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The rate of a reaction quadruples when temperature changes from $27^\circ C$ to $57^\circ C$. Calculate the energy of activation.
Given $R = 8.314$ J $K^{-1} mol^{-1}$, $\log 4 = 0.6021$
A
3.80 kJ/mol
B
3804 kJ/mol
C
38.04 kJ/mol
D
380.4 kJ/mol
Detailed Solution
$\log\frac{k_2}{k_1} = \frac{E_a}{2.303R}\left(\frac{T_2 - T_1}{T_1T_2}\right)$, with $T_1 = 300$ K, $T_2 = 330$ K, $\frac{k_2}{k_1} = 4$
$\log 4 = \frac{E_a}{2.303 \times 8.314} \times \frac{330 - 300}{330 \times 300}$
$E_a = \frac{0.6021 \times 2.303 \times 8.314 \times 330 \times 300}{30}$
$E_a \approx 38040$ J/mol $= 38.04$ kJ/mol