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

`{:(,2SO_2(g) + , O_2(g) hArr , 2SO_3(g)),("Initial conc.", 0,2,1),("At equi.",2x,2+x, 1-2x):}`
`K_p=(p_(SO_3))^2/((pSO_2)^2(pO_2))=(1-2x)^2/((2x)^2(2+x)) ="900 atm"^(-1)`
Assuming x to be very small
`1/((4x^2)xx2)="900 atm"^(-1)`
Solving x=0.0118 atm
`therefore p(SO_2)` =2 x 0.0118 =0.0236 atm , `p(O_2)`= 2 + 0.0118 =2.0118 atm , `p(SO_3)` =1-2 x 0.0118 = 0.976 atm