For the reaction $CH_{3}CH_{2}CH_{2}I + OH^{-} \rightarrow CH_{3}CH_{2}CH_{2}OH + I^{-}$,the rate constant is $1.84 \ (mol \ L^{-1})^{-1} \ min^{-1}$ at $27^{\circ}C$ $(300 \ K)$ and $38.84 \ (mol \ L^{-1})^{-1} \ min^{-1}$ at $327 \ K$. Calculate the activation energy $(E_{a})$ in $cal \ mol^{-1}$.

(A) Given: $T_{1} = 300 \ K$,$k_{1} = 1.84 \ (mol \ L^{-1})^{-1} \ min^{-1}$,$T_{2} = 327 \ K$,$k_{2} = 38.84 \ (mol \ L^{-1})^{-1} \ min^{-1}$,$R = 1.987 \ cal \ K^{-1} \ mol^{-1}$.
Using the Arrhenius equation: $\log(\frac{k_{2}}{k_{1}}) = \frac{E_{a}}{2.303 \ R} \times (\frac{T_{2} - T_{1}}{T_{1} \times T_{2}})$.
Substituting the values: $\log(\frac{38.84}{1.84}) = \frac{E_{a}}{2.303 \times 1.987} \times (\frac{327 - 300}{300 \times 327})$.
$\log(21.108) = \frac{E_{a}}{4.575} \times (\frac{27}{98100})$.
$1.3244 = \frac{E_{a}}{4.575} \times 0.0002752$.
$E_{a} = \frac{1.3244 \times 4.575}{0.0002752} \approx 22000 \ cal \ mol^{-1}$.