Question

In a hypothetical reaction, $A\left(aq\right)\rightleftharpoons 2B\left(q\right)+C\left(aq\right)(1^{st}$ order decomposition)
${}^{\prime }{A}^{\prime }$ is optically active (dextro-rototory) while ${}^{\prime }{B}^{\prime }$ and ${}^{\prime }{C}^{\prime }$ are optically inactive but ${}^{\prime }{B}^{\prime }$ 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 polarised light or by measuring volume of $H_2{O}_2$ consumed in titration.
In an experiment the optical rotation was found to be $\theta ={40}^{\circ }$ at $t=20min$ and $\theta ={10}^{\circ }$ at $t=50min$. from start of the reaction. If the progress would have been monitored by titration method, volume of $H_2{O}_2$ consumed at $t=15min$. (from start) is $40ml$ then volume of $H_2{O}_2$ consumed at $t=60$ min will be:

• A. $60ml$
• B. $75ml$
• C. $52.5ml$
• D. $90ml$

Answer 1

Answer: (A)

$60ml$

$\begin{array}{c}\text{As only}A\text{is optically acitive. So conc. of}A\text{at}t=20\text{min}\\ \text{while concentration of}A\text{at}t=50\text{min}\propto {10}^{\circ }\\ \text{so,}{t}_{1/2}=15\text{min.}\end{array}$

$\begin{array}{c}\text{So, volume consumed of}{H}_2{O}_2\text{at}t=15\text{min}=t_{1/2}\text{,}\\ \text{is according to}50\%\text{production of}B\text{at}t=60\text{min.}\\ \text{Production of}B=94.75\%\text{(four half lives)}\end{array}$

$\begin{array}{cc}\text{So volume consumed}& =40+\frac{40}{2}+\frac{40}{4}+\frac{40}{8}\\ & =40+20+10+5\\ & =75mL\end{array}$