(B) The Arrhenius equation is given by $K = Ae^{-E_a / RT}$.Comparing the given equation $K = (4.5 \times 10^{11} \, s^{-1}) e^{-28000 \, K / T}$ with

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(B) The Arrhenius equation is given by $K = Ae^{-E_a / RT}$.Comparing the given equation $K = (4.5 \times 10^{11} \, s^{-1}) e^{-28000 \, K / T}$ with the Arrhenius equation,we have:$\frac{E_a}{R} = 28000 \, K$$E_a = 28000 \, K \times R$Using $R = 8.314 \, J \, K^{-1} \, mol^{-1}$:$E_a = 28000 \times 8.314 \, J \, mol^{-1} = 232792 \, J \, mol^{-1}$Converting to $KJ \, mol^{-1}$:$E_a = \frac{232792}{1000} \, KJ \, mol^{-1} = 232.79 \, KJ \, mol^{-1}$.

Class 12 Chemistry Chemical Kinetics Collision theory, Energy of activation and Arrhenius equation

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The decomposition of a hydrocarbon follows the equation $K = (4.5 \times 10^{11} \, s^{-1}) e^{-28000 \, K/T}$. What will be the value of activation energy in $KJ \, mol^{-1}$?

A
$669$
B
$232.79$
C
$4.5 \times 10^{11}$
D
$28000$

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Chemical Kinetics

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Collision theory, Energy of activation and Arrhenius equation

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The decomposition of a hydrocarbon follows the equation $K = (4.5 \times 10^{11} \, s^{-1}) e^{-28000 \, K/T}$. What will be the value of activation energy in $KJ \, mol^{-1}$?

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