(A) Using the Arrhenius equation: $\ln(\frac{k_2}{k_1}) = \frac{E_a}{R} (\frac{T_2 - T_1}{T_1 T_2})$.Given that the rate doubles, $\frac{k_2}{k_1} = 2

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

(A) Using the Arrhenius equation: $\ln(\frac{k_2}{k_1}) = \frac{E_a}{R} (\frac{T_2 - T_1}{T_1 T_2})$.Given that the rate doubles, $\frac{k_2}{k_1} = 2$, $T_1 = 298 \text{ K}$, and $T_2 = 308 \text{ K}$.Substituting the values: $\ln(2) = \frac{E_a}{8.314} (\frac{308 - 298}{298 \times 308})$.$0.693 = \frac{E_a}{8.314} (\frac{10}{91784})$.$E_a = \frac{0.693 \times 8.314 \times 91784}{10}$.$E_a \approx 52897 \text{ J mol}^{-1} = 52.897 \text{ kJ mol}^{-1}$.

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

Difficult

The rate of the chemical reaction doubles for an increase of $10 \text{ K}$ in absolute temperature from $298 \text{ K}$. What will be the activation energy?

GUJCET 2026 All GUJCET

A
$52.897 \text{ kJ mol}^{-1}$
B
$51.897 \text{ kJ mol}^{-1}$
C
$42.897 \text{ kJ mol}^{-1}$
D
$41.897 \text{ kJ mol}^{-1}$

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

Previous Year

GUJCET 2026 Paper All GUJCET years →

The rate of reaction increases with a rise in temperature because of:

EasyKCET 2013

View Solution

For an endothermic reaction,where $\Delta H$ represents the enthalpy of the reaction in $kJ/mol$,the minimum value for the energy of activation will be

MediumIIT 1992

View Solution

Define the following terms:
$1.$ Collision
$2.$ Frequency of collision

Medium

View Solution

Find True $(T)$ and False $(F)$ statements among the following:
$1.$ All collisions in a reaction result in the formation of products.
$2.$ All collisions in a reaction are effective.
$3.$ The number of collisions depends on the rate of reaction.

Difficult

View Solution

For a first order decomposition of a certain reaction,rate constant is given by the equation $\log k \left( s^{-1} \right) = 7.14 - \frac{1 \times 10^4 \ K}{T}$. The activation energy of the reaction (in $kJ \ mol^{-1}$) is $(R = 8.3 \ J \ K^{-1} \ mol^{-1})$ (in $.1$)

MediumTS EAMCET 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

The rate of the chemical reaction doubles for an increase of $10 \text{ K}$ in absolute temperature from $298 \text{ K}$. What will be the activation energy?

Similar Questions

The rate of reaction increases with a rise in temperature because of:

For an endothermic reaction,where $\Delta H$ represents the enthalpy of the reaction in $kJ/mol$,the minimum value for the energy of activation will be

Define the following terms:$1.$ Collision$2.$ Frequency of collision

Find True $(T)$ and False $(F)$ statements among the following:$1.$ All collisions in a reaction result in the formation of products.$2.$ All collisions in a reaction are effective.$3.$ The number of collisions depends on the rate of reaction.

For a first order decomposition of a certain reaction,rate constant is given by the equation $\log k \left( s^{-1} \right) = 7.14 - \frac{1 \times 10^4 \ K}{T}$. The activation energy of the reaction (in $kJ \ mol^{-1}$) is $(R = 8.3 \ J \ K^{-1} \ mol^{-1})$ (in $.1$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module

Define the following terms:
$1.$ Collision
$2.$ Frequency of collision

Find True $(T)$ and False $(F)$ statements among the following:
$1.$ All collisions in a reaction result in the formation of products.
$2.$ All collisions in a reaction are effective.
$3.$ The number of collisions depends on the rate of reaction.