1 mole of N_(2) and 3 mole of H_(2) are mixed in a closed vessel of 1 dm^(3) capacity. At equilibrium if the vessel contains a total of 2.4 moles, calculate equilibrium constant K_(c)" for "N_(2)+3H_(2)hArr2NH_(3).

1 mole of N_(2) and 3 mole of H_(2) are mixed in a closed vessel of 1

1.	1 mole of N_(2) and 3 mole of H_(2) are mixed in a closed vessel of 1 dm^(3) capacity. At equilibrium if the vessel contains a total of 2.4 moles, calculate equilibrium constant K_(c)" for "N_(2)+3H_(2)hArr2NH_(3).
Answer» <html><body><p></p>Solution :Initial no. of moles: `underset(<a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a>)(N_(2))+underset(3)(3H_(2))hArrunderset(0)(2NH_(3))` <br/> `{:("No. of moles reacting"),("at equilibrium"):}}x""3x""-` <br/> `{:("No. of moles present"),("at equilibrium"):}}1-x""3-3x""2x` <br/> `therefore` <a href="https://interviewquestions.tuteehub.com/tag/total-711110" style="font-weight:bold;" target="_blank" title="Click to know more about TOTAL">TOTAL</a> no. of moles present at equilibrium = (1 - x) + (3 - 3x) + 2x = 4 - 2x <br/> Given 4 - 2x = 2.4 `""therefore2x=4-2.4orx=1.6/2=0.8` <br/> Since the volume of the vessel is 1 `<a href="https://interviewquestions.tuteehub.com/tag/dm-432223" style="font-weight:bold;" target="_blank" title="Click to know more about DM">DM</a>^(3)`, <br/> `[N_(2)]=(1-x)/1=(1-0.8)/1=0.2" mol"//dm^(3),[H_(2)]=(3-3x)/1=(3-2.4)/1=0.6"mol"//dm^(3)` <br/> `[NH_(3)]=(2x)/1=1.6/1=1.6"mol"//dm^(3)` <br/> `K_(c)=([NH_(3)]^(2))/([N_(2)][H_(2)]^(3))=((1.6)^(2))/((0.2)xx(0.6)^(3))=59.26("mol"//dm^(3))^(-2)`.</body></html>