Question

In a hypothetical reaction, $A\left(aq\right)\rightleftharpoons 2B\left(aq\right)+C\left(aq\right)$ (1${}^{st}$ 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 $\theta ={30}^o$ at $t=20$ min. and $\theta ={15}^o$ 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
• B. 45 ml
• C. 52.5 ml
• D. 90 ml

Answer 1

The correct option is C 52.5 ml

Only A is optically active.
Hence, the concentration of A at $t=20min\propto {30}^o$ and the concentration of A at $t=50min\propto {15}^o$

Thus in 30 min, the concentration decreases to half. So 30 min is the half-life period.

Thus, at $t=30min=t_{1/2}$, 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 $\left(30ml\right)+\left(\frac{30}{2}ml\right)+\left(\frac{30}{4}\right)ml=52.5ml$