Question

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 1

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.