(A) The Arrhenius equation is given by $k = A e^{-E^*/RT}$.Taking the logarithm on both sides,we get $\log \, k = \log \, A - \frac{E^*}{2.303 R} \cdo

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(A) The Arrhenius equation is given by $k = A e^{-E^*/RT}$.Taking the logarithm on both sides,we get $\log \, k = \log \, A - \frac{E^*}{2.303 R} \cdot \frac{1}{T}$.This equation follows the linear form $y = mx + c$,where $y = \log \, k$ and $x = \frac{1}{T}$.Plotting $\log \, k$ versus $\frac{1}{T}$ yields a straight line with a slope equal to $-\frac{E^*}{2.303 R}$.Thus,the activation energy $E^*$ can be determined from the slope of this plot.

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

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The temperature dependence of the rate constant $(k)$ of a chemical reaction is expressed by the Arrhenius equation,$k = A \cdot e^{-E^/RT}$. The activation energy $(E^)$ of the reaction can be calculated by plotting:

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A
$\log \, k$ vs $\frac{1}{T}$
B
$k$ vs $T$
C
$k$ vs $\frac{1}{\log \, T}$
D
$\log \, k$ vs $T$

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Class 12 Questions

Class 12

Chemistry Questions

Chapter

Chemical Kinetics

Subtopic

Collision theory, Energy of activation and Arrhenius equation

Previous Year

AIPMT 2003 Paper All AIPMT years →

Define the following terms:
$1.$ Effective collision
$2.$ Probability or steric factor $(P)$

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How does the graph of the fraction of molecules versus kinetic energy change at different temperatures?

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The rate of the chemical reaction doubles for an increase of $10 \, K$ in absolute temperature from $298 \, K$. Calculate $E_{a}$.

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The rate constant of a first-order reaction is $1.5 \times 10^7 \text{ s}^{-1}$ at $300 \text{ K}$ and $3.0 \times 10^7 \text{ s}^{-1}$ at $330 \text{ K}$. What is the activation energy $(E_a)$ for the reaction? $[R \times 2.303 = 19.15 \text{ J K}^{-1} \text{mol}^{-1}]$

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The activation energy of a first-order reaction at $25\,^{\circ}C$ is $30\,kJ/mol$. In the presence of a catalyst,the activation energy of the same reaction at $25\,^{\circ}C$ becomes $24\,kJ/mol$. The rate of the reaction in the presence of the catalyst will be how many times the original rate?

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The temperature dependence of the rate constant $(k)$ of a chemical reaction is expressed by the Arrhenius equation,$k = A \cdot e^{-E^*/RT}$. The activation energy $(E^*)$ of the reaction can be calculated by plotting:

Similar Questions

Define the following terms:$1.$ Effective collision$2.$ Probability or steric factor $(P)$

How does the graph of the fraction of molecules versus kinetic energy change at different temperatures?

The rate of the chemical reaction doubles for an increase of $10 \, K$ in absolute temperature from $298 \, K$. Calculate $E_{a}$.

The rate constant of a first-order reaction is $1.5 \times 10^7 \text{ s}^{-1}$ at $300 \text{ K}$ and $3.0 \times 10^7 \text{ s}^{-1}$ at $330 \text{ K}$. What is the activation energy $(E_a)$ for the reaction? $[R \times 2.303 = 19.15 \text{ J K}^{-1} \text{mol}^{-1}]$

The activation energy of a first-order reaction at $25\,^{\circ}C$ is $30\,kJ/mol$. In the presence of a catalyst,the activation energy of the same reaction at $25\,^{\circ}C$ becomes $24\,kJ/mol$. The rate of the reaction in the presence of the catalyst will be how many times the original rate?

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The temperature dependence of the rate constant $(k)$ of a chemical reaction is expressed by the Arrhenius equation,$k = A \cdot e^{-E^/RT}$. The activation energy $(E^)$ of the reaction can be calculated by plotting:

Define the following terms:
$1.$ Effective collision
$2.$ Probability or steric factor $(P)$