A pure dextro rotatory substance A under suitable condition converts into an optically inactive product P following 1st order kinetics as
A $⟶$ P (optically inactive)
The following data was recorded for solution of pure A
Time (min) 0 60 $\infty$
Optical rotation +82º +77º +2º

Which of the following is/are correct options?

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optical rotation of solution after 5h is 80°

Half life time of reaction is 8h

Half life time of reaction is 10.75 h

optical rotation of solution after 5h is 60°

answer is B, C.

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Detailed Solution

$K = \frac{2.303}{60} \log_{10} (\frac{80}{75}) = \frac{0.693}{t_{1 / 2}}; t_{1 / 2} = 10.75 h$
$After 5 h K = \frac{2.303}{5} \log_{10} (\frac{80}{α_{t}}) = \frac{0.693}{10.75}$
$∴ \propto_{t} = 58^{0} o.r after 5 h = 58 + 2 = 60^{\circ}$

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