Chapter 5 Phases and Solutions

Chapter 5 Phases and Solutions 5.1 Phase Recognition Homogeneous: A single phase is uniform throughout both in chemical composition and physical state. Note that subdivision does not produce new phase; a block of ice reduced to crushed ice still consists of only one phase. Mixtures of compounds or elements may also appears as a single phase as long as no boundaries exist that allow one region to be distinguished from another region. Heterogeneous: A system consists of more than one phase; the phases are distinguished from each other through separation by distinct boundaries. ice cubes in water two phases: one solid and one liquid

Phase Distinctions in the Water System The curve TTc gives the vapor pressure of water up to the critical point Tc. solid and liquid phase are in equilibrium liquid and vapor phase are in equilibrium ~The curve TB gives the equilibrium conditions for supercooled water, an example of a substance in a metastable state (i.e. not the thermodynamically most stable state). ~This state can be achieved under certain conditions because, in this case, the rate of formation of ice has been reduced by using a very clean sample of water, thereby reducing the number of nucleation sites. ~Formation of metastable states often occurs when the entropy change during the phase transition is negative, as in this case. Note that water is not truly representative of most materials due to that the slope of the line TC is negative. Only a few materials behave as water does in this respect, among them bismuth (鉍) and gallium (鎵). solid and vapor phase are in equilibrium showing the vapor pressure of the solid as a function of temperature At the triple point T, all three phases coexist at the same vapor pressure (0.611 kPa or 4.58 Torr) and temperature (273.16 K or 0.01 C).

Problem 5.12 The variation of the equilibrium vapor pressure with temperature for liquid and solid chlorine in the vicinity of the triple point is given by Use P/Pa in the equations. Calculate the triple point pressure and temperature. Solution At the triple point, the two vapor pressure must be equal since the liquid, solid, and vapor are in all in equilibrium with each other. Therefore, we set

Phase Equilibria in a One-Component System: Water Two phases of the same pure substance to be in equilibrium under given conditions of temperature and pressure, the Gibbs energy of one phase must be equal to that of the other. In other words, the change in Gibbs energy between the two phases is zero. Specially, when equilibrium occurs between ice and water and when there is equilibrium between water and steam At the tripe point, Figure 5.3 ~If only one phase is present under the particular P-T conditions, it will be the phase with the lowest value of G. ~The curvatures for the solid, liquid, vapor regions are all different, but the solid-liquid curves intersect at 0 C, the normal melting point, and the liquid-vapor curves intersect at 100 C, the normal boiling point. ~The effect of reducing the pressure from 1 atm is indicated by a narrowing of the liquid range as shown by the vertical dashed line.

Example 5.1 From the position of the curves representing Gsolid, Gliquid, and Gvapor, predict the phases that are present (more stable) at 1 atm pressure as the temperature increases in Figure 5.3. Solution At temperature less than 0 C, the curve representing Gsolid has the lowest value of Gibbs energy. The solid is the most stable because this phase has the lowest Gibbs energy in this temperature range. Between 0 C and 100 C, the curve for Gliquid is lowest and the liquid is the most stable phase; and above 100 C, the curve for Gvapor is lowest and the vapor is the most stable phase.

We can utilize two of the Maxwell relations from Chapter 3, namely Eq. 3.119 written for 1 mol of substance: ~Since the second of these equations has a negative sign and the curves in Figure 5.3 have negative slopes, the entropy is positive. ~The relative slopes show that (Sm)g>(Sm)l>(Sm)s. ~G must decrease as the pressure is decreased, since Vm is always positive. ~The decrease in Gibbs energy for the gas phase is much greater than for the other two phase due to that the molar volume of the gas Vm(g) is much larger compared to that of the liquid or solid. ~The reduction of pressure reduces the boiling point of all liquids. ~There is an increase in the melting point for water as the pressure is reduced, that is different for most other substances, for which a reduction in pressure normally decreases the melting point of the substances.

Problem 5.1 Diamonds have successfully been prepared by submitting graphite to high pressure. Calculate the approximate minimum pressure needed using fG =0 for graphite and fG =2.90103 J mol-1 for diamond. The densities of the two forms may be taken as independently of pressure and are 2.25 and 3.51 g cm-3, respectively. Solution At equilibrium, G (graphite) =G (diamond); i.e., 2G =0. We are given 1G =2900 J mol-1.

Taking the second derivative of G with respect to T, ~The curves depicting the variation of G with T for each phase do not have the same curvature because CP,m for each phase is different. ~The curvature given by the second derivative increases as T increase since CP,m does not vary greatly with T.

Considering an increase in pressure, we expect the Gibbs energy of a phase to increase. Figure 5.4 In a normal substance, Vm(l)>Vm(s), and thus the increase in pressure will cause a larger increase in Gibbs Energy for the liquid than for the solid. In order to have equilibrium between the two phases at a higher pressure, G must equal 0 and, to accomplish this, the melting point must shift to a higher value as shown in Figure 5.4a. Just the opposite result is obtained for water due to that the molar volume of ice is larger than that of water, the increase in G will be greater for the solid form, thus the melting point is reduced as shown in Figure 5.4b.

5.2 Vaporization and Vapor Pressure Thermodynamics of Vapor Pressure: The Clapeyron Equation If the pressure and temperature are changed infinitesimally in such a way that equilibrium is maintained, Since this expression depends only on T and P, the total derivates may be written as If the molar enthalpy change at constant pressure for the phase transformation is Hm, the term Sm may be written as Hm/T and the above equation becomes Clapeyron Equation It may be applied to vaporization, sublimation, fusion, or solid phase transition of a pure substance. Its derivation involves no assumptions and is thus valid for the general process of equilibrium between any two phases of the same substance.

Example 5.2 What is the expected boiling point of water at 98.7 kPa (approximately 740 Torr, a typical barometric pressure at 275 m altitude)? The heat of vaporization is 2258 J g-1, the molar volume of liquid water is 18.78 cm3 mol-1, and the molar volume of steam is 30.199 dm3 mol-1, all values referring to 373.1 K and 101.325 kPa (1 atm). Solution where Tb refers to the temperature at boiling The new boiling point is 372.42 K.

Example 5.3 Determine the change in the freezing pint of ice with increasing pressure. The molar volume of liquid water is 18.02 cm3 mol-1, and the molar volume of ice is 19.63 cm3 mol-1 at 273.15 K. The molar heat of fusion fusHm=6.009103J mol-1. Solution where the subscript fus refers to the value of the variables at the melting point. Here, since the pressure is not the equilibrium value, P refers to the applied pressure maintained mechanically or through an inert gas. Also [Vm(l)-Vm(s)] is assumed to be approximately constant over a moderate pressure range. This is a 0.74 K decrease in temperature per 100 bar increase in pressure.

The Clausius-Clapeyron Equation When one of the phases in equilibrium is a vapor phase, we assume that Vm(v) is so much larger than Vm(l) that we may neglect Vm(l) in comparison to Vm(v) when the pressure is near 1 bar. (For water, the volume of vapor is at least a thousand times that of a liquid.) Clausius-Clapeyron Equation In order to make use of the above equation, integration is performed assuming vapHm to be independent of temperature and pressure. We thus obtain integration constant

A plot of lnP against 1/T should be linear. The slope of the line is -vapHm/R. A plot of RlnP against 1/T could also be made in which case the intercept on the RlnP axis (i.e. where the temperature is infinitely high) will give the value of vapSm. Figure 5.5 As might be expected over a wide range of temperature, deviation from linearity occur as a result of temperature variation of vapHmand of nonideal gas behavior of the vapor. Even if curvature exists in the plot, the enthalpy change may be calculated by drawing a tangent to the curve at the temperature of interest.

Example 5.4 Benzene has a normal boiling point at 760 Torr of 353.25 K and vapHm=30.76 kJ mol-1. If benzene is to boiled at 30.00 C in a vacuum distillation, to what the value of P must the pressure be lowered? Solution In order to boil benzene its vapor pressure must equal the pressure on the system. The problem is thus a matter of finding the vapor pressure of benzene at 30.00 C. We may use the Clausius-Clapeyron equation considering vapHm constant over the temperature range. T2/K=30.00+273.15=303.15 K.

Problems 5.3 Calculate the heat of vaporization of water at 373.15 K and 101.325 kPa using the Clausius-Clapeyron equation. The vapor pressure of water is 3.17 kPa at 298.15 K. Compare your answer to the CRC Handbook value. Solution The CRC handbook gives 40.6 kJ mol-1.

Problems 5.8 The compound 2-hydroxybiphenyl (o-phenylphenol) boils at 286 C under 101.325 kPa and at 145 C under a reduced pressure of 14 Torr. Calculate the value of the molar enthalpy of vaporization. Compare this value to that given in the CRC Handbook. Solution The CRC handbook gives 71.0 kJ mol-1.

Enthalpy and Entropy of Vaporization: Trouton’s Rule In Section 3.4, page 109, we saw that entropy values may be obtained from the expression where trs specifies a particular transition The enthalpies of vaporization vapSm of most non-hydrogen-bond compounds have values of vapSm in the neighborhood of 88 J K-1 mol-1.This generalization is known as Trouton’s rule and was pointed out in 1884: The intersection of lines on the RlnP axis in Figure 5.5 at approximately 88 J K-1 mol-1 is know an Trouton’s focus.

The effect of hydrogen bonding is seen in the case of water. The abnormally low value for acetic acid may be explained by its appreciable association in the vapor state. Allowing for its apparent molecular weight of about 100 in the vapor state, acetic acid will have a value of vapSm of approximately 100 J K-1 mol-1, in line with other associated liquids. The average of vapSm for a large number of liquids that are not appreciably hydrogen- bonded bears out the value (88 J K-1 mol-1 ). Hildebrand Rule: The entropies of vaporization of unassociated liquids are equal, not at their boiling points, but at temperature at which the vapors occupy equal volumes. Craft’s Rule For associated liquids, a numerical coefficient of 7.510-4 gives better results than 9.310-4, which is used for normal liquids.

Example 5.5 Estimate the enthalpy of vaporization of CS2 if its boiling point is 319.40 K. Solution The experimental value is 28.40 kJ mol-1.

Problems 5.8 The normal boiling point of toluene is 110.62 C. Estimate its vapor pressure at 80.00 C assuming that toluene obeys Trouton’s rule. Solution Applying Trouton’s rule,

Variation of Vapor Pressure with External Pressure For a closed system, with T constant and the total pressure Pt equal to the pressure exerted on the liquid, the following equation may be obtained in a form known as the Gibbs equations: where P is the pressure of the vapor under the total pressure Pt. If we assume that the vapor phase behaves ideally, we may substitute Vm(v)=RT/P, obtaining Since the molar volume of a liquid does not change significantly with pressure, we may assume Vm(l) to be constant through the range of pressure and obtain, by integrating, where Pv is the saturated vapor pressure of the pure liquid without an external pressure. Note that in this case the vapor pressure is now dependent on the total pressure as well as on the temperature. This is because the two phases exist under different pressures.

Example 5.6 The vapor pressure of water without the presence of any other gas, such as air, at 25 C is 3167 Pa. Calculate the vapor pressure of water when the head space, that is, the enclosed volume above the water, already contains an insoluble gas at a pressure of 1 bar. This is an important problem in the bottling industry in the design of beverage containers. Solution This shows that the pressure of the insoluble (inert) gas has very little effect on the water vapor pressure.

Problems 5.5 Estimate the vapor pressure of iodine under an external pressure of 101.3106 Pa at 313.15 K. The density of iodine is 4.93 g cm-3. The vapor pressure at 101.3 kPa is 133 Pa. Solution The molar volume of iodine, I2, is

5.3 Classification of Transitions in Single-Component Systems In each of the transition involving vaporization, fusion, and sublimation occurs without a change in temperature as the enthalpy changes. Other types of phase transitions also occur, such as polymorphic transition in solids (i.e., changes from one crystalline structure to another). The character of the enthalpy and other changes is quite different in these transformations. According to Ehrenfest’s classification, phase transitions are said to be first-order phase transitions if the molar Gibbs energy is continuous from one phase to another, but the derivatives of the Gibbs energy are discontinuous.

For example, at the melting point of ice, even through the Gibbs energies of the liquid water and ice are equal, as required for equilibrium, there is a change in the slope of the G against T curve. There is consequently a break in the S against T curve and also in the V against T curve, since first-order phase transitions There is an increase in heat, while the change in temperature is zero. As a result, since Cp=(H/T)p, its value is infinity at the transition temperature.

According to Tiza’s theory, a second-order phase transition is one in which the discontinuity appears in the thermodynamic properties that are expressed as the second derivative of the Gibbs energy. For second-order transitions, there is no latent heat although there is a finite discontinuity in the heat capacity. In addition the enthalpy and entropy are continuous functions, although their first derivatives, CP=(H/T)P andCP=T(S/T)P,are not. Thus, CP shows a finite change at the transition temperature and fulfills Tisza’s definition since second-order phase transitions An example of this type of transition occurs in the change from the conducting to superconducting state in some metals at temperature near 20 K.

Another common transition, called the lambda transition, is one in which S and V are zero but the heat capacity becomes infinite in the form of the Greek letter lambda (). a lambda transition experienced by liquid helium Unlike the case in a first-order transition, the heat capacity can be measured as close to the transition temperature as possible and is found still to be rising. The enthalpy is continuous but has a vertical inflection point. The transition between He I and He II is an example of a lambda transition that occurs at about 2.2 K. He II has very unusual properties including a practically zero viscosity, leading it to be called a superfluid.

5.4 Ideal Solutions: Raoult’s and Henry’s Laws two-component system x, T, P three-dimensional plot  two-dimension plot ( x, T or x, P) one-component system T, P two-dimensional plot A solution is any homogeneous phase that contains more than one component. Although there is no fundamental difference between components in a solution, we call the component that constitutes the larger proportion of the solution the solvent. The component in lesser proportion is called the solute. ideal gas  ideal solution Considering a solution of molecules of A and B, if the molecules sizes are the same, and the intermolecular attractions of A for A and of B for B are the same as the attraction of A for B, then we may expect the most simple behavior possible from a solution. Thus the solution is considered to be ideal when there is a complete uniformity of intermolecular forces, arising from similarity in molecular size and structure.

Raoult’s Law The vapor pressure P1 of solvent 1 is equal to its mole fraction in the solution multiplied by the vapor pressure P1* of the pure solvent 1. A similar expression holds for substance 2 in the mixture. If the solution has partial vapor pressure that follow above equation, we say that the solution obeys Raoult’s law and behaves ideally. The solutions obeying Raoult’s law are benzene-toluene, ethylene bromide-ethylene chloride, carbon tetrachloride- trichloroethylene, and acetic acid-isobornyl acetate.

2-2 1-1 1-2 If the strength of the interaction between like molecules, 1-1 or 2-2 is greater than that between 1 and 2, the tendency will be to force both components into the vapor phase. This increases the pressure above what is predicted by Raoult’s law and is know as a positive deviation. Negative deviations occurs when the attractions between components 1 and 2 are strong. This may be visualized as a holding back of molecules that would otherwise go into the vapor state. acetone-chloroform system at 328 K chloroform-ethanol system at 318 K

Henry’s Law Henry found that the mass of gas m2 dissolved by a given volume of solvent at constant temperature is proportional to the pressure of the gas in equilibrium with the solution. where k is the Henry’s law constant, c2 is the concentration of the dissolved substance and x2 is the mole fraction of dissolved substance. In the limit of the dilution most liquid solvents obey Raoult’s law but under the same conditions the solute obeys Henry’s law.

Example 5.7 Dry air contains 78.084 mol % N2 and 20.946 mol % O2. Calculate the relative proportion of N2 and O2 dissolved in water under a total pressure of 1.000 bar. Henry’s law constant, k’, for N2 and O2 are 6.51107 Torr and 3.30107 Torr, respectively, at 25 C. Solution First find the partial pressure and from these, the mole fraction. The partial pressures are determined from Dalton’s law. From Henry’s law The relative proportions are This is approximately a ratio of 2 N2 to 1 O2.

Problem 5.35 The vapor pressure of pure ethylene dibromide is 172 Torr and that of pure propylene dibromide is 128 Torr both at 358 K and 1 atm pressure. If these two components follow Raoult’s law, estimate the total vapor pressure in kPa and the vapor composition in equilibrium with a solution that is 0.600 mol fraction propylene dibromide. Solution From Raoult’s law, we have for propylene dibromide (p), For ethylene dibromide (e), The total pressure is The mole fraction of propylene dibromide in the vapor phase is The mole fraction of ethylene dibromide in the vapor phase is

Problem 5.38 Methane dissolves in benzene with a Henry’s constant of 4.27105 Torr. Calculate methane’s molal solubility in benzene at 25 C if the pressure above benzene is 750 Torr. The vapor pressure of benzene is 94.6 Torr at 25 C. Solution Using Raoult’s law Assumption is valid. In 1000g of benzene,

5.5 Partial Molar Quantities Since V=V(n1, n2, n3, ….), at constant temperature and pressure, the increase in V for the general case is given by The increase in volume per mole of component 1 is know as the partial molar volume of component 1. It is given the symbol V1 and is written as Note: The molar volume of pure component 1 will be designated as V1*. This expression may be integrated under the condition of constant composition so that the Vi’s are constant. =dV =0

The mole fractions, xi is This expression is one form of the Gibbs-Duhem equation. For a two-component system by integration Note: Although these expressions are written in terms of volume, they apply equally well to any partial molar quantity.

Problem 5.29 The volume of a solution of NaCl in water is given by the expression where m is the molality. Assume that m nNaCl and that nH2O=55.508 mol, where VH2O*=18.068 cm3 mol-1. Derive an analytical expression for the partial molar volume of H2O in the solution. Solution

Relation of Partial Molar Quantities to Normal Thermodynamic Properties differentiating with respect to n1 Similarly, differentiating with respect to n1 Similarly,

The partial molar Gibbs energy is still a state function and may be represented as a function of temperature and pressure at constant composition. Therefore, upon differentiation, total differential Note: Every relation developed for a system of fixed composition expressed in molar quantities is applicable to each component of the system expressed in terms of the partial molar quantities.

5.6 The Chemical Potential A total differential of the Gibbs function G=G(T, P, nj) is written as partial molar Gibbs energy or chemical potential μi, for the ith component Similarly, the total differential of the internal energy may be written as

A similar treatment of the other thermodynamic functions shows that Note that the constant quantities are the natural variables for each function. The chemical potential is most commonly associated with the Gibbs energy because we most often work with systems at constant temperature and pressure. Under conditions of constant temperature and pressure, a single phase is closed dG=0 for a closed system Suppose that dni mol of component i are transferred from phase α to phase β in the closed system, without any mass crossing the system boundaries. The equilibrium condition would require that For a one-component system, the requirement for equilibrium is that the chemical potential of the substance i is the same in the two phases.

5.7 Thermodynamics of Solutions Raoult’s Law Revisited For any component i of a solution in equilibrium with its vapor, we may write If the vapor behaves ideally, the Gibbs energy for each component is given by Since μi=Gi/ni, we may write where is the chemical potential of the vapor when Pi=1 bar at the temperature T of the system. For a solution in equilibrium with its vapor, For pure liquid in equilibrium with its vapor, where the superscript* represents the value for the pure material.

An expression for the difference between the chemical potentials of the solution and the pure material The relative activity ai of a component in solution is just the ratio of the partial pressure of component i above its solution compared to the vapor pressure of pure component i at the temperature of the system. Based on Raoult’s law, a definition of an ideal solution

Example 5.8 Calculate the activities and activity coefficients for an acetone-chloroform solution in which x2=0.6. The vapor pressure of pure chloroform at 323 K is P2*=98.6 kPa and the vapor pressure above the solution is P2=53.3 kPa. For acetone, the corresponding values are P1*=84.0 kPa and P1=26.6 kPa. Solution The activities are The activity coefficients are determined from the definition fi=ai/xi where fi is the activity coefficient. Therefore, Note that these values are for a solution exhibiting negative behavior from Raoult’s law.

Raoult’s law, If the addition of component 2 lowers the vapor pressure of component 1, then the difference P*-P is the lowering of the vapor pressure. Dividing by P* gives a relative vapor pressure lowering, which is equal to the mole fraction of the solute (component 2). For a two-component system, This form of Raoult’s law is especially useful for solutions of relatively involatile solutes in a volatile solvent. where Wi is the mass and Mi is the molar mass. For a dilute solution, n2 may be neglected and we obtain

Problem 5.39 In a molar determination, 18.04 g of the sugar mannitol was dissolved in 100.0 g of water. The vapor pressure of the solution at 298 K was 2.291 kPa, having been lowered by 0.0410 kPa from the value for pure water. Calculate the molar mass of mannitol. Solution The correct value is 182.18 g mol-1.

Ideal Solutions mixing ideal solution ai=xi The partial molar volume of a component in an ideal solution is equal to the molar volume of the pure component. There is therefore no change in volume on mixing (ΔmixVid=0); that is, the volume of the solution is equal to the sum of the molar volumes of the pure components. Similarly, ideal solution ai=xi As with the volume, the change in enthalpy of mixing is the same as though the components in an ideal solution were pure. Thus there is no heat of solution (ΔmixHid=0 ) when an ideal solution is formed from its components.

The Gibbs energy of mixing ΔmixG for any solution is ideal solution ai=xi <0 For a two-component mixture, Figure 5.12 A plot of this function is shown in Figure 5.12, where it is seen that the curve is symmetrical about x1=1/2=0.5. For more complex systems, the lowest value will occur when the mole fraction is equal to 1/nwhere n is the number of components.

differentiating with respect to temperature The entropy of mixing for an ideal solution is independent of temperature and pressure. The driving force for mixing is purely an entropy effect. Since the mixed state is a more random state, it is therefore a more probable state. If ΔmixG were to be positive, the material would stay as distinct phase; this explains the lack of miscibility found in some liquid systems.

Chapter 5 Phases and Solutions