Calculating Q from a particle diagram
AP often uses particle diagrams to show the mixture in a container. To read one:
- identify what substance each kind of particle represents, and count each;
- convert to concentrations: multiply by the moles each particle represents, then divide by the volume;
- substitute into the expression for Q and compare with K.
If both sides of the equation have the same number of gas moles (Δn = 0), the volume and the “moles per particle” cancel out of the expression, so Q can be calculated straight from the particle counts; otherwise, always convert to concentrations first.
Decide what is happening in each diagram:
What is happening in this container?
A series of diagrams: when is equilibrium reached?
When diagrams of the same container at different times are shown side by side, the sign of equilibrium is that the particle counts stop changing, not that two kinds of particle are equally numerous. A concentration–time graph works the same way: equilibrium is where the curves level off.
Also, the changes in amount are in the ratio of the coefficients, so the equation can be worked out from how the particle counts (or concentrations) change.
Example 1: A series of particle diagrams
2NO₂(g) ⇌ N₂O₄(g) takes place in a container. The numbers of molecules at four times are below (each molecule represents 0.10 mol; the container is 1.0 L):
| Time | t₁ | t₂ | t₃ | t₄ |
|---|---|---|---|---|
| NO₂ | 12 | 6 | 4 | 4 |
| N₂O₄ | 0 | 3 | 4 | 4 |
When is equilibrium reached? Find K.
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Check the data first: conservation of atoms. Each NO₂ contains 1 N and each N₂O₄ contains 2 N. N atoms at each time:
There are always 12 N atoms, so the data are consistent.
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Find when nothing changes. The counts at t₃ and t₄ are identical, so equilibrium has been reached by t₃.
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Convert molecule counts to concentrations. Each molecule represents 0.10 mol; dividing by the 1.0 L volume gives concentrations in M:
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Write K and substitute. NO₂ is the reactant, and its coefficient 2 becomes an exponent:
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Note. The equal numbers of the two molecules at equilibrium are a coincidence; what shows equilibrium is that nothing changes any more.
Example 2: Working out the equation from the changes
In a reaction, [A] falls from 0.80 M to 0.50 M and then stays constant, while [B] rises from 0 to 0.60 M and then stays constant. Write the equation and find K.
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Decide when equilibrium is reached. Both concentrations eventually stop changing, so the forward and reverse rates are equal and the system is at equilibrium. The constant values are the equilibrium concentrations: [A] = 0.50 M and [B] = 0.60 M.
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Calculate the changes.
A is used up, so it is a reactant; B increases, so it is a product.
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Get the coefficients from the ratio of the changes. In one container, the changes in concentration are in the same ratio as the coefficients in the equation:
In lowest whole numbers, the equation is A ⇌ 2B.
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Check with an ICE table. If A decreases by x, B increases by 2x (concentrations in M):
A 2B I 0.80 0 C −x +2x E 0.80 − x 2x From [A] = 0.80 − x = 0.50, x = 0.30, so [B] = 2x = 0.60 M. This matches the data, so A ⇌ 2B is right.
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Write the K expression. The coefficient 2 of B becomes an exponent:
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Substitute the equilibrium concentrations. Only equilibrium concentrations go into K, not starting concentrations or changes:
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Check the direction. At the start [B] = 0, so Q = 0 < K = 0.72. The reaction runs forward, with A decreasing and B increasing, which matches the description.
- A decreases by 0.20 M and B increases by 0.40 M
1 : 2.
- A decreases by 0.30 M and B increases by 0.30 M
1 : 1.
- A decreases by 0.60 M and B increases by 0.30 M
2 : 1.
- A decreases by 0.10 M and B increases by 0.30 M
1 : 3.
- In a particle diagram, A loses 4 particles and B gains 2
2 : 1.
- In a particle diagram, A loses 3 particles and B gains 6
1 : 2.
- B decreases by 0.40 M (reaction in reverse) and A increases by 0.20 M
The changes in A and B are in a 1 : 2 ratio.
- A’s curve drops by 0.05 and B’s curve rises by 0.05
1 : 1.
Common mistakes
Mistake 1: “A diagram with equal numbers of reactant and product particles shows equilibrium.” The sign of equilibrium is that the counts stop changing, that is, Q = K. Equal counts are just one possible coincidence.
Mistake 2: “The diagram with the most product is the one at equilibrium.” How much product there is at equilibrium depends on K. When K is small, there is very little product at equilibrium.
Mistake 3: “Q can always be calculated straight from particle counts.” Only when both sides have the same number of gas moles. Otherwise convert to concentrations (or partial pressures) first.
Quick check
X₂(g) + Y₂(g) ⇌ 2XY(g) has K = 4. A container holds 3 X₂, 3 Y₂, and 4 XY. What happens?
With Δn = 0, the particle counts can be used directly: Q = 4² ÷ (3 × 3) = 1.8 < 4, so the reaction proceeds forward. D: the volume cancels.
Particle diagrams of the same container are drawn at t₁, t₂, t₃, and t₄. The diagrams at t₂, t₃, and t₄ have exactly the same numbers of each particle. Which statement is correct?
From t₂ on the counts don’t change, so the system is at equilibrium; the reaction hasn’t stopped, the forward and reverse rates are just equal.
On a concentration–time graph, [C] rises from 0 to 0.40 M and [D] falls from 0.80 to 0.60 M, after which both curves level off. Which statement is correct?
D decreases by 0.20 and C increases by 0.40, a 1 : 2 ratio, so the equation is D ⇌ 2C. C increases, so it is a product; at equilibrium [C] = 0.40 M and [D] = 0.60 M, which are not equal.
2Z(g) ⇌ Z₂(g) in a 2.0 L container, with each particle representing 0.20 mol. A diagram shows 4 Z and 2 Z₂. What is Q?
[Z] = 4 × 0.20 ÷ 2.0 = 0.40 M and [Z₂] = 2 × 0.20 ÷ 2.0 = 0.20 M, so Q = 0.20 ÷ 0.40² = 1.25. A forgets the square; B uses the particle counts directly (not allowed when Δn ≠ 0); D inverts the expression.
Which statement about particle diagrams of a system at equilibrium is correct?
At equilibrium, reactants and products of a reversible reaction are both present (though one may be very scarce when K is very large or very small).
X₂ + Y₂ ⇌ 2XY has K = 4. Which diagram shows a mixture at equilibrium?
Q = 4² ÷ (1 × 4) = 4 = K. A has Q = 1, C has Q = 1, and D has an infinite Q.
AP-style free response
A₂(g) + B₂(g) ⇌ 2AB(g) takes place in a 1.0 L container, with each molecule representing 0.010 mol. The numbers of molecules at three times are:
| Time | A₂ | B₂ | AB |
|---|---|---|---|
| t₁ | 6 | 4 | 0 |
| t₂ | 4 | 2 | 4 |
| t₃ | 3 | 1 | 6 |
After t₃, the numbers no longer change.
(a) Use the data at t₃ to find K.
(b) Calculate Q at t₂, and state which way the reaction is proceeding at t₂.
(c) Explain why K for this reaction can be calculated directly from the numbers of molecules.
(d) At t₃, 2 more AB molecules are added. Calculate Q at that moment and predict the direction of reaction.
(e) A student says, “At t₃ neither A₂ nor B₂ is used up, so the reaction has stopped.” Identify the problem with this statement.
Model answer and scoring
(a) K = 6² ÷ (3 × 1) = 12 (Δn = 0, so the counts can be used directly).
(b) Q = 4² ÷ (4 × 2) = 2 < 12, so it proceeds forward.
(c) Both sides have the same number of gas moles (2 = 2). Each concentration equals “count × 0.010 mol ÷ 1.0 L,” and the conversion factors in the numerator and denominator cancel exactly.
(d) Q = 8² ÷ (3 × 1) = 21 > 12, so it proceeds in reverse.
(e) The reaction hasn’t stopped; the forward and reverse reactions continue at the same rate, so the counts don’t change. In a reversible reaction at equilibrium, the reactants are never used up.
Scoring: (a) K; (b) Q and direction; (c) Δn = 0; (d) Q > K; (e) dynamic equilibrium.
Summary
- Particle diagram → count particles → convert to concentrations (or use the counts directly when Δn = 0) → calculate Q → compare with K.
- In a series of diagrams or a concentration–time graph, equilibrium starts when the counts (concentrations) stop changing.
- The changes in amount are in the ratio of the coefficients, which lets you work out the equation.
AP tip: When interpreting a particle diagram, write the numbers: “the diagram shows … particles, so Q = ….” Judge equilibrium by comparing Q with K, or by “the counts no longer change,” not by “the two kinds of particles are equally numerous.”