The reaction A(g) + 2B(g) $\rightleftharpoons$ 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 molB. After the reaction reached equilibrium, 1.0 mol of B was found. Calculate K_c for this reaction.

Question

Quizplus · Accepted Answer

First, let's write the equilibrium expression:

Kc = [C]/([A][B]^2)

We know that initially we had:

[A] = 1.0 mol/5.00 L = 0.2 M
[B] = 1.8 mol/5.00 L = 0.36 M

And at equilibrium:

[B] = 1.0 mol/5.00 L = 0.2 M

Since we don't know the equilibrium concentration of C, let's call it x.

Using the stoichiometry of the balanced equation, we can write:

0.2 mol/L = (1.8 mol/L - x) / 2

Solving for x, we get:

x = 1.4 mol/L

Now we can calculate the equilibrium constant:

Kc = (1.4 mol/L) / [(0.2 mol/L)(0.2 mol/L)] = 17

Therefore, the answer is C: 17.

The Reaction A(g) + 2B(g) ⇌\rightleftharpoons⇌ C(g) Was Allowed to Come to Equilibrium

The Reaction A(g) + 2B(g) $\rightleftharpoons$ C(g) Was Allowed to Come to Equilibrium