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 the optical rotation per molar concentration of A, B and C are 60°, 40° and – 80° respectively and the observed rotation of a compound is directly proportional to the concentration of that compound. (Take optical rotation in degree)

$\begin{array}{cc}\text{List -I}& \text{List -II}\\ \text{(I) Rate constant of the reaction (in}hou{r}^{-1})& \left(P\right)1.0\\ \text{(II) Initial concentration of compound A (in M)}& \left(Q\right)2.08\\ \text{(III) If the initial concentratrion of A is 0.8 M then}& \left(R\right)0.5\\ \text{the observed optical rotation of mixture after 1 hour}& \\ \text{(IV) If initially along with A;B and C are also present with}& \left(S\right)-22\\ \text{initial concentration [A] = 0.4M, [B] = 0.1 M and [C] = 0.2 M}& \\ \text{then net observed optical rotation of the mixture after 40 min.}& \left(T\right)22\\ & \left(U\right)-18\end{array}$

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

• A. (I),(Q)
• B. (II),(P)
• C. (II),(R)
• D. (I),(P)

Answer 1

Answer: (A), (C)

The correct options are
A (I),(Q)
C (II),(R)

$A\stackrel{H^{+}}{\to }B+C\mathrm{At}t=0a00\mathrm{At}t=20\min a-xxx\mathrm{At}t=\mathrm{completion}0aa$
According to the question,
60(a-x) + 40x + (-80)x = 5 [since the observed rotation is directly proportional to the concentration] .....(1)
40a - 80a = -20 .....(2)
$\Rightarrow$a = 0.5 M
From equation (1) ,putting the value a=0.5, we get
x = 0.25 M
Since 0.25 is half of 0.5, 20 min is the half life of the reaction.
For a first order reaction, $t_{\frac{1}{2}}=\frac{0.693}{k}$
$\Rightarrow k=2.08{\mathrm{hour}}^{-1}$
So the initial concentration of compound A = 0.5M
Rate constant of the reaction k = 2.08$hou{r}^{-1}$