At 29°C, Kp = 5.3 ✕ 102 atm−2 for the following reaction. N2(g) + 3 H2(g) equilibrium reaction arrow 2 NH3(g) When a certain partial pressure of NH3(g) is put into an otherwise empty rigid vessel at 29°C, equilibrium is reached when 55.1% of the original ammonia has decomposed. What was the original partial pressure of ammonia before any decomposition occurred?

Equilibrium occurs when, in a reversible reaction, the velocity of the formation of the products is equal to the velocity of the formation of the reactants, thus, the concentrations of the compounds remain constant.

The equilibrium can be characterized by a constant, which can be measured by the molar concentration, Kc, or by the partial pressure, Kp. Kc depends only on the liquids, gases, and aqueous solutions, except water, and Kp only on the gases.

They can be calculated by the multiplication of the concentration or partial pressure of the products elevated by their coefficients, divided by the multiplication of the concentration or partial pressure of the reactants elevated by their coefficients.

For the reaction given, let's do an equilibrium chart:

N₂(g) + 3H₂(g) ⇄ 2NH₃(g)

0 0 P Initial

+x +3x -2x Reacts (stoichiometry is 1:3:2)

x 3x P-2x Equilibrium

If 55.1% was decomposed,then, left 44.9% of the initial, thus:

P - 2x = 0.449P

2x = 0.551P

x = 0.2755P

Then at the equilibrium:

pN₂ = 0.2755P

pH₂ = 3*0.2755P = 0.8265P

pNH₃ = P - 2*0.2755P = 0.449P

Kp = (pNH₃)²/[(pN₂)*(pH₂)³]

5.3x10² = (0.449P)²/[0.2755P*(0.8265P)³]

5.3x10² = 0.201601P²/0.15543P⁴

1.29705/P² = 5.3x10²

P² = 1.29705/5.3x10²

P² = 2.447x10⁻³

P = √2.447x10⁻³

P = 0.05 atm