Chemistry question on equilibrium

Question

Consider the following reaction at equilibrium at 135oC: 2AB3(g) ⇌ A2(g) + 3B2(g)

If 2.00 mol of AB3(g) were placed into a 10.0 dm3 container and the concentration of B2(g) at equilibrium was 0.0150 mol.dm─3, then the value of the equilibrium constant (Kc) is:

Answer 1

2AB₃(g) ⇌ A₂(g) + 3B₂(g)

K = [A₂][B₂]³ / [AB₃]²

Set up an ICE table:

2AB₃(g) ⇌ A₂(g) + 3B₂(g)

0.2M...........0...........0........Initial

-2x.............+x........+3x......Change

0.2-2x.........x...........3x.....Equilibrium

At equilibrium 3x = 0.0150 M

Thus, at equilibrium x = 0.005 M

Summary at equilibrium:

[AB3] = 0.2-2x = 0.2 - 0.01 = 0.19

[A2] = x = 0.005

[B2] = 0.0150

Substituting these concentrations into the Kc expression we have...

Kc = [A₂][B₂]³ / [AB₃]² = (0.005)(0.0150)³ / (0.19)²

Kc = 4.67x10^-7

Answer 2

Using the Kp expression for the equation as given K_C = C³_B2C_A2/C²_AB3

At equilibrium (since the volume is constant we can do stoichiometry with concentrations instead of moles):

C_B2 = .015 M

C_A2 = 1/3(.015)M

C_AB3 = (2 moles/10l) - 2/3(.015)M

Sub into K expression.

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