log10k=logA−2.303RTEa
log10k=20.35−T(2.47×103)
2.303REa=2.47×103
Ea=10002.47×103×2.303×8.314=47.29KJ/mole
JEE Main 2021 — Chemistry Physical Chemistry
For the reaction A→B, the rate constant k (in s−1) is given by
log10k=20.35−T(2.47×103)
The energy of activation in kJmol−1 is ________ . (Nearest integer)
[Given : R=8.314JK−1mol−1]
Held on 31 Aug 2021 · Verified 6 Jul 2026.
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