Calculate the ratio of pressure of `CO_(2)" gas and "CO` gas at equilibrium in the reaction, `CO_(2) (g)+C(s) Leftrightarrow 2CO(g)`, If `K_(p)` is 3 bar at 900K and initial pressure of `CO_(2)` is 0.48 bar.

Explanation

`CO_(2)(g)+C(s) Leftrightarrow 2CO(g)`
`"0.48 bar 0 0 at start"`
`"0.48-x 0 2x at equilibrium"`
`K_(P)=(P^(2) co)/(P_(CO_(2))=((2x)^(2))/((0.48-x))=3`
`4x^(2)+3x-1.44=0, x=0.33` (or -1.083, but negative value is neglected)
`P_(CO_(2))` = 0.48-0.33=0.15 bar,
`P_(CO)=2 xx 0.33 =0.66" bar"`
Ratio of pressure `=p_(CO_(2))/(P_(CO))=(0.15)/(0.66)=(15)/(66)=0.227`