(B) The Arrhenius equation is given by $\ln k = \ln A - \frac{E_A}{RT}$.Comparing this with the given equation $\ln k = 33.24 - \frac{2.0 \times 10^{4

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(B) The Arrhenius equation is given by $\ln k = \ln A - \frac{E_A}{RT}$.Comparing this with the given equation $\ln k = 33.24 - \frac{2.0 \times 10^{4}}{T}$,we get:$\frac{E_A}{R} = 2.0 \times 10^{4} \, K$.Therefore,$E_A = 2.0 \times 10^{4} \times R$.Substituting $R = 8.3 \, J \, K^{-1} \, mol^{-1}$:$E_A = 2.0 \times 10^{4} \times 8.3 \, J \, mol^{-1} = 16.6 \times 10^{4} \, J \, mol^{-1}$.Converting to $kJ \, mol^{-1}$:$E_A = \frac{16.6 \times 10^{4}}{1000} \, kJ \, mol^{-1} = 166 \, kJ \, mol^{-1}$.

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

Medium

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

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$15$
B
$166$
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$961$
D
$247$

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Class 12

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Chemical Kinetics

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Collision theory, Energy of activation and Arrhenius equation

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

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For the reaction $2NO + Cl_2 \rightarrow 2NOCl$,the rate is given by $Rate = K[NO]^2[Cl_2]$. How can the rate constant $(K)$ of the reaction be increased?

For the reaction $C_2H_5I + OH^- \rightarrow C_2H_5OH + I^-$,the rate constants at $30^\circ C$ and $60^\circ C$ are $0.325$ and $6.735 \ L \ mol^{-1} \ s^{-1}$ respectively. The value of activation energy $(E_a)$ is .......... calories.

For an effective collision to occur,the colliding molecules must possess:

The rate of reaction becomes double when the temperature of the reaction is increased from $27\,^{\circ}C$ to $57\,^{\circ}C$. The activation energy of the reaction will be ....... $k\,cal$.

Identify the intermediate formed in the reaction $H_2 + I_2 \rightarrow 2 HI$.

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