In the reaction `C(s)+CO_(2)(g) hArr 2CO(g)`, the equilibrium pressure is `12` atm. If `50%` of `CO_(2)` reacts, calculate `K_(p)`.

Options

A. `12 atm`
B. `16 atm`
C. `20 atm`
D. `6 atm`

Explanation

`{:(,,C_((s)),+,CO,hArr,2CO_((g))),(,"Gaseous moles before dissociation",-,,1,,0),(,"Gaseous moles after dissociation",-,,(1-(50)/(100)),,(2xx50)/(100)),(,,-,,0.5,,1):}`
Total moles `=1.5` and `Deltan=1`
Total pressure given at equilibrium `=12 atm`
`K_(p)=((n_(CO))^(2))/((n_(CO_(2)))xx[(P)/(sumn)]^(Deltan)=((1)^(2))/(0.5)xx((12)/(1.5))^(1)`
`K_(p)=(12)/(1.5xx0.5)=16 atm`