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Question

An optically active compound A upon acid catalysed hydrolysis yields two optically active compounds B and C by pseudo first order kinetics with respect to A. The observed rotation of the mixture after 20 min was 5° while after completion of the reaction it was – 20°. If optical rotation per molar concentration of A, B and C are 60°, 40° and – 80° respectively and observed rotation of a compound is directly proportional to concentration of that compound. (Take optical rotation in degree)

List -IList -II(I) Rate constant of the reaction (in hour1)(P)1.0(II) Initial concentration of compound A (in M)(Q)2.08(III) If the initial concentratrion of A is 0.8 M then(R)0.5the observed optical rotation of mixture after 1 hour(IV) If initially along with A;B and C are also present with(S)22initial concentration [A] = 0.4M, [B] = 0.1 M and [C] = 0.2 Mthen net observed optical rotation of the mixture after 40 min.(T)22(U)18

Which of the following options has the correct combination considering List-I and List-II

A
(III),(U)
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B
(IV),(U)
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C
(III),(S)
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D
(IV),(S)
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Solution

The correct option is C (III),(S)
AH+B+CAt t=0 a 0 0At t=20min ax x xAt t=completion 0 a a
So we have calculated,
half life of the reaction t12= 20 min
initial concentration of the compound A = 0.5 M
Rate constant of the reaction k = 2.08hour1
Now if the initial conc of A is 0.8 M then after 1 hour i.e. 3 half lives conc of A = 0.1 M, so B and C will have concentration 0.7 M . Hence, final conc [A] = 0.1 M, [B] = 0.7 M, [C] = 0.7 M
So, net optical rotation =
0.1×60+0.7×40+0.7×(80) = -22
If the initial conc of A is 0.4 M along with [B] = 0.1 M and [C] = 0.2 M then after 40 min i.e. 2 half lives
conc of A = 0.1 M, so 0.3 M get reacted. Hence, final conc [A] = 0.1 M, [B] = 0.4 M, [C] = 0.5 M
So, net optical rotation =
0.1×60+0.4×40+0.5×(80)= -18

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