At a certain temperature, equilibrium constant `(K_(c))` is 16 for the reaction,
`SO_(2)(g)+NO_(2)(g)hArrSO_(3)(g)+NO_(g)`
If we take one mole each of all the four gases in a one litre container, what would be the equilibrium concentrations of `NO` and `NO_(2)`?

Explanation

`{:(,SO_2(g) +, NO_2(g) hArr , SO_3(g)+, NO(g)),("Mole at t=0", 1,1,1,1),("Mole at equilibrium" ,(1-x), (1-x) , (1+x) , (1+x)):}`
`therefore K_C=(("1+x"/V)("1+x"/V))/(("1-x"/V)("1-x"/V))=((1+x)^2)/((1-x)^2)`
`therefore ((1+x)^2)/((1-x)^2)=16 ( therefore K_C=16), ((1+x))/((1-x))=4, x=3/5=0.6`
`therefore [SO_2]=[NO_2]=1-x=1-0.6=0.4 "mol litre"^(-1)` (Volume = 1 litre)
`[SO_3]=[NO]=1+x=1+0.6 = 1.6 "mol litre"^(-1)`