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First Order Reaction

In a hypothet...

Question

In a hypothetical reaction, $A (a q) ⇌ 2 B (a q) + C (a q)$ (1 $^{s t}$ order decomposition)
'A' is optically active (dextro-rotatory) while 'B' and 'C' are optically inactive but 'B' takes part in a titration reaction (fast reaction) with $H_{2} O_{2}$ . Hence the progress of reaction can be monitored by measuring rotation of plane of polarised light or by measuring volume of $H_{2} O_{2}$ consumed in titration.
In an experiment, the optical rotation was found to be $θ = 30^{o}$ at $t = 20$ min. and $θ = 15^{o}$ at $t = 50$ min. from start of the reaction. If the progress would have been monitored by titration method, volume of $H_{2} O_{2}$ consumed at $t = 30$ . If the progress would have been monitored by titration method, volume of $H_{2} O_{2}$ consumed at $t = 30$ min. (from start) is 30 ml then volume of $H_{2} O_{2}$ consumed at $t = 90$ min. will be:

A

60 ml

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B

45 ml

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C

52.5 ml

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D

90 ml

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Solution

The correct option is C 52.5 ml
Only A is optically active.
Hence, the concentration of A at $t = 20 m i n \propto 30^{o}$ and the concentration of A at $t = 50 m i n \propto 15^{o}$
Thus in 30 min, the concentration decreases to half. So 30 min is the half-life period.
Thus, at $t = 30 m i n = t_{1 / 2}$ , the volume of hydrogen peroxide consumed corresponds to 50% production of B.
$t = 90$ min corresponds to three 30 min. The production of B is 87.5%.
Thus volume consumed $(30 m l) + (\frac{30}{2} m l) + (\frac{30}{4}) m l = 52.5 m l$

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