Topic 12 · AS Level · Inorganic chemistry
What this chapter covers12.1
Nitrogen makes up almost four-fifths of the air, and for most purposes it does nothing at all. You breathe it in and out unchanged; food is packed under it because it will not react with anything in the packet. Yet the same element, once it has been forced to react, turns up in fertilisers, in the brown haze over a city at rush hour and in the acid that eats limestone buildings. This chapter is about that contrast: why the N2 molecule is so reluctant to react, what happens to nitrogen once it is part of ammonia or an oxide, and how the oxides of nitrogen made in car engines and lightning go on to cause photochemical smog and acid rain — including the part they play in turning sulfur dioxide into sulfuric acid.
What topic 12 asks you to do
12.1 Nitrogen and sulfur
12.1.1 explain the lack of reactivity of nitrogen, with reference to
triple bond strength and lack of polarity
12.1.2 describe and explain: (a) the basicity of ammonia, using the Brønsted–Lowry
theory (b) the structure of the ammonium ion and its formation by an acid–base reaction
(c) the displacement of ammonia from ammonium salts by an acid–base reaction
12.1.3 state and explain the natural and man-made occurrences of oxides of nitrogen and their
catalytic removal from the exhaust gases of internal combustion engines
12.1.4 understand that atmospheric oxides of nitrogen (NO and NO2) can react with
unburned hydrocarbons to form peroxyacetyl nitrate, PAN, which is a component of photochemical
smog
12.1.5 describe the role of NO and NO2 in the formation of acid rain both directly
and in their catalytic role in the oxidation of atmospheric sulfur dioxide
Read the verbs, because they decide what an answer must contain. 12.1.1 says explain, and names the two reasons it wants: the strength of the triple bond and the lack of polarity of the molecule. "Nitrogen has a triple bond" is not an explanation until it says why the triple bond matters. 12.1.2 says describe and explain three times and each time names the theory or the reaction type — Brønsted–Lowry, acid–base, acid–base — so every answer in that statement is an argument about protons. 12.1.3 has two halves, where oxides of nitrogen come from and how they are removed. 12.1.5 says both directly and catalytically, which is a signal that an answer naming only one of the two roles is half an answer.
What this page covers, and what it leaves out
This page is topic 12 of the 2025–2027 syllabus and nothing else. The Haber process and the Contact process are not here: they are assessed as examples of equilibrium in 7.1 and are covered with the equilibria chapter. Sulfur appears in this topic only through sulfur dioxide and its oxidation in the atmosphere; the manufacture and reactions of sulfuric acid, the uses of ammonium fertilisers and their effect on rivers, and methods of removing sulfur dioxide from power-station flue gases were in older syllabuses and are not required now. A few ideas from other topics are used where the explanation needs them — bond energies (5.1), activation energy and catalysts (8.2, 8.3), σ and π bonds (13.3) — and each is flagged where it is used.
The page is in three parts. Part one is the nitrogen molecule and why it is unreactive. Part two is ammonia and the ammonium ion, seen entirely through the Brønsted–Lowry theory. Part three follows the oxides of nitrogen from the engine to the atmosphere. Each part ends with examination practice taken from past papers, and the final part collects what examiners have reported about this topic, the mistakes that lose marks, a self-test and a knowledge organiser.
The nitrogen molecule12.1.1
A nitrogen atom has the electron configuration 1s2 2s2 2p3: five outer electrons, three of them unpaired. Two nitrogen atoms share three pairs of electrons, which gives each atom an octet — three bonding pairs and one lone pair — and gives the molecule its triple bond.
:N≡N:
A triple bond is not three identical bonds. It is one σ bond and two π bonds. Topic 13.3 describes the atom forming it as sp hybridised: the 2s orbital mixes with one 2p orbital to give two sp hybrid orbitals pointing in opposite directions along the axis of the molecule. One sp orbital on each atom overlaps end-on (head-on) with its partner to form the σ bond; the other holds the lone pair and points away from the bond. That leaves two unhybridised 2p orbitals on each atom, at right angles to each other and to the bond axis. Each overlaps sideways with the matching p orbital on the other atom, giving two π bonds — one with its electron density above and below the axis, the other in front of and behind it (Figure 12.1).
This arrangement puts six bonding electrons between two small nuclei, and the atoms are pulled closer than in any other nitrogen–nitrogen bond. The numbers show how much the third pair of electrons adds.
| Bond | Example | Bond energy / kJ mol−1 | Bond length / nm |
|---|---|---|---|
| N–N | hydrazine, H2N–NH2 | 160 | 0.145 |
| N=N | azo compounds, –N=N– | 409 | 0.125 |
| N≡N | nitrogen, N2 | 944 | 0.110 |
The triple bond is almost six times as strong as the single bond — far more than three single bonds added together. It is one of the strongest covalent bonds known, and that single number drives most of part one.
Carbon monoxide has the same electrons
CO is isoelectronic with N2: both molecules have 14 electrons and a triple bond made of one σ and two π bonds. In CO, one of the three bonding pairs is a dative covalent bond, with both electrons supplied by oxygen. The comparison between the two molecules is the clearest way to see why lack of polarity is the second half of 12.1.1, and section 3 uses it.
Why nitrogen is so unreactive12.1.1
12.1.1 names two reasons, and a complete answer uses both.
Reason one: the triple bond is very strong
Almost every reaction of nitrogen begins by breaking, or at least weakening, the N≡N bond. With a bond energy of 944 kJ mol−1 this costs a great deal of energy, so the activation energy of the reaction is very high. At room temperature, only a vanishingly small fraction of collisions have enough energy to reach it, so the reaction is too slow to observe.
Notice what this argument is not saying. It is not saying that reactions of nitrogen are endothermic. Some are — making nitrogen monoxide from nitrogen and oxygen absorbs energy — but others are exothermic overall: forming ammonia from nitrogen and hydrogen releases energy, and magnesium burning in nitrogen releases a great deal. Nitrogen is unreactive in these reactions for a kinetic reason, not an energetic one. The energy released by forming new bonds only arrives after the old one has been broken, and the barrier is set by the bond that has to break first (Figure 12.2).
Reason two: the molecule has no polarity
Most reactions start with an attraction. A nucleophile is drawn to an atom carrying a partial positive charge; an electrophile is drawn to a region of high electron density or to an atom carrying a partial negative charge. In N2 the two atoms are identical, so they have the same electronegativity and share the bonding electrons exactly equally. There is no permanent dipole — no δ+ end and no δ− end — so there is nothing for an attacking nucleophile or electrophile to be attracted to. The molecule is also small, with its electrons held tightly, so it is not easily polarised by an approaching species either.
Why the second reason is needed: nitrogen against carbon monoxide
If bond strength were the whole story, carbon monoxide should be even less reactive than nitrogen: C≡O has a bond energy of 1077 kJ mol−1. In fact CO burns readily, is oxidised in every catalytic converter and binds to the iron in haemoglobin, which is why it is poisonous. Two differences explain this. First, CO is polar — oxygen is more electronegative than carbon — so it has sites that other species can be attracted to. Second, CO can react without the triple bond breaking completely: turning CO into CO2 only reduces a triple bond to a double one, C≡O → O=C=O. Nitrogen, being non-polar, offers no such foothold. The model below puts numbers on both points.
What a complete answer contains
Examiners have repeatedly reported that "triple bond" appears in almost every answer but often without the qualification that it is strong, or that it needs a lot of energy to break. The word "triple" earns nothing on its own. A strong answer to "explain the lack of reactivity of nitrogen" reads: the N≡N triple bond is very strong (bond energy 944 kJ mol−1), so a very large amount of energy is needed to break it and the activation energy is high; and the molecule is non-polar, so it does not attract electrophiles or nucleophiles. Note also that "strong triple bond" and "high activation energy" make essentially the same point; where a question carries two marks, the non-polarity is what earns the second.
Common trap: stability is not the same as inertness
Saying nitrogen is unreactive "because it is stable" or "because it has a full outer shell" loses the mark. Every atom in every stable molecule has a full outer shell; the ammonia molecule does, and ammonia is reactive. The syllabus wants the two specific features of this molecule: the strength of its triple bond and its lack of polarity.
What it takes to make nitrogen react12.1.1
Unreactive is not the same as inert. Nitrogen does react when enough energy is supplied, and the conditions in which it reacts tell the same story as section 3: every one of them is a way of getting past a very high activation energy.
| Reaction | Equation | Conditions | Where it matters |
|---|---|---|---|
| with oxygen | N2(g) + O2(g) → 2NO(g) | very high temperature: an electric spark or lightning, or the combustion in an engine cylinder | the source of the oxides of nitrogen in part three |
| with hydrogen | N2(g) + 3H2(g) ⇌ 2NH3(g) | iron catalyst, a few hundred °C, high pressure (the Haber process, topic 7) | ammonia for fertilisers |
| with magnesium | 3Mg(s) + N2(g) → Mg3N2(s) | magnesium heated strongly, or burning | a small amount of nitride forms when magnesium burns in air |
In each case the temperature is high because enough molecules must have an energy at least equal to the activation energy (topic 8.2), or a catalyst provides a route with a lower activation energy (8.3). Nitrogen in the air around you, at around 290 K and with nothing to catalyse its reactions, simply has no way over the barrier.
Worked example: magnesium nitride
A question tells you magnesium burns in nitrogen to form an ionic nitride and asks for the equation. Magnesium forms Mg2+; nitrogen, with five outer electrons, gains three to form N3−. The formula that balances the charges is Mg3N2 (3 × +2 = +6; 2 × −3 = −6). Then balance: three Mg atoms and one N2 molecule.
3Mg(s) + N2(g) → Mg3N2(s)
Examiners report that formulae such as MgN3, MgN or a nitrate appear often, and that some candidates write the correct formula and then do not balance the equation. Check the charges of the ions before writing any formula.
Connection: nitrogen in living things
Plants need nitrogen to make proteins, yet almost none of them can use N2 directly — for exactly the reasons in section 3. They depend on nitrogen that has already been "fixed" into ammonium or nitrate ions, by bacteria, by lightning or by the fertiliser industry. The difficulty of fixing nitrogen is why the Haber process was such an important invention, and why it still needs so much energy.
Ammonia as a Brønsted–Lowry base12.1.2(a)
Ammonia is a colourless gas with a sharp, choking smell, and it is extremely soluble in water — a flask of ammonia gas will suck water up into it in a fountain. The solution turns red litmus blue and has a pH of about 11. Something in the solution is producing hydroxide ions, yet ammonia contains no oxygen at all. The Brønsted–Lowry theory explains where they come from.
Brønsted–Lowry definitions (from topic 7.2)
A Brønsted–Lowry acid is a proton (H+ ion) donor. A Brønsted–Lowry base is a proton acceptor. An acid–base reaction is the transfer of a proton from the acid to the base.
To accept a proton, a species needs a lone pair of electrons with which to form a new bond to it: the proton has no electrons of its own to share. The nitrogen atom in NH3 has exactly that — three bonding pairs to hydrogen and one lone pair. When ammonia dissolves, some of its molecules use that lone pair to take a proton from a water molecule:
NH3(aq) + H2O(l) ⇌ NH4+(aq) + OH−(aq)
Water has given up a proton, so here water is acting as the acid; ammonia has accepted it, so ammonia is the base. The hydroxide ion left behind is what makes the solution alkaline. The reaction is reversible, and the reverse reaction is also a proton transfer — NH4+ giving its proton back to OH− — so the two sides contain two conjugate acid–base pairs:
| acid (proton donor) | its conjugate base | |
|---|---|---|
| pair 1 | NH4+ | NH3 |
| pair 2 | H2O | OH− |
The members of a conjugate pair differ by exactly one H+. That is the test to apply to any equation: look for two species whose formulae differ by one hydrogen and one unit of charge.
Ammonia reacts in the same way with acids, but much more completely, because an acid such as hydrochloric acid is a far better proton donor than water. With hydrogen chloride gas, the product is a white smoke of solid ammonium chloride — the classic test for either gas:
NH3(g) + HCl(g) → NH4Cl(s)
and in solution the reaction is between ammonia and the oxonium ion, H3O+, which is how a proton exists in water:
NH3(aq) + H3O+(aq) → NH4+(aq) + H2O(l)
In each case the product contains the ammonium ion, and in each case the reaction is a Brønsted–Lowry acid–base reaction: a proton moves from the acid to the lone pair on nitrogen.
How weak a base is ammonia?12.1.2(a)
The equilibrium in aqueous ammonia lies well to the left. In a typical bench solution only about one ammonia molecule in a hundred has accepted a proton at any moment; the rest are still NH3 molecules, hydrogen-bonded to water. That is what makes ammonia a weak base: it is only partially converted into ions in solution. Sodium hydroxide, by contrast, is a strong base: it is completely ionic, so every formula unit supplies a hydroxide ion.
The difference shows up in the pH. The model below solves the equilibrium exactly for any concentration you choose, using the base dissociation constant of ammonia, and sets the result beside sodium hydroxide of the same concentration.
How to think about it: the particle picture
Picture a litre of 0.10 mol dm−3 aqueous ammonia. At the macroscopic level, it smells of ammonia, turns universal indicator blue-purple and conducts electricity weakly. At the particle level, it is mostly NH3 molecules among water molecules, with roughly one in seventy-five ammonia molecules converted to an NH4+ ion with an OH− ion nearby. The symbolic equation, with its ⇌ sign, is the summary of both. The smell is evidence of the unreacted molecules, and the weak conductivity is evidence that ions are present but in small numbers.
Common trap: "partially dissolves"
Ammonia is extremely soluble in water — hundreds of volumes of gas dissolve in one volume of water. It is weak because it is only partially ionised once dissolved. Examiners regularly report answers describing a weak acid or base as one that "partially dissolves", which earns nothing. Dissolving and ionising are different processes; the word the definition needs is ionises or dissociates.
Forming the ammonium ion12.1.2(b)
When an ammonia molecule accepts a proton, the new N–H bond is formed from the nitrogen's lone pair. Both electrons in that bond come from the same atom, so it is a dative covalent (co-ordinate) bond. The proton has nothing to contribute but its positive charge, and the ion as a whole carries that charge.
NH3 + H+ → NH4+
Once the ion has formed, the dative bond cannot be told apart from the other three. All four N–H bonds are identical in length and strength, because once a pair of electrons is shared it makes no difference which atom it came from. The positive charge is not sitting on the new hydrogen: it belongs to the ion as a whole.
Exam alert: drawing the ammonium ion
When an examination asks for the dot-and-cross diagram of NH4+, the expected answer shows three ordinary shared pairs and one pair in which both electrons carry nitrogen's symbol, with brackets and the + charge. Examiners have reported that most candidates draw this correctly, but also that some answers show a nitrogen atom with five bonds — an "NH5" species — or fail to make the dative bond distinct from the others, or leave out the charge. Nitrogen in period 2 cannot hold more than eight outer electrons.
The shape of the ammonium ion12.1.2(b)
Electron-pair repulsion (topic 3.5) predicts both shapes. Nitrogen in ammonia has four pairs of electrons in its outer shell — three bonding pairs and one lone pair. Four pairs arrange themselves tetrahedrally, but only three of them have atoms at the end, so the molecule is trigonal pyramidal. The lone pair is held closer to the nitrogen than the bonding pairs are and repels them more strongly, squeezing the H–N–H angle down from 109.5° to about 107°.
Protonation turns the lone pair into a bonding pair. The ammonium ion has four bonding pairs and no lone pairs, all repelling equally, so it is a regular tetrahedron with every H–N–H angle 109.5° — the same shape as methane, with which it is isoelectronic. Rotate both below.
| NH3 | NH4+ | |
|---|---|---|
| electron pairs round N | 3 bonding, 1 lone | 4 bonding, 0 lone |
| shape | trigonal pyramidal | tetrahedral |
| H–N–H bond angle | 107° | 109.5° |
| dative bond present? | no | yes — one, indistinguishable once formed |
| Brønsted–Lowry behaviour | base (uses its lone pair) | weak acid (can give a proton back) |
| polarity | polar molecule | an ion, symmetrical |
The ammonium ion as a weak acid12.1.2
The ammonium ion is the conjugate acid of ammonia, and it can give its proton away again. In water, a small proportion of ammonium ions donate a proton to water molecules:
NH4+(aq) + H2O(l) ⇌ NH3(aq) + H3O+(aq)
so a solution of ammonium chloride is slightly acidic, with a pH a little above 5 at 0.1 mol dm−3. The chloride ion plays no part: it is the conjugate base of a strong acid and has no tendency to take a proton back. The ammonium ion is a weak acid because only a very small fraction of the ions donate their proton — the position of equilibrium lies far to the left.
Worked example: writing the equation examiners ask for
Question. In aqueous solution the ammonium ion acts as a weak Brønsted–Lowry acid. Write an equation to show this behaviour. Include state symbols.
Reasoning. "Acid" means the ammonium ion must lose a proton; "in aqueous solution" means the proton goes to water, not to a hydroxide ion; "weak" means the reaction is an equilibrium, so it needs ⇌. The proton on a water molecule makes the oxonium ion, which is aqueous.
NH4+(aq) + H2O(l) ⇌ NH3(aq) + H3O+(aq)
What goes wrong. Examiners reported that the equilibrium sign was rarely used, charges were often missing from the ions, H3O+ was given the state symbol (l), and some answers added OH− instead of water. The last one describes a different reaction — the one in section 10 — not the behaviour of the ion in water. Writing NH4+(aq) ⇌ NH3(aq) + H+(aq) is an acceptable shorthand in many contexts, but it does not show water acting as the base.
Displacing ammonia from ammonium salts12.1.2(c)
Warm any ammonium salt with a solution of sodium hydroxide and a sharp smell appears almost at once: ammonia gas is given off. A piece of damp red litmus paper held at the mouth of the tube turns blue. Ammonia is the only common gas that is alkaline, so this is the standard test for the ammonium ion.
Why it happens: an acid–base reaction
12.1.2(c) says by an acid–base reaction, and that is the key to the explanation. The ammonium ion is a proton donor; the hydroxide ion is a much stronger proton acceptor than ammonia. The hydroxide ion takes a proton from the ammonium ion:
NH4+(aq) + OH−(aq) → NH3(g) + H2O(l)
This is the reverse of the reaction in section 5, and it goes almost to completion for two reasons: hydroxide is a stronger base than ammonia, and warming drives the ammonia out of solution as a gas, so it is continuously removed from the equilibrium. The same ionic equation describes every ammonium salt with every soluble hydroxide, because the anion of the salt and the cation of the base are spectator ions. A basic oxide does the same job, since the oxide ion is an even stronger base than hydroxide:
2NH4+ + O2− → 2NH3 + H2O
The full equation depends on the salt and the base. The model builds it for any pair, balances it, and gives the ionic equation that sits underneath.
Think like a chemist: why farmers do not spread lime and ammonium fertiliser together
Acidic soils are treated with lime, calcium hydroxide, and ammonium salts are used as fertilisers. Applied together, the hydroxide ions displace ammonia from the ammonium ions, and the ammonia escapes into the air as a gas. The nitrogen that the crop was meant to take up is lost, and the lime is used up neutralising the fertiliser instead of the soil. The chemistry is exactly the equation above; only the setting is new.
Practical link: making ammonia in the laboratory
Heating an ammonium salt with a base is also how ammonia is prepared in the laboratory, since nitrogen cannot be persuaded to react with hydrogen on the bench. Ammonium chloride with calcium hydroxide is the usual pair. Ammonia is less dense than air and very soluble in water, so it is collected by upward delivery (downward displacement of air), never over water. It is toxic and irritating to the eyes and lungs, so the preparation is done in a fume cupboard.
Proton transfers, practised12.1.2
Every reaction in this part of the chapter is one proton moving from one species to another. The trainer gives you eight equations, several of them unfamiliar. For each, pick the species that acts as the Brønsted–Lowry acid; the page then identifies both conjugate pairs by finding the two pairs of species that differ by one H+. The answer is worked out from the formulae, not stored.
Common trap: water's role changes
In NH3 + H2O, water is the acid. In NH4+ + H2O, water is the base. Water is amphoteric in the Brønsted–Lowry sense, and its role is decided by its partner. Never label a species acid or base from memory; decide from whether it gains or loses a proton in the equation in front of you.
Examination practice: nitrogen and ammonia12.1.1–12.1.2
These questions are taken from past examination papers. Attempt each one before opening its worked answer. Where the examiners' report commented on a question, the insight is given with the answer. Structured-question answers are worked solutions that show the chemistry each mark needs.
Multiple choice
Structured questions
Past-paper question A1 · 1 mark
Compound P, C4N2(l), burns completely in oxygen: C4N2(l) + 4O2(g) → 4CO2(g) + N2(g). One of the products is nitrogen gas, N2(g).
Explain the lack of reactivity of nitrogen. [1]
Worked answer to A1
The N≡N triple bond is very strong (it needs a lot of energy to break), so the activation energy for its reactions is high. A further correct point is that the molecule is non-polar.
Examiner insight. A common error was not to state that the triple bond is strong. Naming the bond is not an explanation.
Past-paper question A2 · 8 marks
Ammonium iron(II) sulfate, (NH4)2Fe(SO4)2, contains the ammonium ion, NH4+.
(b)(i) Draw a 'dot-and-cross' diagram of an ammonium ion. Show outer shell electrons only. Use × to show electrons from nitrogen. Use ● to show electrons from hydrogen. [2]
(b)(ii) Suggest the shape of an ammonium ion and predict the bond angle. [2]
(c) In aqueous solution the ammonium ion acts as a weak Brønsted–Lowry acid.
(c)(i) Explain the meaning of the term weak Brønsted–Lowry acid. [2]
(c)(ii) Write an equation to show this behaviour of the ammonium ion in water. Include state symbols. [2]
Worked answer to A2
(b)(i) As Figure 12.3: nitrogen in the centre with four pairs of electrons, each pair shared with a hydrogen. Three pairs are one × and one ●; the fourth pair is two × (the dative bond, from nitrogen's lone pair). The whole ion in square brackets with the + charge outside. One mark for the three ordinary shared pairs, one for the dative pair shown as ×× and the charge.
(b)(ii) Tetrahedral; 109.5°. (Four bonding pairs, no lone pairs, equal repulsion.)
(c)(i) A Brønsted–Lowry acid is a proton donor; weak means it is only partially dissociated (ionised) in solution. Two ideas, one mark each.
(c)(ii) NH4+(aq) + H2O(l) ⇌ NH3(aq) + H3O+(aq). One mark for the correct species and charges, one for the ⇌ sign and correct state symbols.
Examiner insight. The dot-and-cross diagram with three covalent bonds and a dative bond, and the shape and bond angle, were answered correctly in many responses. Explaining why the acid was weak was harder, and "partially dissolving" was confused with partially dissociating. The equation in (c)(ii) proved challenging: ⇌ was rare, charges were omitted, H3O+(l) was written instead of H3O+(aq), and hydroxide ions were added instead of water.
Past-paper question A3 · 7 marks
Nitrogen molecules, N2(g), contain two atoms attracted to each other by a triple covalent bond.
(a) Describe how the triple covalent bond forms in a N2(g) molecule. Refer to orbital overlap and hybridisation in your answer. [3]
(c) N2(g) is very unreactive. It is difficult to make ammonia, NH3(g), directly from its elements but it can be made from NH4Cl(s). Identify a reagent and the conditions required to make NH3(g) from NH4Cl(s). [1]
(d) 25 cm3 of 0.10 mol dm−3 HCl(aq) is titrated with 0.10 mol dm−3 NH3(aq). HCl is a strong Brønsted–Lowry acid.
(d)(i) Describe what is meant by a strong Brønsted–Lowry acid. [2]
(d)(ii) NH3 is a weak base. Construct an equation that shows the behaviour of NH3 as a weak Brønsted–Lowry base when dissolved in water. [1]
Part (b) of this question is B4 in the practice at the end of part three.
Worked answer to A3
(a) Each nitrogen atom is sp hybridised. One sp orbital on each atom overlaps end-on (along the axis) to form a σ bond. The two remaining (unhybridised) p orbitals on each atom overlap sideways with those on the other atom to form two π bonds, at right angles to each other. Three marks: sp hybridisation; end-on overlap giving σ; sideways p–p overlap giving two π.
(c) Warm/heat with a base such as sodium hydroxide solution (or calcium hydroxide).
(d)(i) A proton (H+) donor that is fully dissociated (completely ionised) in solution.
(d)(ii) NH3(aq) + H2O(l) ⇌ NH4+(aq) + OH−(aq)
Examiner insight. This nitrogen question was the least confidently answered on its paper. In (a), few answers identified the sp hybrid orbitals of nitrogen or described the modes of overlap precisely, though most recalled one σ and two π bonds. Part (c) and (d)(i) were answered well. In (d)(ii), better-performing candidates included a suitable base and its conjugate acid in their answers. In practice that means an equation with water as the proton donor, producing NH4+ and OH−, and the ⇌ sign that "weak" calls for.
Where oxides of nitrogen come from12.1.3
Two oxides of nitrogen matter in this topic. Nitrogen monoxide, NO (also called nitrogen(II) oxide or nitric oxide), is a colourless gas. Nitrogen dioxide, NO2 (nitrogen(IV) oxide), is a brown, toxic gas and is the colour of the haze over a polluted city. Together they are often written NOx.
Both begin with the reaction section 3 said was so difficult: nitrogen and oxygen from the air combining.
N2(g) + O2(g) → 2NO(g) ΔH = +181 kJ mol−1
The reaction happens only at very high temperatures. There are two important places where air is heated that much.
| Source | What happens | Why the reaction can occur there |
|---|---|---|
| natural: lightning | a lightning discharge passes through air; N2 and O2 combine to NO | the air in the discharge channel reaches temperatures of thousands of kelvin, supplying the activation energy |
| man-made: internal combustion engines | air, not pure oxygen, is drawn into the cylinder with the fuel; some of its nitrogen combines with oxygen during combustion | the burning fuel–air mixture is very hot and at high pressure; in a petrol engine the spark also ignites it |
| man-made: furnaces and power stations | any high-temperature combustion in air | the same reason: high temperature |
Two points in the engine case are worth making explicit, because examiners report they are often missed. The nitrogen does not come from the fuel: petrol is a mixture of hydrocarbons. It comes from the air that is burned with the fuel. And the reaction does not need a catalyst: it is the high temperature in the cylinder that does it.
Once in the atmosphere, nitrogen monoxide is oxidised further by oxygen to nitrogen dioxide. This reaction is slow at the concentrations found in air, but it happens without any special conditions.
2NO(g) + O2(g) → 2NO2(g)
Why high temperature? A model of the equilibrium
The formation of NO is endothermic and, strictly, reversible. The model below computes how much NO air would contain if it were held at a given temperature until equilibrium was reached. It uses ideas from A Level topic 23 (entropy and Gibbs free energy) that you do not need for this topic; what matters here is the result, which shows why only lightning and combustion make significant amounts.
Think like a chemist: why the NO survives the cooling
If equilibrium held at every temperature, the NO made in an engine would turn back into nitrogen and oxygen as the exhaust cooled, because at 300 K the equilibrium lies almost entirely on the left. It does not, and the reason is section 3 again. The decomposition of NO also has a high activation energy, and the exhaust cools within a fraction of a second, so the reverse reaction becomes too slow to happen. The NO is "frozen" at the amount made in the hot cylinder. The high temperature is needed for the forward reaction to go at all; the rapid cooling is why its product is released.
Exam alert: answer the question that was asked
Examiners report two recurring errors here. When asked for the conditions under which oxides of nitrogen are formed in an engine, many candidates wrote about catalytic converters, which remove them. And when asked for a man-made source, "catalytic converter" was a common wrong answer. The formation answer needs: high temperature (in the engine), nitrogen and oxygen from the air, and the equation N2 + O2 → 2NO — with N2 as the reactant, not NO. Lightning is well known as the natural source.
Removing NOx: the catalytic converter12.1.3
The exhaust from a petrol engine contains three pollutants that a catalytic converter is designed to remove: carbon monoxide from incomplete combustion, oxides of nitrogen from the reaction in section 13, and unburnt hydrocarbons. The converter is a stainless-steel canister in the exhaust pipe, containing a ceramic honeycomb coated with a very thin layer of precious metals — platinum, palladium and rhodium. The honeycomb gives the metal a very large surface area for a very small mass of expensive catalyst, so the gases pass over as much catalyst as possible in the short time they spend in the converter.
The central reaction for 12.1.3 is a redox reaction between two of the pollutants. Carbon monoxide is a reducing agent and nitrogen monoxide is an oxidising agent, so on the catalyst surface each gets rid of the other:
2CO(g) + 2NO(g) → 2CO2(g) + N2(g)
Carbon is oxidised from +2 to +4; nitrogen is reduced from +2 to 0. Both products are harmless components of air. Nitrogen dioxide is removed in the same way, and the unburnt hydrocarbons are oxidised too, either by the oxides of nitrogen or by oxygen:
4CO(g) + 2NO2(g) → 4CO2(g) + N2(g)
C8H18(g) + 25NO(g) → 8CO2(g) + 9H2O(g) + 12½N2(g)
These equations are not to be memorised one by one. Every one of them is built the same way: the electrons lost by the carbon-containing species equal the electrons gained by the nitrogen oxide. The model does this for any pair.
What examiners are looking for
Examiners report that many equations for the catalytic converter show only the nitrogen half — NO turning into N2 — and miss the cross-redox reaction in which CO becomes CO2 at the same time. Other common errors: NO written as a reactant when the question names NO2, O2 or NO shown as a product, and equations with the right species that do not balance. When a question names the two pollutants, use exactly those two, and check the oxygen atoms last.
Common trap: the converter's own product
A catalytic converter turns carbon monoxide and hydrocarbons into carbon dioxide, which is a greenhouse gas. It does not reduce carbon dioxide emissions; it increases them slightly. A question asking which pollutant a converter removes is never answered by "carbon dioxide", and "reduced to carbon" is not a reaction it carries out.
Photochemical smog and PAN12.1.4
In a sunny city with heavy traffic, the air can fill with a brownish haze that irritates the eyes and lungs: photochemical smog. The word "photochemical" says that sunlight drives the reactions that make it. It forms from two groups of primary pollutants released from vehicle exhausts — oxides of nitrogen and unburnt hydrocarbons (often called volatile organic compounds, VOCs) — reacting in sunlight with oxygen to form secondary pollutants. One of those secondary pollutants is named by the syllabus: peroxyacetyl nitrate, PAN.
The name can be read off the structure: acetyl is the old name for the ethanoyl group, CH3CO–, the one found in ethanoic (acetic) acid; peroxy refers to the –O–O– link; and the –NO2 group on the end, joined through an oxygen atom, makes the nitrate part. The acetyl part comes from the hydrocarbon, and the nitrogen and extra oxygen come from nitrogen dioxide and the air. That is the whole of what 12.1.4 asks you to understand:
NO / NO2 + unburnt hydrocarbons + O2 —(sunlight)→ PAN (and other secondary pollutants, including ozone)
No equation is required. The reactions that make PAN involve free radicals and many steps, and the syllabus deliberately asks only that you understand that oxides of nitrogen react with unburned hydrocarbons to form PAN and that PAN is a component of photochemical smog. PAN is harmful because it irritates the eyes and respiratory system and damages plants.
Exam alert: name the type of pollution, and name PAN
Examiners report that recall of VOCs and PAN is generally good, but that descriptions of how PAN is produced are often imprecise — an answer needs both the oxides of nitrogen and the unburnt hydrocarbons. When a question asks for the product that contributes to smog, VOCs and carbon monoxide are common wrong answers: they are primary pollutants, not the product. When a question asks for the type of air pollution caused by oxides of nitrogen, the answer is "photochemical smog" (or "acid rain"), not the name of a gas. And unburnt hydrocarbons do not damage the ozone layer — that answer has been reported as a recurring error.
Acid rain: the direct role of NO and NO212.1.5
Rain is always slightly acidic, because carbon dioxide dissolves in it. Acid rain is rain made considerably more acidic by oxides of sulfur and nitrogen. Its effects include damage to limestone and marble buildings and statues, acidification of lakes and rivers so that fish and other organisms die, damage to trees and forests, and leaching of nutrients from soils.
Oxides of nitrogen contribute to it directly, by forming nitric acid. Nitrogen monoxide is first oxidised to nitrogen dioxide (section 13), and nitrogen dioxide then reacts with water and more oxygen:
4NO2(g) + O2(g) + 2H2O(l) → 4HNO3(aq)
Nitrogen dioxide also reacts with water on its own, giving a mixture of two acids — nitric acid and nitrous acid, HNO2:
2NO2(g) + H2O(l) → HNO3(aq) + HNO2(aq)
The second equation is a disproportionation: nitrogen starts at +4 and ends at +5 in HNO3 and +3 in HNO2. Examiners have reported that very few answers mention both acids, and that HNO2 is rarely seen.
Where the sulfur dioxide comes from
Fossil fuels — coal especially, and crude oil before it is refined — contain sulfur compounds as impurities. When the fuel is burned, the sulfur is oxidised to sulfur dioxide:
S + O2 → SO2
Sulfur dioxide dissolves in rain to form sulfurous acid (sulfuric(IV) acid). No oxidation number changes in this step — sulfur stays at +4 — so it is an acid–base reaction, not a redox one:
SO2(g) + H2O(l) → H2SO3(aq)
But the main acid in acid rain is sulfuric acid, H2SO4, in which sulfur is +6. Getting from +4 to +6 means oxidising SO2 to SO3, and that is the reaction in which the oxides of nitrogen play their second, catalytic role.
NO2 as a catalyst for oxidising SO212.1.5
Sulfur dioxide reacts with oxygen only very slowly in clean air. Nitrogen dioxide speeds this up by a two-step cycle. In the first step it oxidises sulfur dioxide to sulfur trioxide, and is itself reduced to nitrogen monoxide:
SO2(g) + NO2(g) → SO3(g) + NO(g)
In the second step, oxygen in the air oxidises the nitrogen monoxide back to nitrogen dioxide:
NO(g) + ½O2(g) → NO2(g) or 2NO(g) + O2(g) → 2NO2(g)
Add the two steps and the NO2 and NO cancel: the nitrogen dioxide used in step 1 is regenerated in step 2. The overall change is simply
SO2(g) + ½O2(g) → SO3(g)
and the sulfur trioxide then reacts with water in the rain:
SO3(g) + H2O(l) → H2SO4(aq)
Nitrogen dioxide meets every part of the definition of a catalyst. It increases the rate of the oxidation of sulfur dioxide, by providing an alternative route with a lower activation energy; it takes part in the reaction; and it is regenerated, so it is not used up and one molecule can oxidise many molecules of sulfur dioxide. Watch that happen in the model.
Worked example: a complete answer to "describe the role of oxides of nitrogen in acid rain"
A full answer to 12.1.5 has both halves.
Directly: NO is oxidised to NO2; NO2 reacts with water (and oxygen) to form nitric acid, HNO3 (and nitrous acid, HNO2): 2NO2 + H2O → HNO3 + HNO2.
Catalytically: SO2 + NO2 → SO3 + NO, then NO + ½O2 → NO2, so NO2 is regenerated and acts as a catalyst; then SO3 + H2O → H2SO4.
Examiners have stated that full credit for the catalytic part needed three equations — the two showing the catalysis and the one converting SO3 into sulfuric acid — together with a statement of the role of the oxides of nitrogen.
Common trap: "it is not used up" is not enough
Examiners report many statements that NO2 is "not used up", with no evidence. The evidence is the regeneration equation: NO2 is a reactant in step 1 and a product in step 2. They also report that many candidates did not know that the prime function of a catalyst is to increase the rate; that the regeneration equation is the one most often wrongly balanced; and that some answers write step 1 backwards, which turns the catalyst into a product of sulfur trioxide. Balance step 2 by counting oxygen atoms: NO has one, NO2 has two, so each NO needs one extra oxygen atom, which is ½O2.
Nitrogen's oxidation numbers12.1
Every nitrogen species in this chapter can be placed on a single scale. Nitrogen has five outer electrons, so its oxidation number runs from −3, when it has in effect gained three (as in ammonia), to +5, when it has in effect lost all five (as in nitric acid). Seeing them together makes the redox chemistry of part three easy to check: in the catalytic converter nitrogen moves down the scale, in the atmosphere it moves up, and in the ammonia chemistry of part two it does not move at all — which is exactly why those reactions are acid–base reactions rather than redox.
Cross-topic connection
The oxidation-number method used here is the one from topic 6. It is worth using on every equation in this chapter as a check. If an equation is meant to be acid–base, no oxidation number should change; if it is meant to be redox, the electrons lost and gained must balance.
Examination practice: oxides of nitrogen12.1.3–12.1.5
These questions are taken from past examination papers. Attempt each one before opening its worked answer; where the examiners' report commented on a question, the insight is given with the answer.
Multiple choice
Structured questions
Past-paper question B1 · 11 marks
Nitric acid, HNO3, can be made by reacting nitrogen dioxide with water. The enthalpy change for the reaction can be measured indirectly using a Hess' cycle.
3NO2(g) + H2O(l) → 2HNO3(l) + NO(g) ΔHr
| substance | ΔHf / kJ mol−1 |
|---|---|
| NO2(g) | 34.0 |
| H2O(l) | −286 |
| HNO3(l) | −173 |
| NO(g) | 91.1 |
(b) Complete the Hess' cycle using the values given in the table and hence calculate the enthalpy change, ΔHr, for this reaction. Show your working. [3]
(c) Nitrogen and oxygen do not react at normal atmospheric temperatures. Explain why. [2]
Nitrogen oxides can be formed naturally in the Earth's atmosphere from nitrogen and oxygen in the air.
(d) State one way that nitrogen oxides are produced naturally. [1]
(e) Nitrogen dioxide, NO2, acts as a homogeneous catalyst in the oxidation of atmospheric sulfur dioxide.
(e)(i) Explain why NO2 is described as a homogeneous catalyst. [3]
(e)(ii) Write equations which describe the two reactions occurring when NO2 acts as a catalyst in the formation of sulfur trioxide from sulfur dioxide. [2]
Worked answer to B1
(b) Both sides of the equation can be formed from the same elements, so ΔHr = ΣΔHf(products) − ΣΔHf(reactants).
Products: 2(−173) + 91.1 = −254.9 kJ mol−1. Reactants: 3(34.0) + (−286) = −184.0 kJ mol−1.
ΔHr = −254.9 − (−184.0) = −70.9 kJ mol−1.
Reasonableness: the reaction makes a strong acid from a gas and water and is known to be exothermic; a value of tens of kJ mol−1 is the right size. The commonest error is to forget to multiply by the coefficients 3 and 2. Note that the question's values (NO 91.1, NO2 34.0) are slightly different from those in this page's data table; always use the data given in the question.
(c) The N≡N triple bond is very strong / needs a lot of energy to break (1), so the activation energy is very high and at normal temperatures very few molecules have enough energy to react (1).
(d) Lightning (electrical storms).
(e)(i) Homogeneous: it is in the same phase (gas) as the reactants (1). Catalyst: it increases the rate of the reaction by providing an alternative route of lower activation energy (1), and it is regenerated / not used up at the end of the reaction (1).
(e)(ii) SO2 + NO2 → SO3 + NO (1); NO + ½O2 → NO2 (1).
Examiner insight. The examiners' report for another paper that asked the same thing noted that many candidates did not know that the prime function of a catalyst is to increase the rate, and that statements that NO2 is "not used up" were frequent but unsupported by the regeneration equation.
Past-paper question B2 · 9 marks
Gases produced in internal combustion engines include carbon monoxide, oxides of nitrogen such as NO2, and unburnt hydrocarbons. These gases are removed from the exhaust before they can enter the atmosphere.
(c)(i) State what is used to remove these gases from the exhaust. [1]
(c)(ii) Write one equation to show how both carbon monoxide, CO, and nitrogen dioxide, NO2, are removed from the exhaust. [1]
(c)(iii) State the environmental consequence of allowing unburnt hydrocarbons to enter the atmosphere. [1]
(d) Vehicle fuels are treated to remove sulfur. If sulfur is present in a fuel when it is burned, SO2 is produced and may be released into the atmosphere where it can form acid rain.
(d)(i) Acid rain can contribute to breathing difficulties. Identify two other consequences of acid rain in the atmosphere. [2]
(d)(ii) NO2 is involved in the production of acid rain from SO2. Give two equations which describe how acid rain is formed by the action of NO2 with SO2. [2]
(d)(iii) NO2 is described as a catalyst during this process. Explain, with the use of an appropriate equation, why NO2 is described as a catalyst. [2]
Worked answer to B2
(c)(i) A catalytic converter.
(c)(ii) 4CO + 2NO2 → 4CO2 + N2 (or 2CO + NO2 → 2CO2 + ½N2). The electrons check: four CO lose 2 each (8); two NO2 gain 4 each (8).
(c)(iii) Formation of photochemical smog (unburnt hydrocarbons react with oxides of nitrogen in sunlight to form PAN).
(d)(i) Any two of: erosion/corrosion of buildings and statues made of limestone or marble; acidification of lakes and rivers, killing fish and other aquatic life; damage to trees, forests and crops; leaching of nutrients (or toxic metal ions) from soils.
(d)(ii) SO2 + NO2 → SO3 + NO and SO3 + H2O → H2SO4.
(d)(iii) NO2 is regenerated: NO + ½O2 → NO2 (1), so it is not used up overall — it is a reactant in the first step and a product in the second (1).
Examiner insight. Most candidates named the catalytic converter. The equation in (c)(ii) proved challenging: some showed NO rather than NO2 as a reagent, others showed O2 or NO as a product, and some had the right species but did not balance. In (c)(iii) very few answers referred to photochemical smog, and some wrongly said unburnt hydrocarbons damage the ozone layer. The consequences of acid rain and the acid-rain equations were generally well known; explaining why NO2 is a catalyst was harder, and not all regeneration equations were balanced.
Past-paper question B3 · 2 marks
SO2 and SO3 are found in the atmosphere. The oxidation of SO2 to SO3 in the atmosphere is catalysed by NO2. The first step of the catalytic oxidation is shown in equation 1.
equation 1 SO2(g) + NO2(g) → SO3(g) + NO(g)
(d)(i) Construct an equation to show how NO2 is regenerated in the catalytic oxidation of SO2. [1]
(d)(ii) NO2 can also react with unburned hydrocarbons to form photochemical smog. State the product of this reaction that contributes to photochemical smog. [1]
Worked answer to B3
(d)(i) NO(g) + ½O2(g) → NO2(g) (or 2NO + O2 → 2NO2).
(d)(ii) Peroxyacetyl nitrate, PAN.
Examiner insight. Some candidates reversed equation 1, ignoring the catalytic role of NO2 in the overall process SO2 + ½O2 → SO3. In (d)(ii) many varied incorrect answers were given, including VOCs and, commonly, CO.
Past-paper question B4 · 3 marks
Nitrogen oxides, NO2 and NO, are produced in internal combustion engines. Release of these gases into the atmosphere leads to the formation of photochemical smog.
(b)(i) Outline how nitrogen oxides are involved in the formation of photochemical smog. [2]
(b)(ii) Construct an equation to demonstrate how a catalytic converter reduces the amount of nitrogen oxide gases released into the atmosphere. [1]
Parts (a), (c) and (d) of this question are A3 in part two.
Worked answer to B4
(b)(i) Nitrogen oxides react with unburnt hydrocarbons (VOCs) in sunlight (1) to form PAN, peroxyacetyl nitrate, a component of photochemical smog (1).
(b)(ii) 2CO + 2NO → 2CO2 + N2 (or 4CO + 2NO2 → 4CO2 + N2).
Examiner insight. Recall of VOCs and of PAN was good, with many able to link the creation of one from the other. In (b)(ii), many answers focused solely on nitrogen and did not include the cross redox process involving CO/CO2 and NOx/N2.
Past-paper question B5 · 6 marks
Molecule M is present in petrol. M is a saturated, non-cyclic hydrocarbon containing eight carbon atoms. When petrol is burned in an internal combustion engine, oxides of nitrogen are released into the atmosphere. Oxides of nitrogen are responsible for the formation of acid rain.
(b)(i) Suggest the conditions required for the production of oxides of nitrogen during combustion of M in an internal combustion engine. Use an appropriate equation in your answer. [2]
(b)(ii) Describe how acid rain is formed in the atmosphere in the presence of oxides of nitrogen and SO2. Identify the role of the oxides of nitrogen in this process. Include all relevant equations. [3]
(b)(iii) State one other type of air pollution that is caused by the production of oxides of nitrogen in an internal combustion engine. [1]
Worked answer to B5
(b)(i) High temperature (and pressure) in the engine, which allows nitrogen and oxygen from the air to react (1): N2 + O2 → 2NO (1).
(b)(ii) SO2 + NO2 → SO3 + NO; NO + ½O2 → NO2; SO3 + H2O → H2SO4; the oxides of nitrogen act as a catalyst (NO2 is regenerated).
(b)(iii) Photochemical smog.
Examiner insight. In (b)(i) a large proportion of candidates wrongly treated this as a question about catalytic converters, and many answers used NO instead of N2 as the reactant. In (b)(ii) only a small number identified the catalytic role of NOx and gave all the relevant equations; many gave only the equation for forming H2SO4, and a significant number wrote about forming HNO3 instead. (b)(iii) was well answered, although some gave the identity of a gas instead of the type of pollution.
Examiner's overall observation12
Examiners' reports on the structured AS papers have commented on this topic in most recent sessions. Taken together, they give a consistent picture.
What candidates generally do well. Lightning is well known as the natural source of oxides of nitrogen, and the catalytic converter is readily named as the device that removes them. The consequences of acid rain, and the equations for sulfuric acid formation, are usually known. The dot-and-cross diagram of the ammonium ion, with three covalent bonds and one dative bond, is usually correct, as are its tetrahedral shape and 109.5° bond angle. Recall of VOCs and PAN as the ingredients and product of photochemical smog has been good, and in the multiple-choice papers questions on what a catalytic converter removes, and on which pollutant contributes to acid rain, were among those candidates found easiest.
Where marks are lost on nitrogen itself. "Triple bond" appears in almost every answer on the unreactivity of nitrogen, but often without the word strong or any reference to the energy needed to break it. Examiners have pointed out that "strong triple bond" and "high activation energy" are in effect one point, so a two-mark answer needs a different second idea — the lack of polarity. Questions on how the triple bond forms expose weak knowledge of orbitals: few candidates identify sp hybridisation of nitrogen, the modes of overlap are described imprecisely, and diagrams of a π bond often do not show sideways overlap of p orbitals. Formulae for magnesium nitride are frequently wrong or left unbalanced.
Where marks are lost on ammonia and the ammonium ion. Definitions of a weak acid or base confuse "partially dissolving" with partially dissociating. The equation for the ammonium ion acting as a weak acid in water is poorly done: the equilibrium sign is rare, charges are omitted, H3O+ is given the state (l), and hydroxide ions are written in place of water. Some diagrams show nitrogen with five bonds, as though NH5 existed, or fail to distinguish the dative bond and the charge.
Where marks are lost on the oxides of nitrogen. Questions on how oxides of nitrogen are formed in engines are frequently answered as if they were about catalytic converters, and "catalytic converter" is given as a man-made source. The equation for formation is often written with NO rather than N2 as the reactant, and the fact that the nitrogen comes from the air is often missing. Equations for the catalytic converter frequently show only the nitrogen change and leave out the simultaneous oxidation of CO to CO2, use the wrong oxide of nitrogen, or are not balanced. For photochemical smog, descriptions of how PAN forms lack precision, "VOCs" or "CO" is given when the product is asked for, and the name of a gas is given when the type of pollution is asked for; some answers claim hydrocarbons damage the ozone layer.
The weakest area: the catalytic role of NO2. Only a small number of candidates give all the relevant equations and identify the catalytic role. Many give only the formation of sulfuric acid, or write about nitric acid instead. The regeneration equation causes the most difficulty and is often wrongly balanced; some answers write the first step backwards. Statements that NO2 is "not used up" are common but unsupported by an equation, and many candidates do not state that a catalyst increases the rate. On the direct role, very few answers mention both HNO3 and HNO2. Examiners have also noted that dissolving SO2 in water to form H2SO3 involves no change in oxidation number, and that SO3 + H2 and H2SO4 are common wrong products for that step.
What a successful response demonstrates. It answers the question actually asked (formation versus removal; product versus primary pollutant; type of pollution versus a gas); it supports every explanation with the specific chemistry — a strong bond and a high activation energy, a lone pair accepting a proton, a catalyst regenerated by a stated equation; and it writes complete, balanced equations with the right species, charges, state symbols and ⇌ where the reaction is an equilibrium.
Things that lose marks in this topic12
Each row is a misconception reported by examiners or one that leads directly to a reported error. Read across: the incorrect idea, why it fails, the correct model, and what it costs.
| Misconception | Why it is wrong | Correct model | Examination consequence |
|---|---|---|---|
| Nitrogen is unreactive because it has a triple bond. | Carbon monoxide has a triple bond and is reactive. What matters is the strength. | N≡N is very strong (944 kJ mol−1): high Ea. The molecule is also non-polar. | Mark for bond strength withheld; second mark lost if polarity is not given. |
| Nitrogen reactions are all endothermic, so they do not happen. | Forming NH3 and Mg3N2 is exothermic. | The barrier is kinetic: activation energy, not ΔH. | Explanation not credited. |
| A weak base partially dissolves. | Ammonia is extremely soluble. | A weak base is partially ionised (protonated) in solution. | Definition mark lost. |
| NH4+ has one bond that is different from the others. | Once formed, a shared pair is a shared pair. | All four N–H bonds are identical; the ion is a regular tetrahedron, 109.5°. | Shape or angle mark lost if 107° is given. |
| Nitrogen can form five bonds in NH4Cl. | Period 2 atoms cannot exceed eight outer electrons. | NH4+ and Cl− are separate ions. | Diagram not credited. |
| NH4+ + OH− shows the ammonium ion acting as an acid in water. | That is the reaction with an alkali, not with water. | NH4+ + H2O ⇌ NH3 + H3O+. | Equation marks lost. |
| Damp blue litmus turning red shows ammonia. | Ammonia is alkaline. | Damp red litmus turns blue. | Observation mark lost; a multiple-choice distractor. |
| The nitrogen in NOx comes from the fuel, or forms in the catalytic converter. | Petrol is hydrocarbons; the converter removes NOx. | N2 and O2 from the air react at the high temperature in the engine. | Formation answer not credited. |
| A catalytic converter changes NO to N2 on its own. | Reduction needs a reducing agent. | 2CO + 2NO → 2CO2 + N2: cross-redox. | Equation mark lost. |
| A catalytic converter removes CO2. | It produces CO2. | It removes CO, NOx and unburnt hydrocarbons. | Multiple-choice distractor. |
| VOCs or CO are the product that causes photochemical smog. | They are primary pollutants. | NOx + unburnt hydrocarbons + sunlight → PAN. | Mark lost. |
| Unburnt hydrocarbons damage the ozone layer. | A different topic, not caused by hydrocarbons from exhausts. | They cause photochemical smog. | Mark lost. |
| NO2 is a catalyst because it is "not used up". | An assertion, not an explanation. | It increases the rate and is regenerated: NO + ½O2 → NO2. | Explanation mark lost. |
| Acid rain from NO2 is only nitric acid. | NO2 disproportionates in water. | 2NO2 + H2O → HNO3 + HNO2. | Incomplete answer. |
| SO2 + H2O is a redox reaction giving H2SO4. | Sulfur stays at +4. | SO2 + H2O → H2SO3; oxidation to +6 needs the NO2 cycle. | Equation mark lost. |
Self-test12
Twelve questions written for this page, covering the whole unit. Each gives the reasoning behind the answer once you have chosen.
Definitions to learn12
| Term | Definition |
|---|---|
| Brønsted–Lowry acid | a proton (H+) donor |
| Brønsted–Lowry base | a proton (H+) acceptor |
| conjugate acid–base pair | two species that differ by one proton, e.g. NH4+/NH3 |
| weak base | a base that is only partially ionised (protonated) in aqueous solution |
| weak acid | an acid that is only partially dissociated (ionised) in aqueous solution |
| dative covalent (co-ordinate) bond | a covalent bond in which both electrons of the shared pair come from the same atom |
| activation energy, Ea | the minimum energy colliding particles need in order to react |
| catalyst | a substance that increases the rate of a reaction by providing an alternative route with a lower activation energy, and is regenerated at the end of the reaction |
| homogeneous catalyst | a catalyst in the same phase as the reactants |
| heterogeneous catalyst | a catalyst in a different phase from the reactants |
| photochemical smog | air pollution formed when oxides of nitrogen and unburnt hydrocarbons react in sunlight; PAN is a component |
| PAN | peroxyacetyl nitrate, CH3C(O)OONO2, a secondary pollutant in photochemical smog |
| acid rain | rain made more acidic than normal by dissolved oxides of sulfur and nitrogen (sulfuric, nitric and nitrous acids) |
| disproportionation | a redox reaction in which the same element is simultaneously oxidised and reduced |
Summary and examination checklist12
Essential knowledge
- N2 is unreactive because its N≡N bond is very strong (944 kJ mol−1), giving reactions a very high activation energy, and because the molecule is non-polar, offering no site for electrophiles or nucleophiles.
- Ammonia is a weak Brønsted–Lowry base: the lone pair on nitrogen accepts a proton. In water: NH3 + H2O ⇌ NH4+ + OH−, far to the left.
- NH4+ forms when that lone pair makes a dative covalent bond to H+. It is tetrahedral, 109.5°; all four bonds are identical once formed.
- NH4+ is a weak acid; ammonium chloride solution is slightly acidic.
- Warming an ammonium salt with a base releases NH3: NH4+ + OH− → NH3 + H2O. Damp red litmus turns blue.
- NO forms from N2 and O2 at high temperature: naturally in lightning; man-made in internal combustion engines (and other high-temperature combustion). NO is oxidised to NO2 in air.
- Catalytic converters (Pt/Pd/Rh on a honeycomb) remove NOx by cross-redox with CO (and hydrocarbons): 2CO + 2NO → 2CO2 + N2.
- NOx + unburnt hydrocarbons in sunlight → PAN, a component of photochemical smog.
- Directly: NO2 forms HNO3 (and HNO2) in rain. Catalytically: SO2 + NO2 → SO3 + NO; NO + ½O2 → NO2; then SO3 + H2O → H2SO4.
Examination checklist
- Did I say the triple bond is strong, and add non-polarity for the second mark?
- Does every Brønsted–Lowry answer mention a proton being donated or accepted, and the lone pair on N?
- Do my equilibrium equations for NH3 and NH4+ in water have ⇌, charges and (aq)/(l)?
- Is my NH4+ diagram showing a distinct dative pair, brackets and the + charge?
- Did I answer formation questions about the engine and removal questions about the converter?
- Does my converter equation contain both a nitrogen oxide and CO (or a hydrocarbon), and balance?
- For smog, did I name PAN as the product and "photochemical smog" as the type of pollution?
- For the catalytic role, did I give both equations, the regeneration, and the words "increases the rate"?
- Did I use the data given in the question rather than remembered values?
Knowledge organiser12
| Idea | Key equation or fact | Conditions / observations | Must-remember distinction |
|---|---|---|---|
| N2 unreactive | N≡N, 944 kJ mol−1; 1σ + 2π; non-polar | reacts only at high T or with a catalyst | strong bond (kinetic barrier), not "stable" or "full shell" |
| NH3 as a base | NH3 + H2O ⇌ NH4+ + OH− | pH ≈ 11 at 0.1 mol dm−3; ~1% ionised | partially ionised, not partially dissolved |
| NH4+ formation | NH3 + H+ → NH4+ (dative bond) | NH3 + HCl → NH4Cl: white smoke | NH3 107° pyramidal; NH4+ 109.5° tetrahedral |
| NH4+ as an acid | NH4+ + H2O ⇌ NH3 + H3O+ | NH4Cl(aq) slightly acidic | water is the base here |
| Displacement of NH3 | NH4+ + OH− → NH3 + H2O | warm; damp red litmus → blue | acid–base, no oxidation-number change |
| Formation of NOx | N2 + O2 → 2NO; 2NO + O2 → 2NO2 | lightning; engines (high T) | N2 from air, not fuel |
| Catalytic converter | 2CO + 2NO → 2CO2 + N2 | Pt/Pd/Rh on ceramic honeycomb | heterogeneous; cross-redox |
| Photochemical smog | NOx + hydrocarbons + O2 (sunlight) → PAN | PAN = CH3C(O)OONO2 | PAN is the product; VOCs and NOx are primary |
| Acid rain, direct | 2NO2 + H2O → HNO3 + HNO2 | NO2 is brown | two acids, disproportionation |
| Acid rain, catalytic | SO2 + NO2 → SO3 + NO; NO + ½O2 → NO2 | then SO3 + H2O → H2SO4 | homogeneous; regenerated; increases rate |
Data used on this page12
These are the values the models on this page compute with. They are reference values chosen to be consistent with one another; they are not the official data booklet, and values in other sources differ by a few units. In an examination, always use the data printed in the question or the data booklet supplied.
Bond energies
| Bond | Bond energy / kJ mol−1 | Used for |
|---|---|---|
| N≡N | 944 | the unreactivity of nitrogen |
| N=N | 409 | comparison, Table 12.1 |
| N–N | 160 | comparison, Table 12.1 |
| C≡O (in carbon monoxide) | 1077 | the comparison with CO |
| C=O (in carbon dioxide) | 805 | oxidising CO |
| O=O | 496 | reactions with oxygen |
| H–H | 436 | forming ammonia |
| N–H | 390 | forming ammonia |
Thermochemical and equilibrium data at 298 K
| Quantity | Value | Used for |
|---|---|---|
| ΔHf of NH3(g) | −46 kJ mol−1 | checking the bond-energy estimate for ammonia |
| ΔHf of NO(g) | +90.3 kJ mol−1 | the NO equilibrium model |
| ΔHf of CO(g) | −111 kJ mol−1 | checking the CO oxidation estimate |
| ΔHf of CO2(g) | −394 kJ mol−1 | checking the CO oxidation estimate |
| ΔS for N2 + O2 → 2NO | +24.8 J K−1 mol−1 | the NO equilibrium model (taken as constant with temperature) |
| Kb of NH3 | 1.8 × 10−5 mol dm−3 | the pH of aqueous ammonia |
| Ka of NH4+ | 5.6 × 10−10 mol dm−3 | the pH of ammonium chloride solution |
| Kw | 1.0 × 10−14 mol2 dm−6 | converting [OH−] to pH |
| Composition of dry air (by volume) | 78% N2, 21% O2 | the NO equilibrium model (the remaining 1%, mainly argon, is taken as unreactive) |
| Electronegativity (Pauling): H, C, N, O | 2.2, 2.6, 3.0, 3.4 | bond polarity in N2, CO and NH3 |
| Ar: H, C, N, O | 1.0, 12.0, 14.0, 16.0 | masses in the catalytic-converter model |