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

` 2 SO_(3) hArr 2 SO_(2) + O_(2) , " " K_(p) = 1/900 atm `
` 2 SO_(3) hArr 2 SO_(2) + O_(2) , " " " K_(p) = 1/900 atm `
` K_(p) = (p_(SO_(2))^(2) xxp_(O_(2)))/(p_(SO_(3))^(2))=(x^(2) xx(2+x/2))/((1-x)^(2)) = 1/100`
As `K_(p)` for this reaction is very small , `x lt lt 1 " Taking " 2 + x/2 cong 2 and (1-x) cong 1 , ` we get
` x^(2) (2) = 1/900 or x^(2) = 1/1800 or x = 0*0236`
Hence , at equilibrium , ` P_(SO_(3))= 1 - x = 1 - 0*0236 " atm ",p_(SO_(2)) = x = 0*0236 `
` p_(O_(2)) = 2 + x/2 = 2 + (0*0236)/2 = 2*0118 " atm " `