The time required for $$10\% $$  completion of a first order reaction at 298$$\,K$$  is equal to that required for its $$25\% $$  completion at 308$$\,K.$$  If the preexponential factor for the reaction is $$3.56 \times {10^9}{s^{ - 1}},$$   the rate constant at $$318\,K$$  is :

Solution :
Let the initial concentration $$\left( A \right) = 100$$
Final concentration at $$298\,K = 100 - 10 = 90$$
Final concentration at $$308\,K = 100 - 25 = 75$$
Substituting the values in the 1^st order rate reaction
$$\eqalign{ & t = \frac{{2.303}}{{{k_{298}}}}{\text{log}}\frac{{100}}{{90}}\,\,...\left( {\text{i}} \right) \cr & t = \frac{{2.303}}{{{k_{308}}}}{\text{log}}\frac{{100}}{{75}}\,\,...\left( {{\text{ii}}} \right) \cr & {\text{From (i) and (ii)}}\,\frac{{{k_{308}}}}{{{k_{208}}}} = 2.73 \cr} $$
Substituting the value in the following relation
$$\eqalign{ & {E_a} = \frac{{2.303R \times {T_1} \times {T_2}}}{{{T_2} - {T_1}}}{\text{log}}\frac{{{k_2}}}{{{k_1}}} \cr & = \frac{{2.303 \times 8.314 \times 298 \times 308}}{{308 - 298}}{\text{log}}\,2.73 \cr & {E_a} = 76.62\,kJ\,mo{l^{ - 1}} \cr & \,\,\,\,\,\,\,\, = 18.39\,kcal\,mo{l^{ - 1}} \cr} $$