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 \[\frac{k}{k}={{e}^{({{E}_{a}}-{{E}_{a}})/RT}}\] \[\Rightarrow \] \[\frac{k}{k}={{e}^{\left( \frac{2000}{2\times 300} \right)}}={{e}^{3.33}}\] [\[\because \]\[{{E}_{a}}-{{E}_{a}}=2000\](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 \[R=28\,R\]
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