The equilibrium constant `K_(p)` of the reaction: `2SO_(2)+O_(2) hArr 2SO_(3)` is `900 atm^(-1)` at `800 K`. A mixture constaining `SO_(3)` and `O_(2)` having initial pressure of 1 atm and 2 atm respectively, is heated at constant volume to equilibriate. Calculate the partial pressure of each gas at `800 K` at equilibrium.

Explanation

`SO_(2) = 2/87 atm , O_(2) = 125/87 atm , SO_(3) = 85/87 atm`
`2SO_(2)(g) + O_(2)(g) rarr 2SO_(3)(g)`
`Kp = 900 = (1-2p)^(2)/((2+p)(2p)^(2)"[Ignoring in comparision to "2]`
p = 1/87
Partial pressure of `SO_(2) = 2p = 2/87 atm`
Partial pressure of `O_(2) = 2 + p = 2 + 1/87 = 125/87 atm`
Partial pressure of `SO_(3) = 1 - 2p = 1 - 2(1/87) = 85/87 atm`