Chemistry - Chemical Kinetics Question with Solution | TestHub

Explain Question : Calculate ${\mathrm{E}}_{\mathrm{a}}$ from k quadrupling over 300 K to 310 K .

Concept : This question is based on two-point Arrhenius equation (Natural $\log$ ).

Solution : The rate (and k ) quadruples, so $\frac{{\mathrm{k}}_2}{{\mathrm{k}}_1}=4$.

Given ${\mathrm{T}}_1=300\mathrm{K}$.

Use the natural $\log$ form : $\mathrm{In}\left(\frac{k_2}{k_1}\right)=\frac{E_a}{R}\left(\frac{T_2-T_1}{T_1{T}_2}\right)$.

Substituting values : In (4) $=\frac{{\mathrm{E}}_{\mathrm{a}}}{8.314}\left(\frac{10}{300\times 310}\right)$.

Since $\mathrm{In}(4)=\mathrm{In}\left(2^2\right)=2\mathrm{In}(2)=2\times 0.693=1.386$

$\Rightarrow 1.386=\frac{{\mathrm{E}}_{\mathrm{a}}}{8.314}\left(\frac{10}{300\times 310}\right)$.

Rearranging for ${\mathrm{E}}_{\mathrm{a}}$ :

${\mathrm{E}}_{\mathrm{a}}=\frac{1.386\times 8.314\times 93000}{10}\approx 107204{\mathrm{Jmol}}^{-1}$.

Converting to $\mathrm{KJ}{\mathrm{mol}}^{-1}:{\mathrm{E}}_{\mathrm{a}}\approx 107.2\mathrm{kJ}{\mathrm{mol}}^{-1}$.

Final answer : Option (A)