For the reaction A → B , the rate constant k (in s - 1 ) is given by log 10 k = 20 . 35 - 2 . 47 &#215…

For the reaction AB, the rate constant k (in s-1) is given by
log10k=20.35-2.47×103T
The energy of activation in kJ mol-1 is ________ . (Nearest integer)
[Given : R=8.314 J K-1 mol-1]

Solution

log10k=logA-Ea2.303RT
log10k=20.35-2.47×103T
Ea2.303R=2.47×103
Ea=2.47×103×2.303×8.3141000=47.29KJ/mole

Asked in: JEE Main 2021 (31 Aug Shift 2)

Practice more Chemical Kinetics questions on Aicharya