Question

Ammonia was formed at 450 ◦C by passing a mixture of nitrogen gas and hydrogen gas at a 1 : 3 mole ratio over a catalyst. When the total pressure was held constant at 10.13 bar it was found that the product gas contained 2.04% by volume of ammonia. For the reaction
(1/2)N2(g) +(3/2)H2(g) <----> NH3(g)
calculate the value of the equilibrium constant K at 450 ◦C.

Answer 1

N₂(g) +3H₂(g) <----> 2NH₃(g)

Initially say total moles = 4

1 mol N2 and 3 mol H2

total volume = V = nRT/P = 4 x 8.314 x (450+273) / 10.13 x 101325Pa = 0.023425 m³ = 23.425 L

N₂(g) +3H₂(g) <----> 2NH₃(g)

1 3 0 initially

1-x 3-3x 2x at equilibrium

total moles = 1-x + 3-3x + 2x = 4-2x

total volume = nRT/P = (4-2x)x 8.314 x 723 K / 10.13 x 101325

2.04 % is ammonia

(4-2x)x 8.314 x 723 K / 10.13 x 101325 x 2.04 /100 = nRT/P = 2x x 8.314 x 723 / 10.13 x 101325

(4-2x) x 2.04 /100 =  2x

8.16 - 4.08x = 200x

204.08x = 8.16

x = 0.04

total volume after reaction = (4-0.08)x 8.314 x 723 K / 10.13 x 101325 = 0.02296 m³ = 22.96L

in equilibrium all the [ ] represents molar concentration

K = [NH₃]² / [N₂][H₂]³ = [0.04/22.96]² / [0.96/22.96L][2.88/22.96L]³

K = 0.03678