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An optically active compound A upon acid catalysed hydrolysis yield two optically active compound B and C by pseudo first order kinetics. The observed rotation of the mixture after 20 min was 5° while after completion of the reaction it was –20°. If optical rotation per mole of A, B & C are 60°, 40° & –80°. Calculate half life of the reaction.
Correct answer is '20'. Can you explain this answer?
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An optically active compound A upon acid catalysed hydrolysis yield tw...
Optical rotation reaction no. of moles
At, t = 20 min
60 (a – x) + 40 x – 80 x = 5
60a – 100 x = 5
30 – 100 x = 5
No. of moles of A is becoming half after 20 min
∴ t1/2 = 20 min
Most Upvoted Answer
An optically active compound A upon acid catalysed hydrolysis yield tw...
Solution:
Given:
- Optically active compound A upon acid catalysed hydrolysis yield two optically active compound B and C.
- Observed rotation of the mixture after 20 min was 5 while after completion of the reaction it was 20.
- Optical rotation per mole of A, B, and C are 60, 40, and 80 respectively.
To find: Half-life of the reaction.
Approach:
- Use the formula for specific rotation and pseudo first-order kinetics to calculate the concentration of A, B, and C at different times.
- Use the formula for half-life to calculate the half-life of the reaction.
Calculations:
- Let the initial concentration of A be [A]₀ and the rate constant of the reaction be k.
- After 20 min, the observed rotation is given by:
- [α] = ([A]₀ - [A]) × 60 + [B] × 40 + [C] × 80
- Substituting [A] = [A]₀e^(-kt), [B] = [A]₀(1 - e^(-kt)) and [C] = [A]₀(1 - e^(-kt)) in the above equation, we get:
- 5 = [A]₀(60e^(-kt) + 40(1 - e^(-kt)) + 80(1 - e^(-kt)))
- Solving for [A]₀e^(-kt), we get:
- [A]₀e^(-kt) = 0.2
- After completion of the reaction, the observed rotation is given by:
- [α] = [B] × 40 + [C] × 80
- Substituting [B] = [A]₀(1 - e^(-kt)) and [C] = [A]₀(1 - e^(-kt)) in the above equation, we get:
- 20 = [A]₀(40(1 - e^(-kt)) + 80(1 - e^(-kt)))
- Solving for [A]₀e^(-kt), we get:
- [A]₀e^(-kt) = 0.5
- Using the formula for half-life, we get:
- t½ = ln(2)/k
- Substituting the value of [A]₀e^(-kt) from the above equations, we get:
- t½ = ln(2)/(20 ln(5/2))
Therefore, the half-life of the reaction is 20 min.
Given:
- Optically active compound A upon acid catalysed hydrolysis yield two optically active compound B and C.
- Observed rotation of the mixture after 20 min was 5 while after completion of the reaction it was 20.
- Optical rotation per mole of A, B, and C are 60, 40, and 80 respectively.
To find: Half-life of the reaction.
Approach:
- Use the formula for specific rotation and pseudo first-order kinetics to calculate the concentration of A, B, and C at different times.
- Use the formula for half-life to calculate the half-life of the reaction.
Calculations:
- Let the initial concentration of A be [A]₀ and the rate constant of the reaction be k.
- After 20 min, the observed rotation is given by:
- [α] = ([A]₀ - [A]) × 60 + [B] × 40 + [C] × 80
- Substituting [A] = [A]₀e^(-kt), [B] = [A]₀(1 - e^(-kt)) and [C] = [A]₀(1 - e^(-kt)) in the above equation, we get:
- 5 = [A]₀(60e^(-kt) + 40(1 - e^(-kt)) + 80(1 - e^(-kt)))
- Solving for [A]₀e^(-kt), we get:
- [A]₀e^(-kt) = 0.2
- After completion of the reaction, the observed rotation is given by:
- [α] = [B] × 40 + [C] × 80
- Substituting [B] = [A]₀(1 - e^(-kt)) and [C] = [A]₀(1 - e^(-kt)) in the above equation, we get:
- 20 = [A]₀(40(1 - e^(-kt)) + 80(1 - e^(-kt)))
- Solving for [A]₀e^(-kt), we get:
- [A]₀e^(-kt) = 0.5
- Using the formula for half-life, we get:
- t½ = ln(2)/k
- Substituting the value of [A]₀e^(-kt) from the above equations, we get:
- t½ = ln(2)/(20 ln(5/2))
Therefore, the half-life of the reaction is 20 min.
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An optically active compound A upon acid catalysed hydrolysis yield two optically active compound B and C by pseudo first order kinetics. The observed rotation of the mixture after 20 min was 5° while after completion of the reaction it was –20°. If optical rotation per mole of A, B & C are 60°, 40° & –80°. Calculate half life of the reaction.Correct answer is '20'. Can you explain this answer?
Question Description
An optically active compound A upon acid catalysed hydrolysis yield two optically active compound B and C by pseudo first order kinetics. The observed rotation of the mixture after 20 min was 5° while after completion of the reaction it was –20°. If optical rotation per mole of A, B & C are 60°, 40° & –80°. Calculate half life of the reaction.Correct answer is '20'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about An optically active compound A upon acid catalysed hydrolysis yield two optically active compound B and C by pseudo first order kinetics. The observed rotation of the mixture after 20 min was 5° while after completion of the reaction it was –20°. If optical rotation per mole of A, B & C are 60°, 40° & –80°. Calculate half life of the reaction.Correct answer is '20'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for An optically active compound A upon acid catalysed hydrolysis yield two optically active compound B and C by pseudo first order kinetics. The observed rotation of the mixture after 20 min was 5° while after completion of the reaction it was –20°. If optical rotation per mole of A, B & C are 60°, 40° & –80°. Calculate half life of the reaction.Correct answer is '20'. Can you explain this answer?.
An optically active compound A upon acid catalysed hydrolysis yield two optically active compound B and C by pseudo first order kinetics. The observed rotation of the mixture after 20 min was 5° while after completion of the reaction it was –20°. If optical rotation per mole of A, B & C are 60°, 40° & –80°. Calculate half life of the reaction.Correct answer is '20'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about An optically active compound A upon acid catalysed hydrolysis yield two optically active compound B and C by pseudo first order kinetics. The observed rotation of the mixture after 20 min was 5° while after completion of the reaction it was –20°. If optical rotation per mole of A, B & C are 60°, 40° & –80°. Calculate half life of the reaction.Correct answer is '20'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for An optically active compound A upon acid catalysed hydrolysis yield two optically active compound B and C by pseudo first order kinetics. The observed rotation of the mixture after 20 min was 5° while after completion of the reaction it was –20°. If optical rotation per mole of A, B & C are 60°, 40° & –80°. Calculate half life of the reaction.Correct answer is '20'. Can you explain this answer?.
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An optically active compound A upon acid catalysed hydrolysis yield two optically active compound B and C by pseudo first order kinetics. The observed rotation of the mixture after 20 min was 5° while after completion of the reaction it was –20°. If optical rotation per mole of A, B & C are 60°, 40° & –80°. Calculate half life of the reaction.Correct answer is '20'. Can you explain this answer?, a detailed solution for An optically active compound A upon acid catalysed hydrolysis yield two optically active compound B and C by pseudo first order kinetics. The observed rotation of the mixture after 20 min was 5° while after completion of the reaction it was –20°. If optical rotation per mole of A, B & C are 60°, 40° & –80°. Calculate half life of the reaction.Correct answer is '20'. Can you explain this answer? has been provided alongside types of An optically active compound A upon acid catalysed hydrolysis yield two optically active compound B and C by pseudo first order kinetics. The observed rotation of the mixture after 20 min was 5° while after completion of the reaction it was –20°. If optical rotation per mole of A, B & C are 60°, 40° & –80°. Calculate half life of the reaction.Correct answer is '20'. Can you explain this answer? theory, EduRev gives you an
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