(C) Given: $T_1 = 25\,^oC = 298\,K$,$T_2 = 35\,^oC = 308\,K$. Rate constant $k_1 = 1 \times 10^{-3}\,s^{-1}$. Since the rate doubles,$k_2 = 2 \times k

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

(C) Given: $T_1 = 25\,^oC = 298\,K$,$T_2 = 35\,^oC = 308\,K$. Rate constant $k_1 = 1 \times 10^{-3}\,s^{-1}$. Since the rate doubles,$k_2 = 2 \times k_1 = 2 \times 10^{-3}\,s^{-1}$. Using the Arrhenius equation: $\log(\frac{k_2}{k_1}) = \frac{E_a}{2.303R} \times (\frac{T_2 - T_1}{T_1 T_2})$. $\log(2) = \frac{E_a}{2.303 \times 8.314} \times (\frac{308 - 298}{308 \times 298})$. $0.3010 = \frac{E_a}{19.147} \times (\frac{10}{91784})$. $E_a = \frac{0.3010 \times 19.147 \times 91784}{10} \approx 52897\,J\,mol^{-1} \approx 53\,kJ\,mol^{-1}$.

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

Medium

The rate constant of a reaction at $25\,^oC$ is $1 \times 10^{-3}\,s^{-1}$. If the rate of the reaction doubles when the temperature is increased to $35\,^oC$,the activation energy of the reaction is .......... $kJ\, mol^{-1}$.

A
$17$
B
$25$
C
$53$
D
$36$

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

According to the Arrhenius equation, in a graph of $\log k$ versus $\frac{1}{T}$, the slope will be . . . . . . .

MediumGUJCET 2026

View Solution

Subtract $(i)$ $\ln \, k_1 = - \frac{E_a}{R T_1} + \ln A$ and $(ii)$ $\ln \, k_2 = - \frac{E_a}{R T_2} + \ln A$ and write the resulting equation.

Easy

View Solution

The activation energy for a reaction at the temperature $T \ K$ was found to be $2.303 \ RT \ J \ mol^{-1}$. The ratio of the rate constant to the Arrhenius factor is

EasyKCET 2010

View Solution

The graph obtained between $\ln k$ ($k=$ Rate constant) on $y$-axis and $1/T$ on $x$-axis is a straight line. The slope of it is $-4 \times 10^4 \ K$. The activation energy of the reaction (in $kJ \ mol^{-1}$) is $(R=8.3 \ J \ K^{-1} \ mol^{-1})$

MediumTS EAMCET 2023

View Solution

$A$ reaction takes place in three steps with individual rate constant and activation energy,

Step Rate constant and Activation energy

$Step \ 1$ $k_1, E_{a_1} = 180 \ kJ \ mol^{-1}$

$Step \ 2$ $k_2, E_{a_2} = 80 \ kJ \ mol^{-1}$

$Step \ 3$ $k_3, E_{a_3} = 50 \ kJ \ mol^{-1}$

Overall rate constant,$k = (k_1 k_2 / k_3)^{2/3}$. The overall activation energy of the reaction will be ........ $kJ \ mol^{-1}$.

Difficult

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

Step	Rate constant and Activation energy
$Step \ 1$	$k_1, E_{a_1} = 180 \ kJ \ mol^{-1}$
$Step \ 2$	$k_2, E_{a_2} = 80 \ kJ \ mol^{-1}$
$Step \ 3$	$k_3, E_{a_3} = 50 \ kJ \ mol^{-1}$

The rate constant of a reaction at $25\,^oC$ is $1 \times 10^{-3}\,s^{-1}$. If the rate of the reaction doubles when the temperature is increased to $35\,^oC$,the activation energy of the reaction is .......... $kJ\, mol^{-1}$.

Similar Questions

According to the Arrhenius equation, in a graph of $\log k$ versus $\frac{1}{T}$, the slope will be . . . . . . .

Subtract $(i)$ $\ln \, k_1 = - \frac{E_a}{R T_1} + \ln A$ and $(ii)$ $\ln \, k_2 = - \frac{E_a}{R T_2} + \ln A$ and write the resulting equation.

The activation energy for a reaction at the temperature $T \ K$ was found to be $2.303 \ RT \ J \ mol^{-1}$. The ratio of the rate constant to the Arrhenius factor is

The graph obtained between $\ln k$ ($k=$ Rate constant) on $y$-axis and $1/T$ on $x$-axis is a straight line. The slope of it is $-4 \times 10^4 \ K$. The activation energy of the reaction (in $kJ \ mol^{-1}$) is $(R=8.3 \ J \ K^{-1} \ mol^{-1})$

Vedclass Test Series

Exam Paper Generator

Online Exam Module