Thermodynamically, most stable form of carbon is;

The correct answer is graphite.

Thermodynamic stability of a substance is determined by its Gibbs free energy ( G ). At a given temperature and pressure, the phase with the lowest G is the most stable. The relationship is:

$G\; =\; H\; -\; T·S$

where H is enthalpy, T temperature, and S entropy. For the two common allotropes of carbon—diamond and graphite—both have the same elemental composition, but their crystal structures differ, leading to different H and S values.

At standard temperature (≈298 K) and pressure (1 atm), the enthalpy of graphite is slightly lower than that of diamond (ΔH≈ -2 kJ mol⁻¹). Graphite also has a higher entropy because its layers can slide and vibrate more freely compared to the rigid three‑dimensional network of diamond (ΔS≈ +1 J mol⁻¹ K⁻¹). Substituting these values into the Gibbs equation shows that the G of graphite is lower:

$\Delta G\; =\; \Delta H\; -\; T·\Delta S\; \approx \; (‑2 kJ mol⁻¹)\; -\; (298 K)(+0.001 kJ mol⁻¹ K⁻¹)\; \approx \; ‑2.3 kJ mol⁻¹$

The negative ΔG indicates that, under ordinary conditions, graphite is thermodynamically favored over diamond. This is why graphite is the most stable form of carbon, while diamond is metastable and will slowly convert to graphite if given sufficient time or the right catalysts.

Other carbon forms such as peat or coal are mixtures of organic matter and contain many impurities; their Gibbs free energies are higher than that of pure graphite, so they are less stable in the pure elemental sense.

Consequently, the most stable allotrope of carbon thermodynamically is graphite.