Chemistry - Chemical Kinetics Question with Solution | TestHub

Arrhenius equation may be given as $k=A{e}^{-\frac{E_a}{RT}}$.

Let $k_{500}$ and $k_{400}$ be the rate constants at temperatures $500K$ and $400K$ (in the presence of a catalyst), respectively. Let $E_{500}$ and $E_{400}$ be the activation energies at temperatures $500K$ and $400K$, respectively.

$k_{500}=A{e}^{-\frac{E_{500}}{R\times 500}}(i)$

$k_{400}=A{e}^{-\frac{E_{400}}{R\times 400}}(ii)$

Given that $k_{500}=k_{400}$ (same rates in the presence and absence of a catalyst).

On comparing equation (i) with equation (ii)

$\frac{E_{500}}{R\times 500}=\frac{E_{400}}{R\times 400}$

$\frac{E_{500}}{5}=\frac{E_{400}}{4}$

$E_{500}=\frac{E_{400}}{4}\times 5$

Given $E_{500}=E_{400}+25$.

Substituting equation (iii)

$$\begin{gathered} \text{displaylines} \\ {E}_{400}+25=E_{400}\times 1.25 \end{gathered}$$

$E_{400}+\frac{25}{0.25}=100KJmo{l}^{-1}$

So, $E_{500}=100+25=125kJmo{l}^{-1}$.