The rate of a reaction double when its temperature changes from 300 K to 310 K. Activation energy of such a reaction will be (R=8.314 JK−1 mol−1 and log2 = 0.301)
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a
53.6 KJK−1 mol−1
b
48.6 JK−1 mol−1
c
58.5 JK−1 mol−1
d
60.5 JK−1 mol−1
answer is A.
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Detailed Solution
From Arrhenius equation, logk2k1=−Ea2.303R(1T2−1T1)Given, k2k1=2; T2=310 KT1=300 KOn putting values, ⇒log2=−Ea2.303×8.314(1310−1300)⇒ Ea=53603.93 J/mol = 53.6 kJ/mol