CHAPTER FIVE GIBBS ENERGY AND PHASE EQUILIBRIUM

Transcription

1 CHAER FIVE GIBBS ENERGY AND HASE EQUILIBRIUM able of Contents INRODUCION... 1 Learning Objectives...1 GIBBS ENERGY... 1 Basic Relationships...1 Constant emperature, Irreversible rocess... 1 Reversible rocess... 2 Gibbs Energy and Spontaneity...3 Standard Gibbs Energy of Reaction...4 Definition... 4 Example roblem 5.1: Spontaneity of the Reaction of Ammonium Nitrate... 5 Relationships for Constant ressure or Constant emperature...6 Gibbs Energy for Constant ressure... 6 Gibbs Energy for Solids and Liquids at Constant emperature... 7 Gibbs Energy for Ideal Gas for Constant emperature... 7 Gibbs Energy for Gas Mixture... 8 HASES OF SUBSANCES ressure-volume-emperature Surface hase Diagram Overview Application of the 2 nd Law Illustration Using CO Example roblem 5.3. hases at Equilibrium for Carbon Dioxide HERMODYNAMICS OF SAURAION CURVES Clapeyron Equation heory Example roblem 5.4: Density of Solid hase of Carbon Dioxide Clausius-Clapeyron Equation General Formulation Solution for Constant Δ β-αhm Dependency of Δβ-αHm on emperature and ressure Solution for Varying Δβ-αHm Saturated Vapor ressure of Water Background Enthalpy of Vaporization Varies with emperature Inverse of Enthalpy of Vaporization CHEMICAL OENIALS

2 roperties Using artial Derivatives Review Internal Energy Internal Energy, Enthalpy, Gibbs and Helmholtz Energies Chemical otential and Internal Energy Gibbs Energy and Chemical otential General Form of Change in Gibbs Energy Chemical otential Defined Using Gibbs Energy Application to Binary Components More Relationships between Gibbs Energy and Chemical otential Chemical otential for Ideal Gases Derivation Molar Fraction Form Integrated Form for Mixture of Ideal Gases (excluded from student handout) Chemical otential of Ideal Solutions Background More on the Chemical otential of ure Substance Definition of Ideal Solutions ALICAIONS OF CHEMICAL OENIALS Background Raoult s Law Introduction Definition Example roblem 5.7: Saturated Vapor ressure of Salt Water Henry's law heory Example roblem 5.7: Dissolved Oxygen Gas in Water Boiling-oint and Freezing-oint emperatures Boiling-oint emperature Example roblem 5.9: Boiling-oint emperature for Sugar Water and Salt Water Freezing-oint emperature Osmotic ressure of Ideal Solution heory van t Hoff Equation Example roblem 5.10: Osmotic ressure of Sugar Water ROBLEM ASSIGMENS

3 Working Notes 3

4 4

5 CHAER FIVE 1 GIBBS ENERGY AND HASE EQUILIBRIUM he only way to get revolutionary advances in science accepted is to wait for all old scientists to die (reflecting on advances in irreversibility in the late 19 th century). Max lanck, Nobel rize (1918) hysicist (and student of R. Clausius), Learning Objectives INRODUCION he key learning objectives for Chapter 5 are to: Understand the definition of Gibbs energy; Be able to compute changes in Gibbs energy for solid, liquid and gases and for chemical reactions, Be able to define saturation curves using Clapeyron and the Clausisi-Clapeyron relationships; Define chemical potential and apply the definition to ideal gases and ideal solutions; Use chemical potential to solve problems with colligative characteristics. Basic Relationships Constant emperature, Irreversible rocess GIBBS ENERGY Gibbs energy (sometimes called Gibbs free-energy or Gibbs energy function) was introduced in Chapter 4. Here it was used to determine the maximum non-expansion work for a system under constant pressure and temperature. It was also shown to be a thermodynamic potential that can be used to determine of system processes between two equilibrium states. Gibbs energy is invaluable property of systems with phase changes and chemical reactions. Concepts of Chapter 4 are expanded upon in this chapter to address direction and equilibrium conditions for these processes Bruce N. Wilson and the Regents of the University of Minnesota. All Rights Reserved. hese notes have not been professionally reviewed. Last Modified: May, Lecture Handouts

6 As given in Chapter 4, Gibbs energy is defined as G = H S = U + V S where G is Gibbs energy and H is enthalpy y defined in Chapter 2 as H = U+V. Gibbs energy is defined by exact differentials. Gibbs energies evaluated per unit mass and per mole are defined as G m = G n and G s = G m By differential our definition of Gibbs energy using enthalpy, we obtain where the change in entropy has been divided into components corresponding to reversible heat transfer (des) and internal generation (dis) as discussed in Chapter 4. For a constant temperature, the differential change in Gibbs energy can be written as As shown later in the chapter, Eq is used to determine the spontaneity or direction of chemical reactions, and it will be called the chemical-reaction form. Although the above relationships are valid for systems with varying pressures, our application to phase change and chemical reaction will often be limited to conditions of constant pressure. Reversible rocess Additional insight can be obtained by evaluating Gibbs energy in term of internal energy. By using H=U+V, Eq can be written as dg = du + dv + Vd d e S d i S Sd By using First Law of du + dv = δq + δwnon, we obtain dg = δq + δw non + Vd d e S d i S Sd = V d S d + δw non d i S where δq = des of Chapter 4 has been used. For internally reversible processes and only expansion work, dis= δwnon = 0. he infinitesimal change in Gibbs energy can then be written as Last Modified: May, Lecture Handouts

7 where the subscript rev has been used to indicate that this relationship is limited to reversible processes. Irreversibilities are often of primary interest when using Gibbs energy. However since dg is an exact differential, the change in Gibbs energy for irreversible processes can be computed from ΔGrev for the same endpoint conditions. Gibbs Energy and Spontaneity In Chapter 4, we defined the Second Law of thermodynamics as ΔS tot = ΔS + ΔS sur Although the use of change in the total entropy is useful in understanding the role of the Second Law in chemical reactions, it is more convenient to assess reactions using the properties of the system only. In Chapter 4, we have previously evaluated the change in entropy of the surroundings as where q is by the definition of heat transfer into the system that has been evaluated using change in enthalpy of a constant pressure system and where the temperature of the surrounding equal of that at the boundary of the system. he change in total entropy can then be written as ΔS tot = ΔS ΔH ) ΔS tot = ΔS ΔH = ΔG ) where the last equality is valid for a constant temperature system as given by Eq We therefore conclude that for a constant temperature and pressure system: he change in Gibbs energy of a system is proportional to the overall change of the entropy of the system and the surroundings. Evaluation of spontaneity is much easier to evaluate using Gibbs energy than total entropy because it is solely dependent on the change in the Last Modified: May, Lecture Handouts

8 system. Let s consider the implications of the above equation for chemical reactions. A general form of the change in entropy involving chemical reactions can be written as: ΔS tot = ΔS sur + ΔS = ΔS sur + (ΔS other + ΔS r ) ) where the change in entropy of the system has a component for chemical reaction (ΔSr) and a component for other possible changes in entropy (ΔSother), such as radiation for photosynthesis. Here we are only interested in changes related to the reaction itself and use the condition that Sother=0. For conditions of constant pressure and temperature, we then obtain ΔS tot,r = ΔS sur + ΔS r = ΔS r ΔH r = ΔG r ) Graphical illustration of spontaneity using Gibbs energy is shown below. he progression of change moves in the direction of ΔG = 0 regardless if the initial condition is located to the left or right side to the final equilibrium value. Figure 5.3. rogression of Change in the System Using Stot or G. We now have the following criteria to determine the spontaneity using Gibbs energy ΔStot,r = 0 ΔGr = 0 Equilibrium ΔStot,r > 0 ΔGr < 0 Spontaneous Reaction ΔStot,r < 0 ΔGr > 0 Not Spontaneous Reaction Standard Gibbs Energy of Reaction Definition As previously discussed, the change in total entropy for a chemical reactions (only) in a system can be written as function of the change in Gibbs energy evaluated for conditions of constant pressure and temperature as Last Modified: May, Lecture Handouts

9 In previous chapters we defined the standard enthalpy and standard entropy of reactions under constant pressure and temperature. We can then define the standard Gibbs energy of reaction corresponding to a pressure of o = 100 ka and temperature of directly as Δ r G m o = ΔG r Δξ = Δ r H m o Δ r S m o ) where we are using the standard enthalpy of reaction defined in Chapter 2 as a) o o o r Hm = ( ν j f Hmj) prod ( ν j f Hmj) react j j and the standard entropy of reaction defined in Chapter 4 as Δ o r S m = υυ jj S o mj pppppppp j υυ jj SS oo mmmm rrrrrrrrrr j b where υj are the stoichiometric coefficients in the chemical reactions,δ ff HH oo mmmm, is the standard enthalpy of formation,ss oo mmmm is the standard molar entropy as given in the thermodynamic data handout and Δξ is the molar extent of reaction as defined and used in Chapters 2 and 4. Example roblem 5.1: Spontaneity of the Reaction of Ammonium Nitrate roblem Definition. Ammonium nitrate (NH4NO3 ) is widely used as a fertilizer. Let s determine if the chemical reaction shown below is spontaneous (NH4 + = Ammonium, NO3 - = Nitrate) for at a constant temperature of 25 C and constant pressure of 100 ka. NH 4 NO 3( S ) H 2O NH + 4 ( aq) + NO 3 ( aq) hermodynamic Data Handout. We obtain for a pressure of 100 ka and temperature of 25 C: Values from our hermodynamics Data Handout. Compound ΔfH o (kj mol-1) S o m (J K -1 mol -1 ) NH4NO3 (s) NH4 + (aq) NO3 - (aq) Last Modified: May, Lecture Handouts

10 Solution- Using Gibbs Energy. Standard reaction Gibbs energy has previously been defined as he standard reaction enthalpy is defined as o o o r Hm = ( ν j f Hmj) prod ( ν j f Hmj) react j j 25C r H = (1)( ) + (1)( 205) (1)( ) = kj mol o m and standard reaction entropy as 25 Δ C o r S m = υυ jj S o mj pppppppp 25C r S o m j υυ jj SS oo mmmm rrrrrrrrrr = (1)(113.4) + (1)(146.4) (1)(151.08) = J K We then obtain the standard reaction Gibbs energy as j -1 mol -1 1 = kj K -1 mol 1 25C r G o m = 25C r H o m 25C r S o m = kj mol ( ) K kj K mol Since the standard reaction Gibbs energy is negative, we conclude that the reaction is spontaneous. Relationships for Constant ressure or Constant emperature Gibbs Energy for Constant ressure We have previously derived dgrev = V d Sd for internally reversible processes with expansion work only (Eq ). For constant pressure, we then obtain Last Modified: May, Lecture Handouts

11 Since Gibbs energy is defined by exact differential, the change for reversible process is identical to that of an irreversible process for the same endpoint conditions. If the entropy is approximately constant over relatively minor changes in temperatures, we then obtain G f ΔG = ΔG rev = dddd G i or in terms of molar Gibbs energy f = S dddd S ( f i ) i G mf ΔG m = dd GG nn = G m ( f ) G m ( i ) S GG mmmm n ( f i ) = S m ( f i ) he approximation of constant entropy over a range of temperature is used later for solids, liquids, and gases to provide insight into the phases of a substance. he above equation can be rearranged as Gibbs Energy for Solids and Liquids at Constant emperature Likewise for dgrev = Vd - Sd, we obtain for a constant temperature If the volume is approximately constant over the pressure of interest, we obtain G f ΔG = ΔG rev = dddd G i or in terms of molar Gibbs energy f = V dddd V ff i i G mf ΔG m = dd GG nn = G m ( f ) G m ( i ) V m ( f i ) GG mmmm he approximation of a constant volume is reasonable for liquids and solids. For these phases, the molar Gibbs increases linearly with pressure, that is, G m ( f ) = G m ( i ) + V m ( f i ) ) Gibbs Energy for Ideal Gas for Constant emperature Last Modified: May, Lecture Handouts

12 For reversible process at constant temperature of a gas, we can write the reversible form of Gibbs energy as where V = nr/ from the ideal gas equation. We now integrate to obtain G f f ΔG = ΔG rev = dddd = nr dddd G i he molar Gibbs energy is then defined as i = nr ln f i G mf ΔG m = dd GG nn GG mmmm = G m ( f ) G m ( i ) = R ln f i he change in molar Gibbs energy is often evaluated relative to the standard-state pressure (i = o = 100 ka). We then obtain G m = G m 0 + R ln ( o ) Gibbs Energy for Gas Mixture In Chapter 3, properties of a gas mixture related to the First Law were determined from the individual components. We are also interested in determining the Gibbs energy for a gas mixture. Once again for air, the mixture consists of N2, O2, H2O, CO2 and other gases. Similar to concepts inherent in Dalton s rule, contributions to Gibbs energy from individual components are assumed to act independently of each other. he Gibbs energy for each of the components are evaluated relative to the standard-state pressure of o =100 ka. his approach is shown in the following schematic. Last Modified: May, Lecture Handouts

13 Figure 5.5. Gibbs Energy for Mixture of Gases. We can easily evaluate the change in Gibbs energy for a jth component with a change of pressure from oo to j by using our definition of molar Gibbs energy developed in the previous section for a constant temperature. For the jth gas, we then have ΔGG mmmm = G mj G 0 mj = R ln ( j o ) where the subscript j refers to the jth gas component, j is the partial pressure of jth gas, and where GG oo mmmm corresponds to the component jth molar Gibbs energy at pressure o. Because our derivation is based on the solution for a single gas, GG oo mmmm also corresponds to a pure substance. If the other components of the mixture can be treated independently of each other, then the subscript j corresponds to any of the components. he total Gibbs energy is defined by adding the components from the individual components acting independently of each other. By multiplying the molar Gibbs energy of the components by the corresponding number of moles, we have kk G t = nn jj GG mmmm jj=1 kk = nn jj jj=1 G 0 mj + R ln j o = nn jj kk jj=1 G 0 mj + R ln ff mmmm tt o where nj is the number of moles for each component and where we have used Dalton s law of Chapter 3 to define the partial pressure as j = fmj t. he molar Gibbs energy for the total mixture of gases is then defined as G mt = GG tt = nn jj nn tt nn tt kk jj=1 G 0 mj + R ln j o = ff mmmm kk jj=1 G 0 mj + R ln j o where nt =n1+n2+ +nk is the total number of moles of the mixture of gases and fmi is the molar fraction as used in Chapter 3. Last Modified: May, Lecture Handouts

14 ressure-volume-emperature Surface HASES OF SUBSANCES hree of the most important properties of thermodynamic systems are pressure, volume, and temperature. For ideal gases, we have a theoretical relationship to determine the third value from two known values. For example, the pressure is defined from a given temperature and specific volume (Vs=V/m) as where the universal gas constant ( R ) and molar mass (mm) are known for a given substance. Clearly, pressure is related to temperature and specific volume as a three dimensional plot as shown below. he general representation of the relationship between temperature, pressure, and specific volume is given by pressure-volume-temperature (-V-) surfaces. A typical surface is shown below for a substance which contracts on freezing (not water which expands). he figure is not to scale. We will first discuss the features in the left-solid figure. Solid, liquid and gas (vapor) phases plot as surfaces. wo phases coexist during phase changes. Singlephase regions are separated by two-phase regions: solid-liquid (small zone), solid-gas (sublimation) and the liquid-vapor. Also shown in the figure is the critical state point. his point corresponds to a pressure of c, specific volume of Vsc and a temperature of c. If a substance is at temperature greater than the critical temperature, then it cannot condense into a liquid, no matter how high the pressure. For > c and > c, there is no clear distinction between what is called a liquid Last Modified: May, Lecture Handouts

15 and what is called a gas. he phase in this region is simply called a fluid. Although the boundaries are not clear in the figure, the liquid phase corresponds to the region where the temperature is less than the isotherm line of c and greater than the triple point temperature. Solid-Liquid Liquid-Saturation Line Fluid Vapor-Saturation Line Critical State c Isobar Solid Liquid Liquid-Gas Gas f d e g h i riple State Isotherm Solid-Gas j a b V s d V s Figure 5.7. hree Dimensional Representation of hases. In the right-sided figure, we have shown an isobar line (f-g-h-i) of equal pressure and isotherm lines (e.g., d-e) of equal temperature. Saturation state refers to any point that lies on a line separating the single-phase from a two-phase region. he a-c line separates the liquid phase from the liquid-vapor region is called the liquid-saturation line corresponding to saturated liquid state. Likewise, the c-b line is the vapor-saturation line and corresponds to the saturated-vapor state. he triple state line corresponds to a temperature and pressure where all three phases exists. his line is parallel to the Vs plane corresponding to a constant, characteristic pressure and varies with the specific volume. For water, the triplestate line corresponds to a temperature of 0.01 C (exactly K because it is used to define the Kelvin scale) and pressure of ka. he solid-liquid region is typically small, where as the liquid-vapor region is considerably larger. Distance a-b is typically 1000 times larger than j-a. Let s consider the region corresponding to both liquid and vapor phases shown by line d-e in the above figures. Discontinuity in isotherms occurs between saturation states of liquid and vapor. hese discontinuities correspond to constant pressure with respect to the specific volume, that is, ( / Vs)=0. Let s consider water liquid that moves along the line h-i because of a temperature increase. Line h-i corresponds to constant pressure that we will select to equal that of the atmosphere. When the water reaches the liquid-saturation line at point i, it begins to boil. he temperature and pressure is constant during the boiling pressure. Last Modified: May, Lecture Handouts

16 he specific volume (and therefore density) at point i is that of liquid. he boiling process will eventually completely convert the liquid to a vapor corresponding to the vapor-saturation line. he vapor specific volume is much larger than that of liquid water. he boiling point is discussed in greater detail in the next section. Shown below is the -V- diagram with a projection upon the - plane and a projection upon the -V plane. he -plane projection is called the phase diagram. Note that the liquid-saturation and vapor-saturation line plot as a single line on this plot. Additional discussion of the phases of substances will be given for these two 2-D diagrams. ressure-volume Diagram Solid-Liquid Liquid-Gas hase Diagram Solid-Liquid V s Vaporization Fusion Liquid-Gas Solid-Gas Sublimation Figure 5.8. rojection onto - and -Vs lanes. V s hase Diagram Overview he projection of the V diagram upon the - plane is called the phase diagram. he liquid-saturation and vapor-saturation line plot as a single line on this plot. Since these two lines are plotted on top of each other it is simply called the liquid-vapor saturation line. he pressure corresponding to this line was called the saturated vapor pressure in Chapter 3. In the terminology of phase diagrams, it is more often referred to as the equilibrium vapor pressure, or even more simply as vapor pressure. he triple state line in the V diagram becomes a point on the phase diagram. A typical curve is shown below for a substance that Last Modified: May, Lecture Handouts

17 contracts with fusion. Also shown is curve indicating the shape of a solid-liquid saturation line for a substance like water that expands with fusion. Let us consider the isobaric path shown in the figure from oint A to oint C. At the temperature at A, the gas is the most thermodynamic stable phase. As the temperature is reduced, we cross the liquid-vapor saturation line to a thermodynamic stable liquid state at B. If the temperature is further reduced we move across the solid-liquid saturation curve to a stable state as a solid identified as C. Figure 5.9. Key Features of a hase Diagram he normal freezing and boiling points are shown in the above figure corresponding to 1 atm ( ka). Boiling point is the temperature at which the equilibrium vapor pressure equals the environmental pressure surrounding the liquid. hysical insight into the boiling point can be obtained by considering a simple description of the impact of internal and external pressures acting on water bubbles. Bubbles quickly collapse in the early stages of formation if the vapor pressure within them is less than the environmental pressure. However, if the internal pressure is greater than the atmospheric pressures, bubbles will form and then burst when the net pressure force exceed surface tension forces at the water-air interface. he saturated vapor pressure of water at 100 C is given by CRC (2009) Handbook of Chemistry and hysics of ka (not 1 atm = ka). Although we can shift the normal boiling point to correspond to this pressure, this adjustment lies within the measurement accuracy of boiling point temperatures (NIS reports a measurement accuracy of ± 0.04 C). he normal melting temperature is very close to 0 C. he standard freezing Last Modified: May, Lecture Handouts

18 and boiling points correspond to a pressure of 1 bar (100 ka). If the boiling point temperature of water is 100 C at = 1 atm, then the standard boiling point is less than 100 C. For purist, the standard boiling point is more accurately given as 99.6 C. his adjustment is small. he standard freezing and boiling points for water are often approximated by 0 C and 100 C, respectively. Also shown in the above figure is the critical point. As previously discussed, the critical point corresponds to the largest temperature and pressure where the liquid phase exists (the smallest temperature and pressure correspond to the triple point). For pressures and temperatures greater than the critical point values, it is difficult to distinguish between liquids and gases (but not solids). Application of the 2 nd Law In this course, we are largely interested in using principles of thermodynamics to understand the observed features of the phase diagram. Let s imagine a system at the triple point pressure and temperature. We now wish to change the system pressure and temperature and then determine the phase as the system reaches equilibrium for this new state. his concept is shown in the schematic below. In Chapter 4, we suggested that a new equilibrium condition was defined when the maximum Stot was obtained. he proper phase is then defined by the phase corresponding to maximum entropy. Insight into why the phase is gas, liquid, or solid for a given pressure and temperature can also be obtained from Gibbs energy. For the transition to a new state under conditions of constant temperature and pressure, Gibbs energy is ΔG = - ΔStot. Since Gibbs energy is an exact differential, we don t need to worry about the details of the non-equilibrium path here. he phase corresponding to the maximum Stot also correspond to the minimum Gibbs energy. Once again, Gibbs energy is defined using system properties and is therefore preferred. Although the minimum Gibbs energy determines the most stable phase state, the rate of Last Modified: May, Lecture Handouts

19 progression is so slow for some substances that time required to achieve this phase may take many years. Illustration Using CO2 We will use carbon dioxide to illustrate the role that Gibbs energy can play in determine the phase of the substance at equilibrium with a given temperature and pressure. he phase diagram for CO2 is shown below. his diagram is based on the analysis of Span and Wagner (1996). For the phase diagram, it is clear that liquid CO2 only occurs at relatively large pressures (approximately 5 times greater than normal atmospheric conditions). Solid CO2 is frequently called dry ice because it transforms itself to a non-solid phase without becoming wet (liquid). he sublimation process occurs under normal atmospheric conditions and is eyecatching. Note that the fluid region does not extend to the solid region. Figure hase Diagram for Carbon Dioxide. he triple point temperature of CO2 is i = C and the triple point pressure is i = 518 ka = 518 kj m -3. he molar Gibbs energies for solid, liquid and gas phases are equal at the triple point and are represented by Gm(i,i) in Fig However, different Gibbs energies are obtained among the phases for other pairs of temperature and pressure values. As previously shown in Fig. 5.10, the phase corresponding to the minimum of these energies is the equilibrium state that is obtained by the system for a particular temperature and pressure pair. his approach is illustrated for carbon dioxide using the example problem given below. Last Modified: May, Lecture Handouts

20 Example roblem 5.3. hases at Equilibrium for Carbon Dioxide roblem Statement. From a stable, initial phase of CO2 corresponding to the triple point, you are interested in determining the phase at a final equilibrium state caused by changes in pressure and temperature. For this problem, you are required to determine the phases for new equilibrium states corresponding to a temperature of -40 C and pressures of (1) 6 bars (600 ka), (2) 50 bars (5000 ka) and (3) 1000 bars (100,000 ka). Solution Approach. You will calculate the molar Gibbs energies for each of the three phases at the final state temperatures and pressures. he substance will move toward the phase corresponding to the minimum Gibbs energy, and therefore the new phase is defined by the smallest of these molar Gibbs energies. At the triple point, all three phases exist together and therefore they all have the same molar Gibbs energy. he progression of change from the triple state point to the temperature and pressures of interest is divided into two reversible paths (any pathway between the two endpoints can be taken because Gibbs energy is defined by an exact differential). he first path is at a constant pressure (triple point pressure of i) from the triple temperature of i= C to f = -40 C. For this constant-pressure path, we can approximate the change in Gibbs energy using Eq , that is, where Gm(i,i) is the molar Gibbs energy at the triple point, Gm(f,i) is the molar Gibbs energy at the temperature of f = -40 C and the triple point pressure of i, and Sm is the molar entropy at i. he second-path computes the change in Gibbs energy for a fixed temperature (=-40 C) at our three different possible pressures. We have previously shown that the change in Gibbs energy with pressure (for constant temperature) is defined for a solid, liquid and gas as G ms ( f, ) = G ms ( f, ii ) + V ms ( i ) = G m ( i, i ) S ms ( f i ) + V ms ( ii ) G ml ( f, ) = G ml ( f, ii ) + V ml ( i ) = G m ( i, i ) S ml ( f i ) + V ml ( ii ) G mg ( f, ) = G mg ( f, ii ) + R f ln ii = G m ( i, i ) S mg ( f i ) + R f ln( i ) where the subscripts of s, l and g correspond to solid, liquid and gas phases. hermodynamic Data. Our relationships for Gibbs energy were derived assuming constant molar entropies and volumes and therefore representative values need to be determined for our idealized paths. hermodynamic data for gas (using = 230 K, = 5 bar) and liquid (using =230 K and = 20 bar) phases were determined directly from Span and Wagner (1996). Molar entropy for solid phase was estimated by adding the molar entropy of freezing Last Modified: May, Lecture Handouts

21 to that of the liquid phase (at =217 K and = 20 bar). he density of dry ice is taken as 1512 kg m -3 and is used to estimate the solid molar volume. Liquid: Sml = 120 J K -1 mol -1 ;Vml = 38.9 cm 3 mol -1 = m 3 mol -1 Gas: Smg = 190 J K -1 mol -1 ; R = (8.314 J K -1 mol -1 )( ) K = J mol -1 We also need to define the molar Gibbs energy corresponding to the triple point, that is, Gm(i,i). Since Gm(i,i) is the same for all three phases, its value only shifts the curve uniformly and doesn t change the location of the minimum value. We will assign Gm(i,i) an arbitrary value of 5000 J mol -1 to simplify the interpretation of results (avoid negative numbers). Solution. he molar Gibbs energy correspond to a change in temperature from i= C = K to = -40 C = K can be computed for each phase using our first path as G ml ( f, i ) = ( ) = 3010 J mol 1 = 3.01 kj mol 1 G mg ( f, i ) = ( ) = 1850 J mol 1 = 1.85 kj mol 1 By using these values, our estimates of specific volumes, and i = ka = kj m -3, we can compute the molar Gibbs energy (kj mol -1 ) for each of three phases at different pressures as: G ml ( f, ) = ( 518.5) G mg ( f, ) = ln Last Modified: May, Lecture Handouts

22 Solutions obtained for these equations are shown below for our three different pressures. he minimum Gibbs energy corresponds to the stable state from the Second Law and are shown in the final column. ressure Gas Liquid Solid Stable Bar kj m -3 kj mol -1 kj mol -1 kj mol -1 hase Liquid Solid Discussion. hese calculations can be repeated for many different pressures. A graphical solution using the three equations directly is given in the figure below. Intersections of the curves correspond to phase change pressures. hese points are also shown on the phase diagram for CO2 of Fig Even with approximations used to estimate molar Gibbs energy (molar volume and entropies vary over the range in temperatures and pressures), the intersection points lie closely to the solid-liquid and liquid-vapor saturation curves. Note that the molar Gibbs energies are equal for the saturation curve values. his concept will be used later to define the slope of the solid-liquid, solid-gas and liquid-gas saturation curves. Figure Molar Gibbs Energy for emperature of -40 C. HERMODYNAMICS OF SAURAION CURVES Last Modified: May, Lecture Handouts

23 Clapeyron Equation heory Let s explore more carefully the lines separating the different phases using the principles of thermodynamics and equilibrium. We are particularly interested in defining the relationship between pressure and temperature for these lines, that is, = (). It is now our task to find the form of the function (). Let s consider the following hypothetical phase diagram with two phases of α and β. he phases α and β are in equilibrium with each other at all points on the phase line. We are defining the β phase as that phase corresponding to greater disorder (an increase in entropy). Based on the example problem in the previous section, we concluded that Gibbs energies of both phases are equal at the solid-vapor, solid-liquid and liquid-vapor saturation curves, that is, Let s now consider a change in and such that we maintain equilibrium as shown in the above figure. We then obtain a change in Gibbs energy along our saturation curve for changes in d and d of dg = dg β α 5.3.2) he phase transitions at our equilibrium condition are reversible. By using our previous derived definition of the change in Gibbs energy (with d and d) for internally reversible processes of dgrev = Vd Sd, the change in Gibbs energies for both phases can be defined as Last Modified: May, Lecture Handouts

24 V β d S d = V d S d 5.3.3) We can rearrange terms as β α α dddd VV ββ dddd VV dddd αα dddd = S β S α = ΔS β α 5.3.4) Since phase transition occurs at the constant temperature and pressure, we have previously shown in Chapter 4 that ΔS β α = Δ β αh where Δ ββ αα HH is the enthalpy of transition from α to β phases. We can now proceed in our derivation using either ΔS or ΔH. Since ΔH varies more slowing with temperature than ΔS, the ΔH form is usually preferred. By using Eq , and by dividing both sides by an equal number of moles for both phases, we obtain 5.3.5) dddd VV mmmm dddd VV dddd mmmm dddd = Δ β αh m = H mβ H mα 5.3.6) By rearranging equations, we obtain the following form of Clapeyron equation dddd dddd = Δ ββ αα HH mm V mβ V mα = Δ ββ αα HH mm ΔV m 5.3.7) We can integrate directly if we assume Δ β-αhm and ΔVm are constant. his is more reasonable for solid liquid than for a transition with a gas phase. For special case of constant ΔVm and a constant Δ β-αhm = Δ fushm, we obtain or fus Hm d d = 5.3.9) V i fus m i Hm = i + ln( ) ) V m i Clapeyron equation is applicable for phase transitions of solid liquid. For transitions involving gases, ΔVm cannot be treated as constant. Solutions for solid gas and liquid gas transitions are given in the next section. In the next section, we will also consider a nonconstant ΔHm. Example roblem 5.4: Density of Solid hase of Carbon Dioxide roblem Definition: From the previous carbon-dioxide phase diagram, we approximated the pressure and temperature for the solid-liquid saturation curve are i =1 = -15 C ( K), Last Modified: May, Lecture Handouts

25 i = 1 = 2340 bar (234,000 kj m -3 ) and =2 = C ( K), =2 = 2573 bar (257,300 kj m -3 ). Estimate the difference in molar volume between the liquid and solid phases. Also estimate the density of dry ice for this molar volume difference. hermodynamic Data: You can use the following additional thermodynamic data: mm = 44 g mol -1 = kg mol -1 ρliq of liquid CO2 (Span and Wagner): 1306 kg m -3 (=2000 bar, = -13 C) 57 CC HH mm (NIS data): 26.1 kj mol -1 Δ ssssss Δ vvvvvv 57 CC HH mm (NIS data): 16.4 kj mol -1 CC mmmm,ss (wiki CO2 Data): 54.6 J K -1 mol -1 (= -85 C) CC mmmm,ll (Span and Wagner): 39.6 J K -1 mol -1 (= -57 C) Solution: Let s first estimate the enthalpy of fusion for dry ice using relationships from Chapter 2 (assuming negligible change with pressure): 13 Δ C 57 fus HH mm = Δ C sub HH mm + CC pppp,ll CC pppp,ss Δ 13 Δ C fus HH mm = ( ) 13 ( 57) = 9.0 kkkk mmmmll 1 By rearranging Clapeyron equation (Eq ) to solve for the change in molar volume, we obtain 9 kkkk ln ΔV m = mmmmmm kkkk = kkkk 5.226xx10 6 mm3 mmmmll 1 = ccmm3 mmmmll 1 mm mm 3 We can then define the density of the solid phase of CO2 as V m,solid = V m,liq ΔV m = 1 ρρ llllll ΔVV mm V m,solid = mm3 kkkk xx10 5 mm 3 mmmmll 1 = 2.846xx10 5 mm 3 mmmmll kkkk mmoooo ρ solid = 1 mol kkkk = V m 2.846xx = 1546 kg m 3 mm3 mmmmmm Last Modified: May, Lecture Handouts

26 he density of the solid phase of CO2 is larger than the density of its liquid phase. Opposite trends are obtained for water as shown in Homework roblem 5.9. Clausius-Clapeyron Equation General Formulation he general framework of the Clapeyron equation is valid for all types of phase transitions. We need to be careful, however, in the evaluation of ΔVm and ΔHm. he assumption of constant ΔVm is not a good approximation for phase transitions involving gases. For solid-gas or liquid-gas phase transitions, we can make two reasonable approximations to obtain the very useful Clausius-Clapeyron equation. he first approximation is that molar volume of solids or liquids is negligible small compared to the much large molar volume of gases. As shown in Chapter 3, the density of air is approximately 1000 smaller than that for liquid water. For the second approximation, we will estimate Vgas using the ideal gas equation, that is, Vgas = nr/. We can then represent ΔVm as, By combining the above approximation with the Clapeyron equation, we obtain for our solidvapor and liquid-vapor saturation curves d d β αhm β αhm = = ( ) ) R ( ) R 2 which is called the Clausius-Clapeyron equation. Once again, Δ β-αhm = Δ vaphm for phase transition of liquid-gas transitions and Δ β-αhm = Δ subhm for phase transitions of solid-gas. Solution for Constant Δ β-αhm he liquid-gas and the solid-gas saturation curves can now be easily defined as a function of temperature for the special case of a constant Δ β-αhm. By rearranging Clausis-Clapeyron equation, we obtain i d = i ( β α R H m d ) 2 = β α R H m i d ) or Last Modified: May, Lecture Handouts

27 β αhm 1 1 ln( ) = ( ) ) i R i We can then obtain for known values at point i and temperature of as = exp( i β α R H m ) 1 1 i ) ) or from a known pressure and values at point i as: R 1 ( )ln( ) = H β α m i 1 i ) or 1 R 1 ( )ln( ) β αhm i i = ) Dependency of Δβ-αHm on emperature and ressure Let s evaluate the change in the molar enthalpy of transition assuming a general form that H = H(,). he change in molar enthalpies of phase transitions can be written as d Δ β α H m = d(h mβ HH mmmm ) = dh mβ ddhh mmmm a) By using multivariate calculus, we can evaluate the change in molar enthalpies as dh m = H m dddd + H m dddd b) which is applicable to both β and α phases. By using the chain rule of calculus of Chapter 1, we obtain H m = H m HH = µ H H m b) where μh = ( / )H is the Joule-homson (also called Joule-Kelvin) coefficient. Since from Chapter 2, Cpm = ( Hm/ ), we obtain the following general relationship for change in enthalpy dh m = µ H CC pppp dddd + CC pppp dddd c) Let s apply Eq c to each of the components in Eq a to obtain Last Modified: May, Lecture Handouts

28 d Δ β α H m = µ H CC pppp dddd + CC pppp dddd ββ µ H CC pppp dddd + CC pppp dddd αα = µ H CC pppp ββ µ H CC pppp αα dddd + (CC pppp,ββ CC pppp,αα ) dddd d) By using d = (d/d) d and by integrating from an initial temperature (o) and to final temperature (), our general relationship for a varying ΔHm is then defined as Δ β α H m = Δ o β α H m + b ( oo ) a) where for vaporization b = CC pppp,ββ CC pppp,αα κκ = (CC pppp,gggggg CC pppp,llllll) κκ b) Variables with an overbar are used to represent average values between and o. he terms in Eq b are defined as C pm = CC oo ppppdddd oo c) dddd μμ HH CC pppp dddd dddd dddd oo μμ HH CC ββ pppp dddd dddd oo αα κ = oo oo d) As discussed in Chapter 2, the change in enthalpy with pressure for an ideal gas or for an incompressible liquid is zero corresponding to μh = 0. herefore κp = 0 for ideal gases and incompressible liquids. We then obtain the adjustment in standard enthalpy of vaporization as a function of temperature as developed in Chapter 2 (see Eq ). Assuming behaviors corresponding to ideal gases and incompressible liquids are often inappropriate for the large pressures given in phase diagrams. Homework roblem 5.9 further explores adjustments in enthalpies for different pressures. Solution for Varying Δβ-αHm Let s now revisit the Clausius-Clapeyron equation for the saturation curves of solid-vapor or liquid-vapor using the linear relationship. We then have d d β αhm β α o Hm + b( o ) = ( ) = ( ) ( R 2 R 2 which can be rearranged as ) ) 1 d = i ( o β α H m + b( o ) d ) R 2 = o β α H m R b o i d + 2 b R i d ) he solution is then obtained as Last Modified: May, Lecture Handouts

29 ln( i ) = ( o β α H m R bo ) 1 = ( o β α 1 H i m R b + ( )ln[( ) R i bo ) 1 1 i + ln[( ) i b / R ) or / ln[ ( / ) i b / R i ] = ( o β α H m R b o ) 1 1 i ) or i = ( ) i b / R exp ( o β α H m R b o ) 1 1 i ) Saturated Vapor ressure of Water Background In Chapter 3, the saturated vapor pressure of water was an important variable in the analysis of mixtures of dry air and water vapor. Saturated vapor pressure was used to compute the vapor pressure from known relative humidity and air temperature data and the vapor pressure from wet bulb and dry bulb temperatures measured with psychrometers. It was also used in Chapter 3 to determine the change in relative humidity for a parcel of air rising in the atmosphere. he saturated vapor pressure corresponds to the liquid-gas saturation curve corresponding to a phase diagram. In terms of the notation in this chapter, the saturation vapor pressure is defined as We are now able to derive relationships between saturated vapor pressure and temperature using the Clausius-Clapeyron equation. he first relationship will be derived assuming a constant molar enthalpy of vaporization. A second relationship is derived assuming a molar enthalpy of vaporization that varies linearly with temperature. A comparison of the two relationships is also given in this section. In Chapter 3, we used the more rigorous relationship based on a non-constant enthalpy of vaporization. Enthalpy of Vaporization Varies with emperature Last Modified: May, Lecture Handouts

30 Let s evaluate the saturated vapor pressure relationship using the temperature dependent enthalpy of vaporization. It is this form that we used in Chapter 3 for determining the saturated vapor pressure of water. he Clausius-Clapeyron equation can be written using the notation for saturated vapor pressure as: e s o vaphm b 1 1 s R o = e ( ) b / i exp[ ( ) ] ) i R i where b is defined by Eq b. We will again select i = 0.01 C and use the corresponding saturated vapor pressure of e s i = ka. We will also evaluate the enthalpy of vaporization as vap Hm = 44 kj mol -1 = 44,000 J mol -1. Although our theoretical framework suggests a b = J K -1 mol -1 at 25 C, the saturated vapor pressure is better represented using b = - 40 J K -1 mol -1. Our framework neglected the molar volume of water, assumed an ideal gas equation for Vm,gas, and neglected changes in enthalpy of phase transition with pressure. We will use b/r = By using these values, we can solve for saturated vapor pressure using 25C e s = ( ) (298.15) exp ( )( ) ) By using = exp( ), we can simplify as or e s = e s = exp( ) exp ) exp ) which is the recommended relationship given in Chapter 3 for determining the saturated vapor pressure at different temperatures. Inverse of Enthalpy of Vaporization If we have a known saturated vapor pressure, we may wish to determine the inverse of our relationship for the enthalpy of vaporization so solve the corresponding temperature. his was important in Chapter 3 to determine the dew-point temperature. Unfortunately, the above equation is too complex to be rearranaged to solve directly for temperature for a given saturated vapor pressure. As an approximation, we will estimate the temperature in the first power term using the solution obtained for constant vaporization enthalpy. We can then rearrange terms as Last Modified: May, Lecture Handouts

31 5292 s s ln(e ) s exp = e ( ) = e ( ) ) s ln(e ) which can be rearranged as or = s s ) ln(e ) ln(19.373) ln[18.88 ln(e )] = ln(e s ) ln(18.88 ln(e s )) ) he above equation was used in Chapter 3 to solve for dewpoint temperatues. Its accuracy is evaluated in the next section. As previously discussed, water boils when the saturated vapor pressure equals the surrounding environmental air pressure. Eq can therefore be used to determine the boiling point temperature using e s = E. roperties Using artial Derivatives Review Internal Energy CHEMICAL OENIALS Useful thermodynamic characteristics can be related to the partial derivatives of the properties of systems. his has already been illustrated in Chapter 1 where we considered internal energy as a function of fundamental properties of mass, temperature, volume, and pressure. It is quickly reviewed here to provide the foundation for additional theoretical work. For a fixed mass (for example, no chemical reactions), internal energy can be taken as a function of two of the remaining three properties that are interrelated by the equation of state (such as the ideal gas equation). By using temperature and volume, we obtained U tttttt = U(, V) 5.4.1) where is temperature and V is volume. By using the definition of total differential, we then obtained where each of the partial derivatives are the change in the internal energy by holding volume or temperature constant. We further defined ( U/ )V as the heat capacity at constant volume (Cv) and then obtained Last Modified: May, Lecture Handouts

32 du tpr = CC v d + U V dv 5.4.3) where we can clearly compute, as previously discussed, the change in internal energy as dutpr = Cv d for a constant volume (dv=0) or non-constant volume if, as for ideal gases, ( U/ V) equals zero. Internal Energy, Enthalpy, Gibbs and Helmholtz Energies Entropy is one of the most important properties of thermodynamic systems. In Chapter 4, we combined the First and Second laws (i.e., Gibbsian equation for reversible processes) to define the change of internal energy (ds = dutpr + dv) and of enthalpy (ds = dhtpr Vd) as du tpr = ds dv dh tpr = ds + V d 5.4.4a) 5.4.4b) Internal energy and enthalpy are more fundamentally tied to entropy rather than temperature and therefore internal energy and enthalpy are represented as Utpr = U(S,V) and Htpr=H(S,), respectively. By evaluating the total differentials for these functions, we obtain du tpr = U S V,n j ds + U V S,n j dv dh tpr = H S p,n j ds + H S,n j d 5.4.5a) 5.4.5b) Where the partial derivatives are evaluated for fixed number of moles in the system for our evaluation of internal energy and enthalpy without phase transitions or chemical reactions. By using the relationships for reversible processes previously derived for Helmholtz energy in Chapter 4 (df=-dv Sd) and Gibbs energy (dgrev = Vd-Sd), we will represent Gibbs energy as G =G(,) and Helmholtz energy as F=F(V,). otal differentials are then defined as dg = G d + G d df = F V dv + F d V 5.4.7a) 5.4.7b) Additional insight into temperature, pressure, volume and entropy can be obtained from the partial derivatives. By comparing terms in the above equations (e.g. Eqs a and 5.4.6a), we conclude that = U S = H V,n j S = U V = F S,n j V 5.4.8a) 5.4.8b) Last Modified: May, Lecture Handouts

33 V = H = G S,n j S = G = F V Chemical otential and Internal Energy 5.4.8c) 5.4.8d) In Chapter 2, the total change in internal energy was divided into possible changes with (1) temperatures of the system, (2) phase transitions and (3) chemical reactions. he relationship in the previous section of dutpr = Cvd + ( U/ V)dV was developed for a constant number of moles and therefore doesn t include possible changes in the internal energy with phase transitions or chemical reactions. hese systems can be quite boring under conditions of constant pressure and temperature (and no non-expansion work); for example the reversible form of dgrev = Vd-Sd = 0. In this section, we develop a theoretical framework that allows phase transitions and chemical reactions to be easily incorporated into representation of the total change in internal energy. Both of these processes change the number of moles of the components within the system. For example, vaporization decreases the number of moles of liquid water and increases the number of vapor moles. Likewise, the number of reactant moles decreases and the number of product moles increases for chemical reactions. A more general formulation for internal energy that includes possible changes in the number of moles is where n1, n2,, nk are the number of moles for each of the k components in the system. A differential change is then defined as du = U S V,n j ds + U V S,n j dv + U n j k j=1 k = ds dv + U n j j=1 S,V,n i j S,V,n i j dn j dn j ) where subscript nj refers to constant number of moles for all components and ni j refers to constant number of moles for all components other than the j th component. Insight from the previous section has been used to evaluate = ( U/ S)V,nj and = - ( U/ V)S,nj (Eqs a and 5.4.8b). he partial derivatives in the summation term are defined as the chemical potential, that is, for the j th component Last Modified: May, Lecture Handouts

34 µ j = U n j S,V,n i j where the chemical potential here is defined for a constant entropy, volume and number of moles of all other components except for the jth component. he chemical potential corresponds to the rate of change in the internal energy of the system corresponding to a change in the number of moles of the jth component. By using the definition of chemical potential, the change in internal energy is defined as he evaluation of chemical potential using the above equations requires conditions of constant entropy. hese conditions are inconvenient to control in an experimental design. We prefer alternative definition that has experimental variables that can be more easily controlled. his definition is given below using Gibbs energy. If we compare the change in internal energy of this section (Eq ) with that obtained directly from the First Law given by Eq , we conclude that the chemical potential term is equal to the irreversible change in entropy. his result is discussed in greater detail in the next section. Gibbs Energy and Chemical otential General Form of Change in Gibbs Energy o incorporate the chemical potentials in the change in Gibbs energy, let s revisit the definition of the infinitesimal change in Gibbs energy as originally given by Eq If we substitute the definition for the change in internal energy defined with chemical reactions (i.e., Eq ), we obtain dg = V d Sd + μμ jj ddnn jj k j=1 = dg rev + μμ jj ddnn jj k j= For the special condition of constant temperature and pressure, we obtain Last Modified: May, Lecture Handouts