Chemical Kinetics Question 121
Question: Consider the reaction $ A\to 2B+C,\Delta H=-15 $ kcal. The energy of activation of backward reaction is $ 20kcalmo{l^{-1}} $ . In presence of catalyst the energy of activation of forward reaction is $ 3kcalmo{l^{-1}} $ . At 400 K. the catalyst causes the rate of the reaction to increase by the number of times equal to
Options:
A) $ {e^{3.5}} $
B) $ {e^{2.5}} $
C) $ {e^{-,2.5}} $
D) $ {e^{2.303}} $
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Answer:
Correct Answer: B
Solution:
[b] $ E_{a}(f)-E_{a}(b)=\Delta H=15,k,cal $
$ \Rightarrow E_{a}(f)=-15+20=5,kcal $
$ \frac{k(catalyst)}{k}={e^{\frac{E_{a}-{E_{a(,catalyst)}}}{RT}}} $
$ ={e^{\frac{(5-3)\times 10^{3}}{2\times 400}}}={e^{2.5}} $
BETA
