(B) The Arrhenius equation is given by $\log K = \log A - \frac{E_a}{2.303 RT}$.Comparing this with the equation of a straight line $y = mx + c$,where

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(B) The Arrhenius equation is given by $\log K = \log A - \frac{E_a}{2.303 RT}$.Comparing this with the equation of a straight line $y = mx + c$,where $y = \log K$ and $x = 1/T$,the slope $m = -\frac{E_a}{2.303 R}$.Given slope $m = -5632$.So,$-5632 = -\frac{E_a}{2.303 \times 8.314 \ J \ K^{-1} \ mol^{-1}}$.$E_a = 5632 \times 2.303 \times 8.314 \ J \ mol^{-1}$.$E_a = 107840 \ J \ mol^{-1} = 107.84 \ kJ \ mol^{-1}$.

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

Difficult

The rate constant of a reaction varies with temperature according to the equation: $\log K = \text{constant} - \frac{E_a}{2.303 RT}$. If a plot of $\log K$ versus $1/T$ yields a straight line with a slope of $-5632$,then the activation energy of the reaction is .......... $kJ \ mol^{-1}$.

A
$127.67$
B
$107.84$
C
$86$
D
$246.8$

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Class 12

Chemistry Questions

Chapter

Chemical Kinetics

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Collision theory, Energy of activation and Arrhenius equation

For a reaction,the value of rate constant at $300 \ K$ is $6.0 \times 10^5 \ s^{-1}$. The value of Arrhenius factor $A$ at infinitely high temperature is

MediumKCET 2023

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Which among the following is correct when energy of activation,$E_a$ of the catalyzed reaction decreases at constant temperature and for same concentration?

EasyMHT CET 2020

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

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Slope of the straight line obtained by plotting $\log_{10} k$ against $\frac{1}{T}$ represents which term?

EasyMHT CET 2018

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For a first order decomposition of a certain reaction,rate constant is given by the equation $\log k \left( s^{-1} \right) = 7.14 - \frac{1 \times 10^4 \ K}{T}$. The activation energy of the reaction (in $kJ \ mol^{-1}$) is $(R = 8.3 \ J \ K^{-1} \ mol^{-1})$ (in $.1$)

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The rate constant of a reaction varies with temperature according to the equation: $\log K = \text{constant} - \frac{E_a}{2.303 RT}$. If a plot of $\log K$ versus $1/T$ yields a straight line with a slope of $-5632$,then the activation energy of the reaction is .......... $kJ \ mol^{-1}$.

Similar Questions

For a reaction,the value of rate constant at $300 \ K$ is $6.0 \times 10^5 \ s^{-1}$. The value of Arrhenius factor $A$ at infinitely high temperature is

Which among the following is correct when energy of activation,$E_a$ of the catalyzed reaction decreases at constant temperature and for same concentration?

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

Slope of the straight line obtained by plotting $\log_{10} k$ against $\frac{1}{T}$ represents which term?

For a first order decomposition of a certain reaction,rate constant is given by the equation $\log k \left( s^{-1} \right) = 7.14 - \frac{1 \times 10^4 \ K}{T}$. The activation energy of the reaction (in $kJ \ mol^{-1}$) is $(R = 8.3 \ J \ K^{-1} \ mol^{-1})$ (in $.1$)

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