(B) The Arrhenius equation is given by: $\log \frac{K_2}{K_1} = \frac{E_a}{2.303R} \left( \frac{T_2 - T_1}{T_1 T_2} \right)$.Given: $T_1 = 25 + 273 =

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$\frac{2.303 \times R \times 298 \times 308}{10} \log 2$

(B) The Arrhenius equation is given by: $\log \frac{K_2}{K_1} = \frac{E_a}{2.303R} \left( \frac{T_2 - T_1}{T_1 T_2} \right)$.Given: $T_1 = 25 + 273 = 298 \, K$,$T_2 = 35 + 273 = 308 \, K$,and $K_2 = 2K_1$.Substituting these values: $\log \frac{2K_1}{K_1} = \frac{E_a}{2.303R} \left( \frac{308 - 298}{298 \times 308} \right)$.$\log 2 = \frac{E_a}{2.303R} \left( \frac{10}{298 \times 308} \right)$.Rearranging for $E_a$: $E_a = \frac{2.303 \times R \times 298 \times 308 \times \log 2}{10}$.

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

Medium

For a certain gaseous reaction,the temperature is increased by $10\,^{\circ}C$ from $25\,^{\circ}C$ to $35\,^{\circ}C$. If the rate of the reaction doubles,what will be the value of the activation energy $(E_a)$?

A
$\frac{10}{2.303R \times 298 \times 308}$
B
$\frac{2.303 \times R \times 298 \times 308}{10} \log 2$
C
$\frac{0.693R \times 10}{298 \times 308}$
D
$\frac{2.303 \times R \times 298 \times 308}{10}$

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Chemical Kinetics

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Collision theory, Energy of activation and Arrhenius equation

The following equation is obtained for a first order reaction at $300 \ K$.
$\log_{10} \frac{k}{A} = 0.00174$
What is the activation energy (in $J \ mol^{-1}$) of the reaction?
$(R = 8.314 \ J \ mol^{-1} \ K^{-1})$

MediumAP EAMCET 2025

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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.

Medium

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The rate constant of a reaction is $2 \times 10^{-3} \ min^{-1}$ at $300 \ K$. By increasing the temperature by $10 \ K$,its value becomes double. Calculate the energy of activation $(E_a)$ and the rate constant at $320 \ K$.

Medium

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The rate constant,the activation energy and the Arrhenius parameter of a chemical reaction at $25\,^oC$ are $3.0 \times 10^{-4}\,s^{-1}$,$104.4\,kJ\,mol^{-1}$ and $6.0 \times 10^{14}\,s^{-1}$ respectively. The value of the rate constant as $T \to \infty$ is

IIT 1996

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Write the Arrhenius equation in the form $\ln \, k = -\frac{E_a}{RT} + \ln \, A$.

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For a certain gaseous reaction,the temperature is increased by $10\,^{\circ}C$ from $25\,^{\circ}C$ to $35\,^{\circ}C$. If the rate of the reaction doubles,what will be the value of the activation energy $(E_a)$?

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The following equation is obtained for a first order reaction at $300 \ K$.$\log_{10} \frac{k}{A} = 0.00174$What is the activation energy (in $J \ mol^{-1}$) of the reaction?$(R = 8.314 \ J \ mol^{-1} \ K^{-1})$

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.

The rate constant of a reaction is $2 \times 10^{-3} \ min^{-1}$ at $300 \ K$. By increasing the temperature by $10 \ K$,its value becomes double. Calculate the energy of activation $(E_a)$ and the rate constant at $320 \ K$.

The rate constant,the activation energy and the Arrhenius parameter of a chemical reaction at $25\,^oC$ are $3.0 \times 10^{-4}\,s^{-1}$,$104.4\,kJ\,mol^{-1}$ and $6.0 \times 10^{14}\,s^{-1}$ respectively. The value of the rate constant as $T \to \infty$ is

Write the Arrhenius equation in the form $\ln \, k = -\frac{E_a}{RT} + \ln \, A$.

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The following equation is obtained for a first order reaction at $300 \ K$.
$\log_{10} \frac{k}{A} = 0.00174$
What is the activation energy (in $J \ mol^{-1}$) of the reaction?
$(R = 8.314 \ J \ mol^{-1} \ K^{-1})$