Chemistry - Chemical Kinetics Question with Solution | TestHub

Given:
Rate constant at $T_1=290\mathrm{K}$ is $k_1=0.026{\mathrm{s}}^{-1}$
Rate constant at $T_2=300\mathrm{K}$ is $k_2=0.58{\mathrm{s}}^{-1}$
Gas constant $R=8.314{\mathrm{J}\mathrm{K}}^{-1}{\mathrm{mol}}^{-1}$

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

Substituting the given values:
$\log \left(\frac{0.58}{0.026}\right)=\frac{E_a}{2.303\times 8.314}\left(\frac{300-290}{290\times 300}\right)$

$\log (22.3)=\frac{E_a}{19.147}\left(\frac{10}{87000}\right)$

$1.348=\frac{E_a}{19.147}\times \frac{1}{8700}$

$E_a=1.348\times 19.147\times 8700$

$E_a=224548{\mathrm{J}\mathrm{mol}}^{-1}$

$E_a=224.55{\mathrm{kJ}\mathrm{mol}}^{-1}$

Answer: $224.55\mathrm{kJ}$