`60 mL` of `H_(2)` and `42` mL of `I_(2)` are heated in a closed vessel. At equilibrium, the vessel contains `20 mL HI`. Calculate degree of dissociation of `HI`.

Answer

A

Explanation

`{:(,H_(2),+,I_(2),hArr,2HI,"Given,"),("Volume at",60,,42,,0,:' 2x=28),(t=0,,,,,,:. x=14),("Volume at",(60-x),,(42-x),,2x,),("equilibrium",(60-14),,(42-14),,28,):}`
Since at constant P and T, moles `prop` Volume of gas )By `PV=nRT`). Thus, volume of gases given can be dorectly, used as concentration. This can be done for reactions leaving `Deltan=0`.
`:. K_(c)=(28xx28)/(46xx28)=28/46`
Now for dissociation of HI,
`{:(,2HI,hArr,H_(2),+,I_(2)),("moles at",1,,0,,0),(t=0,,,,,),("moles at Eq,(1-alpha),,alpha//2,,alpha//2):}`
where `alpha` is degree of dissociation
`K_(c_(1))=alpha^(2)/(4(1-alpha)^(2))=1/K_(c)`
`:. alpha/(2(1-alpha))=sqrt((46/28))`
`:. alpha=0.719` or `71.9%`