At a certain temperature, 0.920 mol SO3 is placed in a 2.00 L container.

Question

2SO3(g)↽−−⇀2SO2(g)+O2(g)At equilibrium, 0.120 mol O2

is present. Calculate 𝐾c.

Answer 1

2 SO3(g) <==> 2 SO2(g) + O2(g)

Step 1 - convert everything to molarity first

Kc is built from concentrations, not moles, so do this before anything else. The container is 2.00 L:

initial [SO3] = 0.920 mol / 2.00 L = 0.460 M
equilibrium [O2] = 0.120 mol / 2.00 L = 0.0600 M

Converting at the end instead of the start is the single most common way this problem is lost, because the number of moles changes as the reaction proceeds.

Step 2 - ICE table, in molarity

Let x be the amount of O2 formed. From the coefficients, forming 1 O2 also forms 2 SO2 and consumes 2 SO3:

[SO3]: 0.460 - 2x
[SO2]: 0 + 2x
[O2]: 0 + x

You are told x = 0.0600 M, so:

[SO3] = 0.460 - 2(0.0600) = 0.460 - 0.120 = 0.340 M
[SO2] = 2(0.0600) = 0.120 M
[O2] = 0.0600 M

Step 3 - plug into Kc

Kc = [SO2]^2 [O2] / [SO3]^2

Kc = (0.120)^2 (0.0600) / (0.340)^2

Kc = (0.0144)(0.0600) / 0.1156 = 8.64 x 10^-4 / 0.1156

Kc = 7.47 x 10^-3

Two checks

Both SO2 and SO3 carry an exponent of 2 because their coefficients are 2, while O2 has an implied 1. Missing one of those squares changes the answer by more than a factor of ten.

And the size makes sense. Kc is well below 1, meaning the equilibrium lies to the left and most of the SO3 has not decomposed - consistent with 0.340 M of the original 0.460 M still sitting there. If you had gotten a Kc greater than 1 with that much starting material left over, something would be wrong.

One note for later in the course: because the moles of gas change here (2 on the left, 3 on the right), Kp does not equal Kc for this reaction. They are related by Kp = Kc(RT)^(delta n) with delta n = +1.