`K_(c)` for `N_(2)O_(4)(g) hArr 2NO_(2)(g)` is `0.00466` at `298 K`. If a `1-L` container initially contained `0.8` mol of `N_(2)O_(4)`, what would be the concentrations of `N_(2)O_(4)` and `NO_(2)` at equilibrium? Also calculate the equilibrium concentration of `N_(2)O_(4)` and `NO_(2)` if the volume is halved at the same temperature.

Answer

A, B, D

Explanation

Suppose x mol of `N_(2)O_(4)` changes to `NO_(2)` at equilibrium.
`{:(N_(2)O_(4),hArr,2NO_(2),,),(0.8-x,,2x,,...("mol" L^(-1) "at equilibrium")):}`
`:. K_(c)=((2x)^(2))/((0.8-x))=0.00466, x=0.03 M`
`:.` At equilibrium, `[N_(2)O_(4)]=0.8-x=0.8-0.03=0.77 M`
`[NO_(2)]=2x=2xx0.03=0.06 M`
Now when the volume is halved, pressure will increase. From Le Chatelier's principle, we know that the equilibrium will shift to left hand side with increase in pressure (or decrease in volume). Further when volume is halved, concentration will be dounled.
Concentration of `N_(2)O_(4)=0.77xx2=1.54 M`
and concentration of `NO_(2)=0.06xx2=0.12 M`
The equilibrium concentration now will be
`{:(N_(2)O_(4),hArr,2NO_(2)),((1.54+y),,(0.12-2y)):}`
`K_(c)=((2x)^(2))/((0.8-x))=0.00466, x=0.03 M`
`y=0.104` and `0.017`
`y=0.104` is unacceptable as in this case `2y gt 0.12`
`:. y=0.017`
`:.` At new equilibrium
`[N_(2)O_(4)]=1.54+0.017=1.557 M`
`[NO_(2)]=0.12-2xx0.017=0.086 M`