Chemical Kinetics Question 122
Question: A reaction takes place in various steps. The rate constant for first, second, third and fifth steps are $ k_1,k_2,k_3 $ and $ k_4 $ respectively The overall rate constant is given by $ k=\frac{k_2}{k_3}{{( \frac{k_1}{k_5} )}^{1/2}} $ . If activation energy are 40,60,50 and 10 kJ/mol respectively, the overall energy of activation (kJ/ mol) is:
Options:
A) 10
B) 20
C) 25
D) none of these
Show Answer
Answer:
Correct Answer: C
Solution:
[c] $ k=A.{e^{-E_{a}/}}^{(RT)} $
$ \therefore $ Effective overall energy of activation $ E_{a}=E_{a}(2)-E_{a}(3)+\frac{1}{2}E_{a}(1)-\frac{1}{2}E_{a}(5) $
$ =60-50+\frac{1}{2}\times 40-\frac{1}{2}\times 10=25,kJ/mol $
BETA
