(D) The Arrhenius equation is given by $K = A \ e^{-E_a / (RT)}$.Comparing this with the given expression $K = A \ e^{-1000}$,we get $E_a / (RT) = 100

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(D) The Arrhenius equation is given by $K = A \ e^{-E_a / (RT)}$.Comparing this with the given expression $K = A \ e^{-1000}$,we get $E_a / (RT) = 1000$.Given $T = 500 \ K$ and the gas constant $R = 2 \ cal \ K^{-1} \ mol^{-1}$.Substituting the values: $E_a = 1000 \times R \times T = 1000 \times 2 \times 500 = 1,000,000 \ cal/mol = 10^6 \ cal/mol$.

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

Medium

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:

A
$100 \ cal/mol$
B
$1000 \ cal/mol$
C
$10^4 \ cal/mol$
D
$10^6 \ cal/mol$

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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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Define the following terms:
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The rate constant is given by the equation $k = p Z e^{-E_a/RT}$. Which factor should register a decrease for the reaction to proceed more rapidly?

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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:

Similar Questions

The activation energy for a simple chemical reaction $A \to B$ is ${E_a}$ in the forward direction. The activation energy for the reverse reaction:

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Define the following terms:
$(a)$ Activation energy
$(b)$ Activated complex