Why doesn't diamond turn into graphite?

♣ Newbie

Take two magnets that are stuck together and pull them apart: it takes effort. Bring them close and they snap together with a clack: joining up gives off energy.

The atoms in a molecule are held together by bonds, and something similar happens with them: breaking a bond costs energy and making one gives it back. A chemical reaction breaks some bonds and makes others. If the new ones grip harder than the old ones, they give back more than the breaking cost, and what’s left over comes out as heat. That’s how the gas on your stove heats things.

If it’s the other way round, the reaction asks for energy and you have to supply it, for example with electricity. You’ll see that at the end, with water.

♦ Intermediate

A bond is the link between two atoms in a molecule. Its bond energy is what it costs to break a mole of those bonds (OpenStax). Some average values:

Hydrogen and oxygen make water vapour: 2 H₂ + O₂ → 2 H₂O. Two H–H and one O=O are broken, which costs 2 × 436 + 498 = 1370 kJ. Four O–H are formed, which give back 4 × 464 = 1856 kJ. 486 kJ is left over, about 243 per mole of water. Measured: 242 kJ (OpenStax; NIST-JANAF gives 241.8).

When energy is left over and comes out as heat, the reaction is exothermic. When you have to put it in, it’s endothermic. That energy left over or missing is called the heat of the reaction.

Bond energies are averages over many molecules, so the sum comes out approximate. For the exact figure, tables give the energy of forming each substance from its elements, and with them you can do the sum without bonds.

♠ Technical

The reaction enthalpy is estimated with ΔᵣH ≈ ΣD(broken) − ΣD(formed) (OpenStax, Chemistry 2e, 7.5). The D values in the table are averages, and the estimate is usually off by tens of kJ: in OpenStax’s example, CO + 2 H₂ → CH₃OH(g) comes out at −107 kJ from bonds and −90.5 kJ from enthalpies of formation.

For water the estimate works because H–H and O=O aren’t averages: each is one specific molecule. With the NIST-JANAF tables at 298.15 K, D(H–H) = 2 ΔfH°(H, g) = 2 × 217.999 = 436.0 kJ/mol, D(O=O) = 2 ΔfH°(O, g) = 498.3 kJ/mol, and the average O–H in water is [2 ΔfH°(H) + ΔfH°(O) − ΔfH°(H₂O, g)] / 2 = 463.5 kJ/mol.

With those D values, the estimate gives exactly ΔfH°(H₂O, g) = −241.826 kJ/mol: it’s the same sum by another route. Hess’s law guarantees it: since H is a state function, ΔᵣH° = Σν ΔfH°(products) − Σν ΔfH°(reactants) doesn’t depend on the intermediate steps (OpenStax, 5.3).

♣ Newbie

Picture a ball in a hollow. Next to it there’s a very long downhill slope, but between the hollow and the slope there’s a small hill. Even though it would be better off at the bottom, the ball goes nowhere until someone pushes it over the hill. After that it rolls down on its own.

Reactions are like that. The gas on your stove and the oxygen in the air can sit together all day without burning, until a spark gives the push. Once it starts, the heat of the flame pushes the molecules next to it, and the reaction keeps going by itself.

What pushes hardest is temperature. When it’s hot, molecules move faster and collide harder, so more collisions make it over the hill. That’s why food keeps longer in the fridge: in the cold, the reactions that spoil it go more slowly.

♦ Intermediate

For two molecules to react, they have to collide with enough energy to start breaking their bonds. That minimum energy is the activation energy (OpenStax). In an energy profile it’s the height of the hill: you start from the reactants, climb to the top and go down to the products. The drop between start and finish is the heat of the reaction, from the previous block. The hill says something else: not how much you gain, but how much it costs to get going.

The energy differs from collision to collision: almost all are weak and a few are very strong. The share of collisions that reaches the top drops enormously if the hill is higher and rises enormously with temperature. With a 50 kJ/mol hill, which is just an example value:

• at 25 °C, roughly 1 in 575 million collisions reaches the top;
• in the fridge, at 4 °C, the reaction runs 4.6 times slower;
• at 35 °C, 1.9 times faster.

These are our own calculations. That’s where the “every 10 degrees, twice as fast” rule comes from, and it only holds for hills about that size: with a 100 kJ/mol hill, the same 10 degrees multiply by 3.7.

A catalyst opens another path with a lower hill. The reaction runs faster, and the start and finish don’t change (OpenStax). Car catalytic converters and the body’s enzymes do this.

♠ Technical

The Arrhenius equation is k = A·e^(−Eₐ/RT) (OpenStax, 12.5). The factor e^(−Eₐ/RT) measures what share of collisions arrives with enough energy; A, the frequency factor, is related to how often molecules collide and how they are oriented. Taking logs, ln k = ln A − (Eₐ/R)(1/T): measure k at several temperatures, plot ln k against 1/T and you get a straight line whose slope gives Eₐ. OpenStax does it with 2 HI → H₂ + I₂ and gets Eₐ = 180 kJ/mol.

Between two temperatures, k₂/k₁ = exp[(Eₐ/R)(1/T₁ − 1/T₂)]. With Eₐ = 50 kJ/mol, from 298.15 to 308.15 K it gives 1.92; with 100 kJ/mol, 3.70 (our own calculation, assuming A stays constant). The rule of doubling every 10 °C isn’t a law: it depends on Eₐ.

A catalyst offers another mechanism, with a lower transition state. Since H is a state function, the start and finish don’t move: in OpenStax’s diagrams with and without a catalyst, «both curves begin and end at the same energies» (12.7).

♣ Newbie

Now picture two valleys. In one, a little higher up, is diamond. In the other, just a little lower, is graphite. Both are pure carbon, arranged in two different ways: you saw it in the atoms lesson.

Going down into the graphite valley would give off a tiny bit of energy, which is why chemists say graphite is more stable. But there’s a huge mountain between the two valleys: to move over to graphite, each carbon atom has to break loose from the diamond, and pulling it loose takes a huge amount of energy. At room temperature, practically no atom manages it.

Diamond isn’t safe because it’s the most stable, but because the mountain is so high. If you heat it above about 1,500 °C in a container with no air, it starts turning into graphite.

♦ Intermediate

To know whether a reaction can go by itself, heat isn’t enough. Chemists use the Gibbs energy, which combines heat and disorder. If the Gibbs energy goes down, the reaction can go by itself. From diamond to graphite, at 25 °C, it goes down by 2.9 kJ/mol (OpenStax; UW–Madison). That’s very little, and the heat it gives off is even less: 1.9 kJ for every 12 grams of carbon, which would warm a glass of water by less than 2 °C (our own calculation).

But being able to doesn’t mean it happens. OpenStax puts it this way: «The spontaneity of a process is not correlated to the speed of the process». To rearrange the carbon, each atom has to break loose from the diamond, and that takes a huge amount of energy.

The hill has been measured: 188 kJ/mol in a powder of tiny diamonds (Butenko and others, 2000) and about 730 kJ/mol on one face of a natural diamond (Davies and Evans, 1972). Each figure comes from a single study. With the 188 one, at 1,600 °C the boundary between diamond and graphite moves 1.5 nanometres per hour. At 25 °C it would take more than 10²⁰ years to move one (our own calculation, stretching their formula far beyond what they measured). The universe is about 10¹⁰ years old.

Diamond is thermodynamically unstable but kinetically stable: it’s said to be metastable (OpenStax). And catalysts work here too: with a layer of nickel, a film of tiny diamonds starts turning into graphite at around 500 °C, and the surface of a diamond crystal above 800 °C (Sedelnikova and others, 2025, who gather it from other studies).

♠ Technical

From OpenStax’s appendix G, diamond → graphite has ΔᵣH° = −1.89 kJ/mol and ΔᵣS° = 5.740 − 2.38 = +3.36 J/(K·mol). So ΔᵣG° = ΔᵣH° − TΔᵣS° = −1.89 − 298.15 × 0.00336 = −2.89 kJ/mol, which matches the −2.90 in the table. Careful: the heat given off is ΔᵣH, −1.89; what decides the direction is ΔᵣG.

The kinetics were measured by Davies and Evans (Proc. R. Soc. Lond. A 328, 413, 1972) between 1850 and 2000 °C, at zero pressure: Eₐ = 728 ± 50 kJ/mol on the 110 faces. For the 111 faces their abstract gives two figures that don’t agree with each other: 253 kcal/mol and 1159 kJ/mol, when 253 kcal is 1059 kJ. They propose that the slow step is pulling a single atom off the surface: one held by two bonds on the 110 faces and by three on the 111.

Butenko and others (J. Appl. Phys. 88, 4380, 2000), with nanodiamond between 1370 and 1870 K, fit v = A·e^(−E/RT) with E = 45 ± 4 kcal/mol (188 kJ/mol) and A = 74 nm/s. They say their formula works «in and around the indicated temperature range» and put the difference from Davies and Evans down to another mechanism at lower temperatures.

If you extrapolate their formula, v(298 K) ≈ 8×10⁻³² nm/s: one nanometre in about 4×10²³ years, between 5×10²⁰ and 4×10²⁶ with their ±4 kcal/mol. With Davies and Evans’s Eₐ, the exponential factor at 25 °C is 10¹¹⁰ times smaller than at 1900 °C. These are extrapolations far beyond what was measured: only the order of magnitude counts. At ambient pressure, diamond is a local minimum of G separated from graphite by a barrier; at very high pressures, diamond becomes the stable one (OpenStax, 16.1).

♣ Newbie

A nail that rusts and a log that burns are doing the same thing: something is joining up with oxygen. The difference is speed: rust takes months and the flame, seconds. That’s why these are called oxidation reactions.

You can make a battery out of water. With electricity, water splits into hydrogen and oxygen: that’s climbing the slope, and it costs energy. Splitting a litre of water costs at least 4.4 kWh, what a kettle uses running for over two hours.

If you then burn that hydrogen, it goes back down the slope and gives back the same energy, as heat. Not one joule more: hydrogen isn’t a source of energy but a way of storing it, like a battery.

♦ Intermediate

In chemistry, oxidation is losing electrons and reduction is gaining them (OpenStax). They always go together, because the electrons one loses, another gains. When hydrogen burns, the hydrogen is oxidised and the oxygen is reduced.

Splitting water with electricity is called electrolysis: 2 H₂O → 2 H₂ + O₂, combustion in reverse. You need at least 1.23 volts across the two electrodes (OpenStax). The energy is the one from block 1, but with liquid water: 285.8 kJ per mole (NIST-JANAF; OpenStax). At 1.23 volts, about 237 kJ of that goes in as electricity and the rest as heat from the surroundings. A litre of water is 55.5 moles: 15.9 MJ, or 4.4 kWh (our own calculation).

Burning the hydrogen gives that energy back, as heat. A fuel cell runs the same reaction without a flame and gives electricity, at about 1.2 volts. Its efficiency is about 50–75 %, against 20–25 % for an engine (OpenStax).

♠ Technical

In electrolysis (OpenStax, 17.7), the anode oxidises water, 2 H₂O(l) → O₂ + 4 H⁺ + 4 e⁻, and oxygen goes from −2 to 0. The cathode reduces, 2 H⁺ + 2 e⁻ → H₂, and hydrogen goes from +1 to 0. E°cell = −1.229 V: you have to impose more than that.

A check with ΔᵣG° = −nFE°: ΔfG°(H₂O, l) = −237.141 kJ/mol (NIST-JANAF) and n = 2, so E° = 237,141 / (2 × 96,485.332) = 1.2289 V. With ΔᵣH instead of ΔᵣG you get 1.481 V, the thermoneutral voltage: above it, the cell needs no heat from outside. The gap between the two voltages is the heat the cell can draw from its surroundings, TΔᵣS = 48.7 kJ/mol, which matches ΔᵣS° = 163.3 J/(K·mol) from appendix G.

The other way round, in a fuel cell, the maximum electrical work is −ΔᵣG = 237.1 kJ/mol, 83 % of what burning the hydrogen would give (our own calculation). OpenStax gives about 1.2 V per cell and a real efficiency of 50–75 %. The energy that comes out never exceeds what it cost to split the water: it’s a chemical battery, not a source.