The `K_(p)` values for the reaction
`H_(2)(g)+I_(2)(g)hArr2HI(g)` at `460^(@)C` is 49. If the initial pressure of `H_(2)` and `I_(2)` are 0.5 atm respectively, determine the partial pressure of each gases at equilibrium.

Explanation

`H_(2_((g)))+I_(2_((g))) hArr 2HI_((g))`
Given `K_(P)=49`
`H_(2_((g)))+I_(2_((g))) hArr 2HI_((g))`
Initial `.P. " " 0.5 " " 0.5 " " 0`
At equilibrium `(0.5-x) (0.5-x) " "2x`
`K_(P)=((2x)^(2))/((0.5-x)(0.5-x)) :. K_(P)=([P_(HI)^(2)])/(P_(H)^(2)P_(2)^(2))`
`4a=((2x)/(0.5-x))^(2)`
`7=(2x)/(0.5-x)`
`3.57x=2x`
`9x=3.5`
`x=(3.5)/(9)`
`=0.3888`
`:. P_(HI)=2xx0.3888`
`=0.778` atm
`P_(H_(2))=0.5-0.388`
`=0.111` atm
`P_(1)=0.5-0.388`
`=0.0111` atm