`K_(c )` for `CO(g)+H_(2)O(g) hArr CO_(2)(g)+H_(2)(g)` at `986^(@)C` is `0.63`. A mixture of `1` mol `H_(2)O(g)` and `3` mol `CO_(2)(g)` is allowed to react to come to an equilibrium. The equilibrium pressure is `2.0` atm.
a. Hoe many moles of `H_(2)` are present at equilibrium ?
b. Calculate partial pressure of each gas at equilibrium.

Explanation

`{:(,CO(g)+,H_(2)O(g),hArr,CO_(2)(g)+,H_(2)(g)),("Initial moles",3,1,,0,0),("moles at",(3-x),(1-x),,x,x),("equilibrium",,,,,),("Total moles at equilibrium"=3-x+1-x+x+x=4,,,,,):}`
Now, `K_(c )=x^(2)/((3-x)(1-x))`
`:. x^(2)/(3+x^(2)-4x)=0.63, ( :. K_(c )=0.63)`
`:. x=0.681`
`:.` moles of `H_(2)` formed `=0.681`
Total pressure at equilibrium`=2 "atm"`,
Total moles at equilibrium=`4`
`p'_(g)=P xx` mole fraction of that gas
`:. P'_(g)=Pxx` mole fraction of that gas
`:. P'_(CO_(2))=P_(H_(2))=(x.P)/(4)=(0.681xx2)/(4)=0.34 "atm"`
`P'_(CO)=((3-x).P)/(4)=1.16 "atm"`
`P'_(H_(2)O)=((1-x).P)/(4)=0.16 "atm"`