Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
0 votes

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?

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

Can you answer this question?

1 Answer

0 votes
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}$
answered Dec 18, 2013 by sreemathi.v
edited Jul 27, 2014 by balaji.thirumalai

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, AIPMT Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App