The rate constant for a first order reaction is given by the following equation :
$lnk = 33.24 - \frac{2.0 \times 10^{4} K}{T}$
The Activation energy for the reaction is given by ${kJmol}^{- 1}$ . (In Nearest integer) (Given : $R = 8.3 J K^{- 1} {mol}^{- 1}$ )

Question

The rate constant for a first order reaction is given by the following equation :

$lnk = 33.24 - \frac{2.0 \times 10^{4} K}{T}$

The Activation energy for the reaction is given by ${kJmol}^{- 1}$ . (In Nearest integer) (Given : $R = 8.3 J K^{- 1} {mol}^{- 1}$ )

TestHub · Accepted Answer

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

Chemistry - Chemical Kinetics Question with Solution | TestHub

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