The rate constants of a reaction at $500 \, K$ and $700 \, K$ are $0.02 \, s^{-1}$ and $0.07 \, s^{-1}$ respectively. Calculate the values of $E_{a}$ and $A$.

Using the Arrhenius equation: $\log \frac{k_{2}}{k_{1}} = \frac{E_{a}}{2.303 \, R} \left[ \frac{T_{2} - T_{1}}{T_{1} T_{2}} \right]$
Given: $k_{1} = 0.02 \, s^{-1}$,$T_{1} = 500 \, K$,$k_{2} = 0.07 \, s^{-1}$,$T_{2} = 700 \, K$,$R = 8.314 \, J \, K^{-1} \, mol^{-1}$.
$\log \frac{0.07}{0.02} = \frac{E_{a}}{2.303 \times 8.314} \left[ \frac{700 - 500}{700 \times 500} \right]$
$\log(3.5) = \frac{E_{a}}{19.147} \times \frac{200}{350000}$
$0.544 = E_{a} \times \frac{200}{6701450}$
$E_{a} = \frac{0.544 \times 6701450}{200} = 18227.9 \, J \, mol^{-1} \approx 18.23 \, kJ \, mol^{-1}$.
Now,calculate $A$ using $k = A e^{-E_{a} / R T}$:
$0.02 = A e^{-18227.9 / (8.314 \times 500)}$
$0.02 = A e^{-4.385}$
$0.02 = A \times 0.01246$
$A = \frac{0.02}{0.01246} = 1.605 \, s^{-1}$.