(A) The Arrhenius equation is given by $K = A \, e^{-E_a / (RT)}$.Comparing this with the given equation $K = 5.0 \times 10^{11} \, e^{-2000/T}$,we ge

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

(A) The Arrhenius equation is given by $K = A \, e^{-E_a / (RT)}$.Comparing this with the given equation $K = 5.0 \times 10^{11} \, e^{-2000/T}$,we get:$E_a / R = 2000 \, K$.Therefore,$E_a = 2000 \, K \times R$.Using the value of the gas constant $R = 8.314 \, J \, K^{-1} \, mol^{-1}$:$E_a = 2000 \times 8.314 = 16628 \, J \, mol^{-1}$.Converting to $kJ \, mol^{-1}$:$E_a = 16.628 \, kJ \, mol^{-1}$.

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

Medium

The decomposition of $N_2O$ into $N_2$ and $O_2$ in the presence of argon follows first-order kinetics. If the rate constant is given by $K = 5.0 \times 10^{11} \, e^{-2000/T}$,what is the activation energy of the reaction?

A
$16.628 \, kJ \, mol^{-1}$
B
$166.28 \, kJ \, mol^{-1}$
C
$166.28 \, J \, mol^{-1}$
D
$16.628 \, J \, mol^{-1}$

Explore More

Browse All

Question Bank

All Subjects

Class 12 Questions

Class 12

Chemistry Questions

Chapter

Chemical Kinetics

Subtopic

Collision theory, Energy of activation and Arrhenius equation

Write the logarithmic form of the Arrhenius equation $k = A e^{-\frac{E_a}{RT}}$.

Easy

View Solution

$A \rightarrow B$
The rate constants of the above reaction at $200 \, K$ and $300 \, K$ are $0.03 \, min^{-1}$ and $0.05 \, min^{-1}$ respectively. The activation energy for the reaction is $.... \, J$ (Nearest integer).
(Given: $\ln 10 = 2.3$,$R = 8.3 \, J \, K^{-1} \, mol^{-1}$,$\log 5 = 0.70$,$\log 3 = 0.48$,$\log 2 = 0.30$)

MediumJEE MAIN 2023

View Solution

For a hypothetical reaction $A \to B$,the activation energies for forward and backward reactions are $19 \ kJ/mol$ and $9 \ kJ/mol$ respectively. The heat of reaction is $.... \ kJ$.

MediumAIPMT 2000

View Solution

The temperature coefficient of a reaction is $2$. When the temperature is increased from $30^{\circ}C$ to $90^{\circ}C$, the rate of reaction is increased by (in $\times$)

DifficultKCET 2018

View Solution

The equation $k = (6.5 \times 10^{12} \, s^{-1}) e^{-26000 \, K / T}$ is followed for the decomposition of compound $A$. The activation energy for the reaction is $..... \, kJ \, mol^{-1}$. [nearest integer] (Given: $R = 8.314 \, J \, K^{-1} \, mol^{-1}$)

EasyJEE MAIN 2022

View Solution

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial

For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free

For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo

The decomposition of $N_2O$ into $N_2$ and $O_2$ in the presence of argon follows first-order kinetics. If the rate constant is given by $K = 5.0 \times 10^{11} \, e^{-2000/T}$,what is the activation energy of the reaction?

Similar Questions

Write the logarithmic form of the Arrhenius equation $k = A e^{-\frac{E_a}{RT}}$.

For a hypothetical reaction $A \to B$,the activation energies for forward and backward reactions are $19 \ kJ/mol$ and $9 \ kJ/mol$ respectively. The heat of reaction is $.... \ kJ$.

The temperature coefficient of a reaction is $2$. When the temperature is increased from $30^{\circ}C$ to $90^{\circ}C$, the rate of reaction is increased by (in $\times$)

The equation $k = (6.5 \times 10^{12} \, s^{-1}) e^{-26000 \, K / T}$ is followed for the decomposition of compound $A$. The activation energy for the reaction is $..... \, kJ \, mol^{-1}$. [nearest integer] (Given: $R = 8.314 \, J \, K^{-1} \, mol^{-1}$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module