Amixture of 1.374 g of h2 and 70.31 g of br2 is heated in a 2.00-l vessel at 700 k. these substances react according to h2(g) + br2(g) ⇜ 2 hbr(g) at equilibrium, the vessel is found to contain 0.566 g of h2. calculate kc.

Answer is:  the equilibrium constant is 57,14.

Chemical reaction: Br₂ + H₂ → 2HBr.

c₀(Br₂) = 70,31 g ÷ 160 g/mol ÷ 2 L.

c₀(Br₂) = 0,22 mol/L.

c₀(H₂) = 1,374 g ÷ 2 g/mol ÷ 2 L.

c₀(H₂) = 0,34 mol/L.

c(H₂) = 0,566 g ÷ 2 g/mol ÷ 2 L = 0,14 mol/L, so 0,2 mol/L reacts.

c(HBr) = 2 · (0,34 mol/L - 0,14 mol/L) = 0,40 mol/L; from chemical reaction n(H₂) : n(HBr) = 1 : 2.

c(Br₂) = 0,22 mol/L - 0,2 mol/L = 0,02 mol/L. n(Br₂) : n(H₂) = 1 : 1.

Kc = c(HBr)² ÷ (c(Br₂) · c(H₂)).

Kc = (0,4 mol/L)² ÷ (0,02 mol/L) · (0,14 mol/L).

Kc = 57,14.