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

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Solution

The Arrhenius equation is,
$k = A e^{- E_{a} / R T}$
In absence of catalyst, $k_{1} = A e^{- 100 / R T}$
In presence of catalyst, $k_{2} = A e^{- 25 / R T}$
So, $\frac{k_{2}}{k_{1}} = e^{75 / R T}$ or $2.303 log \frac{k_{2}}{k_{1}} = \frac{75}{R T}$
or $2.303 log \frac{k_{2}}{k_{1}} = \frac{75}{8.314 \times 10^{- 3} \times 293}$
or $log \frac{k_{2}}{k_{1}} = \frac{75}{8.314 \times 10^{- 3} \times 293 \times 2.303}$
or $\frac{k_{2}}{k_{1}} = 2.34 \times 10^{13}$
As the things being equal in presence or absence of a catalyst,
$\frac{k_{2}}{k_{1}}$ must be $= \frac{r a t e i n p r e s e n c e o f c a t a l y s t}{r a t e i n a b s e n c e o f c a t a l y s t}$
i.e., $\frac{r_{2}}{r_{1}} = \frac{k_{2}}{k_{1}} = 2.34 \times 10^{13}$

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The energy of activation for a reaction is 100 kJ mol−1. Presence of a catalyst lowers the energy of activation by 75%. What will be the effect on rate of reaction at 20oC, other things being equal?

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