At equilibrium, 0.1801 mol of N2 is present. Calculate the equilibrium constant, Kc.

Question

At a certain temperature, 0.3411 mol of N2 and 1.701 mol of H2 are placed in a 5.00-L container.

N2(g)+3H2(g) (double arrows) 2NH3(g)

At equilibrium, 0.1801 mol of N2 is present. Calculate the equilibrium constant, Kc.

Answer 1

N₂ (g) + 3 H₂ (g) <==> 2 NH₃ (g)

K_c = [NH₃]² / [N₂] [H₂]³

Since 0.1801 mol N₂ is present at equilibrium, mol of N₂ that reacts = 0.3411 - 0.1801 = 0.1610 mol N₂

Using stoichiometry, calculate mol of H₂ that reacts with 0.1610 mol N₂.

0.1610 mol N₂ x (3 mol H₂ / 1 mol N₂) = 0.4830 mol H₂

mol H₂ remaining = 1.701 - 0.4830 = 1.218 mol H₂

How much NH₃ is produced from N₂ and H₂? Convert mol of each reactant that "reacts" to mol NH₃. The reactant that produces less NH₃ will be its equilibrium concentration.

0.1610 mol N₂ x (2 mol NH₃ / 1 mol N₂) = 0.3220 mol NH₃

0.4830 mol H₂ x (2 mol NH₃ / 3 mol H₂) = 0.3220 mol NH₃

Both reactants produce the same amount of NH_3, therefore, mol of NH₃ at equilibrium = 0.3220 mol

Next, determine the equilibrium concentration of N₂, H₂ and NH₃.

[N₂]_eq = 0.1801 mol / 5.00 L = 0.0360 M

[H₂]_eq = 1.218 mol / 5.00 L = 0.244 M

[NH₃]_eq = 0.3220 mol / 5.00 L = 0.0644 M

Plug in the above numbers and solve for K_c.

K_c = [NH₃]² / [N₂] [H₂]³

K_c = [0.0644]² / [0.0360] [0.244]³ = 7.93