# An optically active compound A upon acid catalyzed hydrolysis yield two optically active compound B & C by first order kinetics.It is given that $[\alpha_0]$ for $A,B$ & C are $60^o,40^o$ & $-80^o$ respectively.If observed both of the mixture after 20 minute was $5^o$ while after completion of the reaction it was $-20^o$.What is $t_{1/2}$ of reaction?

$\begin{array}{1 1}(a)\;10minutes&(b)\;5minutes\\(c)\;20minutes&(d)\;40minutes\end{array}$

Reaction is $A\rightarrow B+C$
Let initially there are no moles of A are present .
$A\rightarrow B+C$
At $t=0,A\Rightarrow n_0$
At $t=t,A\Rightarrow n_0-\lambda,B\Rightarrow \lambda,C\Rightarrow \lambda$
At $t=\infty,B\Rightarrow n_0,C\Rightarrow n_0$
As $\alpha=n[\alpha]$,at $t=\infty,\alpha_{obs}=-20^o=n_0(40)-n_0(80)$
We get $n_0=1/2$
Now $(\alpha_{\infty}-\alpha_t)\propto n_0-\lambda$ and $(\alpha_{\infty}-\alpha_0)\propto n_0$
We get $k=\large\frac{1}{t}$$ln\large\frac{n_0}{n_0-\lambda}=\large\frac{1}{t}$$ln\big(\large\frac{\alpha_{\infty}-\alpha_0}{\alpha_{\infty}-\alpha_t}\big)$
$\Rightarrow \large\frac{1}{t}$$ln\big(\large\frac{-20-30}{-20-5}\big)=\large\frac{ln\;2}{20}$$mol^{-1}$
$t_{1/2}=\large\frac{ln\;2}{K}$
$\;\;\;\;\;\;=20min$
edited Jul 27, 2014