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 decreases in pressure of `CO_(2)`, then `CO_(2)(g)+C(s) hArr 2CO(g)`
`{:("Initial",,),("Pressusre : 0.48 bar" ,,0),("At equlibrium" ,(0.48-x)"bar","2xbar"):}`
`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+-sqrt((3)^(2))-4(4)(-1.44)]//2xx4`
`=(-3+-5.56)//8`
`=(-3+5.66)//8` (as value of `xx` cannot be negative henc we neglect that value)
`x=2.66//8=0.33`
The equilibrium partial pressure, are,
`P_(CO)=2x=2xx0.33=0.66` bar
`P_(CO_(2))=0.48-x=0.48-0.33=0.15` bar