Question
The rate of a reaction doubles when its temperature changes from 300K to 310K. Activation energy of such a reaction will be (R = 8.314 JK–1 mol–1 and log 2 = 0.301)
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53.6 kJ mol-1
-
48.6 kJ mol-1
-
58.5 kJ mol-1
-
60.5 kJ mol-1
Solution
A.
53.6 kJ mol-1
By using Arrhenius equation,
Given, T2 = 310; T1 = 300K
On putting values in Eq (i), we get
