The rate constant of a reaction is $2 \times 10^{-3} \ min^{-1}$ at $300 \ K$. By increasing the temperature by $10 \ K$,its value becomes double. Calculate the energy of activation $(E_a)$ and the rate constant at $320 \ K$.

(N/A) Given: $k_1 = 2 \times 10^{-3} \ min^{-1}$ at $T_1 = 300 \ K$,$k_2 = 4 \times 10^{-3} \ min^{-1}$ at $T_2 = 310 \ K$.
Using the Arrhenius equation: $\log(k_2/k_1) = \frac{E_a}{2.303R} \times \frac{T_2 - T_1}{T_1 T_2}$.
$\log(2) = \frac{E_a}{2.303 \times 1.987} \times \frac{10}{300 \times 310}$.
$0.3010 = \frac{E_a}{4.575} \times \frac{10}{93000}$.
$E_a = \frac{0.3010 \times 4.575 \times 93000}{10} \approx 12808 \ cal \ mol^{-1}$.
For $T_3 = 320 \ K$: $\log(k_3/k_1) = \frac{E_a}{2.303R} \times \frac{T_3 - T_1}{T_1 T_3}$.
$\log(k_3/2 \times 10^{-3}) = \frac{12808}{2.303 \times 1.987} \times \frac{20}{300 \times 320}$.
$\log(k_3/2 \times 10^{-3}) = 2801.6 \times 0.0002083 = 0.5836$.
$k_3/2 \times 10^{-3} = 10^{0.5836} = 3.833$.
$k_3 = 3.833 \times 2 \times 10^{-3} = 7.666 \times 10^{-3} \ min^{-1}$.