Nitrogen and sulfur

1. Overview

Nitrogen makes up most of the air, yet it hardly reacts. This section explains why, then shows how ammonia acts as a base, forming the ammonium ion, and how a stronger base drives ammonia back out of ammonium salts. It also covers the oxides of nitrogen: how they form in lightning and engines, how catalytic converters remove them, and how they lead to photochemical smog and acid rain, where they also help turn sulfur dioxide into sulfuric acid.

Key Definitions

• Brønsted-Lowry base: a species that accepts a proton (an H⁺ ion) from another species; ammonia does this with the lone pair on its nitrogen atom.
• Coordinate bond (dative covalent bond): a covalent bond in which both electrons of the shared pair come from the same atom, as when the lone pair on nitrogen in NH₃ bonds to an H⁺ ion.
• Ammonium ion: the ion NH₄⁺, formed when ammonia accepts a proton; it has four identical N-H bonds and no lone pair, so it is tetrahedral, with bond angles of 109.5°.
• Non-polar bond: a covalent bond between atoms of equal electronegativity, so the bonding electrons are shared equally and neither atom carries a partial charge, as in N≡N.
• Catalytic converter: a unit in a vehicle's exhaust system in which platinum, palladium and rhodium catalyse reactions that turn NO, NO₂, CO and unburned hydrocarbons into N₂, CO₂ and H₂O.
• Photochemical smog: a haze of air pollutants, including PAN, that forms when sunlight acts on a mixture of oxides of nitrogen and unburned hydrocarbons.
• Peroxyacetyl nitrate (PAN): the compound CH₃C(O)OONO₂, formed in the air when oxides of nitrogen react with unburned hydrocarbons; a component of photochemical smog that irritates the eyes and lungs.
• Acid rain: rain that is more acidic than natural rain (which is slightly acidic from dissolved carbon dioxide), mainly because of sulfuric acid and nitric acid formed from SO₂ and oxides of nitrogen.
• Catalyst: a substance that increases the rate of a reaction by providing a route with a lower activation energy and is chemically unchanged at the end; NO₂ is used up in one step of the oxidation of SO₂ and re-formed in the next.

Content

Why nitrogen is so unreactive

Nitrogen, N₂, is about 78% of the air by volume, yet almost nothing reacts with it at room temperature. Two features of the molecule explain this.

1. A very strong triple bond. The atoms are joined by N≡N: one σ bond and two π bonds (topic 3.4). Its bond energy, 944 kJ mol⁻¹, is the largest for any bond between two atoms of the same element in the Data Booklet. Almost every reaction of N₂ must break at least part of this bond, so the activation energy, $E_{\text{A}}$, is very high. At room temperature, hardly any collisions have energy equal to or greater than $E_{\text{A}}$.

2. No polarity. Both atoms have the same electronegativity (3.0), so the bonding electrons are shared equally and the bond is non-polar. Many reagents react by being drawn to a δ+ or δ− atom: OH⁻ to a δ+ atom, H⁺ to an electron-rich δ− atom. N₂ has neither.

Ammonia is a useful contrast: its N-H bonds are polar (N 3.0, H 2.1), and the lone pair on its nitrogen readily accepts a proton, so it reacts at once with acids.

Nitrogen does react when enough energy is supplied or a catalyst lowers $E_{\text{A}}$: with hydrogen in the Haber process (iron catalyst, high pressure, about 400 to 450 °C, topic 7.1), and with oxygen in lightning and inside hot engines (see below).

Worked example 1: opening up the N≡N bond

Question: Use these Data Booklet bond energies to estimate the energy needed to turn a triple bond into a double bond for nitrogen and for carbon: N≡N 944, N=N 410, C≡C 840 and C=C 610 kJ mol⁻¹. Use the results to explain why molecules with a C≡C bond undergo addition reactions readily but N₂ does not.

Step 1: nitrogen. Going from a triple to a double bond breaks one π bond, so the difference estimates that bond's energy: $$944 - 410 = 534\ \text{kJ mol}^{-1}$$

Step 2: carbon. $$840 - 610 = 230\ \text{kJ mol}^{-1}$$

Step 3: compare. $$\frac{534}{230} = 2.3$$

Answer: about 534 kJ mol⁻¹ for nitrogen and 230 kJ mol⁻¹ for carbon. An addition reaction begins by breaking one π bond, and in N≡N this takes more than twice as much energy, so $E_{\text{A}}$ for adding to N₂ is far higher. N₂ also has no δ+ or δ− atom to attract a reagent. These are estimates: 944 is exact for N₂, but the other three values are averages over many molecules. Topic 3.4 estimates π bonds another way (taking the σ part as a single bond), which gives different numbers, but both methods put nitrogen's π bonds well above carbon's.

Ammonia as a Brønsted-Lowry base

In the Brønsted-Lowry theory (topic 7.2), a base is a proton acceptor. Ammonia, NH₃, has a lone pair on nitrogen that can form a bond to an H⁺ ion, so ammonia is a base.

In water, ammonia takes a proton from a water molecule:

NH₃(aq) + H₂O(l) ⇌ NH₄⁺(aq) + OH⁻(aq)

Water is the acid and ammonia the base. The OH⁻ ions, which come from the water, make the solution alkaline. Only a small fraction of the NH₃ reacts, so ammonia is a weak base and the equation needs ⇌ (topic 7.2).

With acids, the proton transfer goes to completion: NH₃(aq) + H⁺(aq) → NH₄⁺(aq). For example:

• 2NH₃(aq) + H₂SO₄(aq) → (NH₄)₂SO₄(aq)
• NH₃(aq) + HNO₃(aq) → NH₄NO₃(aq)

Both salts are made on a large scale as fertilisers. No water is needed: where ammonia gas meets hydrogen chloride gas, a white smoke of solid ammonium chloride forms, NH₃(g) + HCl(g) → NH₄Cl(s).

The ammonium ion

When NH₃ accepts a proton, its lone pair forms a coordinate (dative covalent) bond to the H⁺, which brings no electrons of its own. The product is the ammonium ion, NH₄⁺. This is an acid-base reaction, not a redox reaction: nitrogen's oxidation number is −3 in both NH₃ and NH₄⁺.

Once formed, the coordinate bond is identical to the other three N-H bonds. Four bonding pairs and no lone pair repel equally, so the ion is tetrahedral (topic 3.5).

	NH₃	NH₄⁺
Bonding pairs on N	3	4
Lone pairs on N	1	0
Shape	pyramidal	tetrahedral
H-N-H angle	107°	109.5°

NH₄⁺ has 10 electrons, like a methane molecule, and the same shape. Its dot-and-cross diagram is in topic 3.7. NH₄⁺ can give its proton back, so it is a weak acid, and solutions of salts such as ammonium chloride and ammonium sulfate are slightly acidic (topic 7.2).

Displacing ammonia from ammonium salts

Warm an ammonium salt with a base such as sodium hydroxide, and ammonia gas is given off:

NH₄⁺(aq) + OH⁻(aq) → NH₃(g) + H₂O(l)

This is also an acid-base reaction: NH₄⁺ donates a proton to OH⁻. Hydroxide is a much stronger base than ammonia, so it takes the proton and ammonia is set free (displaced). Warming drives the very soluble ammonia out of solution as a gas.

• (NH₄)₂SO₄(aq) + 2NaOH(aq) → Na₂SO₄(aq) + 2NH₃(g) + 2H₂O(l)
• 2NH₄Cl(s) + Ca(OH)₂(s) → CaCl₂(s) + 2NH₃(g) + 2H₂O(g), on heating the solids, a common laboratory preparation

Testing for NH₄⁺. Warm the sample with aqueous sodium hydroxide. Hold damp red litmus paper in the mouth of the test-tube, touching neither the liquid nor the glass. If NH₄⁺ is present, the gas given off turns the paper blue.

An acid never releases ammonia. An acid can only offer another proton, and NH₄⁺ has no lone pair to accept one. Releasing NH₃ means removing a proton, which is what a base does. For the same reason, farmers should not spread slaked lime, calcium hydroxide, at the same time as an ammonium fertiliser: the base displaces NH₃, and the nitrogen escapes into the air instead of feeding the crop.

Worked example 2: ammonium sulfate in a fertiliser

Question: A 1.50 g sample of fertiliser contains ammonium sulfate, (NH₄)₂SO₄, mixed with substances that do not release ammonia. It is warmed with excess aqueous sodium hydroxide, and all of the ammonia is passed into 50.0 cm³ of 0.500 mol dm⁻³ hydrochloric acid, which is in excess. Titrating the leftover acid, with methyl orange indicator, needs 25.10 cm³ of 0.200 mol dm⁻³ sodium hydroxide. Calculate the percentage by mass of ammonium sulfate. ($A_{\text{r}}$: H 1.0, N 14.0, O 16.0, S 32.1)

Step 1: ratios. (NH₄)₂SO₄(s) + 2NaOH(aq) → Na₂SO₄(aq) + 2NH₃(g) + 2H₂O(l), so 1 mol of salt gives 2 mol of NH₃. NH₃ reacts with HCl 1 : 1, and so does NaOH.

Step 2: HCl at the start. $$n = 0.500 \times \frac{50.0}{1000} = 0.0250\ \text{mol}$$

Step 3: HCl left over (equal to the NaOH in the titre): $$n = 0.200 \times \frac{25.10}{1000} = 0.005020\ \text{mol}$$

Step 4: HCl used by ammonia, equal to the amount of NH₃: $$0.0250 - 0.005020 = 0.01998\ \text{mol}$$

Step 5: ammonium sulfate, two NH₄⁺ ions per formula unit: $$n = \frac{0.01998}{2} = 0.009990\ \text{mol}$$

Step 6: mass and percentage. $M_{\text{r}}$ = 2(14.0 + 4 × 1.0) + 32.1 + 4 × 16.0 = 132.1 $$m = 0.009990 \times 132.1 = 1.320\ \text{g}$$ $$\frac{1.320}{1.50} \times 100 = 88.0\%$$

Answer: 88.0% ammonium sulfate by mass (3 s.f.). Missing the 2 : 1 ratio gives 176%, impossible for a percentage. Methyl orange changes colour while the solution is still acidic, so the NaOH neutralises only the leftover HCl, not the weakly acidic NH₄⁺ ions (choosing an indicator: topic 7.2).

Where oxides of nitrogen come from

Nitrogen monoxide, NO, is a colourless gas. Nitrogen dioxide, NO₂, is a brown, toxic, acidic gas. Together they are often written NOx. Both start when nitrogen and oxygen from the air react at a very high temperature:

N₂(g) + O₂(g) → 2NO(g)

Once NO has cooled, it reacts with more oxygen in the air, even at ordinary temperatures: 2NO(g) + O₂(g) → 2NO₂(g).

Source	What supplies the high temperature
Natural: lightning	the discharge heats the air in its path to many thousands of °C
Man-made: internal combustion engines	fuel burning in the cylinders heats the air drawn in with it
Man-made: power stations and furnaces	burning fossil fuels

Bacteria in soil also release some NO, but lightning is the natural source to name in an answer. In a car, the nitrogen comes from the air, not the fuel (petrol and diesel are mainly hydrocarbons), and NOx forms inside the engine: the exhaust only carries it out.

Why a high temperature? The reaction must break N≡N, so $E_{\text{A}}$ is very high, and only at a very high temperature do enough molecules have that much energy (topic 8.2). The reaction is also endothermic, so heating moves the position of the equilibrium N₂ + O₂ ⇌ 2NO to the right (topic 7.1). As the gases cool, NO does not change back, because the reverse reaction also has a high activation energy and is far too slow without a catalyst.

Removing NO and NO₂: the catalytic converter

A catalytic converter in the exhaust system holds a ceramic honeycomb with a very large surface area, coated with a thin layer of platinum, palladium and rhodium. Once hot, the metals catalyse reactions between the pollutants. The solid metals and the gases are in different phases, so this is heterogeneous catalysis (topic 8.3):

• 2CO(g) + 2NO(g) → 2CO₂(g) + N₂(g)
• 4CO(g) + 2NO₂(g) → 4CO₂(g) + N₂(g)
• unburned hydrocarbons + NO → CO₂ + H₂O + N₂ (Worked example 3)

These are redox reactions. NO and NO₂ are reduced to N₂ (nitrogen falls from +2 or +4 to 0); CO and hydrocarbons are oxidised (carbon in CO rises from +2 to +4). The reactions are exothermic but very slow on their own. The metals provide a route with a lower activation energy (topic 8.3), so the reactions are fast enough in the short time the gas spends in the converter. How gases attach to the metal surface comes at A Level (topic 26.2), and the harm done by CO and hydrocarbons is in topic 14.1.

Worked example 3: heptane and NO in a catalytic converter

Question: Unburned heptane, C₇H₁₆, reacts with NO on the catalyst to form only CO₂, H₂O and N₂. (a) Write a balanced equation. (b) Use oxidation numbers to show which reactant is oxidised and which is reduced, and check that the electrons balance.

Step 1: carbon and hydrogen. Each C atom ends up in CO₂, and every two H atoms in one H₂O: 7CO₂ and 8H₂O.

Step 2: oxygen. The products hold 7 × 2 + 8 = 22 O atoms, all from NO, so 22NO.

Step 3: nitrogen. 22 N atoms make 11N₂.

Step 4: oxidation numbers. H stays +1 and O stays −2.

• Nitrogen: +2 in NO, 0 in N₂, so NO is reduced: 22 × 2 = 44 electrons gained.
• Carbon: in C₇H₁₆ the seven C atoms add up to −16, balancing 16 H at +1. In 7CO₂ they add up to 7 × (+4) = +28. The change is +28 − (−16) = +44, so heptane is oxidised: 44 electrons lost.

Answer: (a) C₇H₁₆(g) + 22NO(g) → 7CO₂(g) + 8H₂O(g) + 11N₂(g). (b) Heptane is oxidised (its carbon goes from a total of −16 to +28) and NO is reduced (nitrogen +2 to 0); 44 electrons are lost and 44 gained, so the equation balances.

Photochemical smog and PAN

Vehicles give out NO, which soon becomes NO₂ in the air, and unburned hydrocarbons. In strong sunlight, especially in still air over busy cities, they take part in a series of reactions, with oxygen from the air, that form photochemical smog, a haze that irritates the eyes and lungs. "Photochemical" means light drives the first step: sunlight splits NO₂ molecules.

One harmful product is peroxyacetyl nitrate, PAN, CH₃C(O)OONO₂. "Acetyl" is the old name for the ethanoyl group, CH₃CO-, and "peroxy" means an O-O link. PAN irritates the eyes and airways and damages plants.

Vehicles do not give out PAN. It is a secondary pollutant, formed in the air from two pollutants that are given out: oxides of nitrogen and unburned hydrocarbons. Smog also contains other secondary pollutants, such as ozone near the ground.

Acid rain: the direct role of NO and NO₂

Rain is naturally slightly acidic because it dissolves carbon dioxide. Acid rain is more acidic, mainly because of nitric acid and sulfuric acid, and it damages limestone buildings, trees and life in lakes.

NO is only slightly soluble in water, but in the air it is oxidised to NO₂, an acidic oxide that turns damp blue litmus red. With oxygen and water droplets, NO₂ forms nitric acid:

4NO₂(g) + O₂(g) + 2H₂O(l) → 4HNO₃(aq)

With water alone, NO₂ gives nitric acid and nitrous acid, 2NO₂(g) + H₂O(l) → HNO₃(aq) + HNO₂(aq).

Acid rain: NO₂ as a catalyst for oxidising SO₂

Sulfur dioxide, SO₂, comes mainly from burning fossil fuels that contain sulfur compounds, and from volcanoes. To form sulfuric acid, sulfur must be oxidised from +4 to +6. In the route you need to know, SO₂ is first oxidised to sulfur trioxide, SO₃. Oxygen alone does this only very slowly in air, but NO₂ speeds it up:

• Step 1: SO₂(g) + NO₂(g) → SO₃(g) + NO(g)
• Step 2: NO(g) + ½O₂(g) → NO₂(g)
• Overall: SO₂(g) + ½O₂(g) → SO₃(g)
• Then: SO₃(g) + H₂O(l) → H₂SO₄(aq)

In Step 1, sulfur is oxidised (+4 to +6) and nitrogen reduced (+4 to +2); in Step 2, nitrogen is oxidised back to +4. NO₂ is used up in Step 1 and regenerated in Step 2, so it cancels from the overall equation: it is a catalyst. All the species are gases, so it is a homogeneous catalyst (topic 8.3). NO can play the same part, starting with Step 2. Either way, a small amount of NOx can oxidise a much larger amount of SO₂. You meet this type of catalyst again at A Level (topic 26.2).

Key Equations

Ammonia as a base in water: NH₃(aq) + H₂O(l) ⇌ NH₄⁺(aq) + OH⁻(aq)

Forming the ammonium ion: NH₃(aq) + H⁺(aq) → NH₄⁺(aq); as gases, NH₃(g) + HCl(g) → NH₄Cl(s)

Displacing ammonia (warm with a base): NH₄⁺(aq) + OH⁻(aq) → NH₃(g) + H₂O(l)

Forming NOx: N₂(g) + O₂(g) → 2NO(g) at a very high temperature; 2NO(g) + O₂(g) → 2NO₂(g)

Catalytic converter: 2CO(g) + 2NO(g) → 2CO₂(g) + N₂(g). A hydrocarbon with $x$ C atoms and $y$ H atoms needs $2x + \frac{y}{2}$ NO and forms $x + \frac{y}{4}$ N₂.

Nitric acid in rain: 4NO₂(g) + O₂(g) + 2H₂O(l) → 4HNO₃(aq)

NO₂ as a catalyst: SO₂(g) + NO₂(g) → SO₃(g) + NO(g) and NO(g) + ½O₂(g) → NO₂(g); overall SO₂(g) + ½O₂(g) → SO₃(g)

Sulfuric acid in rain: SO₃(g) + H₂O(l) → H₂SO₄(aq)

Amount in solution ($n$ in mol; $c$ in mol dm⁻³; $V$ in dm³, so divide cm³ by 1000): $$n = cV$$

Common Mistakes to Avoid

• Mistake: writing "nitrogen is unreactive because it has a triple bond", or blaming the forces between N₂ molecules. Fix: say the N≡N bond is very strong (944 kJ mol⁻¹), so $E_{\text{A}}$ is very high, and that the molecule is non-polar. A triple bond alone is not enough: C≡C opens far more easily (230 against 534 kJ mol⁻¹ in Worked example 1). The weak forces between N₂ molecules explain its low boiling point, not its lack of reactivity.
• Mistake: saying ammonia is a base because it releases OH⁻, or that its lone pair bonds to OH⁻. Fix: the lone pair accepts H⁺. In NH₃(aq) + H₂O(l) ⇌ NH₄⁺(aq) + OH⁻(aq), the OH⁻ is what remains of the water molecule that gave up the proton.
• Mistake: giving NH₄⁺ a lone pair or a pyramidal or square planar shape. Fix: four bonding pairs and no lone pair make it tetrahedral, 109.5°, with all four N-H bonds identical.
• Mistake: saying an acid releases ammonia from an ammonium salt, or dipping litmus into the solution. Fix: a base releases it. Warm with NaOH(aq) and write "the gas given off turns damp red litmus blue".
• Mistake: naming the catalytic converter, "car exhausts" or acid rain as a source of NOx. Fix: give lightning as the natural source; the man-made one is the high temperature inside the internal combustion engine, where N₂ and O₂ from the air react: N₂(g) + O₂(g) → 2NO(g).
• Mistake: saying the converter oxidises NO, or forming PAN from CO, CO₂ or water. Fix: NO is reduced to N₂ (+2 to 0) while CO is oxidised (+2 to +4); PAN forms from oxides of nitrogen and unburned hydrocarbons in sunlight.
• Mistake: mixing up the two roles: giving the SO₂ cycle when asked how NO₂ forms acid rain directly, giving nitric acid equations for sulfuric acid rain, or leaving out the step that re-forms NO₂. Fix: for the direct role, write 4NO₂(g) + O₂(g) + 2H₂O(l) → 4HNO₃(aq). For the catalytic role, write SO₂(g) + NO₂(g) → SO₃(g) + NO(g), NO(g) + ½O₂(g) → NO₂(g) and SO₃(g) + H₂O(l) → H₂SO₄(aq), then state that NO₂ is regenerated, so it is a catalyst.

Exam Tips

• Explain the lack of reactivity of nitrogen with two separate points: the N≡N bond is very strong, so $E_{\text{A}}$ is very high; and the molecule is non-polar, so it has no δ+ or δ− site to attract reagents. "Strong bond" and "high $E_{\text{A}}$" together make one point, not two.
• Describe the basicity of ammonia with the words "proton acceptor", "lone pair on N" and "coordinate (dative) bond to H⁺". Use ⇌ for ammonia with water and → with an acid, and give state symbols.
• State and explain a source of NOx by naming it and giving the reason, for example "lightning: the very high temperature lets N₂ and O₂ in the air react". For vehicles, write "in the internal combustion engine", not "in the exhaust".
• For a catalytic converter, name the metals, give a balanced equation and use oxidation numbers to say which species is oxidised and which is reduced. Check that electrons lost equal electrons gained (Worked example 3).
• For the catalytic role of NO₂, write the two equations of the cycle and the one forming H₂SO₄, then end with "NO₂ is regenerated, so it is a catalyst".