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» <html><body><p></p>Solution :Total pressure of the mixture = `5xx10^(5) Nm^(-2)` <br/> `therefore` Partial pressure of `NH_(<a href="https://interviewquestions.tuteehub.com/tag/3-301577" style="font-weight:bold;" target="_blank" title="Click to know more about 3">3</a>)=(40xx5xx10^(5))/100=2xx10^(5)Nm^(-2)` <br/> The <a href="https://interviewquestions.tuteehub.com/tag/sum-1234400" style="font-weight:bold;" target="_blank" title="Click to know more about SUM">SUM</a> of the partial pressure of `N_(2)andH_(2)=5xx10^(5)-2xx10^(5)=<a href="https://interviewquestions.tuteehub.com/tag/3xx10-1865443" style="font-weight:bold;" target="_blank" title="Click to know more about 3XX10">3XX10</a>^(5)` <br/> Since `N_(2)andH_(2)` are in the <a href="https://interviewquestions.tuteehub.com/tag/ratio-13379" style="font-weight:bold;" target="_blank" title="Click to know more about RATIO">RATIO</a> 1 : 3 <br/> `P_(N_(2))=1/4xx3xx10^(5)=0.75xx10^(5)Nm^(-2),P_(H_(2))=3/4xx3xx10^(5)=2.25xx10^(5)Nm^(-2)` <br/> `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)`</body></html>