Reaction Quotient and Le Châtelier's Principle: Study Guide (Chemistry 7.10)

From a rule of thumb to a reason

In Topic 7.9 you predicted shifts with Le Châtelier’s principle. That works, but it does not explain why the system must shift. The reaction quotient Q gives the reason.

Recall from Topic 7.3: Q has the same form as the equilibrium expression, but you can calculate it for any mixture, at any moment. For aA + bB ⇌ cC + dD,

Qc = [C]^c [D]^d / ([A]^a [B]^b)

and for gases you can write Qp with partial pressures in place of concentrations. At equilibrium, Q = K.

Here is the key idea of this topic:

1. Before the stress, the system is at equilibrium: Q = K.
2. The stress makes Q ≠ K. The system is no longer at equilibrium.
3. The system reacts (forward or in reverse) until Q = K again. This is the new equilibrium.

A stress can upset the balance in one of two ways. Most stresses change the concentrations in Q but leave K alone. A change in temperature changes the value of K, while the concentrations at that instant stay the same (in a rigid container). Either way, Q and K no longer match, and the concentrations or partial pressures redistribute until they do.

Comparing Q with K

Comparison	What it means	Net reaction	What happens to Q
Q < K	Products too low compared with reactants	Forward (shift right)	Rises to K
Q = K	At equilibrium	None	Stays equal to K
Q > K	Products too high compared with reactants	Reverse (shift left)	Falls to K

Which stresses change Q, and which change K?

Stress	Changes Q?	Changes K?	Example of the effect on Q
Add a reactant	Yes: Q falls	No	Q < K, forward
Add a product	Yes: Q rises	No	Q > K, reverse
Remove a reactant or product	Yes	No	Opposite of adding it
Change the volume of a gas mixture	Yes, if gas coefficients differ	No	See the factor rule below
Dilute a solution	Yes, if dissolved-particle coefficients differ	No	See the factor rule below
Change the temperature (rigid container)	Not at that instant	Yes	K moves away from Q
Add a catalyst, extra solid or pure liquid	No	No	Still Q = K

The factor rule for volume and dilution

When you change the volume, every concentration (or partial pressure) is multiplied by the same factor, s. Each term in Q is raised to its coefficient, so for a system that was at equilibrium (Q = K) before the change,

new Q = s^Δn × K, where Δn = (sum of product coefficients) − (sum of reactant coefficients),

counting only gases (for a volume change) or only dissolved species (for dilution).

• Halving the volume of a gas mixture: s = 2.
• Diluting a solution to twice its volume: s = ½.

If Δn = 0, Q is unchanged and there is no shift. This is the numerical reason behind the “count the molecules” rule from Topic 7.9.

Worked example 1: adding a reactant

Question. For the reaction CO(g) + H₂O(g) ⇌ CO₂(g) + H₂(g), Kc = 4.00 at the temperature of this experiment (an invented value). A 1.00 L flask holds an equilibrium mixture with [CO] = 0.200 M, [H₂O] = 0.200 M, [CO₂] = 0.400 M and [H₂] = 0.400 M. Then 0.200 mol of CO is added at constant temperature and volume. (a) Show that the original mixture was at equilibrium. (b) Calculate Q just after the addition and predict the direction of reaction. (c) Describe how each concentration changes as the new equilibrium is reached.

1. (a) Check K. Qc = (0.400)(0.400) / ((0.200)(0.200)) = 0.160 / 0.0400 = 4.00. This equals Kc, so the mixture was at equilibrium.
2. (b) New concentration. Adding 0.200 mol to 1.00 L raises [CO] to 0.400 M. Nothing else has changed yet.
3. New Q. Qc = (0.400)(0.400) / ((0.400)(0.200)) = 0.160 / 0.0800 = 2.00.
4. Compare. Q = 2.00 < K = 4.00, so there are too few products. The net reaction goes forward.
5. (c) Changes. As the forward reaction runs, [CO] and [H₂O] decrease, and [CO₂] and [H₂] increase, until Q = 4.00 again.

Going further. Solving for the new equilibrium (the method from Topic 7.7) gives [CO] = 0.347 M, [H₂O] = 0.147 M, [CO₂] = [H₂] = 0.453 M. Using the unrounded values, Q = 4.00 again. Only about a quarter of the added CO reacted, and [CO] is still well above its original 0.200 M. This is the “partly undoes the stress” idea from Topic 7.9, now in numbers.

Worked example 2: halving the volume of a gas mixture

Question. For 2SO₂(g) + O₂(g) ⇌ 2SO₃(g), Kp = 128 at a certain temperature (an invented value). An equilibrium mixture has partial pressures P(SO₂) = 0.20 atm, P(O₂) = 0.50 atm and P(SO₃) = 1.6 atm. The volume is halved at constant temperature. Find Qp just after the change and predict the direction of reaction.

1. Check K. Qp = (1.6)² / ((0.20)² × 0.50) = 2.56 / 0.020 = 128. The mixture is at equilibrium.
2. Effect of halving the volume. At constant temperature, each partial pressure doubles: P(SO₂) = 0.40 atm, P(O₂) = 1.00 atm, P(SO₃) = 3.2 atm.
3. New Q. Qp = (3.2)² / ((0.40)² × 1.00) = 10.24 / 0.16 = 64.
4. Compare. Q = 64 < K = 128, so the net reaction goes forward, making more SO₃.

Check with the factor rule. Δn = 2 − (2 + 1) = −1 and s = 2, so new Q = 2⁻¹ × 128 = 64. ✓ This agrees with Le Châtelier’s principle: 3 gas molecules on the left, 2 on the right, so a smaller volume favours the right.

Worked example 3: changing the temperature

Question. Return to the equilibrium mixture of Worked example 1 before any CO was added: [CO] = [H₂O] = 0.200 M and [CO₂] = [H₂] = 0.400 M, with Kc = 4.00. The forward reaction is exothermic. The rigid flask is heated to a higher temperature, where Kc = 2.25 (invented). (a) Explain why the system is no longer at equilibrium. (b) Predict the direction of reaction and how the concentrations change.

1. (a) What changes. Heating a rigid flask does not change the amount of any gas or the volume, so the concentrations, and therefore Q, are unchanged at the instant of heating: Q = 4.00. But K has changed to 2.25. Now Q ≠ K.
2. Why K fell. The forward reaction is exothermic. Raising the temperature favours the endothermic (reverse) direction, so the equilibrium mixture at the higher temperature contains relatively fewer products. That is what a smaller K means.
3. (b) Compare. Q = 4.00 > K = 2.25, so the net reaction goes in reverse. [CO₂] and [H₂] decrease; [CO] and [H₂O] increase.
4. New equilibrium. Because every coefficient is 1, the new concentrations can be found by taking square roots: [CO₂]/[CO] = √2.25 = 1.5. This gives [CO₂] = [H₂] = 0.360 M and [CO] = [H₂O] = 0.240 M. Check: (0.360)² / (0.240)² = 2.25. ✓

When two things change at once

Le Châtelier’s principle struggles when a reactant and a product are added together: one pushes the system forward, the other pushes it back. Q settles the question.

Start again from the original mixture in Worked example 1 ([CO] = [H₂O] = 0.200 M, [CO₂] = [H₂] = 0.400 M, Kc = 4.00). Suppose 0.100 mol of CO and 0.100 mol of CO₂ are added to the 1.00 L flask at the same time. Now [CO] = 0.300 M and [CO₂] = 0.500 M, so

Qc = (0.500)(0.400) / ((0.300)(0.200)) = 0.200 / 0.0600 = 3.33.

Q = 3.33 < K = 4.00, so the net reaction goes forward, even though a product was added too. The added reactant has the bigger effect on Q here because its concentration rose by a larger fraction (from 0.200 M to 0.300 M, an increase of one half) than the product’s did (from 0.400 M to 0.500 M, an increase of one quarter). You do not need to reason about this in words: calculate Q, compare it with K, and the direction follows.

Common misconceptions

• “Adding a reactant changes K.” K depends only on temperature (for a given reaction). Adding a reactant lowers Q; the system then reacts until Q is back to the same K.
• “Q > K means the reaction goes forward because Q is bigger.” It is the reverse. A large Q means too many products, so the system makes reactants.
• “Heating changes Q.” In a rigid container, heating leaves concentrations unchanged at that instant. It is K that moves.
• “At the new equilibrium, every concentration is back to its old value.” Only the ratio in Q returns to K. The individual concentrations are different (Worked example 1).
• “Doubling the pressure doubles Q.” Q changes by s^Δn, which depends on the coefficients. If Δn = 0, Q does not change at all.
• “Raising the temperature always increases K.” Only for an endothermic reaction. For an exothermic reaction K decreases.

Where this leads

Next, in Topic 7.11, Introduction to Solubility Equilibria, you will apply Q and K to a salt dissolving in water, where the equilibrium constant is called Ksp. Previous topic: Introduction to Le Châtelier’s Principle. Now try the practice questions, then use the revision notes and the checklist.