(B) The Arrhenius equation is given by:$k = A e^{-E_a / RT}$Taking the natural logarithm on both sides:$\ln k = \ln A - \frac{E_a}{RT}$Rearranging thi

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(B) The Arrhenius equation is given by:$k = A e^{-E_a / RT}$Taking the natural logarithm on both sides:$\ln k = \ln A - \frac{E_a}{RT}$Rearranging this into the linear equation form $y = mx + c$:$\ln k = -\frac{E_a}{R} \left( \frac{1}{T} \right) + \ln A$Comparing this with $y = mx + c$,where $y = \ln k$,$x = \frac{1}{T}$,and $c = \ln A$,the slope $m$ is equal to $-\frac{E_a}{R}$.

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

Medium

What is the slope of the straight line for the graph drawn between $\ln k$ and $\frac{1}{T}$,where $k$ is the rate constant of a reaction at temperature $T$?

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A
$\frac{-E_a}{2.303 R}$
B
$\frac{-E_a}{R}$
C
$\frac{E_a}{R}$
D
$\frac{R}{E_a}$

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Class 12 Questions

Class 12

Chemistry Questions

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Chemical Kinetics

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Collision theory, Energy of activation and Arrhenius equation

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For a gaseous reaction,a large increase in the rate of reaction with a small increase in temperature indicates:

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In an exothermic reaction $A \to B$,the evolved heat is $280 \ kJ \ mol^{-1}$ and the activation energy is $200 \ kJ \ mol^{-1}$. The activation energy of the reverse reaction $B \to A$ is $.......... \ kJ \ mol^{-1}$.

Medium

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According to the collision theory of reaction rates:

Easy

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The rate constant for the first order decomposition of a certain reaction is described by the equation $\ln k (s^{-1}) = 14.34 - \frac{1.25 \times 10^{4} \ K}{T}$. The energy of activation for this reaction is

MediumAIIMS 2018

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The activation energy for a simple chemical reaction $A \to B$ is ${E_a}$ in the forward direction. The activation energy for the reverse reaction:

EasyAIPMT 2003

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What is the slope of the straight line for the graph drawn between $\ln k$ and $\frac{1}{T}$,where $k$ is the rate constant of a reaction at temperature $T$?

Similar Questions

For a gaseous reaction,a large increase in the rate of reaction with a small increase in temperature indicates:

In an exothermic reaction $A \to B$,the evolved heat is $280 \ kJ \ mol^{-1}$ and the activation energy is $200 \ kJ \ mol^{-1}$. The activation energy of the reverse reaction $B \to A$ is $.......... \ kJ \ mol^{-1}$.

According to the collision theory of reaction rates:

The rate constant for the first order decomposition of a certain reaction is described by the equation $\ln k (s^{-1}) = 14.34 - \frac{1.25 \times 10^{4} \ K}{T}$. The energy of activation for this reaction is

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