If 50% of `CO_(2)` converts to CO at the following equilibrium:
`(1)/(2)C(s)+(1)/(2)CO_(2)(g) iff CO(g)`
and the equilibrium pressure is 12 atm. Calculate `K_(P)`.

Answer

A

Explanation

`{:(,C(s),+,CO_(2),hArr,2CO(g)),("Gaseous moles",-,,1,,0),("before dissociation",,,,,),("Gaseous moles",-,,(1-50/100),,(2xx50)/100),("after dissociation",-,,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/(Sigman)]^(Sigma_(n)) ((1)^(2))/0.5xx(12/1.5)^(1)`
`K_(p)=12/(1.5xx0.5)=16 "atm"`