Question
Consider the following reactions. $A\overset{\quad K_{1} = 6.93 \times 10^{- 2}\text{ mi}\text{n}^{- 1}\quad}{\rightarrow}B\text{ and }A\overset{\quad K_{2} = 13.86 \times 10^{- 2}\text{ mi}\text{n}^{- 1}\quad}{\rightarrow}C$
Here, A, B and C all are optically active compounds. If the optical rotations of A, B and C per unit concentration are 60°, −72° and 42°, respectively, and the initial concentration of A is 2 M, then
Options
Solution
$\frac{\lbrack B\rbrack}{\lbrack C\rbrack} = \frac{K_{1}}{K_{2}} = \frac{1}{2} \Rightarrow \lbrack C\rbrack > \lbrack B\rbrack$
Hence, after long time, the solution will be dextrorotatory.
Now, K= K1 + K2 = 6.93 × 10–2 + 13.86 × 10–2 = 3 × 6.93 × 10–2 min–1
$\therefore t_{1/2} = \frac{\ln 2}{K} = \frac{0.693}{3 \times 6.93 \times 10^{- 2}} = \frac{10}{3}\text{ min}$
A ? B A ? C
t = 0 2M 0 2M 0
t = t 2 – (x + y)M xM 2 – (x + y)M yM
From question, x + y = 1.5 and x/y = 1/2
∴ x = 0.5, y = 1.0
Hence, total rotation = 0.5 × 60° + 0.5 × (–72°) + 1.0 × 42° = 36°
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