Hess's law equation cancellation

Keep the stoichiometry attached to the enthalpy. Use the same abstract $A,B,C$ and suppose each of two entries fixes an amount of substance for every species:Here each unit coefficient represents one mole on the common basis. To obtain $A\to C$, halve $R_1$ and reverse $R_2$. Adding those transformed chemical equations givesThe equal amounts of $B$ on opposite sides cancel, leaving $A\to C$. Doubling this final chemical equation would give $2A\to 2C$ and $\Delta H=-14\mathrm{kJ}$; its change per mole of $A$ would remain $-7\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$.

More generally, write each chemical equation as a stoichiometric vector ${\mathbf{\nu }}_j$, with negative entries for reactants and positive entries for reaction products. Finding coefficients $a_j$ with ${\mathbf{\nu }}_{\mathrm{target}}={\sum }_j{a}_j{\mathbf{\nu }}_j$ also gives $\Delta H_{\mathrm{target}}={\sum }_j{a}_j\Delta H_j$, using compatible state and amount conventions. An identical chemical species cancels only with the same thermodynamic phase and composition conditions. A leftover water or dilution step is a real part of the balance.