If the rate of reaction doubles when the temperature is raised from $20^\circ ext{C}$ to $35^\circ ext{C}$,

Question

If the rate of reaction doubles when the temperature is raised from $20^\circ	ext{C}$ to $35^\circ	ext{C}$, then the activation energy for the reaction will be : ($R = 8.314 	ext{ J mol}^{-1} 	ext{ K}^{-1}$)

Accepted Answer

According to the Arrhenius equation:
$\log \frac{k_2}{k_1} = \frac{E_a}{2.303R} \left[ \frac{T_2 - T_1}{T_1 T_2} ight]$
Given that the rate doubles, $k_2 = 2k_1$.
$T_1 = 20^\circ	ext{C} = 20 + 273 = 293 	ext{ K}$
$T_2 = 35^\circ	ext{C} = 35 + 273 = 308 	ext{ K}$
$R = 8.314 	ext{ J K}^{-1} 	ext{mol}^{-1}$
Substituting the values:
$\log(2) = \frac{E_a}{2.303 	imes 8.314} \left[ \frac{308 - 293}{293 	imes 308} ight]$
$0.3010 = \frac{E_a}{19.147} \left[ \frac{15}{90244} ight]$
$E_a = \frac{0.3010 	imes 19.147 	imes 90244}{15}$
$E_a \approx 34673 	ext{ J mol}^{-1}$
$E_a \approx 34.7 	ext{ kJ mol}^{-1}$

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