Answer:
Option D
Explanation:
According to the Arrhenius equation
K= Ae-Ee/RT
where A= collision number or pre-exponentail factor
R= gas constant
T= absolute temperature
Ea = energy of activation
For reaction R1 ,k = Ae-Ea1/RT ...........(i)
For reaction R2 ,k2 = Ae-Ea2/RT ............(ii)
on dividing Eq (ii) by Eq (i) , we get
k2k1=e−(Ea2−Ea1)RT .........(iii)
(∵ 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