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).

A mixture of N_(2) and H_(2) in the ratio 1 : 3 is allowed to attain e

1.	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).
Answer» Solution :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)`