A sample of `HI(g)` is placed in flask at a pressure of `0.2 atm`. At equilibrium. The partial pressure of `HI(g)` is `0.04 atm`. What is `K_(p)` for the given equilibrium?
`2HI(g) hArr H_(2)(g)+I_(2)(g)`

Answer

`4.0`

Explanation

The initial concentration of HI is 0.2 atm. At equilibrium, it has a partial pressure of 0.04 atm. Therefore, a decrease in the pressure of HI is 0.2 – 0.04 = 0.16. The given reaction is:
`{:(,2HI_((g)),harr,H_(2(g)),+,I_(2(g))),("Initial conc.",0.2 "atm",,0,,0),("At equilibrium",0.04 "atm",,(0.16)/(2),,(2.15)/(2)),(,,,=0.08"atm",,=0.08"atm"):}`
Therefore,
`K_(p) = (p_(H_(2)) xx P_(I_(2)))/(P_(HI)^(2))`
`= (0.08 xx 0.08)/((0.04)^(2))`
`= (0.0064)/(0.0016)`
`= 4.0`
Hence, the value of `K_(p)` for the given equilibrium is `4.0`.