Enthalpy

Definition

Enthalpy is defined by a single, compact combination of state variables:

H = U + PV

where the symbols are:

• H — enthalpy, in joules (J). It is a state function: its value depends only on the current state, not on how the system got there.
• U — internal energy (J): the sum of all microscopic kinetic and potential energies of the particles.
• P — pressure (Pa = N/m²).
• V — volume (m³). Note PV has units Pa·m³ = J, so the sum is dimensionally consistent.

The term PV is the energy the system must expend to push its surroundings aside and occupy its volume V against pressure P. So enthalpy is "internal energy plus the cost of making room." The name comes from the Greek enthalpein ("to warm within"); the symbol H and much of the modern usage trace to Heike Kamerlingh Onnes and the early-20th-century development of physical chemistry, with J. Willard Gibbs having already introduced the "heat function at constant pressure" in the 1870s.

Why enthalpy equals heat at constant pressure

Start from the first law of thermodynamics, dU = δQ − δW. If the only work is expansion against external pressure, δW = P dV, so dU = δQ − P dV. Now differentiate the definition of enthalpy:

dH = dU + P dV + V dP

Substitute dU = δQ − P dV:

dH = (δQ − P dV) + P dV + V dP = δQ + V dP

At constant pressure dP = 0, so the V dP term vanishes and:

dH = δQ  →  ΔH = Q_p

This is the whole point. The heat you measure in a coffee-cup calorimeter or feel from a hand-warmer, in a vessel open to the atmosphere, is exactly ΔH. The awkward expansion work P dV that a gas does on the atmosphere is already absorbed into the bookkeeping — you don't have to track it separately.

Enthalpy vs internal energy: why chemists prefer H

The two heats differ by the work done against the atmosphere:

ΔH = ΔU + Δ(PV)

For reactions involving gases, treating them as ideal gives Δ(PV) = Δn_gas·RT, so:

ΔH = ΔU + Δn_gas · R · T

where Δn_gas is the change in moles of gas (products − reactants), R = 8.314 J/(mol·K), and T is temperature in kelvin. If no gas moles change (Δn_gas = 0), or the reaction is entirely in condensed phases, ΔH ≈ ΔU because the PV term is tiny for solids and liquids.

Condition of process	Heat equals	Apparatus
Constant volume (rigid vessel)	Q_v = ΔU	Bomb calorimeter
Constant pressure (open to atmosphere)	Q_p = ΔH	Coffee-cup / flame calorimeter
Neither (general path)	Q depends on path	—

Because nearly every bench reaction happens open to the room at ~101.325 kPa, the heat you measure is ΔH. That practical fact — not any deep asymmetry between H and U — is why introductory chemistry lives in the language of enthalpy.

Sign conventions: exothermic and endothermic

Signs are always taken from the system's point of view:

• ΔH < 0 (exothermic) — the system releases heat to the surroundings; the surroundings warm up. Combustion, neutralization, most oxidations. Example: CH₄ + 2 O₂ → CO₂ + 2 H₂O, ΔH = −890 kJ/mol.
• ΔH > 0 (endothermic) — the system absorbs heat; the surroundings cool. Melting, boiling, photosynthesis, dissolving ammonium nitrate (instant cold packs). Example: ΔH_vap of water = +40.66 kJ/mol.

A common trap: a negative ΔH does not mean a reaction must occur. Spontaneity is governed by the Gibbs free energy, ΔG = ΔH − TΔS, where S is entropy. An endothermic reaction (ΔH > 0) can still be spontaneous if it increases entropy enough that TΔS outweighs ΔH — which is exactly why ice melts above 0°C.

Hess's law and enthalpy of formation

Because H is a state function, ΔH for a reaction depends only on the initial and final states, never the route. This is Hess's law (Germain Hess, 1840): the enthalpy change of an overall reaction equals the sum of the enthalpy changes of any set of steps that connect the same reactants and products.

The most powerful bookkeeping trick built on Hess's law uses the standard enthalpy of formation ΔH°_f — the enthalpy change to form one mole of a compound from its elements in their standard states (most stable form at 1 bar, usually at 298.15 K). Elements in their standard state are assigned ΔH°_f = 0 by definition. Then for any reaction:

ΔH°_rxn = Σ ΔH°_f(products) − Σ ΔH°_f(reactants)

Substance (standard state, 298.15 K)	ΔH°_f (kJ/mol)
O₂ (g), N₂ (g), C (graphite) — elements	0 (by definition)
H₂O (l)	−285.8
H₂O (g)	−241.8
CO₂ (g)	−393.5
CO (g)	−110.5
CH₄ (g)	−74.6
NH₃ (g)	−45.9
NaCl (s)	−411.2

Worked example: combustion of methane

Compute ΔH° for CH₄(g) + 2 O₂(g) → CO₂(g) + 2 H₂O(l) from formation enthalpies.

ΔH°_rxn = [ΔH°_f(CO₂) + 2·ΔH°_f(H₂O, l)] − [ΔH°_f(CH₄) + 2·ΔH°_f(O₂)]
        = [(−393.5) + 2·(−285.8)] − [(−74.6) + 2·(0)]
        = (−965.1) − (−74.6)
        = −890.5 kJ/mol

The reaction is strongly exothermic — 890 kJ released per mole of methane burned, the basis of natural-gas heating. Notice we never had to burn methane along the tabulated path; Hess's law let us combine formation steps to reach the answer.

To convert this to ΔU we use Δn_gas. Reactants: 1 (CH₄) + 2 (O₂) = 3 mol gas. Products: 1 (CO₂) + 0 (water is liquid) = 1 mol gas. So Δn_gas = 1 − 3 = −2, and at T = 298.15 K:

ΔU = ΔH − Δn_gas·RT = (−890.5 kJ) − (−2)(8.314 J/mol·K)(298.15 K)
   = −890.5 kJ + 4.96 kJ ≈ −885.5 kJ/mol

The difference of ~5 kJ is precisely the expansion work bookkeeping — small but not negligible for gas-changing reactions.

Phase changes: latent heat is an enthalpy change

At a phase transition, temperature stays fixed while heat pours in, so the latent heat is exactly an enthalpy of transition ΔH at constant P. For water:

Transition	ΔH per mole	ΔH per kilogram
Fusion (ice → water, 0°C)	6.01 kJ/mol	334 kJ/kg
Vaporization (water → steam, 100°C)	40.66 kJ/mol	2257 kJ/kg
Sublimation (ice → vapor, 0°C)	≈ 51.0 kJ/mol	≈ 2831 kJ/kg

Vaporization takes about 6.8× more energy than fusion because turning liquid into gas breaks nearly all the hydrogen bonds and does large PV work against the atmosphere. Note also that ΔH_sublimation ≈ ΔH_fusion + ΔH_vaporization — a direct consequence of Hess's law, since H is a state function — provided all three are compared at the same temperature. (At 0°C the vaporization enthalpy is about 45 kJ/mol, so 6.01 + 45 ≈ 51 kJ/mol; the 40.66 kJ/mol in the table is the 100°C value.)

Common mistakes

• Thinking H is "heat content" you can always measure. ΔH equals heat only at constant pressure with no non-expansion work. Change H by doing work and no heat need flow.
• Confusing ΔH sign conventions. Exothermic is ΔH < 0 (system loses energy). Students often flip it because "releasing heat" feels positive.
• Assuming ΔH < 0 means spontaneous. Spontaneity needs ΔG = ΔH − TΔS < 0. Enthalpy is only half the story; entropy and temperature decide.
• Setting ΔH = ΔU always. They differ by Δn_gas·RT whenever the number of gas moles changes. Only equal when Δn_gas = 0 or all species are condensed.
• Forgetting standard states in Hess's law. ΔH°_f is defined for the most stable form (O₂ not O, graphite not diamond) at 1 bar. Mixing states corrupts the sum.
• Using Celsius in Δn_gas·RT. T must be in kelvin (298.15 K, not 25).