What is the activation energy for a reaction, if its rate doubles when the temperature is raised from 20^oC to 35^oC ? $\left(\mathrm{R}=8.314\mathrm{J}{\mathrm{mol}}^{-1}{\mathrm{K}}^{-1}\right)$

detailed solution

Correct option is C

Given, initial temperature T1=20+273=293KFinal temperature, T2=35+273=308KR=8.314Jmol−1K−1Since, rate becomes double on raising temperature r2=2r1⇒r2r1=2As rate constant, k∝r⇒k2k1=2From Arrhenius equation, we know that log k2k1=−Ea2.303RT2−T1T1T2                                                      log 2=Ea2.303×8.314308−293293×308                                                   0.3010=Ea2.303×8.31415293×308                                              Ea=0.3010×2.303×8.314×293×30815                                                  =34673.48Jmol−1=34.7kJmol−1