Question

In the presence of a catalyst, activation energy of a reaction is lowered by $2$ kcal at $27^{\circ }C$. Hence, rate will

• A. 20 times
• B. 28 times
• C. 5 times
• D. 10 times

Answer 1

Answer: (B)

Arrhenius equation is
$k=A{e}^{-E_a/RT}$
In the presence of catalyst rate constant $k^{\prime }=A{e}^{-E_a/RT}$
Eq. (ii ) /Eq. (i), we get
$\frac{k^{\prime }}{k}=e^{(E_a/E_a^{\prime })}/RT$
$\frac{k^{\prime }}{k}=e^{\left(\frac{2000}{2\times 300}\right)}=e^{3.33}$
$(\because E_a-E_a^{\prime }=2000cal$ (given)]
$\because$ On taking log both sides,
$\log \frac{k^{\prime }}{k}=\frac{3.33}{2.303}$
$\frac{k^{\prime }}{k}=$ antilog $1.446=27.92=28$
We know that, rate $(R)\propto k$
$\therefore \frac{k{R}^{\prime }}{R}=\frac{k^{\prime }}{k}=28$
or $R^{\prime }=28R$