`60 mL` of `H_(2)` and `42` mL of `I_(2)` are heated in a closed vessel. At equilibrium, the vessel contains `28 mL HI`. Calculate degree of dissociation of `HI`.

Explanation

`{:(,H_(2),+,I_(2),hArr,2HI,,"(Given)"), ("Volume at t=0", 60,,42,,0,,: 2x=28), ("Volume at equilibrium", (60-x),,(42-x),,2x,, :. x=14), (,(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 directly used as concentration. This can be done only for reactions having `Deltan=0`.
`:. K_(c)=(28xx28)/(46xx28)=(28)/(46)`
`{:("Now for dissociation of" HI,,2HI,hArr,H_(2),+,I_(2)),("Moles at t=0",1,,0,,0),("Moles at equilibrium",(1-alpha),,alpha//2,,alpha//2):}`
(where `alpha` is the 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%`