The equilibrium constant, `K_P`, for the reaction
`2SO_2(g)+O_2(g)hArr2SO_3(g)` is 900 atm at 800K. A mixture containing `SO_3` and `O_2` having initial pressure of 1 and 2 atrn respectively is heated at constant volume to equilibrium. Calculate the partial pressure of each gas at 800K

Explanation

At the beginning, `SO_2(g)` is not present in the reaction system. So, the equilibrium will establish when some amount of `SO_3(g)` decomposes to form `SO_2(g)` & `O_2(g)`. Suppose, the dissociation of some amount of `SO_2(g)` results in a decrease in pressure of `SO_2(g)` by p atm. So, at equilibrium, the pressures of various components will be as follows:
`2SO_2(g)+O_2(g) At the begining (atm): 0 2 1
At equilibrium (atm): `p" "2+p/2" "1-p`
`:.K_p=(p_(SO_3)^2)/(p_(SO_2)^2xxp_(O_2))=900atm`
or `K_p=((1-p)^2)/(p^2xx(2+p/2))=900`
or `((1-p)/p)=30sqrt2=42.42" ":.p=0.023`
`:.` At equilibrium, `p_(SO_2)=0.023atm`
`p_(O_2)=2+(0.023)/(2)=2.0115atm`
and `p_(SO_3)=(1-0.023)atm=0.977atm`