The reaction a (g) + 2b (g) c (g) was allowed to come to equilibrium. the initial amounts of reactants placed into a 5.00 l vessel were 1.0 mol a and 1.8 mol

b. after the reaction reached equilibrium, 1.0 mol of b was found. calculate kc for this reaction.

Answer is: Kc for reaction is 16.66.

Chemical reaction: A + 2B ⇄ C.

c₀ (A) = 1 mol : 5 L; initial concentration of A.

c₀ (A) = 0,2 mol/L.

c₀ (B) = 1,8 mol : 5 L; initial concentration of B.

c₀ (B) = 0,36 mol/L.

c (B) = 1 mol : 5 L.

c (B) = 0,2 mol/L; equilibrium concentration of B.

From chemical reaction: n (A) : n (B) = 1 : 2.

n (A) = 0,8 mol : 2 = 0,4 mol.

c (A) = (1 mol - 0,4 mol) : 5 L.

c (A) = 0,12 mol/L; equilibrium concentration of A.

n (C) : n (B) = 1 : 2.

n (C) = 0,4 mol.

c (C) = 0,4 mol : 5 L.

c (C) = 0,08 mol/L; equilibrium concentration of C.

Kc = 0,08 mol/L : ((0,2 mol/L) ² ·0,12 mol/L).

Kc = 16,66.