SOLUTION THERMODYNAMICS: THEORY • FUNDAMENTAL PROPERTY RELATION • THE CHEMICAL POTENTIAL AND PHASE EQUILIBRIA • PARTIAL PROPERTIES FUNDAMENTAL PROPERTY RELATION The first Tds relation or Gibbs equation: Gibbs energy: Multiplied by n and differentiated eq. (6.3): Enthalpy: Multiplied by n, differentiated and combined with eq. (6.1): Combine eq. (6.3a) and (6.4) to yield: 2 ( ) ( ) ( ) d nU Td nS Pd nV = ÷ (6.1) G H TS ÷ ÷ (6.3) ( ) ( ) ( ) ( ) d nG d nH Td nS nS dT = ÷ ÷ H U PV ÷ + (2.11) ( ) ( ) ( ) d nH Td nS nV dP = + (6.4) (6.3a) ( ) ( ) ( ) = ÷ d nG nV dP nS dT (6.6) Equation (6.6) relates the total Gibbs energy of any closed system to temperature and pressure. An appropriate application is to a single phase fluid in a closed system wherein no chemical reactions occur. For such a system the composition is necessarily constant, and therefore The subscript n indicates that the numbers of moles of all chemical species are held constant. For more general case of a single phase, open system, material may pass into and out of the system, and nG becomes a function of the numbers of moles of the chemical species present, and still a function of T and P. where n i is the number of moles of species i. 3 ( ) ( ) ( ( c c = = ÷ ( ( c c ¸ ¸ ¸ ¸ , , T n P n nG nG nV and nS P T (A) ( ) = 1 2 , , , ,..., ,... i nG g P T n n n • Total differential of nG is The summation is over all species present, and subscript n j indicates that all mole numbers except the ith are held constant. The derivative in the final term is called the chemical potential of species i in the mixture. It is define as With this definition and with the first two partial derivatives [eqn. (A)] replaced by (nV) and –(nS), the preceding equation [eqn. (B)] becomes Equation (11.2) is the fundamental property relation for single phase fluid systems of variable mass and composition. 4 ( ) µ ( c ÷ ( c ¸ ¸ , , i i P T nj nG n (11.1) ( ) ( ) ( ) ( ( c c ( c = + + ( ( ( c c c ¸ ¸ ¸ ¸ ¸ ¸ ¿ , , , , i i i P T nj T n P n nG nG nG d nG dP dT dn P T n (B) ( ) ( ) ( ) µ = ÷ + ¿ i i i d nG nV dP nS dT dn (11.2) For special case of one mole of solution, n = 1 and n i = x i : This equation relates molar Gibbs energy to T, P and {xi}. For special case of a constant composition solution: Although the mole numbers n i of eq. (11.2) are independent variables, the mole fractions x i in eq. (11.3) are not, because ¿ i x i = 1. Eq. (11.3) does imply Other solution properties come from definitions; e.g., the enthalpy, from H = G + TS. Thus, by eq. (11.5), 5 µ = ÷ + ¿ i i i dG VdP SdT dx (11.3) dG VdP SdT = ÷ (6.10) ( ) 1 2 , , , ,..., ,... i G G P T x x x = , P x G S T c | | = ÷ | c \ . , T x G V P c | | = | c \ . (11.5) (11.4) , P x G H G T T c | | = ÷ | c \ . 6 6 When the Gibbs energy is expressed as a function of its canonical variables (T, P and {xi}), it plays the role of a generating function, providing the means for calculation of all other thermodynamic properties by simple mathematical operations (differentiation and elementary algebra), and implicitly represents complete property information. THE CHEMICAL POTENTIAL AND PHASE EQUILIBRIA For a closed system consisting of two phases in equilibrium, each individual phase is open to the other, and mass transfer between phases may occur. Equation (11.2) applies separately to each phase: where superscripts o and | identify the phases. The presumption here is that equilibrium implies uniformity of T and P throughout the entire system. The change in the total Gibbs energy of the two phase system is the sum of these equations. 7 ( ) ( ) ( ) i i i d nG nV dP nS dT dn o o o o o µ = ÷ + ¿ ( ) ( ) ( ) i i i d nG nV dP nS dT dn | | | | | µ = ÷ + ¿ When each total system property is expressed by an equation of the form, the sum is Because the two phase system is closed, eq. (6.6) is also valid. Comparison of the two equations shows that at equilibrium, The changes dn i o and dn i | result from mass transfer between the phases; mass conservation therefore requires 8 ( ) ( ) nM nM nM o | = + ( ) ( ) ( ) i i i i i i d nG nV dP nS dT dn dn o o | | µ µ = ÷ + + ¿ ¿ 0 i i i i i i dn dn o o | | µ µ + = ¿ ¿ ( ) 0 i i i i i dn dn and dn o | o | o µ µ = ÷ ÷ = ¿ ( ) ( ) ( ) = ÷ d nG nV dP nS dT (6.6) ( ) ( ) ( ) ( ) ( ) ( ) i i i i i i d nG d nG nV nV dP nS nS dT dn dn o | o | o | o o | | µ µ ( ( + = + ÷ + + + ¸ ¸ ¸ ¸ ¿ ¿ Quantities dn i o are independent; therefore the only way the left side of the second equation can in general be zero is for each term in parentheses separately to be zero. Hence, where N is the number of species present in the system. For multiple phases (t phases): Example: A glass of liquid water with ice cubes in it. Above 0 o C, the chemical potential of ice is larger than water, so the ice melts. Below 0 o C, the chemical potential of water is larger than ice, so the water freezes. At 0 o C, the water and ice are in equilibrium because their chemical potential are the same. 9 ( ) o | µ µ = =1,2,..., i i i N ( ) o | t µ µ µ = = = = ... 1,2,..., i i i i N (11.6) Multiple phases at the same T and P are in equilibrium when the chemical potential of each species is the same in all phases. A chemical species is transported from a phase of larger potential to a phase of lower potential. Partial molar property of a species i in a solution is define as It is a measure of the response of total property nM to the addition at constant T and P of a differential amount of species i to a finite amount of solution. Three kinds of properties used in solution thermodynamics are distinguished by the following symbolism: ◦ Solution properties M, for example: V, U, H, S, G ◦ Partial properties , for example: ◦ Pure species properties M i , for example: V i , U i , H i , S i , G i Comparison of eq. (11.1) with eq. (11.7) written for the Gibbs energy shows that the chemical potential and the partial molar Gibbs energy are identical; i.e., PARTIAL PROPERTIES ÷ i M 10 ( ) ( c ÷ ( c ¸ ¸ _ , , i i P T nj nM M n (11.7) i M ÷ ÷ ÷ ÷ ÷ ÷ , , , , i i i i i V U H S G µ ÷ ÷ i i G (11.8) 10 ( ) µ ( c ÷ ( c ¸ ¸ , , i i P T nj nG n (11.1) Example When one mole of water is added to a large volume of water at 25ºC, the volume increases by 18cm 3 . The molar volume of pure water would thus be reported as 18cm 3 mol -1 . However, addition of one mole of water to a large volume of pure ethanol results in an increase in volume of only 14cm 3 . The value 14cm 3 is said to be the partial molar volume of water in ethanol. In general, the partial molar volume of a substance i in a mixture is the change in volume per mole of i added to the mixture. 11 ( ) _ , , i i P T nj nV V n ( c ÷ ( c ¸ ¸ Total thermodynamic properties of a homogeneous phase are functions of T, P, and the numbers of moles of the individual species which comprise the phase. Thus, for property M: The total differential of nM is where subscript n indicates that all mole numbers are held constant, and subscript n j that all mole numbers except n i are held constant. Because the first two partial derivatives on the right are evaluated at constant n and because the partial derivative of the last term is given by eq. (11.7), this equation has the simpler form: where subscript x denotes differentiation at constant composition. EQUATIONS RELATING MOLAR AND PARTIAL MOLAR PROPERTIES ( ) ( ) ( ) ÷ | | | | c c = + + | | c c \ . \ . ¿ , , i i i T x P x M M d nM n dP n dT M dn P T 12 ( ) 1 2 , , , ,..., ,... i nM T P n n n = M ( ) ( ) ( ) ( ) ( ( ( c c c = + + ( ( ( c c c ¸ ¸ ¸ ¸ ¸ ¸ ¿ , , , , i i i T n P n P T nj nM nM nM d nM dP dT dn P T n (11.9) 12 Because n i = x i n, Moreover, When dn i and d(nM) are replaced in Eq. (11.9), it becomes The terms containing n are collected and separated from those containing dn to yield The left side of this equation can be zero if each term in brackets be zero too. Therefore, 13 = + i i i dn x dn ndx ( ) ÷ + d nM ndM Mdn ( ) , , i i i i T x P x M M ndM Mdn n dP n dT M x dn ndx P T ÷ c c | | | | + = + + + | | c c \ . \ . ¿ ÷ ÷ ( c c ( | | | | ÷ ÷ ÷ + ÷ = ( | | ( c c \ . \ . ¸ ¸ ( ¸ ¸ ¿ ¿ , , 0 i i i i i i T x P x M M dM dP dT M dx n M x M dn P T ÷ c c | | | | = + + | | c c \ . \ . ¿ , , i i i T x P x M M dM dP dT M dx P T (11.10) and Multiplication of eq.(11.11) by n yields the alternative expression: Equations (11.11) and (11.12) are known as summability relations, they allow calculation of mixture properties from partial properties. Differentiate eq. (11.11) yields: Comparison of this equation with eq. (11.10), yields the Gibbs/Duhem equation: For changes at constant T and P, 14 ÷ = ¿ i i i M x M (11.11) ÷ = ¿ i i i nM n M (11.12) ÷ c c | | | | + ÷ = | | c c \ . \ . ¿ , , 0 i i i T x P x M M dP dT x d M P T ÷ = ¿ 0 i i i x d M ÷ ÷ = + ¿ ¿ i i i i i i dM x d M M dx (11.13) (11.14) A RATIONALE FOR PARTIAL PROPERTIES Partial properties have all characteristics of properties of individual species as they exist in solution. Thus, they may be assigned as property values to the individual species. Partial properties, like solution properties, are functions of composition. In the limit as a solution becomes pure in species i, both M and approach the pure species property M i . For a species that approaches its infinite dilution limit, i.e., the values as its mole fraction approaches zero, no general statements can be made. Values come from experiment or from models of solution behavior. By definition, 15 ÷ i M ÷ ÷ ÷ = = 1 1 lim lim i i i i x x M M M ÷ ÷ · ÷ ÷ 0 lim i i i x M M Equations for partial properties can be summarized as follows: ◦ Definition: which yields partial properties from total properties. ◦ Summability: which yields total properties from partial properties. ◦ Gibbs/Duhem: which shows that the partial properties of species making up solution are not independent of one another. 16 ( ) _ , , i i P T nj nM M n ( c ÷ ( c ¸ ¸ i i i M x M ÷ = ¿ (11.11) , , i i i T x P x M M x d M dP dT P T ÷ c c | | | | = + | | c c \ . \ . ¿ (11.13) (11.7) For binary solution, the summability relation, eq.(11.11) becomes Differentiation of eq. (A) becomes When M is known as a function of x 1 at constant T and P, the appropriate form of the Gibbs/Duhem equation is eq. (11.14), expressed as Because x 1 + x 2 = 1, dx 1 + dx 2 = 0 dx 1 = - dx 2 . Substitute eq. (C) into eq. (B) to eliminate dx 2 gives PARTIAL PROPERTIES IN BINARY SOLUTIONS 1 1 2 2 0 x d M x d M ÷ ÷ + = 17 1 1 2 2 M x M x M ÷ ÷ = + (A) 1 1 1 1 2 2 2 2 dM x d M M dx x d M M dx ÷ ÷ ÷ ÷ = + + + (B) (C) 1 2 1 dM M M dx ÷ ÷ = ÷ (D) 17 Two equivalent forms of eq. (A) result from elimination separately of x 1 and x 2 : In combination with eq. (D) becomes Thus for binary systems, the partial properties are calculated directly from an expression for the solution property as a function of composition at constant T and P. 2 1 1 dM M M x dx ÷ = ÷ 18 ( ) ( ) 1 2 2 1 2 1 2 2 1 1 1 2 1 2 1 2 2 1 1 2 1 2 1 2 1 2 1 1 2 2 1 1 1 1 and x x x x M x M x M M x M x M M M x M x M M x M M x M M M x M M M x M M M ÷ ÷ ÷ ÷ ÷ ÷ ÷ ÷ ÷ ÷ ÷ ÷ ÷ ÷ ÷ ÷ = ÷ = ÷ = ÷ + = + ÷ = ÷ + = + ÷ | | | | = ÷ ÷ = ÷ + | | \ . \ . 1 2 1 dM M M x dx ÷ = + (11.15) (11.16) 18 Eq. (C), the Gibbs/Duhem equation, may be written in derivative forms: When are plotted vs. x 1 , the slopes must be of opposite sign. 19 1 2 1 2 1 1 0 d M d M x x dx dx ÷ ÷ + = 1 2 2 1 1 1 dM x dM dx x dx ÷ ÷ = ÷ (E) (F) 2 1 1 1 2 1 dM x dM dx x dx ÷ ÷ = ÷ (G) 1 2 and M M ÷ ÷ Moreover, Similarly, Thus, plot of vs. x 1 become horizontal as each species approaches purity. 20 1 1 1 2 1 1 1 1 lim 0 Provided lim is finite x x d M d M dx dx ÷ ÷ ÷ ÷ | | | = | \ . 2 2 2 1 1 1 1 1 lim 0 Provided lim is finite x x d M d M dx dx ÷ ÷ ÷ ÷ | | | = | \ . 1 2 and M M ÷ ÷ EXAMPLE 11.2 Figure 11.1 (a) shows a representative plot of M vs. x 1 for a binary system. The tangent line shown extend across the figure, intersecting the edges (at x 1 = 1 and x 1 = 0) at points label I 1 and I 2 . Two equivalent expressions can be written for the slope of this tangent line: The first equation is solved for I 2 ; it combines with the second to give I 1 . Comparison of these expression with eqs. (11.16) and (11.15) show that The tangent intercepts give directly the values of the two partial properties. 21 Describe a graphical interpretation of eqs. (11.15) and (11.16). Solution: 2 1 2 1 1 1 and M I dM dM I I dx x dx ÷ = = ÷ ( ) 2 1 1 1 1 1 and 1 dM dM I M x I M x dx dx = ÷ = + ÷ 1 2 1 2 and I M I M ÷ ÷ = = The limiting values are indicated by Fig. 11.1 (b). For the tangent line drawn at x 1 = 0 (pure species 2), and at the opposite intercept, For the tangent line drawn at x 1 = 1 (pure species 1), and at the opposite intercept, 22 1 1 M M ÷ = 2 2 M M ÷ ÷ · = 2 2 M M ÷ = 1 1 M M ÷ ÷ · = EXAMPLE 11.3 The need arises in a laboratory for 2000 cm 3 of an antifreeze solution consisting of 30 mole % methanol in water. What volumes of pure methanol and of pure water at 25 o C (298.15K) must be mixed to form the 2000 cm 3 of antifreeze, also at 25 o C (298.15K)? Partial molar volumes for methanol and water in a 30 mole % methanol solution and their pure species molar volumes, both at 25 o C (298.15K), are 23 ( ) ( ) 3 -1 3 -1 1 1 3 -1 3 -1 2 2 Methanol 1 : 38.632 cm mol 40.727 cm mol Water 2 : 17.765 cm mol 18.068 cm mol V V V V ÷ ÷ = = = = Solution: Equation (11.11) is written for the molar volume of the binary antifreeze solution, and known values are substituted for the mole fractions and partial volumes: Because the required total volume of solution is V t = 2000 cm 3 , the total number of moles required is Of this, 30% is methanol, and 70% is water: The total volume of each pure species is V i t = n i V i ; thus, i i i V x V ÷ = ¿ 24 ( )( ) ( )( ) 1 1 2 2 3 -1 0.3 38.632 0.7 17.765 24.025 cm mol V x V x V ÷ ÷ = + = + = 2000 83.246 mol 24.025 t V n V = = = 24 ( )( ) ( )( ) 1 2 0.3 83.246 24.974 mol 0.7 83.246 58.272 mol n n = = = = ( )( ) ( )( ) 3 1 3 2 24.974 40.727 1017 cm 58.272 18.068 1053 cm t t V V = = = = i i n x n = 1 2 and V V ÷ ÷ Values of for the binary solution methanol(1)/water(2) at 25 o C (298.15K) are plotted in Fig. 11.2 as functions of x 1 . The line drawn tangent to the V vs x 1 curve at x 1 = 0.3 illustrates the graphical procedure by which values of may be obtained. The curve becomes horizontal at x 1 = 1 and the curve for becomes horizontal at x 1 = 0 or x 2 = 1. The curves for appear to be horizontal at both ends. 25 1 2 , and V V V ÷ ÷ 1 2 and V V ÷ ÷ 1 V ÷ 1 2 and V V ÷ ÷ 2 V ÷ For the tangent line drawn at x 1 = 0 (pure species 2), and at the opposite intercept, For the tangent line drawn at x 1 = 1 (pure species 1), and at the opposite intercept, 1 1 V V ÷ = 2 2 V V ÷ ÷ · = 2 2 V V ÷ = 1 1 V V ÷ ÷ · = EXAMPLE 11.4 The enthalpy of a binary liquid system of species 1 and 2 at fixed T and P is represented by the equation where H is in Jmol -1 . Determine expressions for as a functions of x 1 , numerical values for the pure species enthalpies H 1 and H 2 , and numerical values for the partial enthalpies at infinite dilution 26 ( ) 1 2 1 2 1 2 400 600 40 20 H x x x x x x = + + + 1 2 and H H ÷ ÷ 1 2 and H H ÷ ÷ · · Solution: Replace x 2 by 1 – x 1 in the given equation for H and simplify: By equation (11.15), Then, Replace x 2 by 1 – x 1 and simplify: By eq. (11.16), or 2 3 1 1 1 420 60 40 H x x ÷ = ÷ + 27 3 3 2 1 1 1 1 1 1 600 180 20 180 60 dH H H x x x x x dx ÷ = ÷ = ÷ ÷ + + 3 1 1 600 180 20 H x x = ÷ ÷ (A) 2 1 1 180 60 dH x dx = ÷ ÷ 1 2 1 dH H H x dx ÷ = + 3 2 1 1 1 2 1 2 600 180 20 180 60 H x x x x x ÷ = ÷ ÷ ÷ ÷ (B) 3 2 1 600 40 H x ÷ = + (C) 27 A numerical value for H 1 results by substitution of x 1 = 1 in either eq (A) or (B). Both eqn. yield H 1 = 400 J mol -1 . H 2 is found from either eq. (A) or (C) when x 1 = 0. The result is H 2 = 600 J mol -1 . The infinite dilution are found from eq. (B) and (C) when x 1 = 0 in eq. (B) and x 1 = 1 in eq. (C). The results are: Exercise: Show that the partial properties as given by eqs. (B) and (C) combine by summability to give eq. (A), and conform to all requirements of the Gibbs/Duhem equation. 1 2 and H H ÷ ÷ · · 28 · · - - -1 -1 1 2 H =420 Jmol and H =640 Jmol 28 RELATIONS AMONG PARTIAL PROPERTIES By eq. (11.8), and eq. (11.2) may be written as Application of the criterion of exactness, eq. (6.12), yields the Maxwell relation, 29 i i G µ ÷ ÷ ( ) ( ) ( ) i i i d nG nV dP nS dT G dn ÷ = ÷ + ¿ (11.17) y x M N y x | | c c | | = | | c c \ . \ . (6.12) , , P n T n V S T P c c | | | | = ÷ | | c c \ . \ . (6.16) ( ) ( ) ( ) i i i d nG nV dP nS dT dn µ = ÷ + ¿ (11.2) (11.8) dz Mdx Ndy = + Plus the two additional equations: where subscript n indicates constancy of all n i , and subscript n j indicates that all mole numbers except the ith are held constant. In view of eq. (11.7), the last two equations are most simply expressed: These equations allow calculation of the effects of P and T on the partial Gibbs energy (or chemical potential). They are partial property analogs of eqs. (11.4) and (11.5). , i i T x G V P ÷ ÷ | | c | = | c \ . 30 ( ) , , , j i i P T n T n nV G P n ÷ | | ( c c | = ( | c c ¸ ¸ \ . (11.18) (11.19) , i i P x G S T ÷ ÷ | | c | = ÷ | c \ . 30 Every equation that provides a linear relation among thermodynamic properties of a constant-composition solution has as its counterpart an equation connecting the corresponding partial properties of each species in the solution. ( ) , , , j i i P T n P n nS G T n ÷ | | ( c c | = ÷ ( | c c ¸ ¸ \ . An example is based on the equation that defines enthalpy: H = U + PV For n moles, Differentiation with respect to n i at constant T, P, and n j yields By eq. (11.7) this becomes In a constant-composition solution, is a function of P and T, and therefore: By eqs. (11.18) and (11.19), These examples illustrate the parallelism that exists between equations for a constant-composition solution and the corresponding equations for partial properties of the species in solution. 31 ( ) nH nU P nV = + ( ) ( ) ( ) , , , , , , j j j i i i P T n P T n P T n nH nU nV P n n n ( ( ( c c c = + ( ( ( c c c ¸ ¸ ¸ ¸ ¸ ¸ i i i H U PV ÷ ÷ ÷ = + i G ÷ , , i i i T x P x G G dG dP dT P T ÷ ÷ ÷ | | | | c c | | = + | | c c \ . \ . i i i dG V dP S dT ÷ ÷ ÷ = ÷ Similar to eq. (2.11) H≡U+PV Similar to eq. (6.10) dG=VdP-SdT REFERENCES Smith, J.M., Van Ness, H.C., and Abbott, M.M. 2005. Introduction to Chemical Engineering Thermodynamics. Seventh Edition. Mc Graw-Hill. http://www.chem1.com/acad/webtext/thermeq/TE4.html http://mpdc.mae.cornell.edu/Courses/ENGRD221/LECTURE S/lec26.pdf http://science.csustan.edu/perona/4012/partmolvolsalt_lab20 10.pdf 32 PREPARED BY: MDM. NORASMAH MOHAMMED MANSHOR FACULTY OF CHEMICAL ENGINEERING, UiTM SHAH ALAM.
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