In the reaction `C_((s))+CO_(2(g))hArr2CO_((g))`, the equilibrium pressure is `6.75 atm`. If `50%` of `CO_(2)` reacts, then find the value of `K_(p)`.

Explanation

`{:(,,C_((s)),+,CO_(2(g)),hArr,2CO_((g))),(,"Gaseous mole before dissocitation",-,,1,,0),(,"Gaseous mole",-,,(1-(50)/(100)),,(2xx50)/(100)),(,"after dissociation",,,=0.5,,=1):}`
Total mole `=1.0+0.5=1.5`, `Deltan=1`
Total pressure given at equilibrium `=6.75 atm`
`:. K_(p)=((n_(CO))^(2))/(n_(CO_(2)))xx[(P)/(sumn)]^(n)`
`=((1)^(2))/(0.5)xx[(6.75)/(1.5)]^(1)=9.0 atm`