Direction of Heat Flow

The reaction of powdered aluminum with iron(III) oxide, known as the thermite reaction, generates an enormous amount of heat—enough, in fact, to melt steel (see chapter opening image). The balanced chemical equation for the reaction is as follows:

We can also write this chemical equation as

to indicate that heat is one of the products. Chemical equations in which heat is shown as either a reactant or a product are called thermochemical equations. In this reaction, the system consists of aluminum, iron, and oxygen atoms; everything else, including the container, makes up the surroundings. During the reaction, so much heat is produced that the iron liquefies. Eventually, the system cools; the iron solidifies as heat is transferred to the surroundings. A process in which heat (q) is transferred from a system to its surroundings is described as exothermicA process in which heat $(q)$ is transferred from a system to its surroundings.. By convention, q < 0 for an exothermic reaction.

When you hold an ice cube in your hand, heat from the surroundings (including your hand) is transferred to the system (the ice), causing the ice to melt and your hand to become cold. We can describe this process by the following thermochemical equation:

When heat is transferred to a system from its surroundings, the process is endothermicA process in which heat $(q)$ is transferred to a system from its surroundings.. By convention, q > 0 for an endothermic reaction.

Enthalpy of Reaction

We have stated that the change in energy (ΔE) is equal to the sum of the heat produced and the work performed (Equation 5.3). Work done by an expanding gas is called pressure-volume work, also called PV work. Consider, for example, a reaction that produces a gas, such as dissolving a piece of copper in concentrated nitric acid. The chemical equation for this reaction is as follows:

If the reaction is carried out in a closed system that is maintained at constant pressure by a movable piston, the piston will rise as nitrogen dioxide gas is formed (Figure 5.7 "An Example of Work Performed by a Reaction Carried Out at Constant Pressure"). The system is performing work by lifting the piston against the downward force exerted by the atmosphere (i.e., atmospheric pressure). We find the amount of PV work done by multiplying the external pressure P by the change in volume caused by movement of the piston (ΔV). At a constant external pressure (here, atmospheric pressure)

The negative sign associated with PV work done indicates that the system loses energy. If the volume increases at constant pressure (ΔV > 0), the work done by the system is negative, indicating that a system has lost energy by performing work on its surroundings. Conversely, if the volume decreases (ΔV < 0), the work done by the system is positive, which means that the surroundings have performed work on the system, thereby increasing its energy.

Figure 5.7 An Example of Work Performed by a Reaction Carried Out at Constant Pressure

(a) Initially, the system (a copper penny and concentrated nitric acid) is at atmospheric pressure. (b) When the penny is added to the nitric acid, the volume of NO₂ gas that is formed causes the piston to move upward to maintain the system at atmospheric pressure. In doing so, the system is performing work on its surroundings.

The symbol E in Equation 5.3 represents the internal energyThe sum of the kinetic and potential energies of all of a system’s components. Additionally, ΔE = q + w, where q is the heat produced by the system and w is the work performed by the system. Internal energy is a state function. of a system, which is the sum of the kinetic energy and potential energy of all its components. It is the change in internal energy that produces heat plus work. To measure the energy changes that occur in chemical reactions, chemists usually use a related thermodynamic quantity called enthalpy (H)The sum of a system’s internal energy $E$ and the product of its pressure $P$ and volume $V$: $H=E+PV.$ (from the Greek enthalpein, meaning “to warm”). The enthalpy of a system is defined as the sum of its internal energy E plus the product of its pressure P and volume V:

Because internal energy, pressure, and volume are all state functions, enthalpy is also a state function.

If a chemical change occurs at constant pressure (for a given P, ΔP = 0), the change in enthalpy (ΔH)At constant pressure, the amount of heat transferred from the surroundings to the system or vice versa: .$\Delta H= q_{\text{p}}.$ is

Substituting q + w for ΔE (Equation 5.3) and −w for PΔV (Equation 5.12), we obtain

The subscript p is used here to emphasize that this equation is true only for a process that occurs at constant pressure. From Equation 5.15 we see that at constant pressure the change in enthalpy, ΔH of the system, defined as H_final − H_initial, is equal to the heat gained or lost.

Just as with ΔE, because enthalpy is a state function, the magnitude of ΔH depends on only the initial and final states of the system, not on the path taken. Most important, the enthalpy change is the same even if the process does not occur at constant pressure.

When we study energy changes in chemical reactions, the most important quantity is usually the enthalpy of reaction (ΔH_rxn)The change in enthalpy that occurs during a chemical reaction., the change in enthalpy that occurs during a reaction (such as the dissolution of a piece of copper in nitric acid). If heat flows from a system to its surroundings, the enthalpy of the system decreases, so ΔH_rxn is negative. Conversely, if heat flows from the surroundings to a system, the enthalpy of the system increases, so ΔH_rxn is positive. Thus ΔH_rxn < 0 for an exothermic reaction, and ΔH_rxn > 0 for an endothermic reaction. In chemical reactions, bond breaking requires an input of energy and is therefore an endothermic process, whereas bond making releases energy, which is an exothermic process. The sign conventions for heat flow and enthalpy changes are summarized in the following table:

If ΔH_rxn is negative, then the enthalpy of the products is less than the enthalpy of the reactants; that is, an exothermic reaction is energetically downhill (part (a) in Figure 5.8 "The Enthalpy of Reaction"). Conversely, if ΔH_rxn is positive, then the enthalpy of the products is greater than the enthalpy of the reactants; thus, an endothermic reaction is energetically uphill (part (b) in Figure 5.8 "The Enthalpy of Reaction"). Two important characteristics of enthalpy and changes in enthalpy are summarized in the following discussion.

Figure 5.8 The Enthalpy of Reaction

Energy changes in chemical reactions are usually measured as changes in enthalpy. (a) If heat flows from a system to its surroundings, the enthalpy of the system decreases, ΔH_rxn is negative, and the reaction is exothermic; it is energetically downhill. (b) Conversely, if heat flows from the surroundings to a system, the enthalpy of the system increases, ΔH_rxn is positive, and the reaction is endothermic; it is energetically uphill.

• Reversing a reaction or a process changes the sign of ΔH. Ice absorbs heat when it melts (electrostatic interactions are broken), so liquid water must release heat when it freezes (electrostatic interactions are formed):

  Equation 5.17

  $$\text{heat}+{\text{H}}_{\text{2}}\text{O}\left(\text{s}\right)\to {\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)\Delta H>0$$

  Equation 5.18

  $${\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)\to {\text{H}}_{\text{2}}\text{O}\left(\text{s}\right)+\text{heat}\Delta H<0$$

  In both cases, the magnitude of the enthalpy change is the same; only the sign is different.

• Enthalpy is an extensive property (like mass). The magnitude of ΔH for a reaction is proportional to the amounts of the substances that react. For example, a large fire produces more heat than a single match, even though the chemical reaction—the combustion of wood—is the same in both cases. For this reason, the enthalpy change for a reaction is usually given in kilojoules per mole of a particular reactant or product. Consider Equation 5.19, which describes the reaction of aluminum with iron(III) oxide (Fe₂O₃) at constant pressure. According to the reaction stoichiometry, 2 mol of Fe, 1 mol of Al₂O₃, and 851.5 kJ of heat are produced for every 2 mol of Al and 1 mol of Fe₂O₃ consumed:

  Equation 5.19

  2Al(s) + Fe₂O₃(s) → 2Fe(s) + Al₂O₃(s) + 851.5 kJ

  Thus ΔH = −851.5 kJ/mol of Fe₂O₃. We can also describe ΔH for the reaction as −425.8 kJ/mol of Al: because 2 mol of Al are consumed in the balanced chemical equation, we divide −851.5 kJ by 2. When a value for ΔH, in kilojoules rather than kilojoules per mole, is written after the reaction, as in Equation 5.20, it is the value of ΔH corresponding to the reaction of the molar quantities of reactants as given in the balanced chemical equation:

  Equation 5.20

  $$\text{2Al}\left(\text{s}\right)+{\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}\left(\text{s}\right)\to \text{2Fe}\left(\text{s}\right)+{\text{Al}}_{\text{2}}{\text{O}}_{\text{3}}\left(\text{s}\right)\Delta {\text{H}}_{\text{rxn}}=-\text{851}.\text{5 kJ}$$

  If 4 mol of Al and 2 mol of Fe₂O₃ react, the change in enthalpy is 2 × (−851.5 kJ) = −1703 kJ. We can summarize the relationship between the amount of each substance and the enthalpy change for this reaction as follows:

  Equation 5.21

  $$-\frac{\text{851}\text{.5 kJ}}{\text{2 mol Al}}=-\frac{\text{425}\text{.8 kJ}}{\text{1 mol Al}}=-\frac{\text{1703 kJ}}{\text{4 mol Al}}$$

The relationship between the magnitude of the enthalpy change and the mass of reactants is illustrated in Example 2.

Hess’s Law

Because enthalpy is a state function, the enthalpy change for a reaction depends on only two things: (1) the masses of the reacting substances and (2) the physical states of the reactants and products. It does not depend on the path by which reactants are converted to products. If you climbed a mountain, for example, the altitude change would not depend on whether you climbed the entire way without stopping or you stopped many times to take a break. If you stopped often, the overall change in altitude would be the sum of the changes in altitude for each short stretch climbed. Similarly, when we add two or more balanced chemical equations to obtain a net chemical equation, ΔH for the net reaction is the sum of the ΔH values for the individual reactions. This principle is called Hess’s lawThe enthalpy change $(\Delta H)$ for an overall reaction is the sum of the $\Delta H$ values for the individual reactions., after the Swiss-born Russian chemist Germain Hess (1802–1850), a pioneer in the study of thermochemistry. Hess’s law allows us to calculate ΔH values for reactions that are difficult to carry out directly by adding together the known ΔH values for individual steps that give the overall reaction, even though the overall reaction may not actually occur via those steps.

We can illustrate Hess’s law using the thermite reaction. The overall reaction shown in Equation 5.20 can be viewed as occurring in three distinct steps with known ΔH values. As shown in Figure 5.9 "Energy Changes Accompanying the Thermite Reaction", the first reaction produces 1 mol of solid aluminum oxide (Al₂O₃) and 2 mol of liquid iron at its melting point of 1758°C (part (a) in Equation 5.22); the enthalpy change for this reaction is −732.5 kJ/mol of Fe₂O₃. The second reaction is the conversion of 2 mol of liquid iron at 1758°C to 2 mol of solid iron at 1758°C (part (b) in Equation 5.22); the enthalpy change for this reaction is −13.8 kJ/mol of Fe (−27.6 kJ per 2 mol Fe). In the third reaction, 2 mol of solid iron at 1758°C is converted to 2 mol of solid iron at 25°C (part (c) in Equation 5.22); the enthalpy change for this reaction is −45.5 kJ/mol of Fe (−91.0 kJ per 2 mol Fe). As you can see in Figure 5.9 "Energy Changes Accompanying the Thermite Reaction", the overall reaction is given by the longest arrow (shown on the left), which is the sum of the three shorter arrows (shown on the right). Adding parts (a), (b), and (c) in Equation 5.22 gives the overall reaction, shown in part (d):

The net reaction in part (d) in Equation 5.22 is identical to Equation 5.20. By Hess’s law, the enthalpy change for part (d) is the sum of the enthalpy changes for parts (a), (b), and (c). In essence, Hess’s law enables us to calculate the enthalpy change for the sum of a series of reactions without having to draw a diagram like that in Figure 5.9 "Energy Changes Accompanying the Thermite Reaction".

Figure 5.9 Energy Changes Accompanying the Thermite Reaction

Because enthalpy is a state function, the overall enthalpy change for the reaction of 2 mol of Al(s) with 1 mol of Fe₂O₃(s) is −851.1 kJ, whether the reaction occurs in a single step (ΔH₄, shown on the left) or in three hypothetical steps (shown on the right) that involve the successive formation of solid Al₂O₃ and liquid iron (ΔH₁), solid iron at 1758°C (ΔH₂), and solid iron at 25°C (ΔH₃). Thus ΔH₄ = ΔH₁ + ΔH₂ + ΔH₃, as stated by Hess’s law.

Comparing parts (a) and (d) in Equation 5.22 also illustrates an important point: The magnitude of ΔH for a reaction depends on the physical states of the reactants and the products (gas, liquid, solid, or solution). When the product is liquid iron at its melting point (part (a) in Equation 5.22), only 732.5 kJ of heat are released to the surroundings compared with 852 kJ when the product is solid iron at 25°C (part (d) in Equation 5.22). The difference, 120 kJ, is the amount of energy that is released when 2 mol of liquid iron solidifies and cools to 25°C. It is important to specify the physical state of all reactants and products when writing a thermochemical equation.

When using Hess’s law to calculate the value of ΔH for a reaction, follow this procedure:

1. Identify the equation whose ΔH value is unknown and write individual reactions with known ΔH values that, when added together, will give the desired equation.
2. Arrange the chemical equations so that the reaction of interest is the sum of the individual reactions.
3. If a reaction must be reversed, change the sign of ΔH for that reaction. Additionally, if a reaction must be multiplied by a factor to obtain the correct number of moles of a substance, multiply its ΔH value by that same factor.
4. Add together the individual reactions and their corresponding ΔH values to obtain the reaction of interest and the unknown ΔH.

We illustrate how to use this procedure in Example 3.

Enthalpies of Formation and Reaction

Chapter 2 "Molecules, Ions, and Chemical Formulas", Chapter 3 "Chemical Reactions", and Chapter 4 "Reactions in Aqueous Solution" presented a wide variety of chemical reactions, and you learned how to write balanced chemical equations that include all the reactants and the products except heat. One way to report the heat absorbed or released would be to compile a massive set of reference tables that list the enthalpy changes for all possible chemical reactions, which would require an incredible amount of effort. Fortunately, Hess’s law allows us to calculate the enthalpy change for virtually any conceivable chemical reaction using a relatively small set of tabulated data, such as the following:

• Enthalpy of combustion (ΔH_comb)The change in enthalpy that occurs during a combustion reaction.: Enthalpy changes have been measured for the combustion of virtually any substance that will burn in oxygen; these values are usually reported as the enthalpy of combustion per mole of substance.
• Enthalpy of fusion (ΔH_fus)The enthalpy change that acompanies the melting (fusion) of 1 mol of a substance.: The enthalpy change that accompanies the melting, or fusion, of 1 mol of a substance; these values have been measured for almost all the elements and for most simple compounds.
• Enthalpy of vaporization (ΔH_vap)The enthalpy change that accompanies the vaporization of 1 mol of a substance.: The enthalpy change that accompanies the vaporization of 1 mol of a substance; these values have also been measured for nearly all the elements and for most volatile compounds.
• Enthalpy of solution (ΔH_soln)The change in enthalpy that occurs when a specified amount of solute dissolves in a given quantity of solvent.: The enthalpy change when a specified amount of solute dissolves in a given quantity of solvent.

• Enthalpy of formation (ΔH_f)The enthalpy change for the formation of 1 mol of a compound from its component elements.: The enthalpy change for the formation of 1 mol of a compound from its component elements, such as the formation of carbon dioxide from carbon and oxygen. The corresponding relationship is

  Equation 5.23

  $$\text{elements}\to \text{compound}\Delta H_{\text{rxn}}=\Delta H_{\text{f}}$$

  For example,

  $$\text{C(s)}+{\text{O}}_{\text{2}}\text{(g)}\to {\text{CO}}_{\text{2}}\text{(g)}\Delta H_{\text{rxn}}=\Delta H_{\text{f}}[{\text{CO}}_{\text{2}}\text{(g)}]$$

  The sign convention for ΔH_f is the same as for any enthalpy change: ΔH_f < 0 if heat is released when elements combine to form a compound and ΔH_f > 0 if heat is absorbed. The values of ΔH_vap and ΔH_fus for some common substances are listed in Table 5.1 "Enthalpies of Vaporization and Fusion for Selected Substances at Their Boiling Points and Melting Points". These values are used in enthalpy calculations when any of the substances undergoes a change of physical state during a reaction.

Standard Enthalpies of Formation

The magnitude of ΔH for a reaction depends on the physical states of the reactants and the products (gas, liquid, solid, or solution), the pressure of any gases present, and the temperature at which the reaction is carried out. To avoid confusion caused by differences in reaction conditions and ensure uniformity of data, the scientific community has selected a specific set of conditions under which enthalpy changes are measured. These standard conditions serve as a reference point for measuring differences in enthalpy, much as sea level is the reference point for measuring the height of a mountain or for reporting the altitude of an airplane.

The standard conditionsThe conditions under which most thermochemical data are tabulated: 1 atm for all gases and a concentration of 1.0 M for all species in solution. for which most thermochemical data are tabulated are a pressure of 1 atmosphere (atm) for all gases and a concentration of 1 M for all species in solution (1 mol/L). In addition, each pure substance must be in its standard stateThe most stable form of a pure substance at a pressure of 1 atm at a specified temperature.. This is usually its most stable form at a pressure of 1 atm at a specified temperature. We assume a temperature of 25°C (298 K) for all enthalpy changes given in this text, unless otherwise indicated. Enthalpies of formation measured under these conditions are called standard enthalpies of formation ($\Delta H_{\text{f}}^o$)The enthalpy change for the formation of 1 mol of a compound from its component elements when the component elements are each in their standard states. The standard enthalpy of formation of any element in its most stable form is zero by definition. (which is pronounced “delta H eff naught”). The standard enthalpy of formation of any element in its standard state is zero by definition. For example, although oxygen can exist as ozone (O₃), atomic oxygen (O), and molecular oxygen (O₂), O₂ is the most stable form at 1 atm pressure and 25°C. Similarly, hydrogen is H₂(g), not atomic hydrogen (H). Graphite and diamond are both forms of elemental carbon, but because graphite is more stable at 1 atm pressure and 25°C, the standard state of carbon is graphite (Figure 5.10 "Elemental Carbon"). Therefore, O₂(g), H₂(g), and graphite have $\Delta H_{\text{f}}^o$ values of zero.

Figure 5.10 Elemental Carbon

Although graphite and diamond are both forms of elemental carbon, graphite is more stable at 1 atm pressure and 25°C than diamond is. Given enough time, diamond will revert to graphite under these conditions. Hence graphite is the standard state of carbon.

The standard enthalpy of formation of glucose from the elements at 25°C is the enthalpy change for the following reaction:

$$$$

Equation 5.24

$$\text{6C(s, graphite)}+{\text{6H}}_{\text{2}}\text{(g)}+{\text{3O}}_{\text{2}}\text{(g)}\to {\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}\text{(s)}\Delta H_{\text{f}}^o=-\text{1273}\text{.3 kJ}$$

It is not possible to measure the value of $\Delta H_{\text{f}}^o$ for glucose, −1273.3 kJ/mol, by simply mixing appropriate amounts of graphite, O₂, and H₂ and measuring the heat evolved as glucose is formed; the reaction shown in Equation 5.24 does not occur at a measurable rate under any known conditions. Glucose is not unique; most compounds cannot be prepared by the chemical equations that define their standard enthalpies of formation. Instead, values of $\Delta H_{\text{f}}^o$ are obtained using Hess’s law and standard enthalpy changes that have been measured for other reactions, such as combustion reactions. Values of $\Delta H_{\text{f}}^o$ for an extensive list of compounds are given in Chapter 25 "Appendix A: Standard Thermodynamic Quantities for Chemical Substances at 25°C". Note that $\Delta H_{\text{f}}^o$ values are always reported in kilojoules per mole of the substance of interest. Also notice in Chapter 25 "Appendix A: Standard Thermodynamic Quantities for Chemical Substances at 25°C" that the standard enthalpy of formation of O₂(g) is zero because it is the most stable form of oxygen in its standard state.

Example 4

For the formation of each compound, write a balanced chemical equation corresponding to the standard enthalpy of formation of each compound.

1. HCl(g)
2. MgCO₃(s)
3. CH₃(CH₂)₁₄CO₂H(s) (palmitic acid)

Given: compound

Asked for: balanced chemical equation for its formation from elements in standard states

Strategy:

Use Chapter 25 "Appendix A: Standard Thermodynamic Quantities for Chemical Substances at 25°C" to identify the standard state for each element. Write a chemical equation that describes the formation of the compound from the elements in their standard states and then balance it so that 1 mol of product is made.

Solution:

To calculate the standard enthalpy of formation of a compound, we must start with the elements in their standard states. The standard state of an element can be identified in Chapter 25 "Appendix A: Standard Thermodynamic Quantities for Chemical Substances at 25°C" by a $\Delta H_{\text{f}}^o$ value of 0 kJ/mol.

1. Hydrogen chloride contains one atom of hydrogen and one atom of chlorine. Because the standard states of elemental hydrogen and elemental chlorine are H₂(g) and Cl₂(g), respectively, the unbalanced chemical equation is

   H₂(g) + Cl₂(g) → HCl(g)

   Fractional coefficients are required in this case because $\Delta H_{\text{f}}^o$ values are reported for 1 mol of the product, HCl. Multiplying both H₂(g) and Cl₂(g) by 1/2 balances the equation:

   $$\frac{1}{2}{\text{H}}_{\text{2}}\text{(g)}+\frac{1}{2}{\text{Cl}}_{\text{2}}\text{(g)} \to \text{HCl(g)}$$

2. The standard states of the elements in this compound are Mg(s), C(s, graphite), and O₂(g). The unbalanced chemical equation is thus

   Mg(s) + C(s, graphite) + O₂(g) → MgCO₃(s)

   This equation can be balanced by inspection to give

   $$\text{Mg(s) }+\text{ C(s, graphite)}+\frac{3}{2}{\text{O}}_{\text{2}}\text{(g)}\to {\text{MgCO}}_{\text{3}}\text{(s)}$$

3. Palmitic acid, the major fat in meat and dairy products, contains hydrogen, carbon, and oxygen, so the unbalanced chemical equation for its formation from the elements in their standard states is as follows:

   C(s, graphite) + H₂(g) + O₂(g) → CH₃(CH₂)₁₄CO₂H(s)

   There are 16 carbon atoms and 32 hydrogen atoms in 1 mol of palmitic acid, so the balanced chemical equation is

   16C(s, graphite) + 16H₂(g) + O₂(g) → CH₃(CH₂)₁₄CO₂H(s)

Exercise

For the formation of each compound, write a balanced chemical equation corresponding to the standard enthalpy of formation of each compound.

1. NaCl(s)
2. H₂SO₄(l)
3. CH₃CO₂H(l) (acetic acid)

Answer:

1. $\text{Na(s)}+\frac{1}{2}{\text{Cl}}_{\text{2}}\text{(g)} \to \text{NaCl(s)}$
2. ${\text{H}}_{\text{2}}\text{(g)}+\frac{1}{8}{\text{S}}_{\text{8}}\text{(s)}+{\text{2O}}_{\text{2}}\text{(g)}\to {\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}\text{(l)}$
3. 2C(s) + O₂(g) + 2H₂(g) → CH₃CO₂H(l)

Standard Enthalpies of Reaction

Tabulated values of standard enthalpies of formation can be used to calculate enthalpy changes for any reaction involving substances whose $\Delta H_{\text{f}}^o$ values are known. The standard enthalpy of reaction ($\Delta H_{\text{rxn}}^o$)The enthalpy change that occurs when a reaction is carried out with all reactants and products in their standard state. is the enthalpy change that occurs when a reaction is carried out with all reactants and products in their standard states. Consider the general reaction

Equation 5.25

aA + bB → cC + dD

where A, B, C, and D are chemical substances and a, b, c, and d are their stoichiometric coefficients. The magnitude of $\Delta H_{\text{rxn}}^o$ is the sum of the standard enthalpies of formation of the products, each multiplied by its appropriate coefficient, minus the sum of the standard enthalpies of formation of the reactants, also multiplied by their coefficients:

Equation 5.26

$$\Delta H_{\text{rxn}}^o=\underset{\text{products}}{\text{[}c\Delta H_{\text{f}}^o\text{(C)}+d\Delta H_{\text{f}}^o\text{(D)]}}-\underset{\text{reactants}}{\text{[}a\Delta H_{\text{f}}^o\text{(A)}+b\Delta H_{\text{f}}^o\text{(B)]}}$$

More generally, we can write

Equation 5.27

$$\Delta H_{\text{rxn}}^o=\Sigma m\Delta H_{\text{f}}^o\text{(products)}-\Sigma n\Delta H_{\text{f}}^o\text{(reactants)}$$

where the symbol Σ means “sum of” and m and n are the stoichiometric coefficients of each of the products and the reactants, respectively. “Products minus reactants” summations such as Equation 5.27 arise from the fact that enthalpy is a state function. Because many other thermochemical quantities are also state functions, “products minus reactants” summations are very common in chemistry; we will encounter many others in subsequent chapters.

Note the Pattern

“Products minus reactants” summations are typical of state functions.

To demonstrate the use of tabulated $\Delta H_{\text{f}}^o$ values, we will use them to calculate $\Delta H_{\text{rxn}}^o$ for the combustion of glucose, the reaction that provides energy for your brain:

Equation 5.28

C₆H₁₂O₆(s) + 6O₂(g) → 6CO₂(g) + 6H₂O(l)

Using Equation 5.27, we write

Equation 5.29

$$\Delta H_{\text{rxn}}^o=\text{{6}\Delta H_{\text{f}}^o{\text{[CO}}_{\text{2}}\text{(g)] }+\text{ 6}{H}_{\text{f}}^o{\text{[H}}_{\text{2}}\text{O(l)]}}-\text{{}\Delta H_{\text{f}}^o{\text{[C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}\text{(s)] }+\text{ 6}\Delta H_{\text{f}}^o{\text{[O}}_{\text{2}}\text{(g)]}}$$

From Chapter 25 "Appendix A: Standard Thermodynamic Quantities for Chemical Substances at 25°C", the relevant $\Delta H_{\text{f}}^o$ values are $\Delta H_{\text{f}}^o{\text{[CO}}_{\text{2}}\text{(g)] }=-\text{393}\text{.5 kJ/mol}$, $\Delta H_{\text{f}}^o{\text{[H}}_{\text{2}}\text{O(l)] }=-\text{285}\text{.8 kJ/mol}$, and $\Delta H_{\text{f}}^o{\text{[C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}\text{(s)] }=-\text{1273}\text{.3 kJ/mol}$. Because O₂(g) is a pure element in its standard state, $\Delta H_{\text{f}}^o{\text{[O}}_{\text{2}}\text{(g)] }=0\text{ kJ/mol}\text{.}$ Inserting these values into Equation 5.29 and changing the subscript to indicate that this is a combustion reaction, we obtain

Equation 5.30

$$\begin{array}{cc}\Delta H_{\text{comb}}^o\hfill & =\text{[6(}-\text{393}\text{.5 kJ/mol) }+\text{ 6(}-\text{285}\text{.8 kJ/mol)]}\hfill \\ \hfill & -\text{ [}-\text{1273}\text{.3 kJ/mol }+\text{ 6(0 kJ/mol)] }=-\text{2802}\text{.5 kJ/mol}\hfill \end{array}$$

As illustrated in Figure 5.11 "A Thermochemical Cycle for the Combustion of Glucose", we can use Equation 5.29 to calculate $\Delta H_{\text{comb}}^o$ for glucose because enthalpy is a state function. The figure shows two pathways from reactants (middle left) to products (bottom). The more direct pathway is the downward green arrow labeled $\Delta H_{\text{comb}}^o\text{.}$ The alternative hypothetical pathway consists of four separate reactions that convert the reactants to the elements in their standard states (upward purple arrow at left) and then convert the elements into the desired products (downward purple arrows at right). The reactions that convert the reactants to the elements are the reverse of the equations that define the $\Delta H_{\text{f}}^o$ values of the reactants. Consequently, the enthalpy changes are

$$\begin{array}{c}\Delta H_1^o=\Delta H_{\text{f}}^o\text{[glucose(s)] }=-\text{1}\cancel{\text{mol glucose}}\left(\frac{\text{1273}\text{.3 kJ}}{\text{1}\cancel{\text{mol glucose}}}\right)=+\text{1273}\text{.3 kJ}\\ \Delta H_2^o=\text{ 6}\Delta H_{\text{f}}^o{\text{[O}}_{\text{2}}\text{(g)] }=-\text{6}\cancel{{\text{mol O}}_{\text{2}}}\left(\frac{\text{0 kJ}}{\text{1}\cancel{{\text{mol O}}_{\text{2}}}}\right)=\text{ 0 kJ}\end{array}$$

(Recall that when we reverse a reaction, we must also reverse the sign of the accompanying enthalpy change.) The overall enthalpy change for conversion of the reactants (1 mol of glucose and 6 mol of O₂) to the elements is therefore +1273.3 kJ.

Figure 5.11 A Thermochemical Cycle for the Combustion of Glucose

Two hypothetical pathways are shown from the reactants to the products. The green arrow labeled $\Delta H_{\text{comb}}^o$ indicates the combustion reaction. Alternatively, we could first convert the reactants to the elements via the reverse of the equations that define their standard enthalpies of formation (the upward arrow, labeled $\Delta H_1^o$ and $\Delta H_2^o$). Then we could convert the elements to the products via the equations used to define their standard enthalpies of formation (the downward arrows, labeled $\Delta H_3^o$ and $\Delta H_4^o$). Because enthalpy is a state function, $\Delta H_{\text{comb}}^o$ is equal to the sum of the enthalpy changes $\Delta H_1^o+\Delta H_2^o+\Delta H_3^o+\Delta H_4^o\text{.}$

The reactions that convert the elements to final products (downward purple arrows in Figure 5.11 "A Thermochemical Cycle for the Combustion of Glucose") are identical to those used to define the $\Delta H_{\text{f}}^o$ values of the products. Consequently, the enthalpy changes (from Chapter 25 "Appendix A: Standard Thermodynamic Quantities for Chemical Substances at 25°C") are

$$\begin{array}{c}\Delta H_3^o=\text{ 6}\Delta H_{\text{f}}^o{\text{[CO}}_{\text{2}}\text{(g)] }=6\cancel{{\text{mol CO}}_{\text{2}}}\left(\frac{-393.5\text{ kJ}}{\text{1}\cancel{{\text{mol CO}}_{\text{2}}}}\right)=-\text{2361}\text{.0 kJ}\\ \Delta H_4^o=\text{ 6}\Delta H_{\text{f}}^o{\text{[H}}_{\text{2}}\text{O(l)] }=-\text{6}\cancel{{\text{mol H}}_{\text{2}}\text{O}}\left(\frac{-\text{285}\text{.8 kJ}}{\text{1}\cancel{{\text{mol H}}_{\text{2}}\text{O}}}\right)=-1714.8\text{ kJ}\end{array}$$

The overall enthalpy change for the conversion of the elements to products (6 mol of carbon dioxide and 6 mol of liquid water) is therefore −4075.8 kJ. Because enthalpy is a state function, the difference in enthalpy between an initial state and a final state can be computed using any pathway that connects the two. Thus the enthalpy change for the combustion of glucose to carbon dioxide and water is the sum of the enthalpy changes for the conversion of glucose and oxygen to the elements (+1273.3 kJ) and for the conversion of the elements to carbon dioxide and water (−4075.8 kJ):

Equation 5.31

$$\Delta H_{\text{comb}}^o=\text{ +1273}\text{.3 kJ }+\text{ (}-\text{4075}\text{.8 kJ) }=-\text{2802}\text{.5 kJ}$$

This is the same result we obtained using the “products minus reactants” rule (Equation 5.27) and $\Delta H_{\text{f}}^o$ values. The two results must be the same because Equation 5.29 is just a more compact way of describing the thermochemical cycle shown in Figure 5.11 "A Thermochemical Cycle for the Combustion of Glucose".

Example 5

Long-chain fatty acids such as palmitic acid [CH₃(CH₂)₁₄CO₂H] are one of the two major sources of energy in our diet ($\Delta H_{\text{f}}^o=-\text{891}\text{.5 kJ/mol}$). Use the data in Chapter 25 "Appendix A: Standard Thermodynamic Quantities for Chemical Substances at 25°C" to calculate $\Delta H_{\text{comb}}^o$ for the combustion of palmitic acid. Based on the energy released in combustion per gram, which is the better fuel—glucose or palmitic acid?

Given: compound and $\Delta H_{\text{f}}^o$ values

Asked for: $\Delta H_{\text{comb}}^o$ per mole and per gram

Strategy:

A After writing the balanced chemical equation for the reaction, use Equation 5.27 and the values from Chapter 25 "Appendix A: Standard Thermodynamic Quantities for Chemical Substances at 25°C" to calculate $\Delta H_{\text{comb}}^o\text{,}$ the energy released by the combustion of 1 mol of palmitic acid.

B Divide this value by the molar mass of palmitic acid to find the energy released from the combustion of 1 g of palmitic acid. Compare this value with the value calculated in Equation 5.30 for the combustion of glucose to determine which is the better fuel.

Solution:

A To determine the energy released by the combustion of palmitic acid, we need to calculate its $\Delta H_{\text{comb}}^o\text{.}$ As always, the first requirement is a balanced chemical equation:

C₁₆H₃₂O₂(s) + 23O₂(g) → 16CO₂(g) + 16H₂O(l)

Using Equation 5.27 (“products minus reactants”) with $\Delta H_{\text{f}}^o$ values from Chapter 25 "Appendix A: Standard Thermodynamic Quantities for Chemical Substances at 25°C" (and omitting the physical states of the reactants and products to save space) gives

$$$$ $$\begin{array}{ll}\Delta H_{\text{comb}}^o\hfill & =\Sigma m\Delta H_{\text{f}}^o\text{ (products)}-\Sigma n\Delta H_{\text{f}}^o\text{ (reactants)}\hfill \\ \hfill & =\text{[16(}-\text{393}{\text{.5 kJ/mol CO}}_{\text{2}}\text{) }+\text{ 16(}-\text{285}{\text{.8 kJ/mol H}}_{\text{2}}\text{O}]\hfill \\ \hfill & -\text{[(}-\text{891}{\text{.5 kJ/mol C}}_{\text{16}}{\text{H}}_{\text{32}}{\text{O}}_{\text{2}}\text{)}+{\text{23(0 kJ/mol O}}_{\text{2}}\text{)]}\\ \hfill & =-\text{9977}\text{.3 kJ/mol}\hfill \end{array}$$

This is the energy released by the combustion of 1 mol of palmitic acid.

B The energy released by the combustion of 1 g of palmitic acid is

$$\Delta H_{\text{comb}}^o\text{ per gram }=\left(\frac{\text{9977}\text{.3 kJ}}{\cancel{\text{1 mol}}}\right)\left(\frac{\cancel{\text{1 mol}}}{\text{256}\text{.42 g}}\right)=-\text{38}\text{.910 kJ/g}$$

As calculated in Equation 5.30, $\Delta H_{\text{comb}}^o$ of glucose is −2802.5 kJ/mol. The energy released by the combustion of 1 g of glucose is therefore

$$\Delta H_{\text{comb}}^o\text{ per gram }=\left(\frac{-\text{2802}\text{.5 kJ}}{\cancel{\text{1 mol}}}\right)\left(\frac{\cancel{\text{1 mol}}}{\text{180}\text{.16 g}}\right)=-\text{15}\text{.556 kJ/g}$$

The combustion of fats such as palmitic acid releases more than twice as much energy per gram as the combustion of sugars such as glucose. This is one reason many people try to minimize the fat content in their diets to lose weight.

Exercise

Use the data in Chapter 25 "Appendix A: Standard Thermodynamic Quantities for Chemical Substances at 25°C" to calculate $\Delta H_{\text{rxn}}^o$ for the water–gas shift reaction, which is used industrially on an enormous scale to obtain H₂(g):

$$\underset{\text{water–gas shift reaction}}{{\text{CO(g) + H}}_{\text{2}}\text{O(g)}\to {\text{CO}}_{\text{2}}{\text{(g) + H}}_{\text{2}}\text{(g)}}$$

Answer: −41.2 kJ/mol

We can also measure the enthalpy change for another reaction, such as a combustion reaction, and then use it to calculate a compound’s $\Delta H_{\text{f}}^o\text{,}$ which we cannot obtain otherwise. This procedure is illustrated in Example 6.

Example 6

Beginning in 1923, tetraethyllead [(C₂H₅)₄Pb] was used as an antiknock additive in gasoline in the United States. Its use was completely phased out in 1986 because of the health risks associated with chronic lead exposure. Tetraethyllead is a highly poisonous, colorless liquid that burns in air to give an orange flame with a green halo. The combustion products are CO₂(g), H₂O(l), and red PbO(s). What is the standard enthalpy of formation of tetraethyllead, given that $\Delta H_{\text{comb}}^o$ is −19.29 kJ/g for the combustion of tetraethyllead and $\Delta H_{\text{f}}^o$ of red PbO(s) is −219.0 kJ/mol?

Given: reactant, products, and $\Delta H_{\text{comb}}^o$ values

Asked for: $\Delta H_{\text{f}}^o$ of reactant

Strategy:

A Write the balanced chemical equation for the combustion of tetraethyllead. Then insert the appropriate quantities into Equation 5.27 to get the equation for $\Delta H_{\text{f}}^o$ of tetraethyllead.

B Convert $\Delta H_{\text{comb}}^o$ per gram given in the problem to $\Delta H_{\text{comb}}^o$ per mole by multiplying $\Delta H_{\text{comb}}^o$ per gram by the molar mass of tetraethyllead.

C Use Chapter 25 "Appendix A: Standard Thermodynamic Quantities for Chemical Substances at 25°C" to obtain values of $\Delta H_{\text{f}}^o$ for the other reactants and products. Insert these values into the equation for $\Delta H_{\text{f}}^o$ of tetraethyllead and solve the equation.

Solution:

A The balanced chemical equation for the combustion reaction is as follows:

2(C₂H₅)₄Pb(l) + 27O₂(g) → 2PbO(s) + 16CO₂(g) + 20H₂O(l)

Using Equation 5.27 gives

$$\begin{array}{cc}\Delta H_{\text{comb}}^o& =\text{ [2}\Delta H_{\text{f}}^o\text{(PbO) }+\text{ 16}\Delta H_{\text{f}}^o{\text{(CO}}_{\text{2}}\text{) }+\text{ 20}\Delta H_{\text{f}}^o{\text{(H}}_{\text{2}}\text{O)]}-\text{{2}\Delta H_{\text{f}}^o{\text{[(C}}_{\text{2}}{\text{H}}_{\text{5}}{\text{)}}_{\text{4}}\text{Pb] + 27}\Delta H_{\text{f}}^o{\text{(O}}_{\text{2}}\text{)}}\end{array}$$

Solving for $\Delta H_{\text{f}}^o{\text{[(C}}_{\text{2}}{\text{H}}_{\text{5}}{\text{)}}_{\text{4}}\text{Pb]}$ gives

$$\Delta H_{\text{f}}^o{\text{[(C}}_{\text{2}}{\text{H}}_{\text{5}}{\text{)}}_{\text{4}}\text{Pb] }=\Delta H_{\text{f}}^o\text{(PbO) }+\text{ 8}\Delta H_{\text{f}}^o{\text{(CO}}_{\text{2}}\text{) }+\text{ 10}\Delta H_{\text{f}}^o{\text{(H}}_{\text{2}}\text{O)}-\frac{27}{2}\Delta H_{\text{f}}^o{\text{(O}}_{\text{2}}\text{)}-\frac{\Delta H_{\text{comb}}^o}{2}$$

The values of all terms other than $\Delta H_{\text{f}}^o{\text{[(C}}_{\text{2}}{\text{H}}_{\text{5}}{\text{)}}_{\text{4}}\text{Pb]}$ and $\Delta H_{\text{comb}}^o$ are given in Chapter 25 "Appendix A: Standard Thermodynamic Quantities for Chemical Substances at 25°C".

B The magnitude of $\Delta H_{\text{comb}}^o$ is given in the problem in kilojoules per gram of tetraethyllead. We must therefore multiply this value by the molar mass of tetraethyllead (323.44 g/mol) to get $\Delta H_{\text{comb}}^o$ for 1 mol of tetraethyllead:

$$\Delta H_{\text{comb}}^o=\left(\frac{-\text{19}\text{.29 kJ}}{\cancel{\text{g}}}\right)\left(\frac{323.44\cancel{\text{g}}}{\text{mol}}\right)=-\text{6329 kJ/mol}$$

Because the balanced chemical equation contains 2 mol of tetraethyllead, $\Delta H_{\text{rxn}}^o$ is

$$\Delta H_{\text{rxn}}^o=\text{ 2}\cancel{{\text{mol (C}}_{\text{2}}{\text{H}}_{\text{5}}{\text{)}}_{\text{4}}\text{Pb}}\left(\frac{-\text{6329 kJ}}{\text{1}\cancel{{\text{mol (C}}_{\text{2}}{\text{H}}_{\text{5}}{\text{)}}_{\text{4}}\text{Pb}}}\right)=-\text{12,480 kJ}$$

C Inserting the appropriate values into the equation for $\Delta H_{\text{f}}^o{\text{[(C}}_{\text{2}}{\text{H}}_{\text{5}}{\text{)}}_{\text{4}}\text{Pb]}$ gives

$$\begin{array}{ll}\Delta H_{\text{f}}^o{\text{[(C}}_{\text{2}}{\text{H}}_{\text{5}}{\text{)}}_{\text{4}}\text{Pb]}\hfill & =\text{[(1 }\cancel{\text{mol PbO}}\text{)(}-\text{219}\text{.0 kJ/}\cancel{\text{mol PbO}}\text{)]}\hfill \\ \hfill & +\text{[(8 }\cancel{{\text{mol CO}}_{\text{2}}}\text{)(}-\text{393}\text{.5 kJ/}\cancel{{\text{mol CO}}_{\text{2}}}\text{)]}\hfill \\ \hfill & +[\text{(10 }\cancel{{\text{mol H}}_{\text{2}}\text{O}}\text{)(}-\text{285}\text{.8 kJ/}\cancel{{\text{mol H}}_{\text{2}}\text{O}}\text{)]}\hfill \\ \hfill & -\left[\left(\frac{27}{2}\cancel{{\text{mol O}}_{\text{2}}}\right)\text{(}0\text{ kJ/}\cancel{{\text{mol O}}_{\text{2}}}\text{)}\right]\hfill \\ \hfill & -\left[\frac{-\mathrm{12,480}}{2}{\text{ kJ/mol (C}}_{\text{2}}{\text{H}}_{\text{5}}{\text{)}}_{\text{4}}\text{Pb}\right]\hfill \\ \hfill & =-\text{219}\text{.0 kJ}-\text{3148 kJ}-\text{2858 kJ}-\text{0 kJ }+\text{ 6240 kJ/mol}\hfill \\ \hfill & =\text{15 kJ/mol}\hfill \end{array}$$

Exercise

Ammonium sulfate [(NH₄)₂SO₄] is used as a fire retardant and wood preservative; it is prepared industrially by the highly exothermic reaction of gaseous ammonia with sulfuric acid:

2NH₃(g) + H₂SO₄(aq) → (NH₄)₂SO₄(s)

The value of $\Delta H_{\text{rxn}}^o$ is −2805 kJ/g H₂SO₄. Use the data in Chapter 25 "Appendix A: Standard Thermodynamic Quantities for Chemical Substances at 25°C" to calculate the standard enthalpy of formation of ammonium sulfate (in kilojoules per mole).

Answer: −1181 kJ/mol

Enthalpies of Solution and Dilution

Physical changes, such as melting or vaporization, and chemical reactions, in which one substance is converted to another, are accompanied by changes in enthalpy. Two other kinds of changes that are accompanied by changes in enthalpy are the dissolution of solids and the dilution of concentrated solutions.

The dissolution of a solid can be described as follows:

The values of ΔH_soln for some common substances are given in Table 5.2 "Enthalpies of Solution at 25°C of Selected Ionic Compounds in Water (in kJ/mol)". The sign and the magnitude of ΔH_soln depend on specific attractive and repulsive interactions between the solute and the solvent; these factors will be discussed in Chapter 13 "Solutions". When substances dissolve, the process can be either exothermic (ΔH_soln < 0) or endothermic (ΔH_soln > 0), as you can see from the data in Table 5.2 "Enthalpies of Solution at 25°C of Selected Ionic Compounds in Water (in kJ/mol)".

Substances with large positive or negative enthalpies of solution have commercial applications as instant cold or hot packs. Single-use versions of these products are based on the dissolution of either calcium chloride (CaCl₂, ΔH_soln = −81.3 kJ/mol) or ammonium nitrate (NH₄NO₃, ΔH_soln = +25.7 kJ/mol). Both types consist of a plastic bag that contains about 100 mL of water plus a dry chemical (40 g of CaCl₂ or 30 g of NH₄NO₃) in a separate plastic pouch. When the pack is twisted or struck sharply, the inner plastic bag of water ruptures, and the salt dissolves in the water. If the salt is CaCl₂, heat is released to produce a solution with a temperature of about 90°C; hence the product is an “instant hot compress.” If the salt is NH₄NO₃, heat is absorbed when it dissolves, and the temperature drops to about 0° for an “instant cold pack.”

A similar product based on the precipitation of sodium acetate, not its dissolution, is marketed as a reusable hand warmer (Figure 5.12 "An Instant Hot Pack Based on the Crystallization of Sodium Acetate"). At high temperatures, sodium acetate forms a highly concentrated aqueous solution. With cooling, an unstable supersaturated solution containing excess solute is formed. When the pack is agitated, sodium acetate trihydrate [CH₃CO₂Na·3H₂O] crystallizes, and heat is evolved:

A bag of concentrated sodium acetate solution can be carried until heat is needed, at which time vigorous agitation induces crystallization and heat is released. The pack can be reused after it is immersed in hot water until the sodium acetate redissolves.

Figure 5.12 An Instant Hot Pack Based on the Crystallization of Sodium Acetate

The hot pack is at room temperature prior to agitation (left). Because the sodium acetate is in solution, you can see the metal disc inside the pack. After the hot pack has been agitated, the sodium acetate crystallizes (right) to release heat. Because of the mass of white sodium acetate that has crystallized, the metal disc is no longer visible.

The amount of heat released or absorbed when a substance is dissolved is not a constant; it depends on the final concentration of the solute. The ΔH_soln values given previously and in Table 5.2 "Enthalpies of Solution at 25°C of Selected Ionic Compounds in Water (in kJ/mol)", for example, were obtained by measuring the enthalpy changes at various concentrations and extrapolating the data to infinite dilution.

Because ΔH_soln depends on the concentration of the solute, diluting a solution can produce a change in enthalpy. If the initial dissolution process is exothermic (ΔH < 0), then the dilution process is also exothermic. This phenomenon is particularly relevant for strong acids and bases, which are often sold or stored as concentrated aqueous solutions. If water is added to a concentrated solution of sulfuric acid (which is 98% H₂SO₄ and 2% H₂O) or sodium hydroxide, the heat released by the large negative ΔH can cause the solution to boil. Dangerous spattering of strong acid or base can be avoided if the concentrated acid or base is slowly added to water, so that the heat liberated is largely dissipated by the water. Thus you should never add water to a strong acid or base; a useful way to avoid the danger is to remember: Add water to acid and get blasted!