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What is the activation energy for a reaction if its rate doubles when the temperature is raised from $20^\circ C$ to $35^\circ C$? (R = 8.314 J $mol^{-1}K^{-1}$)
A
15.1 kJ $mol^{-1}$
B
342 kJ $mol^{-1}$
C
269 kJ $mol^{-1}$
D
34.7 kJ $mol^{-1}$
Detailed Solution
$\log\frac{K_2}{K_1} = \frac{E_a}{2.303R}\left[\frac{1}{T_1} - \frac{1}{T_2}\right]$, with $\frac{K_2}{K_1} = 2$
$\log2 = \frac{E_a}{2.303\times8.314\times10^{-3}}\left[\frac{1}{293} - \frac{1}{308}\right]$
$E_a = 34.7$ kJ $mol^{-1}$
$\log2 = \frac{E_a}{2.303\times8.314\times10^{-3}}\left[\frac{1}{293} - \frac{1}{308}\right]$
$E_a = 34.7$ kJ $mol^{-1}$