The value of `K_(p)` for the reaction, `CO_(2)(g)+C(s)hArr2CO(g)` is `3.0` at `1000K` . If initially `P_(CO_(2))=0.48` bar and `P_(CO)=0` bar and pure graphite is present, calculate the equilibrium partial pressures 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)),("Intial",,,,,),("pressure:",0.48"bar",,,,0):}`
At equilibrium
`{:((0.48-x)"bar",2x"bar"):}`
`K_(p)=(p_(CO)^(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+-(3)^(2)-4(4)(-1.44)]//2xx4`
`=(-3+-5.66)//8`
`=(-3+5.66)//8` (as value of x cannot be negative hence we neglect that value)
`x=2.66//8=0.33`
The equilibrium partial pressures are ,
`p_(CO_(2))=2x=2xx0.33=0.66"bar"`
`p_(CO_(2))=0.48-x=0.48-0.33=0.15"bar"`