Chemical Kinetics Question 57
Question: The rate constant of a certain reaction is given by log $ K=A-\frac{B}{T}+C,\log ,T $ . Then activation energy of reaction at 300 K is:
Options:
A) $ \frac{B}{T}+C,\log ,T $
B) $ [2.303,B+CT]R $
C) C log T
D) B + CTR
Show Answer
Answer:
Correct Answer: B
Solution:
[b] $ \log k=A-\frac{B}{T}+C\log T $ Or In $ k=2.303A-\frac{B\times 2.303}{T}+C $ In T $ \frac{d, Ink}{dT}=[ \frac{2.303B}{T^{2}}+\frac{C}{T} ],\frac{d( Ink)}{dt}=\frac{E_{a}}{RT^{2}} $
$ E_{a}=[2.303B+CT]R. $
