The shape of the lesson
By the end of the lesson, learners can
- Inert nitrogen, basic ammonia: Compare the N 2 molecule with NH 3 in the simulator, then add H + to ammonia and watch the shape change.
- Ammonium salts and fertilisers: Neutralise ammonia with different acids in the simulator, then warm an ammonium salt with sodium hydroxide and test the gas.
- N2 inert: strong triple bond, non-polar.
- NH3 pyramidal 107°; NH4+ tetrahedral 109.5°.
- NH4+ + OH− → NH3 + H2O; red litmus blue.
- %N: count every N atom.
One picture every idea returns to
Nitrogen is unreactive because the N≡N triple bond is very strong (944 kJ mol−1) and the molecule is non-polar, so it does not attract electrophiles or nucleophiles. Ammonia is a Brønsted–Lowry base: the lone pair on nitrogen accepts a proton, forming the ammonium ion, NH4+, by a dative bond. NH3 is pyramidal (107°) and NH4+ tetrahedral (109.5°). Ammonium salts react with warm alkali to release ammonia, which turns damp red litmus blue. Ammonium salts are used as fertilisers; their value depends on the percentage of nitrogen by mass.
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, An inert gas that feeds the world.. Nitrogen makes up most of the air yet barely reacts with anything. Put the figure on the board: it is the one picture every idea in the lesson returns to. | Whole class, then pairs |
| 5 to 17 min | Teach and check: Inert nitrogen, basic ammonia | Compare the N 2 molecule with NH 3 in the simulator, then add H + to ammonia and watch the shape change. Project tab 1 of the lab (Nitrogen and sulfur lab, ammonia and NH4+) and let a student drive it while the class predicts. | Whole class, then pairs |
| 17 to 30 min | Teach and check: Ammonium salts and fertilisers | Neutralise ammonia with different acids in the simulator, then warm an ammonium salt with sodium hydroxide and test the gas. Project tab 2 of the lab (Nitrogen and sulfur lab, ammonium salts) and let a student drive it while the class predicts. | Frayer Model |
| 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, 13 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 Ammonium salts and fertilisers, 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.
Frayer Model
Pairs fill a four-box grid for the key term: definition, characteristics, examples and non-examples. The non-examples box is where the misconceptions surface, so give it the most time.
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 |
|---|---|
| Nitrogen is unreactive because of its strong triple bond and lack of polarity, not because it is small or a gas. | When calculating percentage of nitrogen, count every nitrogen atom: NH 4 NO 3 has two, (NH 4) 2 SO 4 has two, urea has two. |
Nitrogen, ammonia and the ammonium ion: worksheet
6 questions, 13 marks. Written in the style of the papers; no past paper question is reproduced.
Explain why nitrogen is very unreactive.
Describe, with an equation, how ammonia behaves as a base, and explain the change in bond angle.
Write an equation for the reaction of ammonium sulfate with sodium hydroxide, and describe a test for the gas produced.
Calculate the percentage by mass of nitrogen in ammonium nitrate, NH4NO3.
A farmer needs 70.0 kg of nitrogen. Calculate the mass of urea, CO(NH2)2 (Mr 60.0), required.
Explain why farmers should not add calcium hydroxide to soil at the same time as an ammonium fertiliser.
Answer key and marking notes
- The N≡N triple bond is very strong (944 kJ mol−1), so a lot of energy is needed to break it.
- The molecule is non-polar, so it does not attract electrophiles or nucleophiles.
- NH3 + H+ → NH4+; the nitrogen lone pair forms a dative bond to H+.
- NH3: 3 bonding pairs + 1 lone pair, 107°.
- NH4+: 4 bonding pairs, no lone pairs, so the repulsions are equal: 109.5°.
- (NH4)2SO4 + 2NaOH → Na2SO4 + 2NH3 + 2H2O.
- Ammonia turns damp red litmus paper blue.
- Mr = 14.0 + 4.0 + 14.0 + 48.0 = 80.0.
- %N = 28.0 ÷ 80.0 × 100 = 35.0%.
- Urea is 28.0 ÷ 60.0 = 46.7% nitrogen.
- Mass = 70.0 ÷ 0.467 = 150 kg.
- Ca(OH)2 is a base, which reacts with ammonium ions: NH4+ + OH− → NH3 + H2O.
- Nitrogen is lost as ammonia gas, so the fertiliser is wasted.
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.
Nitrogen makes up about 78% of the air, yet N2 is very unreactive. What is the main reason?
- Athe N≡N triple bond is very strong (944 kJ mol−1) and the molecule is non-polar, so there is no partial charge to attract electrophiles or nucleophiles
- Bthe N2 molecule is very small, so other molecules cannot collide with it
- Cnitrogen atoms have a full outer shell, like the noble gases
- Dthe molecule is polar, so it is surrounded by a shell of water molecules in the air
The diagram shows ammonia reacting with a hydrogen ion. Which statement correctly describes the role of ammonia?
- Aammonia is a Brønsted-Lowry base: its nitrogen lone pair accepts a proton, forming a dative covalent bond in NH4+
- Bammonia is a Brønsted-Lowry acid, because it donates a proton to form NH4+
- Cammonia is a base because it donates a hydroxide ion directly from the molecule
- Dammonia gains an electron from H+, so an ionic bond forms between N and H
Which row gives the correct shapes and bond angles of NH3 and NH4+?
- ANH3 pyramidal, 107°; NH4+ tetrahedral, 109.5°
- BNH3 trigonal planar, 120°; NH4+ tetrahedral, 109.5°
- CNH3 tetrahedral, 109.5°; NH4+ pyramidal, 107°
- DNH3 pyramidal, 107°; NH4+ square planar, 90°
A white solid is warmed with aqueous sodium hydroxide. A gas is given off that turns damp red litmus paper blue. Which ion is present in the solid, and what is the equation if the solid is ammonium chloride?
- ANH4+; NH4Cl + NaOH → NaCl + NH3 + H2O
- BNH4+; NH4Cl + NaOH → NaCl + N2 + H2 + H2O
- CCl−; NH4Cl + NaOH → NaCl + NH3 + H2O
- DNH4+; NH4Cl + NaOH → NaCl + HCl + NH3
Ammonium sulfate, a common fertiliser, is made by reacting ammonia with sulfuric acid. Which equation is correct?
- A2NH3 + H2SO4 → (NH4)2SO4
- BNH3 + H2SO4 → NH4SO4
- C2NH3 + H2SO4 → (NH4)2SO4 + H2
- DNH3 + H2SO4 → (NH4)2SO4 + H2O
What is the percentage by mass of nitrogen in ammonium sulfate, (NH4)2SO4? (Ar H 1.0, N 14.0, O 16.0, S 32.1)
- A21.2%
- B10.6%
- C42.4%
- D35.0%
The chart shows the percentage of nitrogen by mass in three fertilisers. A farmer needs to supply 100 kg of nitrogen using ammonium sulfate. What mass of ammonium sulfate is needed, to 3 s.f.?
- A472 kg
- B21.2 kg
- C286 kg
- D214 kg
In the Haber process, N2 + 3H2 ⇌ 2NH3 (forward reaction exothermic). The usual conditions are about 450 °C, 200 atm and an iron catalyst. Why are a high temperature and a catalyst used even though the forward reaction is exothermic?
- Athe very strong N≡N bond gives a high activation energy, so without them the rate would be far too slow; 450 °C is a compromise between rate and yield
- Ba high temperature shifts the equilibrium to the right, increasing the yield of ammonia
- Cthe catalyst shifts the equilibrium to the right, so a higher temperature can be used without losing yield
- Dnitrogen is a small molecule, so it needs a high temperature to collide with hydrogen
Quiz key
Breaking the N≡N triple bond needs 944 kJ mol−1, one of the strongest bonds known, so reactions of N2 have a very high activation energy. The two atoms are identical, so the molecule is non-polar and has no δ+ or δ− sites to attract reagents. Being small has nothing to do with it; small molecules such as H2 or F2 can be very reactive, so 'small' is the classic error. A nitrogen atom has five outer electrons, not a full shell; it completes its octet only by sharing three pairs in N2. N2 is non-polar, not polar, and is not hydrated in air.
In NH3 + H+ → NH4+, ammonia accepts the proton, so it is a Brønsted-Lowry base. The lone pair on nitrogen is donated to the empty orbital of H+, forming a dative (coordinate) bond; once formed it is identical to the other three N-H bonds. Calling ammonia an acid is the classic error: it gains H+, it does not donate it. Ammonia contains no OH group; in water it produces OH− by taking H+ from H2O (NH3 + H2O ⇌ NH4+ + OH−). The bond formed is a shared pair donated by N, a covalent bond, not an ionic one, and H+ has no electron to give.
In NH3 nitrogen has three bonding pairs and one lone pair. The lone pair repels more strongly than bonding pairs, squeezing the H-N-H angle from 109.5° to about 107°, and the shape (atoms only) is pyramidal. In NH4+ the lone pair has become a fourth bonding pair, so four equal pairs give a regular tetrahedron at 109.5°. Trigonal planar ignores the lone pair on N. Swapping the two shapes puts the lone pair on the ion that no longer has one. Four electron pairs never adopt a square planar arrangement here; that needs six electron pairs, two of them lone pairs.
Ammonia is the only common alkaline gas, so the solid contains the ammonium ion. The stronger base OH− removes a proton from NH4+ and displaces ammonia: NH4+ + OH− → NH3 + H2O, or in full NH4Cl + NaOH → NaCl + NH3 + H2O. Ammonia is not decomposed into N2 and H2 by gentle warming, and neither of those gases affects litmus. Chloride ions are identified with acidified silver nitrate, not by this test, and do not release a gas with NaOH. HCl could not be formed in excess alkali (it would be neutralised), and the equation is unbalanced.
Sulfuric acid is diprotic, so each H2SO4 donates two protons to two NH3 molecules, giving two NH4+ ions for each SO42− ion: 2NH3 + H2SO4 → (NH4)2SO4. The salt is a soluble source of nitrogen for crops. NH4SO4 has the wrong charge balance: one NH4+ cannot balance SO42−. No hydrogen gas is formed; ammonia neutralises acids by accepting protons, it does not displace hydrogen like a metal. The last equation is unbalanced and no water is produced, because ammonia (unlike a metal hydroxide) contains no OH to form water.
Mr((NH4)2SO4) = 2 × 18.0 + 32.1 + 64.0 = 132.1. The formula contains two N atoms, mass 28.0, so %N = 28.0 / 132.1 × 100 = 21.2%. 10.6% counts only one N atom, the classic error of ignoring the subscript 2 outside the bracket. 42.4% counts four N atoms (56.0 / 132.1), doubling twice. 35.0% is the value for ammonium nitrate, NH4NO3, not ammonium sulfate.
Mass of fertiliser = mass of N / fraction of N = 100 / 0.212 = 472 kg (using the exact fraction 28.0 / 132.1 also gives 472 kg). Urea, with the highest %N (46.7%), would need only 214 kg, which is why it is a cheap fertiliser to transport, but that value is for urea, not ammonium sulfate. 21.2 kg multiplies by the percentage instead of dividing, giving the mass of N in 100 kg of fertiliser. 286 kg uses the 35.0% value for ammonium nitrate.
Because the N≡N bond (944 kJ mol−1) is so strong, the activation energy is high. A low temperature would favour ammonia at equilibrium but the rate would be uneconomically slow, so about 450 °C with an iron catalyst is chosen as a compromise; 200 atm pushes the equilibrium towards the side with fewer gas moles. Raising the temperature shifts an exothermic equilibrium to the LEFT, lowering the yield, so it is not used for that reason. A catalyst speeds up forward and reverse reactions equally and does not shift the position of equilibrium. Molecular size is not the issue; the strength of the triple bond is.
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.