At `700 K, CO_(2)` and `H_(2)` react to form `CO` and `H_(2)O`. For this purpose, `K_(c )` is `0.11`. If a mixture of `0.45` mol of `CO_(2)` and `0.45` mol of `H_(2)` is heated to `700 K`.
(a) Find out amount of each gas at equilibrium.
(b) When equilibrium has been reached, another `0.34` mol of `CO_(2)` and `0.34` mol of `H_(2)` are added to the reaction mixture. Find the composition of of mixture at new equilibrium.

Explanation

(`a`) `{:(,CO_(2),+,H_(2),hArr,CO,+,H_(2)O( :.K_(c)=0.11)),("Moles at t=0",0.45,,0.45,,0,,0),("Moles at equilibrium",(0.45-x),,(0.45-x),,x,,x):}`
`K_(c)=0.11=(x*x)/((0.45-x)^(2)`
(` :' Delta n=0`, :. Volume terms are not needed.)
or `(x)/((0.45-x))=0.3317`
`:. x=0.112`
`:. "Mole of" CO_(2)="Mole of" H_(2)`
`=0.45-0.112=0.338 "mole"`
`"Mole of" CO="Mole of" H_(2)O=0.112 "mole"`
(`b`) `{:(,CO_(2),+,H_(2),hArr,CO,+,H_(2)O),("Initial mole",0.45,,0.45,,0,,0),("Moles further added",(0.45+0.34),,(0.45+0.34),,0,,0),("Mole at new equilibrium",(0.79-x),,(0.79-x),,x,,x):}`
`:. K_(c)=(x^(2))/((0.79-x)^(2))=0.11`
`:. x=0.197`
`:. "Mole of" CO_(2)="Mole of" H_(2)`
`=0.79-0.197=0.593 "mole"`
`"Mole of" CO="Mole of" H_(2)O=0.197 "mole"`