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).

A mixture of 1 mole of N_(2) and 3 moles of H_(2) is allowed to react

1.	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» Solution :INITIAL no. of MOLES: `underset(1)(N_(2))+underset(3)(3H_(2))hArrunderset(0)(2NH_(3))` `{:("No. of moles reacting"),("at equilibrium"):}}x""3X""2x=0.6thereforex=0.3` `{:("No. of moles present"),("at equilibrium"):}}{:(1-0.3,3-0.9),(=0.7,=2.1):}" "0.6` `therefore` Total no. of moles present at equilibrium = 0.7 + 2.1 + 0.6 = 3.4 Partial pressure = MOLE fraction `xx` total pressure Since the total pressure = 10 bar. `P_(N_(2))=0.7/3.4xx100`bar = 20.59 bar `P_(H_(2))=2.1/3.4xx100` bar = 61.76 bar `P_(NH_(3))=0.6/3.4xx100` bar = 17.65 bar `K_(p)=(P_(NH_(3))^(2))/(P_(NH_(2))xxP_(H_(2))^(3))=((17.65)^(2))/(20.59xx(61*76)^(3))=6.420xx10^(-5)`