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.

Answer

`p_(SO_2)` = 0.0236 , `p_(O_2)` = 2.0118 , `p_(SO_3)` = 0.976

Explanation

`{:(,2SO_2(g)+,O_2(g) hArr , 2So_3(g)),("Initial pressure", 0,"2 atm" , " 1 atm"):}`
Since volume of the vessel remains constant, so that the partial pressures are proportional to moles. Let us assume that at equilibrium 2s moles of `SO_3` have decomposed to give 2x moles of `SO_2(g)` and x mole of `O_2`. Thus at equilibrium
`{:(,2SO_2+, O_2 hArr , 2SO_3),("Equi pressure", "2x atm","(2+x) atm" , "(1-2x) atm"):}`
`K_p=([p_(SO_3)]^2)/([p_(SO_2)]^2[p_(O_2)])=900`
or `(1-2x)^2/((2x)^2(2+x))=900`
Since x is very small assuming 1-2x `approx` 1,2+x `approx` 2 and solving for x
x=0.0118
`p_(SO_2)`=2x = 2 x 0.0118 = 0.0236 atm
`p_(O_2)`=2+x =2 + 0.0118 =2.0118 atm
`p_(SO_3)` =1-2x =1-2 x 0.0118 =0.976 atm