Question

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

• A. 20 times
• B. 28 times
• C. 14 times
• D. remain the same

Answer 1

Answer: (B)

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