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 $2.303 \, \text{log}\frac{k_{2}}{k_{1}}=\frac{75}{8.314 \times 1 0^{- 3} \times 293}$

or $\text{log}\frac{k_{2}}{k_{1}}=\frac{75}{8.314 \times 1 0^{- 3} \times 293 \times 2.303}$

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., $\frac{r_{2}}{r_{1}}=\frac{k_{2}}{k_{1}}=2.34\times 10^{13}$