Major Concept Area: Equilibrium. The concept of dynamic equilibrium is an extremely important one in chemistry. In Chapter 9, we took the first step toward an understanding of equilibrium by considering energy changes in chemical and physical processes. In this chapter, we discuss the second major contributing factor to the establishment of equilibrium: the tendency of the universe to seek maximum disorder.
Specific Concepts in this Chapter:
In Chapter 10, we introduced the concept of disorder by discussing the free expansion of an ideal gas. Common experience tells us that this process is spontaneous--it occurs without outside assistance. The change in potential energy for this process is zero, so its spontaneity is not due to the tendency to minimize PE. Instead, the process is driven by the tendency to increase disorder. Our earlier discussion of disorder was vague in two respects. First, we did not distinguish between the disorder of the system and that of the surroundings. Second, disorder was treated only qualitatively. In this chapter, we will put the disorder concept on a much firmer basis by making it quantitative. To do this will require that we differentiate system and surroundings, as we have done for energy changes in Chapter 9.
11-1 The Second Law of Thermodynamics. Consider again the free expansion of an ideal gas, carried out in the apparatus in Figure 11-1. Expansion of the gas (the system) increases its disorder because each gas molecule has a larger volume to explore. Disorder is greater because knowledge of the location of a gas molecule is less. Clearly, then, free expansion leads to an increase in disorder--in entropy--of the system: DSsys > 0. What happens in the surroundings when the gas expands? Other than the small effort expended in opening the connecting valve, absolutely nothing happens. No boundary of the system moves, so the surroundings does not experience work. There is no heat flow to or from the surroundings because the gas is ideal and thus does not change internal energy on expansion. Absolutely nothing is experienced by an observer in the surroundings. It therefore follows that the disorder in the surroundings is unchanged by the free expansion of the gas: DSsurr = 0. The system and surroundings together constitute the universe. Combining our two results gives equation 11-1-1 for the free expansion of an ideal gas:
Stated in words, equation 11-1-1 says that in any spontaneous process, the entropy of the universe increases. Although the equation was developed by consideration of a single simple process, it has proved to have far-reaching and profound generality: it is true for any spontaneous process. Thus for the burning of natural gas; the rusting of iron; the respiration of living organisms; and the light-promoted process of photosynthesis, the disorder (entropy) of the universe increases. The simple and powerful statement in Equation 11-1-1 is called the Second Law of Thermodynamics.
The Second Law is in many respects more subtle and difficult to understand than is the First Law. Yet it is the Second Law that allows prediction of spontaneity, so to gain some understanding of it is crucially important. To facilitate understanding, we will make liberal use of examples in this chapter to illustrate the important ideas; and we will set down a number of simple guidelines for predicting the sign of the entropy changes for system and surroundings in physical and chemical processes. These guidelines are not hard and fast rules--instead, they give the correct result most, but not all, of the time. However, their simplicity makes them worth remembering.
The search for a quantitative measure of entropy was a long and difficult one. The analysis of steam engines (the original genesis of thermodynamics) was the source of one approach, formulated by Rudolf Clausius in the 1880s. The Clausius formulation expresses entropy in terms of heat and temperature. The second approach, that of Ludwig Boltzmann, was based on considerations of probability and statistics. We have already made the connection between disorder and probability in our discussion in Chapter 10. Although the two quantitative formulations of entropy appear different, it has been shown that they are equivalent in any application to which they may both be applied. We will briefly consider these two approaches to entropy, beginning with the Clausius formulation.
11-2 Entropy and Heat. That there is a connection between disorder and heat is somewhat expected. First, heat, the random thermal motion of molecules, is clearly the most disordered and non-useful form of energy. Second, much of our experience indicates that disorder is produced when heat is added to systems. The solid form of a substance (water, sodium chloride, iron) melts, and the liquid form boils, when heat is added. Gases expand when heated. Atoms form a plasma of nuclei and electrons when heated to the temperatures of stars. These qualitative realizations help us to appreciate the Clausius definition of entropy in equation 11-2-1.
Stated in words, the equation says that the change in entropy in a system when a process is carried out may be calculated by dividing the amount of heat added to the system by the temperature of the system. Strictly speaking, equation 11-2-1 is valid only when the temperature of the system remains constant. Most of the processes that we will consider occur at constant temperature, however, and equation 11-2-1 applies. The subscript "max" on the heat term represents a restriction on the equation. It indicates that when calculating the entropy change for a change in the system from a particular initial to a particular final state, we must use the heat that would be added to the system along the maximum heat path. This is known as the reversible path. Our intention here is merely to gain appreciation for the connection of entropy to heat. Consequently, we will not discuss this idea of maximum heat further here. The concepts are discussed more fully in Appendix F, for those who wish to pursue the matter in more depth.
We now proceed to examine the "sense" of equation 11-2-1. First, the Clausius formulation of entropy recognizes the clear connection between disorder and heat. The more heat we add, the more disorder we produce, at a given temperature. Second, the equation says that the disorder produced by addition of a given amount of heat is less when the system is at high temperature than when it is at low temperature. This is also reasonable. In a system at low T (say, near absolute zero) the kinetic energy of random molecular motion is small. At high T (say 1000 K), however, the thermal energy is substantial. Addition of a specified amount of heat to such a system produces a much larger fractional change in disorder at low T than at high T. Equation 11-2-1 recognizes that if a system is already highly disordered, a small amount of heat will not make much difference. For example, when your room has not been cleaned for a month (or longer), the addition of one more dirty sock to the mess will hardly be noticeable. In contrast, if your room has just been cleaned (lowered in entropy), that dirty sock on the floor will be very noticeable. Third, the equation tells us that the units of disorder are those of energy divided by temperature, or J/K. Finally, entropy is represented in equation 11-2-1 and in Chapter 10 by an upper case symbol, which we reserve for state functions. Entropy is a state function. This means that the change in entropy of a system is independent of the path taken between initial and final states. Two simple and useful guidelines can be distilled from equation 11-2-1. Both guidelines refer to systems consisting of a single pure substance.
Solution. To convert Fe(s) at 0 oC to Fe(s) at 25 oC requires the addition of heat. As a result of the addition of heat, the average kinetic energy of the iron atoms increases, with a corresponding increase in disorder in the system. Fe(s) has higher entropy at 25 o than at 0 oC.
To convert H2O(l) at 298 K to H2O(g) at 398 K requires the addition of heat. First, liquid water must be heated from 298 K to the boiling point, 373 K. This increases the thermal motions, and the disorder, of the molecules of the liquid. Second, liquid water must be boiled at 373 K. This frees water molecules from the potential well of the liquid and allows them to move independently of one another; disorder increases substantially. Finally, water vapor at 373 K must be heated to 398 K. Again, average molecular kinetic energy (and disorder) increases with the addition of heat. H2O(g) at 398 K has substantially higher entropy than H2O(l) at 298 K.
To convert Ar(g) at 150 K to Ar(g) at 300 K requires the addition of heat. This increases the random thermal molecular motion, and the disorder of the system. Ar(g) has higher entropy at 300 K than at 150 K.
Solution. Because heat is added to a pure substance in this process, we can say with confidence that the entropy of the system (substance) increases. Thus DSsys > 0. In this case, we can go further by actually applying equation 11-2-1 to calculate the change in entropy that occurs when ethanol is vaporized, because phase changes carried out at the normal phase change temperature satisfy the "maximum heat path" restriction on 11-2-1.
Further, vaporization at the normal boiling point implies a constant pressure of 1 atm. In constant pressure processes, q = DH. Thus
Example 11-2 leads to a third useful entropy guideline:
11-3 Entropy and Probability: the Boltzmann Interpretation of Entropy. The Clausius interpretation of entropy is a pragmatic one, based on engine analyisis. Ludwig Boltzmann took a different aproach, based on the idea that there is a parallel between the amount of disorder in a system in a particular state and the number of ways that the system may be arranged in that state. In a very general way, this can be stated as
where W = the number of arrangements possible for a particular state of a system, and f is a to-be-determined function. We suppose at the outset that a state with only one possible arrangement has no disorder--that is, S = 0--and that S should increase smoothly as W increases. Thus our first constraint on the form of the function, f, is that f(1) = 0. To arrive at a second constraint on f, consider a simple example, illustrated in Figure 11-2. Shown is a row of 10 buckets placed side-by-side. A ping pong ball is tossed at the row of buckets and lands in one of them (we assume for simplicity that there are no misses--the ball always goes in one of the buckets). The state of the resulting system is described by saying that there is one ball in the set of 10 buckets. Because the ball can be in any of the buckets, the number of arrangements for the state, W, is 10. There is thus some disorder associated with this state. An observer asked to view the row of buckets containing one ball would be uncertain as to where the ball is. His uncertainty is a measure of the disorder.
We now double the amount of material in the system by tossing a second ball, which also lands in one of the buckets. The state of the system is described by saying that there are 2 balls in the set of 10 buckets. There are 10 possible locations for the first ball. For each of these possibilities, there are 10 possible locations for the second ball. The total number of arrangements for the state is 10 * 10 = 102. By similar reasoning, if n balls in succession were tossed, the number of arrangements would be 10n. In general, for a state consisting of N balls placed in L locations, the number of arrangements is given by equation 11-3-2.
The entropy of the state is S = f(LN). We now impose the second constraint on the form of the function f: S must double when the amount of material in the system (N) doubles. In equation form,
The function that satisfies both imposed constraints (f(1) = 0 and f(L2N) = 2f(LN)) is the logarithm. Thus
where C is a proportionality constant. Equation 11-3-4 is the Boltzmann interpretation of entropy. According to 11-3-4, the entropy of a system is directly proportional to the number of particles (such as ping pong balls, or molecules), N, in the system and to the natural logarithm of the number of locations each particle can have. Thus the amount of disorder goes up directly with the amount of matter in the system, in agreement with the second constraint. When each particle has only one allowed location, L is 1 and S is 0. The first constraint is therefore satisfied by 11-3-4. Further, we see that as the number of allowed locations for a particle (L) goes up, S increases more and more slowly. This is seen below:
| L | S | Increment in S |
|---|---|---|
| 1 | 0 | |
| 2 | .693 NC | .693 NC |
| 3 | 1.10 NC | .41 NC |
| 4 | 1.39 NC | .29 NC |
| 5 | 1.61 NC | .22 NC |
| . | ||
| . | ||
| 99 | 4.60 NC | |
| 100 | 4.61 NC | 0.01 NC |
When the system has low entropy initially (L is small, say 1), the addition of one more allowable location has a dramatic effect on S. Thus increasing L from 1 to 2 increases S by 0.693 NC. However, when the system has high entropy initially (L is large, say 99), increasing L by 1 increases S by only 0.01 NC, a negligible effect.
Boltzmann expressed equation 11-3-4 in somewhat different form, as shown in 11-3-5.
The constant, k, is called Boltzmann's constant. It has the value 1.38 * 10-23 J/K. A plot of S versus W, the number of arrangements of N particles in L locations, is shown in Figure 11-3. The slope of the plot is steep when L is small, but approaches zero when L becomes large, showing that DS becomes smaller for a given value of DW as S (and W) increases. Thus when W increases from 1 to 200, S increases by about 7 * 10-23 J/K; but when W increases by the same amount from 800 to 1000, S increases by only 0.3 * 10-23 J/K. The following mathematical operations, based on equation 11-3-5, show analytically what Figure 11-3 shows visually.
Because W = LN, where N = the number of particles in the system and L = the number of options per particle,
Doubling N gives S = k ln L2N = 2Nk ln L. Thus S goes up in direct proportion to N. However, doubling L gives S = k ln (2L)N = Nk (ln 2 + ln L) = Nk ln L + Nk ln 2. In this case, S goes up incrementally by N ln 2 each time the number of options per particle increases by a factor of 2.
The Boltzmann and Clausius interpretations of entropy are thus in agreement that the more disordered a system is initially, the less difference an incremental change in the number of arrangements makes. If a state, "a", has more possible arrangements than a state, "b", then state "a" is more probable than state "b". The Boltzmann interpretation of entropy explicitly recognizes the connection between entropy and probability that we developed intuitively in Chapter 10. According to this interpretation, the second law of thermodynamics says that the universe seeks states of higher probability.
Example 11-3. Consider a system consisting of a pair of dice in a closed box. Suppose that we have no information about this state of the system other than that there are 2 dice in a box. How many "particles" constitute this system? How many arrangements are possible for the system in this state? What is the entropy of the system in this state? Now suppose that the system is known to be in a "state" in which the sum of the numbers showing on the top faces of the two dice is 2. What is the entropy of this state? What is the entropy of the state in which the number total is 7? The most probable state of the system is the one for which we know only that there are two dice in a box. What is the next most probable state (numerical total) of the system? Discuss the entropies of the least and most probable numerical totals (states).
Solution. The first state to be considered is the one described by saying that there are 2 dice in a box. We will consider each die to be a particle. Then the system consists of 2 particles. Each particle can be in any of 6 "locations" (a die can have 1, 2, 3, 4, 5, or 6 showing on its top face). The total number of arrangements of the 2 particles in the described state is 6 * 6 = 62, or 36. This is the number of arrangements possible for the system so
To assess the probabilities and entropies of the other described states (numerical totals of 2 and 7), we must figure out the number of ways to obtain each state:
| State (Numerical Total) | Die 1 | Die 2 | # Arrangements |
|---|---|---|---|
| 2 | 1 | 1 | 1 |
| 3 | 1 | 2 | |
| 2 | 1 | 2 | |
| 4 | 1 | 3 | |
| 2 | 2 | ||
| 3 | 1 | 3 | |
| 5 | 1 | 4 | |
| 2 | 3 | ||
| 3 | 2 | ||
| 4 | 1 | 4 | |
| 6 | 1 | 5 | |
| 2 | 4 | ||
| 3 | 3 | ||
| 4 | 2 | ||
| 5 | 1 | 5 | |
| 7 | 1 | 6 | |
| 2 | 5 | ||
| 3 | 4 | ||
| 3 | 4 | ||
| 5 | 2 | ||
| 6 | 1 | 6 | |
| and so on. |
The most probable state is 7, because 6 different arrangements of the system produce it. The least probable states are 2 and 12, because each is produced by only one arrangement. State 7 has highest probability and highest entropy, with value k ln 6; states 2 and 12 have lowest probability and lowest entropy, with value k ln 1 = 0.
The entropy associated with states (numerical totals) of dice may be thought of as follows. A blindfolded person is told that the dice show a total of 2, and is asked what number is showing on each die. S/he can, with certainty, state that each die shows 1. In contrast, when told that the dice show a total of 7, s/he can only guess, with 1 chance in 6 of being correct, at the numbers on individual die. S/he is uncertain. Finally, when told only that there are 2 dice in a box, s/he has only 1 chance in 36 of guessing the numbers on individual die. The extent of uncertainty is a measure of the entropy of the state of the system.
Example 11-4. Consider a system consisting of 1 mole of HCl molecules in the solid state. Suppose that in the HCl crystal all molecules have their bond axes aligned in the same direction, but that the bond dipoles can point randomly in either the "up" or "down" direction, as shown in Figure 11-4a. What is the entropy of this system? What would the entropy of the system be if all of the bond dipoles were oriented the same way, as in Figure 11-4b?
Solution. To apply the Boltzmann entropy equation, we must know the number of particles in the system, and the number of arrangements that each particle may have. For one mole of HCl, there are Avogadro's Number of molecules. Each molecule behaves as a single particle, because the atoms are bonded together. The number of particles is therefore 6.02 * 1023. The number of arrangements available to each particle is 2--the bond dipole can be either "up" or "down". The entropy of the HCl crystal is
This is a much larger entropy than we found for the pair of dice, because the number of particles is huge. The uncertainty in the location of each particle is not severe, because it must be in one arrangement or the other. However, the uncertainty goes up dramatically when Avogadro's number of particles, each with the 2-state choice, is involved.
If instead of one mole of HCl, we consider 3 moles, the number of particles triples, as does the entropy (to 17.28 J/K). However, if we instead triple the number of arrangements possible for each HCl molecule to 6 (now the bond dipole can point "up", "down", "left", "right", and "in" or "out" of the page), the entropy does NOT triple:
The change in entropy when the number of arrangements is tripled from 2 to 6 is 9.13 J/K. When the number of arrangements for an HCl molecule is again tripled, to 18, entropy again changes by 9.13 J/K. This stresses that the more disordered a system is initially, the less difference an increase in freedom of orientation makes.
From the Boltzmann interpretation of entropy, we extract more guidelines:
If each molecule of a sample of gas is provided a larger volume to explore, then the number of possible arrangements, L, of the molecule is increased. Equation 11-3-5 predicts that the entropy must also increase. This is the basis for Guideline 4. This guideline is formulated in terms of gases simply because liquids and solids undergo negligible volume changes in most processes of interest to us. An increase in temperature increases the random thermal motions of molecules, effectively increasing the number of arrangements available to them. This too increases S by Equation 11-3-5, leading to Guideline 5.
11-4 Entropy Changes in Chemical Reactions. We now turn to the problem of entropy changes in chemical reactions. To calculate these will require that we have available quantitative data on the entropy of 1 mole of each pure substance participating in the reaction. Such data are obtained experimentally by judicious combination of the Boltzmann and Clausius interpretations of entropy. Before looking at things quantitatively, though, it is best to develop a qualitative feel for entropy changes in processes. That is where we begin.
Qualitative Considerations. Is it possible to predict simply by examining the equation for a physical or chemical process whether or not the process is spontaneous from left to right? The answer: sometimes. In order to make such a prediction, it is necessary to demonstrate that DSuniv for the process is greater than zero. This in turn requires consideration of both DSsys and DSsurr:
If we can demonstrate that DSuniv > 0 for a process of interest, we may conclude that the process will spontaneously occur. If instead we demonstrate that DSuniv < 0 for the process, we may conclude that the process is not spontaneous, but that the reverse process is. It is important to realize that DSuniv may be positive even when either D Ssys or DSsurr (but not both) is < 0. Only the total of the two contributions matters. Thus entropy may decrease in the system by, say, 100 J/K, as long as it increases in the surroundings by more than 100 J/K, so that DSuniv is > 0. Similarly, S may decrease in the surroundings, as long as it increases more in the system. We have already had quite a lot to say about the entropy of systems consisting of single pure substances, Ssys, and the factors affecting it. We must now extend these ideas to processes in which more than one pure substance may be involved. Several guidelines are useful:
Example 11-5. Predict the sign of DSsys for each process, and explain the prediction.
Solution. In the process, CH4(g) ---> C(g) + 4H(g), bonds are broken and no bonds are formed to replace them. Disorder will surely increase in this process, and DSsys > 0. Note that both the number of independent particles and the number of moles of gas also increase, reinforcing our prediction. Because bonds are broken in this process, we also predict DH > 0; bond breaking is always endothermic. Because this process involves atomization of 1 mole of methane, DH for the process is the enthalpy of atomization of methane.
In the process H2O(g) ---> H2O(l), a highly disordered gas is converted to a relatively organized liquid. Disorder decreases, and DSsys < 0. Based on our work in Chapter 9, we expect condensation of vapor to be exothermic: DH < 0. Energy leaves the system and enters the surroundings in this process.
From the two processes considered in Example 11-5, we can develop an important and useful generalization. For a process in which the bonds of a substance are ruptured, both DS and DH are > 0. When bonds are broken, the system climbs out of a potential energy well. This requires that molecular forces be overcome, a process that is always endothermic: DH > 0. When forces are overcome, increased particle freedom inevitably results: DS > 0.
In the language of Chapter 6,
The process of escaping a potential well is shown in Figure 11-5.
When methane is atomized--that is, deaggregated--covalent bonds are ruptured, and the five atoms are free to move independently; when water condenses (aggregates), bonds are formed as intermolecular forces operate in the liquid phase. The freedom of motion of the molecules is more restricted, and their potential energy is lower than in the gas phase.
Example 11-6. Predict qualitatively the sign of DSsys and DHsys for the reaction,
Solution. Here we have little help. This can not be viewed as a process that involves only the breaking or making of bonds. Although F-F and H-H bonds must be broken to form fluorine and hydrogen atoms, they are replaced with H-F bonds in the products. The moles of gas stays the same in the process (2 moles in the reactants replaced by 2 moles in the products), as does the number of free particles. It is reasonable in such a case to predict that the entropy change will be small: DSsys approximately 0. However, the enthalpy change depends on the relative bond strengths. We cannot qualitatively predict this without considerable experience. We conclude:
Example 11-6 conveys another important lesson. In most chemical reactions, bonds in the reactants are broken; but bonds are formed in the products. In the majority of cases, the number of chemical bonds in reactants and products is the same, because the same atoms, with the same valences, are present. In such processes, the vast majority, it is not possible to use equation 11-4-2 to qualitatively assess the entropy and enthalpy changes. In most chemical reactions, the atoms proceed from the potential well of the reactants to the potential well of the products, as shown in Figure 11-6. Usually the same number of chemical bonds is involved in both states, and a simple qualitative prediction of DS and DH based on 11-4-2 is not possible. Instead, the overall enthalpy change depends upon which of the two potential wells is deeper; and the overall entropy change depends on whether the reactants or the products are more disordered. The process shown in the figure is endothermic; the sign of its entropy change is not discernible from the potential well diagram.
Example 11-7. For each reaction, determine the change in the number of covalent bonds when reactants are converted to products.
Solution. In each case, the number of bonds in each reactant or product molecule can be counted from a correct Lewis structure of the molecule. You should verify each bond number given below, and should review Chapter 3 if you have difficulty.
| CH4(g) + | 2 O2(g) ---> | CO2(g) + | 2 H2O(l) | |
| #bonds | 4 | 4 | 4 | 4 |
| Change in number of bonds = 0 | ||||
| CO(g) + | H2O(g) ---> | CO2(g) + | H2(g) | |
| #bonds | 3 | 2 | 4 | 1 |
| Change in number of bonds = 0 | ||||
| 2 SO2(g) + | O2(g) ---> | 2 SO3(g) | ||
| #bonds | 6 | 2 | 8 | |
| Change in number of bonds = 0 | ||||
| 4 NH3(g) + | 5 O2(g) ---> | 4 NO(g) + | 6 H2O(g) | |
| #bonds | 12 | 10 | 8 | 12 |
| Change in number of bonds = -2 (A more sophisticated treatment of bonding than used in Chapter 3 indicates that NO actually has a bond order of 2.5. Allowing for this, Change in number of bonds = 0). | ||||
| C6H12O6(s) + | 6 O2(g) ---> | 6 CO2(g) + | 6 H2O(l) | |
| #bonds | 24 | 12 | 24 | 12 |
| Change in number of bonds = 0. |
We must now address DSsurr. This turns out to be a fairly simple matter. We offer Guideline 7 as a means of assessing DSsurr:
Heat entering the surroundings increases the random motion of molecules; disorder increases. Heat leaving the surroundings decreases the random thermal motion of molecules; disorder decreases. We can make this guideline even more useful. For processes carried out at constant pressure, q = DHsys. The heat entering the surroundings is thus -DHsys! If we can make a qualitative assessment of DHsys, we can easily make an assessment of DSsurr.
Example 11-8. Assess the spontaneity of each process.
Solution. We qualitatively assess DSsys and DSsurr for each process and attempt to draw a conclusion about the sign of DSuniv.
In this process, gas is produced from reactants in condensed phases. Thus DSsys > 0. In addition, the reaction is exothermic, meaning that heat passes into the surroundings. Thus DSsurr > 0. We conclude that DSuniv must be greater than zero; Zn(s) spontaneously reacts with aqueous HCl to give the indicated products.
In this process, two solids are converted to a third solid, 2 moles of dissolved gas, and 10 moles of liquid water. Thus water molecules previously confined to the crystal lattice of a reactant have enhanced freedom in the products, and two additional water molecules have formed. Ammonia (NH3) that is confined in an ionic lattice in the reactants is free to explore the volume of the aqueous solution in the products. It is reasonable to predict that entropy increases: DSsys > 0. However, the reaction is endothermic, meaning that heat leaves the surroundings. Thus DSsurr < 0. We have no information about relative magnitudes, and can say only that if DSsys > - DSsurr, reaction will be spontaneous.
Three moles of gaseous reactants are consumed in this process. We confidently predict DSsys < 0. However, reaction is strongly exothermic, so heat enters the surroundings, leading to DSsurr > 0. Again, without quantitative information, we cannot make a firm prediction. We can say only that if DSsurr > -DSsys, the process will be spontaneous.
Equation 11-1-1 reveals in general terms whether or not processes of particular types are possible. Figure 11-7 pictures two reactions being carried out in reaction chambers at constant pressure. In 11-7a, the reaction, R ---> P, is exothermic; in 11-7b, the process is endothermic. Can equation 11-1-1 be used to formulate any general statements about the entropy change of the system for these two situations? Consider first the exothermic process in Figure 11-7a. For an exothermic process, DHsys is negative, so heat flows from system to surroundings. Because heat enters the surroundings, DSsurr > 0. Thus
For the endothermic process in Figure 11-7b, DHsys is positive, so heat flows from surroundings to system during the reaction. Because heat leaves the surroundings, DSsurr < 0. It follows that
An endothermic process can be spontaneous only if DSsys > 0.
The Third Law of Thermodynamics. The Boltzmann equation tells us that the entropy of a system is zero when there is only one possible arrangement. This is a situation of perfect order. Under what conditions does one mole of a substance have perfect order? The disorder in a sample of substance is associated with the uncertainty in the positions of the particles--molecules, atoms, or ions--of which the substance is composed. This uncertainty is smallest when the substance is in the solid phase, in which each particle is located at a specific position in a lattice that has long-range order. So we will suppose that we can create from 1 mole of the substance a perfect crystal, one with no lattice imperfections at all. In practice, this is impossible. However, there is nothing to prevent us from imagining such a crystal. The disorder in such a crystal is small because each particle occupies a well-defined site within it. However, the particles are in constant motion (KE = 3kT/2), jiggling against each other, which introduces some uncertainty--motional disorder--in their positions. But we can reduce this to a minimum by reducing T to 0 K. Again, this is not possible in practice, but can be imagined. We have, then, a perfect crystal of the substance at 0 K. The Third Law of Thermodynamics is a statement about this crystal:
Based on the third law, it has proven possible to measure experimentally the total amount of disorder in one mole of a pure substance in its standard state at 298 K. It is not necessary for you to worry at this stage of your education about how this is done. This total entropy (disorder) of 1 mole of a substance in its standard state at 298 K is called its standard, or absolute, entropy. It is symbolized So. This quantity has been measured and tabulated for many substances, and is shown for a few of them in Table 11-1. Appendix G contains a more extensive tabulation.
| Substance | So (J/mole-K) at 298 K |
|---|---|
| Fe(s) | 27.28 |
| Fe2O3(s) | 87.40 |
| C(s) | 5.74 |
| Mg(s) | 23.68 |
| MgO(s) | 26.94 |
| H2O(l) | 69.91 |
| C2H6O(l) | 160.7 |
| H2O(g) | 188.83 |
| N2(g) | 191.61 |
| H2(g) | 130.68 |
| CO2(g) | 213.74 |
Example 11-9. Make sense of the entropy data in Table 11-1.
Solution. There are a number of important statements to be made about the absolute entropy data in the table. First, in contrast to standard enthalpies of formation in Table 9-2,
The absolute entropy of a substance is the actual (not relative) disorder content of 1 mole of the substance at 298 K. It is possible to know the actual disorder content because of the recognition that a perfect crystal at 0 K has no disorder. There is thus a natural zero, or reference point, for entropy. In contrast, there is no way to know the actual enthalpy content of a substance because there are so many contributions to enthalpy (energy) that cannot be absolutely measured. We can measure changes in enthalpy, but not absolute values. Thus, for convenience, we reference enthalpies of compounds to those of the elements in their standard states, which are arbitrarily taken as zero. This is analogous to arbitrarily defining gravitational potential energy equal to zero at the earth's surface. It is a zero-point of convenience, but not of reality.
Second, again in contrast to the case for standard enthalpies of formation (Table 9-2),
This is easy to understand: it is not possible to have less disorder than is present in a perfect crystal at absolute zero, where the disorder is zero (order is perfect).
Third, it is clear from the data that the absolute entropies of solids are small and those of gases are large. The two liquids in the table have entropies that are intermediate in value.
This is primarily due to the increasing uncertainty in molecular location in progressing from the solid to the gas phase.
Now focus on the solids. Some of the data are puzzling. Although the oxide of iron, Fe2O3, has larger absolute entropy than Fe(s), the oxide of magnesium has lower absolute entropy than Mg(s). This is a result of the lattice structures of the solids and is not readily explained at this point; we will not concern ourselves with it. However, we note also that So for solid carbon is very much smaller than that for solid magnesium. The reason for this is that atoms in the carbon lattice are bonded covalently. The bonding electrons are restricted to locations between pairs of carbon atoms, and make only a small contribution to the entropy of the solid. In contrast, magnesium is a typical metal. Its valence electrons are free to move over the entirety of the crystal, making a substantial contribution to the absolute entropy.
An examination of the two liquids in the table shows that the absolute entropy for ethyl alcohol is substantially larger than that for water. In fact, the value for ethyl alcohol is not much less than that for gases. This is attributed to the molecular complexity of ethyl alcohol, which has 9 atoms per molecule, compared with 3 atoms per molecule of water.
This generalization is reinforced when we examine the gases. Of the gases listed, the least complex, H2, has the smallest So, and the most complex, CO2, has the largest value. But what about H2 and N2, which are both diatomic yet differ substantially in So? Here the difference in complexity originates not in the number of atoms per molecule, but in the number of electrons per atom. Generally, it is found that as the number of electrons per molecule increases, so does So. The number of electrons increases as molar mass increases, leading to our last generalization.
Quantitative Entropy Changes in Chemical Reactions. If we know the absolute entropy for each substance in a chemical reaction, it is easy to calculate the entropy change, DSoR, for the reaction. We find the sum of the absolute entropies of the products and subtract from it the sum of the absolute entropies of the reactants. Of course, if a reaction involves more than one mole of a substance, we must multiply its absolute entropy by this number of moles. The entropy change so obtained is automatically a standard entropy change because the absolute entropies are standard values. That is, the absolute entropy is the entropy of 1 mole of a substance in its standard state at 298 K. In equation form,
Example 11-10. Predict the sign of DSoR for the reaction of Mg(s) with CO2(g), then calculate its value.
Solution. Our prediction is that the entropy of the solid products will be lower than that of the reactants, one of which is a gas with presumably large So. Thus DSoR is expected to be < 0. Now we calculate it, using data from Table 11-1:
The system becomes more ordered as a result of reaction, a result that we qualitatively predicted.
Based on the calculated value of DSoR, would we expect the reaction to spontaneously occur? The initial response might be "no, because the disorder in the system decreases." However, we can test this prediction by trying the reaction, using the setup in Figure 11-8. A block of dry ice (solid CO2) is cut in half. A cup-shaped depression is carved out in the center of the bottom half and filled with magnesium turnings. The magnesium is then heated with a propane torch until it is hot enough to start to burn. At this point, the top half of the dry ice block is replaced, covering the cup. The reaction begins slowly, but then escalates until the dry ice block glows white from the light emitted by the magnesium as it reacts violently with the CO2 vapor produced in the cup, and a white smoke of MgO spews forth from the crack between blocks. After watching this spectacular display, we conclude that the reaction is indeed spontaneous, despite the prediction based on the dramatically negative value of DSoR for the reaction.
The problem with our prediction is that it ignores the entropy change of the surroundings. A Second-Law prediction of spontaneity (or lack thereof) must be based on DS for the universe, which includes not only DSsys, but also DSsurr. To rationalize the spontaneity of the Mg-CO2 reaction, we must calculate DSsurr in addition to DSsys. We can do this with the help of one more guideline:
But the heat put into the surroundings is the negative of the heat of reaction, qR. Thus
The reaction is done at constant pressure, because it is open to the atmosphere. Thus qR = DHR. Finally, then
The standard enthalpy change of reaction can be calculated using standard enthalpies of formation for CO2 and MgO from Appendix G. The other two substances are elements in their standard states, and therefore have DHfo = 0.
When the entropy change in its entirety is considered, the reaction is hugely spontaneous.
Standard Entropy of Atomization. The standard entropy of atomization of a substance, DSoatom, is the entropy change that accompanies the conversion of 1 mole of the substance in its standard state to its constituent atoms in the gas phase at a specified temperature, usually 298 K. Several atomization processes and their corresponding standard entropy changes are shown below:
A more complete tabulation is found in Appendix G. We can make several observations about standard entropies of atomization. First, they are entropy changes, because they characterize processes. Thus they must include the symbol D, in contrast to absolute entropies. Second, they are always positive. Bond breaking always increases disorder, because atoms that were previously bound together can now move independently.
The standard entropy (change) for a chemical reaction of any type can be calculated from the standard entropies of atomization of the reactants and products by imagining a Hess's Law cycle in which, first, the reactants are atomized, requiring the input of the total of the standard entropies of atomization of the reactants; and second, the atoms are recombined to form products, with the release of the total of the standard entropies of atomization of the products. We have used this Hess's Law view of a chemical process several times before, and it should be familiar to you now. The overall change in entropy for the reaction, then, is given by 11-4-6:
Standard entropies of atomization provide an alternative to absolute entropies for the determination of entropy changes in chemical reactions.
11-5 The Gibbs Function. Example 11-10 suggests that, at constant temperature and pressure,
The constant pressure restriction is necessary to use DHsys in place of q; the constant temperature restriction is necessary to use DSsurr = qsurr/T, which is an isothermal expression. Equation 11-5-1 is interesting because it expresses the entropy change of the universe solely in terms of system functions! To put both sides of the equation into energy units, we rewrite it as
This suggests that we define a new energy function of the system, usually called the Gibbs Free Energy and symbolized G, such that
A change in G at constant temperature is then given by
and it is clear that DG, the change in a system function, measures TDSuniv. This new function, G, incorporates the influences of both enthalpy and entropy in determining the spontaneity of a process and enables us to reframe the spontaneity criterion of the second law of thermodynamics:
The change in Gibbs Free Energy, DG, reflects a compromise in the effects of DH and DS on the spontaneity of a process. For the atomization of methane, Example 11-5, bonds are broken in converting reactants to products. Because bond breaking is always endothermic, yet leads to increased freedom of motion of the atoms involved, both DH and DS for the process are positive. Reactants are favored by enthalpy; but products are favored by entropy. Whether the process is spontaneous or not is determined by the relative magnitudes of DH and TDS in equation 11-5-3. For the atomization of methane at 298 K, the enthalpy "wins"; the process is not spontaneous from left to right.
Example 11-11. Apply the new criterion of spontaneity to the Mg-CO2 reaction of Example 11-10.
Solution. We use DG = DH - TDS at 298 K. The enthalpy and entropy changes of reaction have already been calculated in the previous example.
The reaction is predicted to be spontaneous. Note that some care is required to express the changes in enthalpy and entropy in the same units.
Standard Free Energy of Formation, DGfo. Application of the free energy criterion to chemical reactions is easy. One approach is to calculate DHRo and DSRo from tabulated enthalpies of formation and absolute entropies, then calculate DGRo for the reaction using equation 11-5-3. A second approach is to use so-called standard free energies of formation, which have been tabulated by chemists for many important substances. The standard free energy (change) of formation of a substance is the free energy change that occurs when 1 mole of the substance is made from its elements in their standard states. The definition is analogous to that of standard enthalpy (change) of formation.
For calcium carbonate, the standard free energy of formation is the free energy change for the following reaction:
Table 11-2 gives standard free energies of formation at 298 K for calcium carbonate and a number of other common substances. A more extensive tablulation is in Appendix G. For an element in its standard state, the standard free energy of formation is zero, because there is no change in either enthalpy or entropy when 1 mole of an element is formed from itself. If we apply Hess's Law to the free energy change in the same way that we did for enthalpy in Chapter 9, we arrive at equation 11-5-4:
| Substance | DGfo |
|---|---|
| CaCO3 | -1128.76 |
| CaO | -604.17 |
| CH4(g) | -50.8 |
| C2H2(g) | 209.2 |
| C2H4(g) | 68.11 |
| C6H12O6(s) | -910.4 |
| CO2(g) | -394.4 |
| H2O(l) | -236.81 |
| MgO(s) | -569.6 |
(11-5-4): DGRo = S[DGfo(prod)] - S[DGfo (react)]
We apply this to an important reaction.
Example 11-12. Is the photosynthesis reaction spontaneous at 25 oC?
Solution. To answer the question, we require standard free energies of formation for all participants in the reaction. These are obtained from Appendix G.
| Substance | DGfo, kJ/mole |
|---|---|
| CO2(g) | -394.4 |
| H2O(l) | -236.8 |
| C6H12O6(s) | -910.4 |
| O2(g) | 0 |
The reaction is predicted to be non-spontaneous (impossible) at ordinary temperature, yet it occurs on a tremendous scale all around us. An interesting question: why does photosynthesis occur?
Standard Free Energy of Atomization. The standard free energy (change) of atomization of a substance, DGoatom, is the free energy change accompanying the conversion of 1 mole of the substance in its standard state to atoms in the gas phase. The standard free energy of atomization is related to the standard enthalpy and entropy changes of atomization via equation 11-5-5, which is just 11-5-3 applied to the process of atomization.
Tabulated values of DGoatom at T = 298 K are given in Appendix G. Equation 11-5-6 shows how they can be used, according to the usual Hess's Law cycle in which reactants are converted to products via the gas phase atoms, to calculate the overall standard free energy change for any chemical reaction of interest.
This equation can be used as an alternative to equation 11-5-4 involving standard free energies of formation.
Because both the standard enthalpy and entropy of atomization are invariably positive numbers, it is clear from 11-5-5 that the standard free energy of atomization can be positive or negative, depending on the relative magnitudes of the two terms. It turns out that the enthalpy term dominates at 298 K for most substances, so most of the values in Appendix G are > 0. At high temperature, the second term becomes dominant, so that many standard free energies of atomization become negative.
The Interpretation of DG as "Free" Energy. For a process carried out at constant temperature and pressure, the value of DG provides a measure of the portion of the total energy produced by the process that is "free" to be used for doing work of some kind. This is the origin of the name, free energy. To clarify this, we consider the reaction of hydrogen and oxygen to form water at 298 K.
By now, we are aware that this process is exothermic, and can use standard enthalpies of formation or atomization to show that DHRo = -572 kJ. It is reasonable to wonder whether all of this released energy can be harnessed to do work; for example, to run an engine. In fact, only a portion of this total released energy can be so used, for reasons that we now consider.
A qualitative assessment of the entropy change for the reaction reveals that DSsys < 0. Using absolute entropies, it is a simple matter to show that DSRo = -326 J/K. For the reaction to be spontaneous, it is necessary that DSuniv be positive; thus the entropy of the surroundings must increase by a minimum of 326 J/K to guarantee spontaneity. At 298 K, this increase in entropy of the surroundings requires that at least TDSsurr = 97.1 kJ of heat be provided to the surroundings during the reaction. This heat must be drawn from the total energy produced by the reaction. As a consequence, only 475 kJ of the total 572 kJ is actually available to do work. The remainder must be paid to the surroundings as heat to guarantee an overall entropy increase.
In contrast, for an exothermic reaction having DSsys > 0, it is not necessary for any energy to be transferred to the surroundings to guarantee that DSuniv > 0. For such a process, not only is the whole of DH free to do work; but the quantity T DSsys is also available. Work in excess of the magnitude of DH is obtainable from such a process.
Figure 11-9 pictures the "free energy" available from processes having varying signs and magnitudes of DHsys and DSsys.
The Temperature Dependence of DG. Experiment shows that neither DH nor DS for chemical reactions is particularly sensitive to temperature. Thus a reaction run at 400 K is expected to have nearly the same standard enthalpy and entropy changes as the standard values at 298 K. Both DHRo and DSRo can therefore be treated as constants within a fairly large range of temperature centered on 298 K. Since DGo = DHo - TDSo, it follows that a plot of DGRo against T should be linear with
Example 11-13. Consider the simple physical process
for which DHRo > 0 and DSRo > 0. Predict the general form of a plot of DGRo versus T for the process, and discuss significant aspects of the plot.
Solution. Since the standard enthalpy and entropy for the process are both positive, the plot should have a positive intercept and a negative slope, as shown in Figure 11-10. We make the following observations from the plot:
From the plot, we form a conclusion that we have already drawn tentatively in Chapter 10. We add it to the list of entropy guidelines.
We have previously stated that the atomization of methane is non-spontaneous at 298 K because the entropy increase is not sufficient to offset the large enthalpy increase that accompanies the rupture of the carbon-hydrogen covalent bonds.
Guideline 9 suggests, however, than an increase in temperature should favor atomization. Indeed, at very high temperature, such that TDS is more positive than DH, atomization becomes spontaneous. This is a manifestation of the general statement that we developed in Chapter 6: Increased kinetic energy (temperature) favors deaggregation.
Example 11-14. The Thermodynamics of a Rubber Band. We carry out a simple experiment with a rubber band, illustrated in Figure 11-11, which produces a surprising result. A rubber band is stretched and allowed to come to air temperature. It is then allowed to spontaneously contract. Its temperature is measured immediately after the process and is found to be lower than the initial temperature; i.e., contraction of the rubber band is endothermic. Why is contraction endothermic? Why does the rubber band spontaneously contract?
Solution. We apply the First Law to the process. First, since the rubber band contracts very quickly (try it!), there is no time for heat flow from the surroundings into the rubber band. Therefore
Second, since the rubber band contracts by itself with no assistance from outside, there is no work done. Thus
The inevitable conclusion from the first law is that
The internal energy of the rubber band does not change when it contracts.
That the temperature of the rubber band decreases on contraction implies that the average kinetic energy of the rubber molecules decreases:
Since DE = DKE + DPE, it follows that DPE > 0. When the rubber band contracts, its potential energy increases in some way. Realize that this conclusion is forced on us by experiment (the observed decrease in T) and application of the First Law. It is possible to reach this conclusion without saying anything about the molecular nature of the rubber band. It is a conclusion arrived at by macroscopic observation. To go further--to understand the manner in which potential energy increases for the rubber band--requires that we switch to the molecular perspective.
Rubber is a polymer, which means that its molecules are like long strings. When the rubber band is stretched, so are the molecules. They align in the direction of stretching and thereby approach each other closely along their entire lengths. The intermolecular forces are strong, and PE is correspondingly low. The alignment of molecules in the stretched state is shown at left in Figure 11-11b. When the rubber band contracts, the molecules relax and become tangled, like a mass of spaghetti (Figure 11-11b, at right). In this state, molecules are in contact only where they cross over each other. Intermolecular forces are weak, and PE is correspondingly high. According to this simple picture of the rubber molecules, intermolecular PE increases when the rubber band contracts. Since there is no time for heat to flow into the rubber band to supply the energy needed to overcome intermolecular forces, the rubber band gets the energy from itself; it converts molecular kinetic energy to potential energy, and cools.
Example 11-15. A device called Hayward's Wheel is shown in Figure 11-12. The spokes of a bicycle wheel are replaced by rubber bands, and the wheel is suspended above the ground as shown, supported by the axle. Initially the wheel is stationary (i.e., not rotating). A bank of heating lamps is placed behind the wheel, close to the rubber bands. When the lamps are turned on, which way will the wheel rotate (clockwise or counterclockwise))?
Solution. The wheel will rotate counterclockwise. See if you can develop a reasonable explanation.
In this chapter, we have discussed both the qualitative and quantitative aspects of entropy and have applied the ideas to chemical and physical reactions. In the course of this development, entropy has emerged as an odd quantity. It is a state function for the system, like energy; yet it is very different from energy. When something happens in the world, whether it is a natural process like a volcanic eruption or a man-driven process like the manufacture of an automobile, energy may be changed in form, but it is not changed in amount: the total energy of the universe is constant (conserved). However, this is not true of entropy. Entropy is created every time something happens! In fact, only events that create entropy are allowed. This gives a preferred direction -- a one-wayness -- to events that has led to the idea that entropy is in some way connected to the unidirectional flow of time. Further, the fact that entropy is continually being created means that it is not a conserved quantity for the universe. Yet we have defined it to be conserved for the system of interest. Therefore, it is not conserved for the surroundings. This is a very odd state function indeed. It is conserved for what we choose to define as the system, but it is not in general conserved outside of that system. The justification for dealing with entropy in this strange way is that it works in both the explanatory and the predictive senses. It is therefore very important to grapple and come to terms with the strangeness of entropy. The interested reader may refer to Appendix F for more on entropy.
Supplement: But What About When DG = 0...? We have seen that when DGo < 0 for a chemical or physical process, the process is spontaneous left to right from a starting point in which all reactants and products are in their standard states. When DGo > 0, the process is spontaneous right to left from a starting point in which all reactants and products are in their standard states. It is logical, then, that when DGo = 0, the process is equally spontaneous in both directions; it is in equilibrium under standard conditions. In this situation, reactants and products exist together indefinitely, each in its standard state, because the driving force for conversion of reactants to products is balanced by the driving force for conversion of products to reactants. In Chapter 10, we learned that this situation is dynamic. At equilibrium, both conversions continue to take place at the molecular level. They take place at equal rates, however, so that from the macroscopic perspective, amounts of reactants and products do not change.
The conversion of liquid water to water vapor is a process with which everyone has some familiarity, so we will use this process to explore the connection between equilibrium and DGo = 0. The process is shown below:
We have used the double arrow to indicate that the process establishes dynamic equilibrium in a closed container. Suppose we are interested in determining the temperature at which liquid water is in equilibrium with water vapor at 1 atm pressure (i.e., the temperature at which the vapor pressure of water is 1 atm). We can estimate this temperature by taking advantage of the fact that DGo = 0 at this temperature. Using data from Appendix G, it can be shown that DHRo = 44.05 kJ and DSRo = 118.7 J/K for this process. At any particular temperature, it is true that DGRo = DHRo - TDSRo. Setting DGRo = 0 and solving for T gives
Thus the temperature at which liquid water exists in equilibrium with water vapor at 1 atm pressure is 370 K. At this temperature, the conversion of liquid to vapor and vapor to liquid are equally spontaneous. We recognize that 370 K is the normal boiling point of water (actually, we do not obtain exactly 373 K because we have used values of the standard enthalpy and entropy for the process that are valid at 298 K. The values are somewhat different at 373 K). For a pure substance, the temperature at which DGo for conversion of liquid to vapor = 0 is the normal boiling point.
Chemical reactions carried out in closed systems also come to dynamic equilibrium, in which the forward reaction rate (conversion of reactants to products) and the reverse reaction rate (conversion of products to reactants) are equal. For a chemical reaction, DGRo = 0 implies that reactants and products exist in equilibrium with each substance in its standard state. The temperature at which this situation exists can be estimated using the same approach used for phase equilibrium above. For example, reaction 11-S-2 has DHRo = 57.2 kJ and DSRo = 175.7 J/K. Note that these values are consistent with bond breaking.
DGRo = 0 at a temperature such that T = DHRo/DSRo, or 326 K. At this temperature, we can have N2O4 and NO2 in the same container, each with a partial pressure of 1 bar, with forward and reverse reaction tendencies exactly the same. In other words, the reaction is at equilibrium.
A very interesting aspect of equation 11-S-1 deserves mention. Unless DHRo and DSRo have the same sign (both negative or both positive), equation 11-S-1 yields a negative Kelvin temperature! The physical significance of this is that processes for which DHRo and DSRo have opposite signs can not have DGRo = 0 at any temperature. They are either spontaneous under standard conditions at all temperatures (exothermic with positive entropy change) or non-spontaneous under standard conditions at all temperatures (endothermic with negative entropy change).
Our study of phase equilibrium in Chapter 10 revealed that two phases of a pure substance can be in equilibrium over a fairly wide range of temperature and pressure conditions. At 298 K, liquid water is in equilibrium with water vapor at 23.8 torr (0.0313 atm) in a closed container.
This process is equally spontaneous in both directions; consequently it must be true that DG = 0. Because the pressure of water vapor is not 1 atm, however, the water vapor is not in its standard state, and DG is not a standard value. It is written without the superscript o to indicate this. The standard free energy change for process 11-S-3 can be calculated in the usual way: DGRo = DHRo - TDSRo = 8.7 kJ. This tells us that the process in which water vapor at 1 atm is formed from liquid water at 25 oC is nonspontaneous; steam at 1 atm pressure spontaneously condenses at 25 oC. Similarly, reaction 11-S-2 establishes equilibrium at temperatures other than 326 K. In general, the pressures of NO2 and N2O4 are non-standard values, though, unless T = 326 K. For an equilibrium situation at all temperatures other than this, DGR = 0, but DGRo not equal to 0.
We will explore the connection of DG to equilibrium in more detail in Chapter 12.
Applications| Substance | DHfo, kJ/mole | So, J/mole-K |
|---|---|---|
| CO2(g) | -395.5 | 213.74 |
| H2O(l) | -285.83 | 69.91 |
| C6H12O6(s) | -1273 | 212.1 |
| O2(g) | 205.14 |
| Substance | DHfo, kJ/mole | DSo, J/mole-K |
|---|---|---|
| C6H6(l) | 49.04 | 172.8 |
| C6H6(s) | 39.12 | 137.7 |