Chemistry - Chemical Kinetics Question with Solution | TestHub

Given:
$T_1=27^{\circ }\mathrm{C}=300\mathrm{K}$
$T_2=37^{\circ }\mathrm{C}=310\mathrm{K}$
$\frac{k_2}{k_1}=2$
$R=8.314\mathrm{J}{\mathrm{K}}^{-1}\mathrm{mo}{\mathrm{l}}^{-1}$

Using the Arrhenius equation:
${\log }_{10}\left(\frac{k_2}{k_1}\right)=\frac{E_a}{2.303R}\left(\frac{T_2-T_1}{T_1{T}_2}\right)$

Substituting the values:
${\log }_{10}(2)=\frac{E_a}{2.303\times 8.314}\left(\frac{310-300}{300\times 310}\right)$

$0.3010=\frac{E_a}{19.147}\left(\frac{10}{93000}\right)$

$E_a=\frac{0.3010\times 19.147\times 93000}{10}$

$E_a=53598.6\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}$

$E_a=53.60\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$

Answer: 53.60