Editing Standard enthalpy of formation (section)

==Organic compounds==
The formation reactions for most organic compounds are hypothetical. For instance, carbon and hydrogen will not directly react to form [[methane]] ({{chem2|CH4}}), so that the standard enthalpy of formation cannot be measured directly. However the [[Heat of combustion|standard enthalpy of combustion]] is readily measurable using [[Calorimeter#Bomb calorimeters|bomb calorimetry]]. The standard enthalpy of formation is then determined using [[Hess's law]]. The combustion of methane:
:<chem>CH4 + 2 O2 -> CO2 + 2 H2O</chem>
is equivalent to the sum of the hypothetical decomposition into elements followed by the combustion of the elements to form [[carbon dioxide]] ({{chem2|CO2}}) and water ({{chem2|H2O}}):
:<chem>CH4 -> C + 2H2</chem>
:<chem>C + O2 -> CO2</chem>
:<chem>2H2 + O2 -> 2H2O</chem>

Applying Hess's law, 
:<math>\Delta_\text{comb} H^\ominus ( \text{CH}_4 ) = [ \Delta_\text{f} H^\ominus (\text{CO}_2) + 2 \Delta_\text{f} H^\ominus ( \text{H}_2 \text{O} ) ] - \Delta_\text{f} H^\ominus (\text{CH}_4).</math>

Solving for the standard of enthalpy of formation,
:<math>\Delta_\text{f} H^\ominus (\text{CH}_4) = [ \Delta_\text{f} H^\ominus (\text{CO}_2) + 2 \Delta_\text{f} H^\ominus (\text{H}_2 \text{O})] - \Delta_\text{comb} H^\ominus (\text{CH}_4).</math>

The value of {{tmath|\Delta_\text{f} H^\ominus (\text{CH}_4)}} is determined to be −74.8&nbsp;kJ/mol. The negative sign shows that the reaction, if it were to proceed, would be [[exothermic]]; that is, methane is enthalpically more stable than hydrogen gas and carbon.

It is possible to predict heats of formation for simple [[Strain (chemistry)|unstrained]] organic compounds with the [[heat of formation group additivity]] method.