Question

At 450°C, ammonia gas will decompose according to the following equation: 2 NH3 (g)  N2 (g) + 3 H2 (g) Kc = 4.50 at 475˚C An unknown quantity of NH3 is placed in a reaction flask (with no N2 or H2) and is allowed to come to equilibrium at 475°C. The equilibrium concentration of H2 is then determined to be 0.252 M. Determine the initial concentration of NH3 placed in the flask

Answer 1

Answer:

0.2024 M

Explanation:

For the decomposition reactio given, let's do an equilibrium chart. Let's call the initial concentration of NH₃ as C:

2NH₃(g) ⇄ N₂(g) + 3H₂(g)

C 0 0 Initial

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

C - 2x x 3x Equilibrium

3x = 0.252

x = 0.084 M

The equilibrium constant (Kc) is the multiplication of the concentrations of the products elevated by their coefficients, divided by the multiplication of reactants concentrations elevated by their coefficients.

Kc = ([H₂]³*[N₂])/([NH₃]²)

4.50 = [(0.252)³*(0.084)]/(C - 2*0.084)²

4.50 = 0.00533/(C - 0.168)²

4.50 = 0.00533/(C² - 0.336C + 0.028224)

4.50C² - 1.512C + 0.127008 = 0.00533

4.50C² - 1.512C + 0.121678 = 0

Solving the equation by a graphic calculator, for C > 0.168

C = 0.2024 M