(D) The Arrhenius equation is given by $\ln k = \ln A - \frac{E_a}{RT}$.Converting to base $10$,we get $\log_{10} k = \log_{10} A - \frac{E_a}{2.303 R

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(D) The Arrhenius equation is given by $\ln k = \ln A - \frac{E_a}{RT}$.Converting to base $10$,we get $\log_{10} k = \log_{10} A - \frac{E_a}{2.303 RT}$.Comparing this with the given equation $\log_{10} k = -2000 \left(\frac{1}{T}\right) + 6$,we identify the slope as $-\frac{E_a}{2.303 R} = -2000$.Thus,$E_a = 2000 \times 2.303 \times R$.Using $R = 8 \, J \, mol^{-1} K^{-1}$ and the approximation $\ln x \approx 2.3 \log_{10} x$ provided in the question:$E_a = 2000 \times 2.3 \times 8 = 36800 \, J \, mol^{-1}$.Converting to $kJ \, mol^{-1}$,we get $E_a = \frac{36800}{1000} = 36.8 \, kJ \, mol^{-1}$.

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

Medium

For a first order reaction $A \rightarrow P$,the temperature $(T)$ dependent rate constant $(k)$ was found to follow the equation $\log_{10} k = -(2000) \frac{1}{T} + 6$. The activation energy $(E_a)$ of the reaction in $kJ \, mol^{-1}$ will be ......... (Given: $\ln x = 2.3 \times \log_{10} x$ and $R = 8 \, J \, mol^{-1} K^{-1}$)

A
$9.2$
B
$16.2$
C
$116.6$
D
$36.8$

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Chemical Kinetics

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Collision theory, Energy of activation and Arrhenius equation

At $25\,^{\circ}C$,the values of rate constant,activation energy and Arrhenius constant of a reaction are $3 \times 10^{-4}\, s^{-1}$,$129\, kJ/mol$ and $2 \times 10^{15}\, s^{-1}$ respectively. The value of rate constant as $T \to \infty$ is

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Assertion $(A)$ : $A$ catalyst increases the rate of a reaction.
Reason $(R)$ : In presence of a catalyst, the activation energy of the reaction increases.
The correct answer is

MediumAP EAMCET 2007

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The temperature at which the rate constants of the two gaseous reactions given below become equal is . . . . . . $K$. (Nearest integer).
$X \longrightarrow Y \quad k_1 = 10^6 e^{\frac{-30000}{T}}$
$P \longrightarrow Q \quad k_2 = 10^4 e^{\frac{-24000}{T}}$
Given: $\ln 10 = 2.303$

DifficultJEE MAIN 2026

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The activation energy of a reaction is $94.14 \, kJ \, mol^{-1}$ and the rate constant at $310 \, K$ is $10^{-2} \, s^{-1}$. What is the value of the frequency factor $(A)$?

Medium

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The following graph is obtained for a reaction $(A \rightarrow P)$. The activation energy ($E_{a}$ in $kJ \ mol^{-1}$) and heat of reaction ($|\Delta H|$ in $kJ \ mol^{-1}$) for this reaction are respectively ($x=$ reaction coordinate; $y=E$ in $kJ \ mol^{-1}$)

MediumAP EAMCET 2025

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At $25\,^{\circ}C$,the values of rate constant,activation energy and Arrhenius constant of a reaction are $3 \times 10^{-4}\, s^{-1}$,$129\, kJ/mol$ and $2 \times 10^{15}\, s^{-1}$ respectively. The value of rate constant as $T \to \infty$ is

Assertion $(A)$ : $A$ catalyst increases the rate of a reaction.Reason $(R)$ : In presence of a catalyst, the activation energy of the reaction increases.The correct answer is

The temperature at which the rate constants of the two gaseous reactions given below become equal is . . . . . . $K$. (Nearest integer).$X \longrightarrow Y \quad k_1 = 10^6 e^{\frac{-30000}{T}}$$P \longrightarrow Q \quad k_2 = 10^4 e^{\frac{-24000}{T}}$Given: $\ln 10 = 2.303$

The activation energy of a reaction is $94.14 \, kJ \, mol^{-1}$ and the rate constant at $310 \, K$ is $10^{-2} \, s^{-1}$. What is the value of the frequency factor $(A)$?

The following graph is obtained for a reaction $(A \rightarrow P)$. The activation energy ($E_{a}$ in $kJ \ mol^{-1}$) and heat of reaction ($|\Delta H|$ in $kJ \ mol^{-1}$) for this reaction are respectively ($x=$ reaction coordinate; $y=E$ in $kJ \ mol^{-1}$)

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Assertion $(A)$ : $A$ catalyst increases the rate of a reaction.
Reason $(R)$ : In presence of a catalyst, the activation energy of the reaction increases.
The correct answer is

The temperature at which the rate constants of the two gaseous reactions given below become equal is . . . . . . $K$. (Nearest integer).
$X \longrightarrow Y \quad k_1 = 10^6 e^{\frac{-30000}{T}}$
$P \longrightarrow Q \quad k_2 = 10^4 e^{\frac{-24000}{T}}$
Given: $\ln 10 = 2.303$