Chemical Engineering - Chemical Reaction Engineering - Discussion

Discussion Forum : Chemical Reaction Engineering - Section 2 (Q.No. 3)
3.
At a given value of E/R (ratio of activation energy and gas constant), the ratio of the rate constants at 500°K and 400°K is 2, if Arrhenious law is used. What will be this ratio, if transition state theory is used with the same value of E/R?
1.6
2
2.24
2.5
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
4 comments Page 1 of 1.

Niki said:   1 decade ago
According to Arrhenious law,
ln (K2/K1)= E/R (1/T1 -1/T2)
ln (2)= E/R ( 1/400 - 1/500)
E/R = 1386
Now applying Transition state theory,
ln (K2/K1)= E/R (1/T1 -1/T2) + ln (T2/T1)
ln (K2/K1) =1386( 1/400 - 1/500) + ln (500/400)
K2/K1 = 2.4999
~2.5
(3)

Arun said:   6 years ago
Explain the term in transition state formula.

Ram said:   3 years ago
500/400 * 2 = 2.5.
(3)

Sunil Sharma said:   2 years ago
Explain the answer.

Post your comments here:

Your comments will be displayed after verification.