`TOPIC 3IDEAL SOLUTIONS, FUGACITY, ACTIVITY, AND STANDARD STATES1I. PARTIAL AND APPARENT MOLAR PROPERTIES2MOLAR VS. PARTIAL MOLAR QUANTITIESMolar values of state functions are defined as follows: U =U / n S = S/n V =V /n etc. These are useful only in the case of singlecomponent systems and dependent only on pressure and temperature, not composition. Partial molar quantities are defined according to: U    Ui =     ni  T ,P ,n j S  Si =   n    i T ,P ,n jPHYSICAL INTERPRETATION OF PARTIAL MOLAR VOLUMESThe partial molar volume of component i in a system is equal to the infinitesimal increase or decrease in the volume, divided by the infinitesimal number of moles of the substance which is added, while maintaining T, P and quantities of all other components constant. Another way to visualize this is to think of the change in volume on adding a small amount of component i to an infinite total volume of the system. Note: partial molar quantities can be positive or negative! 4 V    Vi =    n i T ,P ,n jThese are dependent on T, P, and composition.3SUMMING PARTIAL MOLAR QUANTITIESThe total value for a state function of a system is obtained by summing the partial molar volumes of its components according to: nWe can also express the summations in terms of molar state functions and mole fractions:U=X Ui i =1niH =  X i Hii=1nV =  X iVii=1nU = n1U 1 + n2U 2 + n3U 3 + n4U 4 + K + nnU n =  niU iH =  ni H ii=1 nXi =niV =  niVii= 1ni=1ni= 1niWe can manipulate partial molar quantities in a manner identical to the way we manipulate total quantities. As with total state functions, we cannot know absolute values, only differences (except for V and S)!5In the case of the volume of a two-component system, e.g., NaCl-H2O, we can write:V = X NaCl NaCl + X H 2OVH 2O V61Schematic plot of the molar volume of aqueous NaCl solutions as a function of mole fraction of NaCl .HOW TO DETERMINE PARTIAL MOLAR VOLUMERefer to the previous diagram. Triangles A and B are similar, so it is true that:X NaCl /(V solution - x ) = X H 2O /( y - Vsolution )VyX NaCl y - X NaClV solution = X H 2 OV solution - X H 2O xAx(XH 2O+ X NaCl ) Vsolution = X NaCl y + X H 2O xsoBXNaClXHVsolution2ObutX NaCl + X H 2 O = 1VNaCl = yV solution = X NaCl y + X H 2O xVH 2O = xcomparison with shows thatV = X NaClVNaCl + X H2OVH2O0.00.20.40.60.81.07X NaClSo the partial molar volumes can be determined from the intercepts of a line tangent to the plot of volume vs. mole fraction. 8 G  Gi =   µi  n    i T , P,n jAbsolute Molar Gibbs Free EnergyPARTIAL MOLAR FREE ENERGY THE CHEMICAL POTENTIALChemical potentialSchematic plot of chemical potential vs. mole fraction for a binary systemµBoo + X Aµ A o X BµBµAoG mixingThe previous relationships also apply:µBG =  ni µ ii= 1nG =  X i µii=1nIt can also be shown that: U µi =   n  i   A   H  =  =     S V ,n j  n i T ,V ,n j  ni ,    S ,P ,n jµA0.0 0.2 0.4 0.6 0.8 1.0109XBCOMPOSITIONAL CHANGESThe Master equations that we developed previously for one-component systems can now be written as:dU = TdS - PdV +  µ i dnii =1 n nAN ADDITIONAL REQUIREMENT FOR EQUILIBRIUMConsider a system with components i, j, k, l, ... distributed among phases , , , , ... At equilibrium it must be true that: µi  = µi  = µ i = µ i = ... µj  = µj  = µ j = µ j = ... µk = µk = µk = µk = ... µl  = µl  = µ l = µ l = ... etc.12dH = TdS + VdP +  µ i dnii =1dG = - SdT + VdP +  µ i dnii =1 nndA = - SdT - PdV +  µ i dnii =1112Chemical potentials represent the slope of the Gibbs free energy surface in compositional space. Thus, a component will move from a phase in which it has a high chemical potential, to one in which it has a low chemical potential, until its chemical potential in all phases is the same. Specific example: Consider a silicate melt in equilibrium with forsterite (Mg 2 SiO4), and enstatite (Mg 2 Si2O6 ). At equilibrium the following must be true: µ Mgmelt = µ Mg Fo = µ Mg En µSi melt = µ SiFo = µ SiEn µO melt = µ OFo = µOEn13GIBBS-DUHEM EQUATIONFor a homogenous phase of two components, A and B, the Master Equation becomes:dG = - SdT + VdP + µ Adn A + µ B dnBIf we now specify equilibrium at constant T and PdG = µ Adn A + µ B dnB = 0Now, we have shown above thatG = µ A nA + µ B nBDifferentiating this we obtaindG = µ AdnA + µ B dnB + nA dµ A + nB dµ B14GIBBS-DUHEM EQUATION CONTINUEDAt equilibriumdG = µ A dnA + µ B dnB + nAdµ A + nB dµ B = 0GIBBS-DUHEM EQUATION CONTINUEDStarting with the expression:n Adµ A + n B dµ B = 0Substituting the previous expressiondG = µ Adn A + µ B dnB = 0If we divide through both sides by dXA we get µ   µ  n A A  + nB  B  =0  X   X   A  T , P ,nB  A  T , P ,nBwe obtainn Adµ A + n B dµ B = 0And now dividing by nA + n B we get: µ   µ  X A A  + (1 - X A ) B  =0  X   X   A T , P ,nB  A T , P ,nB15 16In the general case we get the Gibbs-Duhem equation n n dµ = 0i =1 i iAPPARENT MOLAR QUANTITIESAlthough in principle, partial molar quantities can be measured from intercepts of lines tangent to a plot of state functions vs. mole fraction as outlined previously, they are not determined this way in practice. In practice, apparent molar quantities are determined. For a state function like volume, the apparent molar volume,  V, is given by  V = (V - n 1V1 °)/n 2 where n 1 , and n2 are the number of moles of solvent and solute, respectively, and V 1 ° is the molar volume of pure solvent.17Total volume of a solution as a function of solute concentration. Illustrates the difference between partial and apparent molar volume.Volume cm3 (V)Volume of pure solventn2 VV2Volume attributed to solute2n1 V1oVolume attributed to solvent0n2Moles of solute n2183Comparison of apparent molar and partial molar volumesThe apparent molar volume is the volume that would be attributed to one mole of solute in solution if it is assumed that the solvent contributes the same volume it has in the pure state. Starting with the definition of apparent molar volume:  V = (V - n 1V1 °)/n 2 we can rearrange to get V = n1 V1° + n 2  V and dividing by (n1 + n2 ), V = X1 V1 ° + X2  V Thus, the volume of solution can be calculated knowing  V instead of the partial molar volume.19V 2o V1o V1VV2V20.00.20.40.60.81.020X2PARTIAL MOLAR VOLUMES FROM APPARENT MOLAR VOLUMES V   ( n V o + n2V )   V2 =  = 1 1  n     n2  2 T ,P , n1  T , P ,n1  V2 = n2  V  n  2   + V   n1or   V2 = m V  + V  m n1II. IDEAL SOLUTIONSIf V measurements are fit by an equation of the type: V = a + bm + cm 2 then we have V 2 = m(b + 2cm) + a + bm + cm2 or V2 = a + 2bm + 3cm221 22THERMODYNAMICS OF IDEAL SOLUTIONSAn ideal solution is one that satisfies the following equation: µi - µ i° = RT ln Xi where µ i is the chemical potential of some component i in a solution and µ i° is the chemical potential of that component in the pure form. Recall that G =  X i µiiG ideals o l' n =  X iµi + RT  X i ln X io i iG mech mix =  X i µiioGidealsol'n = Gmech mix + RT  X i ln X iiGideal mix = Gideal s o l' n - Gmech mix = RT  X i ln X iisubstituting we getGidealsol' n =  X i ( µio + RT ln X i )i23These equations tell us that the free energy of an ideal solution is the sum of two terms: the free energy of a mechanical mixture, and a free energy of ideal mixing.244ENTHALPY AND VOLUME OF AN IDEAL SOLUTIONH ideal s o l'n = H mech mix =ENTROPY OF MIXINGGideal mix = RT  X i ln X iiX Hi i io i=  XiH iioVideal s o l'n = Vmech mix =  X iVi =o XVioIf i is a pure substance.i ibut we have and soGideal mix = Hideal mix -TS ideal mix Hideal mix = 0iThere is no volume or enthalpy change upon ideal mixing. In other words:Gideal mix = - TSideal mix = RT  X i ln X i Sideal mix = - R  X i ln X iiVideal mix = H ideal mix = 0However, there is a change in entropy upon ideal mixing. Because the solution is more disordered, 25 entropy increases!Thus, the only contribution to Gideal mix is an entropy contribution!26µB oMolar Gibbs Free EnergyG mech mixo µ + = XA Ao X Bµ BMolar VolumeµA oV sol'n=o X AV A+ XB VBoVB oGideal sol'non soluti idealV Ao0.00.20.40.60.81.0270.00.20.40.60.81.028XBXBMolar EnthalpyH sol'nHAoH = XA A lution eal so ido+ X BHBoHB oIII. FUGACITY AND ACTIVITY0.00.20.40.60.81.0XB29 305FUGACITYStarting with dG = VdP - SdT at constant T this becomes dG = VdP For an ideal gas dG = (RT/P)dP = RT dln P This is true for ideal gases only, but it would be nice to have a similar form for real fluids. dG = RT dln  where  is the fugacity  = /P  1 as P  0  is the fugacity coefficient  = P Fugacity may be thought of as a thermodynamic pressure; it has units of pressure.31MEASUREMENT OF FUGACITY G    =V  P  T   (RT ln f )   RT ( ln f )    =  =V P P  T  T V  ( ln f )    =  P T RT V   ln f =  P  RT V   ln f -  ln P =   P -  ln P  RT 32 V  1  ln  f  =  ln  =   P  P -   P    P  RT  1 V  ln  =  -  P RT P   V 1 ln  =   -  P RT P  0PZ  ( ln f )    = P T P  ln f = Z P  P = Z  ln P ln f -  ln P = Z  ln P -  ln P  f  =  ln  = ( Z - 1)  ln P =  Z - 1  P  ln      P  P ( )Alternatively, we can begin again with: V  ( ln f )    =  P T RT But we now define the compressibility factor Z: PV Z= RT  ln = P 0P0 Z - 1   P  P P  Z -1  ln  =    dP 0  P 33 34The above equation is the basis of the experimental determination of fugacities from P-V-T data. We can substitute into the integral (Z-1)/P calculated from any equation of state, or we can integrate graphically.CALCULATION OF EQUILIBRIUM BOUNDARIES INVOLVING FLUIDSA number of mineral reactions involve only solid minerals and either H 2 O or CO2, e.g. KAl3 Si3O10 (OH)2  KAlSi3 O8 + Al2 O3 + H 2 O or CaCO3 + SiO2  CaSiO3 + CO2 or in general A(s)  B(s) + C(fluid) rV = VB + VC - VA = (VB - VA) + V C rV = sV + Vfluid36(Z-1)/Parea = ln 0PP356VdP =   VdP +  Vr sfluiddPThe pressure integral for the solids is then evaluated using the constant sV approximation and that for the fluid is evaluated using fugacities.rV° = V°Kspar + V°cor + V°H 2 O - V°musc rV° = (V°Kspar + V° cor - V° musc) + V°H2 O rV° = sV° + V°H2 OPf  P=1  rVdP =  sV ( P - 1) + RT ln P f P=1   PVr P roP P  f  o dP =  sV odP + VH 2O dP =  sV o ( P - 1) + RT ln H2O , P   fH O, 1bar  P Pr r  2 For the muscovite breakdown reaction above, we can start with the equation:o o     o o r GT , P = r GTr , Pr +    r G dT +   r G dP T P = P  P T = T   r   r Tr Pr T P o = r GT r, Pr -  r S odT +  rV odP Tr Pr37  T  o o 0 = r GTr -  r STr (T - Tr ) +  r aT - Tr - T ln  T   r   2 2  rb r c (T + Tr - 2TTr ) 2 2 o + (2TTr - T - Tr ) + +  sV ( P - 1) 2 2TTr2  fH O , P  2  + RT ln f   H 2O, 1bar A function of pressure and temperature!38TPFUGACITIES IN GASEOUS SOLUTIONSStarting with the following, in terms of partial molar volumes: Vi  ( ln f i )    =  P T RT We obtain the expression for the fugacity coefficient of a component in a solution:P 1  V ln i =   i -  P P 0  RT39ALL CONSTITUENTS HAVE A FUGACITYThe expression dG = RT dln  may be integrated between two states 1 and 2 to give: G2 - G1 = RT ln ( 2 /1 ) This equation applies to a pure one-component system. For a solution we must use chemical potentials and we write: µi&quot; - µi' = RT ln (i&quot;/ i') This equation makes no stipulation as to the state of component i, and can therefore refer to solid, liquid or gas.40· Solids and liquids therefore are also associated with a fugacity. In some cases, this fugacity can be thought of as a vapor pressure. Fugacity can also be thought of as an escaping tendency. · However, in some cases, a vapor phase may not exist, but a fugacity always exists. One must realize that the fugacity is a thermodynamic model parameter, not always an approximation to a real pressure. · Fugacities of solid phases or individual components of solid solutions are not generally known. · Fugacities are absolute physical properties.41ACTIVITIESThe absolute values of the fugacities of solids and liquids cannot always be determined, but their ratios can be. Consider µ i&quot; - µi' = RT ln (i&quot;/ i') If we let one of these states be a reference state, this can be rewritten: µi - µ i° = RT ln ( i/i°) We now define the activity of constituent i to be ai =  i/i° Thus µi - µ i° = RT ln ai427DALTON'S LAWDalton (1811) discovered that, at low total pressures, a mixture of gases exerts a pressure equal to the sum of the pressures that each constituent gas would exert if each alone occupied the same volume. Strictly true only for ideal gases, but is approximately true at low total pressure where real gases approach ideality. For each gas we have: P1 V = n 1 RT P2 V = n 2 RT etc. For the gas mixture we have: PtotalV =  ni RTi43If we divide the expression for each constituent by the expression for the mixture we obtain:P n 1 = 1 = X1 P  niiP2 n = 2 = X2 P  niietc.or P1 = X 1 · Ptotal P2 = X 2 · Ptotal etc. Ptotal = P1 + P 2 + P3 + ... P1 , P 2 , etc. are called the partial pressures.44HENRY'S LAWHenry (1803) was studying the solubility of gases in liquids. He found that the amount of gas dissolved in a liquid in contact with it was directly proportional to the pressure on the gas, i.e., Pi = Kh,i· Xi Kh,i is a constant called the Henry's Law constant. In practice, this law hold's only at relatively low values of P i.RAOULT'S LAWRaoult (1887) studied vapor-liquid systems in which two or more liquid components were mixed in known proportions and the liquid was equilibrated with its own vapor. The composition of the vapor was then determined. The total vapor pressure of the system was low, so the vapor behaved ideally and conformed to Dalton's law. In such systems, the partial pressures of the gaseous components were found to be a linear function of the their mole fraction in the liquid.4645Thus, for a binary system A-B, he obtained: PA = XA·PA º and PB = X B ·PB º where P Aº and PB º are the vapor pressures of pure components A and B, respectively.P A 108oPtota lP6PB4o20 0.0 0.2 0.4 0.6 0.8 1.0The only way that such a simple relationship as Raoult's law can hold is if the intermolecular forces between A-A, B-B, and A-B are identical. Solutions in which this is the case are called ideal solutions. The most general way of expressing Raoult's Law is: Pi = Xi·Piº Very few systems follow Raoult's Law over the entire range of composition from Xi = 0 to Xi = 1. However, Raoult's Law often applies to the solvent in dilute solutions, whereas the solute in dilute solutions follows Henry's Law.47 48AXBB8Partial pressure in the mixture acetone-chloroform at 35.2°C. This mixture exhibits negative deviations from Raoult's LawPartial pressure in the mixture carbon disulfide-acetone at 35.2°C. This mixture exhibits positive deviations from Raoult's Law4950THE GIBBS-DUHEM EQUATION REVISITEDPreviously we derived the Gibbs-Duhem equation for a binary solution:(1 - X B ) µA      X B T ,P , nA µ  + XB  B  =0  X   B T , P ,n AThis equation shows that the slopes of tangents to curves of chemical potential vs. mole fraction for binary solutions are not independent of one another.  µ A  For example, if XB = 0, and  X  has a finite  BT ,P , nAThis can be rearranged to give: µ A    X    B  T , P , nA XB =-  µ B  (1 - X B )   X    B  T , P , nA51 µ  =. value, then  B   X   B T , P ,nAIf XA = 0.5, then   µ A   µ   = - B  etc.   X   X B T , P ,n A  B T , P ,n A52Chemical potentials in solutions of carbon disulfide and acetone.THE DUHEM-MARGULES EQUATIONStarting with the Gibbs-Duhem equation:(1 - X B ) µ A     Gibbs- Duhem Equation  X B  T , P, n A µ  + X B B  =0  X   B  T ,P , n A µ A    X    B T , P ,nA XB =-  µ B  (1 - X B )    X   B T , P ,nAIf we recall that dµi = RT dln  i the substitution and obtain:we can make(1 - X B )  ln f A       X B  T ,P , n A  ln f B  = -X B   X    B  T ,P ,n A53  ln f A    X    B T ,P , n XB A =-   ln fB  (1- X B )    X   B T ,P , nA549When the vapors are nearly perfect gases, we may substitute partial pressures for fugacities to obtain the approximate relation:  ln PA     X   B  T , P ,n A XB =-   ln PB  (1 - X B )    X   B  T , P ,nAApplication of the Duhem -Margules equation. Partial pressure is plotted on the Y-axis.Duhem -Margules EquationRealizing that dX B = -dX A, and that d ln P = d P/P we can rewrite this as: PA    PA  X   A T , P ,nB XA = PB  P   B  XB  X   B T , P ,nA PA   PA  X    A T , P ,nB XA = PB  P  B   XB  X   B T , P ,nA55565758THE LEWIS FUGACITY RULEThis is a variation on Raoult's Law:IDEAL MIXING AND ACTIVITYIf we compare the definition of the activity: ai =  i/i° and a rearrangement of the Lewis Fugacity Rule: Xi = fimixture /fipure we see that for solutions that obey the Lewis Fugacity Rule ai = Xi We can also now write: µi - µ i° = RT ln Xi which is considered another form of Raoult's Law.60fimixture = Xi· fipureThis states that the fugacity of a constituent in a mixture is equal to it's mole fraction times its fugacity in the pure state. Many substances that do not obey Raoult's Law do in fact obey the Lewis Fugacity Rule.fApureFugacityp f B ure0.00.20.40.60.81.0AXBB5910Activity relations for an ideal binary system.1.0NON-IDEAL MIXINGaA aBActivity0.00.20.40.60.81.0XBIt turns out that these relations hold not only for liquid and gaseous solutions, but also for solid solutions.· As already discussed, most real solutions do not conform to Raoult's Law over the entire compositional range. · However, whether solid, liquid or gas, in many solutions, the component in excess (solvent) follows Raoult's Law and the minor component (solute) follows Henry's Law over a limited range at low mole fractions.61 62Positive deviation from Raoult's Law1.0ANOTHER NIFTY APPLICATION OF THE GIBBS-DUHEM EQUATIONTask: Prove that, if the solute in a binary solution obeys Henry's Law, then the solvent obeys Raoult's Law. Starting with the Gibbs-Duhem equation for a binary systemn Ad µ A + n B d µ B = 00.8aAaBRaoultian activity0.60.4and dividing through both sides by nA + n B we get0.2X Adµ A + X B dµ B = 0X Ad µ A = - X B d µ B0.0 0.0 0.2 0.4XB0.60.81.063At low pressures we havedµi = RT dln Pi64So we can now write: X A d ln PA = - X B d ln PB Now if component A is the solute and obeys Henry's Law we have: PA = Kh,A · XA Taking the natural logarithm of both sides we have: ln PA = ln Kh,A + ln XA Now differentiating we get: d ln P A = d ln XA so now X Ad ln X A = - X B d ln PBXA dX A = - X B d ln PB XANow-dXB = dXAso- dX B = - X B d ln PB dX B = d ln PB d ln X B = d ln P B XBNow we integrate from X B = 1 to X B = XBlnwhere P B° is the partial pressure of B when XB = 1, i.e., the partial pressure of pure B.ln X B = ln PB PBo XB = PB o PB66XB P = ln B 1 Po BdX A = - X B d ln P B65PB = X B  PBoRaoult's Law!11ACTIVITY COEFFICIENTSThe ideal solution is useful as a model with which real solutions are compared. This comparison is effected by taking the ratio of the activity of the real solution relative to that of the ideal solution. This ratio is called the activity coefficient. Deviations from Raoult's Law are expressed by the Raoultian activity coefficient  R ai =  R,iXiDeviations from Henry's Law are expressed by the Henryian activity coefficient  H ai =  H,iXi Activity coefficients, because they are ratios of activities, are unitless. A major difference between the two types of activity coefficients is that: R  1 as X  1, but H  1 as X  0 Thus,  H is usually more useful for solutes in dilute solutions.6867STANDARD STATESBecause the activity is the ratio of two fugacities, i.e., ai =  i/i° the value of the activity depends on the reference state chosen for i°. This state we usually refer to as the standard state. The choice of the standard state is completely arbitrary. The standard state need not be a real state. It is only necessary that we be able to calculate or measure the ratio of the fugacity of the constituent in the real state to that in the standard state. 70IV. STANDARD STATES69A STANDARD STATE HAS FOUR ATTRIBUTES· · · · Temperature Pressure Composition A particular, well-defined state (e.g., ideal gas, ideal solution, solid, liquid, etc.)STANDARD STATES FOR GASESSingle Ideal Gas Starting with the relationship: µ2 - µ1 = RT ln (P2 /P1 ) we can assign our standard state to be the ideal gas at 1 bar and any temperature. In this case we can write: µ - µ° = RT ln P and µ - µ° is the difference in chemical potential between an ideal gas at T and P, and an ideal gas at T and 1 bar.If desirable, we can permit T or P to be variable, i.e., on a sliding scale.717212Ideal mixture of ideal gases For such a mixture we can write: µ1 - µ1 ° = RT ln [(X 1 P)/(X1 P)°] If we choose our standard state to be the pure ideal gas 1 at any temperature and 1 bar, then X1° = 1 and P° = 1, so µ1 - µ1 ° = RT ln (X1 P) = RT ln P 1 Non-ideal gases For non-ideal gases we would write: µ1 - µ1 ° = RT ln (f1 /f1°) but recall that lim (fi/Pi) Pi 0 = 1. So if we chose our standard state to be the pure, ideal gas at any temperature and P = 1 bar, we getµ1 - µ1 ° = RT ln f1 This equation is frequently written, but rarely understood. It only has meaning if the standard state is specified to be the pure ideal gas at any temperature and 1 bar. This is the most commonly chosen standard state for gases and supercritical fluids. However, there is no reason why this particular standard state has to be chosen. We could just as easily choose: 1) the pure ideal gas at any T and 10 bars; 2) a pure real gas at 25°C and 1 bar; 3) a specific mixture of gases at any T and  bars; or 4) any other welldefined standard state.7473LIQUIDS AND SOLIDSThe following equation applies to liquids and solids as well: µi - µ i° = RT ln (fi /fi°) Fixed pressure standard state The standard state is chosen to be the pure phase at the temperature of interest and 1 bar. Then fi° = 1, so a i = fi. In this case, it is necessary to know fi at each and every set of P-T conditions of interest. Variable pressure standard state The standard state is the pure phase at the pressure and temperature of interest.75Under these conditions, fi = fi°, so a i = 1. The only way the activity of a solid deviates from unity under this standard state is when the solid is not pure, but is a solid solution. It may seem that the second standard state is easier to deal with in terms of pressure corrections. However, with the first standard state, the pressure correction is applied to fi, whereas in the second standard state, the correction is applied to µ i°. In either case, volume data for the constituent are required to make the correction.76AQUEOUS SOLUTIONSThe activities of solutes in dilute solutions are more closely approximated with Henry's Law than Raoult's Law. Thus, a somewhat different standard state is applied. We start with the equation expressing the difference in chemical potentials between two solutions with different molalities: µi&quot; - µ i' = RT ln (H&quot;m&quot;/ H'm') If we let one solution be the standard state, we can write: µi - µ i° = RT ln ( Hm)/(Hm)°77We then define the standard state to be the hypothetically ideal one-molal solution at the temperature and pressure of interest. Under these conditions  H = 1 and m = 1, so ( Hm)° = 1, and we write: µi - µ i° = RT ln H m This somewhat strange standard state is necessary, because if we let the standard state be the infinitely dilute solution, we would have  H = 1 and m = 0, so ( Hm)° = 0, which would result in an undefined value of: µi - µ i° = RT ln ( Hm)/ ( Hm)°7813`

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