CHEMISTRY THE CENTRAL SCIENCE

5 THERMOCHEMISTRY

5.7 ENTHALPIES OF FORMATION

We can use the methods just discussed to calculate enthalpy changes for a great many reactions from tabulated ΔH values. For example, extensive tables exist of enthalpies of vaporization (ΔH for converting liquids to gases), enthalpies of fusion (ΔH for melting solids), enthalpies of combustion (ΔH 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), ΔHf, where the subscript f indicates that the substance has been formed from its constituent elements.

The magnitude of any enthalpy change depends on the temperature, pressure, and state (gas, liquid, or solid crystalline form) of the reactants and products. To compare enthalpies of different reactions, we must 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) and the temperature of interest, which we usually choose to be 298 K (25 °C).* The standard enthalpy change of a reaction is defined as the enthalpy change when all reactants and products are in their standard states. We denote a standard enthalpy change as ΔH°, where the superscript ° indicates standard-state conditions.

The standard enthalpy of formation of a compound, Δf, is the change in enthalpy for the reaction that forms one mole of the compound from its elements with all substances in their standard states:

If:     elements (in standard state) → compound (1 mol in standard state)

Then:   ΔH = 

TABLE 5.3 • Standard Enthalpies of Formation, ΔH°f, at 298 K

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 usually used for the formation reaction. For example, the standard enthalpy of formation for ethanol, C2H5OH, is the enthalpy change for the reaction

The elemental source of oxygen is O2, not O or O3, because O2 is the stable form of oxygen at 298 K and atmospheric pressure. Similarly, the elemental source of carbon is graphite and not diamond because graphite is the more stable (lower-energy) form at 298 K and atmospheric pressure (see Practice Exercise 5.8). Likewise, the most stable form of hydrogen under standard conditions is H2(g), so this is used as the source of hydrogen in Equation 5.25.

The stoichiometry of formation reactions always indicates that one mole of the desired substance is produced, as in Equation 5.25. As a result, standard enthalpies of formation are reported in kJ/mol of the substance being formed. Some values are given in TABLE 5.3, and a more extensive 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.

GIVE IT SOME THOUGHT

Ozone, O3(g), is a form of elemental oxygen produced during electrical discharge. Is  for O3(g) necessarily zero?

SAMPLE EXERCISE 5.10 Equations Associated with Enthalpies of Formation

For which of these reactions at 25 °C does the enthalpy change represent a standard enthalpy of formation? For each that does not, what changes are needed to make it an equation whose ΔH is an enthalpy of formation?

SOLUTION

Analyze The standard enthalpy of formation is represented by a reaction in which each reactant is an element in its standard state and the product is one mole of the compound.

Plan We need to examine each equation to determine (1) whether the reaction is one in which one mole of substance is formed from the elements, and (2) whether the reactant elements are in their standard states.

Solve In (a) 1 mol Na2O is formed from the elements sodium and oxygen in their proper states, solid Na and O2 gas, respectively. Therefore, the enthalpy change for reaction (a) corresponds to a standard enthalpy of formation.

In (b) potassium is given as a liquid. It must be changed to the solid form, its standard state at room temperature. Furthermore, two moles of product are formed, so the enthalpy change for the reaction as written is twice the standard enthalpy of formation of KCl(s). The equation for the formation reaction of 1 mol of KCl(s) is

Reaction (c) does not form a substance from its elements. Instead, a substance decomposes to its elements, so this reaction must be reversed. Next, the element carbon is given as diamond, whereas graphite is the standard state of carbon at room temperature and 1 atm pressure. The equation that correctly represents the enthalpy of formation of glucose from its elements is

PRACTICE EXERCISE

Write the equation corresponding to the standard enthalpy of formation of liquid carbon tetrachloride (CCl4).

Answer: C(graphite) + 2 Cl2(g) → CCl4(l)

Using Enthalpies of Formation to Calculate Enthalpies of Reaction

We can use Hess's law and tabulations of  values, such as those in Table 5.3 and Appendix C, 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), to CO2(g) and H2O(l) under standard conditions:

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

(Note that it is sometimes useful to add subscripts to the enthalpy changes, as we have done here, to keep track of the associations between reactions and their ΔH values.)

Notice that we have used Hess's law to write the standard enthalpy change for Equation 5.29 as the sum of the enthalpy changes for Equations 5.26 through 5.28. We can use values from Table 5.3 to calculate :

FIGURE 5.23, an enthalpy diagram for Equation 5.29, shows our propane combustion reaction broken down to the three reactions. Several aspects of this calculation depend on the guidelines we discussed in Section 5.4.

1. Decomposition. Equation 5.26 is the reverse of the formation reaction for C3H8(g), so the enthalpy change for this decomposition reaction is the negative of the  value for the propane formation reaction: .

FIGURE 5.23 Enthalpy diagram for propane combustion.

2. Formation of CO2. Equation 5.27 is the formation reaction for 3 mol of CO2(g). Because enthalpy is an extensive property, the enthalpy change for this step is .

3. Formation of H2O. The enthalpy change for Equation 5.28, formation of 4 mol of H2O, is . The reaction specifies that H2O(l) is produced, so be careful to use the value of  for H2O(l) and not the value for H2O(g).

Note that in this analysis we assume that the stoichiometric coefficients in the balanced equation represent moles. For Equation 5.29, therefore, ΔH°rxn = –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) × (ΔH°f in kJ/mol) = kJ. We therefore report  in kJ.

We can break down any reaction into formation reactions as we have done here. When we do, we obtain the general result that the standard enthalpy change of a reaction is the sum of the standard enthalpies of formation of the products minus the standard enthalpies of formation of the reactants:

The symbol Σ (sigma) means “the sum of,” and n and m are the stoichiometric coefficients of the relevant chemical equation. The first term on the right in Equation 5.31 represents the formation reactions of the products, which are written in the “forward” direction in the chemical equation, that is, elements reacting to form products. This term is analogous to Equations 5.27 and 5.28. The second term on the right in Equation 5.31 represents the reverse of the formation reactions of the reactants, analogous to Equation 5.26, which is why this term is preceded by a minus sign.

SAMPLE EXERCISE 5.11 Calculating an Enthalpy of Reaction from Enthalpies of Formation

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

SOLUTION

Analyze (a) We are given a reaction [combustion of C6H6(l) to form CO2(g) and H2O(l)] and asked to calculate its standard enthalpy change, ΔH°. (b) We then need to compare the quantity of heat produced by combustion of 1.00 g C6H6 with that produced by 1.00 g C3H8, whose combustion was treated previously in the text. (See Equations 5.29 and 5.30.)

Plan (a) We need to write the balanced equation for the combustion of C6H6. We then look up  values in Appendix C or in Table 5.3 and apply Equation 5.31 to calculate the enthalpy change the text previously to calculate the enthalpy change per gram of that for the reaction. (b) We use the molar mass of C6H6 to change the enthalpy change per mole to that per gram. We similarly use the molar mass of C3H8 and the enthalpy change per mole calculated in substance.

Solve

(a) We know that a combustion reaction involves O2(g) as a reactant. Thus, the balanced equation for the combustion reaction of 1 mol C6H6(l) is

We can calculate ΔH° for this reaction by using Equation 5.31 and data in Table 5.3. Remember to multiply the  value for each substance in the reaction by that substance's stoichiometric coefficient. Recall also that  for any element in its most stable form under standard conditions, so :

(b) From the example worked in the text, ΔH° = –2220 kJ for the combustion of 1 mol of propane. In part (a) of this ex ercise we determined that ΔH° = –3267 kJ for the combustion of 1 mol benzene. To determine the heat of com bustion per gram of each substance, we use the molar masses to convert moles to grams:

C3H8(g): (–2220 kJ/mol)(1 mol/44.1 g) = –50.3 kJ/g

C6H6(l): (–3267 kJ/mol)(1 mol/78.1 g) = –41.8 kJ/g

Comment 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

Use Table 5.3 to calculate the enthalpy change for the combustion of 1 mol of ethanol:

C2H5OH(l) + 3 O2(g) → 2 CO2(g) + 3 H2O(l)

Answer: –1367 kJ

SAMPLE EXERCISE 5.12 Calculating an Enthalpy of Formation Using an Enthalpy of Reaction

The standard enthalpy change for the reaction CaCO3(s) → CaO(s) + CO2(g) is 178.1 kJ. Use Table 5.3 to calculate the standard enthalpy of formation of CaCO3(s).

SOLUTION

Analyze Our goal is to obtain .

Plan We begin by writing the expression for the standard enthalpy change for the reaction:

Solve Inserting the given  and the known  values from Table 5.3 or Appendix C, we have

Solving for  gives

Check We expect the enthalpy of formation of a stable solid such as calcium carbonate to be negative, as obtained.

PRACTICE EXERCISE

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

CuO(s) + H2(g) → Cu(s) + H2O(l) ΔH° = –129.7 kJ

Answer: –156.1 kJ/mol