The value of `K_(p)` for the reaction
`CO_(2)(g)+C(s)hArr2CO(g)`
is `3.0` bar at `1000 K`. If initially `P_(CO_(2)) = 0.48` bar, `P_(CO) = 0` bar and pure graphite is present then determine equilibrium partial pressue of `CO` and `CO_(2)` .

Explanation

For the reaction
Let 'x' be the decrease in pressure of `CO_(2)`.
then
`CO_(2)(g) + C(s) hArr 2CO(g)`
`{:("Initial",,),("pressure":,0.48 "bar",0):}`
At equilibrium
`(0.48 -x)`bar , `2x`bar
`K_(p) = (p_(CO_(2))^(2))/(p_(CO_(2)))`
`K_(p) = (2x)^(2)//(0.48-x) = 3`
`4x^(2) = 3(0.48 - x)`
`4x^(2) = 1.44 -x`
`4x^(2) + 3x - 1.44 = 0`
`a=4,b=3,c=-1.44`
`x=((-b+-sqrt(b^(2)-4ac)))/(2a)`
`=[-3+-sqrt((3)^(2))-4(4)(-1.44)]//2xx4`
`=(-3+-5.66)//8`
`=(-3+5.66)//8` (as value of x cannot be nagative hence we neglect that value)
`x = 2.66//8 = 0.33`
The equilibrium particle pressures are,
`p_(CO_(2)) = 2x=2 xx 0.33 = 0.66 "bar"`
`p_(CO_(2)) = 0.48 - x = 0.48 - x = 0.48 - 0.33 = 0.15 "bar"`