Hess’s law allows us to construct new chemical reactions and predict what their enthalpies of reaction will be. This is a very useful tool because now we don’t have to measure the enthalpy changes of every possible reaction. We need measure only the enthalpy changes of certain benchmark reactions and then use these reactions to algebraically construct any possible reaction and combine the enthalpies of the benchmark reactions accordingly.

But what are the benchmark reactions? We need to have some agreed-on sets of reactions that provide the central data for any thermochemical equation.

Formation reactions are chemical reactions that form one mole of a substance from its constituent elements in their standard states. By standard states we mean as a diatomic molecule if that is how the element exists and the proper phase at normal temperatures (typically room temperature). The product is one mole of substance, which may require that coefficients on the reactant side be fractional (a change from our normal insistence that all coefficients be whole numbers). For example, the formation reaction for methane (CH₄) is

C(s) + 2 H₂(g) → CH₄(g)

The formation reaction for carbon dioxide (CO₂) is

C(s) + O₂(g) → CO₂(g)

In both cases, one of the elements is a diatomic molecule because that is the standard state for that particular element. The formation reaction for H₂O—

2 H₂(g) + O₂(g) → 2 H₂O(ℓ)

—is not in a standard state because the coefficient on the product is 2; for a proper formation reaction, only one mole of product is formed. Thus, we have to divide all coefficients by 2:

H₂(g) + 1/2 O₂(g) → H₂O(ℓ)

On a molecular scale, we are using half of an oxygen molecule, which may be problematic to visualize. However, on a molar level, it implies that we are reacting only half of a mole of oxygen molecules, which should be an easy concept for us to understand.

Given the formula of any substance, you should be able to write the proper formation reaction for that substance.

The enthalpy change for a formation reaction is called the enthalpy of formation The subscript f is the clue that the reaction of interest is a formation reaction. Thus, for the formation of FeO(s),

Note that now we are using kJ/mol as the unit because it is understood that the enthalpy change is for one mole of substance. Note, too, by definition, that the enthalpy of formation of an element is exactly zero because making an element from an element is no change. For example,

H₂(g) → H₂(g) ΔH_f = 0

Formation reactions and their enthalpies are important because these are the thermochemical data that are tabulated for any chemical reaction. Table 7.2 “Enthalpies of Formation for Various Substances” lists some enthalpies of formation for a variety of substances; in some cases, however, phases can be important (e.g., for H₂O).

It is easy to show that any general chemical equation can be written in terms of the formation reactions of its reactants and products, some of them reversed (which means the sign must change in accordance with Hess’s law). For example, consider

2 NO₂(g) → N₂O₄(g)

We can write it in terms of the (reverse) formation reaction of NO₂ and the formation reaction of N₂O₄:

We must multiply the first reaction by 2 to get the correct overall balanced equation. We are simply using Hess’s law in combining the ΔH_f values of the formation reactions.

Example 13

Show that the reaction

Fe₂O₃(s) + 3 SO₃(g) → Fe₂(SO₄)₃(s)

can be written as a combination of formation reactions.

Solution

There will be three formation reactions. The one for the products will be written as a formation reaction, while the ones for the reactants will be written in reverse. Furthermore, the formation reaction for SO₃ will be multiplied by 3 because there are three moles of SO₃ in the balanced chemical equation. The formation reactions are as follows:

2 Fe(s) + 3 S(s) + 6 O₂(g) → Fe₂(SO₄)₃(s)

When these three equations are combined and simplified, the overall reaction is

Fe₂O₃(s) + 3 SO₃(s) → Fe₂(SO₄)₃(s)

Test Yourself

Write the formation reactions that will yield

2 SO₂(g) + O₂(g) → 2 SO₃(g).

Answer

Now that we have established formation reactions as the major type of thermochemical reaction we will be interested in, do we always need to write all the formation reactions when we want to determine the enthalpy change of any random chemical reaction? No. There is an easier way. You may have noticed in all our examples that we change the signs on all the enthalpies of formation of the reactants, and we don’t change the signs on the enthalpies of formation of the products. We also multiply the enthalpies of formation of any substance by its coefficient—technically, even when it is just 1. This allows us to make the following statement: the enthalpy change of any chemical reaction is equal to the sum of the enthalpies of formation of the products minus the sum of the enthalpies of formation of the reactants. In mathematical terms,

where n_p and n_r are the number of moles of products and reactants, respectively (even if they are just 1 mol), and ΔH_f,p and ΔH_f,r are the enthalpies of formation of the product and reactant species, respectively. This products-minus-reactants scheme is very useful in determining the enthalpy change of any chemical reaction, if the enthalpy of formation data are available. Because the mol units cancel when multiplying the amount by the enthalpy of formation, the enthalpy change of the chemical reaction has units of energy (joules or kilojoules) only.

Example 14

Use the products-minus-reactants approach to determine the enthalpy of reaction for

Solution

The enthalpies of formation are multiplied by the number of moles of each substance in the chemical equation, and the total enthalpy of formation for reactants is subtracted from the total enthalpy of formation of the products:

ΔHrxn=[(2 mol)(−92.3 kJ/ mol)+(1 mol)(0 kJ/ mol)]−[(2 mol)(−36.3 kJ/ mol)+(1 mol)(0 kJ/ mol)]

All the mol units cancel. Multiplying and combining all the values, we get

ΔH_rxn = −112.0 kJ

Test Yourself

What is the enthalpy of reaction for this chemical equation?

Answer

+2.8 kJ