8.314 \times 300 \times 320}{20}$
$\ \ = 43.84\ kJ$