At a certain temperature, 0.3611 mol of N2 and 1.741 mol of H2 are placed in a 3.50 L container.

Question

At a certain temperature, 0.3611N2 and 1.741 mol of H2 are placed in a 3.50 L container.

N2(g)+3H2(g)↽−−⇀2NH3(g)

At equilibrium, 0.1401 mol of N2 is present. Calculate the equilibrium constant, 𝐾c

Answer 1

0.3611 mol N₂ / 3.50 L = 0.1032 M N₂

1.741 mol H₂ / 3.50 L = 0.4974 M H₂

@ equilibrium [N₂] = 0.1401 mol / 3.5 L = 0.0400 M N₂

N₂(g) + 3H₂(g) ==> 2NH₃(g)

0.1032......0.4974............0............Initial

-x...............-3x...............+2x.........Change

0.1032-x....0.4974-3x.....2x.........Equilibrium

Since we know that 0.1032-x = 0.0400 M, we can solve for x, and thus find all concentrations @ equilibrium

0.1032 - x = 0.0400

x = 0.0632

Concentrations @ equilibrium:

[N₂] = 0.0400 M

[H₂] = 0.4974 - (3x0.0632) = 0.3078 M

[NH₃] = 2x = 0.1264 M

Kc = [NH₃]² / [N₂][H₂]³

Kc = (0.1264)² / (0.0400)(0.3078)³ = 0.01598 / 1.17x10-3

Kc = 13.66

(be sure to check all of the math)