question_answer

The degree of dissociation of HI at a particular temperature is 0.8. The volume of 2 M \[N{{a}_{2}}{{S}_{2}}{{O}_{3}}\] solution required to neutralise the iodine present in an equilibrium mixture of a reaction when 2 mole each of \[{{H}_{2}}\] and \[{{I}_{2}}\], are heated in a closed vessel of 2 litre capacity is

A) 1.6

B) 0.25

C) 0.4       

D) 0.16

Correct Answer: A

Solution :

Given, \[\alpha \]for \[\operatorname{HI}=0.8\]

\[\therefore \]\[{{K}_{e}}~=\frac{{{\alpha }^{2}}}{4{{(1-\alpha )}^{2}}}=\frac{{{(0.8)}^{2}}}{4{{(1-0.8)}^{2}}}=4\]

Now

\[\therefore K_{e}^{'}=\frac{1}{{{K}_{e}}}=\frac{4{{x}^{2}}}{{{\left( 2-x \right)}^{2}}}=\frac{1}{4}\]

\[\operatorname{x}=2/5\]

Thus, \[{{\operatorname{I}}_{2}}\]left =\[2-\frac{2}{5}=\frac{8}{5}\]mole \[=\frac{8}{5}\times 2\]equivalent

Or \[{{\operatorname{Na}}_{2}}{{S}_{2}}{{O}_{3}}\]required can be calculate as;

\[\operatorname{Meq}.\,Of\,N{{a}_{2}}{{S}_{2}}{{O}_{3}}=-Meq.Of\,{{I}_{2}}\,left\]

\[2\times V=\frac{8}{5}\times 2\times 1000\]

\[\therefore \]\[V=1600mL=1.6\,litre\]