For the reaction,
`0.5 C(s)+0.5CO_(2)(g) hArr CO(g)`
the equilibrium pressure is 12 atm. If `CO_(2)` conversion is 50%, the value of `K_(p)`, in atom is

Options

A. 4
B. 1
C. 0.5
D. 2

Answer

A

Explanation

For the reaction,
`0.5 C(s) +0.5 CO_(2) (g) hArr CO(g)`
`0.5 CO_(2) (g)` can give `CO(g)`.
If 50% conversion of `CO_(2)(g)` takes place.
We have concentration of `CO_(2)(g)=[0.25]`
`:' K_(p)` is calculated only for gaseous items, thus
`K_(p)=(["Product"]_((g)))/(["Reactant"]_((g)))=([1])/([0.25])=4`
Hence, option (a) is the correct answer.