Question

Analysis of the gases in a sealed reaction vessel containing NH₃, N₂, and H₂ at equilibrium at 400 °C established the concentration of N₂ to be 1.2 M and the concentration of H₂ to be 0.24 M.

a) True
b) False

Answer 1

The equilibrium molar concentration of NH3 is 0.48 M.

Step-by-step explanation:

To calculate the equilibrium molar concentration of NH3, we need to use the equilibrium constant, Kc. The balanced equation for the reaction is N2(g) + 3H2(g) ⇒ 2NH3(g). The equilibrium constant expression can be written as [NH3]^2 / [N2][H2]^3. Since we are given the concentrations of N2 and H2, we can substitute them into the equation and solve for [NH3].

Kc = [NH3]^2 / ([N2][H2]^3) = 0.50

Substituting the given values, we have:

[NH3]^2 = (1.2 M)^2 = 1.44 M

1.44 M = 0.50 * (1.2 M)(0.24 M)^3

[NH3] = √(1.44 M * (0.50 / (1.2 M)(0.24 M)^3)) = 0.48 M

Therefore, the equilibrium molar concentration of NH3 is 0.48 M.