A mixture of `N_(2) and H_(2)` in the ratio 1 : 3 is allowed to attain equilibrium. At equilibrium, the total pressure is `5xx10^(-5)Nm^(-2)` and the mixture contains 40% by volume of `NH_(3)`. Calculate `K_(p)`.

Explanation

Total pressure of the mixture = `5xx10^(5) Nm^(-2)`
`therefore` Partial pressure of `NH_(3)=(40xx5xx10^(5))/100=2xx10^(5)Nm^(-2)`
The sum of the partial pressure of `N_(2)andH_(2)=5xx10^(5)-2xx10^(5)=3xx10^(5)`
Since `N_(2)andH_(2)` are in the ratio 1 : 3
`P_(N_(2))=1/4xx3xx10^(5)=0.75xx10^(5)Nm^(-2),P_(H_(2))=3/4xx3xx10^(5)=2.25xx10^(5)Nm^(-2)`
`K_(p)=(P_(NH_(3))^(2))/(P_(N_(2))xxP_(H_(2))^(3))=((2xx10^(5))^(2))/((0.75xx10^(5))xx(2.25xx10^(5))^(3))=0.468xx10^(-10)(Nm^(-2))^(-2)`