Question

For a first order reaction $A\to 2B+C$. It was found that at the end of 10 minutes from the start,the total optical rotation of the system was ${60}^0$ and when the reaction is complete.It was ${180}^0$.The B and C are only optically active and initially only A was taken.
i) What is the rate constant of the above reaction (in hour)?
ii) At what time (in minute) from the start, total optical rotation will be ${90}^0$.
Take log2=0.3,log3=0.48,log7=0.85, in 10 =2.3

Answer 1

The given reaction is:-

$A\to 2B+C$

Given that only $B$ and $C$ are optically active.

Now, at $t=0$, only $A$ is present which is optically inactive. So there will not be any observed optical rotation. Now,

Amount of $A$ present initially $\propto$ total optical rotation when reaction is complete

$\Rightarrow [A{]}_0\propto {180}^{\circ }$

Amount of $A$ present at any time, $t\propto$ (optical rotation of the system at time $t=0$)-(optical rotation at $t=t$)

$\Rightarrow [A{]}_t\propto {120}^{\circ }$ (after $10$ min)

For first order reaction,

$K=\frac{1}{t}ln\frac{[A{]}_0}{[A{]}_t}$

where, $K$= rate constant, $t$= time

$[A{]}_0$= concentration of $A$ at $t=0$

$[A{]}_t$= concentration of $A$ at $t=t$

$\Rightarrow K=\frac{1}{10}\times ln\left(\frac{180}{120}\right)$

$\Rightarrow \frac{1}{10}\times 0.41=0.041$ $mi{n}^{-1}$

Now, for $[A{]}_t\propto {90}^{\circ }$ we have,

$t=\frac{1}{K}ln\left(\frac{[A{]}_0}{[A{]}_t}\right)$

$=\frac{1}{0.041}ln\left(\frac{180}{90}\right)$

$\Rightarrow t=16.9$ $min$