On introducing a catalyst at 500 K the rate of a first order reaction increases by 1.718 times. The activation energy in presence of catalyst is 4.15 kJ ${\text{mol}}^{-1}$. The slope of the plot of $\ln \mathrm{k}\left({\text{in sec}}^{-1}\right)$ versus $\frac{1}{\mathrm{T}}$ (T in Kelvin) in absence of catalyst is $(\mathrm{R}=8.3{\text{ J mol}}^{-1}{\text{ K}}^{-1})$

$\begin{array}{l}\mathrm{k}={\mathrm{Ae}}^{-{\mathrm{E}}_{\mathrm{a}}/\mathrm{RT}}\\ \mathrm{k}\text{'}={\mathrm{Ae}}^{-{\mathrm{E}}_{\mathrm{a}}\text{'}/\mathrm{RT}}\end{array}$

(where k = rate constant for non catalysed reaction and k' = rate constant for catalysed reaction. $E_a=$activation energy for non-catalysed reaction and $E_a\text{'}=$activation energy for catalysed reaction) 

$\frac{k\text{'}}{k}=e^{E_a-E{\text{'}}_a/RT}$

Also given $k\text{'}=k+1.718k=2.718k$

$\therefore 2.718=e^{\frac{E_a-E_a\text{'}}{RT}}$

$\begin{array}{l}{\log }_{\mathrm{e}}2.718=\frac{{\mathrm{E}}_{\mathrm{a}}-{\mathrm{E}}_{\mathrm{a}}\text{'}}{8.314\times {10}^{-3}\times 500}\\ {\mathrm{E}}_{\mathrm{a}}-{\mathrm{E}}_{\mathrm{a}\text{'}}=4.11\\ \therefore {\mathrm{E}}_{\mathrm{a}}=8.3{\text{ kJ/mol}}^{-1}\\ \mathrm{k}={\mathrm{Ae}}^{-{\mathrm{E}}_{\mathrm{a}}/\mathrm{RT}}\\ \therefore {\log }_{\mathrm{e}}\mathrm{k}={\log }_{\mathrm{e}}\mathrm{A}-\frac{{\mathrm{E}}_{\mathrm{a}}}{\mathrm{RT}}\end{array}$

This is an equation for straight line with slope $=-\frac{{\mathrm{E}}_{\mathrm{a}}}{\mathrm{R}}=-\frac{8.3}{8.3\times {10}^{-3}}$

                                           $=-1000$