(A) The Arrhenius equation is given by $\ln k = \ln A - \frac{E_{a}}{RT}$.Comparing this with the equation of a straight line $y = mx + c$,where $y =

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

(A) The Arrhenius equation is given by $\ln k = \ln A - \frac{E_{a}}{RT}$.Comparing this with the equation of a straight line $y = mx + c$,where $y = \ln k$ and $x = \frac{1}{T}$,the slope $m = -\frac{E_{a}}{R}$.Given that the slope is $-5 \times 10^{3} \, K$,we have $-\frac{E_{a}}{R} = -5 \times 10^{3} \, K$.Therefore,$E_{a} = 5 \times 10^{3} \times 8.314 \, J \, mol^{-1} = 41570 \, J \, mol^{-1}$.Converting to $kJ \, mol^{-1}$,we get $E_{a} = 41.57 \, kJ \, mol^{-1} \approx 41.5 \, kJ \, mol^{-1}$.

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

Medium

The slope of the Arrhenius plot $(\ln k \, vs \, \frac{1}{T})$ of a first-order reaction is $-5 \times 10^{3} \, K$. The value of $E_{a}$ of the reaction is: (in $kJ \, mol^{-1}$)
$[\text{Given } R = 8.314 \, J \, K^{-1} \, mol^{-1}]$

NEET 2021 All NEET

A
$41.5$
B
$83.0$
C
$166$
D
$-83$

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

Previous Year

NEET 2021 Paper All NEET years →

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

Medium

View Solution

If the rate constants of a reaction at $500 \ K$ and $700 \ K$ are $0.002 \ s^{-1}$ and $0.06 \ s^{-1}$,respectively,the value of activation energy is $(R=8.314 \ J \ mol^{-1} \ K^{-1}, \log 3=0.477)$.

MediumTS EAMCET 2018

View Solution

The activation energy of a reaction is $0$. The rate constant of the reaction:

Medium

View Solution

The relation between the fraction of molecules $\left( \frac{N_E}{N_T} \right)$ and temperature is explained by the Boltzmann and Maxwell distribution graph.

Difficult

View Solution

Reactant $A$ converts to product $D$ through the given mechanism (with the net evolution of heat) :
$A \rightarrow B$ $slow ; \Delta H=+ve$
$B \rightarrow C$ $fast ; \Delta H=-ve$
$C \rightarrow D$ $fast ; \Delta H=-ve$

Which of the following represents the above reaction mechanism ?

DifficultJEE MAIN 2025

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

$A \rightarrow B$	$slow ; \Delta H=+ve$
$B \rightarrow C$	$fast ; \Delta H=-ve$
$C \rightarrow D$	$fast ; \Delta H=-ve$

The slope of the Arrhenius plot $(\ln k \, vs \, \frac{1}{T})$ of a first-order reaction is $-5 \times 10^{3} \, K$. The value of $E_{a}$ of the reaction is: (in $kJ \, mol^{-1}$)$[\text{Given } R = 8.314 \, J \, K^{-1} \, mol^{-1}]$

Similar Questions

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

If the rate constants of a reaction at $500 \ K$ and $700 \ K$ are $0.002 \ s^{-1}$ and $0.06 \ s^{-1}$,respectively,the value of activation energy is $(R=8.314 \ J \ mol^{-1} \ K^{-1}, \log 3=0.477)$.

The activation energy of a reaction is $0$. The rate constant of the reaction:

The relation between the fraction of molecules $\left( \frac{N_E}{N_T} \right)$ and temperature is explained by the Boltzmann and Maxwell distribution graph.

Reactant $A$ converts to product $D$ through the given mechanism (with the net evolution of heat) :$A \rightarrow B$$slow ; \Delta H=+ve$$B \rightarrow C$$fast ; \Delta H=-ve$$C \rightarrow D$$fast ; \Delta H=-ve$Which of the following represents the above reaction mechanism ?

Vedclass Test Series

Exam Paper Generator

Online Exam Module

The slope of the Arrhenius plot $(\ln k \, vs \, \frac{1}{T})$ of a first-order reaction is $-5 \times 10^{3} \, K$. The value of $E_{a}$ of the reaction is: (in $kJ \, mol^{-1}$)
$[\text{Given } R = 8.314 \, J \, K^{-1} \, mol^{-1}]$

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

Reactant $A$ converts to product $D$ through the given mechanism (with the net evolution of heat) :
$A \rightarrow B$ $slow ; \Delta H=+ve$
$B \rightarrow C$ $fast ; \Delta H=-ve$
$C \rightarrow D$ $fast ; \Delta H=-ve$

Which of the following represents the above reaction mechanism ?