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

`{:(,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 Eq",(0.45-x),,(0.4-x),,x,,x):}`
`:. K_(c )=0.11=(x.x)/((0.45-x)^(2)) :' K_(c )=([H_(2)O][CO])/([CO_(2)][H_(2)])`
`( :. Deltan=0, :.` volume terms are not needed)
or `(x)/((0.45-x))=0.3317` and `x=0.112`
`:.` moles of `CO_(2)`= moles of `H_(2)`
`=0.46-0.112 =0.338` moles
moles of `CO`= moles of `H_(2)O=0.112`
(b). `{:(,CO_(2),+,H_(2),hArr,CO,+,H_(2)O),("Initial moles",0.45,,0.45,,0,,0),("moles further added",(0.45+0.34),,(0.45+0.34),,0,,0),("moles of new equilibrium",(0.79-x),,(0.79-x),,x,,x):}`
`:. K_(c )=x^(2)/((0.79-x)^(2))=0.11 and x=0.197`
`:.` "moles" of `CO_(2)`= "moles" of `H_(2)=0.79-0.197=0.593 "mol"`
moles of `CO`= moles of `H_(2)O=0.197 "mol"`