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

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 this reaction reach equilibrium. The molar ratio of NH3 is 0.0385. Calculate (1) the standard equilibrium constant under this condition. (2) At 673 K, the total pressure of this system if the molar ratio of NH3 is 0.05.

Expert Solution

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

N2 + 3H2 -------------> 2NH3

initial 1 3 0

change -0.01925 -0.05775 0.03885

equilibrium (1-0.01925) (3-0.05775) 0.03885

Feed in the equation,

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

= 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) =

queen_honey_blossom answered 10 months ago

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