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Lesson 4 · 30 min

Nitrogen and phosphorus

The N≡N triple bond makes dinitrogen nearly inert, so fixing nitrogen is a problem for life and industry alike. Walk the nitrogen cycle across its oxidation-state map, then see why phosphorus prefers P₄ tetrahedra.

In this lesson

  • Explain the inertness of N₂ and the different structures of elemental nitrogen and phosphorus from bond enthalpies
  • Describe the three ways of fixing nitrogen (Haber process, nitrogenase, lightning) and judge their direction from thermodynamic data
  • Read the energy changes of the nitrogen cycle from nitrogen’s oxidation-state map (Frost diagram)
  • Write the three steps of the Ostwald process for nitric acid, and meet the allotropes of phosphorus and the phosphates

Every breath you take is about 78 % nitrogen, yet crops are so often short of nitrogen that, a century and more ago, people shipped bird guano and saltpetre from South America to fertilise their fields. From 1909 Haber and Bosch turned the nitrogen of the air into ammonia — by some estimates, about half the nitrogen atoms in our bodies today have passed through a Haber–Bosch plant. Why is dinitrogen so hard to use?

Predict first

In water at pH 7, with no other oxidant or reductant, which nitrogen species is thermodynamically the most stable (lowest on the Frost diagram)?

Pick an answer and lock it in before you look. A wrong guess teaches more than no guess.

One of the strongest bonds there is

From the enthalpy of formation of nitrogen atoms the N≡N triple bond works out at 945 kJ/mol, while an N–N single bond is only about 163 kJ/mol — three single bonds add up to only about half of one triple bond. Two NX2\ce{N2} molecules (two triple bonds, about 1890 kJ) are far more stable than a hypothetical tetrahedral NX4\ce{N4} (six single bonds, about 980 kJ), so nitrogen is diatomic. NX2\ce{N2} is also non-polar, with a very low highest occupied orbital and a very high lowest empty one, so it neither gives nor takes electrons easily: breaking into the triple bond means climbing a high activation barrier.

Phosphorus is the opposite. π overlap between 3p orbitals is poor, so P≡P is a bad bargain compared with three P–P single bonds, and white phosphorus consists of PX4\ce{P4} tetrahedra. Their 60° bond angles are badly strained: white phosphorus ignites in air at about 30 °C, must be stored under water and is highly toxic. Heating it gives polymeric red phosphorus, much more stable (it is on the side of matchboxes); high pressure gives black phosphorus, a layered semiconductor rather like graphite.

Fixing nitrogen: three routes

  • The Haber–Bosch process: NX2+3 HX2⇌2 NHX3\ce{N2} + \ce{3H2} \mathrel{\ce{<=>}} \ce{2NH3} has ΔH∘=−92\Delta H^\circ = -92 kJ and, at 298 K, ΔG∘=−33\Delta G^\circ = -33 kJ — thermodynamically favourable but imperceptibly slow at room temperature. Industry uses an iron catalyst at 400–500 °C and 150–300 bar as a compromise between rate and yield (see the Haber process).
  • Nitrogenase: microbes such as root-nodule bacteria reduce NX2\ce{N2} to NHX3\ce{NH3} at room temperature and pressure with an iron–molybdenum active site, spending 16 ATP per NX2\ce{N2} fixed.
  • Lightning: NX2+OX2→2 NO\ce{N2} + \ce{O2} \mathrel{\ce{->}} \ce{2NO} has ΔG∘\Delta G^\circ = +175 kJ at 298 K, strongly unfavourable. Even at the ~3000 K of a lightning channel it is still +108 kJ, but the equilibrium constant has risen to about 0.01 — enough to make useful amounts of NO.

Once fixed, nitrogen keeps changing oxidation state in soil and water: bacteria oxidise ammonia to nitrite and then nitrate (nitrification), and bacteria in oxygen-poor places reduce nitrate back to NX2\ce{N2} (denitrification). The energy of every step can be read straight off the Frost diagram.

ΔG=F[∑iνi(NE∘)i−(∑iνiNi)Epartner]\Delta G = F\Big[\sum_i \nu_i (NE^\circ)_i - \Big(\sum_i \nu_i N_i\Big) E_\mathrm{partner}\Big]
The Gibbs energy of a nitrogen-cycle step: the height difference on the Frost diagram, minus the part carried by the electrons handed to (or taken from) the partner couple.
(NE∘)i(NE^\circ)_i
Frost ordinate of species i at the pH (per N atom)
νi\nu_i
stoichiometric coefficient in N atoms (negative for reactants)
NiN_i
oxidation state of nitrogen in species i
EpartnerE_\mathrm{partner}
potential of the partner couple (O₂/H₂O or H⁺/H₂) at that pH
FF
Faraday constant
A Frost diagram from standard potentials, with the nitrogen cycle

The oxidation-state map of nitrogen

NH₄⁺N₂H₅⁺NH₂OHN₂N₂ONONO₂⁻N₂O₄NO₃⁻−3−2−1012345Oxidation state of nitrogen N−1012345NE° / V (relative to N₂)

Lowest point N₂: the thermodynamically most stable nitrogen species at this pH.

A step of the nitrogen cycle

2 NHX4X++3 OX2→2 NOX2X−+4 HX++2 HX2O\ce{2NH4+ + 3O2 -> 2NO2- + 4H+ + 2H2O}

ΔG (as written)
−275 kJ
per N atom transformed
−275 kJ
Electrons lost per N atom (negative = gained)
6
Electron partner
O₂/H₂O

Exergonic: it can run on its own — organisms harvest this energy.

Ammonia-oxidising bacteria such as Nitrosomonas.

ammonium · N = −3 · NE° = 0.83 V

Ammonium fertilisers; a product of protein breakdown.

The ordinate NE° = −ΔG°f/F per N atom relative to N₂ is computed along a chain of CRC standard potentials (Vanýsek) with pKa values from Harris — the same numbers as the redox unit’s Frost diagrams. Each step’s ΔG = F·[Σν·NE° − (Σν·N)·E(partner)], with unit activities except H⁺ and gases at 1 bar. H₂ merely stands in for an electron donor: real denitrifying bacteria use organic matter, and nitrogenase also spends 16 ATP per N₂ fixed. The map shows where energy flows, not how fast.
Step through the nitrogen cycle: the two nitrification steps release about 275 kJ and 75 kJ per nitrogen atom — the energy nitrifying bacteria live on — while anammox comproportionates −3 and +3 nitrogen to N₂. Then change the pH to 0 or 14 and see which points move with pH and which stay put.

Oxides of nitrogen and nitric acid

Nitrogen forms oxides in every state from +1 to +5: laughing gas NX2O\ce{N2O} (+1), NO with one unpaired electron (+2), red-brown NOX2\ce{NO2} and its colourless dimer NX2OX4\ce{N2O4} (+4, 2 NOX2⇌NX2OX4\ce{2NO2} \mathrel{\ce{<=>}} \ce{N2O4}), and NX2OX5\ce{N2O5} (+5), the anhydride of nitric acid. NO turns into red-brown NOX2\ce{NO2} the moment it meets air, and NOX2\ce{NO2} is a leading actor in photochemical smog.

The Ostwald process makes nitric acid from ammonia:

  1. 4 NHX3+5 OX2→4 NO+6 HX2O\ce{4NH3} + \ce{5O2} \mathrel{\ce{->}} \ce{4NO} + \ce{6H2O}, over a platinum–rhodium gauze at about 900 °C with a contact time of about a thousandth of a second;
  2. after the gas is cooled, 2 NO+OX2→2 NOX2\ce{2NO} + \ce{O2} \mathrel{\ce{->}} \ce{2NO2};
  3. 3 NOX2+HX2O→2 HNOX3+NO\ce{3NO2} + \ce{H2O} \mathrel{\ce{->}} \ce{2HNO3} + \ce{NO}, with the NO recycled to step 2.

Nitric acid is a strong acid and a strong oxidant: with copper the concentrated acid gives NOX2\ce{NO2} and the dilute acid NO — not hydrogen.

Worked example

How much heat does the first Ostwald step release?

From ΔfH∘\Delta_\mathrm{f}H^\circ (kJ/mol): NHX3\ce{NH3}(g) −45.9, NO(g) +91.3, HX2O\ce{H2O}(g) −241.8, find ΔrH∘\Delta_\mathrm{r}H^\circ for 4 NHX3+5 OX2→4 NO+6 HX2O(g)\ce{4NH3} + \ce{5O2} \mathrel{\ce{->}} \ce{4NO} + \ce{6H2O(g)}.

Phosphoric acid and the phosphates

Almost all phosphorus in nature occurs as phosphate. Phosphoric acid HX3POX4\ce{H3PO4} is triprotic (pKa\mathrm{p}K_\mathrm{a} = 2.15, 7.20, 12.38); its speciation is in polyprotic acids. Bone and tooth enamel are mostly hydroxyapatite CaX5(POX4)X3OH\ce{Ca5(PO4)3OH}; fluoride replaces its OHX−\ce{OH-} to give the more acid-resistant fluorapatite, which is why toothpaste contains fluoride. Insoluble phosphate rock must be treated with sulfuric acid to make soluble “superphosphate” fertiliser, and the backbone of DNA and the energy currency ATP both rely on phosphate ester bonds.

Concept check

In the Ostwald process, why is the NO–air mixture cooled before NO is converted to NOX2\ce{NO2}?

Try the calculation

The standard enthalpy of formation of nitrogen atoms is ΔfH∘(N,g)=472.68\Delta_\mathrm{f}H^\circ(\ce{N}, g) = 472.68 kJ/mol. What is the N≡N bond enthalpy?

kJ/mol

Key takeaways

  • The N≡N bond enthalpy is 945 kJ/mol and N₂ is non-polar with a huge activation barrier, so it is kinetically inert; phosphorus has weak π bonds and prefers P₄ tetrahedra and polymers.
  • Nitrogen is fixed by the Haber process, nitrogenase and lightning — favourable or not thermodynamically, all need a catalyst or extreme conditions to get over the barrier.
  • N₂ is the lowest point of the Frost diagram at pH 7; the energies of nitrification, denitrification and anammox can be read straight from the map.
  • The Ostwald process oxidises NH₃ to NO over Pt–Rh, oxidises NO to NO₂ after cooling, and absorbs NO₂ in water as HNO₃.
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Data and references

  1. J. D. Cox, D. D. Wagman, V. A. Medvedev, CODATA Key Values for Thermodynamics (1989); CRC Handbook of Chemistry and Physics, 97th ed. (2016), “Standard thermodynamic properties of chemical substances”.
  2. P. Vanýsek, “Electrochemical series”, CRC Handbook of Chemistry and Physics, 91st ed. (2010).
  3. N. N. Greenwood, A. Earnshaw, Chemistry of the Elements, 2nd ed. (1997), ch. 11–12 (nitrogen, phosphorus).
  4. V. Smil, Enriching the Earth: Fritz Haber, Carl Bosch, and the Transformation of World Food Production, MIT Press (2001).
  5. M. Strous et al., “Missing lithotroph identified as new planctomycete”, Nature 400, 446 (1999) (anammox).