= Enthalpy cycle
{title2=$\sum_{\rm cycle}\Delta H=0$}
An <enthalpy cycle> is a collection of compatible <thermodynamic processes> that returns the entire material inventory to its initial <thermodynamic state>. Its signed <enthalpy> changes sum to zero. This supplies a consistency check and lets a missing change be recovered from the others.
Consider three abstract <thermodynamic states> $A,B,C$ of the same conserved material inventory at common <temperature> and <pressure>. Choose their <molar enthalpies>, on the same one-<mole> basis, as
$$
h_A=0,\qquad h_B=18\,\mathrm{kJ\,mol^{-1}},\qquad h_C=-7\,\mathrm{kJ\,mol^{-1}}.
$$
These are constructed values, not material measurements. Endpoint subtraction gives
$$
\begin{aligned}
\Delta h_{AB}&=h_B-h_A=+18\,\mathrm{kJ\,mol^{-1}},\\
\Delta h_{BC}&=h_C-h_B=-25\,\mathrm{kJ\,mol^{-1}},\\
\Delta h_{AC}&=h_C-h_A=-7\,\mathrm{kJ\,mol^{-1}}.
\end{aligned}
$$
Thus $+18-25=-7$: the two-step route and the direct route have the same <molar enthalpy> change. Closing the route with $C\to A$ gives $+18-25+7=0$. Shifting every $h$ by the same arbitrary reference constant leaves all these differences unchanged.
\Image[/hess-law-enthalpy-state-paths.png]
{title=Two routes between the same <thermodynamic states> give the same <enthalpy> change}
{id=image-hess-law-enthalpy-state-paths}
{description=Original abstract states with <molar enthalpies> $0$, $18$, and $-7\,\mathrm{kJ\,mol^{-1}}$ on one common basis. Arrows show signed endpoint differences, not an <activation energy> curve, a <reaction mechanism>, or elapsed time.}
The short proof is cancellation:
$$
(H_B-H_A)+(H_C-H_B)=H_C-H_A.
$$
Adding any finite number of consecutive endpoint differences works the same way. For a closed <enthalpy cycle> the final $H_A$ cancels the initial $H_A$. A nonzero experimental residual can therefore expose mismatched states or measurement error; it does not identify which individual input is responsible.
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