(B) . According to the Arrhenius equation,$K = A e^{-E_a / RT}$,the rate constant $K$ depends on temperature. As temperature increases,the fraction of

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the fraction of molecules having energy in excess of the activation energy increases with temperature

(B) . According to the Arrhenius equation,$K = A e^{-E_a / RT}$,the rate constant $K$ depends on temperature. As temperature increases,the fraction of molecules having kinetic energy greater than the activation energy $(E_a)$ increases significantly. This factor,represented by $e^{-E_a / RT}$,is the primary reason for the rapid increase in the reaction rate.

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

Medium

The rate of gas phase chemical reactions generally increases rapidly with a rise in temperature. This is mainly because

KVPY 2011 All KVPY

A
the collision frequency increases with temperature
B
the fraction of molecules having energy in excess of the activation energy increases with temperature
C
the activation energy decreases with temperature
D
the average kinetic energy of molecules increases with temperature

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Class 12

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Chemical Kinetics

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Collision theory, Energy of activation and Arrhenius equation

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Activation energy of any reaction depends on

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The decomposition of a hydrocarbon follows the equation $k = (4.5 \times 10^{11} \ s^{-1}) e^{-28000 \ K / T}$. Calculate the activation energy $E_a$.

Medium

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The reason for almost doubling the rate of reaction on increasing the temperature of the reaction system by $10 \, ^\circ C$ is

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According to the collision theory of reaction rates,the rate of reaction increases with temperature due to:

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The rate constant for a chemical reaction taking place at $500 \ K$ is expressed as $K = A \ e^{-1000}$. The activation energy of the reaction is:

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The rate of gas phase chemical reactions generally increases rapidly with a rise in temperature. This is mainly because

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Activation energy of any reaction depends on

The decomposition of a hydrocarbon follows the equation $k = (4.5 \times 10^{11} \ s^{-1}) e^{-28000 \ K / T}$. Calculate the activation energy $E_a$.

The reason for almost doubling the rate of reaction on increasing the temperature of the reaction system by $10 \, ^\circ C$ is

According to the collision theory of reaction rates,the rate of reaction increases with temperature due to:

The rate constant for a chemical reaction taking place at $500 \ K$ is expressed as $K = A \ e^{-1000}$. The activation energy of the reaction is:

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