The rate constant for a first order reaction is given by the equation:lnk=33.24-2.0×104 K TThe Activation energy for the reaction is given by _________kJmol-1. (In Nearest integer)(Given: R=8.3 J K-1 mol-1 )

lnk=lnA-EART Given: lnk=33.24-2.0×104 T ∴ on comparing EAR=2.0×104 ∴EA=2.0×104×R ⇒EA=2.0×104×8.3 J ⇒EA=166×103 J=166 kJ

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The rate constant for a first order reaction is given by the 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: $R = 8.3 J K^{- 1} {mol}^{- 1}$ )

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answer is 166.

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Detailed Solution

$lnk = lnA - \frac{E_{A}}{RT} Given: lnk = 33.24 - \frac{2.0 \times 10^{4}}{T} ∴ on comparing \frac{E_{A}}{R} = 2.0 \times 10^{4} ∴ E_{A} = 2.0 \times 10^{4} \times R \Rightarrow E_{A} = 2.0 \times 10^{4} \times 8.3 J \Rightarrow E_{A} = 166 \times 10^{3} J = 166 kJ$