`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 equilibrium",(3-x),,(1-x),,x,,x):}`
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`)
Solving this quadratic equation by the formula:
`x=(-b+-sqrt(b^(2)-4ac))/(2a)`
`:. x=0.681`
`:. "Moles of" H_(2) formed=0.681`
Total pressure at equilibrium `=2 atm`
Total moles at equilibrium `=4`
`P'_(8)=P_(M)xx"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`