The degree of dissociation of `HI` at a particualr temperature is `0.8`. Calculate the volume of `2 M Na_(2)S_(2)O_(3)` solution required to neutralise the iodine present in an equilibrium mixture of a reaction when 2 mol each of `H_(2)` and `I_(2)` are heated in a closed vessel of `2 L` capacity and the equilibrium mixture is freezed.

Explanation

`2HI(g)+H_(2)(g)+I_(2)(g)`
`{:("Before dissociation",,1,0,0),("After dissociation",,(1-alpha),alpha/2,alpha/2):}`
`:. K_(c )=alpha^(2)/(4(1-alpha)^(2))=((0.8)^(2))/(4(1-0.8)^(2))=4 (since, alpha=0.8)`
Now, `H_(2)(g)+I_(2)(g)hArr2HI(g)`
`{:("Initial moles",,2,2,0),("moles After reaction",,(2-x),(2=x),2x):}`
`:. K_(c )^(')=1/K_(c )=((2x)^(2))/((2-x)^(2))`
or `((2x)^(2))/((2-x)^(2))=1/4 rArr (2x)/(2-x)=1/2`
or `x=2/5`
Thus, moles of `I_(2)` left `=2-2/5=8/5`
Equivalents of `Na_(2)S_(2)O_(3)`= Equivalents of `I_(2)` left at equilibrium (where `V` is the volume in `L`)
`2xxV=8/5xx2`
`V=1.6 L`