Enthalpy of Reaction, Formation, & Combustion

symbol for enthalpy of reaction

Core Concepts

In this article, you will learn the basics of enthalpy, as well as how to use enthalpy of formation to calculate enthalpies of reaction and enthalpies of combustion.

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Chemists and physicists tend to define change in enthalpy as the heat exchange of a system at constant pressure.

Enthalpy is an important thermodynamic concept because it informs whether a process is likely to occur, including chemical reactions. Specifically, chemists often use Gibbs free energy to represent the favorability, or spontaneity, of a reaction. Enthalpy has a direct relationship with Gibbs free energy, as indicated by the equation:

    \begin{gather*} {dG = dH - TdS} \end{gather*}

Since negative changes in Gibbs indicate spontaneous reactions, many “exothermic” reactions, involving negative changes in enthalpy, tend to be spontaneous. The opposite goes for “endothermic” reactions that have positive changes in enthalpy. Due to the importance of enthalpy in thermodynamically describing chemical reactions, chemists have determined many ways of measuring and calculating enthalpy.

Enthalpy of Formation

One of the most common ways of calculating the enthalpy of a reaction (or “heat of reaction”) involves using what chemists call enthalpies of formation (or “heat of formation”). For context, each molecule has a characteristic enthalpy of formation. This enthalpy essentially represents the sum total energy of each bond in the molecule. Expressed differently, a molecule’s enthalpy of formation is the heat associated with forming it from its most basic components.

For example, take the reaction forming hexane (C6H14) from six moles of elemental carbon and seven moles of H2. The heat given off by this reaction, under constant pressure, is equal to the enthalpy of formation of hexane.

    \begin{gather*} {6C + 7H_{2} \rightarrow C_{6}H_{14}} \\ {\Delta H_{f}\degree = -199\text{kJ/mol}} \end{gather*}

It’s important to note that the above value for enthalpy of formation only applies under “standard conditions”. Specifically, a temperature of 20 degrees Celsius (or 298.15K) and a pressure of 1 atm. In context, you will often see standard conditions abbreviated to “STP” for “standard temperature and pressure”. If the reaction takes place under nonstandard conditions, the enthalpy of formation would change. All enthalpies that assume standard conditions have a small circle in superscript, similar to the notation for degrees.

Importantly, all enthalpies are what chemists call state functions, in that the change in enthalpy between two states is equal to the difference between the states’ enthalpies, regardless of the intermediate steps taken between the two states. Thus, regardless of mechanism between the carbon and hydrogen, an enthalpic change of -199kJ/mol always occurs under standard conditions.

For More Help, Watch out Interactive Video Explaining Enthalpy of Formation!

Enthalpy of Reaction

Using enthalpies of formation, you can calculate the enthalpic change of any chemical reaction at a given temperature. Importantly, there are many ways of calculating an enthalpy of reaction. Some examples include using bond enthalpies or the temperature change to the surroundings. However, if your reaction involves well known reactants under familiar conditions, then the necessary enthalpies of formation exist online. In such cases, with a fairly simple formula, enthalpies of formation most easily yield the reaction enthalpy.

To calculate enthalpy of reaction, you need to multiply the enthalpies of formation of each of your reactants by the stoichiometric coefficients of those reactants in the balanced chemical equation. Then, you need to add the multiplied enthalpies of the products and reactants separately. Finally, you subtract the combined enthalpy of the reactants from the products to yield the overall enthalpy of reaction.

    \begin{gather*} {aA + bB \rightarrow cC + dD} \\ {\Delta H_{rxn}\degree = \left( \left( c*\Delta H_{f,C}\degree\right) + \left( d*\Delta H_{f,D}\degree\right) \right) - \left( \left( a*\Delta H_{f,A}\degree\right) + \left( b*\Delta H_{f,B}\degree\right) \right)} \end{gather*}

Interestingly, chemists also use this process of subtracting the combined values of the reactants from the combined values of the products to calculate an overall reaction value for many other state variables, such as entropy of reaction and Gibbs free energy of reaction. Broadly, chemists call this method of calculating change in a state variable Hess’s Law, after Swiss chemist Germain Hess. Many chemistry students memorize the phrase “products minus reactants” to remember the formula associated with Hess’s Law.

    \begin{align*} {\Delta H_{rxn}\degree &= \sum \Delta H_{f,products}\degree - \sum \Delta H_{f,reactants}\degree} \\ {\Delta S_{rxn}\degree &= \sum S_{products}\degree - \sum S_{reactants}\degree} \\ {\Delta G_{rxn}\degree &= \sum \Delta G_{f,products}\degree - \sum \Delta G_{f,reactants}\degree} \end{align*}

Let’s look at a worked example.

Enthalpy of Reaction Example: Nitrogen Dioxide Decomposition

Nitrogen dioxide sometimes decomposes into nitrogen monoxide and diatomic oxygen according to the following chemical reaction:

    \begin{gather*} {2NO_{2} \rightarrow 2NO + O_{2}} \end{gather*}

Under standard conditions, nitrogen dioxide and nitrogen monoxide have enthalpies of formation of 33.18\text{kJ/mol} and 90.25\text{kJ/mol}, respectively. Because elemental oxygen occurs naturally as diatomic oxygen, O2 has an enthalpy of formation of zero.

    \begin{align*} {\Delta H_{f, NO_{2}}\degree &= 33.18\text{kJ/mol}} \\ {\Delta H_{f, NO}\degree &= 90.25\text{kJ/mol}} \\ {\Delta H_{f, O_{2}}\degree &= 0\text{kJ/mol}} \end{align*}

To calculate the enthalpy of reaction, we need to multiply the enthalpies of formation of both nitrogen dioxide and nitrogen monoxide by 2, because both have a stoichiometric coefficient of 2 in the balanced chemical equation. Then, we take the multiplied enthalpy of nitrogen dioxide (the “products”) and subtract the multiplied enthalpy of nitrogen monoxide (the “reactants”) to get our overall enthalpy of reaction of 114.14 kilojoules per “one mole of reaction”. This positive reaction enthalpy reveals that the decomposition of nitrogen dioxide is endothermic.

    \begin{align*} {\Delta H_{rxn}\degree&=\sum\Delta H_{f,products}\degree -\sum\Delta H_{f,reactants}\degree}\\{\Delta H_{rxn}\degree&=\left( 2*90.25 \right) -\left(2*33.18\right) =114.14\text{kJ/mol}} \end{align*}

Importantly, “one mole of reaction” here refers to the decomposition of two moles of nitrogen dioxide. This is because we calculated the reaction enthalpy using a balanced equation which gave nitrogen dioxide a coefficient of 2. If you instead wanted to know the reaction enthalpy of the decomposition of one mole of nitrogen dioxide, you can simply divide the enthalpy we calculated by two, since that would be a “half mole of reaction.” Similarly, if you wanted to do the same for four moles of nitrogen dioxide, or “two moles of reaction,” you would multiply our value by 2.

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Enthalpy of Combustion

File:Midsummer bonfire closeup.jpg
An example of a combustion reaction. Source.

Combustion reactions provide one of the most common reaction types for which chemists use Hess’s Law to then calculate enthalpy of reaction from enthalpies of formation. Importantly, the term “enthalpy of combustion” is used for such enthalpies of reaction, specifically concerning the molecule being combusted. For instance, chemists would use the phrase “hexane’s enthalpy of combustion” to describe the standard reaction enthalpy associated with the combustion reaction of hexane.

The same rules as enthalpy of reaction apply when calculating enthalpy of combustion, with the added benefit that different combustion reactions often have the same products. For hydrocarbons, the products generally only involve carbon dioxide and water, though the quantities may differ depending on the number of carbons and oxygens in the molecule. Combustion may also produce nitrogen dioxide and hydrogen sulfide if the combusted molecule has nitrogens or sulfurs.

\Delta H_{f} \degree \text{(kJ/mol)}
Table of enthalpies of formation for common combustion products

Let’s look at another example, this time a combustion reaction.

Enthalpy of Combustion Reaction Example: Ethanol Combustion

When one molecule of ethanol combusts under standard conditions, the reaction then produces two carbon dioxides and three water molecules according to the following equation:

    \begin{gather*} {H_{3}CCH_{2}OH + 3O_{3} \rightarrow 2CO_{2} + 3H_{2}O} \end{gather*}

Thus, to yield the enthalpy of the combustion reaction, we then sum the enthalpies of formation of the products, weighted by their stoichiometric coefficients, and subtract the enthalpy of formation of ethanol (-277.69\text{kJ/mol}). The result is ethanol’s standard enthalpy of combustion of -1234.79\text{kJ/mol}.

    \begin{align*} {\Delta H_{rxn}\degree &= \left( \left( 2 *\Delta H_{f,CO_{2}}\degreeright) + \left( 3*\Delta H_{f,H_{2}O}\degree\right) \right) - \left(\Delta H_{f,ethanol}\degree\right)} \end{align*}

    \begin{align*} {\Delta H_{rxn}\degree &= \left( \left(2 *-393.51\right) + \left( 3*-241.82\right) \right) - \left( -277.69\right)} \end{align*}

Enthalpy of Reaction Practice Problems

Problem 1

What is the enthalpy of reaction?

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    \begin{gather*} {CaO + CO_{2} \rightarrow CaCO_{3}} \end{gather*}

Problem 2

What is the enthalpy of reaction?

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    \begin{gather*} {2Ba + O_{2} \rightarrow 2BaO} \end{gather*}

Problem 3

What is the enthalpy of combustion of cysteine?

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    \begin{gather*} {C_{3}H_{7}NO_{2}S + 3O_{2} \rightarrow 3CO_{2} + H_{2}S + NO_{2}} \end{gather*}

Enthalpy of Reaction Practice Problem Solutions

1: -258.0 \text{kJ/mol}

2: -1116 \text{kJ/mol}

3: -600.6 \text{kJ/mol}