Major Concept Area: Equilibrium. The concept of dynamic equilibrium is an extremely important one in chemistry. In this chapter, we introduce the second major contributing factor in the establishment of equilibrium--entropy. We then apply energy and entropy ideas to the physical equilibria established among the phases of one or more pure substances under varying conditions of temperature and pressure. The concept of dynamic equilibrium is then extended to a variety of situations involving a balance of opposing rates across a boundary between two phases. In all cases we will find that the rate of passage from one phase to another depends on the surface area of contact between the two phases, and the concentration of substance in the phase of origin. This idea will help us to understand equilibria involving 2 (or 3) phases of a single pure substance. Such equilibria occur all around us in nature and are important in weather and the deterioration of infrastructure. We will then apply the idea to equilibria in which a substance distributes itself between 2 other substances that are in contact with each other; and to equilibria involving the dissolution of molecular and ionic solids and gases in liquids. Solubility of solids and gases is extremely important. For example, fluoridation of water is done to convert the principal structural component of teeth, apatite (Ca5(OH)(PO4)3), to fluoroapatite (Ca5(F)(PO4)3), which is both less soluble and more resistant to chemical attack by food components (particularly acids!) than is apatite. The dissolution of gases such as CO2 and O2 is essential to the aquatic life cycle.
Specific Concepts in this Chapter:
10-1 Qualitative Aspects of Disorder and Entropy . We have seen in previous chapters that of the three phases of matter, the solid has the lowest potential energy. We have also seen that a major driving force for physical and chemical systems is the tendency toward minimum potential energy. This leaves us with a question: why do solids melt to form liquids, and why do these vaporize to gases? If the solid is the lowest-energy phase, and if all systems seek lowest energy, it seems that all substances should be solids at all temperatures and pressures. But this is clearly wrong. To understand the way pure substances behave requires an additional concept.
Consider the system shown at left in Figure 10-1. An ideal gas is confined in bulb A, connected via a closed stopcock to evacuated bulb B. What will happen when the stopcock is opened? Experience tells us that the gas will expand to occupy bulb B until the pressure in both bulbs is uniform. Why does this happen? An initial response might be that by expanding, the gas decreases in potential energy. But closer examination reveals that this is not true. First, if the gas expands at constant temperature (which we can arrange), the average molecular kinetic energy remains the same. Thus DKE = 0. Second, the gas is ideal, so there are no forces between molecules. There is therefore no potential energy involved, and DPE = 0. (If the gas is real rather than ideal, there are weak attractive forces; expansion results in an increase in distance of separation of the molecules and a correponding increase in PE. For a real gas, DPE is small but > 0.) Thus the drive to minimum energy cannot be responsible for the tendency of a gas to expand. The only change on expansion is that each molecule has a larger volume available to it. The location of a molecule is less certain than it was before expansion. Expansion has the effect of increasing the disorder, or randomness, of the gas molecules.
Now consider the reverse process, for which the initial state is the expanded one, at the right of Figure 10-1. What is the chance that the molecules in bulb B will spontaneously move through the stopcock into bulb A to give the state at the left of the figure? Such a process would lead to a decrease in the disorder of the system (as well as a decrease in potential energy for the real gas). But such a process has never been observed. Thus experience tells us that it is very improbable that the reverse process will occur. Expansion is spontaneous, meaning it occurs without outside assistance; compression is not spontaneous. There is a natural direction for this process that leads to an increase in the randomness, or disorder, in the system. The amount of disorder in a system is measured by a quantity called entropy, symbolized S. The higher the disorder in a system, the higher the entropy. The change in entropy in a process is the difference between the entropies of the final and initial states, and is symbolized DS. It seems to be a law of nature that entropy increases (DS > 0) in any spontaneous process.
A valid question at this point is this: "Why is there a tendency toward increased disorder?" A simple macroscopic example, based on the puzzle toy called Rubik's cube, will shed some light on this question. The toy consist of a cube with each face a different color. Each face can be independently rotated in either direction (clockwise or counterclockwise); rotation causes scrambling of the colors. We now consider what happens when we begin with the cube in the most ordered state -- each face showing only one color -- and make a few random 90o face rotations. Begin with a 90o clockwise rotation of the top face of the cube. This produces a situation in which four of the faces show 2 colors. If we agree that any configuration of the cube in which 2 or more faces show more than one color is a disordered state, then our single rotation has produced a disordered state. Now give the cube to a blindfolded person. Inform the person that by making a single 90o rotation of one face, it is possible for him/her to return the cube to the ordered state. Tell the person to make a single random move. What is the probability that this move will produce the ordered state? There are 12 possible 90o moves (6 faces, 2 directions per face). Of these, 1 will produce the ordered state, but 11 will produce other disordered states. The probability that the person's move will produce the ordered state is 1 in 12. The probability that a random move will produce another disordered state is 11 times larger than the probability of producing the ordered state. The odds are very unfavorable, even when the cube is only one move away from the ordered state. If the cube were 6 moves away from the ordered state, the odds against restoring the ordered state by a sequence of 6 random moves would be 2,985,983 to 1. Disordered states are more probable than ordered ones. Collections of molecules, like Rubik's cube, tend toward the most probable situation, because molecular motions are completely random. A gas expands because there is a high probability that some of the molecules in bulb A of Figure 10-1 will wander through the stopcock into bulb B. This requires only that some molecules in bulb A be moving toward the right. On the other hand, once the gas has expanded, it can spontaneously contract only if, completely by chance, all of the »1023 molecules in bulb B happen to move toward the stopcock at the same time. The probability of this happening is essentially zero. The drive to maximum entropy is a consequence of the fact that a collection of particles, each moving randomly, is governed by the laws of statistics and probability.
We are now equipped with two natural laws of great power:
We can extend the usefulness of these laws by considering a second simple and familiar process, the transformation of a pure substance from the solid to the vapor phase:
We have indicated that the process is accompanied by increases in enthalpy and disorder. It is common knowledge that for any pure substance, the gas phase is favored at relatively high temperature and the solid phase at relatively low temperature. This simple, familiar fact allows us to conclude that law 2 is more important than law 1 at high T; and 1 is more important than 2 at low T. Although this statement has been based on a single simple process, it is generally true, for all systems. The question posed at the beginning of the section can now be addressed. Although the solid is the phase of minimum energy, the gas is the phase of maximum disorder. At low T, the drive to minimum energy predominates, and the solid is favored. At high T, the drive to maximum S predominates, and the gas is favored.
Example 10-1. Predict the sign of DS for each process:
Solution.
b) A bond is formed between two chlorine atoms. This aggregation restricts the freedom of the atoms, which must now travel together as a molecule. Disorder decreases, meaning
DS < 0.c) A given amount of Ne(g) is allowed to occupy a larger volume. Its disorder increases:
DS > 0.d) The temperature of a given amount of solid carbon is increased. The thermal motions of the atoms will increase in speed and in amplitude, causing more disorder.
DS > 0.10-2 Liquid-Vapor Equilibrium. In Chapter 7 we explored the concept of vapor pressure by means of a thought experiment, pictured in Figures 7-20 and 7-22. Our observation following injection of water into the box was that the pressure rose gradually from zero to 23.8 torr at a temperature of 25oC, where it levelled off and remained constant indefinitely. The pressure resulted from formation of water vapor by evaporation of some of the liquid water. We address two interesting questions: 1) Why does evaporation occur? 2) Why does the pressure rise to 23.8 torr, then stop changing? We take first a macroscopic, then a molecular, view. Consider the process in equation 10-2-1:
The drive to minimum energy favors the liquid phase. However, the drive to increased disorder favors the gas phase. Two natural tendencies try to drive the process in equation 10-2-1 in opposite directions. The result is a compromise in which some water is present as liquid and some as vapor. This is our macroscopic interpretation of the vapor pressure phenomenon. We now seek an understanding at the molecular level.
An important result from our models of the gas, liquid, and solid phases can be stated succinctly as follows: at a given temperature, all molecules have the same average kinetic energy, whether they are present in the solid, liquid, or gas phase. Further, the distribution of kinetic energies follows the Maxwell-Boltzmann distribution, regardless of phase. The general form of the Maxwell-Boltzmann distribution is reproduced in Figure 10-2. In the ensuing discussion, we will apply this plot to the liquid phase of a pure substance. As the curve indicates, at any moment there are some molecules moving very slowly, a large number moving with intermediate (near-average) kinetic energies, and a few molecules with very high kinetic energy. We focus now on the molecules near the surface of the liquid, because it is these that have a chance to escape from the liquid into the space above. Molecules near the surface possess a range of kinetic energies, like those in the bulk. A molecule near the surface and moving toward the surface will escape the liquid if its kinetic energy is sufficient to overcome the attractive forces of nearby molecules in the liquid. In other words, the kinetic energy must be at least equal in magnitude to the depth of the liquid potential well. We will call this minimum required kinetic energy the escape kinetic energy. This quantity is indicated on the abscissa of Figure 10-2. The number of liquid molecules having at least this kinetic energy is proportional to the area under the curve to the right of a vertical line passing through the escape KE.
When liquid water is injected into an evacuated box, it spreads out on the bottom and molecules begin to evaporate at a rate depending on two quantities. First, the number of molecules that evaporate per unit time is proportional to the number having at least the escape kinetic energy, given by the shaded area in Figure 10-2. Second, the evaporation rate is proportional to the number of molecules at the surface -- i.e., to the exposed liquid surface area. In equation form,
(The constant, kE, is called a rate constant. It is a proportionality constant relating the evaporation rate to macroscopic variables on which rate depends. In this case, surface area is the only variable of importance, as long as temperature is constant. The size of the rate constant depends upon the fraction of molecules with KE greater than the escape KE. Because this depends on temperature, kE is temperature-dependent. This is indicated in the equation.) Evaporation leads to a buildup of molecules in the vapor phase over time. These exert pressure on the container walls, which increases directly as the number of molecules. Further, they collide occasionally with the liquid surface, where they may once again be captured by the intermolecular forces of the closely packed molecules at the liquid surface -- in other words, they condense. The rate of condensation is proportional to the exposed surface area of liquid and to the number of molecules in the vapor phase. This number is directly proportional to the pressure exerted by the vapor. The rate of condensation, in equation form, is
As the number of molecules in the vapor phase continues to increase as a result of a steady rate of evaporation, a point is eventually reached at which
Although both evaporation and condensation still occur, they occur at equal rates. There will be no further change in the number of molecules in the vapor phase, hence no further change in pressure. We describe this situation as a state of phase equilibrium. Attainment of phase equilibrium is shown graphically in Figure 10-3. Equation 10-2-5 is used to represent the equilibrium between liquid and gas phases:
The double arrow indicates that the two processes, evaporation and condensation, occur simultaneously and at equal rates. Phase equilibrium is dynamic, rather than static, because although it appears at the macroscopic level that nothing is happening (vapor pressure is constant), there is a lot of action at the molecular level. This can be easily substantiated by starting with a system consisting of H2O in the liquid phase and D2O (deuterium oxide) in the vapor phase. After some time has elapsed, both forms of water will be found (by, for example, mass spectrometry) in both phases, indicating that evaporation and condensation are continuously occurring.
The pressure exerted by the vapor at equilibrium is called the equilibrium vapor pressure of the liquid, symbolized Pvap. Equations 10-2-2 and 10-2-3 can be used to obtain an expression for Pvap in terms of the constants, kE and kC. Since at equilibrium, RE and RC are equal, we write
Solving for Pvap and cancelling the surface area terms in numerator and denominator gives equation 10-2-7.
(10-2-7): Pvap = kE(T)/kC
This equation shows that for a particular liquid at a fixed temperature, Pvap is constant, in agreement with experiment. The vapor pressure of a pure liquid provides the first example of an equilibrium constant, denoted Keq. For a liquid in equilibrium with vapor, Pvap = Keq. Keq is a ratio of rate constants.
Each liquid has a characteristic vapor pressure at a given T whose value depends primarily on kE. The greater the intermolecular forces, the greater the escape KE, and the smaller the value of kE. Small kE means smaller Pvap. Since the constant kC does not depend on the potential well depth, it is almost the same for all liquids. Equilibrium vapor pressures for several common laboratory liquids at 25oC are given in Table 10-1.
| Substance | Pvap (torr) |
|---|---|
| water (H2O) | 24 |
| ethanol (H2O) | 65 |
| chloroform (H2O) | 215 |
| diethylether (H2O) | 545 |
Example 10-2. A sample of liquid of molecular weight 46.0 g/mole and density 1.04 g/mL is injected into a 150-mL closed container and spreads out to provide a surface area of 25 cm2. Over a period of 1.0 minute, the pressure due to vapor rises to an equilibrium value of 45 torr at 297 K. Calculate the average net rate of evaporation of liquid in moles/s, molecules/s, and mL/s.
Solution. Use the ideal gas law to calculate the moles of gas present in the gas phase at the end of the 1 minute period:
This corresponds to nNo = 2.20 * 1020 molecules. The volume of liquid converted to vapor is calculated by multiplying the moles of liquid by the molar mass, then dividing by density to convert to mL:
The average evaporation rates in moles/s, molecules/s, and mL/s can now be calculated by dividing the amount evaporated by the total time required, which is 60 s:
10-3 The Temperature Dependence of Vapor Pressure. When the temperature of a liquid-vapor equilibrium system is increased, the equilibrium vapor pressure increases. There are two reasons for this.
We now attempt to justify these statements at the molecular level. An increase in temperature means that the kinetic energy of the molecules increases in both phases. Molecules at the liquid surface jiggle more vigorously, and a greater number of them have sufficient KE to escape the potential well of the liquid and enter the vapor phase. The increase in the number of molecules in the vapor phase results in an increased vapor pressure. Figure 10-2 shows the effect of increased temperature on the Maxwell-Boltzmann kinetic energy distribution. As the curve shifts right with increasing T, the area under the curve to the right of the escape KE increases exponentially. Pvap therefore increases exponentially with T. Mathematically, the T-dependence of Pvap can be understood in terms of equation 10-2-2. The minimum KE requirement for escape from the liquid is imbedded in the constant kE, which depends markedly on T. kC has almost no temperature dependence because no minimum KE is required for a molecule to fall into a potential well. The temperature dependence of Pvap therefore parallels that of kE , which directly reflects the shift in the Maxwell-Boltzmann distribution.
It can be shown, both experimentally and theoretically, that vapor pressure varies with temperature according to equation 10-3-1:
This is usually written in logarithmic form as the Clausius-Clapeyron equation, equation 10-3-2:
D
Hovap and DSovap are the standard molar enthalpy and entropy of vaporization, respectively. (We will say more about standard entropy changes in Chapter 11.) Equation 10-3-2 suggests that a plot of lnPvap vs. 1/T should yield the enthalpy of vaporization, DHovap, from the slope, and the entropy of vaporization, DSovap, from the intercept. This is a convenient experimental approach to the measurement of these quantities. The form of the Clausius-Clapeyron equation is general in physical science; all molecular processes exhibit temperature dependence of this form.The vapor pressure is plotted against temperature for several common laboratory liquids in Figure 10-4. The magnitude of Pvap for a liquid at a particular temperature and the rate of change of Pvap with temperature depend on the magnitude of the intermolecular forces in the liquid phase. The stronger the forces, the lower Pvap at a given T and the steeper the rise in the curve. Of the liquids plotted in Figure 10-4, water clearly has the largest intermolecular forces, due to strong hydrogen bonding.
Having seen how Pvap depends on T, we consider a problem involving conceptual aspects of l-v phase equilibrium.
Example 10-3. In the system in Figure 10-5a, the liquid and vapor of a pure substance are in equilibrium at some temperature, in a cylinder fitted with a moveable piston. Under these conditions, rates of evaporation and condensation are equal. We carry out two sequential processes that perturb the equilibrium and examine how the system reattains equilibrium.
At time t1, T is instantly increased. The system returns to equilibrium while the piston is held at position A throughout.
What effect will the T increase have on the equilibrium? Increasing T will cause RE to immediately increase (the Maxwell-Boltzmann distribution). However, because kC is independent of temperature, RC will be unaffected by the increase of T. Therefore, immediately following the T increase, RE will exceed RC; the system is no longer at equilibrium. There will be a net flow of molecules from liquid to vapor. As the number of molecules in the vapor phase increases, the vapor pressure increases, causing RC to increase, until RE and RC are once again equal. The system eventually reaches a new equilibrium, with a higher Pvap. It is important to realize that it takes time for the population of molecules in the gas phase to build up. Thus RC adjusts much more slowly to the T change than does RE.
At time t2, the piston is instantly pulled up to position B. The system returns to equilibrium. T is constant throughout.
Moving the piston has no effect on the rate of evaporation, which depends only on temperature. However, the increase in volume of the vapor space causes an instantaneous decrease in the vapor pressure and RC. There is again an imbalance of rates that results in a net flow of molecules from liquid to vapor. As the number of molecules in the vapor builds up, RC increases correspondingly. This change is gradual, but eventually RC again becomes equal to RE. Equilibrium is reestablished with the original values of RE and RC, but with a larger amount of vapor.
Figure 10-5b shows a graphical interpretation of the effects of the two processes on RE and RC. Note that the liquid (RE) is unaffected by movement of the piston and adjusts instantly to a temperature change; the vapor (RC) is affected by both processes, and is slow to adjust after they are carried out. The shape of the Pvap-time curve during adjustment is a typical kinetics curve, or dynamics curve. The slope of the curve is steep at first, far from equilibrium, and gradually decreases as the Pvap tapers smoothly and continuously to its equilibrium value.
Boiling -- The Boiling Point. The boiling point of a liquid is the temperature at which the vapor pressure of the liquid is equal to the pressure above the liquid. The normal boiling point is the temperature at which the vapor pressure is 1 atm, the "normal" value of pressure above a liquid when it stands open to the atmosphere.
When a liquid boils, pockets of vapor form in the liquid and rise to its surface, where the vapor escapes to the atmosphere. The bubbles are not bubbles of air -- they are pockets of the gaseous form of the substance. At temperatures where the vapor pressure of liquid is less than the pressure above the liquid, pockets cannot form because they are immediately collapsed by the greater outside pressure. Only when the vapor pressure becomes equal to the outside pressure do the pockets become self-supporting. Under these conditions, boiling can occur. Figure 10-6 illustrates the relationship of the pressure inside a vapor pocket to the outside pressure.
The temperature at which a liquid boils can be lowered by reducing the pressure above it. Under these conditions the pockets become self-supporting at a lower value of Pvap , hence at a lower temperature. A plot of Pvap versus T such as those in Figure 10-4 shows how the boiling point changes with external pressure. A practical manifestation of this phenomenon is the longer cooking time required at high altitude, where atmospheric pressure is substantially less than 1 atm. The boiling point of water is less than 100oC, so food takes longer to cook than at lower altitude. If you wanted to boil water at room temperature (25oC), to what value would you have to lower the pressure above it?
A thought experiment in which a liquid is boiled under controlled conditions will illustrate the volume-temperature characteristics of the process. We begin with a sample of a pure liquid in a cylinder fitted with a piston, at an initial temperature Ti which is below the normal boiling point, Tb, of the liquid. The outside pressure on the piston is 1 atm. This situation is pictured in Figure 10-7. At Ti, Pvap of the liquid is less than 1 atm so vapor cannot exist in the cylinder. We justify this statement as follows. Suppose there were some vapor above the liquid in the cylinder. It would exert a pressure Pvap on the piston. Because Pvap < 1 atm, the piston would begin to move in. In response, vapor would condense to maintain the equilibrium value of Pvap. This would cause the piston to move further in. This process of vapor condensation and inward movement of the piston would continue until there was no vapor left. Thus in the beginning, there can be no vapor present. We now slowly heat the liquid to its normal boiling point, and boil it. As long as T < Tb, Pvap < 1 atm and the piston will not move. But at the T where Pvap becomes equal to 1 atm, vapor begins to form in the cylinder, and the piston begins to recede. More and more liquid evaporates in order to maintain Pvap at 1 atm until all of the liquid disappears. At this point only vapor will be present in the cylinder. Finally, we continue heating the vapor. The volume of vapor increases with T according to the ideal gas law. A plot of the volume of the system as a function of T for the boiling process is shown in Figure 10-8. Several features of the plot are important:
10-4 Solid-Vapor Equilibrium. Just as liquid and vapor can coexist in equilibrium under certain temperature-pressure conditions, so too can solid and vapor coexist, without any liquid present. This situation can be directly observed by placing some solid iodine (I2) in a closed Erlenmeyer flask. Purple iodine vapor is visible in the space above the solid when the flask is held against a white background. Solid-vapor phase equilibrium is represented as in equation 10-4-1.
The double arrow means that conversion of solid to vapor (sublimation) and vapor to solid (deposition) occur at equal rates, so that there is no macroscopic change in the system.
The temperature dependence of the pressure of vapor in equilibrium with solid is similar to that of the liquid vapor pressure. It is governed by a form of the Clausius-Clapeyron equation in which the enthalpy and entropy of vaporization are replaced by the corresponding quantities for the sublimation process,
DHosub and DSosub:Plots of ln Pvap versus 1/T for both liquid-vapor and solid-vapor situations are given in Figure 10-9. As shown in the figure, the straight line for the solid has a more negative slope and a more positive intercept than the liquid line. This is true because
The increase in potential energy for conversion of solid to vapor must exceed that for conversion of liquid to vapor, and similarly for the increase in disorder, because the solid is the phase of lowest PE and disorder. The inequalities in 10-4-3 require that the two lines cross at some temperature, Tt. At T > Tt, the vapor pressure of the solid exceeds that of the liquid; at T < Tt, the vapor pressure of the solid is less than that of the liquid.
Ultimately we would like to develop a graphical presentation of the system that shows directly the vapor pressure of the system as a function of temperature. To do this, we translate the graph of ln Pvap versus 1/T to one in which Pvap is plotted directly against temperature. Each possible state of the system is represented by a (P,T) point on this plot. To perform the translation, we make use of 3 facts:
Translation of the Clausius-Clapeyron plots to Pvap versus T plots gives the result in Figure 10-10. The curve labelled "l-v" represents all P-T combinations where liquid and vapor can coexist in equilibrium. The curve labelled "s-v" represents all P-T combinations at which solid and vapor can coexist in equilibrium. Because the point at which these two curves intersect lies on both curves simultaneously, all three phases must coexist together here. This is called the triple point, with temperature Tt.
Solid-Liquid Equilibrium. Completion of our graphical presentation of the states of the system requires only the addition of a curve of all P,T combinations at which solid and liquid coexist in equilibrium. One point of this curve -- the triple point -- is already plotted. For most substances, the "s-l" line slopes sharply up from the triple point, with a positive slope, as indicated in Figure 10-11. The steep slope of this line reflects the fact that solid and liquid are both condensed phases, with properties that depend very little on pressure. A big change in pressure results in only a small change in the temperature of the equilibrium system. The intersection of the horizontal dotted line at P = 1 atm in Figure 10-11 with the s-l curve is called the normal melting point of the substance. This is the temperature at which solid and liquid exist at equilibrium under 1 atm pressure. For most substances, the vapor pressure at the triple point is less than 1 atm. It follows that the melting temperature, Tm, is somewhat greater than the triple point temperature, Tt (although the difference is usually slight). The intersection with the l-v curve is the normal boiling point of the substance. As we have seen, this is the temperature at which the vapor pressure of the liquid reaches 1 atm.
10-5 The Phase Diagram for a Pure Substance. The plot in Figure 10-11 is called a phase diagram. Although details of temperature and pressure vary from one substance to another, all pure substances have phase diagrams that resemble the generalized diagram in the figure. A phase diagram is a superposition of l-v, s-v, and l-s curves for a substance, and shows the phase behavior of the substance as a function of temperature and the pressure to which the substance is subjected. It is important to realize that the diagram is strictly applicable only to a sample of substance contained in a closed system in which no other substance is present. A convenient apparatus in which to study the phase behavior of a substance under these restrictive conditions is a cylinder fitted with a piston (e.g., a syringe). Such an apparatus allows the system to be maintained at any desired temperature, and the desired pressure may be exerted on the substance via the piston. In discussing phase diagrams, we assume that the pure substance being studied is enclosed in such an apparatus.
Points that lie on the lines in a phase diagram represent those P-T conditions at which two phases coexist in equilibrium. What is the significance of (P,T) points that lie off the lines? To answer this, we consider the situation in Figure 10-12. As agreed above, the substance is confined in a cylinder/piston apparatus, shown at top. The phase diagram for the substance is shown below. Initially the system is at point A in the diagram. Consistent with this, the cylinder contains the liquid and vapor phases in equilibrium at TA and vapor pressure PA. The external pressure on the piston is also kept at PA, so the piston is stationary. Starting at A, we carry out two processes.
Process 1. Increase the external pressure to PB, while keeping temperature constant at TA. This process is represented by the vertical arrow from A to B in the figure.
Since Pext (equal to PB) now exceeds the internal pressure, PA, the piston moves in. Vapor in the system condenses to maintain the equilibrium vapor pressure, PA. The piston continues to move in until all vapor condenses and the piston rests directly on top of the liquid. When the vapor is completely condensed, the internal pressure rises to PB, and the piston stops. The system is now at point B on the diagram. At this point, only liquid exists in the system. We conclude that only liquid can exist at points between the s-l and l-v curves.
Process 2. Again starting at A, decrease Pext to PC at constant T. This process is represented by the vertical arrow from A to C.
Since Pext is now less than the internal pressure, the piston experiences an imbalance of forces and moves out. With the increased volume of the system, liquid must vaporize in order to maintain the equilibrium vapor pressure PA. The outward movement continues until all liquid in the cylinder has vaporized. At this point, only gas exists in the cylinder. The piston expands a bit further until the gas pressure drops to PC (ideal gas law), then stops. We conclude that at point C and at all points below the l-v and s-v lines, only gas exists in the cylinder. Generalizing, (P,T) points lying off the lines in the phase diagram represent conditions under which only one phase may be present at equilibrium. Regions in which only one phase can exist are indicated by the appropriate symbols in the diagram. Notice that the solid phase is stable at relatively low temperatures and high pressures; the gas phase is stable at relatively high temperatures and low pressures; and the liquid is stable at intermediate T and P.
Example 10-4. Consider the phase diagram in Figure 10-13a. Plot the volume of the system as a function of temperature as the system is taken from point A to point D on the diagram.
Solution. From A to B, the solid is heated. It's temperature increases but volume remains essentially constant, as indicated in Figure 10-13b. When point B is reached, solid begins to melt. T remains constant as long as both phases are present, but volume increases slightly as solid converts to liquid. This is shown as the small vertical segment at TB in Figure 10-13b. When all solid has melted, only liquid is present at point B on the diagram. With continued addition of heat, the temperature of the liquid increases but, as with the solid, there is little change in volume. This is indicated by the horizontal segment between TB and TC in Figure 10-13b. When point C is reached, liquid begins to vaporize. T remains constant while both phases are present, but volume increases dramatically (approximately 1000-fold). This is represented by the long vertical rise at TC in Figure 10-13b. When all liquid has vaporized, the volume of the vapor increases linearly with temperature according to the ideal gas law. This is represented by the segment between TC and TD in Figure 10-13b.
10-6 Le Chatelier's Principle. Le Chatelier's principle is an important tool for predicting the effects of pressure and temperature changes on phase equilibria:
A system in equilibrium subjected to an external stress responds to reduce the effect of the stress.
There are two stresses that are important for phase equilibria. We examine them in terms of the stress principle.
Change in pressure. Suppose we increase the pressure on a system in phase equilibrium, say a liquid-vapor equilibrium, shown in equation 10-6-1.
Le Chatelier's Principle predicts that the equilibrium will shift to offset the effect of the stress. Intuitively, we expect the system to decrease its volume, to retreat from the increased pressure. This will happen in a liquid-vapor equilibrium system if some vapor converts to liquid. The system shifts toward the denser phase.
A molecular interpretation of this situation can be made with the aid of Figure 10-14. A liquid is in equilibrium with vapor at left. The piston is at position A, and the external pressure is equal to the equilibrium vapor pressure of the system. The piston is then suddenly moved down to a new position, B. The instantaneous effect of this stress is to increase the pressure exerted by the gas inside the cylinder. The increase in pressure causes the rate of condensation, RC, to become greater than the rate of evaporation, RE, which depends only on temperature. A net flow of molecules from vapor to liquid results -- i.e., the system shifts toward the denser liquid phase -- to reduce the internal pressure to the equilibrium value. Once this value is reached, RE and RC will again be equal, and the system will be in equilibrium again. The difference between the final and initial equilibrium states is that there is less vapor and more liquid in the final state. The equilibrium in Equation 10-6-1 shifts to the left as a result of the pressure increase.
Change in temperature. Le Chatelier's Principle predicts that increasing the temperature of a system in phase equilibrium will cause it to shift to use heat. Conversely, a temperature decrease will result in a shift that produces heat. For the equilibrium in equation 10-6-1, an increase in T will cause a shift toward the vapor phase (to the right), since conversion of liquid to vapor is endothermic and consumes heat. This will increase the vapor pressure of the liquid, as we already know. As a second example, consider the liquid-solid equilibrium in equation 10-6-2.
For the process as written, DH < 0. An increase in the temperature of the system causes a shift in the endothermic direction; i.e., to the left.
The Phase Diagram for Water. Water, a common substance upon which we depend for life, has unusual phase behavior. If water initially at 25oC is cooled to its freezing point of 0oC, it becomes gradually denser until a temperature of 4oC is reached. At this temperature, water has its maximum density. With further cooling, the density of the liquid decreases again until solid (ice) begins to form. Surprisingly, liquid water near 0oC is denser than solid water. Thus water is unusual, since for most pure substances, the solid phase is denser than the liquid. The low density of ice is a consequence of a very open lattice structure that results from extensive hydrogen bonding between neighboring molecules. Each water molecule in the lattice is engaged in 4 hydrogen bonds, 2 using the O atom, and 1 for each of the H atoms. Breakdown of this hydrogen bonded network upon melting results in a collapse of the open lattice, with an increase in density. What will happen if pressure is applied to a block of ice in equilibrium with liquid at 0oC? According to Le Chatelier, the solid-liquid equilibrium will shift toward the denser phase, in this case the liquid. Application of pressure to ice melts it. In terms of the phase diagram, this can happen only if the l-s equilibrium line has a negative slope, rather than the positive slope of most substances. The phase diagram for water is shown in Figure 10-15. The slope of the s-l line therefore gives information about the relative densities of solid and liquid phases.
There are a number of practical manifestations of the unusual phase diagram for water. Perhaps the most important is that lakes and other natural bodies of water freeze from the top down, because as water cools from 4 to 0oC , it becomes less dense and rises to the surface. The layer of ice that forms on the surface then insulates the water below and prevents the entire lake from freezing. Aquatic life is thus enabled to survive the cold of winter. A second, and costly, consequence of the reversed s-l line is the damage done to roads by the freezing of water in surface cracks. The water freezes, expands, and pushes the pavement out of the way. The result is frost heaves and potholes. The same process is responsible for the erosion of natural rock formations and for the gradual recovery by nature of man-made structures. In the latter case, water freezes in cracks and fissures, opening them up to the growth of vegetation. The root systems of the vegetation cause further breakup of the man-made surface, until eventually nature has taken over. Finally, the phase diagram for water shows why snow and ice melt on roads when we drive on them. Were water a typical substance, tire pressure would serve to compress the solid phase, making winter driving less slippery.
10-7 Phase Equilibrium in Solutions. Many of the ideas presented in the discussion of phase equilibrium of pure substances are also applicable to mixtures of pure substances. In this section we discuss the so-called colligative properties of solutions (homogeneous mixtures) in terms of phase equilibrium.
Solutions and Concentration Units. We begin with definitions of some terms that are used in discussing solutions:
If we add 1.0 g of sugar to 10 mL of water in a beaker, the sugar falls to the bottom of the beaker where it forms a pile. Over the period of a few minutes, the pile becomes smaller until finally no solid sugar remains visible. The sugar has dissolved in the water. The result is a solution of sugar in water. Because water is present in major amount, it is the solvent, and sugar is the solute. The solution is clear and displays all of the usual properties of a liquid. There is no cloudiness (opacity) to indicate the presence of the sugar.
In preparing a solution, it is usually important to specify, as precisely as possible, the amount of solute present per unit amount of solvent or per unit amount of solution. This is called the concentration of the solution. The concentration is an intensive property of the solution, independent of the amount of solution present. In this respect, concentration is similar to density. There are several ways to express concentration, discussed below.
If A is the only solute, then XA = moles A/(moles A + moles solvent)
There are several other ways to express concentration, but those above are most common. We will use molarity most frequently.
Example 10-5. How many mL of 0.26 M H2SO4 are required to react completely with 26.72 mL of 0.18 M NaOH according to
Solution.
Moles NaOH = 26.72 mL NaOH * 0.18 mole NaOH/1000 mL NaOH = 4.810 * 10-3 moles NaOH
From the stoichiometry of the equation,
Suppose that we require 100 mL of a 0.10 M sulfuric acid solution, but all we have available is commercial concentrated sulfuric acid, which is 18 M in H2SO4. We can prepare the required solution by dilution. Dilution is the process of adding a known volume of a concentrated solution of a reagent to water (or another solvent) to give a less concentrated solution. It is a very frequently-used laboratory technique. The key to dilution is that the number of moles of H2SO4 in the required volume of 18 M H2SO4 is the same as the number of moles in 100 mL of 0.1 M H2SO4. The number of moles is the product of solution volume and molarity both before and after dilution. Using subscript "i" for initial quantities and "f" for final quantities,
Here Vi is the volume of concentrated H2SO4, Mi is its molarity, Vf is the volume of dilute H2SO4, and Mf is its molarity. Vi can easily be calculated from the known values of the other 3 quantities. The result is 0.556 mL.
Solutions have some interesting properties called colligative properties. They depend on the nature of the solvent and the amount (but not the identity) of the solute. These properties may be understood in terms of phase equilibrium. We begin with the effect of a solute on the vapor pressure of a liquid solvent.
Raoult's Law. Consider a solution of a non-volatile solute, B, in a volatile solvent, A. A volatile substance has an appreciable vapor pressure at room temperature; a non-volatile substance has essentially zero vapor pressure. How is Pvap of the solvent in the solution related to Pvap of the pure solvent? Raoult answered this question experimentally in 1886 when he showed that equation 10-7-2 is valid for ideal solutions:
PA = the vapor pressure above the solution due to solvent, XA = the mole fraction of solvent, and PAo = the vapor pressure of pure solvent. This equation is called Raoult's Law. We can make several statements about this law.
1) XA < 1 so the vapor pressure of the solution is less than that of pure solvent.
2) No term in the equation involves any chemical or physical property of the solute. It doesn't matter what the solute is,
only that it is non-volatile.
3) The equation is ideal because it holds only for ideal solutions. An ideal solution is one in which intermolecular forces
between A molecules, between B molecules, and between A and B molecules are the same. If these
forces are the same, there is no enthalpy change when solute dissolves in solvent. It is important to realize that Raoult's
Law has very limited applicability. Very few real solutions even approach the ideal, and most are definitely NOT ideal.
These do not obey Raoult's Law. Some solutions are nearly ideal when very dilute; thus Raoult's law approximately describes
the vapor pressures above most dilute solutions.
4) The law is purely statistical. It says that the tendency for solvent molecules to escape the liquid is in proportion to
the relative abundance (mole fraction) of solvent molecules at the liquid surface.
Raoult's Law can be understood physically by considering the liquid surfaces of pure solvent and solution, shown in Figure 10-16. In pure solvent, all surface molecules are solvent. However, in the solution, a fraction of molecules at the surface are solute molecules, which are non-volatile (cannot escape to the vapor phase). These reduce the surface area available for escape of solvent molecules, so that the evaporation rate of solvent is reduced from that in pure solvent. The fraction of surface area occupied by solvent molecules is directly related to the mole fraction of solvent. The evaporation rate, and consequently the vapor pressure, is reduced by a factor equal to the ratio of surface areas, which is the mole fraction of solvent. Raoult's law, equation 10-7-2, follows. (A more detailed development of equation 10-7-2 is found in Appendix E).
Colligative Properties.The extent to which the vapor pressure is lowered depends quantitatively on the amount of solute present. Defining DP = PAo - PA, and substituting for PA using Raoult's Law, gives
The vapor pressure lowering, DP, is the first of four colligative properties. The remaining three are consequences of DP, and may be examined in terms of the phase diagrams for the pure solvent and the solution. These are shown superimposed in Figure 10-17. The l-v line for the solution (dashed line) is lower at all temperatures than that for the pure solvent (solid line). It therefore intersects the P = 1 atm line at a higher temperature. The boiling point of the solution is higher than that of pure solvent by an amount DTb, which is called the boiling point elevation, and is indicated on the diagram. The boiling point elevation is the second colligative property. DTb is defined as Tb(solution) - Tb(solvent), which makes it a positive number. The boiling point elevation is directly proportional to the molality of the solute:
The value of the constant, Kb, called the boiling point elevation constant for the solvent, depends only on properties of the solvent. Thus equation 10-7-4 is valid for a particular solvent no matter what the solute is.
Because the l-v line for the solution is lower, it intersects the s-v line at lower T than does the l-v curve for the pure solvent. The s-l line for the solution (dashed in the figure) therefore lies to the left of that for the pure solvent. Consequently the triple point for the solution is lower than that for the solvent by an amount DTf, indicated on the diagram. Over small ranges of T and P, the s-l lines are linear and roughly parallel. It follows that the freezing point for the solution is lower than that for the solvent by about the same amount, DTf. DTf is called the freezing point depression, and is defined to be a positive number: DTf = Tf - Tf(solution). It is the third colligative property. The freezing point depression is similar to the boiling point elevation in being directly proportional to the solute molality:
Kf is the freezing point depression constant and depends only on properties of the solvent. Table 10-2 gives values of Kb and Kf for several common solvents.
Table 10-2: Molal Boiling Point and Freezing Point Constants
| Solvent | Tb(oC) | Kb | Tf(oC) | Kf |
|---|---|---|---|---|
| acetic acid | 118.2 | 2.93 | 17 | 3.90 |
| chloroform | 61.2 | 3.63 | ||
| naphthalene | 80 | 6.8 | ||
| water | 100 | 0.52 | 0 | 1.86 |
| camphor | 179 | 40 | ||
| carbon tetrachloride | 76.6 | 5.03 | ||
| ethanol | 78.6 | 1.22 | ||
| n-octane | 125.7 | 4.02 |
Unlike the vapor pressure lowering, which is usually small and difficult to measure accurately, DTb and DTf are relatively easily and accurately measured. The freezing point depression used to be applied in the determination of molar masses of unknown substances. However, because molar masses can now be determined much more accurately using mass spectrometry, the freezing point method is seldom used.
There are a number of practical uses of the freezing point depression and boiling point elevation colligative properties. For example, sodium chloride is often added to water for cooking, not only for flavor, but to raise the boiling temperature and accelerate the cooking chemistry. Calcium chloride is spread on icy roads in winter to lower the freezing point of the water. And antifreeze (ethylene glycol) is added to the cooling systems of automobiles to protect against boilover in the summer and coolant freeze in winter.
The fourth colligative property is called osmotic pressure. It can be understood in terms of the apparatus in Figure 10-18, which consists of two arms, separated by a semi-permeable membrane. This is a membrane with pores that allow passage of small molecules (solvent) but block larger molecules (solute). In the left arm of the apparatus is placed a pure solvent, usually water. In the right arm is placed a solution of the same solvent. The dots in the figure represent molecules of solute present in the solution. Solvent molecules can pass through the membrane in either direction. However, their passage from right to left (from solution to pure solvent) is impeded somewhat because solute molecules block their access to some of the pores. More solvent molecules per unit time pass from solvent to solution than in the reverse direction. The result is that the amount of solvent in the right arm increases with time and the solution becomes more dilute. It is possible to prevent the net flow of solvent from left to right by applying pressure to the top of the liquid column in the right arm of the apparatus. The applied pressure increases the rate of passage of solvent from solution to solvent until it is equal to the rate of passage in the other direction. Under these conditions, the system is in equilibrium and the concentration of the solution does not change with time. The pressure that must be applied to stop the net flow of solvent is called the osmotic pressure, symbolized P. Its magnitude is directly proportional to the temperature and the molarity of the solution. The ideal gas constant, R, is the proportionality constant:
Realizing that molarity is moles of solute per unit volume of solution, expressed in liters, we can convert equation 10-7-6 to a form analogous to the ideal gas law:
It may seem odd that the ideal gas constant appears in an equation that describes a property of a liquid solution. However, a non-volatile solute dissolved in a solvent behaves much like a gas, because the solute molecules are far apart and are free to roam over the entire solution volume. This similarity between a gas and a dissolved solute causes osmotic pressure to obey a gas-law type equation. The similarity in fact extends further. The osmotic pressure calculated from 10-7-6 has exactly the same value as the pressure of a sample of gas occupying the same volume at the same temperature. We know that the gas will expand if the pressure exerted on it is less than the pressure that it exerts. In the same way, the solution in Figure 10-18 tends to "expand" (increase in volume) by dilution when in contact with pure solvent unless an external pressure equal to its osmotic (internal) pressure is applied.
A simple calculation using equation 10-7-6 shows that concentrations as low as 10-4 M give readily measureable osmotic pressure. The osmotic pressure is therefore much more sensitive to the presence of solute than the other colligative properties. It can be used to determine the molar masses of huge molecules like proteins, which dissolve in water to give solutions that may be concentrated in terms of weight percent, but are of very small molarity due to the large molar masses (on the order of 10000 m).
Example 10-6. A solution of 0.0720 g of the blood protein, hemoglobin, in 100 mL water has an osmotic pressure of 2.10 torr at 300 K. What is the molar mass of hemoglobin?
Solution. Calculate the concentration in moles hemoglobin per liter using the osmotic pressure. Then equate the moles per liter with the grams per liter to get molar mass.
Since 100 mL of solution contains 0.0720 g, a liter would contain 0.720 g. Thus
You might think about whether freezing point depression or boiling point elevation measurements would be useful in determining this MM.
It may not be clear that osmotic pressure, the tendency for solvent to flow spontaneously from solvent to solution across the barrier, is a consequence of the vapor pressure lowering of the solvent in the solution. Figure 10-19 shows two beakers, one containing pure solvent, the other the same solution that is in the right arm of the apparatus in Figure 10-18. The beakers are placed in a closed container. Solvent begins to evaporate from each beaker in an attempt to establish the equilibrium vapor pressure. However, this pressure is less for the solution than for the pure solvent. As pure solvent evaporates from the left beaker in an attempt to establish the vapor pressure, Povap, it condenses in the right beaker as the solution tries to lower the vapor pressure to its equilibrium value, Pvap. This process has the same result as the process in Figure 10-18: the solution gradually becomes more dilute due to spontaneous "flow" of solvent into it. The origin of the osmotic pressure phenomenon in the solvent vapor pressure lowering is clear.
Mixtures of Volatile Liquids. We continue this chapter with a discussion of the vapor pressure over a mixture of two volatile liquids, A and B. What should the total vapor pressure, PTvap, be over this solution, assuming Raoult's Law to be valid? Each liquid should independently follow Raoult's Law:
The total vapor pressure should then be the sum of the two partial vapor pressures, according to Dalton's partial pressure law:
Plots of Pvap(A), Pvap(B), and PTvap versus XA should be linear, as shown in Figure 10-20. If the intermolecular force between an A and a B molecule is identical to that between two A molecules or two B molecules, the solution is ideal and produces plots like those in Figure 10-20. A mixture of hexane (a chain of 6 carbon atoms with attached hydrogens) and heptane (a chain of 7 carbon atoms with attached hydrogens), which have very similar molecular structures and compatible intermolecular forces, is very nearly ideal. The same is true of a mixture of carbon tetrachloride and carbon tetrabromide. For a pair of ideal liquids, there is essentially no change in the intermolecular forces on mixing. Thus there is no change in potential energy; the enthalpy change accompanying mixing, DHsoln, is zero. The volume change on mixing, DVsoln, is nearly zero as well, because intermolecular separation distances will not change unless there is an increase or decrease in forces of attraction.
Ideal liquid mixtures are rare. It is much more common for two liquids to form a non-ideal mixture, for which the vapor pressure plots are non-linear. Consider, for example, a mixture of acetone and heptane, with the vapor pressure plots in Figure 10-21. Mixtures producing plots like this are said to exhibit positive deviations from Raoult's Law, because the vapor pressures are larger than expected for an ideal mixture. Larger vapor pressures imply that the molecules more readily escape from the liquid phase than they would if the intermolecular forces were ideally compatible. Addition of heptane to acetone disrupts the relatively strong dipole-dipole forces between acetone molecules. The weaker dipole-induced dipole forces between acetone and heptane molecules are insufficient to make up for the disruption. Acetone molecules therefore escape the liquid more easily when surrounded by heptane molecules than when surrounded by other acetone molecules. When intermolecular forces between unlike molecules are weaker than those between like molecules, positive deviations are observed. DHsoln and DVsoln for such mixtures are both greater than zero.
In contrast, mixtures of acetone and chloroform produce the vapor pressure plots in Figure 10-22. In this case, the vapor pressures are smaller than the ideal values, implying that the molecules escape the liquid phase less readily in the mixtures than they would if the intermolecular forces were ideally compatible. Plots like those in Figure 10-22 exhibit negative deviations from Raoult's Law. Intermolecular forces between acetone and chloroform molecules are relatively strong, involving some hydrogen bonding in addition to the expected dipole-dipole contribution. A molecule of chloroform therefore escapes the liquid less readily when surrounded by acetone molecules than when surrounded by other chloroform molecules. When intermolecular forces between unlike molecules are stronger than those between like molecules, negative deviations are observed. DHsoln and DVsoln for such mixtures are both less than zero.
To some extent it is possible to predict whether or not two liquids will show deviations from Raoult's Law based on the compatibility of their molecular structures. Two liquids with very similar, non-polar molecular structures tend to form nearly ideal solutions, with at most only small deviations from linearity. The hexane/heptane and CCl4/CBr4 examples cited above illustrate this. Solutions of liquids with similar polar structures often exhibit negative deviations from Raoult's Law. In contrast, two liquids with quite different structures, one polar and one non-polar, are incompatible, and therefore show positive deviations. When the molecular structures of the liquids are very different, as is the case for H2O (polar) and CCl4 (non-polar), the molecules are so eager to escape one another that they do so even in the liquid phase. Such liquids do not dissolve in one another, and are said to be immiscible (i.e., do not mix). Immiscible liquids form two layers when mixed, with the denser liquid on the bottom.
10-8 Additional Equilibria Involving Two Phases.
In our discussion of the phases of a single pure substance, we found that equilibrium involving two phases is achieved when the rate of passage from phase 1 to phase 2 is balanced by an equal rate of passage from phase 2 to phase 1:
The rate of passage from phase 1 to phase 2 is directly proportional to the concentration of substance in phase 1 and to the surface area, SA, of contact between the two phases. A similar statement holds for the rate of passage from phase 2 to 1:
When these two rates become equal, equilibrium exists, and the following relationship characterizes it:
Keq is an equilibrium constant, and is seen to be the ratio of rate constants for opposing processes. We now consider a number of other two-phase equilibrium situations for which the above considerations apply.
Distribution of a Solute Between Two Immiscible Liquids in Contact. As we have discussed in Chapter 6, when two immiscible liquids such as water and chloroform (CHCl3) are placed in contact, they mutually exclude each other, forming two liquid layers. The less dense liquid floats on the denser one, and a boundary surface is clearly visible. It is interesting to think about what will happen if a third substance, solid or liquid, is added to the system consisting of two immiscible liquids, A and B. If the third substance, C, is soluble in one of the liquids (say A) and insoluble in the other (B), it will dissolve exclusively in the A layer, as in example 6-14. More interesting is the situation in which C is soluble in both liquids to at least some extent. In this case, we find that C partitions, or distributes, itself between the two liquid layers. We further find that the concentration of the third substance is higher in the liquid with which its intermolecular forces are most compatible. The ratio of the concentration of C in liquid B to that in liquid A is found to attain a constant value Kd, defined as in (10-8-4):
This constant ratio is established no matter how the 3-component mixture is prepared. Three such ways follow:
A and B are first brought into contact. C is then added, and the container is shaken.
C is first dissolved in A. A (containing C) and B are then brought into contact, and the container is shaken.
C is first dissolved in B. A and B (containing C) are then brought into contact, and the container is shaken.
In addition, the ratio is reestablished if the mixture is perturbed in any of the following ways:
We can understand this behavior in terms of the diagram in Figure 10-23, which shows layers of A and B in contact at a boundary, and molecules of C initially dissolved in liquid A. The boundary between the two liquids provides a "doorway" from one to the other through which C molecules, which have reasonable interactions with both A and B molecules, can pass. Thus molecules of C will tend to move across the boundary from liquid A, where they are initially dissolved, into liquid B, in which there are initially no molecules of C. In other words, they begin to distribute themselves between the two liquids. After some time, enough C molecules will have passed into liquid B that the rates of passage of C from A to B and back will be the same. This is a situation of dynamic equilibrium, represented as in (10-8-5a). The equality of rates is shown in (10-8-5b).
The rate of passage from A to B is directly proportional to the concentration of C in liquid A, and to the surface area over which A and B are in contact.
Equating the two rates at equilibrium gives
Once the equilibrium situation is attained, there will be no further change in the concentration of C in the two layers. In other words, the constant ratio in (10-8-8) will be attained. From the magnitude of Kd, we may easily deduce which liquid C prefers. If Kd < 1, C prefers A; if > 1, C prefers B.
In summary, distribution of a substance between two liquid phases is an example of dynamic equilibrium, a situation in which two opposing processes (here, movement of C both ways across the boundary) occur at equal rates.
Example 10-7. A vial contains 1 mL of water and 1 mL of 1,1,2,2-tetrachloroethane, which are immiscible. To the vial is added 0.1 g of acetone, which is soluble in both water and TCE. When the equilibrium distribution of acetone between the two solvents has been achieved, it is found that Kd = [acetone]H2O/[acetone]TCE = 2.8.
To the vial is now added 1 mL of TCE containing 0.05 g of acetone, and the vial is shaken until equilibrium is once again attained.
Solution.
1) At equilibrium, the ratio of the concentration of acetone in water to that in TCE must be 2.8. Because the two solvents are present in equal volumes, the ratio of the moles acetone in water to moles acetone in TCE must also be 2.8. Further, the total moles of acetone present is 0.1 g/58.08 g/mole = 0.00172 moles.
Masses may now be calculated from moles:
2) Molarities may be easily calculated from the moles acetone and the volume of each layer:
3) To address this, we introduce an approach that is generally useful for situations in which a system proceeds from specified concentration conditions to equilibrium. The approach begins with an equation representing the process:
Beneath each substance in the equation we write the moles of that substance that we start with. Thus we put moles of acetone present in each phase prior to addition of 0.05 g of acetone in 1 mL TCE. These are equilibrium values, so we write "original equilibrium" to the left of these amounts:
| AcetoneTCE | <==> | AcetoneH2O | |
| original equilibrium: | 4.53*10-4 moles | 1.27*10-3 moles |
Beneath the "original equilibrium" line of information, we enter the amount of acetone added to the TCE. This is 0.05 g, or 8.60*10-4 moles.
| AcetoneTCE | <==> | AcetoneH2O | |
| original equilibrium: | 4.53*10-4 moles | 1.27*10-3 moles | |
| add: | 8.60*10-4 moles | 0 moles |
The amounts in the start and add lines are now added to give the "initial" values, from which the system will proceed toward a new equilibrium:
| AcetoneTCE | <==> | AcetoneH2O | |
| original equilibrium: | 4.53*10-4 moles | 1.27*10-3 moles | |
| add: | 8.60*10-4 moles | 0 moles | |
| initial: | 13.13*10-4 moles | 1.27*10-3 moles |
It is unlikely that the system is still in equilibrium following addition of extra acetone and TCE. Assuming this, there must be movement of acetone from one solvent to the other in order to restore the equilibrium concentration ratio to 2.8. We will suppose that "x" moles of acetone leaves the TCE layer and enters the water layer as equilibrium is restored. This information is entered in a "change" line:
| AcetoneTCE | <==> | AcetoneH2O | |
| original equilibrium: | 4.53*10-4 moles | 1.27*10-3 moles | |
| add: | 8.60*10-4 moles | 0 moles | |
| initial: | 13.13*10-4 moles | 1.27*10-3 moles | |
| change: | -x moles | x moles |
Note that the change in amount of acetone in TCE is taken as negative, consistent with our assumption that acetone will move from the TCE layer to the water layer. Finally, the amounts present at the new equilibrium are calculated by adding the change to the initial value:
| AcetoneTCE | <==> | AcetoneH2O | |
| original equilibrium: | 4.53*10-4 moles | 1.27*10-3 moles | |
| add: | 8.60*10-4 moles | 0 moles | |
| initial: | 13.13*10-4 moles | 1.27*10-3 moles | |
| change: | -x moles | x moles | |
| new equilibrium: | 1.313*10-3 - x moles | 1.27*10-3 + x moles |
We now have expressions for the moles of acetone in each solvent after the new equilibrium is established, and can convert these to concentrations by dividing by the volumes of the solvents:
Substituting these into the Kd expression produces
Substituting this value into the expressions in the "new equilibrium" line gives
4) Initially, the vial contains 1 mL of water and 1 mL of TCE, with acetone concentrations of 1.27 and 0.453 M, respectively. Then 1 mL of TCE containing 0.05 g acetone is added, and the system returns to equilibrium. At equilibrium, the vial contains 1 mL of water and 2 mL of TCE, with acetone concentrations of 1.50 and 0.538 M, respectively (these values are calculated from known moles acetone in each layer, and known volume of each layer). Thus from time zero to the time at which the addition of acetone in TCE takes place, concentrations are level at 1.27 and 0.453 M in the water and TCE layers respectively. At a time sufficiently long following the addition, concentrations are again level at 1.50 and 0.538 M, respectively. These two set of conditions are shown in Figure 10-24a. It remains for us to determine what happens to concentrations during the time interval between t1, when TCE addition takes place, and t2, by which equilibrium has once again been attained.
We can easily determine the concentrations of acetone in the two layers immediately following TCE addition, but before any redistribution of acetone takes place. First, the concentration of acetone in the water layer is the same as before TCE addition, at 1.27 M. The concentration of acetone in the TCE layer jumps from 0.453 M just before addition to 0.657 M immediately following addition. The latter value is calculated by dividing the total number of moles of acetone in the TCE layer by the volume of the TCE layer after addition, 0.002 L. From the post-addition values, the concentrations in the two layers must smoothly evolve according to typical dynamics curves as the acetone redistributes itself to reattain equilibrium. Thus the concentration-time curves are completed as shown in Figure 10-24b.
Solutions of Molecular Solids in Liquids. A molecular solid is one in which covalent molecules are arranged in a lattice held together by intermolecular forces exerted by one molecule on its neighbors. When such a solid is placed in contact with a solvent in which it dissolves, molecules at the surfaces of crystals in contact with the solvent can escape the potential well of the crystal if they possess sufficient kinetic energy, and can enter the potential well of the solvent. The rate of this escape from the solid phase is proportional to the surface area of the solid and to the "concentration" of substance in the molecular solid, which is constant. Thus
Similarly, dissolved molecules collide with the surface of the solid and return to the solid phase at a rate given in 10-8-10.
At equilibrium, when the two opposing rates are equal,
Thus the equilibrium constant is simply the concentration of substance dissolved in the solvent.
As we might expect from our work with phase diagrams, a change in pressure has little effect on the equilibrium established between a pure solid and a saturated liquid solution of the solid (equation 10-8-3).
Both the solid and solution (liquid) phases are condensed, hence incompressible--their volumes are essentially unaffected by pressure. Unless there is a change in volume with pressure, there can be no change in concentration (the number of molecules per unit volume); and unless there is a concentration change in a phase, there can be no change in the rate at which molecules or ions leave the phase as long as T is constant (we have seen that rate processes are related directly to concentration). All phase equilibria are dynamic equilibria involving a balance of opposing rates. Unless one of the rates is affected by pressure, a change in P will have no effect on the position of equilibrium. Since neither the forward nor the reverse rate of an equilibrium involving two condensed phases is affected by pressure, this type of equilibrium is unperturbed when P is varied. The equilibrium between a solid and its saturated solution is of this type, where in this case the two balanced rates are the rate of dissolution and the rate of crystallization. The equilibrium between s and l phases of a pure substance is very similar. The temperature at which such an equilibrium exists is unaffected by even huge changes in P. Thus the solubility of a solid in a liquid should be independent of pressure.
Example 10-8. The solubility of a molecular solid, A(s), in water is 9.0 g per 100 mL of water at 20 oC. 6.4 g of A(s) is added to 23.6 mL H2O. What fraction of A(s) dissolves?
To the resulting solution is added 14.4 mL of water. What is the concentration of A(s) when equilibrium is reached, in grams A(s) per 100 mL water?
Solution. The mass of A that will dissolve in 23.6 mL H2O is
The fraction of A that dissolves is 2.12/6.4 = 0.33.
The equilibrium constant for the process
is simply the concentration of A(aq) in a saturated solution. Thus Keq = 9.0 g/100 mL. Before addition of extra water, the system is at equilibrium and satisfies this Keq. Immediately after addition of water, the concentration of dissolved A is 2.12 g A/(23.6 + 14.4) mL H2O = 5.58 g A/100 mL water. According to Le Chatelier, the process above must shift to the right to offset the dilution of A caused by addition of water. Shift will continue either until all solid A is dissolved, or until the solution is once again saturated. To saturate 38 mL water requires (9.0 g A/100 mL water) * 38 mL = 3.42 g A, which is less than the total initial amount of A present. Thus the solution will achieve equilibrium with 3.42 g A dissolved and 6.4 - 3.42 = 3.0 g remaining in the solid phase.
Solutions of Gases in Liquids. The solubility of a gas in a liquid is markedly affected by the pressure at which the gas is maintained over the liquid. This is a consequence of the effect of pressure on the volume, hence the concentration, of a gas. Consider the experiment shown in Figure 10-25. We begin with a quantity of water at 25 oC in a cylinder fitted with a piston and a gas inlet (Figure 10-25a). Initially, the space above the water is free of all gas except water vapor, which exerts the characteristic vapor pressure at 25 oC. We now fill the space above the water with a gas, B(g), to some pressure PB1 by opening the gas inlet valve, while holding the piston fixed in place, until the desired pressure is reached (Figure 10-25b). As soon as the gas is admitted to the cylinder, gas molecules begin to strike the walls of the cylinder and the surface of the water with a frequency that depends on the concentration (pressure) of gas molecules in the space above the liquid. Some of the molecules that strike the water surface enter the liquid--i.e., they dissolve. The rate of dissolution is directly proportional to the pressure of the gas and to the surface area, SA, of the water, as indicated in equation 10-8-13.
Here k1 is a proportionality constant (a rate constant). Gas molecules that dissolve move randomly in the liquid phase, and at any given time, some of them will be moving toward the surface. Those with sufficient energy will escape the liquid and reenter the gas phase. The rate of escape is directly proportional to the surface area of the liquid, and to the number of molecules travelling toward the surface, which in turn is directly related to the concentration of dissolved gas. Thus
where k2 is also a rate constant. Immediately after the gas is introduced to the cylinder, the rate of escape is 0 because there is no dissolved gas, but the rate of dissolution is greater than 0 because P > 0. The initial imbalance of rates leads to a buildup of dissolved molecules, an increase in [B], and a corresponding increase in Rateescape, until Ratedissolution = Rateescape. At this point, dynamic equilibrium between the gas and solution phases has been established. The dynamic equilibrium between a gas and its solution is expressed as in 10-8-15.
where the double arrow shows the opposing equal rates. Equating the right sides of 10-8-13 and 10-8-14 and rearranging gives
Equilibrium is characterized by a particular ratio of the concentration of dissolved B to the pressure of B above the solution.
Now consider quickly moving the piston down (Figure 10-25c) so that the volume above the liquid is halved. This doubles the pressure of the gas above the solution, and immediately doubles Rdissolution (10-8-13) because the number of collisions of gas molecules with the surface is now twice as great. However, Rescape is initially unaffected. The immediate effect of the pressure increase is thus to make Rdissolution > Rescape. This imbalance results in a net flow of molecules from the gas to the liquid phase, causing a gradual buildup in [B], until the two opposing rates again become equal. The new equilibrium is established with higher values of both P and [B], but their ratio, given by k1/k2, is the same as before the pressure increase. Thus the concentration of dissolved gas--its solubility--is directly proportional to the pressure of gas above the solution. We express this in mathematical terms by writing equation 10-8-16 somewhat differently.
Equation 10-8-17, and the verbal statement that it summarizes, is known as Henry's Law. K is called the Henry's Law constant. The values of K for dissolution of several gases in water are given in Table 10-3. For a gas pressure of 1 atm, these constants are numerically the same as the molar solubilities of the gases.
Table 10-3: Henrys Law Constants (Solubilities) of several gases in water at 20 oC, with 1 atm gas pressure
| Gas | K(M/atm), x 103 |
|---|---|
| N2 | 0.69 |
| CO | 1.04 |
| O2 | 1.38 |
| Ar | 1.50 |
| Kr | 2.79 |
Example 10-9. A corked bottle of champagne is allowed to warm from 0 oC to 25 oC over a 1-hour period. The pressure of CO2 in the bottle was 2.5 atm at 0 oC. What is the concentration of dissolved CO2 at 0 oC? What is the pressure of CO2 in the bottle at 25 oC? You may assume that
Solution. We start at 0 oC because we know the pressure of CO2 and can calculate the Henry's Law constant at this temperature. The concentration of dissolved CO2 at 0 oC can be calculated from Henry's Law as follows:
We have multiplied by 1.5 to obtain the Henry's Law constant appropriate at the lower temperature. Thus at 0 oC we have 11.6*10-2 moles/L * 1.00 L = 0.116 moles CO2 dissolved in the liquid, and n = PV/RT = (2.5)(0.050)/(0.08206)(273) = 0.00558 moles CO2 in the gas space above the liquid. The total moles CO2 in the bottle is therefore 0.116 + 0.00558 = 0.1216 moles. This total must remain constant as the champagne warms.
At 25 oC the ratio of pressure to concentration must satisfy the Henry's Law equilibrium constant, which is 3.1 * 10-2 at the higher temperature. Thus
So that we can make use of conservation of total moles CO2 during the warming process, we should convert the ratio on the left to a mole ratio. This is easy using the volumes of the two phases:
Substituting for [CO2] and PCO2 and rearranging gives
Knowing both the ratio and the sum of the moles dissolved and in the gas allows us to calculate both:
Finally we convert back to pressure and concentration:
In summary, warming causes a slight decrease in concentration of dissolved gas from 0.116 to 0.114 M, and an increase in pressure of gas in the head space from 2.5 to 3.68 atm.
Solutions of Ionic Solids in Liquids; The Solubility Product, Ksp. The Solubility Rules. Almost all salts are strong electrolytes; they dissolve as separate positive and negative ions. However, it is easy to show experimentally that although some salts are very soluble (dissolve readily) in water, others dissolve to only a small extent. For example 34.7 g of sodium chloride, NaCl, dissolves in 100 mL of water at 10oC, compared with only 0.000089 g of silver chloride, AgCl. Chemists have developed empirical rules that can be used to decide whether or not a salt is soluble. These are presented below. For application of these rules, we consider a salt to be soluble if its solubility exceeds 0.01 moles per liter of solution. Salts that dissolve to a lesser extent than 0.01 M are insoluble.
Solubility Rules
A salt is soluble if it contains:
A salt is insoluble if it contains:
Example 10-10. Use the solubility rules to predict solubilities of NaCl, Fe2S3, (NH4)2S, and CaCO3.
Solution.
NaCl. According to rule 1, this is soluble.
Fe2S3. According to rule 7, this is insoluble.
(NH4)2S. According to rule 7, this is soluble.
CaCO3. According to rule 7, this is insoluble.
The solubility rules can be used to predict whether a precipitate (solid) will form when solutions of two electrolytes are mixed.
Example 10-11. A solution of sodium chloride is mixed with a solution of silver nitrate. Will a precipitate form?
Solution. According to the solubility rules, NaCl and AgNO3 are soluble salts. This makes it possible to prepare a fairly concentrated solution of each salt. Because salts are strong electrolytes, these solutions contain the constitutent ions of the salts: Na+(aq), Cl-(aq) and Ag+(aq), NO3-, respectively. Mixing the solutions produces a solution containing all four ions. Is there a cation-anion combination that will produce an insoluble salt? Rule 3 says that AgCl is insoluble. A precipitate of AgCl will form when the two solutions are mixed, as shown below.
Na+(aq) + Cl-(aq) + Ag+(aq) + NO3-(aq) --> Na+(aq) + NO3-(aq) + AgCl(s)
Eliminating spectator ions gives the net ionic equation:
Ag+(aq) + Cl-(aq) --> AgCl(s)
The solubility rules, based on empirical observation, are useful for predicting the outcome of mixing solutions of electrolytes. This is applied in chemical analysis, in which selective precipitation of an ion from solution can enable us to determine the amount of the ion that was initially present by weighing the precipitate.
Although many ionic salts show substantial solubility in water, a number of simple and common salts fail to dissolve. These are called slightly soluble salts. Combinations of OH-, CO32-, and S2- with many cations, particularly those of the transition metals, are examples of such salts. We now want to examine the quantitative aspects of their dissolution, using the principles of equilibrium learned thus far in this chapter. Consider adding a quantity of the salt MX(s) (Mp+ is the cation, Xp- is the anion) to a beaker of water. Assume that we know how much of the salt will dissolve in the water, and that the amount added is more than this. What happens at the molecular level after the salt is placed in contact with the water? The situation is shown in Figure 10-26. Figure 10-26a shows the initial situation, immediately after addition of the salt to the water, but before dissolution of the solid. Upon contact with water, ions on the surface of the crystals begin to escape into the liquid at a rate that depends upon the forces of attraction between these ions and other ions in the crystal on one hand, and the ions and water molecules on the other. The rate of dissolution is proportional to the surface area, SA, of the solid salt, and the concentration of salt, which is constant in the solid.
The rate of dissolution remains essentially constant as long as solid is present and the temperature is maintained constant. As dissolution proceeds, the concentrations of the ions in the liquid phase gradually build up. These ions move randomly in the liquid phase, occasionally colliding with the surface of the solid, where some of them stick--i.e., they crystallize. As their number builds, so does the rate at which they collide with the surface of the solid and crystallize. The rate of crystallization is proportional to the exposed surface area of solid, and to the product of the concentrations of positive and negative ions. The product is required because positive and negative ions must collide at the surface in order for crystallization to occur.
A situation in which dissolution and crystallization occur with unequal rates is shown in Figure 10-26b. Eventually, rates of dissolution and crystallization become equal, and there is no further net dissolution of solid. At this point, pictured in 10-26c, dynamic equilibrium has been established between the solid and its ions in solution. This is similar to other phase equilibria discussed earlier in this chapter. The equilibrium is represented in equation 10-8-20, where the double arrow shows the two opposing rates.
Figure 10-26d shows graphically the equalization of the rates of dissolution and precipitation over time. The equilibrium in 10-8-20 is certainly a phase equilibrium, because it involves two phases: a solid phase, represented by the salt, and a liquid phase, the solution. Equating the expressions for the rates of dissolution and crystallization, the expression for the equilibrium constant is obtained:
We give a special symbol to the equilibrium constant for dissolution of a slightly soluble salt: Ksp. The subscript, sp, stands for "solubility product' because it is the PRODUCT of ion concentrations, which are related to the molar SOLUBILITY of the salt. The molar solubility is the number of moles of salt that dissolve per liter of solution. The solubility product expression for the slightly soluble salt, bismuth sulfide, which dissolves according to
is Ksp = [Bi3+]2[S2-]3. The value of Ksp is 2 x 10-70 at 298 K, indicating that the concentrations of ions in equilibrium with solid bismuth sulfide are very small.
Before considering calculations using solubility product equilibria, we define a few terms used to describe solutions of salts. Consider that we have a beaker containing one liter of pure water. To this we add a quantity of a salt, MnXm(s). Suppose first that the quantity of solid added is less than the amount that will dissolve in one liter of water. We will see a gradual disappearance of solid salt as dissolution occurs, until all of the solid is gone. The resulting solution, containing less than the maximum possible amount of dissolved salt, is said to be unsaturated. In such a solution, the product [Mp+]n[Xq-]m is less than Ksp for the salt. Now suppose that we add more solid to the beaker. Since the solution is unsaturated, we expect that some of the added solid will dissolve, until the solution contains the maximum possible amount. This amount depends on the solubility of the salt in water at the water temperature. We can identify this situation visually, because we will see a quantity of undissolved solid on the bottom of the beaker that does not change in size with time. This indicates that there is no longer any net dissolution of solid. At the molecular level, of course, a dynamic conversion between solid salt and dissolved ions is taking place, in which salt is dissolving and ions are combining to form solid at equal rates. At the macroscopic level, as is true for any equilibrium, the dynamic aspects of the microscopic balance are not apparent, and it appears as if nothing is happening. A solution containing the maximum possible amount of dissolved solid at a particular temperature is said to be saturated. In a saturated solution, it must be true that the product of ion concentrations, each concentration raised to the appropriate power, is equal to Ksp. Finally, suppose that we are able somehow to cause more than the maximum possible amount of salt to dissolve. In such a solution, which is said to be supersaturated, the product of ion concentrations exceeds Ksp, and the system is not at equilibrium. Eventually, we expect precipitation (solid formation) to occur until the product of ion concentrations becomes equal to Ksp, at which point equilibrium will have been established.
How do we prepare a supersaturated solution of a salt? Suppose that the salt becomes more soluble in water as the temperature of the water increases. Both sodium acetate, NaC2H3O2, and alum, KAl(SO4)2, qualify, as do many other salts. We consider sodium acetate. The solubility of sodium acetate in water at 25 oC is 125 g per 100 mL of water. At 100 oC (the boiling point of water) the solubility increases to 170 g per 100 mL of water. If we weigh an amount of sodium acetate that is greater than 125 g but less than 170 g, add it to 100 mL of water, and warm the mixture with stirring to near 100 oC, all of the sodium acetate should dissolve. (In order to observe supersaturation, it is necessary to remove all solid material from the solution while it is hot. This can be done by careful filtration of the hot solution.) We now carefully cover the solution to exclude dust and dirt, and allow it to slowly cool to room temperature. With luck, all of the sodium acetate stays dissolved, even though the equilibrium solubility at 25 oC is exceeded. This is a supersaturated solution. How can a substance remain dissolved when it should precipitate? Why does the salt not crystallize from solution as the system cools, so as to always maintain the solubility product equilibrium? The answer is interesting--and simple. Formation of crystals of solid from ions or molecules dissolved in a solvent is similar to the freezing process, in which solid forms from liquid:
For this process, DPE < 0, favoring formation of solid; but DS < 0 (because solid is more ordered than liquid), opposing formation of solid. Supersaturation is thus analogous to supercooling of liquids. For crystallization of a salt to begin, it is necessary that positive and negative ions, moving randomly in solution, coalesce in the proper arrangement for a crystal to begin to form. Because this process is opposed by entropy, it often takes some time, during which the solution can hold more dissolved material than it should. Once the beginnings of a crystal form, however, this serves as a nucleus, or template, on which other ions can deposit, and crystallization proceeds rapidly. Thus we can initiate crystallization of sodium acetate from the supersaturated solution by adding a small crystal of sodium acetate as a template. Crystallization then proceeds until the Ksp is satisfied (i.e., until the solution is saturated). (Often, addition of any small solid particles to a supersaturated solution promotes crystallization. That is why filtration to remove dust is necessary.) Generally, the more complicated and/or unsymmetrical the structures of the ions of the salt, the more difficult it is for the ions to coalesce in the correct way to initiate crystallization. Thus we expect salts of complex ions to more readily form supersaturated solutions than salts of simple ions. In agreement with this guideline, it is very difficult to prepare a supersaturated solution of NaCl; whereas preparation of supersaturated sodium acetate, which contains an anion of complex structure, is readily accomplished.
Example 10-12. Determine the molar solubility of CaCO3 in water at 25 oC.
Solution. In this example, we illustrate what might be called the standard approach to the determination of salt solubility using Ksp. After carrying this through in the usual way, we will compare the result with the known solubility of CaCO3 to demonstrate that Ksp calculations must in most cases be taken with a grain of salt (no pun intended).
We can approach this problem using a general method for treating chemical equilibria that will be extensively used in Chapter 12. We write the equation for dissolution of calcium carbonate, and underneath the species we write the initial concentrations, the changes in concentrations that take place during reaction, and the concentrations that prevail at equilibrium. This is done below.
| (10-8-23): | CaCO3(s) ---> | Ca2+(aq) + | CO32- | Ksp = [Ca2+][CO32-] = 5 * 10-9 M2 | |
| initial | xs | 0 | 0 | ||
| change | -x | x | x | ||
| equilibrium | xs | x | x |
(Note: because CaCO3 is a pure solid, it is omitted from the Ksp expression. In order to establish equilibrium with its ions, it is necessary only that a sufficient amount of it be present. Thus we have written under its formula that we have an excess (xs) initially, and still have an excess at equilibrium.)
Substituting, x2 = 5 * 10-9, giving x = 7 * 10-5 M.
Thus according to our calculation, 7 * 10-5 moles (0.007 g) of CaCO3 will dissolve in one liter of water. According to the Handbook of Chemistry and Physics, the solubility of calcium carbonate is 0.0015 g per 100 mL of water, or 0.015 g per liter, for one crystalline form; and 0.014 g per liter for another. The solubility thus depends on the crystalline form adopted by the solid, so already we have a complication not addressed by our standard calculation above. Further, we see that at best our calculated result is wrong by a factor of about 2! The solubility of CaCO3 is actually two times greater than we would expect. The major reason for the discrepancy is that the carbonate anion is substantially basic, and reacts with water in the hydrolysis process below:
This process causes the concentration of the carbonate dianion to decrease, shifting the solubility equilibrium, 10-8-23, to the right by Le Chatelier's Principle. Reaction 10-8-24 therefore enhances the solubility of CaCO3. It is the rule rather than the exception that the solubilities of slightly soluble salts are influenced by reactions that the ions produced by dissolution may subsequently undergo. For this reason, the standard approach to Ksp calculations gives at best only a gross estimate of concentrations, and at worst, a completely inaccurate result.
Additional examples of Ksp calculations are presented in Appendix H for completeness. However, be aware that the calculations illustrated there are no more accurate than the one shown here.
In this chapter we have explored the concept of phase equilibrium in detail. In the discussion of the equilibria among phases of a single pure substance, the equilibrium vapor pressure has emerged as a quantity of great importance. It measures the "escaping tendency" of the molecules in the liquid (or solid) phase by directly reflecting at the macroscopic scale the magnitudes of the forces of attraction between molecules. By measuring its temperature dependence, we can obtain the depth of the liquid potential well and can measure the change in disorder that occurs when the liquid vaporizes. By measuring its decrease, or consequences thereof, in solutions, we can determine the amount and even the molecular formula of a dissolved solute. This very simple macroscopic property of liquids, then, provides us with an abundance of information about happenings in the liquid at the molecular level.
More generally, we have seen that dynamic equilibrium, a concept of immense importance in chemistry, has a number of defining characteristics:
The word, "equilibrium," is used somewhat casually by scientists. It is useful to distinguish between equilibrium in general and dynamic equilibrium as we have characterized it here. A pure substance subjected to a pressure and a temperature such that only the liquid phase of the substance may exist is certainly at equilibrium in the sense that no change in the system will occur in the absence of a perturbation. However, this equilibrium is not dynamic, because only one state of the system is present, so opposing processes do not occur. At least 2 states of a system must coexist in order for a situation of balanced opposing processes to be possible.
Supplement: Journey Around a Phase Diagram. The phase diagram for a typical pure substance is the superposition of l-v, s-v, and s-l P versus T curves for the substance. At points located on the lines, two phases coexist in equilibrium; at points located off the lines, only one phase exists.
An understanding of phase diagrams requires an understanding of several aspects of the equilibrium vapor pressure. First, the vapor pressure, Pvap, of a pure liquid increases exponentially with temperature. This may be rationalized by partitioning the vapor pressure increase into two contributions. Say that there are on average N1 molecules in the vapor phase prior to the temperature increase. When T is raised, the pressure due to these N1 molecules increases linearly by the ideal gas law. However, at the higher temperature a larger fraction of molecules in the liquid phase possesses the minimum kinetic energy needed to escape the potential well of the liquid, resulting in an increase in the number of molecules in the vapor phase. Say that an additional N2 molecules enter the vapor phase as T is raised. Then at the higher temperature, there are on average N1 + N2 molecules in the gas phase. The total increase in Pvap is the result of contributions from the average number of molecules already present in the gas phase and from the additional molecules that enter the gas phase as a result of the T increase. According to this rationale, Pvap increases more rapidly than linearly. The strict mathematical dependence is governed by the Clausius-Clapeyron equation:
The temperature dependence of the pressure of vapor in equilibrium with a solid is governed by the same equation. The only difference is that the enthalpy and entropy terms must be appropriate to sublimation rather than vaporization, as in Equation 10-S-2.
As discussed earlier in the chapter, a plot of lnPvap versus 1/T for both the liquid-vapor and solid-vapor equilibria on the same set of axes must have the general appearance of Figure 10-9. The lines must cross at some temperature T = Tt. The phase diagram results from translation of the lnPvap versus 1/T plots for liquid and solid to Pvap versus T plots, as shown in Figure 10-10. The curve labelled l-v represents all P-T combinations at which liquid and vapor coexist in equilibrium. Similarly, the curve labelled s-v represents P-T combinations for solid-vapor equilibrium. All three phases coexist at the triple point.
The locus of P,T combinations at which solid and liquid coexist in equilibrium must include the triple point, and slopes sharply and positively up from it. The line is steeply sloped because solid and liquid are both condensed phases, with the consequence that their properties are largely pressure insensitive. Normally, the solid phase has higher density than the liquid phase. For most substances, then, the line has positive slope, reflecting the tendency for the substance to convert to the denser phase under increased pressure. For a few substances, water among them, liquid is denser than solid near the melting point, and increased pressure converts solid to liquid. These substances have a negatively-sloped s-l line.
Figure S-1 shows a phase diagram for a pure substance. It is instructive to make a cyclic "journey" around this phase diagram, to improve understanding of the changes that take place in the system as the pressure and temperature are varied. The pure substance is considered to be in a cylinder/piston container, with no other substances present. The pressure on the system is exerted through the piston, and may be changed at will. The system temperature is maintained by immersing the cylinder in a bath at the desired T. Phase diagrams apply strictly only when the pure substance is contained by itself in such an apparatus.
The journey begins and ends at point A on the diagram. The system is taken by numbered paths to various lettered states and is ultimately returned to state A. The appearance and state of the system are pictured in Figure S-2a. State A of the substance is characterized by pressure PA and temperature TA. Since point A is on the l-v line, the situation is one of liquid-vapor equilibrium. The total moles of substance in the system, nT, is given by the sum of the moles of substance in the liquid and vapor phases, nlA + nvA (the symbol nlA signifies the moles of substance in the liquid phase in state A). Note that the piston rests against the mechanical stops of the cylinder because the external pressure, Pext, is kept somewhat below PA.
Operation 1: The external pressure is increased to PB, maintaining temperature constant at TA. What happens in the system? All vapor condenses until only liquid exists in the cylinder, because Pext exceeds the value of the equilibrium vapor pressure at TA. At pressure, temperature points between the s-l and l-v lines, only liquid exists in the system. The system now looks as in Figure S-2b.
Operation 2: Pext is decreased by an infinitesimal (i.e., very small) amount to PB - dP, and T is increased to TC. The external pressure must be decreased so that the piston will move. However, it is decreased only enough to
