MME 2010 METALLURGICAL THERMODYNAMICS II. Partial Properties of Solutions

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1 MME 2010 METALLURGICAL THERMODYNAMICS II Partial Properties of Solutions

2 A total property of a system consisting of multiple substances is represented as nm = n i M i If the system consists of a liquid mixture comprised of molecules of the same size and equal forces between, like a solid mixture, the total property is the same: nm = n i M i Ideal solution is such a liquid behaving like a solid mixture In other words, the components of the ideal solution avoid intimate mixing In contrast, real solutions consist of interacting chemical species and their total system property is apportioned or shared by individual components as their partial properties nm = n i M i

The partial molar volume assigned to each component in a real solution is more or less than

3 Apportioning of system properties as a puzzle analogy Consider the surface area of puzzle pieces and relate them to the volume of individual components in solutions. The partial molar volume assigned to each component in a real solution is more or less than their volumes at pure state depending on their interactions Ideal solution nm = n i V i V i > V i Real solution nv = n i V i

Although partial properties are shared system properties, in other words they are obtained from the system property, they may be needed in calculation of the total property of another system they

4 Although partial properties are shared system properties, in other words they are obtained from the system property, they may be needed in calculation of the total property of another system they become a part of Partial property of each species in the solution is required in order to obtain the total property of the solution Partial molar volume V i of a species is a function of its molar ratio x i x i = n i So both total solution property and individual partial property approach the pure species property as a solution becomes pure in species i: n i lim M = lim M i = M i x i 1 x i 1

Consider adding a drop of water to a beaker containing an equimolar mixture of alcohol and water at room temperature T and atmospheric pressure P The drop that is also at T and P, contains Δn w moles

5 Consider adding a drop of water to a beaker containing an equimolar mixture of alcohol and water at room temperature T and atmospheric pressure P The drop that is also at T and P, contains Δn w moles Allowing enough time for thermal equilibrium, you measure the volume change and see that nv = V w n w < V w n w where V w is the molar volume of pure water and V w represents the effective molar volume of water in the solution V w = (nv) n w The effective molar volume of water is considered a property of the original equimolar solution when the process is taken to the limit of an infinitesimal drop V w = (nv) lim n w 0 n w = d(nv) dn w or V w = nv n w T,P,na

6 Hence V i = nv n i T,P,nj Total molar volume of a solution is equal to the sum of the product of the partial molar volume of each species and its mole number nv = n 1 V 1 + n 2 V 2 + n 3 V 3 + +n m V m Total differential equation of solution volume: d nv = nv P T,n dp + nv T P,n dt + i nv n i P,T,n j dn i At constant temperature and pressure d nv = V i dn i since n i = x i n, i dn i = x i dn + ndx i

7 And since the complete differential of total volume d(nv) = ndv + Vdn, ndv + Vdn = Seperating the terms n and dn, i V i (x i dn + ndx i ) dv i V i dx i n + V i V i x i dn = 0 The solution may contain any moles n, and its size dn may vary by any value Thus V = i x i V i or nv = i n i V i

8 Example How much water must be added to 1000 ml of NiSO 4 -water solution containing 96% NiSO 4.6H 2 O by volume in order to make a plating solution containing 64% NiSO 4.6H 2 O by volume? Partial molar volume of NiSO 4.6H 2 O = 98.1 ml for 96% solution, 96.6 ml for 64% solution Partial molar volume of H 2 O = 14.5 ml for 96% solution, 17.1 ml for 64% solution

9 For a system of constant total composition of n=1, total differential of volume is represented as dv = V P T,n dp + V T P,n dt + i V i dx i Comparing with the general expression for dv = i x i dv i + i V i dx i and substitution yields: V dp + V dt P T,n T P,n i x i dv i = 0 The equation must hold for all changes in P, T, and V i caused by changes of state in a homogeneous phase. For constant T and P, i x i dv i = 0 Partial properties of species making up a solution are dependent on each other

10 So far relationships between total and partial volume of solutions have been discussed The same relationships hold for other extensive properties including enthalpy, entropy, internal energy and Gibbs free energy V = x 1 V 1 + x 2 V 2 + +x m V m dv = dx 1 V 1 + dx 2 V 2 + +dx m V m x 1 dv 1 + x 2 dv 2 + +x m dv m = 0 S = x 1 S 1 + x 2 S 2 + +x m S m ds = dx 1 S 1 + dx 2 S 2 + +dx m S m x 1 ds 1 + x 2 ds 2 + +x m ds m = 0 H = x 1 H 1 + x 2 H 2 + +x m H m dh = dx 1 H 1 + dx 2 H 2 + +dx m H m x 1 dh 1 + x 2 dh 2 + +x m dh m = 0 G = x 1 G 1 + x 2 G 2 + +x m G m dg = dx 1 G 1 + dx 2 G 2 + +dx m G m x 1 dg 1 + x 2 dg 2 + +x m dg m = 0 U = x 1 U 1 + x 2 U 2 + +x m U m du = dx 1 U 1 + dx 2 U 2 + +dx m U m x 1 du 1 + x 2 du 2 + +x m du m = 0

11 Partial properties in binary solutions Equations covered so far enable the calculation of a partial property as a function of composition. An alternative graphical way is more convenient for binary solutions Since x A + x B = 1, dx A = dx B Partial property changes total to 0 M = x A M A + x B M B dm = x A dm A + x B dm B + M A dx A + M B dx B dm = x A dm A + x B dm B + (M A M B )dx A dm dx A = (M A M B ) M = M A + x B (M B M A ) M = x A M A M B + M B M A = M x B dm dx B M B = M x A dm dx A

12 The relationships just derived enable determination of partial molar properties of components when the total solution property and its variation with concentration is known dv V A = V x B dx B dv V B = V x A dx

13 Example Derive expressions for the partial molar volumes of the components of a binary solution as a function of composition. The total volume change of mixing is V M = 2.7x A x B cm 3 /mol V A = V M x B d V M dx B V B = V M x A d V M dx A

14 The relationships just derived enable determination of partial molar properties of components when the total solution property and its variation with concentration is known G A = μ A = Gx B dg dx B Recall that μ A id = G A o + RT ln x A

15 Example Variation of total molar properties of the Cd-Sn alloy system with composition at 773 K are given in table form. Obtain the partial molar properties of the components graphically for 45% Sn and calculate the activities of the components at 773 K X Sn dg, cal dh, cal TdS, cal G Sn = G Sn o + RT ln a Sn dg, cal dh, cal TdS, cal

16 Relationships between partial molar properties The sound basis of thermodynamic relationships among pure substances with constant composition allows us to make use of all of them for analysis of solutions with variable composition by substituting molar properties of pure substances with the partial molar properties G i = H i T S i G i T P,nj = S i (G i T) = H ( 1 T) i P,nj G i P T,nj = V i du = TdS PdV dg = VdP SdT dh = TdS + VdP da = PdV SdT

17 Energy, Joules/mol Example Determine expressions for partial molar enthalpies and the pure species enthalpies for an arbitrary binary solution system of A and B. The total molar enthalpy variation by concentration is given as: M A = M x B dm dx B H = 400x A + 600x B + x A x B (40x A + 20x B ) J/mol dh 300 pha phb

18 Example The free energy of a binary ideal solution of species 1 and 2 at ambient T and P is represented by the equation G id = x 1 G 1 + x 2 G 2 + RT(x 1 ln x 1 + x 2 ln x 2 ) J/mole Determine expressions for G 1 and G 2 at equimolar concentration G 1 = G x 2 dg dx 2 G 2 = G x 1 dg dx 1

19 Obtaining total solution property from one of the components Total solution property in terms of partial properties is simplified by virtue of the equation X A + = 1 M = x A M A + x B M B M A = M + (1 X A )( dm dx A ) Multiplying both sides by dx A and dividing by 2 M A dx A 2 = MdX A + 1 X A dm 2 Replacing (1-X A )= and dx A =-d M A dx A 2 = Md + dm 2 = d M

20 Integrating both sides gives M d M = M (at =1) M M B o = X A =0 X A =0 Setting partial mixing or excess properties ΔM M and M E equal to zero at =1 ΔM M = X A =0 constant X A X A X A M A dx A 2 M A AdX A 2 M A dx A 2 M E X A = X A =0 M E AdX A 2 constant

21 M X A M B o = X A =0 M A dx A 2 Values of integrals can be obtained either analytically or graphically The following steps are involved in analytical integration: Replacement of the property M A, M M A or M E A into the integral Organization of the function inside the integral in such a way to leave only one variable Integration and replacement of the limits Graphical integration involves determination of the area under the curve M A /2, M M A/B2 or M E A/2 vs. X A between the limits

Example Partial volume of mixing of A is given as V A = 10x B 2 + 40x B 3 Find the total volume of the solution at x A = 0.669 X V A =0.669 V A dx A = X 2 = B X A =0 X A =0 X A =0.

22 Example Partial volume of mixing of A is given as V A = 10x B x B 3 Find the total volume of the solution at x A = X V A =0.669 V A dx A = X 2 = B X A =0 X A =0 X A = x B x B 3 dx A 2 V = X A =0 X A = x B dx A = X A =0.669 X A =0 (50 40 X A )dx A V = = 24.5 V = 8.11cm 3 To check the solution, V = x A x B 30x A + 50x B V = ( (0.331) = 8.11 cm 3

23 Homework question The activity of a component for binary systems comprised of species that are chemically similar is represented as lnα 1 = (1-2x 1 ) Obtain G/RT value for x 1 = X G 1 =0.398 RT(X 2 ) = G 1 dx 1 2 X 1 =0 RTX 2

24 Further derivation of the fundamental partial property equations enable determination of the partial property of one component, M A if the variation of the other s, M B as a function of composition is known substituting by x A = 1 x B yields x A dm A + x B dm B = 0 dm B = 1 x B x B dm A Equation can be integrated between limits M B and M B o M BdMB = M B M B o = M B o (M A ) XB=x (M A ) XB=1 1 x B x B dm A Because M B is constant at x B = 1 and equals M B o as 1 x B x B equals 0

25 M B is obtained from the integration by adding M o B directly if M A = f x B known is M B M B o = (M A ) XB=x (M A ) XB=1 1 x B x B dm A Analytical determination involves the following steps: Organize M A as a function of only one composition variable Take the derivative of M A and replace it into the integral Organize the function inside the integral in such a way to leave only one variable and change the limits of the integral if necessary, to make it in accord with the derivative of the integral. Make use of the relationship dx A =-d Integrate the function, replace the limits to determine the value of the integral

26 Graphical determination of M B M B o is possible by plotting a x A x B - M A graph The area under the curve between the limits (V A ) XB=1 and (V A ) XB=a gives V B V B o V B V B o = (V A ) XB=a (V A ) XB=1 V B V B o = x A x B U 2ydU U 1 dv A

27 However there are some limitations with graphical determination: The value of M A becomes - if M A has logarithmic composition terms X A / becomes when becomes zero These problems may be resolved for excess properties and properties of mixing using the α-function: For any component i, α i is defined as α i = ME i (1 X i ) 2 α-function = 0/0 when X i becomes 1 so that its integral gives a finite value

28 Derivation of the α-function M E B at = (M E A ) XB=a (M E A) XB=1 X A dm E A Let M E A = α A 2, Then dm E A = dα A 2 + 2α A d Replacing into integral yields M E B at = M E B at = (α A ) XB=a X XB=a A (α A ) XB=1 2 dα A XB=1 (α A ) XB=aXA dα A (α A ) XB=1 XB=a XB=1 The first integral is expanded by virtue of the identity d(xy) = ydx + xdy X A 2α A X A d (α A ) XB=aXA dα A = d(x A α A ) α A d(x A ) 2α A d (α A ) XB=1

29 M E B at = d X A α A + α A d(x A ) 2α A X A d M E B at = X A α A + α A X A d + α A dx A 2α A X A d Grouping the last terms, M E B at = X A α A α A 2X A X A + d M E B at = X A α A α A d α A = ME A 2 Numerical value for X A α A is readily calculated, the value of the integral can be determined graphically or analytically Analytical integration with α-function is done by replacing α A into the equation and integrating between the limits Graphical determination is done by obtaining the area under vs. α A graph

30 X Ni/X Fe Example Consider the binary system Fe-Ni at 1600 C X Ni a Ni ΔGm Ni X Ni/X Fe M Fe = ( M Ni ) XFe=a ( M Ni ) XFe=1 1 x Fe x Fe d M Ni The value of partial Gibbs energy of mixing -ΔGm Ni goes to infinity at the lower limit and results in an unbounded area This problem is overcome by using partial excess Gibbs energy which reaches a finite value of RT lnγ o Ni at X Ni = Using partial excess values provides good 2.00 results as long as X Fe >0 since X Ni /X Fe = 1.00 Area at X Fe = α-function enables obtaining finite integral over all X values Area

3. APLICATION OF GIBBS-DUHEM EQUATION

3. APLICATION OF GIBBS-DUHEM EQUATION Gibbs-Duhem Equation When extensive property of a solution is given by; Q' = Q'(T,P,n,n...) The change in extensive property with composition was; dq Q n dn Q n dn