(B) According to the Arrhenius equation,$\ln \ k = \ln \ A - \frac{E_a}{RT}$.We are plotting $\ln \ k$ versus $\frac{10^{3}}{T}$.Rewriting the equatio

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(B) According to the Arrhenius equation,$\ln \ k = \ln \ A - \frac{E_a}{RT}$.We are plotting $\ln \ k$ versus $\frac{10^{3}}{T}$.Rewriting the equation: $\ln \ k = \ln \ A - \left( \frac{E_a}{R \times 10^{3}} \right) \times \left( \frac{10^{3}}{T} \right)$.Comparing this with the equation of a straight line $y = mx + c$,the slope $m = -\frac{E_a}{R \times 10^{3}}$.Given the slope is $-18.5$,we have: $-18.5 = -\frac{E_a}{8.31 \times 10^{3}}$.$E_a = 18.5 \times 8.31 \times 10^{3} \ J \ mol^{-1} = 153735 \ J \ mol^{-1}$.Converting to $kJ \ mol^{-1}$: $E_a = 153.735 \ kJ \ mol^{-1}$.Rounding to the nearest integer,we get $154 \ kJ \ mol^{-1}$.

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

Medium

The rate constants for the decomposition of acetaldehyde have been measured over the temperature range $700-1000 \ K$. The data has been analysed by plotting a $\ln \ k \ vs \ \frac{10^{3}}{T}$ graph,which gives a slope of $-18.5$. The value of activation energy for the reaction is $...... \ kJ \ mol^{-1}$. (Nearest integer) (Given: $R = 8.31 \ J \ K^{-1} \ mol^{-1}$)

JEE MAIN 2022 All JEE MAIN

A
$234$
B
$154$
C
$701$
D
$185$

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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 slope of the Arrhenius plot $(\ln k \, vs \, \frac{1}{T})$ of a first-order reaction is $-5 \times 10^{3} \, K$. The value of $E_{a}$ of the reaction is: (in $kJ \, mol^{-1}$)
$[\text{Given } R = 8.314 \, J \, K^{-1} \, mol^{-1}]$

MediumNEET 2021

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The decomposition of formic acid on a gold surface follows first-order kinetics. If the rate constant at $300 \ K$ is $1.0 \times 10^{-3} \ s^{-1}$ and the activation energy $E_a = 11.488 \ kJ \ mol^{-1}$,the rate constant at $200 \ K$ is ............ $\times 10^{-5} \ s^{-1}$.
(Round off to the Nearest Integer).
(Given: $R = 8.314 \ J \ mol^{-1} K^{-1}$)

DifficultJEE MAIN 2021

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For $A + B \longrightarrow C + D$; $\Delta H = -20 \ kJ \ mol^{-1}$,the activation energy of the forward reaction is $85 \ kJ \ mol^{-1}$. The activation energy for the backward reaction is.....$kJ \ mol^{-1}$

Medium

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For $N_2 + 3H_2 \rightarrow 2NH_3$,$\Delta H = -22 \ kcal$,and $E_a = 70 \ kcal$. Hence $E_a$ for $2NH_3 \rightarrow N_2 + 3H_2$ is $.....$ $kcal$.

Medium

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If the half-lives of a first-order reaction at $350 \ K$ and $300 \ K$ are $2 \ s$ and $20 \ s$ respectively,the activation energy of the reaction in $kJ \ mol^{-1}$ is:

MediumAP EAMCET 2020

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The slope of the Arrhenius plot $(\ln k \, vs \, \frac{1}{T})$ of a first-order reaction is $-5 \times 10^{3} \, K$. The value of $E_{a}$ of the reaction is: (in $kJ \, mol^{-1}$)$[\text{Given } R = 8.314 \, J \, K^{-1} \, mol^{-1}]$

For $A + B \longrightarrow C + D$; $\Delta H = -20 \ kJ \ mol^{-1}$,the activation energy of the forward reaction is $85 \ kJ \ mol^{-1}$. The activation energy for the backward reaction is.....$kJ \ mol^{-1}$

For $N_2 + 3H_2 \rightarrow 2NH_3$,$\Delta H = -22 \ kcal$,and $E_a = 70 \ kcal$. Hence $E_a$ for $2NH_3 \rightarrow N_2 + 3H_2$ is $.....$ $kcal$.

If the half-lives of a first-order reaction at $350 \ K$ and $300 \ K$ are $2 \ s$ and $20 \ s$ respectively,the activation energy of the reaction in $kJ \ mol^{-1}$ is:

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The slope of the Arrhenius plot $(\ln k \, vs \, \frac{1}{T})$ of a first-order reaction is $-5 \times 10^{3} \, K$. The value of $E_{a}$ of the reaction is: (in $kJ \, mol^{-1}$)
$[\text{Given } R = 8.314 \, J \, K^{-1} \, mol^{-1}]$