Discussion Forum

Two reactions R₁ and  R₂  have identical pre-exponential factors. The activation energy of R₁ exceeds that of R₂ by 10 kj mol^-1 . If k₁ and k₂ are rate constants for reactions R₁ and R2, respectively at 300 K. then ln $(\frac{k_{2}}{k_{1}})$ is equal to   (R=8.314 J mol^-1 K^-1)

Answer:

Explanation:

According to the Arrhenius equation

                   K= Ae^-E_e/RT 

   where A= collision number or pre-exponentail factor

     R= gas constant

      T= absolute temperature

     E_a = energy of activation

    For reaction R₁ ,k = Ae^-E_a1/RT           ...........(i)

    For reaction R₂ ,k₂ =  Ae^-E_a2/RT   ............(ii)

on dividing Eq  (ii) by Eq (i) , we get

          $\frac{k_{2}}{k_{1}}=e^-{\frac{(E_{a_{2}}-E_{a_{1}})}{RT}}$                .........(iii)

($\because$ Pre -exponental factor 'A' is same for both reactions)

      Taking In on both the sides of Eq  (iii) 

    We get

             $ln(\frac{k_{2}}{k_{1}})=\frac{E_{a_{1}}-E_{a_{2}}}{RT}$

     Given,   $E_{a_{1}}=E_{a_{2}}+10 kJ mol^{-1}$

                                       $E_{a_{1}}=E_{a_{2}}+10000 J mol^{-1}$

   $\therefore ln\frac{k_{2}}{k_{1}}=\frac{10000 J mol^{-1}}{8.314 Jmol^{-1}K^{-1}\times 300K}=4$