Chemistry - Chemical Kinetics Question with Solution | TestHub

Using the Arrhenius equation:

$\log \left(\frac{k_2}{k_1}\right)=\frac{E_a}{2.303R}\left(\frac{T_2-T_1}{T_1{T}_2}\right)$

Given $T_1=298\text{ K}$, $T_2=308\text{ K}$, $\frac{k_2}{k_1}=2$, and $2.303R=19.15{\text{ J K}}^{-1}{\text{ mol}}^{-1}$.

Substituting the values:

$\log (2)=\frac{E_a}{19.15}\left(\frac{308-298}{298\times 308}\right)$

$0.301=\frac{E_a}{19.15}\left(\frac{10}{91784}\right)$

$E_a=\frac{0.301\times 19.15\times 91784}{10}$

$E_a\approx 52897{\text{ J mol}}^{-1}=52.89{\text{ kJ mol}}^{-1}$

Answer: $52.89{\text{ KJ mol}}^{-1}$