For Ammonia synthesis reaction: 3H2(g)+N2(g)?2NH3, Under 673 K, 1000 kPa, the initial molar ratio of H2...

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For Ammonia synthesis reaction: 3H2(g)+N2(g)?2NH3, Under 673 K,1000 kPa, the initial molar ratio of H2 and N2 is 3:1 and then thisreaction reach equilibrium. The molar ratio of NH3 is 0.0385.Calculate (1) the standard equilibrium constant under thiscondition. (2) At 673 K, the total pressure of this system if themolar ratio of NH3 is 0.05.

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(1) Keq = [NH3]^2/[N2][H2]^3

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â N2Â Â Â Â Â Â Â Â Â +Â Â Â Â Â Â Â Â Â 3H2Â Â Â Â ------------->Â Â Â 2NH3

initialÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 1Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 3Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 0

changeÂ Â Â Â Â Â Â Â Â Â Â Â -0.01925Â Â Â Â Â Â Â Â Â Â -0.05775Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 0.03885

equilibriumÂ Â Â Â Â (1-0.01925)Â Â Â Â Â (3-0.05775)Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 0.03885

Feed in the equation,

Keq = (0.03885)^2/(1-0.01925)(3-0.05775)^3

Â Â Â Â Â Â Â = 6.04 x 10^-5

(2) Total moles = 1 + 3 + 0.5 = 4.5

total presure of the system = 4.5 x 0.0821 x 673/22.4 = 11.1 atm

P(N2) =

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For Ammonia synthesis reaction: 3H2(g)+N2(g)?2NH3, Under 673 K, 1000 kPa, the initial molar ratio of H2...

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