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

Explanation

`{:(,,H_(2),+,I_(2),hArr,2HI),(,"Initial pressure",0.5,,0.5,,0),(,"At equilibrium",(0.5-x),,(0.5-x),,2x):}`
`:. K_(p)=((2x)^(2))/((0.5-x)^(2))=49`
or `(2x)/(0.5-x)=7`
or `2x=3.5-7x`
or `9x=3.5`
`x=(3.5)/(9)=0.389`
At equilibrium,
`P'_(H_(2))=0.5-0.389=0.111 atm`
`P'_(I_(2))=0.5-0.389=0.111 atm`
`P'_(HI)=0.389xx2=0.778 atm`