Major Concept Area: Equilibrium. The concept of dynamic equilibrium is an extremely important one in chemistry. In this chapter, we take the first step toward an understanding of equilibrium by considering energy changes in chemical and physical processes. As we will see in later chapters, the tendency of a system to seek minimum energy is one of two major contributing factors to the establishment of equilibrium.
Specific Concepts in this Chapter:9-1 Terminology. We stated in Chapter 1 that chemical reactions usually either produce or consume energy. Similarly, as you know from Chapter 7 and from experience, phase changes of pure substances involve energy changes. For example, input of heat is necessary to melt ice, or to boil water. Energy is released as heat when steam condenses or water freezes. In this chapter we will lay the foundation for quantitative discussion of energy changes in chemical reactions and in physical processes, such as changes of phase. The formal structure for discussion of energy is a portion of the discipline of Thermodynamics. Thermodynamics began as the study of the exchange of heat and work between an engine and its surroundings. From these humble beginnings has grown a branch of science so powerful that it can tell us whether or not a process of interest can occur, and to what extent. We now begin a study of thermodynamics. In Chapter 10 we will return to a discussion of phase changes and build upon the energy concepts developed here.
Scientists are concerned with the interaction between a system and its surroundings, which occurs across a boundary between them. A system may be changed from one state to another by a process, which usually involves an exchange of energy between system and surroundings. The emphasized words are important. They occur frequently in any discussion of science, and should mean the same thing to all who use them. We define each of these words and a few others below:
We have discussed energy in Chapters 2, 6, and 7. Systems frequently change energy in processes, and it is these changes that we actually measure. The change in energy of a system in a process will be written DE, where the symbol, D , means "change." The change in energy is the difference between the energy after the process (final value) and the energy before the process (initial value):
Any process involves two energy changes, one for the system, the other for the surroundings. These are always equal and opposite in sign:
Energy lost by the system must be gained by the surroundings, and vice versa, because energy is a conserved quantity: the total energy of the universe is constant. Generally, we will be more interested in DEsys than in DEsurr; unless otherwise stated, the symbol DE will refer to the energy change of the system.
As we have seen, energy is of only two types: kinetic (motion) and potential (position). The energy, E, of a system is the sum of its kinetic and potential energies: E = KE + PE; the change in E is the sum of changes in KE and PE: DE = DKE + DPE. Kinetic energy is possessed by a moving baseball (translation), a spinning top (rotation) and a spring that is repeatedly stretching and contracting (vibration). Molecules undergo analogous motions, as explored in Chapter 4. Potential energy is possessed by a rock on a mountain, a compressed spring, a positive and a negative charge separated by some finite distance (as in a chemical bond) and by mass (through the Einstein equation, E = mc2). Potential energy results from forces. Important for chemists is the PE associated with chemical bonds. When bonds are broken, the Coulomb force of attraction of two nuclei for the bonding electrons must be overcome, and PE increases. When bonds are formed, force of attraction is increased and PE decreases. Thus
Energy changes in chemical reactions are primarily potential energy changes that result from breaking bonds and making new ones.
9-2 Heat, Work, and the First Law. Suppose we desire to change the internal energy of a system. How can we do it? Surprisingly, there are only two ways: we can transfer heat to or from the system; or we can arrange for work to be done on or by the system. All of thermodynamics is built upon the measurement of heat and work. The quantity of heat involved in a process is represented by the letter, q. We will adopt the convention that q is positive when heat is added to a system, and negative when it is removed from a system. The amount of work involved in a process is represented by the letter, w. We will adopt the convention that w is positive when work is done on a system by its surroundings, and negative when work is done by the system. If an amount of heat q is added to a system, and an amount of work w is done on it, the (internal) energy of the system changes by the amount DE. This is expressed succinctly in equation 9-2-1. This very simple and important equation is called the First Law of Thermodynamics. It is a statement of the law of energy conservation.
Before proceeding further, we discuss a point that is probably obvious but that is nonetheless important. Although our focus will normally be on the heat added to the system (whether positive or negative) and the work done on the system by the surroundings (whether positive or negative), we will on occasion prefer to view things from the perspective of the surroundings. That is, we will be interested in the heat added to the surroundings, or the work done on the surroundings by the system. In all processes, the heat added to the system and to the surroundings are equal and opposite (that is, if heat enters the system, it must leave the surroundings, and vice versa). Similarly, in all processes, the work done on the system and by the system are equal and opposite (that is, if the system does work on the surroundings, the surroundings does an equal and opposite work on the system). Thus
Heat and work are familiar yet elusive concepts. We may define heat as energy that "flows" due to a temperature difference. The spontaneous movement of heat is always from high to low temperature. For example, heat flows from a hot cup of coffee to the cooler surroundings, but never in the reverse direction. The quantity of heat added to a system can be readily determined from the temperature change of the system and its heat capacity, C:
The heat capacity of a system is defined as the quantity of heat that must be added to the system to increase its temperature by 1oC (or 1 K). For a pure substance, the value of C depends on the amount of substance being heated (or cooled). Values of the heat capacity per gram (the specific heat, Cs) or heat capacity per mole (the molar heat capacity, Cm) are therefore tabulated. Table 9-1 gives the specific heats and molar heat capacities for a number of common substances.
| Substance | Cs, J/g-K | Cm, J/mole-K |
|---|---|---|
| H2O(l) | 4.184 | 75.376 |
| Fe(s) | 0.449 | 25.1 |
| O2(g) | 0.9172 | 29.35 |
| NaCl(s) | 0.8641 | 50.50 |
| Al(s) | 0.901 | 24.3 |
The water will not boil; in fact, it's temperature barely changes, because there is so much more water than iron, and because the heat capacity of water is 10 times that of iron.
From the molecular viewpoint, heat is related to the changes in the random thermal motions of molecules. The molecules of a system that is hotter than its surroundings have higher average kinetic energy than the molecules of the surroundings. These "hot" molecules collide with the boundary of the system and transfer some of their kinetic energy to the molecules of the boundary. Molecules in the surroundings in turn receive kinetic energy by colliding with the boundary molecules. In this way, the excess thermal (kinetic) energy of the system molecules is transferred across the boundary to the molecules of the surroundings. The average KE of the system molecules decreases and that of the molecules in the surroundings increases in this way, until system and surroundings are at the same temperature. Once this situation has been reached, there will be no further tendency for kinetic energy to be transferred one way or the other; a state of thermal equilibrium has been reached. Thus heat flow is the result of a difference of molecular kinetic energies on two sides of a boundary.
We define work as the energy expended when a force moves matter through a distance. Mathematically, it is the product of the force, F, exerted and the distance, r, moved, as in equation 9-2-3.
This equation is correct only when the force remains at the same value over the entire distance. Since this is frequently and generally not the case, we replace equation 9-2-3 with the differential equation, 9-2-4.
Here dw is the infinitesimal amount of work done when force F moves a mass through the infinitesimal distance, dr. The distance is short enough that F may be considered constant. To obtain the total work done over the full distance, r, which is the sum of the infinitesimal distances, dr, we add up the infinitesimal amounts of work. This is done by integration of equation 9-2-4.
Equation 9-2-5 is the general expression for work in any situation. We now wish to tailor it to a form that is more useful for chemical applications. Since the work done when chemical reactions take place is the result of expansion or contraction of the system volume against a pressure exerted by the surroundings, we must translate equation 9-2-5 into pressure-volume terms. Figure 9-1 enables us to do this.
The figure shows a gas confined in a cylinder by a piston of area A. The surroundings exerts a pressure, Pext, on the piston. We will calculate the work done on the system when the external pressure contracts it an amount DVsys, from Vi to Vf (Vf < Vi), assuming that the external pressure remains constant throughout the contraction. During the contraction, the piston moves in a linear distance, d. Since the piston moves in the same direction that the external pressure is pushing, the work done by the surroundings (on the system) is positive. The volume change of the surroundings during the process is D Vsurr = Ad. But the volume change of the system is the negative of this, so that Ad = -DVsys. The force exerted by the surroundings is F = PextA, since pressure is force per unit area. It follows that the work done in the process is
We have successfully expressed the work in terms of the outside pressure and the volume change of the system. There are two things to note about this equation. First, the system has been contracted by the surroundings, so work has been done on it. According to our convention, w should be positive. However, DVsys is < 0. The - sign on the right side insures that a positive work is calculated when the system is compressed. Second, Pext need not be constant throughout the contraction process. We must therefore sum over infinitesimal amounts of work as we did to obtain equation 9-2-5. The final result for the mechanical work of expansion, important for chemical processes, is equation 9-2-6.
Two things are noteworthy about this equation. First, it says that there can be no work done unless the system changes volume; that is, a boundary of the system must move. Second, if Pext is constant throughout a process, then integration is simple and w = -PextDVsys.
At the molecular level, work results because random molecular collisions with a movable system boundary transfer momentum to the boundary and push it back against an external force. The motions of the system molecules are completely random, as always. But their random motion is converted to an organized and macroscopically observable motion of the boundary that can be used to do something useful in the surroundings. Work is quite different from heat, which "flows" because random molecular motions on one side of the boundary are converted to random molecular motions on the other side. Heat never leaves the realm of randomness; work does.
Work is done on a system by the surroundings when a moveable system boundary is pushed in by the surroundings. The unidirectional boundary movement transfers energy to the system molecules that collide with it, much as a ping-pong ball is accelerated by a moving paddle. This tends to increase the average kinetic energy of the system molecules, and the system temperature goes up; the system warms when work is done on it if no heat is transferred.
We now apply the first law to gas expansions that are important in chemical processes.
D E: Because the gas is ideal, there are no intermolecular forces and no potential energy. The internal energy of the gas is all kinetic energy, and we may write:
Further, T is constant. Since KE depends only on temperature, KE is constant. Therefore
Work: Since Pext is constant throughout the process, w = -PextDVsys:
Heat: We obtain this from the First Law of Thermodynamics:
Work flows out of the system (the system does work on the surroundings by pushing the surroundings back), an equal amount of heat flows in, and the internal energy (all kinetic) of the gas does not change. This is shown in Figure 9-2b. We arrived at this result by considering only macroscopic properties -- the volume of the gas, and the pressure exerted on the system by the surroundings. It is instructive, though, to think about what the gas molecules are doing during the process. The piston is pushed back because the number of molecular collisions on the inside exceeds the number on the outside. Molecules of the gas transfer energy to the piston as they push it back, which tends to slow them down. Thus the gas molecules tend to lose kinetic energy as they push back the surroundings. This kinetic energy is replenished, however, by the influx of heat from the surroundings. This energy is transferred to the gas via collisions of the gas molecules with the walls. The net result is that the average molecular speed, hence the kinetic energy, remains constant.
DE: Since the gas is ideal and T is constant, DE = 0.
Work: Some thought is required to realize that there is no work done in this process. In almost all situations in which we will be interested, we must be careful to define the system so that it is closed; that is, so that no matter crosses the system boundary. We can have a closed system in this problem only by including the right chamber as a part of the system, even when it is empty. Otherwise, when the stopcock is opened, matter (the gas) will flow across the system boundary, and the system will be open. If we define the system to include the right box, then the system volume does not change in the process, and
It is important to distinguish here between the volume of the system and the volume occupied by the gas. The latter changes; the former does not.
Although the molecules spontaneously move into the right chamber when the stopcock is opened, they do not lose kinetic energy because they do not have to push back the surroundings when they expand. Therefore no influx of heat occurs because there is no temperature (kinetic energy) difference between the gas molecules and the molecules of the surroundings.
We have examined gas expansions in some detail because they are extremely important in considerations of chemical reactions involving gases, and in practical areas as well. For example, the operation of the internal combustion engine is based on the forceful expansion of gases produced in the controlled explosion of gasoline. These gases push back pistons, which turn a driveshaft and propel an automobile. Constricted veins and arteries are opened up non-surgically by inflation of a balloon in the constricted area. The work done by the balloon in pushing back the wall of the blood vessel restores free blood flow. Air pressure in the air locks on submarines is used to push back the sea water that would otherwise come pouring in. The process of breathing involves inflation and deflation of the lungs; the energy required to supply the work done in this process comes from controlled oxidation of food by the cells of the body. Finally, and very importantly, warming and cooling of the atmosphere by the sun expands and contracts atmospheric gases, causing weather patterns and storms. Appendix J presents first law analyses of somewhat more complex gas expansion processes.
State and Path Functions. Examples 9-2 and 9-3 illustrate two isothermal gas expansions, with the initial and final states of the system the same in both cases. It is instructive to compare the results:
Isothermal expansion against constant Pext:The amounts of heat and work that "flow" during a process connecting specified initial and final states depend on how the process is carried out. In contrast, the change in the energy of the system is the same in both expansions, independent of how the process is performed. Quantities that depend on the path followed between states are called path functions. Quantities whose values are independent of path are called state functions. The total change in the value of a state function for any cyclic process (a process in which the initital and final states of a system are the same) is zero. We say that state functions are conserved during any complete cycle. Path functions are not conserved. In thermodynamics, state functions are represented by upper case (capital) letters; e.g., P, V, T, and E. Path functions are represented by lower case letters; e.g., q and w.
In light of this discussion, the first law may seem odd, because it expresses the change in a state function as the sum of two path functions. Although heat and work are both path-dependent quantities, their sum is not. That this should be true is not obvious. That it is true has been verified in all known applications of the first law of thermodynamics.
9-3 Enthalpy. Enthalpy, symbolized H, is the most important energy quantity for discussion of chemical reactions. It is defined by equation 9-3-1.
Enthalpy is a state function of the system because E, P, and V are state functions. It is obviously an energy quantity because both terms on the right side of equation 9-3-1 have energy units. At this stage of our development, though, it is not clear why we need a new state function, or why it should have the form of equation 9-3-1. We attempt to justify the form before moving on.
From the first law of thermodynamics it is obvious that if a process is carried out at constant volume (so that DVsys = 0), no work is done and DE = q. This is an important result; it tells us that if we measure the heat added to a system in a constant volume process, we immediately know the change in energy of the system. To emphasize the constant volume restriction on the equation, it is usually written
Unfortunately, although the equation is useful, it is not convenient to carry out processes at constant volume. This requires using completely closed containers, sealed off from the outside, which makes isolation and study of the product of a chemical reaction difficult. In practice, chemists do most of their work under conditions of constant pressure, and let the volume of the system change as it will. In most cases, chemical reactions are carried out in containers that are open to the atmosphere, and the constant pressure is atmospheric pressure. An energy state function whose change is measured by the heat added to the system in a constant pressure process would therefore be more useful than E for the analysis of chemical reactions. Such a state function, which we might call H to suggest its connection with heat, would then satisfy equation 9-3-3:
What form should the new function H have? We can answer this by applying the first law to the constant pressure process:
for a constant pressure process, Pext = Psys = constant. We may therefore replace Pext with Psys and take the pressure out from under the integral sign to obtain equation 9-3-4.
Expanding the terms and solving for q gives 9-3-5
Now notice that in equation 9-3-5 the difference between the final and initial values of the quantity, E + PV appears. We can make equation 9-3-5 look like equation 9-3-3 by substituting H = E + PV. The final result is equation 9-3-6.
The subscript P is added to remind us of the restrictions on the truth of the equation. This explains why enthalpy is defined as it is in equation 9-3-1, and suggests the following statement: the enthalpy change of a system for a process carried out at constant pressure is the amount of heat absorbed by the system during the process. Based on equation 9-3-1, we may think of enthalpy as the energy of the system plus the energy of interaction of the system with the surroundings, represented by the PV term. When P is constant, this interaction energy is the work done by the system on the surroundings.
The substitution of nRT for PV follows from the ideal gas law. Since n and T are both constant throughout the process, D(nRT) is zero. Thus
Molecular Energy, Revisited. The internal energy of a system is the total of the energy possessed by its atoms and molecules. We have discussed the types of molecular energy in terms of equation 6-1-3, Chapter 6. We will now apply that equation to three systems of increasing molecular complexity, to show ultimately that the most important contribution to molecular energy is the potential energy of chemical bonds.
First, consider a system consisting of an ideal gas. There are no intermolecular forces between molecules of an ideal gas, so there is no intermolecular potential energy; and there is no internal molecular structure to worry about. The only energy that such a system has is the translational kinetic energy of the molecules as they move about in the container. For 1 mole of an ideal gas at 298 K, the total energy is
We now increase the complexity of our system slightly by considering a real monatomic gas. Monatomic means that the "molecules" are single atoms. Examples of real monatomic gases are the noble gases, He, Ne, Kr, Xe, and Rn. Atoms of these gases exert weak (but non-zero) intermolecular forces of the instantaneous dipole-induced dipole type on one another, so there will be a small potential energy in such a system in addition to the translational kinetic energy at 298 K. In addition, real atoms have internal structure (i.e., nuclei and electrons). We will ignore the energy associated with this structure. For real monatomic gases, then,
The value of potential energy used here is for the noble gas, Ar. It is negative because we follow the convention established in our discussion of the potential well in Chapter 6: potential energy is 0 when the atoms or molecules are far enough apart that the forces fall essentially to 0, and decreases from zero at smaller distances as the attractive forces take effect.
We now make one more increase in complexity and consider a real polyatomic gas. In this case, each gas molecule contains 2 or more atoms. Examples are HCl (hydrogen chloride), NH3 (ammonia), and CH4 (methane). The total energy of such a molecule, again ignoring the internal atomic structure, is in equation 9-3-9.
This equation bears some resemblance to equation 4-1-2. Espin and Enuclei have been omitted because they are not involved in chemical reactions, and therefore will not change. Etrans and Erot have been explicitly acknowledged as kinetic energies. Evib has been divided into kinetic and potential portions (the energy of vibration interconverts between potential and kinetic as the vibration moves from either extreme (stretched or compressed), where there is no motion, to the point where there is no tension in the spring). PEinter acknowledges that molecules exert attractive forces on one another, just as do the atoms of argon. Finally, PEbonds is the potential energy resulting from the chemical bonds within the molecules, holding the atoms together. Equation 9-3-9 is evaluated for HCl in equation 9-3-10:
The first 5 terms of 9-3-10 are similar in magnitude and small. The dominant term is clearly the potential energy of chemical bond formation. We will generalize from this one specific example: the internal energy of a system consisting of real polyatomic gases is primarily the potential energy of chemical bonds. When we consider E for chemical reactions in the next section, we will equate it to the change in bond potential energy, and assume that the remaining (small) terms in equation 9-3-10 are very nearly the same for reactants and products.
9-4 Applications of Energy Ideas to Chemical Reactions. A chemical reaction in its most general form is presented in equation 9-4-1.
In thermodynamic terms, the reaction is a process in which a system proceeds from an initial state (the reactants) to a final state (the products). During the process, an amount of heat, q, is added to the system. The system undergoes a change in internal energy, DE, and a change in enthalpy, DH. As we have seen, both DE and DH can be related to q under certain conditions. A reaction for which q > 0 (i.e., in which heat is transferred from surroundings to system) is endothermic ("endo" means "in"; thus, heat in); one for which q < 0, exothermic ("exo" means "out"; hence, heat out). If the reaction is done at constant pressure, q = DH. If the reaction is done at constant volume, q = DE = DPEbonds. We state without proof that DH and DE are approximately equal for chemical reactions, and both provide a good measure of the change in potential energy that occurs when bonds in the reactants break and bonds in the products form. In this text, we will not worry about the small difference between DH and DE. In more advanced courses, the distinction between the two quantities will be made clear.
The exchange of energy between system and surroundings is analogous to the transfer of money from a bank account to ready cash. When funds (energy) are withdrawn from the account (system), the account balance (internal energy) goes down by the same amount that the ready cash increases outside of the account (heat appears in the surroundings).
Conservation of Enthalpy -- Hess's Law. It should come as no surprise that enthalpy, a form of energy, is conserved in chemical reactions. This is something that is now well-established and that we rely upon in chemistry. However, in the early application of thermodynamics to chemical reactions, this was not known. It was put on a firm experimental basis by Hess. We will introduce the ideas generally, then apply them to a particular example.
Suppose that we are interested in the enthalpy change, DHunk for the chemical reaction
but for some reason are unable to determine it experimentally (by carrying out the reaction and measuring the heat produced or absorbed). Suppose, however, that we can measure the enthalpy changes for two related reactions:
The reactions are related to 9-4-2, because when they are added together and substances common to both sides cancelled, equation 9-4-2 results. It would be useful if the three enthalpy changes were also related in this way; i.e., DHunk = DH1 + DH2. We can use an energy level diagram to show that this should be true. Figure 9-5 shows three states plotted on such a diagram. The initial state for the unknown reaction is A, 2B; the final state for the unknown reaction is state C. The intermediate state on the diagram is the initial state for reaction (2). DH1, DH2, and DH3 are shown as arrows on the diagram (strictly speaking, the reaction in the diagram is
However, cancellation of B gives reaction (1).) It is clear that the arrow representing DHunk is the vector sum of the arrows representing DH1 and DH2. This demonstrates what we wanted to show.
Hess measured the enthalpies for many series of related reactions and found that in all cases, enthalpy was conserved. He summarized his work in a simple statement, which is paraphrased below:
Hess's Law: if several reactions can be added or subtracted to give a desired reaction, their enthalpy changes can be combined in the same way to give the enthalpy change of the desired reaction.
Stated another way the law says that no matter what pathway we take from one state to another, the net change in enthalpy will be the same. Enthalpy is a state function.
A generally useful thought process will be applied to this simple example. Our goal is to decide how to combine equations (a) and (b) to obtain the equation with unknown enthalpy change. The desired equation has 1 mole of C(s) on the left. Equation (b) has C(s) on the left, just as we need. To obtain 1 mole C(s) on the left, we take equation (b) as is:
The next substance in the desired equation is O2(g). Both (a) and (b) contain O2(g). We thus have no idea whether to add in equation (a), equation (b), or some combination of them. We solve this problem by ignoring the O2(g). The next substance in the desired equation is CO(g). 1 mole of it occurs on the right side. Equation (a) involves 2 moles of CO(g), on the left. We can get 1 mole CO(g) by taking 1/2 of equation (a). We can get CO on the right by subtracting equation (a). Thus
(At this point you should carry out this combination of equations (a) and (b) to see that the desired equation is obtained.) According to Hess's Law, the unknown enthalpy change is obtained with the same combination of known enthalpies:
Try plotting the 3 reactions on an energy level diagram to obtain the unknown enthalpy diagrammatically.
We look now at a more complex example.| Reaction | DH, kJ |
|---|---|
| (a) C(s) + O2(g) ---> CO2(g) | -393.5 |
| (b) H2(g) + 1/2 O2(g) ---> H2O(l) | -285.85 |
| (c) 6C(s) + 6H2(g) + 3O2(g) ---> C6H12O6(s) | -1273.02 |
Since O2 appears in all 3 given equations, it cannot give us any information about which equation to take, or what factor to multiply by. We ignore it. Here is the desired combination of (a), (b), and (c):
(See if this combination gives the desired equation.) The enthalpies then combine in the same way:
Photosynthesis is strongly endothermic. This means that a large amount of energy must be supplied to carry out the reaction.
For the chemical reaction
the standard enthalpy of reaction, DHRo (the subscript R stands for "reaction"), is defined as the enthalpy change that occurs when the reaction is carried out under standard conditions -- i.e., with each reactant and product in its standard state. The standard state of a pure substance is the form (solid, liquid, or gas) in which the substance is most stable when subjected to 1 atm pressure at a specified temperature. Solutions are in the standard state when all solutes are present at 1.0 M concentration, again at a specified temperature. In most cases, the specified temperature is 298 K (for convenience), but it need not be. For example, hydrogen is most stable as a diatomic gas at 1 atm pressure and 298 K; its standard state at 298 K is therefore H2(g). The standard state of water at 298 K is H2O(l), but at 398 K is H2O(g). That of iron(III) oxide is the solid ionic lattice, Fe2O3(s), at 298 K. Each substance has only one standard state at a specified temperature, but many non-standard states. Several non-standard states for hydrogen at 298 K are listed below:
The symbol, DHRo, is used for the enthalpy change of any reaction carried out under standard conditions.
9-5 Special Processes and their Associated Standard Enthalpy Changes. Standard Enthalpy of Formation. A vast number of reaction enthalpies has been measured, so many that to tabulate them all would require a compendium of unmanageable size. In addition, such a compendium would be immediately out of date as new reaction enthalpies became known. Clearly some efficient means of tabulation is desirable, and has been achieved using the so-called standard enthalpy of formation, symbolized DHfo. The standard enthalpy of formation of a substance is defined as the enthalpy change that occurs when 1 mole of the substance is made from its constituent elements in their standard states.
The elements of ethyl alcohol are carbon, hydrogen, and oxygen. Their standard states are C(graphite), H2(g), and O2(g). We write these elemental standard states on the left of the equation:
Finally, we balance the equation by changing only the coefficients on the left (not the right, because we must form 1 mole of the desired product):
The standard enthalpy change for this reaction is by definition DHfo(C2H5OH). In similar fashion we obtain the following formation reactions for the other 3 substances:
The last reaction is interesting because nothing happens! Elemental bromine is made from itself. The enthalpy change must be zero. The same will be true when any other element in its standard state is made from itself. Thus, the standard enthalpy of formation of an element in its standard state is zero.
As Example 9-8 indicates, standard enthalpies of formation may be viewed as chemical potential energies of substances relative to the chemical potential energies of the elements in their standard states.
Rather than keep a huge, running compendium of standard enthalpy changes for arbitrary reactions, chemists tabulate DHfo values for individual substances at 298 K. From the standard enthalpies of formation, it is possible to calculate the standard enthalpy changes for many reactions involving these substances. The latter enthalpy changes then need not be tabulated. To see how this works, consider the diagram in Figure 9-6. The figure shows a general chemical reaction in which some set of reactants, R, is converted to products, P. We can construct a second pathway for this process involving two steps. In the first step, Reactants are decomposed into their constituent elements in the standard state. The associated enthalpy change is simply the negative of the sum of the standard enthalpies of formation of the reactants. The negative sign is required because reactants are being decomposed, not formed. In the second step, these elements are recombined to form the Products. The enthalpy change for this step is the sum of the standard enthalpies of formation of the products. Hess's Law tells us that the net enthalpy changes for the two paths must be the same. It therefore follows that the enthalpy change for the overall reaction is given by equation 9-5-1:
Standard enthalpies of formation for a few common substances are given in Table 9-2. A more extensive compilation is found in Appendix G.
| Substance | DHfo |
|---|---|
| CaCO3 | -1207.1 |
| CaO | -635.5 |
| CH4(g) | -74.8 |
| C2H2(g) | 226.7 |
| C2H4(g) | 52.3 |
| C6H12O6(s) | -1273.02 |
| CO2(g) | -393.5 |
| H2O(l) | -285.85 |
| MgO(s) | -601.8 |
Figure 9-7 shows several standard enthalpies of formation plotted on an energy diagram. The energies are plotted relative to those of the elements in their standard states, all found at zero on the diagram. Using a diagram like this, it is possible to use vector addition to obtain the value for DHRo of a specified reaction. Such a procedure is the graphical analogy to equation 9-5-1.
| Substance | DHfo |
|---|---|
| N2H4(g) | 95.40 |
| N2O4(g) | 9.66 |
| N2(g) | ---- |
| H2O(l) | -285.85 |
Equation 9-5-1 is now applied:
The reaction is strongly exothermic.
Enthalpies of Phase Transformation. In Chapter 7 we described a thought experiment in which we began with ice in a syringe at -10oC. We gradually added heat to the system, watching the changes within it, until the final temperature was 110oC and the water was present as vapor. All pure substances undergo temperature-induced conversion from the solid state to the gas state in essentially the same way. The range of temperature involved will vary from substance to substance, but the phase change characteristics do not. Figure 9-8 shows a plot of the heat added per mole of substance versus the temperature of the substance. It is a graphical summary of the thought experiment in Chapter 7. The plot is called a heating curve. In region 1 (T1 to the melting point, Tf) the solid is heated. The added heat increases the temperature (kinetic energy) of the molecules of the solid. The amount of heat added per mole of substance is linearly related to temperature according to equation 9-5-2
The slope of the plot in region 1 is the heat capacity of the solid, Csolid. Since the addition of energy to the system is carried out at constant pressure, q = DH. Thus
When the melting temperature is reached, it is maintained until all solid has melted. Region 2 (Tf) shows that the temperature of the system remains constant despite the addition of heat. In this region, the kinetic energy of the system is constant but the potential energy increases as molecules pull apart against the intermolecular forces; the solid melts. The energy added in this region is called the heat (or enthalpy) of fusion, DHfus. Region 3 (Tf to Tb) is like region 1 except that liquid, not solid, is present, so the slope is Cliquid. Region 4 (Tb) shows vaporization at the boiling point, Tb, with addition of the enthalpy of vaporization, DHvap. Finally, region 5 shows heating of the vapor at a rate Cgas. Regions 2 and 4 clearly show the system climbing out of the potential well of the solid phase, first melting to liquid, then vaporizing to gas. For most substances, DHvap is larger than DHfus because the increase in distance of molecular separation from liquid to vapor is much larger. The values of DHfus and DHvap in Figure 9-8 apply at the temperatures Tf and Tb, respectively. Just as for chemical reactions, however, it is convenient to tabulate phase change enthalpies for 1 mole of substance under standard conditions at 298 K. Table 9-3 shows standard molar enthalpies of fusion, DHofus, and vaporization, DHovap, for a number of common substances.
| Substance | DHfuso | DHvapo | |
|---|---|---|---|
| water | 6.01 | 40.7 | |
| carbon tetrachloride | 2.47 | 30.00 | |
| carbon dioxide | 8.33 | 25.23(s) | |
| methyl alcohol | 3.16 | 35.27 | |
| ethyl alcohol | 4.60 | 43.5 | |
| ammonia | 5.65 | 23.35 |
Since the process is done at constant pressure, DHvap = q = 40.67 kJ. The definition of enthalpy can be used to obtain DE:
The pressure is constant, so D(PV) = PDV = P(Vgas - Vliquid). The volume of 1 mole of liquid water is approximately 18 mL (18.0 g divided by the density of water, about 1 g/mL). The volume of a mole of water vapor is about 31.01 L at 373 K, from the ideal gas law. Clearly the liquid volume is neglible. It follows that D(PV) = PVgas = 31.01 L-atm. Conversion to joules is accomplished using the ratio of the gas constant in joules to the gas constant in L-atm:
Solving equation 9-5-4 for DE gives
DH and DE have similar values. The small difference of 3.1 kJ is a measure of the work done on the surroundings during the vaporization process, as the gas pushes back the atmosphere. This calculation substantiates our earlier decision to ignore the usually small difference between DE and DH.
20.94 kJ of heat will be released to the surroundings when the vapor condenses.
In Chapter 7, and thus far in this chapter, we have discussed phase changes carried out by heating at constant pressure. We have referred to these as temperature-induced phase changes. We now discuss the energetics of pressure-induced phase changes. These are accomplished by varying the pressure exerted on the substance while keeping its temperature constant. Consider a sample of a substance in the gas phase, confined in a cylinder/piston apparatus as in Figure 9-9a. The temperature of the cylinder and the gas is maintained constant at T1. While keeping the temperature fixed, we gradually increase the pressure on the gas by pushing on the piston. This causes the molecules to crowd more closely together. However, it does not affect the average speeds of the molecules. They will collide with other molecules more frequently and change speed more often, but as long as temperature does not change, the average kinetic energy will not either. When pressure gets high enough, the gas will liquefy, and the potential energy of the molecules will correspondingly decrease. This energy will be released as heat, and the molecules will temporarily speed up. However, as this heat is transferred to the constant temperature bath through the walls of the cylinder, the gas temperature will return to T1 and the average molecular speed will return to the original value. Further increase in pressure, still maintaining T constant, causes the liquid to solidify. Again there will be a temporary increase in molecular speed when the molecular potential energy decreases, but this will return to the original value as the heat of fusion is transferred to the constant temperature surroundings. A plot of the potential energy (enthalpy) of the system as a function of applied pressure is shown in Figure 9-9b. The plot is idealized in showing no change in potential energy with pressure when the substance is present as a single phase. In actuality, slight decreases in potential energy occur with increasing pressure in the gas and liquid phases, and a slight increase occurs in the solid phase. However, these changes are small compared with the changes that occur during phase transformation. As long as temperature remains constant, the kinetic energy of an average molecule remains the same at 3kT/2, regardless of phase, and the speeds obey the Maxwell-Boltzmann Distribution.
The thought experiment described above can be done in the laboratory using carbon dioxide, CO2. Starting with a sample of CO2 gas at -50 oC and 1 atm pressure, an increase of pressure to about 5.3 atm causes liquefaction at a constant temperature of -50 oC. A further increase in pressure to about 20 atm causes the liquid to solidify at -50 oC.
Application of Hess's Law to the process of conversion of solid to gas directly (path 1) or indirectly via the liquid phase (path 2) leads to the conclusion that for any pure substance
This is indicated in terms of the potential well in Figure 9-10.
Enthalpy of Atomization as an Alternative to Enthalpy of Formation. Tabulations of standard thermodynamic data for chemical compounds are invariably based on the standard enthalpy of formation, DHof, discussed earlier. We have seen that this quantity is useful for calculating the overall enthalpy changes for a huge number of chemical reactions. In this section, we will explore an alternative approach to the tabulation of thermodynamic enthalpy data, one that more directly emphasizes the importance of chemical bonding in chemical reactions.
Spontaneous conversion of a particular set of reactants to a particular set of products either does or does not tend to occur. As we will learn, there are two major factors involved in whether a reaction occurs. One of these is an energy factor, which we can summarize as follows:
A reaction tends to occur if the bonds between atoms in the products are stronger than the bonds between atoms in the reactants.
The enthalpy change measures directly the relative strengths of bonds in reactants and products. If DH is positive, we interpret this to mean that more energy is required to break the bonds in the reactants than is released in forming the bonds of the products. The reverse is true for negative DH. It would be nice to express this H, which is so clearly related to relative strengths of bonds, in terms of a quantity for each chemical species which directly represents the strengths of the bonds in the species. Unfortunately, DHof does not do this. Instead, it represents the difference in bond strengths between the species and its elements in their standard states, which in most cases also contain bonds.
The quantity that we seek is the standard enthalpy of atomization, DHoatom, for the species. This may be defined as the energy required to convert 1 mole of the species completely to its constituent atoms in the gas phase. For example, the standard enthalpy of atomization of CaCO3 is the enthalpy change for the following process:
There are no chemical bonds between the isolated atoms in the gas phase. Thus DHoatom represents the total enthalpy change required to break all of the bonds in 1 mole of calcium carbonate. It directly measures the strengths of the bonds in this material. Breaking chemical bonds is always endothermic; that is, it always requires the input of energy:
Consequently, enthalpies of atomization are invariably positive numbers.
Let's apply this idea to the general chemical reaction in equation 9-5-6.
In Figure 9-6, we applied Hess's Law to this process by first converting the reactants to elements in their standard states. These were then allowed to form products in their standard states. An alternate Hess's Law path is shown in Figure 9-11. Here the chemical and physical bonds in the reactants are broken by the input of an amount of energy equal to the total of the atomization enthalpies of all of the reactants. This converts the reactants to atoms, in the gas phase. These atoms are then allowed to come together in the configurations corresponding to the products. The bonds that form in this process release an amount of energy equal to the negative of the total atomization enthalpy of the products. The enthalpy of reaction is thus seen as the net result of breaking the reactant bonds and replacing them with product bonds. Since reaction 9-5-6 is perfectly general, we may write
The importance of bonds in determining the energy change of a reaction is thus directly seen. Table 9-4 contains standard enthalpies of atomization for a number of elements and compounds.
The requisite enthalpies of atomization are found in Table 9-4.
Solution.An output of 9648 + 5827 = 15475 kJ is produced in forming bonds in the products. Overall, the bonds in the products are stronger; the reaction is exothermic. DHR = -2799 kJ
9-6 Experimental Measurement of DH: Calorimetry. The measurement of enthalpy changes of chemical and physical processes is accomplished by calorimetry. The chemical or physical process is carried out on a precisely known scale in an insulated container called a calorimeter. The insulation prevents the flow of heat from the surroundings to the system (inside the calorimeter) or vice versa. The accompanying diagram summarizes the calorimetric experiment pictorially. Inside the calorimeter, the heat produced or consumed by the process is absorbed or provided by the calorimeter, which will cause the temperature of the calorimeter and its contents to change. From the measured temperature change and the known heat capacity of the calorimeter, the heat absorbed or provided by the calorimeter can be calculated using equation 9-4 (qcal = CcalDT). For an exothermic process, for which qprocess < 0, the heat produced is absorbed by the calorimeter. Therefore qcal is > 0 and the temperature increases. For an endothermic process (qprocess > 0), qcal < 0 and temperature decreases. In either case, qcal = -qprocess. Substituting -qprocess for qcal gives the fundamental equation of calorimetry:
qprocess can then be adjusted for the scale of the reaction to give a value per mole of process, qprocess,m. If the process is carried out at constant pressure, qprocess,m = DHprocess (approximately = DEprocess); if at constant volume, qprocess,m = DEprocess (approximately = DHprocess).
This is the heat of reaction for burning 1.765 g ethyl alcohol. We can convert this to a standard heat (a heat per mole) by dividing by the number of moles of ethyl alcohol.
Since the reaction was done at constant volume, q = DERo = -1366 kJ. As we expect by now, DHRo is very close to this (-1368 kJ).
It was anticipated that the enthalpy of reaction for sulfuric acid should be twice that for hydrochloric acid, because the former contains two hydrogen ions, while the former contains only one.
The experiments were carried out as follows. First, the reaction of HCl with NaOH was investigated using 2.0 M solutions of acid and base. Several volumes of acid solution (between 2 and 40 mL) were reacted with the same volume of base solution (50 mL). In all cases, the total volume of solution in the calorimeter was adjusted to 100 mL by addition of the necessary volume of water (the difference between 50 mL and the volume of acid to be used) to the calorimeter along with the acid solution. The acid solution in the calorimeter and the base solution outside the calorimeter were adjusted to the same temperature, and the base was quickly added to the calorimeter. The resulting temperature change was measured. Data are presented in Table 9-5; observed temperature change is plotted against the millimoles of acid used (volume used multiplied by molarity of the acid solution) in Figure 9-12. The same procedure was then followed using 2.0 M H2SO4 and 2.0 M NaOH. The results of these experiments are also shown in the table and figure.
The heat capacity of the calorimeter containing 100 mL of aqueous solution is known to be 350 J/ oC. What is the relationship between the enthalpies of reaction of the two acids with NaOH? What are the values of the reaction enthalpies?
Discussion. Several conclusions can be drawn from the figure. First, for both acids, the observed temperature change is directly proportional to the moles of acid used. Because the acid is the limiting reagent, this is equivalent to saying that the temperature change is proportional to the amount of reaction. This substantiates that reaction enthalpy is an extensive property--its value depends on the amounts of substances reacted. The slopes of the two linear plots give the temperature change per millimole of acid. The values obtained from the best linear fit of the data are
Second, it is clear that the value for sulfuric acid is somewhat more than twice as large as the HCl value. The ratio of slopes is 2.41. The data plotted in Figure 9-12 are quite good; there is minimal scatter around the best fit straight lines. It is thus hard to justify rounding 2.41 to 2.00, the anticipated value. The enthalpy of reaction of H2SO4 with NaOH is about 20% larger than expected. This requires explanation.
As we will see in detail in Chapter 13, strong acids such as HCl react with water as follows:
Reaction of such an acid with a base such as NaOH actually involves the following reaction:
In anticipating that the enthalpy of reaction for sulfuric acid would be twice that for hydrochloric acid, we assumed that sulfuric acid readily gives up both hydrogen ions to water:
If this were true, however, the temperature change per mole of sulfuric acid would be exactly twice the value for hydrochloric acid. The fact that it is not reveals to us that sulfuric acid releases the first hydrogen ion readily, but not the second one. When dissolved in water, sulfuric acid releases the first hydrogen ion as follows:
The second hydrogen ion, however, remains attached to the sulfate anion. When base is added, H3O+ produced by the first hydrogen ion reacts with OH- as expected. In order for the second proton to react with OH-, it must be removed from the sulfate ion:
Hess's law allows us to formulate this in terms of the following two steps:
The extra 20% of heat produced in the sulfuric acid-sodium hydroxide reaction is DH1.
Finally, the enthalpy per mole of reaction can be obtained by multiplying the temperature change per mole by the heat capacity of the calorimeter:
From these values, we obtain DH1 = -22.8 kJ/mole of HSO4-.
9-7 The Relationship Between Energy and Chemical Bonds. Bond energy. Structures that result from covalent bonding between atoms were discussed in detail in Chapter 3. There we developed methods for understanding and predicting the number of bonds in a molecule or polyatomic ion, and the three-dimensional distribution of atoms in space that results from the bonding. In this chapter, we have stressed that chemical reactions involve bonding rearrangements, and that the energy changes that accompany reactions reflect the relative strengths of bonds in reactants and products. In this section, we discuss the relationship between the enthalpy change for a reaction and the strengths of specific covalent bonds in reactants and products.
The total bond energy of a gaseous molecule is the energy required to decompose the molecule to isolated, gaseous atoms. For example, the bond energy for CO2(g) is the enthalpy change for the following reaction:
This is simply the enthalpy of atomization of CO2(g). We would now like to partition this total enthalpy among the individual bonds in the CO2 molecule. The structure of CO2, developed using the principles of Chapter 3, is shown in Figure 9-13. The rupture of two moles of C=O bonds in CO2 to form gaseous atoms requires 1609 kJ of energy; thus the energy required to break one C=O double bond is 804.5 kJ/mole. Similarly, one fourth of the total bond energy (enthalpy of atomization) of methane, CH4, is the average strength of a C-H single bond in methane. Total bond energies for a great many molecules have been measured. The results indicate that the strength of, say, an I-F bond is very much (although not exactly) the same no matter what molecule it occurs in. The data below support this statement:
| Compound | I-F Bond Strength |
|---|---|
| IF | 278 |
| IF3 | 272 |
| IF5 | 268 |
| IF7 | 231 |
It has thus been possible to assign an average bond energy to each common bond type. Table 9-6 summarizes bond energy data for a large number of chemical bonds. Note that all of the entries in the table are positive numbers; bond breaking is always an endothermic process.
Using the bond energy information in the table, it is possible to estimate the value of DH for a large number of gas phase chemical reactions, using the simple Hess's Law cycle illustrated in Figure 9-14. It is important to stress that this is an estimate, because the tabulated bond energies are average values. According to the figure, we visualize a pathway for the reaction in which reactants are first converted to gas phase atoms, then the atoms are allowed to recombine to form products. Disassembling the reactants requires the input of an energy equal to the total energy of all of the bonds in the reactants. This quantity, S[BE(reactants)] is invariably positive. Assembling the products releases the total bond energy of the products, S[BE(products)]. This step is exothermic, as bonds are formed. If the total bond energy of reactants exceeds that of products, the reaction is endothermic (i.e., more energy must be put in to break the reactant bonds than is produced in forming product bonds); if the bond energy of products exceeds that of reactants, the reaction is exothermic. The value of DH for the reaction may thus be estimated using equation 9-7-1
| Reactants | BE | Products | BE |
|---|---|---|---|
| C(3)C | 835 | C=O (4) | 4 x 799 |
| C-H (2) | 2 x 411 | O-H (2) | 2 x 459 |
| O=O (2.5) | 2.5 x 494 | ||
| total: | 2892 | total: | 4114 |
Bonds in the products are stronger than in the reactants. The reaction is exothermic in the amount 1222 kJ. Note that this value is similar to, but not exactly the same as, the value calculated using standard enthalpies of formation (you should work this out). The standard enthalpy of formation is a very accurate quantity, which applies specifically to a particular compound. In contrast, bond energies are average values, and thus cannot be expected to reproduce exactly the specific bond strengths in a particular compound. Enthalpy changes calculated using average bond energies are approximations.
Solution. Average bond energies can not be used because the reaction is not entirely in the gas phase; one of the products, H2O, is formed as a liquid. The total bond energy of a substance in the liquid phase includes not only the energies of the covalent bonds, but also the potential energy due to the intermolecular forces operative in the liquid. To break the bonds in liquid H2O requires, first, that the liquid be vaporized, then that the two covalent bonds be broken. The average bond energies in Table 9-6 apply to covalent bond breaking only, and have nothing to say about phase changes.
Of course enthalpies of atomization can be used. These measure the energy required to break all of the bonds in a substance, which includes the phase change enthalpies.
Chemical Reactions: Driving Force/Spontaneity. Exothermic processes produce heat, and in the process adopt a final atomic arrangement that has lower potential energy (enthalpy) than the initial arrangement. Many such processes are spontaneous -- that is, they occur naturally, without assistance from outside the system. For example, the rusting of iron is a spontaneous process that releases 825 kJ of heat per mole of iron(III) oxide produced; natural gas spontaneously burns in oxygen, releasing 890 kJ per mole of CH4; and hydrogen gas reacts spontaneously with oxygen gas to form water, with the release of 286 kJ of energy per mole of water in the form of heat and some light. It was at one time thought that all spontaneous chemical processes were exothermic, based on the well-known mechanical tendency to minimize potential energy (e.g., water flows downhill). Thus the drive to decrease the potential energy of the system was thought to be the driving force for chemical reactions. In fact, although many spontaneous chemical processes are exothermic, some are endothermic. For example, ice melts spontaneously at 25oC, even though the process is endothermic. Ammonium nitrate, NH4NO3, spontaneously dissolves in water, despite the fact that a substantial quantity of heat is transferred to the system from the surroundings in the process (the water becomes cold as the salt dissolves). The drive to minimize potential energy (enthalpy) is clearly an important factor in determining spontaneity, because so many spontaneous processes are exothermic. However, it is not the only factor. In the next chapter, we will discuss the second factor involved in spontaneity, in the context of phase equilibrium.
Supplement: Hess's Law and Biological Glucose Oxidation.
Applications| DHo, kJ/mole | |
|---|---|
| a) C(s) + O2(g) ---> CO2(g) | -393.5 |
| b) H2(g) + 1/2 O2(g) ---> H2O(l) | -285.85 |
| c) 6 C(s) + 6 H2(g) + 3 O2(g) ---> C6H12O6(s) | -1273.02 |
| DHof, kJ | |
|---|---|
| C(s) + O2(g) ---> CO2(g) | -393.5 |
| CO(g) + 1/2 O2(g) ---> CO2(g) | -110.5 |
| C(s) + 1/2 O2(g) ---> CO(g) | unknown |
| DHo, kJ | |
|---|---|
| (a) C2H6(g) + 7/2 O2(g) ---> 2CO2(g) + 3H2O(l) | -1560.09 |
| (b) H2(g) + 1/2 O2(g) ---> H2O(l) | -285.85 |
| (c) C(graph) + O2(g) ---> CO2(g) | -393.51 |
| (d) 2C(graph) + 3H2(g) ---> C2H6(g) | ? |
| DHo | |
|---|---|
| N2(g) + 3H2(g) ---> 2NH3(g) | -92.05 kJ |
| H2(g) ---> 2H(g) | 435.97 |
| N2(g) ---> 2N(g) | 941.4 |
| Bond | BE, kJ/mole |
|---|---|
| H-C | 411 |
| H-O | 459 |
| C=O | 799 |
| C(3)O | 1071 |
| C-O | 358 |