Standard enthalpy change of combustion

The standard enthalpy of combustion is the enthalpy change when one mole of a reactant completely burns in excess oxygen under standard thermodynamic conditions (although experimental values are usually obtained under different conditions and subsequently adjusted). By definition, the combustion reactions are always particularly exothermic and so enthalpies of combustion are always negative, although the values for individual combustions may vary.

The most common way of calculating the enthalpy change of combustion (or formation) is by using a Hess cycle or by using numerical based bond enthalpies. It is commonly denoted as $\Delta H_{{{\mathrm {comb}}}}^{{\circ }}$ or $\Delta H_{{{\mathrm {c}}}}^{{\circ }}$. When the enthalpy required is not a combustion, it can be denoted as $\Delta H_{{{\mathrm {total}}}}^{{\circ }}$. Enthalpies of combustion are typically measured using bomb calorimetry, and have units of energy (typically kJ); strictly speaking, the enthalpy change per mole of substance combusted is the standard molar enthalpy of combustion (which typically would have units of kJ mol−1).

Alcohols and alkanes

For alcohols and alkanes containing the same number of carbon atoms, e.g. methane (CH4) and methanol (CH3OH), for which the standard enthalpy change of the alkane would be more negative than the alcohol ($\Delta H_{{{\mathrm {c}}}}^{{\circ }}[{\mathrm {CH_{{4(g)}}}}]$ = −890.3 kJ mol−1, $\Delta H_{{{\mathrm {c}}}}^{{\circ }}[{\mathrm {CH_{3}OH_{{(l)}}}}]$ = −726.0 kJ mol−1).

Since the complete combustion of both of these carbon compounds produce carbon dioxide (CO2) and water (H2O), there is more bond-breaking and bond-making when methane is burnt. The presence of an -OH bond on methanol means that there is less bond-breaking and bond-making to produce water compared to methane.[1]

References

1. ^ Edexcel AS Chemistry by Ann Fullick and Bob McDuell