(A) The Arrhenius equation is given by $K = A \, e^{-E_a / (RT)}$.Comparing this with the given equation $K = 5.0 \times 10^{11} \, e^{-2000/T}$,we ge

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(A) The Arrhenius equation is given by $K = A \, e^{-E_a / (RT)}$.Comparing this with the given equation $K = 5.0 \times 10^{11} \, e^{-2000/T}$,we get:$E_a / R = 2000 \, K$.Therefore,$E_a = 2000 \, K \times R$.Using the value of the gas constant $R = 8.314 \, J \, K^{-1} \, mol^{-1}$:$E_a = 2000 \times 8.314 = 16628 \, J \, mol^{-1}$.Converting to $kJ \, mol^{-1}$:$E_a = 16.628 \, kJ \, mol^{-1}$.

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

Medium

The decomposition of $N_2O$ into $N_2$ and $O_2$ in the presence of argon follows first-order kinetics. If the rate constant is given by $K = 5.0 \times 10^{11} \, e^{-2000/T}$,what is the activation energy of the reaction?

A
$16.628 \, kJ \, mol^{-1}$
B
$166.28 \, kJ \, mol^{-1}$
C
$166.28 \, J \, mol^{-1}$
D
$16.628 \, J \, mol^{-1}$

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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 $N_2O$ into $N_2$ and $O_2$ in the presence of argon follows first-order kinetics. If the rate constant is given by $K = 5.0 \times 10^{11} \, e^{-2000/T}$,what is the activation energy of the reaction?

Similar Questions

Which of the following options is incorrect?

For the reaction $A \rightleftharpoons B$,if $E_a$ is the activation energy of the forward reaction,then the activation energy of the backward reaction is:

For a reaction,the temperature coefficient is $2$. Calculate the activation energy $(E_a)$ in $kJ/mol$. (Assume the temperature range is $25^{\circ}C$ to $35^{\circ}C$)

The rate of a chemical reaction doubles for every $10^\circ \text{C}$ rise in temperature. If the temperature is increased by $50^\circ \text{C}$,the rate of reaction will increase by a factor of ........

For a reaction,the threshold energy is $75 \ kJ/mol$. If the internal energy of the reactants is $20 \ kJ/mol$,the activation energy (in $kJ/mol$) is

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