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.

Answer

`p_(H_(2)) = p_(I_(2)) = 0*11 " atm ", p_(HI)= 0.78 " atm"`

Explanation

`{:(,H_(2),+,I_(2),hArr,2 HI),("Intial ",0*5,,0*5,,0),("At. eqm.",0*5-x,,0*5-x,,2" "x):}`
`K_(p) = (2x)^(2)/(0*5-x)^(2) = 49 or (2 x)/(0*5 - x)=7 or 2 x = 3*5- 7 x`
or ` 9 x = 3*5 or x= (3*5)/9=0*39`
`:. " Pressure of " H_(2) and I_(2) " at eqm."= 0*5 - 0*39= 0*11 " atm"`
Pressure of HI at eqm. `= 2 xx 0*39 = 0*78 " atm "`