A) 20 times
B) 28 times
C) 14 times
D) remain the same
Correct Answer: B
Solution :
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 \[\frac{k'}{k}={{e}^{({{E}_{a}}-E_{a}^{'})/RT}}\] \[\Rightarrow \] \[\frac{k'}{k}={{e}^{\left( \frac{2000}{2\times 300} \right)}}={{e}^{3.33}}\] [\[\because \]\[{{E}_{a}}-{{E}_{a'}}=2000\,\,\text{cal}\](given)] On taking log both sides, \[\log \frac{k'}{k}=\frac{3.33}{2.303}\] \[\frac{k'}{k}=\text{antilog}\,\text{1}\text{.446}\,\,\text{=}\,\text{27}\text{.92}\approx 28\] We know that, rate \[(R)\propto k\] \[\therefore \] \[\frac{R'}{R}=\frac{k'}{k}=28\] or \[R'=28\,R\]
You need to login to perform this action.
You will be redirected in
3 sec
