(D) The Arrhenius equation is given by $k = A e^{-E_a / (RT)}$.At infinitely high temperature,i.e.,$T \to \infty$,the term $\frac{E_a}{RT} \to 0$.Ther

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(D) The Arrhenius equation is given by $k = A e^{-E_a / (RT)}$.At infinitely high temperature,i.e.,$T \to \infty$,the term $\frac{E_a}{RT} \to 0$.Therefore,the equation becomes $k = A e^0 = A \times 1 = A$.Since the Arrhenius factor $A$ is a constant independent of temperature,its value remains the same as the rate constant $k$ at $T \to \infty$.Given that $k = 6.0 \times 10^5 \ s^{-1}$ at $300 \ K$,and $A$ is independent of temperature,the value of $A$ is $6.0 \times 10^5 \ s^{-1}$.

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

Medium

For a reaction,the value of rate constant at $300 \ K$ is $6.0 \times 10^5 \ s^{-1}$. The value of Arrhenius factor $A$ at infinitely high temperature is

KCET 2023 All KCET

A
$6 \times 10^5 \times e^{-E_a / (300 R)}$
B
$e^{-E_a / (300 R)}$
C
$\frac{6 \times 10^{-5}}{300}$
D
$6 \times 10^5$

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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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For a reaction,the value of rate constant at $300 \ K$ is $6.0 \times 10^5 \ s^{-1}$. The value of Arrhenius factor $A$ at infinitely high temperature is

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$A$ catalyst lowers the activation energy for a certain reaction from $83.314 \, kJ \, mol^{-1}$ to $75 \, kJ \, mol^{-1}$ at $500 \, K$. What will be the rate of reaction as compared to the uncatalysed reaction? Assume other things being equal.

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