Chemical Kinetics and Nuclear ChemistryHard
Question
Rate constant k varies with temperature by equation , log k(min-1) = 5 - 
. We can conclude:
. We can conclude:Options
A.pre-exponential factor A is 5
B.Ea is 2000 kcal
C.pre-exponential factor A is 105
D.Ea is 9.212 kcal
Solution
(C) Given, log k (min-1) = 5 - 
Compare this with
log K = log A -
we find A = 1 × 105
(D)
= - 2000
Ea = 9.212 k cal.
Compare this with
log K = log A -
we find A = 1 × 105
(D)
= - 2000Ea = 9.212 k cal.
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