For the equilibrium `2SO_3(g) leftrightarrow 2SO_2(g)+O_2(g)` the partial pressure `SO_3,SO_2 and O_2` gases , at 650 K are respectively 0.3 bar ,0.6 bar and 0.4 bar. IF the moles of both the oxides of sulphur are so adjusted as equal, what will be the partial pressure of `O_2`.

Explanation

The equilibrium given as `2SO_3(g) leftrightarrow 2SO_2(g)+O_2(g)`
Equilibrium constant `K_p= (p_(so_2)^2 po)/(p_(so_3)^p) =(0.6 times 0.6 times 0.4)/(0.3 times 0.3)=1.6` bar
Upon adjustment, `K_p` does not change 1.6 bar = `K_p=(x^2 P_O_2)/x^2`
Partial pressure of oxygen =1.6 bar.