(B) According to the Arrhenius equation,$k = A e^{-E_a / RT}$.Taking the natural logarithm on both sides,$\ln k = \ln A - \frac{E_a}{RT}$.Differentiat

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$A$ high value of activation energy

(B) According to the Arrhenius equation,$k = A e^{-E_a / RT}$.Taking the natural logarithm on both sides,$\ln k = \ln A - \frac{E_a}{RT}$.Differentiating with respect to temperature,$\frac{d(\ln k)}{dT} = \frac{E_a}{RT^2}$.This equation shows that the rate constant $k$ is highly sensitive to temperature changes when the activation energy $E_a$ is high.Therefore,a small increase in temperature leads to a significant increase in the rate of reaction if the activation energy is high.

Class 12 Chemistry Chemical Kinetics 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:

A
$A$ high value of the pre-exponential factor
B
$A$ high value of activation energy
C
$A$ large number of collisions at high temperature
D
An increase in the average energy of molecules

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Class 12

Chemistry Questions

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Chemical Kinetics

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Collision theory, Energy of activation and Arrhenius equation

For a chemical reaction,if the temperature is increased from $25\,^{\circ}C$ to $55\,^{\circ}C$,the rate of reaction will change by a factor of (Assume $\mu = 3$).

Medium

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$A$ catalyst:

MediumIIT 1984

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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 taken for half of the initial amount of $N_2O_5$ to decompose is $12 \ min$ at $310 \ K$ and $2 \ hrs$ at $300 \ K$. The activation energy of the reaction in $kJ \ mol^{-1}$ is $\left(R=8.3 \ J \ K^{-1} \ mol^{-1}\right)$

EasyAP EAMCET 2019

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For an elementary chemical reaction, the Arrhenius plot is given below. If the energy of activation is $6.64 \ kJ \ mol^{-1}$ and $R = 8.3 \ J \ K^{-1} \ mol^{-1}$, the temperature at which the rate constant becomes $e^2 \ min^{-1}$, is (in $K$)

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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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For a chemical reaction,if the temperature is increased from $25\,^{\circ}C$ to $55\,^{\circ}C$,the rate of reaction will change by a factor of (Assume $\mu = 3$).

$A$ catalyst:

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 taken for half of the initial amount of $N_2O_5$ to decompose is $12 \ min$ at $310 \ K$ and $2 \ hrs$ at $300 \ K$. The activation energy of the reaction in $kJ \ mol^{-1}$ is $\left(R=8.3 \ J \ K^{-1} \ mol^{-1}\right)$

For an elementary chemical reaction, the Arrhenius plot is given below. If the energy of activation is $6.64 \ kJ \ mol^{-1}$ and $R = 8.3 \ J \ K^{-1} \ mol^{-1}$, the temperature at which the rate constant becomes $e^2 \ min^{-1}$, is (in $K$)

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