If 0.25 mol of br2 and 0.55 mol of cl2 are introduced into a 3.0-l container at 400 k, what will be the equilibrium concentration of br2

Answer is:  the equilibrium concentration of Br₂ is 0,02 mol/L.
Chemical reaction: Br₂ + Cl₂ → 2BrCl.
Kc = 7,0.
c₀(Br₂) = 0,25 mol ÷ 3 L.
c₀(Br₂) = 0,083 mol/L.
c₀(Cl₂) = 0,55 mol ÷ 3 L.
c₀(Cl₂) = 0,183 mol/L.
Kc = c(BrCl)² ÷ c(Br₂) · c(Cl₂).
7 = (2x)² ÷ (0,083 mol/L - x) · (0,183 mol/L - x).
7 = 4x² ÷ (0,083 mol/L - x) · (0,183 mol/L - x).
Solve quadratic equation: x = 0,063 mol/L.
c(Br₂) = 0,083 mol/L - 0,063 mol/L = 0,02 mol/L.

Answer (b) supersaturation;