JEE Exam > JEE Questions > The decomposition of a substance follows firs...
Start Learning for Free
The decomposition of a substance follows first order kinetics. If its conc. is reduced to 1/8 th of its initial value, in 24 minutes, the rate constant of decomposition process is
- a)1/24 min⁻1
- b)0.692/24 min⁻1
- c)2.303/24 log (1/8) min⁻1
- d)2.303/24 log (8/1) min⁻1
Correct answer is option 'D'. Can you explain this answer?
Most Upvoted Answer
The decomposition of a substance follows first order kinetics. If its ...
Solution:
Given that, the decomposition of a substance follows first order kinetics and its concentration is reduced to 1/8th of its initial value in 24 minutes.
Let us write the first-order rate equation for the decomposition of the substance as follows:
Rate of decomposition = k [A]
where k is the rate constant and [A] is the concentration of the substance.
We know that, for a first-order reaction, the half-life is given by the expression:
t1/2 = 0.693/k
where t1/2 is the time taken for the concentration of the substance to reduce to half its initial value.
Let us assume that the initial concentration of the substance is [A]0 and its concentration after 24 minutes is [A]1.
Given that, [A]1 = 1/8 [A]0
Therefore, the concentration of the substance that is decomposed in 24 minutes is:
[A]0 - [A]1 = [A]0 - 1/8 [A]0 = 7/8 [A]0
Let us substitute the given values in the first-order rate equation:
Rate of decomposition = k [A]0
= k (7/8 [A]0) / 24 min
= (7/8) k [A]0 / 24 min
= k/3.4286
Let us substitute the value of t1/2 in the above equation:
k/3.4286 = 0.693/t1/2
= 0.693/ln2 * k
=> k = 2.303/24 * ln(8/1) min^-1
Hence, the rate constant of the decomposition process is 2.303/24 * ln(8/1) min^-1. Therefore, the correct option is D.
Given that, the decomposition of a substance follows first order kinetics and its concentration is reduced to 1/8th of its initial value in 24 minutes.
Let us write the first-order rate equation for the decomposition of the substance as follows:
Rate of decomposition = k [A]
where k is the rate constant and [A] is the concentration of the substance.
We know that, for a first-order reaction, the half-life is given by the expression:
t1/2 = 0.693/k
where t1/2 is the time taken for the concentration of the substance to reduce to half its initial value.
Let us assume that the initial concentration of the substance is [A]0 and its concentration after 24 minutes is [A]1.
Given that, [A]1 = 1/8 [A]0
Therefore, the concentration of the substance that is decomposed in 24 minutes is:
[A]0 - [A]1 = [A]0 - 1/8 [A]0 = 7/8 [A]0
Let us substitute the given values in the first-order rate equation:
Rate of decomposition = k [A]0
= k (7/8 [A]0) / 24 min
= (7/8) k [A]0 / 24 min
= k/3.4286
Let us substitute the value of t1/2 in the above equation:
k/3.4286 = 0.693/t1/2
= 0.693/ln2 * k
=> k = 2.303/24 * ln(8/1) min^-1
Hence, the rate constant of the decomposition process is 2.303/24 * ln(8/1) min^-1. Therefore, the correct option is D.
|
Explore Courses for JEE exam
|
|
Similar JEE Doubts
Top Courses for JEEView all
The decomposition of a substance follows first order kinetics. If its conc. is reduced to 1/8 th of its initial value, in 24 minutes, the rate constant of decomposition process isa)1/24 min1b)0.692/24 min1c)2.303/24 log (1/8) min1d)2.303/24 log (8/1) min1Correct answer is option 'D'. Can you explain this answer?
Question Description
The decomposition of a substance follows first order kinetics. If its conc. is reduced to 1/8 th of its initial value, in 24 minutes, the rate constant of decomposition process isa)1/24 min1b)0.692/24 min1c)2.303/24 log (1/8) min1d)2.303/24 log (8/1) min1Correct answer is option 'D'. Can you explain this answer? for JEE 2025 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about The decomposition of a substance follows first order kinetics. If its conc. is reduced to 1/8 th of its initial value, in 24 minutes, the rate constant of decomposition process isa)1/24 min1b)0.692/24 min1c)2.303/24 log (1/8) min1d)2.303/24 log (8/1) min1Correct answer is option 'D'. Can you explain this answer? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The decomposition of a substance follows first order kinetics. If its conc. is reduced to 1/8 th of its initial value, in 24 minutes, the rate constant of decomposition process isa)1/24 min1b)0.692/24 min1c)2.303/24 log (1/8) min1d)2.303/24 log (8/1) min1Correct answer is option 'D'. Can you explain this answer?.
The decomposition of a substance follows first order kinetics. If its conc. is reduced to 1/8 th of its initial value, in 24 minutes, the rate constant of decomposition process isa)1/24 min1b)0.692/24 min1c)2.303/24 log (1/8) min1d)2.303/24 log (8/1) min1Correct answer is option 'D'. Can you explain this answer? for JEE 2025 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about The decomposition of a substance follows first order kinetics. If its conc. is reduced to 1/8 th of its initial value, in 24 minutes, the rate constant of decomposition process isa)1/24 min1b)0.692/24 min1c)2.303/24 log (1/8) min1d)2.303/24 log (8/1) min1Correct answer is option 'D'. Can you explain this answer? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The decomposition of a substance follows first order kinetics. If its conc. is reduced to 1/8 th of its initial value, in 24 minutes, the rate constant of decomposition process isa)1/24 min1b)0.692/24 min1c)2.303/24 log (1/8) min1d)2.303/24 log (8/1) min1Correct answer is option 'D'. Can you explain this answer?.
Solutions for The decomposition of a substance follows first order kinetics. If its conc. is reduced to 1/8 th of its initial value, in 24 minutes, the rate constant of decomposition process isa)1/24 min1b)0.692/24 min1c)2.303/24 log (1/8) min1d)2.303/24 log (8/1) min1Correct answer is option 'D'. Can you explain this answer? in English & in Hindi are available as part of our courses for JEE.
Download more important topics, notes, lectures and mock test series for JEE Exam by signing up for free.
Here you can find the meaning of The decomposition of a substance follows first order kinetics. If its conc. is reduced to 1/8 th of its initial value, in 24 minutes, the rate constant of decomposition process isa)1/24 min1b)0.692/24 min1c)2.303/24 log (1/8) min1d)2.303/24 log (8/1) min1Correct answer is option 'D'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of
The decomposition of a substance follows first order kinetics. If its conc. is reduced to 1/8 th of its initial value, in 24 minutes, the rate constant of decomposition process isa)1/24 min1b)0.692/24 min1c)2.303/24 log (1/8) min1d)2.303/24 log (8/1) min1Correct answer is option 'D'. Can you explain this answer?, a detailed solution for The decomposition of a substance follows first order kinetics. If its conc. is reduced to 1/8 th of its initial value, in 24 minutes, the rate constant of decomposition process isa)1/24 min1b)0.692/24 min1c)2.303/24 log (1/8) min1d)2.303/24 log (8/1) min1Correct answer is option 'D'. Can you explain this answer? has been provided alongside types of The decomposition of a substance follows first order kinetics. If its conc. is reduced to 1/8 th of its initial value, in 24 minutes, the rate constant of decomposition process isa)1/24 min1b)0.692/24 min1c)2.303/24 log (1/8) min1d)2.303/24 log (8/1) min1Correct answer is option 'D'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice The decomposition of a substance follows first order kinetics. If its conc. is reduced to 1/8 th of its initial value, in 24 minutes, the rate constant of decomposition process isa)1/24 min1b)0.692/24 min1c)2.303/24 log (1/8) min1d)2.303/24 log (8/1) min1Correct answer is option 'D'. Can you explain this answer? tests, examples and also practice JEE tests.
|
|
Explore Courses for JEE 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