The rate of the chemical reaction doubles for an increase of $10 \, K$ in absolute temperature from $298 \, K$. Calculate $E_{a}$.

(N/A) Given: $T_{1} = 298 \, K$,$T_{2} = 308 \, K$,$k_{2} = 2k_{1}$,$R = 8.314 \, J \, K^{-1} \, mol^{-1}$.
Using the Arrhenius equation: $\log \frac{k_{2}}{k_{1}} = \frac{E_{a}}{2.303 \, R} \left[ \frac{T_{2} - T_{1}}{T_{1} T_{2}} \right]$.
Substituting the values: $\log 2 = \frac{E_{a}}{2.303 \times 8.314} \left[ \frac{10}{298 \times 308} \right]$.
Solving for $E_{a}$: $E_{a} = \frac{0.3010 \times 2.303 \times 8.314 \times 298 \times 308}{10}$.
$E_{a} \approx 52897.78 \, J \, mol^{-1} = 52.9 \, kJ \, mol^{-1}$.