Q.Calculate the standard heat of formation of propane, if its heat of combustion is $-2220.2$ kJ mol$^{-1}$. The heats of formation of $CO_2(g)$ and $H_2O(l)$ are $-393.5$ and $-285.8$ kJ mol$^{-1}$ respectively.

Chemists name a whole family of specific enthalpy changes, each tied to a precisely-defined process. Standard heat of formation $\Delta H_f^0$: enthalpy change forming 1 mole of compound from its elements in standard states (298 K, 1 bar), with every pure element fixed at zero by convention. Standard enthalpy of reaction from formation data: $\Delta H_r^0=\sum\Delta H_f^0(\text{products})-\sum\Delta H_f^0(\text{reactants})$. Heat of combustion $\Delta H_C$: enthalpy change burning 1 mole completely in excess oxygen -- ALWAYS negative, since combustion is always exothermic. Heat of solution: enthalpy change dissolving 1 mole of solute in a specified amount of solvent. Heat of neutralisation: enthalpy change neutralising 1 gram-equivalent of acid by 1 gram-equivalent of base in dilute solution -- remarkably constant at $-57.32$ kJ for ANY strong acid + strong base combination, because Arrhenius's theory says both fully ionise, so the reaction is always really just $H^+(aq)+OH^-(aq)\rightarrow H_2O(l)$ underneath the spectator ions. Molar heat of fusion/vapourisation/sublimation: enthalpy change melting/boiling/sublimi …