Chemical Kinetics Question 148
Question: A catalyst lowers the activation energy of a certain reaction from 83.314 to $ 75kJmo{l^{-1}} $ at 500 K. What will be the rate of reaction as compared to uncatalysed reaction- Assume other things being equal.
Options:
A) Double
B) 28 times
C) 7.38 times
D) $ 7.38\times 10^{3} $ times
Show Answer
Answer:
Correct Answer: C
Solution:
[c] $ \frac{k_2}{k_1}=\frac{A{e^{-{E_{a_2}}/RT}}}{A{e^{-{E_{a_1}}/RT}}}={e^{({E_{a_1}}-{E_{a_2}})/RT}} $
$ 2.303\log \frac{k_2}{k_1}=\frac{{E_{a_1}}-E_{a}}{RT} $
$ =\frac{( 83.314-75 )\times 10^{3}}{8.314\times 500}=2 $
$ \log k_2=\frac{2}{2.303}=0.868 $ Taking Antilog $ k_2=7.38 $
BETA
