Plzz solve this
Q.5.25 The rate constant of a first order reaction at 25$°$C is 0.24 $s^{-1}$. If the energy of activation of the reaction is 88 kJ $mo{l}^{-1}$, at what temperature would this reaction have rate constant of $4\times {10}^{-2} s^{-1}$?
(283.7 K)

Dear student,

According to Arrhenius equation : $\log \frac{K_1}{K_2} = \frac{E_a}{2.303R}(\frac{1}{T_2} -\frac{1}{T_1})$
                                              K₁ = 0.24 s^-1 and T₁ = 298 K
                                             K₂  = 4 $\times$10^-2 s^-1   T₂ = ?  E_a = 88 kj/mol = 88000 j /mol
Now on putting the values ;  $\log \frac{0.24}{4\times {10}^{-2}} = \frac{88000}{2.303\times 8.314} (\frac{1}{T_2} - \frac{1}{298})$
                                              0.77815    =   4596.02 $(\frac{1}{T_2} - \frac{1}{298})$
                                  0.000169        = $(\frac{1}{T_2} - \frac{1}{298})$
                                  T₂  =283.7 K
  Regards