Question

The equilibrium constant for the following reaction is 6.30 at 723K. 2NH3 <=> N2(g) + 3H2(g) If an equilibrium mixture of the three gases in a 10.2 L container at 723 K contains 0.221 mol of NH3(g) and 0.315 mol of N2(g), the equilibrium concentration of H2 is _________ M. (Enter numbers as numbers, no units. For example, 300 minutes would be 300. For letters, enter A, B, or C. Enter numbers in scientific notation using e# format. For example 1.43×10-4 would be 1.43e-4.)

Answer 1

Answer:

0.458 M

Explanation:

Let's consider the following reaction at equilibrium.

2 NH₃(g) ⇄ N₂(g) + 3 H₂(g)

The concentrations at equilbrium are:

[NH₃] = 0.221 mol / 10.2 L = 0.0217 M

[N₂] = 0.315 mol / 10.2 L = 0.0309 M

[H₂] = ?

We can find the concentration of H₂ at equilibrium using the equilibrium constant (K).

K = [N₂] × [H₂]³ / [NH₃]²

[H₂] = ∛(K × [NH₃]²/ [N₂])

[H₂] = ∛(6.30 × 0.0217²/ 0.0309) = 0.458 M

Answer 2

Answer:

The equilibrium concentration of H2 is 0.4579 M

Explanation:

Step 1: Data given

The equilibrium constant for the following reaction is 6.30 at 723K

Volume = 10.2 L

Number of moles NH3 = 0.221 moles

Number of moles N2 = 0.315 moles

Step 2: The balanced equation

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

Step 3: Calculate concentration

Concentration = moles / volume

Concentration NH3= 0.221 moles / 10.2 L

Concentration NH3 = 0.0217 M

Concentration N2 = 0.315 moles / 10.2 L

Concentration N2 = 0.0309 M

Step 4: Define Kc

Kc = [H2]³[N2] / [NH3]²

6.30 = [H2]³ * (0.0309) / (0.0217)²

[H2]³ = 0.09601 M

[H2] = 0.4579 M

The equilibrium concentration of H2 is 0.4579 M