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Solution :

First determine the equilibrium constant. <br> `K_(c )` for `H_(2)(g)+I_(2)(g) hArr 2HI(g)` <br> `K_(c )=([HI]^(2))/([H_(2)][I_(2)])=(0.8)^(2)/(0.08xx0.08)=100` <br> When `0.4 "mol"` of `HI` are added, equilibrium is disturbed At that instant, `[HI]=0.8+0.4=1.2 M` <br> `rArr Q gt K_(c )` since `Q=(1.2)^(2)/(0.08xx0.08)=225` <br> `rArr` Backward reaction dominates and the equilibrium shifts to the left. <br> Let `2x`= concentration of HI consumed (while going left) Then concentration of each of `H_(2)` and `I_(2)` formed = `x` <br> `rArr [HI]=1.2-2x, [H_(2)]=0.08+x`, <br> `[I_(2)]=0.08+x and K_(c )=100` <br> `rArr K_(c )=((1.2-2x)^(2))/((0.08+x)(0.08+x))=100` <br> `rArr` Take square root on the side to get <br> `x=0.033` <br> Finally, the equilibrium concentrations are: <br> `[HI]=1.2-2x=1.2-0.033xx2=1.13 M` <br> `[H_(2)]=0.08+x=0.08+0.033=0.11 M` <br> `[I_(2)]=0.08+x=0.08+0.033=0.11 M`