The exam rarely hands you a table of equilibrium concentrations. It shows you pictures: boxes of colored circles, or graphs of concentration against time. This page shows how to read both, how to tell when a picture shows equilibrium, and how to get Q or K from it. Readers report that students lose points by counting carelessly, by using limiting-reactant thinking where an equilibrium is shown, and by putting raw counts into K when the volume does not cancel.
Is it at equilibrium?
An equilibrium particle diagram shows a mixture whose composition no longer changes. On its own, one box cannot prove equilibrium: you need either two snapshots with the same counts, or a statement that the system has reached equilibrium. In a time series, the first time the counts stop changing is when equilibrium is reached. Expect to see reactants and products together; an equilibrium mixture is never "all used up".
From particles to concentrations
![A container at equilibrium with 4 W atoms and 4 W2 molecules in 2.0 liters, each particle standing for 0.10 mol. Step 1, count: 4 W and 4 W2. Step 2, convert: 0.40 mol divided by 2.0 L is 0.20 M for each. Step 3, substitute: Kc = [W2] over [W] squared = 0.20 over 0.20 squared = 5.0. A note warns that putting counts straight into K would give 0.25, which is wrong because the volume does not cancel when gas moles differ on the two sides.](../figures/particles-to-k.svg)
Kc uses molarity. To get molarity from a diagram, count each kind of particle, turn the count into moles (the question tells you how many moles each particle stands for), and divide by the volume.
Worked example. For 2 W(g) ⇌ W2(g) in a 2.0 L container at equilibrium, a diagram shows 4 W and 4 W2, and each particle stands for 0.10 mol. Find Kc.
1. Count: 4 W, 4 W2.
2. Convert: 4 × 0.10 mol = 0.40 mol of each; 0.40 mol / 2.0 L = 0.20 M of each.
3. Substitute: Kc = [W2] / [W]2 = 0.20 / (0.20)2 = 5.0.
Putting the counts straight in would give 4/42 = 0.25, which is wrong.
When can you use counts directly? When the total gas coefficients are the same on both sides, as in X2(g) + Y2(g) ⇌ 2 XY(g) (2 and 2). Then every V and every "moles per particle" factor cancels:
Q = (nXY/V)2 / ((nX₂/V)(nY₂/V)) = nXY2 / (nX₂ nY₂)
If the gas coefficients differ, convert first. When in doubt, convert; it is never wrong.
Using diagrams to predict direction
Worked example. For X2(g) + Y2(g) ⇌ 2 XY(g), Kc = 4.0. A box holds 3 X2, 3 Y2 and 2 XY. Which way will it react, and what could it look like at equilibrium?
1. Q = 22 / (3 × 3) = 0.44. (Gas moles equal on both sides, so counts are fine.)
2. Q < K, so the net reaction goes forward: X2 and Y2 decrease, XY increases.
3. Try one more reaction event: 2 X2, 2 Y2, 4 XY. Atoms are conserved (X atoms: 2 × 2 + 4 = 8, the same as 2 × 3 + 2 = 8), and Q = 16/4 = 4.0 = K. That is the equilibrium picture.
Notice what this is not: it is not a limiting-reactant problem. The reaction does not run until X2 or Y2 is gone; it stops changing when Q reaches K.
Reading concentration-time graphs
- Equilibrium is where every curve becomes horizontal. Where two curves cross means nothing special: equal concentrations at one instant.
- Changes follow the coefficients. For A ⇌ 2 B, B rises by twice as much as A falls.
- K comes from the plateau values. Read them after the curves flatten and substitute.
- Q at an earlier time uses values read before the plateau; it should be less than K if the curves show a net forward reaction.
Worked example. A flask starts with 0.80 M A: A(g) ⇌ 2 B(g). The curves level off at [A] = 0.40 M and [B] = 0.80 M. Find Kc.
1. Check the ratio: A fell by 0.40 M and B rose by 0.80 M, ratio 1 : 2, matching the coefficients.
2. Kc = [B]2 / [A] = (0.80)2 / 0.40 = 1.6.