Lesson 11: The Equilibrium ConstantAP

Put a reversible reaction in a sealed container and something remarkable happens: the concentrations stop changing, even though the reaction has not stopped. Forward and reverse reactions continue at equal rates — this is dynamic equilibrium. For the general reaction aA + bB ⇌ cC + dD, the concentrations at equilibrium always satisfy one relationship, the law of mass action: Kc = [C]^c[D]^d ÷ ([A]^a[B]^b), where the brackets mean equilibrium molarities and the exponents are the stoichiometric coefficients. Kc is constant at a given temperature no matter what concentrations you start from; it changes only with temperature. Two construction rules are absolute. Pure solids and pure liquids are left out of the expression (their "concentration" is constant and is absorbed into K), and every concentration is an equilibrium value — putting initial concentrations into the expression gives the reaction quotient Q, not K (CED Topics 7.1–7.5).

What the size of K tells you

K is a scoreboard for the equilibrium position. K >> 1 (say 10³ or more) means the equilibrium mixture is mostly products — the reaction "goes essentially to completion". K << 1 (10⁻³ or less) means it barely proceeds at all. K near 1 means significant amounts of everything. For gas-phase reactions you can equally well write Kp using partial pressures; the two are linked by Kp = Kc(RT)^Δn, where Δn is (moles of gaseous products − moles of gaseous reactants) and R = 0.08206 L·atm/mol·K. When Δn = 0 the two constants are equal — a favourite quick-check on the exam.

Worked example 1: converting between Kp and Kc

For 2 SO₃(g) ⇌ 2 SO₂(g) + O₂(g) at 1000 K, Kc = 0.280. Calculate Kp. Count gas moles: products 2 + 1 = 3; reactants 2; Δn = +1. Kp = Kc(RT)^Δn = 0.280 × (0.08206 × 1000)¹ = 0.280 × 82.06 = 23.0. The pressure-based constant is much larger only because of the units conversion — Kp and Kc describe the same equilibrium, never two different positions. If Δn had been negative, Kp would have come out smaller than Kc.

ICE tables: the equilibrium calculator

Equilibrium problems are solved with an ICE table — Initial, Change, Equilibrium — which turns the algebra of "some unknown amount x reacts" into routine. Write the initial concentrations, let the change be ±x multiplied by each coefficient, and express every equilibrium concentration in terms of x. Substitute into the K expression and solve. Two solution patterns recur. If K is very small, x is negligible next to the initial concentrations, and you may drop the "−x" in the denominator — but you must verify afterwards that x is less than about 5% of the value it was subtracted from. And sometimes the expression is a perfect square in disguise, and taking the square root of both sides avoids the quadratic entirely.

Worked example 2: solving an ICE problem by the square-root shortcut

For H₂(g) + I₂(g) ⇌ 2 HI(g) at 430 °C, Kc = 54.3. A flask starts with [H₂] = [I₂] = 0.500 M and no HI. Find all equilibrium concentrations. ICE: [H₂] = 0.500 − x; [I₂] = 0.500 − x; [HI] = 2x. Set up: 54.3 = (2x)² ÷ (0.500 − x)². The numerator and denominator are both perfect squares, so take the square root of both sides: 7.37 = 2x ÷ (0.500 − x). Solve: 2x = 7.37(0.500 − x) = 3.685 − 7.37x, so 9.37x = 3.685 and x = 0.393 M. Therefore [H₂] = [I₂] = 0.500 − 0.393 = 0.107 M and [HI] = 2 × 0.393 = 0.787 M. Check: 0.787² ÷ (0.107 × 0.107) = 0.619 ÷ 0.0114 ≈ 54 ✓. Always run this check: it costs ten seconds and catches most algebra slips.

Worked example 3: when the quadratic is unavoidable

For N₂O₄(g) ⇌ 2 NO₂(g) at a temperature where Kc = 0.360, a flask starts with [N₂O₄] = 0.500 M. Find the equilibrium concentrations. ICE: [N₂O₄] = 0.500 − x; [NO₂] = 2x. Set up: 0.360 = (2x)² ÷ (0.500 − x) = 4x² ÷ (0.500 − x). Multiply out: 4x² = 0.180 − 0.360x, so 4x² + 0.360x − 0.180 = 0. Quadratic formula: x = [−0.360 + √(0.360² − 4 × 4 × (−0.180))] ÷ (2 × 4) = [−0.360 + √(0.1296 + 2.88)] ÷ 8 = (−0.360 + 1.735) ÷ 8 = 0.172 M. (Take the positive root; the negative root would give negative concentrations.) Therefore [N₂O₄] = 0.500 − 0.172 = 0.328 M and [NO₂] = 0.344 M. Check: 0.344² ÷ 0.328 = 0.118 ÷ 0.328 = 0.360 ✓. Note that x here is 34% of the starting value — far too big for the small-x approximation, which is why the quadratic was required.

Common mistakes

• Inserting initial concentrations into the K expression and calling the result K. That calculation gives Q, the reaction quotient — useful next lesson, wrong here.
• Including solids and pure liquids in the expression. For CaCO₃(s) ⇌ CaO(s) + CO₂(g), Kc = [CO₂], full stop; the two solids never appear.
• Forgetting the coefficients as exponents. For N₂ + 3 H₂ ⇌ 2 NH₃ the denominator contains [H₂]³; omitting the cube is the single commonest setup error.
• Applying the small-x approximation without the 5% check, or applying it when K is large. If the check fails, the answer needs the quadratic.
• Taking the negative quadratic root or reporting a concentration larger than the starting amount of its source. Sanity-check every ICE answer: no equilibrium concentration can be negative, and a reactant can only decrease.
• Treating Kc and Kp as different equilibria instead of the same equilibrium in different units.

AP Exam Experience

Question 1. What is the equilibrium constant expression for the reaction CaCO₃(s) ⇌ CaO(s) + CO₂(g)?

• (A) Kc = [CaO][CO₂] ÷ [CaCO₃]
• (B) Kc = [CO₂]
• (C) Kc = [CaCO₃] ÷ ([CaO][CO₂])
• (D) Kc = [CaO] ÷ [CaCO₃]

Question 2. For a reaction with Kc = 2.5 × 10⁻¹² at 25 °C, the equilibrium mixture contains

• (A) mostly products, with very little reactant
• (B) roughly equal amounts of reactants and products
• (C) mostly reactants, with very little product
• (D) only reactants, because the reaction does not occur at all

Question 3. For the reaction H₂(g) + I₂(g) ⇌ 2 HI(g), Kc = 54 at a certain temperature. A mixture has [H₂] = [I₂] = 0.100 M and [HI] = 0.735 M. Which statement is true?

• (A) The mixture is at equilibrium
• (B) The mixture is not at equilibrium and the reaction proceeds to the right
• (C) The mixture is not at equilibrium and the reaction proceeds to the left
• (D) Nothing can be concluded without knowing the volume

Answer key. 1. (B) — pure solids are omitted, leaving only the gas. 2. (C) — a tiny K means the equilibrium lies far to the left; "mostly reactants" but not (D), because equilibrium is dynamic and some product always forms. 3. (A) — Q = (0.735)² ÷ (0.100 × 0.100) = 0.540 ÷ 0.0100 = 54 = Kc, so the mixture is at equilibrium.

Free-response question. Phosphorus pentachloride decomposes: PCl₅(g) ⇌ PCl₃(g) + Cl₂(g), with Kc = 1.20 at 250 °C. A flask is charged with PCl₅ at an initial concentration of 0.100 M and no products. (a) Write the equilibrium constant expression for the reaction. (b) Calculate the equilibrium concentrations of all three gases. (c) Calculate the percent dissociation of PCl₅ (the fraction of the original PCl₅ that has decomposed, as a percent). (d) The equilibrium mixture is transferred to a flask of half the volume at the same temperature. Without calculating, state whether the percent dissociation increases, decreases or stays the same, and justify your answer using Q. Scoring guideline.

• (a) 1 point for Kc = [PCl₃][Cl₂] ÷ [PCl₅].
• (b) 1 point for setting up x² ÷ (0.100 − x) = 1.20 (with [PCl₃] = [Cl₂] = x from an ICE table); 1 point for the quadratic solution x = 0.0928 M, giving [PCl₃] = [Cl₂] = 0.0928 M and [PCl₅] = 0.0072 M. (Reject the negative root explicitly or implicitly by positive concentrations.)
• (c) 1 point for 0.0928 ÷ 0.100 × 100 = 92.8% dissociated.
• (d) 1 point for "decreases", justified by noting that halving the volume doubles every concentration, making Q = (0.186 × 0.186) ÷ 0.0144 ≈ 2.4, which exceeds Kc = 1.20, so the reaction shifts left, re-forming PCl₅ — consistent with Le Châtelier's principle, since the left side has fewer moles of gas.

Equilibrium in the real world: why K runs industry and oceans

Equilibrium constants are operating instructions wherever chemistry runs at scale. In the Haber process the modest value of K at working temperature is the entire business problem: Lesson 16 will show K for ammonia synthesis falls from about 6 × 10⁵ at 25 °C to below 1 at 470 °C, which is why plants run hot for speed, squeeze to 200 atmospheres to push the position with Le Châtelier's principle, and recycle the unconverted gas rather than accepting a low single-pass yield — the ICE table from this lesson, run on a loop. The same mathematics governs the oceans: the equilibrium CO₂ + H₂O ⇌ H⁺ + HCO₃⁻ has a fixed K, so as atmospheric CO₂ rises, Q exceeds K and the reaction shifts right, lowering the pH and consuming the carbonate that shell-building organisms need — ocean acidification is an ICE problem with the atmosphere as the initial concentration. And every buffer in the next units, every slightly soluble salt, every weak acid in this course is this same law of mass action wearing different clothes: write the expression, build the table, solve for x.

Deeper: K under algebraic surgery

Equilibrium constants transform predictably when equations are manipulated, and short exam questions exploit all three rules. Reverse the equation and K inverts: if N₂O₄ ⇌ 2 NO₂ has Kc = 0.360, then 2 NO₂ ⇌ N₂O₄ has Kc = 1 ÷ 0.360 = 2.78. Multiply the equation by a factor n and K is raised to the power n: doubling the N₂O₄ equation gives Kc = 0.360² = 0.130; halving it (½ N₂O₄ ⇌ NO₂) gives Kc = √0.360 = 0.600. Add two equations and their constants multiply: if A ⇌ B has K₁ = 4.0 and B ⇌ C has K₂ = 0.25, the combined A ⇌ C has K = 4.0 × 0.25 = 1.0. The rules are just the mass-action expression following the algebra: reversing swaps numerator and denominator, scaling raises every concentration's exponent. They also explain a favourite experimental design: an unfavourable step (small K) can be dragged along by coupling it to a very favourable one (large K) whose product swamps the combined constant — the chemical version of paying for an uphill purchase with a downhill windfall.