Chemical Kinetics Question 285

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}} $

Show Answer

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}} $



NCERT Chapter Video Solution

Dual Pane