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.

Answer

A, C

Explanation

`{:(CO_(2)(g),+,H_(2)(g),hArr,CO(g),+,H_(2)(g)),(1,,1,,0,,0),(,,,,,,"moles at t=0"),((1-x),,(1-x),,x,,x),(,,,,,,"moles at equilibrium"):}`
Given that volume `%` of `CO=0.16`
`:.` moles of `CO=x`
The moles at equilibrium `=1-x+1-x+x+x=2`
`:. x/2=0.6/100 rArr :. 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)^(2))=1.03xx10^(-5)`