(B) According to the Arrhenius equation,the rate constant $k$ is given by $k = A e^{-E_a/RT}$.As the temperature $T$ increases,the fraction of molecul

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Increase in the number of activated molecules

(B) According to the Arrhenius equation,the rate constant $k$ is given by $k = A e^{-E_a/RT}$.As the temperature $T$ increases,the fraction of molecules having energy equal to or greater than the activation energy $(E_a)$ increases significantly.This leads to a larger number of effective collisions,resulting in a substantial increase in the reaction rate.Therefore,the correct reason is the increase in the number of activated molecules.

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

Easy

$A$ large increase in the rate of reaction for a small increase in temperature is due to:

A
Increase in the number of collisions
B
Increase in the number of activated molecules
C
Decrease in the activation energy
D
Decrease in the mean free path

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Class 12 Questions

Class 12

Chemistry Questions

Chapter

Chemical Kinetics

Subtopic

Collision theory, Energy of activation and Arrhenius equation

Give an example of a collision in the proper direction that results in the formation of a product.

Easy

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For a first-order reaction,if the rate constant is $k_1$ at temperature $T_1$ and $k_2$ at temperature $T_2$,which of the following relations is correct? ($E_a$ = activation energy)

Easy

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The rate constant $(K')$ of one reaction is double the rate constant $(K'')$ of another reaction. What is the relationship between the corresponding activation energies of the two reactions (${E_a}'$ and ${E_a}''$)?

Medium

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The time required for $10 \%$ completion of a first order reaction at $298 \,K$ is equal to that required for its $25 \%$ completion at $308 \,K$. If the value of $A$ is $4 \times 10^{10} \,s^{-1}$,calculate $k$ at $318 \,K$ and $E_a$.

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For two reactions,the values of the pre-exponential factor are the same. However,the difference between their activation energy values is $24.9 \, kJ \, mol^{-1}$. The ratio of their rate constants at $300 \, K$ is ....

Medium

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$A$ large increase in the rate of reaction for a small increase in temperature is due to:

Similar Questions

Give an example of a collision in the proper direction that results in the formation of a product.

For a first-order reaction,if the rate constant is $k_1$ at temperature $T_1$ and $k_2$ at temperature $T_2$,which of the following relations is correct? ($E_a$ = activation energy)

The rate constant $(K')$ of one reaction is double the rate constant $(K'')$ of another reaction. What is the relationship between the corresponding activation energies of the two reactions (${E_a}'$ and ${E_a}''$)?

The time required for $10 \%$ completion of a first order reaction at $298 \,K$ is equal to that required for its $25 \%$ completion at $308 \,K$. If the value of $A$ is $4 \times 10^{10} \,s^{-1}$,calculate $k$ at $318 \,K$ and $E_a$.

For two reactions,the values of the pre-exponential factor are the same. However,the difference between their activation energy values is $24.9 \, kJ \, mol^{-1}$. The ratio of their rate constants at $300 \, K$ is ....

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