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JEE Main202227 Jun 2022Morning ShiftChemistryChemical KineticsActual

The rate constant for a first order reaction is given by the following equation : lnk = 33 . 24 - 2 . 0 × 10 4 K T The Activation energy for the reaction is given by kJmol - 1 . (In Nearest integer) (Given : R = 8 . 3 J K - 1 mol - 1 )

Correct answer

0

Step-by-step solution

lnk = lnA - E A RT Given: lnk = 33 . 24 - 2 . 0 × 10 4   T ∴ on comparing E A R = 2 . 0 × 10 4 ∴   E A = 2 . 0 × 10 4 × R ⇒ E A = 2 . 0 × 10 4 × 8 . 3   J ⇒ E A = 16 . 6 × 10 4   J = 166 kJ

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