Reaction Quotient and Equilibrium Constant: Study Guide (Chemistry 7.3)

One number that describes a reacting mixture

In Topic 7.2 you saw that a reversible reaction runs in whichever direction is faster until the forward and reverse rates become equal. While that happens, the amounts of reactants and products keep changing. Chemists want a single number that sums up “how far towards products” a mixture is at any moment. That number is the reaction quotient, Q.

For a general reaction

aA + bB ⇌ cC + dD

the reaction quotient in concentrations is

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

Three rules build it:

1. Products on top, reactants underneath. The species on the right of the arrow go in the numerator.
2. Each concentration is raised to the power of its coefficient in the balanced equation. A coefficient of 2 means “squared”, not “times 2”.
3. Terms are multiplied, never added. [C]^c[D]^d means [C]^c × [D]^d.

This way of building the expression from the balanced equation is called the law of mass action. Square brackets mean concentration in mol L⁻¹ at the moment you are describing.

Q at any moment, K at equilibrium

Q can be worked out for any mixture: at the start, halfway, or at the end. Its value changes as the reaction proceeds. When the system reaches equilibrium, the concentrations stop changing, so Q stops changing too. The value it settles at is the equilibrium constant, K:

• at equilibrium, Qc = Kc
• at equilibrium, Qp = Kp (for the pressure version, below)

The key idea is that the reaction always changes the mixture so that Q moves towards K. Two flasks of the same reaction at the same temperature can start with very different mixtures, but both end with the same value of Q, because K is a constant for that reaction at that temperature. Figure 1 shows this.

Read the graph in three steps:

• Axes. Q has no unit; time is in minutes. The scale goes up in even steps (2 on Q, 5 min on time).
• Shape. Q changes fastest at the start, then more slowly as the forward and reverse rates get closer.
• Flat part. Once the line is flat, the system is at equilibrium and the flat value is K. Both runs give 4.0.

A Q-time graph is a good way to show K because it separates the one constant value (K) from the many values Q passes through on the way. Which direction a reaction runs for a given Q and K is developed in Topic 7.7.

Gases: Qp and Kp

For a reaction involving gases you can measure partial pressures instead of concentrations. The expression has exactly the same shape, with P for the partial pressure of each gas:

Qp = (P_C)^c (P_D)^d / ((P_A)^a (P_B)^b), and at equilibrium Qp = Kp.

Partial pressures are usually in atm. Two points matter:

• Kc and Kp are different numbers for most reactions, because one uses mol L⁻¹ and the other atm. You will not be asked to convert one into the other, but you must notice which one a question gives or asks for. Never put a pressure into a Kc expression or a concentration into a Kp expression.
• Only gases have partial pressures. So Kp is written for reactions in which every species that appears in the expression is a gas. Reactions in solution use Kc. Calculations where a dissolved substance is in equilibrium with the same substance as a gas are outside this course.

What to leave out: pure solids and pure liquids

The concentration of a gas or a dissolved substance depends on how much of it is in a given volume. A pure solid or a pure liquid is different. Its “concentration” is fixed by its density: a lump of carbon has the same number of moles per litre of carbon whether the lump is large or small. Adding more solid does not change that value, so it cannot shift the mixture. That is why:

Pure solids (s) and pure liquids (l) are left out of Q and K.

Species in the equation	In Qc?	In Qp?
Gas, (g)	yes, as [X]	yes, as P_X
Dissolved, (aq)	yes, as [X]	no
Pure solid, (s)	no	no
Pure liquid, (l), including water as the solvent	no	no

The solid or liquid still has to be present for the equilibrium to exist. It is only its amount that does not matter.

Worked example 1: writing expressions

Question. Write Kc for each reaction. Where it is possible, also write Kp.

(a) 4NH₃(g) + 5O₂(g) ⇌ 4NO(g) + 6H₂O(g) (b) Fe₃O₄(s) + 4H₂(g) ⇌ 3Fe(s) + 4H₂O(g) (c) Ag⁺(aq) + 2NH₃(aq) ⇌ Ag(NH₃)₂⁺(aq) (d) NH₃(aq) + H₂O(l) ⇌ NH₄⁺(aq) + OH⁻(aq)

(a) Every species is a gas, so all four appear, each raised to its coefficient. Water here is a gas, so it stays in.

Kc = [NO]⁴[H₂O]⁶ / ([NH₃]⁴[O₂]⁵) Kp = (P_NO)⁴(P_H₂O)⁶ / ((P_NH₃)⁴(P_O₂)⁵)

(b) Fe₃O₄ and Fe are solids, so they are left out. Only the two gases remain.

Kc = [H₂O]⁴ / [H₂]⁴ Kp = (P_H₂O)⁴ / (P_H₂)⁴

(c) All species are dissolved, so this reaction has a Kc but no Kp.

Kc = [Ag(NH₃)₂⁺] / ([Ag⁺][NH₃]²)

The complex ion is one species, so its whole formula sits inside one pair of brackets.

(d) Water is the liquid solvent, so it is left out.

Kc = [NH₄⁺][OH⁻] / [NH₃]

Check. In each answer, the powers match the coefficients, products are on top, and no (s) or (l) species appears.

Worked example 2: calculating Q for a snapshot

Question. A flask holds a mixture for the reaction N₂(g) + 3H₂(g) ⇌ 2NH₃(g). At one moment, [N₂] = 0.50 mol L⁻¹, [H₂] = 0.20 mol L⁻¹ and [NH₃] = 0.10 mol L⁻¹. For this example, take Kc = 0.50 at the temperature of the flask (a value chosen for practice). Calculate Qc and decide whether the mixture is at equilibrium.

1. Write the expression: Qc = [NH₃]² / ([N₂][H₂]³).
2. Substitute: Qc = (0.10)² / ((0.50)(0.20)³).
3. Work out the pieces: (0.10)² = 0.010; (0.20)³ = 0.0080; 0.50 × 0.0080 = 0.0040.
4. Divide: Qc = 0.010 ÷ 0.0040 = 2.5.

Interpretation. Qc = 2.5 is not equal to Kc = 0.50, so the mixture is not at equilibrium. The concentrations will keep changing until Qc falls to 0.50 (five times smaller than now).

Watch out. Leaving out the powers gives 0.10 ÷ (0.50 × 0.20) = 1.0. Multiplying by the coefficients instead, (2 × 0.10) ÷ (0.50 × 3 × 0.20), gives 0.67. Both are wrong.

Worked example 3: a solid in a gas equilibrium

Question. Hot carbon reacts with carbon dioxide: C(s) + CO₂(g) ⇌ 2CO(g). In a closed vessel, P_CO = 0.40 atm and P_CO₂ = 0.80 atm. In this question, take Kp = 0.20 at the temperature of the vessel.

(a) Write Kp. (b) Calculate Qp and decide whether the system is at equilibrium. (c) A student adds another 5 g of carbon. Does Qp change?

(a) Carbon is a solid, so Kp = (P_CO)² / P_CO₂.

(b) Qp = (0.40)² / 0.80 = 0.16 / 0.80 = 0.20. This equals Kp, so the system is at equilibrium.

(c) No. Carbon does not appear in Qp. Extra solid does not change P_CO or P_CO₂, so Qp stays at 0.20 and the system stays at equilibrium.

Common misconceptions

• “Q and K are the same thing.” Q is for any moment; K is the single value Q reaches at equilibrium. Q = K only at equilibrium.
• Multiplying by coefficients. Coefficients become powers. 2NH₃ gives [NH₃]², not 2[NH₃].
• Reactants on top. The expression is always products over reactants for the reaction as written.
• Adding terms. Concentrations in the numerator (and in the denominator) are multiplied together.
• Including solids, liquids and the solvent. Leave out every (s) and (l) species, but keep H₂O(g).
• Putting pressures in Kc, or concentrations in Kp. Check the subscript and use matching quantities.
• Thinking K changes when you add more of a substance. Adding a reactant changes Q, not K. At a fixed temperature, K is constant.
• Using starting amounts in K. K must be calculated from equilibrium amounts. Starting amounts give a value of Q.

Where this leads

Next, in Topic 7.4, you will put measured equilibrium concentrations and pressures into these expressions to calculate K. Later topics use K to judge how far a reaction goes (7.5), to combine reactions (7.6) and to predict which way a mixture moves when Q ≠ K (7.7 and 7.10). If you need a reminder of why forward and reverse rates become equal, go back to Topic 7.2.

Try the practice questions now, then use the revision notes and the checklist to consolidate.