Question

The energy of activation for a reaction is $100\text{kJ mo}{\text{l}}^{-1}$ . Presence of a catalyst lowers the activation energy by 75%. What will be effect on rate of reaction at $20^{o}C$ , other things being equal?

• A. increases by a factor of
  $2.34\times 10^{13}$
• B. increases by a factor of
  $2.34\times 10^{10}$
• C. increases by a factor of $10^{15}$
• D. no effect

Answer 1

Answer: (A)

The Arrhenius equation is

$k=\text{A}{\text{e}}^{-E_a/\text{RT}}$

In absence of catalyst, $k_1=\text{A}{\text{e}}^{-100/\text{RT}}$

In presence of catalyst, $k_2=\text{A}{\text{e}}^{-25/\text{RT}}$

So $\frac{k_2}{k_1}=e^{-75/\text{RT}}$ or 2.303 $\text{lo}{\text{g}}^{\frac{k_2}{k_1}}=\frac{75}{\text{RT}}$

or $2.303\text{log}\frac{k_2}{k_1}=\frac{75}{8.314\times 10^{-3}\times 293}$

or $\text{log}\frac{k_2}{k_1}=\frac{75}{8.314\times 10^{-3}\times 293\times 2.303}$

or $\frac{k_2}{k_1}=2.34\times 10^{13}$

As the things being equal in presence or absence of a catalyst,

${\text{k}}_{\text{2}}=$ rate in presence of catalyst

${\text{k}}_{\text{1}}\text{=}$ rate in absence of catalyst

i.e., $\frac{r_2}{r_1}=\frac{k_2}{k_1}=2.34\times 10^{13}$