The shape of the lesson
By the end of the lesson, learners can
- Writing the expression: Build Kc expressions in the simulator, then vary the starting amounts.
- Calculating K c: If you are given equilibrium concentrations, substitute directly.
- What changes K c?: Only temperature.
- Equilibrium concentrations only.
- Table: initial, change, equilibrium, then ÷ volume.
- Units: cancel mol dm−3 each time.
- Only temperature changes Kc.
One picture every idea returns to
For aA + bB ⇌ cC + dD, Kc = [C]c[D]d ÷ [A]a[B]b, using equilibrium concentrations in mol dm−3. Products go on top, and each concentration is raised to the power of its coefficient. The units come from cancelling the concentration units, and may be none at all. A large Kc means the position lies to the right. Kc changes only with temperature; concentration, pressure and catalysts leave it unchanged.
Forty-five minutes, phase by phase
| Time | Phase | What happens in the room | Grouping |
|---|---|---|---|
| 0 to 5 min | Hook: predict | Open with the question in the title, A number for the position.. Le Chatelier tells you which way an equilibrium moves. Put the figure on the board: it is the one picture every idea in the lesson returns to. | Whole class, then pairs |
| 5 to 13 min | Teach and check: Writing the expression | Build Kc expressions in the simulator, then vary the starting amounts. Project tab 3 of the lab (Equilibria lab, Kc) and let a student drive it while the class predicts. | Whole class, then pairs |
| 13 to 21 min | Teach and check: Calculating K c | If you are given equilibrium concentrations, substitute directly. | Pairs |
| 21 to 30 min | Teach and check: What changes K c? | Only temperature. | Rally Coach |
| 30 to 36 min | Check | The four Stage 1 questions from the student lesson, answered on mini whiteboards or by show of fingers. Re-teach the one most get wrong. | Individual, then whole class |
| 36 to 43 min | Plenary | The exam-style questions from Stage 2, one at a time: thirty seconds alone, then a random call. The feedback lines in the key give the examiner's wording. | Individual, random call |
| 43 to 45 min | Exit and homework | Set the worksheet below (6 questions, 15 marks). Learners self-assess against the objectives before they leave. | Individual |
Protect the parts that carry the learning
The sequence carries about fifty minutes of material, so the lab is driven by one student while the class predicts rather than explored freely. Times are a guide.
Protected: the check at 30 minutes. It is the only individual read on whether the key idea landed, and it is not cut.
Cut first if the build overruns: the last check in What changes K c?, never the check or the plenary.
In a 60 minute block: run worksheet questions 1 to 3 in class as a Rally Coach set and model both the hardest worked example and its check live.
Rally Coach
Pairs share one worksheet. One partner solves the first question aloud while the other coaches, checking each line and offering a prompt rather than the answer; they swap roles for the next question.
The full structure, with its variations and the other twelve, is on the cooperative-learning page.
What to head off, and how
| Trap learners fall into | Teaching move that pre-empts it |
|---|---|
| Use equilibrium concentrations, not initial amounts, and convert moles to concentrations by dividing by the volume unless the volumes cancel. | Products always go on top. Work out the units separately each time by cancelling, rather than guessing. |
The equilibrium constant Kc: worksheet
6 questions, 15 marks. Written in the style of the papers; no past paper question is reproduced.
Write the Kc expression and deduce its units for N2O4(g) ⇌ 2NO2(g).
1.00 mol of N2O4 is placed in a 1.00 dm3 flask; at equilibrium 20.0% has dissociated to NO2. Calculate Kc.
1.00 mol of ethanoic acid and 1.00 mol of ethanol reach equilibrium with 0.667 mol of ethyl ethanoate: CH3COOH + C2H5OH ⇌ CH3COOC2H5 + H2O. Calculate Kc and explain why the volume is not needed.
0.500 mol each of H2 and I2 are mixed in 1.00 dm3. Kc = 49.0 for H2 + I2 ⇌ 2HI. Calculate the equilibrium amount of HI.
For 2SO2 + O2 ⇌ 2SO3, ΔH = −197 kJ mol−1. State and explain the effect on Kc of increasing the temperature.
Explain why adding more O2 to the SO2/O2/SO3 equilibrium mixture at constant temperature does not change Kc.
Answer key and marking notes
- Kc = [NO2]2 ÷ [N2O4].
- Units: (mol dm−3)2 ÷ mol dm−3 = mol dm−3.
- 0.200 mol N2O4 reacts, forming 0.400 mol NO2; 0.800 mol N2O4 remains.
- In 1.00 dm3: [N2O4] = 0.800, [NO2] = 0.400 mol dm−3.
- Kc = 0.4002 ÷ 0.800 = 0.200 mol dm−3.
- Equilibrium amounts: acid 0.333, alcohol 0.333, ester 0.667, water 0.667 mol.
- Kc = (0.667 × 0.667) ÷ (0.333 × 0.333) = 4.0.
- There are two concentration terms on top and bottom, so V cancels; Kc has no units.
- Let x mol of H2 react: [HI] = 2x, [H2] = [I2] = 0.500 − x.
- (2x)2 ÷ (0.500 − x)2 = 49.0, so 2x ÷ (0.500 − x) = 7.00.
- 2x = 3.50 − 7.00x, so x = 0.389 and HI = 0.778 mol.
- The forward reaction is exothermic, so increasing the temperature shifts the position to the left.
- [SO3] falls while [SO2] and [O2] rise, so Kc decreases.
- Adding O2 momentarily makes the concentration ratio smaller than Kc.
- The position shifts right, forming SO3, until the ratio again equals the same Kc; only temperature changes Kc.
8 multiple-choice questions on this lesson
From the topic question bank. Assign them in the app for a marked score in your gradebook, or run the topic as a class game.
Which is the correct expression for Kc for the reaction N2(g) + 3H2(g) ⇌ 2NH3(g)?
- AKc = [NH3]2 / ([N2] [H2]3)
- BKc = [N2] [H2]3 / [NH3]2
- CKc = [NH3] / ([N2] [H2])
- DKc = 2[NH3] / ([N2] + 3[H2])
What are the units of Kc for N2(g) + 3H2(g) ⇌ 2NH3(g), with concentrations in mol dm−3?
- Amol−2 dm36
- Bmol2 dm3−6
- Cno units
- Dmol dm−3
For H2(g) + I2(g) ⇌ 2HI(g) at a certain temperature, the equilibrium concentrations are [H2] = 0.0200 mol dm−3, [I2] = 0.0300 mol dm−3 and [HI] = 0.150 mol dm−3. What is Kc?
- A37.5
- B0.0267
- C250
- D500
Which change alters the value of Kc for a reaction at equilibrium?
- Achanging the temperature
- Badding a catalyst
- Cincreasing the total pressure
- Dadding more of one reactant
The graph shows the reaction N2O4(g) ⇌ 2NO2(g) reaching equilibrium in a sealed 1.00 dm3 flask. Using the equilibrium concentrations read from the graph, what is Kc?
- A0.360 mol dm−3
- B3.00 mol dm−3
- C2.78 mol−1 dm33
- D0.360 mol−1 dm33
1.00 mol of ethanoic acid and 2.00 mol of ethanol are mixed with a little acid catalyst and left to reach equilibrium: CH3COOH + C2H5OH ⇌ CH3COOC2H5 + H2O. At equilibrium, 0.845 mol of ethyl ethanoate is present. What is Kc?
- A3.99
- B0.357
- C0.251
- D4.72
0.200 mol of N2O4 is placed in a sealed 2.00 dm3 flask: N2O4(g) ⇌ 2NO2(g). At equilibrium, 0.160 mol of N2O4 remains. What is Kc?
- A0.0200 mol dm−3
- B0.0400 mol dm−3
- C0.00500 mol dm−3
- D0.500 mol dm−3
For 2SO2(g) + O2(g) ⇌ 2SO3(g), ΔH = −197 kJ mol−1. What happens when the temperature of the equilibrium mixture is raised?
- AKc decreases and the equilibrium moves to the left
- BKc stays the same, but the equilibrium moves to the left
- CKc increases, because the rates of reaction increase
- DKc stays the same, because only changes in concentration affect equilibrium
Quiz key
For aA + bB ⇌ cC, Kc = [C]c / ([A]a[B]b). Here that gives [NH3]2 / ([N2] [H2]3). Putting the reactants on top gives the expression for the reverse reaction, which is 1/Kc (upside down). Leaving out the powers ignores the 2 and the 3 in the equation. The coefficients are powers, not multipliers, and the concentrations in the denominator are multiplied, not added.
Units = (mol dm−3)2 / [(mol dm−3)(mol dm−3)3] = (mol dm−3)2 − 4 = (mol dm−3)−2 = mol−2 dm36. mol2 dm3−6 comes from the upside-down expression. Kc has no units only when the total powers on the top and bottom are equal, which is not the case here (2 on top, 4 below). mol dm−3 would need one more power on top than below.
Kc = 0.1502 / (0.0200 × 0.0300) = 0.0225 / 6.00 × 10−4 = 37.5. The units cancel (mol2 dm3−6 on top and bottom), so Kc has no units. 0.0267 is the expression upside down (1/37.5). 250 forgets to square [HI]. 500 multiplies [HI] by 2 instead of squaring it.
Kc is constant at a given temperature, so only a change in temperature changes its value. A catalyst speeds up the forward and reverse reactions equally and changes neither the position nor Kc. Increasing pressure can shift the position of a gas equilibrium, but the concentrations rearrange so that the ratio in the expression returns to the same Kc. Adding a reactant also shifts the position right until the ratio again equals the unchanged Kc.
At equilibrium [N2O4] = 0.040 and [NO2] = 0.120 mol dm−3. Kc = 0.1202 / 0.040 = 0.0144 / 0.040 = 0.360, with units (mol dm−3)2 / mol dm−3 = mol dm−3. 3.00 forgets to square [NO2]. 2.78 mol−1 dm33 is the expression upside down. 0.360 mol−1 dm33 has the right number but inverted units.
Equilibrium amounts: ester = water = 0.845 mol; acid = 1.00 − 0.845 = 0.155 mol; ethanol = 2.00 − 0.845 = 1.155 mol. The volume V cancels (two concentrations on top and two below), so Kc = (0.845 × 0.845) / (0.155 × 1.155) = 0.714 / 0.179 = 3.99, with no units. 0.357 uses the initial amounts of acid and ethanol instead of the equilibrium amounts. 0.251 is the expression upside down. 4.72 leaves out the water, which is a product in the same (liquid) mixture and must be included.
0.200 − 0.160 = 0.0400 mol N2O4 reacts, forming 2 × 0.0400 = 0.0800 mol NO2. Concentrations: [N2O4] = 0.160 / 2.00 = 0.0800 and [NO2] = 0.0800 / 2.00 = 0.0400 mol dm−3. Kc = 0.04002 / 0.0800 = 0.0200 mol dm−3. 0.0400 uses moles instead of concentrations; here the volume does not cancel, because the powers on top and bottom differ. 0.00500 forgets the 1 : 2 ratio and takes 0.0400 mol of NO2. 0.500 forgets to square [NO2].
Raising the temperature favours the endothermic direction, here the reverse reaction, so the equilibrium moves left: [SO3] falls while [SO2] and [O2] rise, and Kc (products over reactants) decreases. Kc does not stay the same when the temperature changes; only concentration and pressure changes leave it fixed. Faster rates do not change Kc; for an exothermic reaction Kc falls with rising temperature. Temperature, not only concentration, affects the equilibrium.
What to project
The student lesson embeds these tabs. Open them full screen on the projector; every slider has a drag handle on the stage.
Slides for this lesson
A projectable deck for every pack is on the way. Until then, present the student lesson full screen: its sections, figure and lab tabs follow the same order as the plan above.