(A) Activation energy $(E_a)$ is the minimum extra amount of energy that must be supplied to the reactant molecules so that their total energy becomes

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

The amount of energy to be added to the actual energy of a molecule so that the threshold energy is reached

(A) Activation energy $(E_a)$ is the minimum extra amount of energy that must be supplied to the reactant molecules so that their total energy becomes equal to the threshold energy $(E_T)$.Mathematically,$E_a = E_T - E_{\text{average}}$,where $E_{\text{average}}$ is the average kinetic energy of the reactant molecules.Therefore,option $(A)$ is the correct definition.

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

Easy

Activation energy is defined as:

A
The amount of energy to be added to the actual energy of a molecule so that the threshold energy is reached
B
The amount of energy the molecule must contain so that it reacts
C
The energy which a molecule should have in order to enter into an effective collision
D
The average kinetic energy of the molecule

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

Choose the incorrect statement -

Medium

View Solution

If the activation energy of a reaction is $80.9 \, kJ \, mol^{-1}$,the fraction of molecules at $700 \, K$,having enough energy to react to form products is $e^{-x}$. The value of $x$ is ....... . (Rounded off to the nearest integer) $[$Use $R = 8.31 \, J \, K^{-1} \, mol^{-1}]$

EasyJEE MAIN 2021

View Solution

The following figure shows a graph of $\log_{10}K$ vs $\frac{1}{T}$,where $K$ is the rate constant and $T$ is the temperature. The straight line $BC$ has a slope,$\tan \theta = -\frac{1}{2.303}$,and an intercept of $5$ on the $Y$-axis. Thus,$E_a$,the energy of activation,is ....... $cal$.

Medium

View Solution

For a first order reaction $A \rightarrow P$,the temperature $(T)$ dependent rate constant $(k)$ was found to follow the equation $\log k = -(2000) \frac{1}{T} + 6.0$. The pre-exponential factor $A$ and the activation energy $E_{a}$,respectively,are

DifficultIIT 2009

View Solution

For a reaction $E_a = 0$ and $k = 3.2 \times 10^8 \ s^{-1}$ at $300 \ K$. The value of frequency factor at $400 \ K$ would be

Medium

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

Activation energy is defined as:

Similar Questions

Choose the incorrect statement -

If the activation energy of a reaction is $80.9 \, kJ \, mol^{-1}$,the fraction of molecules at $700 \, K$,having enough energy to react to form products is $e^{-x}$. The value of $x$ is ....... . (Rounded off to the nearest integer) $[$Use $R = 8.31 \, J \, K^{-1} \, mol^{-1}]$

The following figure shows a graph of $\log_{10}K$ vs $\frac{1}{T}$,where $K$ is the rate constant and $T$ is the temperature. The straight line $BC$ has a slope,$\tan \theta = -\frac{1}{2.303}$,and an intercept of $5$ on the $Y$-axis. Thus,$E_a$,the energy of activation,is ....... $cal$.

For a first order reaction $A \rightarrow P$,the temperature $(T)$ dependent rate constant $(k)$ was found to follow the equation $\log k = -(2000) \frac{1}{T} + 6.0$. The pre-exponential factor $A$ and the activation energy $E_{a}$,respectively,are

For a reaction $E_a = 0$ and $k = 3.2 \times 10^8 \ s^{-1}$ at $300 \ K$. The value of frequency factor at $400 \ K$ would be

Vedclass Test Series

Exam Paper Generator

Online Exam Module