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A mixture of 1 mole of N_(2) and 3 moles of H_(2) is allowed to react at a constant pressure of 100 bar. At equilibrium, 0.6 mole of ammonia is formed. Calculate the equilibrium constant for the reaction N_(2)+3H_(2)hArr2NH_(3).

Answer» <html><body><p></p>Solution :<a href="https://interviewquestions.tuteehub.com/tag/initial-499269" style="font-weight:bold;" target="_blank" title="Click to know more about INITIAL">INITIAL</a> no. of <a href="https://interviewquestions.tuteehub.com/tag/moles-1100459" style="font-weight:bold;" target="_blank" title="Click to know more about MOLES">MOLES</a>: `underset(1)(N_(2))+underset(3)(3H_(2))hArrunderset(0)(2NH_(3))` <br/> `{:("No. of moles reacting"),("at equilibrium"):}}x""<a href="https://interviewquestions.tuteehub.com/tag/3x-310805" style="font-weight:bold;" target="_blank" title="Click to know more about 3X">3X</a>""2x=0.6thereforex=0.3` <br/> `{:("No. of moles present"),("at equilibrium"):}}{:(1-0.3,3-0.9),(=0.7,=2.1):}" "0.6` <br/> `therefore` Total no. of moles present at equilibrium = 0.7 + 2.1 + 0.6 = 3.4 <br/> Partial pressure = <a href="https://interviewquestions.tuteehub.com/tag/mole-1100299" style="font-weight:bold;" target="_blank" title="Click to know more about MOLE">MOLE</a> fraction `xx` total pressure <br/> Since the total pressure = 10 bar. <br/> `P_(N_(2))=0.7/3.4xx100`bar = 20.59 bar <br/> `P_(H_(2))=2.1/3.4xx100` bar = 61.76 bar <br/> `P_(NH_(3))=0.6/3.4xx100` bar = 17.65 bar <br/> `K_(p)=(P_(NH_(3))^(2))/(P_(NH_(2))xxP_(H_(2))^(3))=((17.65)^(2))/(20.59xx(61*76)^(3))=6.420xx10^(-5)`</body></html>


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