Class 12 Exam > Class 12 Questions > The activation energy of the reaction at a gi...
Start Learning for Free
The activation energy of the reaction at a given temperature is found to (2.303 RT) J mol-1. The ratio of rate constant to the Arrhenius factor is
- a)0.01
- b)0.1
- c)0.02
- d)0.001
Correct answer is option 'B'. Can you explain this answer?
| FREE This question is part of | Download PDF Attempt this Test |
Verified Answer
The activation energy of the reaction at a given temperature is found ...
Most Upvoted Answer
The activation energy of the reaction at a given temperature is found ...
The activation energy of a reaction is the minimum amount of energy required for a chemical reaction to occur. It can be determined experimentally by measuring the rate constant of the reaction at different temperatures and using the Arrhenius equation. The Arrhenius equation relates the rate constant (k) of a reaction to the temperature (T) and the activation energy (Ea) of the reaction.
The Arrhenius equation is given by:
k = A * exp(-Ea/RT)
Where:
k is the rate constant
A is the pre-exponential factor or the frequency factor
Ea is the activation energy
R is the gas constant (8.314 J/mol·K)
T is the temperature in Kelvin
The ratio of the rate constant to the Arrhenius factor is given by:
k/A = exp(-Ea/RT)
To find the ratio of the rate constant to the Arrhenius factor, we need to express the activation energy in terms of RT. Given that the activation energy is (2.303 RT) J/mol, we can rewrite it as:
Ea = 2.303 RT
Substituting this value into the equation, we have:
k/A = exp(-(2.303 RT)/RT)
Simplifying the equation, we get:
k/A = exp(-2.303)
Since the exponential function e^(-2.303) is approximately equal to 0.1, we can conclude that:
k/A ≈ 0.1
Therefore, the correct answer is option 'B' - 0.1. The ratio of the rate constant to the Arrhenius factor is approximately 0.1.
The Arrhenius equation is given by:
k = A * exp(-Ea/RT)
Where:
k is the rate constant
A is the pre-exponential factor or the frequency factor
Ea is the activation energy
R is the gas constant (8.314 J/mol·K)
T is the temperature in Kelvin
The ratio of the rate constant to the Arrhenius factor is given by:
k/A = exp(-Ea/RT)
To find the ratio of the rate constant to the Arrhenius factor, we need to express the activation energy in terms of RT. Given that the activation energy is (2.303 RT) J/mol, we can rewrite it as:
Ea = 2.303 RT
Substituting this value into the equation, we have:
k/A = exp(-(2.303 RT)/RT)
Simplifying the equation, we get:
k/A = exp(-2.303)
Since the exponential function e^(-2.303) is approximately equal to 0.1, we can conclude that:
k/A ≈ 0.1
Therefore, the correct answer is option 'B' - 0.1. The ratio of the rate constant to the Arrhenius factor is approximately 0.1.
Free Test
FREE
| Start Free Test |
Community Answer
The activation energy of the reaction at a given temperature is found ...
Option B : 0.1
|
Explore Courses for Class 12 exam
|
|
Similar Class 12 Doubts
The activation energy of the reaction at a given temperature is found to (2.303 RT) J mol-1. The ratio of rate constant to the Arrhenius factor isa)0.01b)0.1c)0.02d)0.001Correct answer is option 'B'. Can you explain this answer?
Question Description
The activation energy of the reaction at a given temperature is found to (2.303 RT) J mol-1. The ratio of rate constant to the Arrhenius factor isa)0.01b)0.1c)0.02d)0.001Correct answer is option 'B'. Can you explain this answer? for Class 12 2024 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about The activation energy of the reaction at a given temperature is found to (2.303 RT) J mol-1. The ratio of rate constant to the Arrhenius factor isa)0.01b)0.1c)0.02d)0.001Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for Class 12 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The activation energy of the reaction at a given temperature is found to (2.303 RT) J mol-1. The ratio of rate constant to the Arrhenius factor isa)0.01b)0.1c)0.02d)0.001Correct answer is option 'B'. Can you explain this answer?.
The activation energy of the reaction at a given temperature is found to (2.303 RT) J mol-1. The ratio of rate constant to the Arrhenius factor isa)0.01b)0.1c)0.02d)0.001Correct answer is option 'B'. Can you explain this answer? for Class 12 2024 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about The activation energy of the reaction at a given temperature is found to (2.303 RT) J mol-1. The ratio of rate constant to the Arrhenius factor isa)0.01b)0.1c)0.02d)0.001Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for Class 12 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The activation energy of the reaction at a given temperature is found to (2.303 RT) J mol-1. The ratio of rate constant to the Arrhenius factor isa)0.01b)0.1c)0.02d)0.001Correct answer is option 'B'. Can you explain this answer?.
Solutions for The activation energy of the reaction at a given temperature is found to (2.303 RT) J mol-1. The ratio of rate constant to the Arrhenius factor isa)0.01b)0.1c)0.02d)0.001Correct answer is option 'B'. Can you explain this answer? in English & in Hindi are available as part of our courses for Class 12.
Download more important topics, notes, lectures and mock test series for Class 12 Exam by signing up for free.
Here you can find the meaning of The activation energy of the reaction at a given temperature is found to (2.303 RT) J mol-1. The ratio of rate constant to the Arrhenius factor isa)0.01b)0.1c)0.02d)0.001Correct answer is option 'B'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of
The activation energy of the reaction at a given temperature is found to (2.303 RT) J mol-1. The ratio of rate constant to the Arrhenius factor isa)0.01b)0.1c)0.02d)0.001Correct answer is option 'B'. Can you explain this answer?, a detailed solution for The activation energy of the reaction at a given temperature is found to (2.303 RT) J mol-1. The ratio of rate constant to the Arrhenius factor isa)0.01b)0.1c)0.02d)0.001Correct answer is option 'B'. Can you explain this answer? has been provided alongside types of The activation energy of the reaction at a given temperature is found to (2.303 RT) J mol-1. The ratio of rate constant to the Arrhenius factor isa)0.01b)0.1c)0.02d)0.001Correct answer is option 'B'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice The activation energy of the reaction at a given temperature is found to (2.303 RT) J mol-1. The ratio of rate constant to the Arrhenius factor isa)0.01b)0.1c)0.02d)0.001Correct answer is option 'B'. Can you explain this answer? tests, examples and also practice Class 12 tests.
|
|
Explore Courses for Class 12 exam
|
|
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
|
© EduRev
|
Education Revolution
|
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup