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. `K_(p)` will be equal to 4
B. `K_(p)` will be equal to 16
C. The initial pressure =8
D. The partial pressure of CO is 8 atm at equlibrium

Answer

B, C, D

Explanation

`K_(p)=(P_(CO)^(2))/(P_(CO_(2))),P_(CO_(2))+P_(CO)=12implies1/2P_(CO)+P_(CO)=12`
`[thereforeP_(CO_(2))=1/2P_(CO)]`
`3/2P_(CO)=12impliesP_(CO)=(2xx12)/(3)=8`
`impliesK_(p)=((P_(CO))^(2))/((P_(CO_(2))))=(8xx8)/(4)=16" "["Initially only" CO_(2)` was present, therefore initial pressure =8]
Hence choices (b), (c), (d) are correct while (a) is incorrect.