Question

In the presence of a catalyst, activation energy of a reaction is lowered by 2 kcal 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)

28 times

Arrhenius equation is $K=A{e}^{-EQ/RT}$------------------$\left(i\right)$

In the presence of catalyst$.$

rate constant

$K=A{e}^{-EA/RT}$------------------------------$\left(ii\right)$

Dividing eq $\left(i\right)$ by $\left(ii\right)$

We get $\frac{K}{K}=e^{\left(EQ-EA\right)/RT}$

$\Rightarrow \frac{K}{K}=e^{\left(\frac{2000}{2\times 300}\right)}=e^{3.33}$

$\because E_Q-E_A=2000$ $($ 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 $\left(R\right)\alpha K$

$\therefore \frac{R}{R}=\frac{K}{K}=28$

or $R=28R$

Hence$,$ option $\left(B\right)$ is correct$.$