Question

The energy of activation for a reaction at $300K$ is $100kJmo{l}^{-1}$. For same concentration, presence of a catalyst lowers the energy of activation by $75\%$. Calculate the value of
${\log }_{10}\left(\frac{r_2}{r_1}\right)$
$r_2=\text{rate in presence of catalyst}$
$r_1=\text{rate in absence of catalyst}$

• A. 13.06
• B. 7.94
• C. 10.02
• D. 3.21

Answer 1

The correct option is A 13.06

The Arrhenius equation is,
$k=A{e}^{-E_a/RT}$
In absence of catalyst, $k_1=A{e}^{-100/RT}$
In presence of catalyst, $k_2=A{e}^{-25/RT}$
So, $\frac{k_2}{k_1}=e^{75/RT}$ or $2.303log\frac{k_2}{k_1}=\frac{75}{RT}$
or $2.303log\frac{k_2}{k_1}=\frac{75}{8.314\times {10}^{-3}\times 300}$
or
$log\frac{k_2}{k_1}=\frac{75}{8.314\times {10}^{-3}\times 300\times 2.303}\approx 13.06$
Since
$\text{Rate}=k[\text{Conc}{]}^n$
At a particular concentration,
$\text{Rate}\propto k$
So,
$\frac{k_2}{k_1}=\frac{\text{rate in presence of catalyst}}{\text{rate in absence of catalyst}}$
i.e., $\frac{r_2}{r_1}=\frac{k_2}{k_1}$
Hence,
${\log }_{10}\left(\frac{r_2}{r_1}\right)=13.06$
Hence, option (a) is correct.