Question

1. Nitrogen and hydrogen gases are mixed in a closed 3500 mL flask. They react to form ammonia gas. At equilibrium, the mixture contains 0.25 mol of ammonia gas and 0.080 mol hydrogen gas. The equilibrium constant for this reaction is 5.81 x 10^5. What amount of nitrogen gas is present at equilibrium? All the entities are in gaseous form.

Answer 1

The amount of nitrogen gas present at equilibrium is 0.0039 moles.

Volume of the system =  3500 mL or 3.5 L

Concentration of ammonia at equilibrium =  0.25 mol/3.5 L = 0.07 M

Concentration of hydrogen gas at equilibrium =  0.080 mol//3.5 L = 0.02 M

Equilibrium constant for this reaction = 5.81 x 10^5

The reaction for the formation of ammonia is;

N2 + 3H2 ⇄ 2NH3

So;

K = [NH3]^2/[N2] [H2]^3

K [N2] [H2]^3 =  [NH3]^2

[N2] =  [NH3]^2/K  [H2]^3

[N2] = [ 0.07]^2/ 5.81 x 10^5 × [0.02]^3

[N2] = 0.0011 M

Amount of N2 =  0.0011 M ×  3.5 L = 0.0039 moles

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