# 5.7 Enthalpies of Formation

By using the methods we have just discussed, we can calculate the enthalpy changes for a great many reactions from tabulated H values. Many experimental data are tabulated according to the type of process. For example, extensive tables exist of enthalpies of vaporization (H for converting liquids to gases), enthalpies of fusion (H for melting solids), enthalpies of combustion (H for combusting a substance in oxygen), and so forth. A particularly important process used for tabulating thermochemical data is the formation of a compound from its constituent elements. The enthalpy change associated with this process is called the enthalpy of formation (or heat of formation) and is labeled Hf, where the subscript f indicates that the substance has been formed from its elements.

The magnitude of any enthalpy change depends on the conditions of temperature, pressure, and state (gas, liquid, or solid, crystalline form) of the reactants and products. In order to compare the enthalpies of different reactions, it is convenient to define a set of conditions, called a standard state, at which most enthalpies are tabulated. The standard state of a substance is its pure form at atmospheric pressure (1 atm; see Section 10.2) and the temperature of interest, which we usually choose to be 298 K (25°C). The standard enthalpy of a reaction is defined as the enthalpy change when all reactants and products are in their standard states. We denote a standard enthalpy as , where the superscript ° indicates standard-state conditions.

The standard enthalpy of formation of a compound, , is the change in enthalpy for the reaction that forms 1 mol of the compound from its elements, with all substances in their standard states. We usually report values at 298 K. If an element exists in more than one form under standard conditions, the most stable form of the element is used for the formation reaction. For example, the standard enthalpy of formation for ethanol, C2H5OH, is the enthalpy change for the following reaction:

[5.22]

The elemental source of oxygen is O2, not O or O3, because O2 is the stable form of oxygen at 298 K and standard atmospheric pressure. Similarly, the elemental source of carbon is graphite and not diamond, because the former is the more stable (lower energy) form at 298 K and standard atmospheric pressure (see Practice Exercise 5.7). Likewise, the most stable form of hydrogen under standard conditions is H2(g), so this is used as the source of hydrogen in Equation 5.22.

The stoichiometry of formation reactions always indicates that 1 mol of the desired substance is produced, as in Equation 5.22. As a result, enthalpies of formation are reported in kJ/mol of the substance. Several standard enthalpies of formation are given in Table 5.3. A more complete table is provided in Appendix C. By definition, the standard enthalpy of formation of the most stable form of any element is zero because there is no formation reaction needed when the element is already in its standard state. Thus, the values of for C(graphite), H2(g), O2(g), and the standard states of other elements are zero by definition.

## Using Enthalpies of Formation to Calculate Enthalpies of Reaction

Why are tabulations of so useful? As we will see in this section, we can use Hess's law to calculate the standard enthalpy change for any reaction for which we know the values for all reactants and products. For example, consider the combustion of propane gas, C3H8(g), with oxygen to form CO2(g) and H2O(l) under standard conditions:

We can write this equation as the sum of three formation reactions:

[5.23]

[5.24]

[5.25]

[5.26]

From Hess's law we can write the standard enthalpy change for the overall reaction, Equation 5.26, as the sum of the enthalpy changes for the processes in Equations 5.23 through 5.25. We can then use values from Table 5.3 to compute a numerical value for :

[5.27]

We notice that several aspects of the above calculation depend on the guidelines we discussed in Section 5.4:

1. Equation 5.23 is the reverse of the formation reaction for C3H8(g). As a result, the enthalpy change for this reaction is -[C3H8(g)].
2. Equation 5.24 is the formation reaction for 3 mol of CO2(g). Because enthalpy is an extensive property, the enthalpy change for this step is 3[CO2(g)]. Similarly, the enthalpy change for Equation 5.25 is 4[H2O(l)]. Notice that the reaction specifies that H2O(l) was produced; we were careful to use the value of for H2O(l) rather than that for H2O(g).
3. We assume that the stoichiometric coefficients in the balanced equation represent moles. Thus, for Equation 5.27, the value = -2220 kJ represents the enthalpy change for the reaction of 1 mol C3H8 and 5 mol O2 to form 3 mol CO2 and 4 mol H2O. The product of the number of moles and the enthalpy change in kJ/mol has the units kJ: (number of moles) ( in kJ/mol) = kJ. We therefore report in kJ.

Figure 5.20 presents an enthalpy diagram for Equation 5.26, showing how it can be broken into steps involving formation reactions.

Figure 5.20 Enthalpy diagram for the combustion of 1 mol of propane gas, C3H8(g). The overall reaction is C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(l). We can imagine this reaction as occurring in three steps. First, C3H8(g) is decomposed to its elements, so H1 = -[C3H8(g)]. Second, 3 mol CO2(g) are formed, so H2 = 3[CO2(g)]. Finally, 4 mol H2O(l) are formed; thus, H3 = 4[H2O(l)]. Hess's law tells us that = H1 + H2 + H3. This same result is given by Equation 5.28 because [O2(g)] = 0.

We can break down any reaction into formation reactions as we have done here. When we do, we obtain the following general result for the calculation of from the values of the reactants and products:

[5.28]

The symbol (sigma) means "the sum of," and n and m are the stoichiometric coefficients of the reaction. The first term in Equation 5.28 represents the formation reactions of the products, which are written in the "forward" direction; that is, elements reacting to form products. This term is analogous to Equations 5.24 and 5.25 in the previous example. The second term represents the reverse of the formation reactions of the reactants, as in Equation 5.23, which is why the values have a minus sign in front of them.

#### SAMPLE EXERCISE 5.9

(a) Calculate the standard enthalpy change for the combustion of 1 mol of benzene, C6H6(l), to CO2(g) and H2O(l). (b) Compare the quantity of heat produced by combustion of 1.00 g propane to that produced by 1.00 g benzene.

SOLUTION (a) We know that a combustion reaction involves O2(g) as a reactant. Our first step is to write a balanced reaction for the combustion reaction of 1 mol C6H6(l):

We can calculate for the reaction by using Equation 5.28 and data in Table 5.3. We remember to multiply the value for each substance in the reaction by that substance's stoichiometric coefficient. We also recall that = 0 for any element in its most stable form under standard conditions, so [O2(g)] = 0:

(b) From the example worked in the text, we have = -2220 kJ for the combustion of 1 mol of propane. In part (a) of this exercise we determined that = -3267 kJ for the combustion of 1 mol benzene. To determine the heat of combustion per gram of each substance, we use the molecular weights to convert moles to grams:

Both propane and benzene are hydrocarbons. As a rule, the energy obtained from the combustion of a gram of hydrocarbon is between 40 and 50 kJ.

#### PRACTICE EXERCISE

Using the standard enthalpies of formation listed in Table 5.3, calculate the enthalpy change for the combustion of 1 mol of ethanol:

#### SAMPLE EXERCISE 5.10

The standard enthalpy change for the reaction

is 178.1 kJ. From the values for the standard enthalpies of formation of CaO(s) and CO2(g) given in Table 5.3, calculate the standard enthalpy of formation of CaCO3(s).

SOLUTION The standard enthalpy change in the reaction is

Inserting the known values, we have

Solving for (CaCO3) gives

#### PRACTICE EXERCISE

Given this standard enthalpy of reaction, use the standard enthalpies of formation in Table 5.3 to calculate the standard enthalpy of formation of CuO(s):