Chapter 5 Energy Changes in Chemical Reactions

Forms of Energy

The forms of energy include thermal energy, radiant energy, electrical energy, nuclear energy, and chemical energy (Figure 5.1 "Forms of Energy"). Thermal energyEnergy that results from atomic and molecular motion; the faster the motion, the higher the thermal energy. results from atomic and molecular motion; the faster the motion, the greater the thermal energy. The temperatureA measure of an object’s thermal energy content. of an object is a measure of its thermal energy content. Radiant energyOne of the five forms of energy, radiant energy is carried by light, microwaves, and radio waves (the other forms of energy are thermal, chemical, nuclear, and electrical). Objects left in bright sunshine or exposed to microwaves become warm because much of the radiant energy they absorb is converted to thermal energy. is the energy carried by light, microwaves, and radio waves. Objects left in bright sunshine or exposed to microwaves become warm because much of the radiant energy they absorb is converted to thermal energy. Electrical energyOne of the five forms of energy, electrical energy results from the flow of electrically charged particles. The other four forms of energy are radiant, thermal, chemical, and nuclear. results from the flow of electrically charged particles. When the ground and a cloud develop a separation of charge, for example, the resulting flow of electrons from one to the other produces lightning, a natural form of electrical energy. Nuclear energyOne of the five forms of energy, nuclear energy is stored in the nucleus of an atom. The other four forms of energy are radiant, thermal, chemical, and electrical. is stored in the nucleus of an atom, and chemical energyOne of the five forms of energy, chemical energy is stored within a chemical compound because of a particular arrangement of atoms. The other four forms of energy are radiant, thermal, nuclear, and electrical. is stored within a chemical compound because of a particular arrangement of atoms.

Figure 5.1 Forms of Energy

(a) Thermal energy results from atomic and molecular motion; molten steel at 2000°C has a very high thermal energy content. (b) Radiant energy (e.g., from the sun) is the energy in light, microwaves, and radio waves. (c) Lightning is an example of electrical energy, which is due to the flow of electrically charged particles. (d) Nuclear energy is released when particles in the nucleus of the atom are rearranged. (e) Chemical energy results from the particular arrangement of atoms in a chemical compound; the heat and light produced in this reaction are due to energy released during the breaking and reforming of chemical bonds.

Electrical energy, nuclear energy, and chemical energy are different forms of potential energy (PE)Energy stored in an object because of its relative position or orientation., which is energy stored in an object because of the relative positions or orientations of its components. A brick lying on the windowsill of a 10th-floor office has a great deal of potential energy, but until its position changes by falling, the energy is contained. In contrast, kinetic energy (KE)Energy due to the motion of an object: $KE=\frac{1}{2}m{v}^{\text{2}}\text{,}$ where $m$ is the mass of the object and $v$ is its velocity. is energy due to the motion of an object. When the brick falls, its potential energy is transformed to kinetic energy, which is then transferred to the object on the ground that it strikes. The electrostatic attraction between oppositely charged particles is a form of potential energy, which is converted to kinetic energy when the charged particles move toward each other.

Energy can be converted from one form to another (Figure 5.2 "Interconversion of Forms of Energy") or, as we saw with the brick, transferred from one object to another. For example, when you climb a ladder to a high diving board, your body uses chemical energy produced by the combustion of organic molecules. As you climb, the chemical energy is converted to mechanical work to overcome the force of gravity. When you stand on the end of the diving board, your potential energy is greater than it was before you climbed the ladder: the greater the distance from the water, the greater the potential energy. When you then dive into the water, your potential energy is converted to kinetic energy as you fall, and when you hit the surface, some of that energy is transferred to the water, causing it to splash into the air. Chemical energy can also be converted to radiant energy; one common example is the light emitted by fireflies, which is produced from a chemical reaction.

Figure 5.2 Interconversion of Forms of Energy

When a swimmer steps off the platform to dive into the water, potential energy is converted to kinetic energy. As the swimmer climbs back up to the top of the diving platform, chemical energy is converted to mechanical work.

Although energy can be converted from one form to another, the total amount of energy in the universe remains constant. This is known as the law of conservation of energyThe total amount of energy in the universe remains constant. Energy can be neither created nor destroyed, but it can be converted from one form to another..As you will learn in Chapter 18 "Chemical Thermodynamics", the law of conservation of energy is also known as the first law of thermodynamics. Energy cannot be created or destroyed.

Energy, Heat, and Work

One definition of energyThe capacity to do work. is the capacity to do work. The easiest form of work to visualize is mechanical workThe energy required to move an object a distance $d$ when opposed by a force $F$: $w=F\times d.$ (Figure 5.3 "An Example of Mechanical Work"), which is the energy required to move an object a distance d when opposed by a force F, such as gravity:

Because the force (F) that opposes the action is equal to the mass (m) of the object times its acceleration (a), we can also write Equation 5.1 as follows:Recall from Chapter 1 "Introduction to Chemistry" that weight is a force caused by the gravitational attraction between two masses, such as you and Earth.

Figure 5.3 An Example of Mechanical Work

One form of energy is mechanical work, the energy required to move an object of mass m a distance d when opposed by a force F, such as gravity.

Consider the mechanical work required for you to travel from the first floor of a building to the second. Whether you take an elevator or an escalator, trudge upstairs, or leap up the stairs two at a time, energy is expended to overcome the force of gravity. The amount of work done (w) and thus the energy required depends on three things: (1) the height of the second floor (the distance d); (2) your mass, which must be raised that distance against the downward acceleration due to gravity; and (3) your path, as you will learn in Section 5.2 "Enthalpy".

In contrast, heat (q)Thermal energy that can be transformed from an object at one temperature to an object at another temperature. is thermal energy that can be transferred from an object at one temperature to an object at another temperature. The net transfer of thermal energy stops when the two objects reach the same temperature.

The energy of an object can be changed only by the transfer of energy to or from another object in the form of heat,As you will learn in Chapter 6 "The Structure of Atoms", hot objects can also lose energy as radiant energy, such as heat or light. This energy is converted to heat when it is absorbed by another object. Hence radiant energy is equivalent to heat. work performed on or by the object, or some combination of heat and work. Consider, for example, the energy stored in a fully charged battery. As shown in Figure 5.4 "Energy Transfer", this energy can be used primarily to perform work (e.g., running an electric fan) or to generate light and heat (e.g., illuminating a light bulb). When the battery is fully discharged in either case, the total change in energy is the same, even though the fraction released as work or heat varies greatly. The sum of the heat produced and the work performed equals the change in energy (ΔE):

Figure 5.4 Energy Transfer

Discharging a fully charged battery releases the same amount of energy whether the battery is used to run a fan (a) or illuminate a light bulb (b). In (a), most of the energy is used to perform work, which turns the blades of the fan and thus moves the air; only a small portion of the energy is released as heat by the electric motor. In (b), all the energy is released as heat and light; no work is done.

Energy is an extensive property of matter—for example, the amount of thermal energy in an object is proportional to both its mass and its temperature. (For more information on the properties of matter, see Chapter 1 "Introduction to Chemistry".) A water heater that holds 150 L of water at 50°C contains much more thermal energy than does a 1 L pan of water at 50°C. Similarly, a bomb contains much more chemical energy than does a firecracker. We now present a more detailed description of kinetic and potential energy.

Kinetic and Potential Energy

The kinetic energy of an object is related to its mass m and velocity v:

For example, the kinetic energy of a 1360 kg (approximately 3000 lb) automobile traveling at a velocity of 26.8 m/s (approximately 60 mi/h) is

Because all forms of energy can be interconverted, energy in any form can be expressed using the same units as kinetic energy. The SI unit of energy, the joule (J)The SI unit of energy: $\text{1 J }=\text{ 1 kg}•{\text{m}}^{\text{2}}/{\text{s}}^{\text{2}}.$,The joule is named after the British physicist James Joule (1818–1889), an early worker in the field of energy. is defined as 1 kilogram·meter²/second² (kg·m²/s²). Because a joule is such a small quantity of energy, chemists usually express energy in kilojoules (1 kJ = 10³ J). For example, the kinetic energy of the 1360 kg car traveling at 26.8 m/s is 4.88 × 10⁵ J or 4.88 × 10² kJ. It is important to remember that the units of energy are the same regardless of the form of energy, whether thermal, radiant, chemical, or any other form. Because heat and work result in changes in energy, their units must also be the same.

To demonstrate, let’s calculate the potential energy of the same 1360 kg automobile if it were parked on the top level of a parking garage 36.6 m (120 ft) high. Its potential energy is equivalent to the amount of work required to raise the vehicle from street level to the top level of the parking garage, which is given by Equation 5.1 (w = Fd). According to Equation 5.2, the force (F) exerted by gravity on any object is equal to its mass (m, in this case, 1360 kg) times the acceleration (a) due to gravity (g, 9.81 m/s² at Earth’s surface). The distance (d) is the height (h) above street level (in this case, 36.6 m). Thus the potential energy of the car is as follows:

The units of potential energy are the same as the units of kinetic energy. Notice that in this case the potential energy of the stationary automobile at the top of a 36.6 m high parking garage is the same as its kinetic energy at 60 mi/h. If the vehicle fell from the roof of the parking garage, its potential energy would be converted to kinetic energy, and it is reasonable to infer that the vehicle would be traveling at 60 mi/h just before it hit the ground, neglecting air resistance. After the car hit the ground, its potential and kinetic energy would both be zero.

Potential energy is usually defined relative to an arbitrary standard position (in this case, the street was assigned an elevation of zero). As a result, we usually calculate only differences in potential energy: in this case, the difference between the potential energy of the car on the top level of the parking garage and the potential energy of the same car on the street at the base of the garage.

A recent and spectacular example of the conversion of potential energy to kinetic energy was seen by the earthquake near the east coast of Honshu, Japan, on March 11, 2011. The magnitude 9.0 earthquake occurred along the Japan Trench subduction zone, the interface boundary between the Pacific and North American geological plates. During its westward movement, the Pacific plate became trapped under the North American plate, and its further movement was prevented. When there was sufficient potential energy to allow the Pacific plate to break free, approximately 7.1 × 10¹⁵ kJ of potential energy was released as kinetic energy, the equivalent of 4.75 × 10⁸ tn of TNT (trinitrotoluene) or 25,003 nuclear bombs. The island of Japan experienced the worst devastation in its history from the earthquake, resulting tsunami, and aftershocks. Historical records indicate that an earthquake of such force occurs in some region of the globe approximately every 1000 years. One such earthquake and resulting tsunami is speculated to have caused the destruction of the lost city of Atlantis, referred to by the ancient Greek philosopher Plato.

Energy can also be expressed in the non-SI units of calories (cal)A non-SI unit of energy: 1 cal = 4.184 J exactly., where 1 cal was originally defined as the amount of energy needed to raise the temperature of exactly 1 g of water from 14.5°C to 15.5°C.We specify the exact temperatures because the amount of energy needed to raise the temperature of 1 g of water 1°C varies slightly with elevation. To three significant figures, however, this amount is 1.00 cal over the temperature range 0°C–100°C. The name is derived from the Latin calor, meaning “heat.” Although energy may be expressed as either calories or joules, calories were defined in terms of heat, whereas joules were defined in terms of motion. Because calories and joules are both units of energy, however, the calorie is now defined in terms of the joule:

In this text, we will use the SI units—joules (J) and kilojoules (kJ)—exclusively, except when we deal with nutritional information, addressed in Section 5.4 "Thermochemistry and Nutrition".

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!

Heat Capacity

We have seen that the temperature of an object changes when it absorbs or loses thermal energy. The magnitude of the temperature change depends on both the amount of thermal energy transferred (q) and the heat capacity of the object. Its heat capacity (C)The amount of energy needed to raise the temperature of an object 1°C. The units of heat capacity are joules per degree Celsius $\left(\text{J}/°\text{C}\right).$ is the amount of energy needed to raise the temperature of the object exactly 1°C; the units of C are joules per degree Celsius (J/°C). The change in temperature (ΔT) is

where q is the amount of heat (in joules), C is the heat capacity (in joules per degree Celsius), and ΔT is T_final − T_initial (in degrees Celsius). Note that ΔT is always written as the final temperature minus the initial temperature. The value of C is intrinsically a positive number, but ΔT and q can be either positive or negative, and they both must have the same sign. If ΔT and q are positive, then heat flows from the surroundings into an object. If ΔT and q are negative, then heat flows from an object into its surroundings.

The heat capacity of an object depends on both its mass and its composition. For example, doubling the mass of an object doubles its heat capacity. Consequently, the amount of substance must be indicated when the heat capacity of the substance is reported. The molar heat capacity (C_p)The amount of energy needed to increase the temperature of 1 mol of a substance by 1°C. The units of $C_{\text{p}}$ are $\text{J}/\left(\text{mol}•°\text{C}\right).$ is the amount of energy needed to increase the temperature of 1 mol of a substance by 1°C; the units of C_p are thus J/(mol·°C).The subscript p indicates that the value was measured at constant pressure. The specific heat (C_s)The amount of energy needed to increase the temperature of 1 g of a substance by 1°C. The units of $C_{\text{s}}$ are $\text{J}/\left(\text{g}•°\text{C}\right).$ is the amount of energy needed to increase the temperature of 1 g of a substance by 1°C; its units are thus J/(g·°C). We can relate the quantity of a substance, the amount of heat transferred, its heat capacity, and the temperature change in two ways:

The specific heats of some common substances are given in Table 5.3 "Specific Heats of Selected Substances at 25°C". Note that the specific heat values of most solids are less than 1 J/(g·°C), whereas those of most liquids are about 2 J/(g·°C). Water in its solid and liquid states is an exception. The heat capacity of ice is twice as high as that of most solids; the heat capacity of liquid water, 4.184 J/(g·°C), is one of the highest known.

The high specific heat of liquid water has important implications for life on Earth. A given mass of water releases more than five times as much heat for a 1°C temperature change as does the same mass of limestone or granite. Consequently, coastal regions of our planet tend to have less variable climates than regions in the center of a continent. After absorbing large amounts of thermal energy from the sun in summer, the water slowly releases the energy during the winter, thus keeping coastal areas warmer than otherwise would be expected (Figure 5.13 "The High Specific Heat of Liquid Water Has Major Effects on Climate"). Water’s capacity to absorb large amounts of energy without undergoing a large increase in temperature also explains why swimming pools and waterbeds are usually heated. Heat must be applied to raise the temperature of the water to a comfortable level for swimming or sleeping and to maintain that level as heat is exchanged with the surroundings. Moreover, because the human body is about 70% water by mass, a great deal of energy is required to change its temperature by even 1°C. Consequently, the mechanism for maintaining our body temperature at about 37°C does not have to be as finely tuned as would be necessary if our bodies were primarily composed of a substance with a lower specific heat.

Figure 5.13 The High Specific Heat of Liquid Water Has Major Effects on Climate

Regions that are near very large bodies of water, such as oceans or lakes, tend to have smaller temperature differences between summer and winter months than regions in the center of a continent. The contours on this map show the difference between January and July monthly mean surface temperatures (in degrees Celsius).

Example 7

A home solar energy storage unit uses 400 L of water for storing thermal energy. On a sunny day, the initial temperature of the water is 22.0°C. During the course of the day, the temperature of the water rises to 38.0°C as it circulates through the water wall. How much energy has been stored in the water? (The density of water at 22.0°C is 0.998 g/mL.)

Passive solar system. During the day (a), sunlight is absorbed by water circulating in the water wall. At night (b), heat stored in the water wall continues to warm the air inside the house.

Given: volume and density of water and initial and final temperatures

Asked for: amount of energy stored

Strategy:

A Use the density of water at 22.0°C to obtain the mass of water (m) that corresponds to 400 L of water. Then compute ΔT for the water.

B Determine the amount of heat absorbed by substituting values for m, C_s, and ΔT into Equation 5.36.

Solution:

A The mass of water is

$${\text{mass of H}}_{\text{2}}\text{O }=\text{ 400 }\cancel{\text{L}}\left(\frac{\text{1000 }\cancel{\text{mL}}}{\text{1 }\cancel{\text{L}}}\right)\left(\frac{\text{0}\text{.998 g}}{\text{1 }\cancel{\text{mL}}}\right)=\text{ 3}\text{.99}\times {\text{10}}^{\text{5}}{\text{ g H}}_{\text{2}}\text{O}$$

The temperature change (ΔT) is 38.0°C − 22.0°C = +16.0°C.

B From Table 5.3 "Specific Heats of Selected Substances at 25°C", the specific heat of water is 4.184 J/(g·°C). From Equation 5.36, the heat absorbed by the water is thus

$$q=m{C}_{\text{s}}\Delta T=\text{ (3}\text{.99}\times {\text{10}}^{\text{5}}\cancel{\text{g}}\text{)}\left(\frac{\text{4}\text{.184 J}}{\cancel{\text{g}}\text{•}\cancel{\text{°C}}}\right)(\text{16}\text{.0}\cancel{\text{°C}})=\text{ 2}\text{.67}\times {\text{10}}^{\text{7}}\text{ J }=\text{ 2}\text{.67}\times {\text{10}}^{\text{4}}\text{ kJ}$$

Both q and ΔT are positive, consistent with the fact that the water has absorbed energy.

Exercise

Some solar energy devices used in homes circulate air over a bed of rocks that absorb thermal energy from the sun. If a house uses a solar heating system that contains 2500 kg of sandstone rocks, what amount of energy is stored if the temperature of the rocks increases from 20.0°C to 34.5°C during the day? Assume that the specific heat of sandstone is the same as that of quartz (SiO₂) in Table 5.3 "Specific Heats of Selected Substances at 25°C".

Answer: 2.7 × 10⁴ kJ (Even though the mass of sandstone is more than six times the mass of the water in Example 7, the amount of thermal energy stored is the same to two significant figures.)

When two objects at different temperatures are placed in contact, heat flows from the warmer object to the cooler one until the temperature of both objects is the same. The law of conservation of energy says that the total energy cannot change during this process:

The equation implies that the amount of heat that flows from a warmer object is the same as the amount of heat that flows into a cooler object. Because the direction of heat flow is opposite for the two objects, the sign of the heat flow values must be opposite:

Thus heat is conserved in any such process, consistent with the law of conservation of energy.

Substituting for q from Equation 5.36 gives

which can be rearranged to give

When two objects initially at different temperatures are placed in contact, we can use Equation 5.40 to calculate the final temperature if we know the chemical composition and mass of the objects.

Measuring Heat Flow

In Example 7, radiant energy from the sun was used to raise the temperature of water. A calorimetric experiment uses essentially the same procedure, except that the thermal energy change accompanying a chemical reaction is responsible for the change in temperature that takes place in a calorimeter. If the reaction releases heat (q_rxn < 0), then heat is absorbed by the calorimeter (q_calorimeter > 0) and its temperature increases. Conversely, if the reaction absorbs heat (q_rxn > 0), then heat is transferred from the calorimeter to the system (q_calorimeter < 0) and the temperature of the calorimeter decreases. In both cases, the amount of heat absorbed or released by the calorimeter is equal in magnitude and opposite in sign to the amount of heat produced or consumed by the reaction. The heat capacity of the calorimeter or of the reaction mixture may be used to calculate the amount of heat released or absorbed by the chemical reaction. The amount of heat released or absorbed per gram or mole of reactant can then be calculated from the mass of the reactants.

Constant-Pressure Calorimetry

Because ΔH is defined as the heat flow at constant pressure, measurements made using a constant-pressure calorimeterA device used to measure enthalpy changes in chemical processes at constant pressure. give ΔH values directly. This device is particularly well suited to studying reactions carried out in solution at a constant atmospheric pressure. A “student” version, called a coffee-cup calorimeter (Figure 5.14 "A Coffee-Cup Calorimeter"), is often encountered in general chemistry laboratories. Commercial calorimeters operate on the same principle, but they can be used with smaller volumes of solution, have better thermal insulation, and can detect a change in temperature as small as several millionths of a degree (10⁻⁶°C). Because the heat released or absorbed at constant pressure is equal to ΔH, the relationship between heat and ΔH_rxn is

Equation 5.41

ΔH_rxn = q_rxn = −q_calorimeter = −mC_sΔT

The use of a constant-pressure calorimeter is illustrated in Example 9.

Figure 5.14 A Coffee-Cup Calorimeter

This simplified version of a constant-pressure calorimeter consists of two Styrofoam cups nested and sealed with an insulated stopper to thermally isolate the system (the solution being studied) from the surroundings (the air and the laboratory bench). Two holes in the stopper allow the use of a thermometer to measure the temperature and a stirrer to mix the reactants.

Example 9

When 5.03 g of solid potassium hydroxide are dissolved in 100.0 mL of distilled water in a coffee-cup calorimeter, the temperature of the liquid increases from 23.0°C to 34.7°C. The density of water in this temperature range averages 0.9969 g/cm³. What is ΔH_soln (in kilojoules per mole)? Assume that the calorimeter absorbs a negligible amount of heat and, because of the large volume of water, the specific heat of the solution is the same as the specific heat of pure water.

Given: mass of substance, volume of solvent, and initial and final temperatures

Asked for: ΔH_soln

Strategy:

A Calculate the mass of the solution from its volume and density and calculate the temperature change of the solution.

B Find the heat flow that accompanies the dissolution reaction by substituting the appropriate values into Equation 5.41.

C Use the molar mass of KOH to calculate ΔH_soln.

Solution:

A To calculate ΔH_soln, we must first determine the amount of heat released in the calorimetry experiment. The mass of the solution is

$$\text{(100}\text{.0 }\cancel{\text{mL}}{\text{ H}}_{\text{2}}\text{O)(0}\text{.9969 g/}\cancel{\text{mL}}\text{) }+\text{ 5}\text{.03 g KOH }=\text{ 104}\text{.72 g}$$

The temperature change is (34.7°C − 23.0°C) = +11.7°C.

B Because the solution is not very concentrated (approximately 0.9 M), we assume that the specific heat of the solution is the same as that of water. The heat flow that accompanies dissolution is thus

$$q_{\text{calorimeter}}=m{C}_{\text{s}}\Delta T=\text{ (104}\text{.72 }\cancel{\text{g}}\text{)}\left(\frac{\text{4}\text{.184 J}}{\cancel{\text{g}}·\cancel{\text{°C}}}\right)(\text{11}\text{.7}\cancel{\text{°C}}\text{) }=\text{ 5130 J }=\text{ 5}\text{.13 kJ}$$

The temperature of the solution increased because heat was absorbed by the solution (q > 0). Where did this heat come from? It was released by KOH dissolving in water. From Equation 5.41, we see that

ΔH_rxn = −q_calorimeter = −5.13 kJ

This experiment tells us that dissolving 5.03 g of KOH in water is accompanied by the release of 5.13 kJ of energy. Because the temperature of the solution increased, the dissolution of KOH in water must be exothermic.

C The last step is to use the molar mass of KOH to calculate ΔH_soln—the heat released when dissolving 1 mol of KOH:

$$\Delta H_{\text{soln}}=\left(\frac{-\text{5}\text{.13 kJ}}{\text{5}\text{.03 }\cancel{\text{g}}}\right)\left(\frac{\text{56}\text{.11 }\cancel{\text{g}}}{\text{1 mol}}\right)=-\text{57}\text{.2 kJ/mol}$$

Exercise

A coffee-cup calorimeter contains 50.0 mL of distilled water at 22.7°C. Solid ammonium bromide (3.14 g) is added and the solution is stirred, giving a final temperature of 20.3°C. Using the same assumptions as in Example 9, find ΔH_soln for NH₄Br (in kilojoules per mole).

Answer: 16.6 kJ/mol

Constant-Volume Calorimetry

Constant-pressure calorimeters are not very well suited for studying reactions in which one or more of the reactants is a gas, such as a combustion reaction. The enthalpy changes that accompany combustion reactions are therefore measured using a constant-volume calorimeter, such as the bomb calorimeterA device used to measure energy changes in chemical processes. shown schematically in Figure 5.15 "A Bomb Calorimeter". The reactant is placed in a steel cup inside a steel vessel with a fixed volume (the “bomb”). The bomb is then sealed, filled with excess oxygen gas, and placed inside an insulated container that holds a known amount of water. Because combustion reactions are exothermic, the temperature of the bath and the calorimeter increases during combustion. If the heat capacity of the bomb and the mass of water are known, the heat released can be calculated.

Figure 5.15 A Bomb Calorimeter

After the temperature of the water in the insulated container has reached a constant value, the combustion reaction is initiated by passing an electric current through a wire embedded in the sample. Because this calorimeter operates at constant volume, the heat released is not precisely the same as the enthalpy change for the reaction.

Because the volume of the system (the inside of the bomb) is fixed, the combustion reaction occurs under conditions in which the volume, but not the pressure, is constant. As you will learn in Chapter 18 "Chemical Thermodynamics", the heat released by a reaction carried out at constant volume is identical to the change in internal energy (ΔE) rather than the enthalpy change (ΔH); ΔE is related to ΔH by an expression that depends on the change in the number of moles of gas during the reaction. The difference between the heat flow measured at constant volume and the enthalpy change is usually quite small, however (on the order of a few percent). Assuming that ΔE < ΔH, the relationship between the measured temperature change and ΔH_comb is given in Equation 5.42, where C_bomb is the total heat capacity of the steel bomb and the water surrounding it:

Equation 5.42

ΔH_comb < q_comb = −q_calorimeter = −C_bombΔT

To measure the heat capacity of the calorimeter, we first burn a carefully weighed mass of a standard compound whose enthalpy of combustion is accurately known. Benzoic acid (C₆H₅CO₂H) is often used for this purpose because it is a crystalline solid that can be obtained in high purity. The combustion of benzoic acid in a bomb calorimeter releases 26.38 kJ of heat per gram (i.e., its ΔH_comb = −26.38 kJ/g). This value and the measured increase in temperature of the calorimeter can be used in Equation 5.42 to determine C_bomb. The use of a bomb calorimeter to measure the ΔH_comb of a substance is illustrated in Example 10.

Example 10

The combustion of 0.579 g of benzoic acid in a bomb calorimeter caused a 2.08°C increase in the temperature of the calorimeter. The chamber was then emptied and recharged with 1.732 g of glucose and excess oxygen. Ignition of the glucose resulted in a temperature increase of 3.64°C. What is the ΔH_comb of glucose?

Given: mass and ΔT for combustion of standard and sample

Asked for: ΔH_comb of glucose

Strategy:

A Calculate the value of q_rxn for benzoic acid by multiplying the mass of benzoic acid by its ΔH_comb. Then use Equation 5.42 to determine the heat capacity of the calorimeter (C_bomb) from q_comb and ΔT.

B Calculate the amount of heat released during the combustion of glucose by multiplying the heat capacity of the bomb by the temperature change. Determine the ΔH_comb of glucose by multiplying the amount of heat released per gram by the molar mass of glucose.

Solution:

The first step is to use Equation 5.42 and the information obtained from the combustion of benzoic acid to calculate C_bomb. We are given ΔT, and we can calculate q_comb from the mass of benzoic acid:

$$q_{\text{comb}}=\text{ (0}\text{.579 }\cancel{\text{g}}\text{)(}-\text{26}\text{.38 kJ/}\cancel{\text{g}}\text{) }=-\text{15}\text{.3 kJ}$$

From Equation 5.42,

$$-C_{\text{bomb}}=\frac{q_{\text{comb}}}{\Delta T}=\frac{-\text{15}\text{.3 kJ}}{\text{2}\text{.08°C}}=-\text{7}\text{.34 kJ/°C}$$

B According to the strategy, we can now use the heat capacity of the bomb to calculate the amount of heat released during the combustion of glucose:

$$q_{\text{comb}}=-C_{\text{bomb}}\Delta T=\text{ (}-\text{7}\text{.34 kJ/}\cancel{\text{°C}}\text{)(3}\text{.64 }\cancel{\text{°C}}\text{) }=-\text{26}\text{.7 kJ}$$

Because the combustion of 1.732 g of glucose released 26.7 kJ of energy, the ΔH_comb of glucose is

$$\Delta H_{\text{comb}}=\left(\frac{-\text{26}\text{.7 kJ}}{\text{1}\text{.732 }\cancel{\text{g}}}\right)\left(\frac{\text{180}\text{.16 }\cancel{\text{g}}}{\text{mol}}\right)=-\text{2780 kJ/mol }=-\text{2}\text{.78}\times {\text{10}}^{\text{3}}\text{ kJ/mol}$$

This result is in good agreement (< 1% error) with the value of ΔH_comb = −2803 kJ/mol that we calculated in Section 5.2 "Enthalpy" using enthalpies of formation.

Exercise

When 2.123 g of benzoic acid is ignited in a bomb calorimeter, a temperature increase of 4.75°C is observed. When 1.932 g of methylhydrazine (CH₃NHNH₂) is ignited in the same calorimeter, the temperature increase is 4.64°C. Calculate the ΔH_comb of methylhydrazine, the fuel used in the maneuvering jets of the US space shuttle.

Answer: −1.30 × 10³ kJ/mol

Key Equations

relationship of quantity of a substance, heat capacity, heat flow, and temperature change

Equation 5.35: q = nC_pΔT

Equation 5.36: q = mC_sΔT

constant-pressure calorimetry

Equation 5.41: ΔH_rxn = q_rxn = −q_calorimeter = −mC_sΔT

constant-volume calorimetry

Equation 5.42: ΔH_comb < q_comb = −q_calorimeter = −C_bombΔT

Summary

Calorimetry is the set of techniques used to measure enthalpy changes during chemical processes. It uses devices called calorimeters, which measure the change in temperature when a chemical reaction is carried out. The magnitude of the temperature change depends on the amount of heat released or absorbed and on the heat capacity of the system. The heat capacity (C) of an object is the amount of energy needed to raise its temperature by 1°C; its units are joules per degree Celsius. The specific heat (C_s) of a substance is the amount of energy needed to raise the temperature of 1 g of the substance by 1°C, and the molar heat capacity (C_p) is the amount of energy needed to raise the temperature of 1 mol of a substance by 1°C. Liquid water has one of the highest specific heats known. Heat flow measurements can be made with either a constant-pressure calorimeter, which gives ΔH values directly, or a bomb calorimeter, which operates at constant volume and is particularly useful for measuring enthalpies of combustion.

Key Takeaway

• Calorimetry measures enthalpy changes during chemical processes, where the magnitude of the temperature change depends on the amount of heat released or absorbed and on the heat capacity of the system.

Conceptual Problems

1. Can an object have a negative heat capacity? Why or why not?

2. What two factors determine the heat capacity of an object? Does the specific heat also depend on these two factors? Explain your answer.

3. Explain why regions along seacoasts have a more moderate climate than inland regions do.

4. Although soapstone is more expensive than brick, soapstone is frequently the building material of choice for fireplaces, particularly in northern climates with harsh winters. Propose an explanation for this.

Numerical Problems

Please be sure you are familiar with the topics discussed in Essential Skills 4 (Section 5.6 "Essential Skills 4") before proceeding to the Numerical Problems.

1. Using Equation 5.35 and Equation 5.36, derive a mathematical relationship between C_s and C_p.

2. Complete the following table for 28.0 g of each element at an initial temperature of 22.0°C.

   Element	q (J)	Cp [J/(mol·K)]	Final T (°C)
   nickel	137	26.07
   silicon		19.789	3.0
   zinc	603		77.5
   mercury	137		57

3. Using Table 5.3 "Specific Heats of Selected Substances at 25°C", how much heat is needed to raise the temperature of a 2.5 g piece of copper wire from 20°C to 80°C? How much heat is needed to increase the temperature of an equivalent mass of aluminum by the same amount? If you were using one of these metals to channel heat away from electrical components, which metal would you use? Once heated, which metal will cool faster? Give the specific heat for each metal.

4. Gold has a molar heat capacity of 25.418 J/(mol·K), and silver has a molar heat capacity of 23.350 J/(mol·K).

   1. If you put silver and gold spoons of equal mass into a cup of hot liquid and wait until the temperature of the liquid is constant, which spoon will take longer to cool down when removed from the hot liquid?
   2. If 8.00 g spoons of each metal at 20.0°C are placed in an insulated mug with 50.0 g of water at 97.0°C, what will be the final temperature of the water after the system has equilibrated? (Assume that no heat is transferred to the surroundings.)

5. In an exothermic reaction, how much heat would need to be evolved to raise the temperature of 150 mL of water 7.5°C? Explain how this process illustrates the law of conservation of energy.

6. How much heat must be evolved by a reaction to raise the temperature of 8.0 oz of water 5.0°C? What mass of lithium iodide would need to be dissolved in this volume of water to produce this temperature change?

7. A solution is made by dissolving 3.35 g of an unknown salt in 150 mL of water, and the temperature of the water rises 3.0°C. The addition of a silver nitrate solution results in a precipitate. Assuming that the heat capacity of the solution is the same as that of pure water, use the information in Table 5.2 "Enthalpies of Solution at 25°C of Selected Ionic Compounds in Water (in kJ/mol)" and solubility rules to identify the salt.

8. Using the data in Table 5.8 "Enthalpies of Combustion of Common Fuels and Selected Organic Compounds", calculate the change in temperature of a calorimeter with a heat capacity of 1.78 kJ/°C when 3.0 g of charcoal is burned in the calorimeter. If the calorimeter is in a 2 L bath of water at an initial temperature of 21.5°C, what will be the final temperature of the water after the combustion reaction (assuming no heat is lost to the surroundings)?

9. A 3.00 g sample of TNT (trinitrotoluene, C₇H₅N₃O₆) is placed in a bomb calorimeter with a heat capacity of 1.93 kJ/°C; the ΔH_comb of TNT is −3403.5 kJ/mol. If the initial temperature of the calorimeter is 19.8°C, what will be the final temperature of the calorimeter after the combustion reaction (assuming no heat is lost to the surroundings)? What is the ΔH_f of TNT?

Answers

1. C_p = C_s × (molar mass)

3. For Cu: q = 58 J; For Al: q = 130 J; Even though the values of the molar heat capacities are very similar for the two metals, the specific heat of Cu is only about half as large as that of Al, due to the greater molar mass of Cu versus Al: C_s = 0.385 and 0.897 J/(g·K) for Cu and Al, respectively. Thus loss of one joule of heat will cause almost twice as large a decrease in temperature of Cu versus Al.

5. 4.7 kJ

7. ΔH_soln = −0.56 kJ/g; based on reaction with AgNO₃, salt contains halide; dividing ΔH_soln values in Table 5.2 "Enthalpies of Solution at 25°C of Selected Ionic Compounds in Water (in kJ/mol)" by molar mass of salts gives lithium bromide as best match, with −0.56 kJ/g.

9. T_final = 43.1°C; the combustion reaction is 4C₇H₅N₃O₆(s) + 21O₂(g) → 28CO₂(g) + 10H₂O(g) + 6N₂(g); $\Delta H_{\text{f}}^o$ (TNT) = −65.5 kJ/mol