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JEE Main 202131 AugustEvening Shift
Exam
JEE Main
Year
2021
Date
31 August
Shift
Evening Shift
Subject
Chemistry
Chapter
Chemical Kinetics

JEE Main 2021 · 31 August · Evening Shift · Chemistry · Chemical Kinetics

For the reaction A → B, the rate constant k (in s ^ - 1 ) is given...

For the reaction A→B,A \to B, the rate constant kk (in s−1{s}^{- 1}) is given by
log⁡10k=20.35−(2.47×103)T\log_{10}k = 20 . 35 - \dfrac{\left(2 . 47 \times {10}^{3}\right)}{T}
The energy of activation in kJmol−1kJ{mol}^{- 1} is ________ . (Nearest integer)
[Given : R=8.314JK−1mol−1R = 8 . 314J{K}^{- 1}{mol}^{- 1}]

Correct answer

0

Step-by-step solution

log⁡10k=logA−Ea2.303RT\log_{10}k = logA - \dfrac{{E}_{a}}{2 . 303RT}
log⁡10k=20.35−(2.47×103)T\log_{10}k = 20 . 35 - \dfrac{\left(2 . 47 \times {10}^{3}\right)}{T}
Ea2.303R=2.47×103\dfrac{{E}_{a}}{2 . 303R} = 2 . 47 \times {10}^{3}
Ea=2.47×103×2.303×8.3141000=47.29KJ/mole{E}_{a} = \dfrac{2 . 47 \times 103 \times 2 . 303 \times 8 . 314}{1000} = 47 . 29KJ / mole
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