91Ó°ÊÓ

From the Arrhenius Equation, the rate of reaction at two different temperatures is given as

\({\bf{log}}\frac{{{{\bf{k}}_{\bf{2}}}}}{{{{\bf{k}}_{\bf{1}}}}}{\bf{ = }}\frac{{{{\bf{E}}_{\bf{a}}}}}{{{\bf{2}}{\bf{.303*R}}}}\left( {\frac{{\bf{1}}}{{{{\bf{T}}_{\bf{1}}}}}{\bf{ - }}\frac{{\bf{1}}}{{{{\bf{T}}_{\bf{2}}}}}} \right)\)

where \({{\bf{k}}_{\bf{1}}}\)and \({{\bf{k}}_{\bf{2}}}\)are the rate constants at \({{\bf{T}}_{\bf{1}}}\) and \({{\bf{T}}_{\bf{2}}}\) where \({{\bf{T}}_{\bf{1}}}{\bf{ < }}{{\bf{T}}_{\bf{2}}}\). \({{\bf{E}}_{\bf{a}}}\)is the activation energy (in J) and R is the gas constant.

Given,

The ratio of \(\frac{{{{\bf{k}}_{\bf{2}}}}}{{{{\bf{k}}_{\bf{1}}}}}{\bf{ = 1}}{\bf{.47}}\)

\({{\bf{T}}_{\bf{1}}} = 298\,K\)

\({{\bf{T}}_{\bf{2}}} = 293K\)

Replacing the values in the Arrhenius equation,

\(\begin{align}\log \frac{{0.1}}{{0.054}} &= \frac{{{E_a}}}{{2.303 \times 8.314}}\left( {\frac{1}{{293}} - \frac{1}{{298}}} \right)\\{E_a} &= 0.267 \times 2.303 \times 8.314 \times \left( {\frac{{293 \times 298}}{{298 - 293}}} \right)\\{E_a} &= 89477.5J/mol\\{E_a} &= 89.48kJ/mol\end{align}\)