Question

Ammonia has been studied as an alternative "clean" fuel for internal combustion engines, since its reaction with oxygen produces only nitrogen and water vapor, and in the liquid form it is easily transported. An industrial chemist studying this reaction fills a 500 mL flask with 4.8 atm of ammonia gas and 1.9 atm of oxygen gas at 33 degress Celsius. He then raises the temperature, and when the mixture has come to equilibrium measures the partial pressure of nitrogen gas to be 0.63 atm. Calculate the pressure equilibrium constant for the combustion of ammonia at the final temperature of the mixture.

Answer 1

Answer:

0.364

Explanation:

Let's do an equilibrium chart for the reaction of combustion of ammonia:

2NH₃(g) + (3/2)O₂(g) ⇄ N₂(g) + 3H₂O(g)

4.8atm 1.9atm 0 0 Initial

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

4.8-2x 1.9-(3/2)x x 3x Equilibrium

At equilibrium the velocity of formation of the products is equal to the velocity of the formation of the reactants, thus the partial pressures remain constant.

If pN₂ = 0.63 atm, x = 0.63 atm, thus, at equilibrium

pNH₃ = 4.8 - 2*0.63 = 3.54 atm

pO₂ = 1.9 -(3/2)*0.63 = 0.955 atm

pH₂O = 3*0.63 = 1.89 atm

The pressure equilibrium constant (Kp) is calculated with the partial pressure of the gases substances:

Kp = [(pN₂)*(pH₂O)³]/[(pNH₃)²*$(pO_2)^{3/2}$]

Kp = [0.63*(1.89)³]/[(3.54)²*$(0.955)^{3/2}$]

Kp = 4.2533/11.6953

Kp = 0.364