(C) According to the Arrhenius equation,$k = Ae^{-E_a / RT}$.Let $E_{a_1}$ be the initial activation energy and $E_{a_2}$ be the activation energy aft

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(C) According to the Arrhenius equation,$k = Ae^{-E_a / RT}$.Let $E_{a_1}$ be the initial activation energy and $E_{a_2}$ be the activation energy after adding the catalyst.$k_1 = Ae^{-E_{a_1} / RT} \dots (1)$$k_2 = 4k_1 = Ae^{-E_{a_2} / RT} \dots (2)$Dividing $(2)$ by $(1)$:$4 = e^{(E_{a_1} - E_{a_2}) / RT}$Taking natural logarithm on both sides:$\ln 4 = \frac{E_{a_1} - E_{a_2}}{RT}$$E_{a_2} - E_{a_1} = -RT \ln 4$Given $T = 27 + 273 = 300 \ K$ and $\ln 4 = 1.386$.$\Delta E_a = E_{a_2} - E_{a_1} = -(8.314 \ J \ K^{-1} \ mol^{-1} \times 300 \ K \times 1.386)$$\Delta E_a = -3457.3 \ J / mol \approx -3.45 \ kJ / mol$.

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

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The rate constant of a reaction is increased $4$ times after the addition of a catalyst to the reaction mixture at the same temperature of $27^{\circ} C$. The change in the activation energy of this reaction is (Take $\ln(1/4) = -1.386, R = 8.314 \ J \ K^{-1} \ mol^{-1}$)

TS EAMCET 2022 All TS EAMCET

A
$-15 \ kJ / mol$
B
$-1.5 \ kJ / mol$
C
$-3.45 \ kJ / mol$
D
$-34.5 \ kJ / mol$

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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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For the decomposition of $N_2O_5,$ the reaction is $2N_2O_{5(g)} \rightarrow 4NO_{2(g)} + O_{2(g)}$ with activation energy $E_a.$ If the reaction is written as $N_2O_{5(g)} \rightarrow 2NO_{2(g)} + 1/2 O_{2(g)}$ with activation energy $E_a',$ what is the relationship between $E_a$ and $E_a'$?

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The following equation is obtained for a first order reaction at $300 \ K$.
$\log_{10} \frac{k}{A} = 0.00174$
What is the activation energy (in $J \ mol^{-1}$) of the reaction?
$(R = 8.314 \ J \ mol^{-1} \ K^{-1})$

MediumAP EAMCET 2025

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The rate of a reaction quadruples when temperature changes from $27^{\circ} C$ to $57^{\circ} C$. Calculate the energy of activation.
Given $R=8.314 \ J \ K^{-1} \ mol^{-1}, \log 4=0.6021$

MediumNEET 2024

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For a chemical reaction,the rate constant,activation energy,and Arrhenius factor at $25 \, ^\circ C$ are $3.0 \times 10^{-4} \, s^{-1}$,$104.4 \, kJ \, mol^{-1}$,and $6.0 \times 10^{14} \, s^{-1}$ respectively. Find the value of the rate constant as $T \rightarrow \infty$.

Medium

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$A$ reaction takes place in three steps with individual rate constant and activation energy as follows:
$Step$ $Rate \ constant$ $Activation \ energy$
$Step-1$ $k_1$ $E_{a1} = 180 \ kJ/mol$
$Step-2$ $k_2$ $E_{a2} = 80 \ kJ/mol$
$Step-3$ $k_3$ $E_{a3} = 50 \ kJ/mol$
If overall rate constant,$k = (\frac{k_1 k_2}{k_3})^{2/3}$,then overall activation energy of the reaction will be .......... $kJ/mol$.

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$Step$	$Rate \ constant$	$Activation \ energy$
$Step-1$	$k_1$	$E_{a1} = 180 \ kJ/mol$
$Step-2$	$k_2$	$E_{a2} = 80 \ kJ/mol$
$Step-3$	$k_3$	$E_{a3} = 50 \ kJ/mol$

The rate constant of a reaction is increased $4$ times after the addition of a catalyst to the reaction mixture at the same temperature of $27^{\circ} C$. The change in the activation energy of this reaction is (Take $\ln(1/4) = -1.386, R = 8.314 \ J \ K^{-1} \ mol^{-1}$)

Similar Questions

For the decomposition of $N_2O_5,$ the reaction is $2N_2O_{5(g)} \rightarrow 4NO_{2(g)} + O_{2(g)}$ with activation energy $E_a.$ If the reaction is written as $N_2O_{5(g)} \rightarrow 2NO_{2(g)} + 1/2 O_{2(g)}$ with activation energy $E_a',$ what is the relationship between $E_a$ and $E_a'$?

The following equation is obtained for a first order reaction at $300 \ K$.$\log_{10} \frac{k}{A} = 0.00174$What is the activation energy (in $J \ mol^{-1}$) of the reaction?$(R = 8.314 \ J \ mol^{-1} \ K^{-1})$

The rate of a reaction quadruples when temperature changes from $27^{\circ} C$ to $57^{\circ} C$. Calculate the energy of activation.Given $R=8.314 \ J \ K^{-1} \ mol^{-1}, \log 4=0.6021$

For a chemical reaction,the rate constant,activation energy,and Arrhenius factor at $25 \, ^\circ C$ are $3.0 \times 10^{-4} \, s^{-1}$,$104.4 \, kJ \, mol^{-1}$,and $6.0 \times 10^{14} \, s^{-1}$ respectively. Find the value of the rate constant as $T \rightarrow \infty$.

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The following equation is obtained for a first order reaction at $300 \ K$.
$\log_{10} \frac{k}{A} = 0.00174$
What is the activation energy (in $J \ mol^{-1}$) of the reaction?
$(R = 8.314 \ J \ mol^{-1} \ K^{-1})$

The rate of a reaction quadruples when temperature changes from $27^{\circ} C$ to $57^{\circ} C$. Calculate the energy of activation.
Given $R=8.314 \ J \ K^{-1} \ mol^{-1}, \log 4=0.6021$