Finding the equilibrium constant

Question

At a certain temperature, 0.5011 mol of N2 and 1.621 mol of H2 are placed in a 2.50-L container.

N2(g) + 3H2(g) --><---2NH3(g)



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

Answer 1

N₂(g) + 3H₂(g) <====>2NH₃(g)

0.5011..1.621..............0......Initial....

-.361.... -1.083..........+1.0022...Change

0.1401...0.538.........1.0022....Equilibrium

Final Concentrations at equilibrium...

[N₂] = 0.1401 mol/2.50 L = 0.0560 M

[H₂] = 0.538 mol/2.50 L = 0.2152 M

[NH₃] = 1.0022/2.5 L = 0.4088 M

Kc = [NH₃]²/[N₂][H₂]³

Kc = (0.4088)²/(0.0560)(0.2152)³

Kc = 299.4