Question

The following data were obtained for a given reaction at $300K$.

Reaction		Energy of activation
		$(kJmo{l}^{-1})$
(i)	uncatalysed	$76$
(ii)	catalysed	$57$

The factor by which rate of catalysed reaction is increased, is

• A. 21
• B. 2100
• C. 2000
• D. 1200

Answer 1

Answer: (C)

Using Arrhenius equation:
$K=Ae\frac{E_a}{Rt}$, we get-
$\log k=\log A-\frac{E_a}{2.303RT}$
$\log k_1=\log A-\frac{E_{a_{(1)}}}{2.303RT}\dots (i)$
and $\log k_2=\log A-\frac{E_{a_{(2)}}}{2.303RT}$
or $\log \frac{k_2}{k_1}=\frac{1}{2.303RT}\left[E_{a_{(1)}}-E_{a_{(2)}}\right]$ (from (i) and (ii))
$=\frac{1}{2.303\times 8.314\times 300}(76000-57000)$
or $\log \frac{k_2}{k_1}=\frac{19000}{2.303\times 8.314\times 300}=\frac{190}{6.9\times 8.314}$
On taking antilog-
or $\frac{k_2}{k_1}=2000$