Two reactions $${R_1}$$ and $${R_2}$$ have identical pre-exponential factors. Activation energy of $${R_1}$$ exceeds that of $${R_2}$$ by $$10\,kJ\,mo{l^{ - 1}}.$$   If $${k_1}$$  and $${k_2}$$  are rate constants for reactions $${R_1}$$ and $${R_2}$$ respectively at $$300 K$$ , then In $$\left( {{k_2}/{k_1}} \right)$$   is equal to :
$$\left( {R = 8.314\,J\,mo{l^{ - 1}}{K^{ - 1}}} \right)$$

Solution :
$$\eqalign{ & {\text{From arrhenius equation,}} \cr & k = A.e{\,^{\frac{{ - Ea}}{{RT}}}} \cr & so,\,\,\,{k_1} = A.e{\,^{\frac{{ - {E_{{a_1}}}}}{{RT}}}}\,\,\,\,.....\left( 1 \right) \cr & {k_2} = A.{e^{\,\,\frac{{ - {E_{{a_2}}}}}{{RT}}}}\,\,\,\,\,\,.....\left( 2 \right) \cr & {\text{On dividing equation (2)/(1)}} \Rightarrow \frac{{{k_2}}}{{{k_1}}} = {e^{\frac{{\left( {{E_{{a_1}}} - {E_{{a_2}}}} \right)}}{{RT}}}} \cr & {\text{ln}}\left( {\frac{{{k_2}}}{{{k_1}}}} \right) \cr & = \frac{{{E_{{a_1}}} - {E_{{a_2}}}}}{{RT}} \cr & = \frac{{10,000}}{{8.314 \times 300}} \cr & = 4 \cr} $$