The shape of the lesson
By the end of the lesson, learners can
- Partial pressures: In the simulator, change the total pressure of an equilibrium mixture and watch each gas's share of the pressure.
- Calculating K p: Follow the chain in Fig. 1: equilibrium moles, then mole fractions, then partial pressures, then substitute.
- What changes K p?: Exactly the same rule as for Kc: temperature only.
- Mole fraction = n ÷ total equilibrium moles.
- Partial pressure = mole fraction × total pressure.
- Units from the pressure unit, cancelled like Kc.
- Only temperature changes Kp.
One picture every idea returns to
The partial pressure of a gas is the pressure it would exert alone in the container: p = mole fraction × total pressure, where mole fraction = moles of that gas ÷ total moles of gas. The partial pressures add up to the total pressure. Kp is written like Kc but with partial pressures: for N2O4 ⇌ 2NO2, Kp = p(NO2)2 ÷ p(N2O4). Units come from the pressure units (Pa, kPa or atm). Like Kc, Kp changes only with temperature.
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, Pressures instead of concentrations.. For reactions between gases it is easier to measure pressures than concentrations. 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: Partial pressures | In the simulator, change the total pressure of an equilibrium mixture and watch each gas's share of the pressure. Project tab 4 of the lab (Equilibria lab, Kp) and let a student drive it while the class predicts. | Whole class, then pairs |
| 13 to 21 min | Teach and check: Calculating K p | Follow the chain in Fig. 1: equilibrium moles, then mole fractions, then partial pressures, then substitute. | Pairs |
| 21 to 30 min | Teach and check: What changes K p? | Exactly the same rule as for Kc: temperature only. | Connect-Extend-Challenge |
| 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, 14 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 p?, 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.
Connect-Extend-Challenge
After the new idea, each student writes three lines: how it connects to what they knew, how it extends their thinking, and what still challenges them. Pairs compare, then the challenges become the plenary.
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 |
|---|---|
| Partial pressure uses the mole fraction of the equilibrium mixture, so first find the equilibrium moles and the new total. | A common slip is dividing by the initial number of moles. Also remember that increasing the total pressure changes the position of equilibrium but not the value of K p. |
The equilibrium constant Kp: worksheet
6 questions, 14 marks. Written in the style of the papers; no past paper question is reproduced.
Define partial pressure and state how it is calculated.
Write the Kp expression and units (in Pa) for N2(g) + 3H2(g) ⇌ 2NH3(g).
1.00 mol of N2O4 is allowed to reach equilibrium at a total pressure of 100 kPa; 20.0% dissociates into NO2. Calculate Kp.
At equilibrium p(SO2) = 20 kPa, p(O2) = 10 kPa and p(SO3) = 70 kPa. Calculate Kp for 2SO2 + O2 ⇌ 2SO3 and give its units.
For N2O4 ⇌ 2NO2, ΔH = +58 kJ mol−1. State and explain the effect of increasing the temperature on Kp.
Explain why increasing the total pressure on the N2O4/NO2 equilibrium changes the composition of the mixture but not Kp.
Answer key and marking notes
- The pressure a gas in a mixture would exert if it alone occupied the whole container.
- Partial pressure = mole fraction × total pressure.
- Kp = p(NH3)2 ÷ [p(N2) × p(H2)3].
- Units: Pa2 ÷ Pa4 = Pa−2.
- Equilibrium moles: N2O4 0.800, NO2 0.400; total 1.20.
- Mole fractions: 0.667 and 0.333.
- Partial pressures: 66.7 kPa and 33.3 kPa.
- Kp = 33.32 ÷ 66.7 = 16.7 kPa.
- Kp = 702 ÷ (202 × 10) = 4900 ÷ 4000.
- Kp = 1.23 kPa−1.
- The forward reaction is endothermic, so the position shifts right on heating.
- p(NO2) increases and p(N2O4) decreases, so Kp increases.
- Higher pressure shifts the position to the left, towards fewer gas molecules (1 mol rather than 2).
- The partial pressures readjust until p(NO2)2 ÷ p(N2O4) equals the same Kp, which depends only on temperature.
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.
An equilibrium mixture contains 1.00 mol of N2, 3.00 mol of H2 and 1.00 mol of NH3 at a total pressure of 500 kPa. What is the partial pressure of NH3?
- A100 kPa
- B167 kPa
- C125 kPa
- D500 kPa
Which is the correct expression for Kp for N2(g) + 3H2(g) ⇌ 2NH3(g)?
- AKp = p(NH3)2 / (p(N2) × p(H2)3)
- BKp = p(N2) × p(H2)3 / p(NH3)2
- CKp = p(NH3) / (p(N2) × p(H2))
- DKp = [NH3]2 / ([N2] [H2]3)
What are the units of Kp for N2O4(g) ⇌ 2NO2(g), with partial pressures in kPa?
- AkPa
- BkPa−1
- Cno units
- DkPa2
The total pressure on the equilibrium N2(g) + 3H2(g) ⇌ 2NH3(g) is increased at constant temperature. What happens to the value of Kp?
- Ait stays the same; the partial pressures change until their ratio again equals Kp
- Bit increases, because the equilibrium moves to the right
- Cit decreases, because there are fewer moles of gas on the right
- Dit increases, because all the partial pressures increase
For N2O4(g) ⇌ 2NO2(g) at equilibrium, p(N2O4) = 60.0 kPa and p(NO2) = 40.0 kPa. What is Kp?
- A26.7 kPa
- B0.667 kPa
- C0.0375 kPa−1
- D26.7 kPa−1
An equilibrium mixture for 2SO2(g) + O2(g) ⇌ 2SO3(g) contains 0.300 mol SO2, 0.100 mol O2 and 1.60 mol SO3 at a total pressure of 200 kPa. What is Kp?
- A2.84 kPa−1
- B0.533 kPa−1
- C284 kPa−1
- D2.84 kPa
For N2O4(g) ⇌ 2NO2(g), ΔH = +58 kJ mol−1. What is the effect of increasing the temperature on Kp?
- AKp increases, because the equilibrium moves to the right in the endothermic direction
- BKp decreases, because the equilibrium moves to the left
- CKp does not change, because it is a constant
- DKp does not change, because the total pressure is unchanged
1.00 mol of N2O4 is heated and 20.0% of it dissociates: N2O4(g) ⇌ 2NO2(g). The total pressure at equilibrium is 100 kPa. What is Kp?
- A16.7 kPa
- B20.0 kPa
- C5.00 kPa
- D0.500 kPa
Quiz key
Total moles = 1.00 + 3.00 + 1.00 = 5.00 mol, so the mole fraction of NH3 = 1.00 / 5.00 = 0.200 and p(NH3) = 0.200 × 500 = 100 kPa. 167 kPa divides the total pressure by the number of different gases (3), ignoring how many moles of each there are. 125 kPa divides by the moles of the other gases (4) instead of the total (5). 500 kPa is the total pressure; each gas contributes only its share.
Kp uses equilibrium partial pressures, products over reactants, each raised to the power of its coefficient: p(NH3)2 / (p(N2) × p(H2)3). The reactants-over-products version is upside down (it is 1/Kp). Leaving out the powers ignores the coefficients 2 and 3. The expression in square brackets is Kc, which uses concentrations, not partial pressures.
Units = kPa2 / kPa = kPa. kPa−1 comes from writing the expression upside down. There would be no units only if the total powers on the top and bottom were equal (here 2 on top and 1 below). kPa2 forgets to divide by the partial pressure of N2O4.
Kp depends only on temperature. Increasing the pressure does move the position to the right (fewer gas moles), but that shift is exactly what brings the ratio of partial pressures back to the same Kp value. The shift in position is a consequence of the unchanged Kp, not a sign that Kp has grown. The number of gas moles decides the direction of the shift, not a change in Kp. All partial pressures rise at first, but the expression is a ratio, and the system adjusts until it returns to Kp.
Kp = 40.02 / 60.0 = 1600 / 60.0 = 26.7 kPa (kPa2 / kPa = kPa). 0.667 kPa forgets to square p(NO2). 0.0375 kPa−1 is the expression upside down. 26.7 kPa−1 has the correct value but the units of the inverted expression.
Total moles = 0.300 + 0.100 + 1.60 = 2.00, so the mole fractions are 0.150, 0.0500 and 0.800 and the partial pressures are p(SO2) = 30.0, p(O2) = 10.0 and p(SO3) = 160 kPa. Kp = 1602 / (30.02 × 10.0) = 25600 / 9000 = 2.84 kPa−1 (kPa2 / kPa3). 0.533 leaves out the squares. 284 substitutes the moles instead of the partial pressures. 2.84 kPa has the right number but the wrong units; there are more pressure terms below than above.
Increasing the temperature favours the endothermic direction, here the forward reaction, so more NO2 forms: p(NO2) rises, p(N2O4) falls and Kp increases. Moving left would be the effect for an exothermic forward reaction. Kp is constant only at a fixed temperature, so a change in temperature changes it. Keeping the total pressure constant does not keep Kp constant; the temperature change alters the equilibrium constant itself.
At equilibrium: N2O4 = 0.800 mol and NO2 = 2 × 0.200 = 0.400 mol; total = 1.20 mol. Partial pressures: p(N2O4) = (0.800 / 1.20) × 100 = 66.7 kPa and p(NO2) = (0.400 / 1.20) × 100 = 33.3 kPa. Kp = 33.32 / 66.7 = 16.7 kPa. 20.0 kPa uses 1.00 mol as the total, forgetting that the number of moles rises on dissociation. 5.00 kPa forgets the 1 : 2 ratio and takes 0.200 mol of NO2. 0.500 kPa forgets to square p(NO2).
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.