The rate of reaction doubles when its temperature changes from 300 K to 310 K. The activation energy of the reaction will be: (R = 8.314 JK⁻¹ mol⁻¹ and log 2 = 0.301)

53.6 kJ mol⁻¹
48.6 kJ mol⁻¹
58.5 kJ mol⁻¹
60.5 kJ mol⁻¹

Correct answer: 53.6 kJ mol⁻¹

Solution

log(k2/k1) = Ea/(2.303 R) x (1/T1 - 1/T2). 0.301 = Ea/(2.303 x 8.314) x (10/(300x310)); Ea = 0.301 x 19.15 / 1.075e-4 ~ 53.6 kJ/mol (index 0).

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