The first order rate constant for the decomposition of ethyl iodide by the reaction $C_{2}H_{5}I_{(g)} \rightarrow C_{2}H_{4(g)} + HI_{(g)}$ at $600 \ K$ is $1.60 \times 10^{-5} \ s^{-1}$. Its energy of activation is $209 \ kJ/mol$. Calculate the rate constant of the reaction at $700 \ K$.

(N/A) We use the Arrhenius equation in the form: $\log \frac{k_{2}}{k_{1}} = \frac{E_{a}}{2.303 \ R} \left[ \frac{T_{2} - T_{1}}{T_{1} T_{2}} \right]$
Given: $k_{1} = 1.60 \times 10^{-5} \ s^{-1}$,$T_{1} = 600 \ K$,$T_{2} = 700 \ K$,$E_{a} = 209000 \ J \ mol^{-1}$,$R = 8.314 \ J \ K^{-1} \ mol^{-1}$.
Substituting the values:
$\log \frac{k_{2}}{1.60 \times 10^{-5}} = \frac{209000}{2.303 \times 8.314} \left[ \frac{700 - 600}{600 \times 700} \right]$
$\log \frac{k_{2}}{1.60 \times 10^{-5}} = 10921.6 \times \frac{100}{420000} = 2.599$
$\frac{k_{2}}{1.60 \times 10^{-5}} = \text{antilog}(2.599) \approx 397.19$
$k_{2} = 397.19 \times 1.60 \times 10^{-5} \approx 6.36 \times 10^{-3} \ s^{-1}$