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 K_c for this reaction.

Question

Quizplus · Accepted Answer

We can set up an ICE table to help us solve for the equilibrium constant.

Reaction: A(g) + 2B(g) ⇌ C(g)

Initial:              1.0 mol        1.8 mol              0 mol
Change:         -x                -2x                    +x
Equilibrium:  1.0 - x         1.8 - 2x           x

We know that at equilibrium, 1.0 mol of B is present. Therefore:

1.8 - 2x + x = 1.0

x = 0.4 mol

Substituting this value into the equilibrium expression:

Kc = [C]/([A][B]^2) = (0.4 mol) / [(1.0 mol)(1.8 mol)^2] = 0.060 M^-2

Therefore, the answer is choice C.

The Reaction A(g)+ 2B(g) C(g)was Allowed