(C) According to the Arrhenius equation,$K = A \, e^{-Ea/RT}$.Assuming the pre-exponential factor $A$ is the same for both reactions:$K_1 = A \, e^{-E

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(C) According to the Arrhenius equation,$K = A \, e^{-Ea/RT}$.Assuming the pre-exponential factor $A$ is the same for both reactions:$K_1 = A \, e^{-Ea_1/RT}$$K_2 = A \, e^{-Ea_2/RT}$Given $Ea_1 = \frac{Ea_2}{3}$,we have $Ea_2 = 3Ea_1$.Substituting $Ea_2$ into the expression for $K_2$:$K_2 = A \, e^{-3Ea_1/RT}$Now,divide $K_1$ by $K_2$:$\frac{K_1}{K_2} = \frac{A \, e^{-Ea_1/RT}}{A \, e^{-3Ea_1/RT}} = e^{(-Ea_1 + 3Ea_1)/RT} = e^{2Ea_1/RT}$Therefore,$K_1 = K_2 \, e^{2Ea_1/RT}$.

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

Medium

Reactant $A$ shows two reactions:
$A \xrightarrow{K_1} B$,activation energy $= Ea_1$
$A \xrightarrow{K_2} C$,activation energy $= Ea_2$
If $Ea_1 = \frac{Ea_2}{3}$,then the relation between $K_1$ and $K_2$ is:

A
$K_2 = K_1 \, e^{Ea_1/RT}$
B
$K_2 = K_1 \, e^{2Ea_1/RT}$
C
$K_1 = K_2 \, e^{2Ea_1/RT}$
D
$K_1 = 2K_2 \, e^{Ea_1/RT}$

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Class 12

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Chemical Kinetics

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Collision theory, Energy of activation and Arrhenius equation

In a reversible reaction,the catalyst

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The rate constant for a first order reaction at $300\, ^\circ C$ for which $E_a$ is $35\, kcal\, mol^{-1}$ and frequency constant $(A)$ is $1.45 \times 10^{11} s^{-1}$ will be:

Medium

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For a given reaction,the overall enthalpy change is $+100 \ kJ/mol$ and the activation energy of the reverse reaction is $+200 \ kJ/mol$. The activation energy for the forward reaction would be......$kJ/mol$.

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An increase in temperature by $10\,^{\circ}C$,generally increases the rate of a reaction by .......... times.

Easy

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The rate constant for the decomposition of $N_{2}O_{5}$ at various temperatures is given below:
$T / ^{\circ}C$ $0$ $20$ $40$ $60$ $80$
$10^{5} \times k / s^{-1}$ $0.0787$ $1.70$ $25.7$ $178$ $2140$

Draw a graph between $\ln k$ and $1 / T$ and calculate the values of $A$ and $E_{a}.$ Predict the rate constant at $30^{\circ}C$ and $50^{\circ}C$.

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$T / ^{\circ}C$	$0$	$20$	$40$	$60$	$80$
$10^{5} \times k / s^{-1}$	$0.0787$	$1.70$	$25.7$	$178$	$2140$

Reactant $A$ shows two reactions:$A \xrightarrow{K_1} B$,activation energy $= Ea_1$$A \xrightarrow{K_2} C$,activation energy $= Ea_2$If $Ea_1 = \frac{Ea_2}{3}$,then the relation between $K_1$ and $K_2$ is:

Similar Questions

In a reversible reaction,the catalyst

The rate constant for a first order reaction at $300\, ^\circ C$ for which $E_a$ is $35\, kcal\, mol^{-1}$ and frequency constant $(A)$ is $1.45 \times 10^{11} s^{-1}$ will be:

For a given reaction,the overall enthalpy change is $+100 \ kJ/mol$ and the activation energy of the reverse reaction is $+200 \ kJ/mol$. The activation energy for the forward reaction would be......$kJ/mol$.

An increase in temperature by $10\,^{\circ}C$,generally increases the rate of a reaction by .......... times.

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Reactant $A$ shows two reactions:
$A \xrightarrow{K_1} B$,activation energy $= Ea_1$
$A \xrightarrow{K_2} C$,activation energy $= Ea_2$
If $Ea_1 = \frac{Ea_2}{3}$,then the relation between $K_1$ and $K_2$ is: