Question

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

Reaction		Energy of activation
		$ \, \, \, \, \, \, \, (k J \, mol^{-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-
$\log k =\log A -\frac{ E _{ a }}{2.303 RT }$
$\log k _{1}=\log A -\frac{ E _{ a _{(1)}}}{2.303 RT } \ldots ( i )$
and $\log k _{2}=\log A -\frac{ E _{ a _{(2)}}}{2.303 RT }$
or $\log \frac{ k _{2}}{ k _{1}}=\frac{1}{2.303 RT }\left[ E _{ a _{(1)}}- E _{ a _{(2)}}\right]$ (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$