Wikipedia has two different pages for enthalpy of combustion and heat of combustion. Their definition is the same and they use the same symbol $\Delta H_c^\circ $

It says that enthalpies of combustion are always negative as these reactions are exothermic. The heat of combustion tables list positive values for all substances. Why do these values differ in sign?

  • $|\Delta H_c^\circ|$ =$\text{heat of combustion}$ – Satwik Pasani Nov 30 '13 at 14:50

Looking at Wikipedia for the definitions:

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).

The heat of combustion $(\Delta H_c^\circ)$ is the energy released as heat when a compound undergoes complete combustion with oxygen under standard conditions. The chemical reaction is typically a hydrocarbon reacting with oxygen to form carbon dioxide, water and heat.

On first inspection, the two seem the same, however, there is a very slight different use in terminology and wording.

1.) What are standard conditions?

Referring back to Wikipedia:

Standard conditions for temperature and pressure are standard sets of conditions for experimental measurements established to allow comparisons to be made between different sets of data.

In chemistry, IUPAC established standard temperature and pressure (informally abbreviated as STP) as a temperature of 273.15 K (0 °C, 32 °F) and an absolute pressure of 100 kPa (14.504 psi, 0.987 atm, 1 bar)

As you know, the standard enthalpy of combustion can be referred to as as $\Delta H ^{\circ} _{\mathrm{total}}$ whose units are in kJ. The molar enthalpy of combustion is how much energy is released per kJ/mol. How is the standard enthalpy of combustion found?

Using the individual enthalpies of formation, and Hess's law, we can calculate the standard enthalpy of combustion, or $\Delta H ^{\circ} _{\mathrm{total}}$.

$\Delta H_{reaction}^\ominus = \sum \Delta H_{\mathrm f \,(products)}^{\ominus} - \sum \Delta H_{\mathrm f \,(reactants)}^{\ominus}$

The standard enthalpies of formation, or the negative values are located here https://en.wikipedia.org/wiki/Standard_enthalpy_change_of_formation_(data_table) and as well, generally in your textbook in an appendix along with the standard entropy of formation, and Gibbs free energy of formation, as all three are state functions.

2.) Why do the signs differ in the values from your textbook compared to the table in Wikipedia?

For the heat of combustion, it is generally expressed in units of higher heating value, lower heating value, and gross heating value. As the article states in Wikipedia, the heating value (or energy value or calorific value) of a substance, usually a fuel or food (see food energy), is the amount of heat released during the combustion of a specified amount of it.

In summary, the sign difference that I see is due to different terminology, they are specifically referring to units of HHV, LHV, and GHV and expressing those in the table below. Although, heat of combustion may also be calculated as the difference between the heat of formation $\big( \Delta H_f^\circ \big)$ of the products and reactants just as the standard enthalpy of combustion, it seems to me that the heat of combustion is specialized in a way to be applied to fuels, and as such, different from the standard enthalpy of combustion that we are accustomed to.

Perhaps someone could clarify a little more on this?

The lower heating value is the measured, or empirical heat of combustion. It is lower because some of the internal energy (enthalpy) that is released as heat goes into the production of water in vapor form.

Since standard enthalpy values require compounds in their standard states (water in liquid form), the higher heating value must be calculated. This is done by adjusting for the heat of vaporization of water for the given mass. Typically this done by adding the adjustment quantity to the measured value.

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.