At a certain temperature, 0.4011 mol of N2 and 1.621 mol of H2 are placed in a 3.50 L container. N2 (g) + 3H2 (g) ↽--⇀2NH3 (g) At equilibrium, 0.1001 mol of N2 is present. Calculate the equilibrium constant, Kc.

The equilibrium constant ≈120.07

Explanation:

Step 1: The balanced equation

N2 (g) + 3H2 (g) ↽--⇀2NH3 (g)

This means that for 1 mole of N2 consumed, there is needed 3 moles of H2 to produce 2 moles of NH3

Step 2: Calculate initial moles

Reactants:

⇒N2: 0.4011 mol es

⇒H2: 1.621 mol es

Products

⇒NH3: 0 mol

Step 3: Calculate number of moles that changed

N2: 0.4011 mol - 0.1001 mol = 0.3010 mol

H2: 3 * 0.3010 mol = 0.9030 mol

NH3: 2 * 0.3010 mol = 0.6020 mol

Step 4: Calculate moles at the equilibrium

N2: 0.1001 moles

⇒[N2] = 0.1001moles / 3.5L = 0.0286 M

H2: 1.621 moles - 0.9030 moles = 0.718 moles

⇒[H2] = 0.718 moles / 3.5L : 0.205 M

NH3: 0.6020 mol es

⇒[NH3] = 0.6020 / 3.5 = 0.172M

Step 5: Calculate Kc

Kc = [NH3]² / ([N2] [H2]³)

Kc = 0.172² / (0.0286*0.205³)

Kc = 0.029584‬ / (0.0286 * 0.008615125‬)

Kc ≈120.07

The Kcis a high value, this shows the position of the equilibrium is to the right.

The equilibrium constant ≈120.07