Enthalpy of Reaction and Thermochemical Equations

Enthalpy depends only on the current state of a system, not on how it got there. That is what makes indirect routes legitimate: any path from the same reactants to the same products gives the same ΔH.

A state function is fixed only when the endpoints are actually the same, including temperature, pressure, composition and phases. Heating liquid water from 20 to 30 °C is not the same final state as heating it to 40 °C. To compare two routes, write complete initial and final states first. Heat itself is not a state function: changing the work path can change q even when ΔU is unchanged.

The ΔH quoted applies to the molar amounts written in that equation, with those exact physical states. Changing water from liquid to gas changes ΔH, because vaporisation costs energy.

For 2 H₂(g) + O₂(g) → 2 H₂O(l), ΔH = −572 kJ means the stoichiometric batch containing 2 mol H₂ releases 572 kJ. It is −286 kJ per mole of H₂ consumed or water formed, and −572 kJ per mole of O₂ consumed. A bare “kJ/mol” is ambiguous unless the equation or species basis is supplied. For an actual amount, reaction extent ξ = nconsumed/its coefficient gives q = ξΔrH at constant pressure under the stated work assumptions.

Because ΔH is proportional to the amount reacting, it behaves exactly like a coefficient. Convert grams to moles, then use the ratio of kilojoules to moles from the thermochemical equation.

If only 0.300 mol H₂ reacts completely with excess O₂, ξ = 0.300/2 = 0.150 mol of reaction as written, so q = (0.150)(−572) = −85.8 kJ. In words, 85.8 kJ is released. The released-energy magnitude is positive, while the system's enthalpy change is negative. If both reactants have finite amounts, identify the limiting one before choosing ξ.

A thermochemical equation includes phases because changing the product from liquid to gas changes the heat released. If liquid water is formed and then vaporized, vaporization absorbs energy, making the net combustion enthalpy less negative. Adjust the written reaction using Hess's law and the number of water molecules produced.

For CH₄ combustion producing two waters, changing H₂O(l) to H₂O(g) adds two vaporization steps. With +44 kJ/mol at the stated temperature, ΔHgas-products = −890 + 2(44) = −802 kJ per mole CH₄. The combustion still releases heat, but less is collected if water remains vapor. Reversing the phase correction gives condensation, −44 kJ/mol. Never cancel H₂O(l) against H₂O(g) in a Hess sum without including that phase-change energy.