A mixture of one mole of `CO_(2)` and "mole" of `H_(2)` attains equilibrium at a temperature of `250^(@)C` and a total pressure of `0.1` atm for the change `CO_(2)(g)+H_(2)(g) hArr CO(g)+H_(2)O(g)`. Calculate `K_(p)` if the analysis of final reaction mixture shows `0.16` volume percent of CO.

Explanation

`{:(,CO_2(g)+, H_2(g) hArr , CO(g)+, H_2O(g)),("Mole at t=0", 1,1,0,0),("Mole of equilibrium" ,(1-x), (1-x), x, x):}`
Given that Vol. % of CO=0.16
Mole of CO= `pi`
Total mole at equilibrium = 1 – x + 1 – x + x + x = 2
`therefore X/2=0.16/100`
`therefore` x=0.0032
Now, `K_C=K_p=X^2/((1-X)^2)` (? `Deltan=0` , volume terms are not needed )
`K_P=((0.0032)^2)/((1-0.0032))=1.03xx10^(-5)`