The rate constant is doubled when temperature increases from 27oC to 37oC. Activation energy in kJ is:

A. 34.2

B. 58.8

C. 53.6

D. 100.8

Answer:- Correct option is C) 53.6

We are specified that:

When T1​=27+273=300K

Let k1​=k

When T2​=37+273=310K

k2​=2k

Substituting these values, the equation:

log(k1​k2​​) =2.303Ea​​×(T1​T2​T2​–T1​​)

We will get:

log(k2k​) =2.303×8.314Ea​​ (300×310310−300​)

log (2) =2.303×8.314Ea​​ (300×31010​)

Ea​=53598.6 Jmol−1

Ea​=53.6 kJmol−1

Henceforth, the energy of activation of the reaction is 53.6 kJmol−1