Week 11 · Chapter 3 (B): surface area and concentration, the law of mass action, the equilibrium constant Kc and what its size means (short, calculation-light text: two and a half book pages, add worked examples and prediction tasks) · 3 lessons
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The law of mass action and the equilibrium constant
Chemistry Sec 3 Term 1, Chapter 3: pp. 59–61
GoalState the law of mass action for a one-step reaction, write Kc for a reversible reaction, and calculate Kc from equilibrium concentrations.
New words
law of mass action · at a constant temperature the rate of a one-step reaction is proportional to the product of the reactant concentrations, each raised to its coefficientrate constant k · the number that links the rate to the concentrations; it changes with temperature but not with concentrationequilibrium constant Kc · the product concentrations multiplied together, divided by the reactant concentrations multiplied together, each raised to its coefficient, at equilibrium
Explanation
The law of mass action is named after two Norwegian scientists, Waage and Guldberg. For a reaction that happens in one step, A + B → C + D, the rate is r₁ = k₁[A][B]: k₁ is the rate constant and square brackets mean concentration in mol/L. For a reaction with several steps the powers must be found by experiment, so use the law as written only for one-step reactions; the book’s iron(III) thiocyanate illustration is written as if it were one step, which is a simplification. In a reversible reaction A + B ⇌ C + D the backward rate is r₂ = k₂[C][D]. At equilibrium r₁ = r₂, so k₁[A][B] = k₂[C][D] and k₁ ÷ k₂ = [C][D] ÷ ([A][B]). This ratio is a constant at a given temperature and is called Kc. In general Kc has the products on top and the reactants below, and every concentration is raised to its coefficient in the balanced equation. Put in the equilibrium concentrations, never the starting ones. At this level Kc is written without units (extra: the units really depend on the equation).
Reversible reaction
Kc
N₂O₄ ⇌ 2NO₂
[NO₂]² ÷ [N₂O₄]
2SO₂ + O₂ ⇌ 2SO₃
[SO₃]² ÷ ([SO₂]² × [O₂])
CO + H₂O ⇌ CO₂ + H₂
([CO₂] × [H₂]) ÷ ([CO] × [H₂O])
N₂ + 3H₂ ⇌ 2NH₃
[NH₃]² ÷ ([N₂] × [H₂]³)
Worked examples
ƒKc of N₂O₄ ⇌ 2NO₂
Kc = [NO₂]² ÷ [N₂O₄] = 0.120² ÷ 0.040 = 0.36
Kc from concentrations. For N₂O₄ ⇌ 2NO₂ the equilibrium concentrations are [N₂O₄] = 0.040 mol/L and [NO₂] = 0.120 mol/L. Kc = [NO₂]² ÷ [N₂O₄] = 0.120² ÷ 0.040 = 0.0144 ÷ 0.040 = 0.36. The coefficient 2 becomes the power 2, and the answer is less than 1, so there is more N₂O₄ than the NO₂ squared.
Three substances. For 2SO₂ + O₂ ⇌ 2SO₃ the concentrations are [SO₂] = 0.20, [O₂] = 0.10 and [SO₃] = 0.40 mol/L. Kc = [SO₃]² ÷ ([SO₂]² × [O₂]) = 0.16 ÷ (0.04 × 0.10) = 0.16 ÷ 0.004 = 40. Square the numbers first, then multiply the lower ones. For CO + H₂O ⇌ CO₂ + H₂ with 0.20, 0.30, 0.60 and 0.40 mol/L the powers are all 1: Kc = (0.60 × 0.40) ÷ (0.20 × 0.30) = 0.24 ÷ 0.06 = 4.0.
Predicting a rate. Suppose a one-step reaction has rate = k[A][B]². Doubling [B] multiplies the rate by 2² = 4. Doubling both [A] and [B] multiplies it by 2 × 2² = 8. Tripling [A] alone multiplies it by 3. The prediction is valid only because the reaction is one step, and the powers are the coefficients. A scientist would check the rate law by experiment.
Class activity
Pair counting with tokens, paper only. Cut out 3 red paper tokens for A and 2 blue tokens for B. Draw a line from every red token to every blue token: you get 3 × 2 = 6 pairs, a picture of the number of possible meetings, which is what the rate is proportional to. Now use 6 red tokens and 2 blue: how many pairs? Then 6 red and 4 blue? Write the three counts in a table and compare them with the idea rate = k[A][B]. Keep your tokens for the next lesson.
Practice
1
Write the Kc expression for 2NO + O₂ ⇌ 2NO₂.
Kc = [NO₂]² ÷ ([NO]² × [O₂]).
2
For PCl₅ ⇌ PCl₃ + Cl₂ the equilibrium concentrations are [PCl₅] = 0.50, [PCl₃] = 0.10 and [Cl₂] = 0.10 mol/L. Find Kc.
