Chapter 5. Simple mixtures

1. Chapter 5. Simple mixtures At equilibrium the chemical potential of a species is the same in every phase. How to express the chemical potential of a substance in terms of its mole fraction in a mixture in Raoult’s and Henry’s laws. The effect of a solute on certain thermodynamic properties of a solution (boiling point; melting point, osmosis ). To express the chemical potential of a substance in a real mixture in terms of a property is known as the activity. Chemistry deals with mixtures, including mixtures of substances that can react together.  at first stage, a binary mixtures (a mixture of two components) that do not react together is considered.

2. The thermodynamic description of mixtures 5.1 Partial molar quantities: (a) partial molar volume: At 25oC, 1 mol H2O was added to a huge volume of water, the volume increases by 18 cm3  18 cm3 mol-1 is the molar volume of pure water. When 1 mol H2O was added to a huge volume of ethanol, the volume increases by only 14 cm3  14 cm3 mol-1 is the molar volume in pure ethanol.  In general, the partial molar volume of a substance A in a mixture is the change in volume per mole of A added to a large volume of the mixture. The partial molar volumes of the components of a mixturevary with composition because the environment of each type of molecule changes as the compositions from pure A to pure B.  modification of the mutual forces.

3. VJ = (V/nJ)p,T,n’ the subscript n’ signifies the amount of all other substances present are constant. Slope of a plot of V vs. nJ.  VJ

4. VJ value is dependent on the composition. When the composition of the mixture is changed by addition of dnA and dnB  dV = (V/nA)p,T,nBdnA + (V/nB)p,T,nA dnB= VA dnA + VB dnB When the composition is held constant, as the amount of A and B are increased, the final volume of a mixture can be calculated by integration. V =  VA dnA+  VB dnB = VA  dnA + VB  dnB =VAnA + VBnB Because V is a state function the final result is valid however the solution is in fact prepared.

5. Illustration 5.1 A polynomial fit to measurements of the total volume of a water/ethanol mixture at 25oC that contains 1.00 kg of water. V = 1002.93 + 54.6664 x – .36394 x2 + 0.028256 x3 VE = (V/nE) = (v/x) dv/dx = 54.66664 – 2(0.36394) x + 0.084768 x2 Molar volumes are always positive but partial molar volume quantities need not be. Exp. The limiting partial molar volume of MgSO4 in water is – 1.4 cm3 mol-1.  The salts break up the open structure of water as ions become hydrated.

6. (b) Partial molar Gibbs energies The chemical potential energy is defined as the partial molar Gibbs energy. J = (G/nJ)p,T,n’  The chemical potential is the slope of the Gibbs energy against the amount of the component J The total Gibbs energy of a mixture is G = nAA + nBB. The fundamental equation of chemical thermodynamics relates the change in Gibbs energy to changes in pressure, temperature, and composition: dG = Vdp – SdT + AdnA + BdnB + .

7. At constant pressure and temperature dG = AdnA + BdnB + . dG = dwadd, max  dwadd,max = AdnA + BdnB + . The additional work (non-expansion, electrical work) can arise from the changing composition of a system (or electrochemical cell). (c) The wider significance of the chemical potential: G = U + pV – TS ; U = – pV + TS + G  dU = – pdV – v dp + SdT + TdS + Vdp – SdT + AdnA + BdnB + . = – pdV + TdS + AdnA + BdnB + . At constant S and V; dU = AdnA + BdnB + .  J = (U/nJ)S,V,n’ J = (H/nJ)S,p,n’ ; J = (U/nJ)V,T,n’

8. (d) The Gibbs-Duhem equation: At constant pressure and temperature, the total Gibbs energy of a binary mixture is G = nAA + nBB dG = AdnA + BdnB + nAdA + nBdB  dG = AdnA + BdnB  nAdA + nBdB = 0 The Gibbs–Duhem equation is J nJdμJ = 0. In a system in which the compositions have mutual interaction or are mutually exchangeable, the chemical potential of one component can not change independently of the chemical potentials of the other components. In a binary mixture, dB = – (nA/nB) dA  The same line of reasoning applies to all partial molar quantities. Exp. Where the Vwater increases, the Vethanol decreases. In practice, the Gibbs-Duhem equation is used to determine the partial molar properties of one component in a binary mixture from the second component.

9. Example 5.1 Using the Gibbs-Duhem equation: The experimental values of the partial molar volume of K2SO4(aq) at 298 K are found to fit the expression: vB = 32.280 + 18.216 x1/2 x is the numerical value of the morality of K2SO4 (x =b/b). Use the Gibbs-Duhem equation to derive an equation for molar volume of water in the solution. vwater = 18.079 cm3 mol-1 nA dvA + nB dvB = 0  dvA = – (nA/nB) dvB vA = vA* –  (nA/nB) dvB ; vA*: vA/cm3 mol-1 is the numerical value of the molar volume of pure A. dvB/dx = 9.108 x–1/2 vA = vA* – 9.108 o (nA/nB) x–1/2 dx nB/nA = nB/(1 kg)/MA = nBMA/1kg = bMA = x bMA  vA = vA* – 9.108 MAbo x1/2 dx = vA*– 2/3 {9.108 MAb(b/b)3/2} b/b b/b

10. Example 5.1 Using the Gibbs-Duhem equation:  vA /(cm3 mol-1) = 18.079 – 0.1094 (b/b)3/2 5.2 The thermodynamics of mixing: At constant temperature and pressure, systems tend toward lower Gibbs energy. A spontaneous mixing process is that of two gases introduced into the same container. The mixing is spontaneous, so it must correspond to a decrease in G.

11. (a) The Gibbs energy of mixing of perfect gases:  = Gm  =  + RT ln p/p : standard chemical potential, the  of the pure gas at 1.0 bar. For simpler notation, the p/p is replaced by p p: the numerical value of p in bars.  =  + RT ln p The Gibbs energy of the total system: Gi = nAA + nBB = nA (A + RT ln p) + nB (B + RT ln p) After mixing the partial pressure of the gases are pA and pB with pA + pB = p. Gf = nA (A + RT ln pA) + nB (B + RT ln pB) Gmix = nA RT ln (pA/p) + nB RT ln (pB/p)

12. Use the relation between partial pressure and mole fraction; pJ/p = J  Gmix = nRT (A ln A + B ln B) Because J < 1, Gmix < 0 Gmix is negative for all compositions.

13. Exp. 5.2 Calculate a Gibbs energy of mixing Gi = (3.0 mol) ((H2) + RT ln 3p) + (1.0 mol) ((N2) + RT ln p) Gf = (3.0 mol) ((H2) + RT ln 3/2 p) + (1.0 mol) ((N2) + RT ln ½ p) Gmix = (3.0 mol) RT ln (3/2 p/3p) + (1.0 mol) RT ln (½ p/p) = – 6.9 kJ Gmix = the mixing itself and the changes in pressure of the two gases. When 3.0 mol H2 mixes with 1.0 mol N2at same pressure, with the volumes of the vessels adjusted accordingly, Gmix = – 5.6 kJ

14. (b) Other thermodynamic mixing functions: (G/T)p,n = – S  The entropy of mixing of two perfect gases is given by ΔmixS = (mixG/T)p,nA,nB = –nR(xA ln xA + xB ln xB) > 0 For perfect gases, the enthalpy of mixing ΔmixH = 0. The enthalpy of mixing is zero because there are no interactions between molecules forming in the gas mixture.

15. 5.3 The chemical potentials of liquids: We should denoted quantities to pure substances by a superscript *. A* is the chemical potential of the liquid A*(l). A* =A0 + RT ln pA* (relative pressure pA* = pA/p0) If another substance (solute) is also present in the liquid, the chemical potential is changed to A. A = A0 + RT ln pA (pA = pA/p0)  A0 = A* – RT ln pA*  A = A* – RT ln pA* + RT ln pA = A*+ RT ln pA/pA*

16. In a series of experiments on mixtures of closely related liquids, the partial vapor pressure of each component to its vapor pressure of pure liquid, pA/pA* is approximately equal to the mole fraction of A in the liquid mixture. Raoult’s law: pA = ApA*

17. Mixtures that obey the law throughout the composition range from pure A to pure B are called ideal solution.  A = A* + RT ln A Molecular interpretation 5.1 The presence of a second component reduces the rate at which A molecules leave the surface of the liquid but does not inhibit the return rate. rate of vaporization = k A rate of condensation = k’pA At equilibrium, k A = k’pA pA = (k/k’) A For pure liquid, A = 1  pA* = (k/k’)

18. Some solutions depart significantly from Raoult’s law. The law is therefore a good approximation for the properties of the solventif the solution is dilute.

19. (b) Ideal-dilute solution: William Henry: for real solutions at low concentrations, although the vapor pressure is proportional to its mole fraction, the constant of proportionality is not the vapor pressure of the pure substance. pB = BKB Mixtures for which the solute obeys Henry’s law and the solvent obeys Raoult’s law, are called ideal-dilute solutions.

20. In a dilute solution: The solvent molecules are in an environment very much like the one they have in the pure liquid. In contrast, the solute molecules are surrounded by the solvent molecules, which is entirely different from their environment when pure. Unless the solvent and solute molecules happen to very similar.  Raoult’s law.

21. Example. Investigate the validity of Raoult’s and Henry’s laws: Rault’s law: pJ = JpJ* Henry’s law: pJ = J KJ KJ is an empirical constant chosen so that the plot of the vapor pressure of J against its molar fraction is tangent to the experimental curve at J = 0. K(propane) = 23.3 kPa K(chloroform) = 22.0 kPa

22. In practice, Henry’s law is expressed in terms of mobility, b, of the solute. pB = bB KB  respiration: oxygen, diving, mountaineering, anesthetics.

23. I5.1 Gas solubility and breathing: We inhale about 500 cm3 of air with each breath we take. As the diaphragm (橫隔膜) is depressed and chest expands  a decrease in pressure of 100 Pa related to atmospheric pressure. The total volume in the lungs is about 6 dm3 (L). Some air remains in the lungs at all time to prevent the collapse of the alveoli (肺泡). Alveolar gas is in fact a mixture of newly inhaled air and air about to be exhaled. pO2 in arterial blood is about 40 Torr  pO2 in freshly inhaled air is about 104 Torr. Arterial blood remains in the capillary passing through the wall of the alveolus for about 0.75 s, but such is the steepness of the pressure gradient that it becomes fully saturated with oxygen in about 0.25 s. If the lungs collect fluid, then the respiratory membrane thickens, diffusion is greatly slow, and body tissue begin to suffer from oxygen starvation.

24. CO2 moves in the opposite direction across the respiratory tissue, but the pCO2 gradient is much less, corresponding to about 5 Torr in blood and 40 Torr in air at equilibrium. Because the CO2 is much more soluble in the alveolar fluid than oxygen is, equal amount of oxygen and carbon dioxide are exchanged in each breath. A hyperbaric oxygen chamber, high pO2, is used to treat certain types of disease, such as CO poisoning, disease caused by anaerobic bacteria. In scuba diving, one unfortunate consequence of breathing air at high pressure is that N2 is much more soluble in fatty tissue than in water, so it tends to dissolve in central nervous system, bone marrow, and fat reserves.  nitrogen narcosis (氮醉).

25. The Properties of Solutions: 5.4 Liquid mixtures: (a) Ideal solution The Gibbs energy of mixing of two liquids to form an ideal solution is calculated as in exactly the same way as for two gases. Before mixing: Gi = nAA* + nBB* After mixing: Gf = nA {A* + RTln A} + nB {B* + RTln B}  mixG = nRT {A RTln A + B RTln B} ; n = nA + nB mixS = nR {A RTln A + B RTln B}  mixH = mixG + TmixS = 0 The mixV = 0 (G/p)T = V  ( mixG/p)T = mixV = 0 mixG is independent of pressure.

26. In a perfect gas there are no forces acting between molecules  In an ideal solutions, there are interactions, but the average energy of A–B interactions in the mixture is the same as the average energy of A–A and B–B interactions in the pure liquids. Real solution: No only may there be enthalpy and volume changes when liquids mix, but there may also be an additional contribution to the entropy arising from the way in which the molecules of one type might cluster together instead of mingling freely with the others. Large and positive mixH or if mixS << 0, then the mixG might be positive for mixing.  separation is spontaneous and the liquid may be immiscible. When they are miscible only over a certain range compositions, the liquid might be partially miscible.

27. (b) Excess functions and regular solutions: An excess function (XE) is the difference between the observed thermodynamic function of mixing and the function for an ideal solution. SE = mixS – mixSideal A regular solution: as one in which two kinds of molecules are distributed randomly but have different energies of interactions with each other.  HE  0 but SE = 0. Suppose the HE depends on compositions:  HE = nRTAB

28. HE = nRTAB  is a dimensionless parameter that is a measure of the energy AB interactions related to that of the AA and BB interactions. If  < 0, mixing is exothermic and solute-solvent interactions are more favourable than solvent-solvent and solute-solute interactions. If  > 0, mixing is endothermic. For a regular solution: mixG = nRT {AlnA + BlnB + AB}

29. mixG = nRT {AlnA + BlnB + AB} The important feature is that  > 2 the graph shows two minima separated by a maximum. At  >2, the system will separate spontaneously into two phases with compositions corresponding to the two minima.

30. 5.5 Colligative properties: (眾數性質) A colligative property is a property that depends only on the number of solute particles present, not their identity.  the elevation of boiling point, the depression of freezing point, osmotic pressure. Throughout the following: (1). The solute is not volatile. (2). The solute does not dissolve in the solid solvent. (a) The common feature of colligative properties: For an ideal-dilute solution, the reduction of the chemical potential of the solvent is from A* for pure solvent to A* + RT lnA when a solute is present.

31. There is no direct influence of the solute on the chemical potential of the solvent vapor and the solid solvent because the solute appears in neither the vapor nor the solid. A reduction in chemical potential of solvent  boiling point is raised and freezing point is lowered.

32. Molecular interpretation 5.2 The lowering of the chemical potential is not the energy of interaction of the solute and solvent particles, because it occurs even in an ideal solution. If it is not an enthalpy effect, it must be an entropy effect. S = kB ln The vapor pressure reflects the tendency of the solution towards greater entropy. When a solute is present, there is an additional contribution to the entropy of the liquid, even in an ideal solution.  S(solution  gas) < S(solvent  gas)  lower vapor pressure and higher boiling point.

33. (b) The elevation of boiling point: The heterogeneous equilibrium between the solvent vapor and the solvent in solution A*(g) = A*(l) + RT ln A The presence of a solute at a mole fraction B causes an increase in normal boiling point from T* to T* + T, T = KB ; K = RT*2/ vapH

34. Justification 5.1 ln A = (A*(g) – A*(l))/RT = vapG/RT  Gibbs-Helmholtz equation: ((G/T)/T)p = – H/T2  d lnA /dT = 1/R d (vapG/T)/dT = – vapH/RT2 0 d lnA = – (1/R) T* vapH/T2 dT  ln A = ln(1 – B ) = vapH/R (1/T – 1/T*) When B << 1, ln(1 – B)  – B B = vapH/R (1/T* – 1/T) 1/T* – 1/T = (T – T*)/TT*  T/T*2 lnA T

35. For practical applications, we note that the molar fraction of B is proportional to its molality, b, in the solution.  T = Kb b

36. (c) The depression of freezing point: The heterogeneous equilibrium between pure solid solvent and solution with solute present at a mole fractionB: A*(s) = A*(l) + RT ln A  T = K’B ; K’ = RT*2/ fusH  T = Kf b Cryoscopy (Greek "freeze-viewing"): A technique for determining the molecular weight of a solute by dissolving a known quantity of it in a solvent and recording the amount by which the freezing point of the solvent drops.

37. (d) Solubility: Solubility is not strictly a colligative property (identity of the solute). Saturation is a state of equilibrium, with the undissolved solute in equilibrium with the dissolved solute. In solution B* = B*(l) + RT ln B B*(s) = B*(l) + RT ln B RT ln B = {B*(s) – B*(l)}/RT = – fusG/RT 0 d lnB = – (1/R) Tf fusH/T2 dT  ln B = fusH/R (1/Tf – 1/T)

38. The solubility of B decreases exponentially as the temperature is lowered from its melting point. High melting point, large enthalpies  low solubility. This predict fails to predict that solutes will have different solubilities in different solvents, for no solvent properties appear in the expression.

39. (e) Osmosis: Osmosis is the spontaneous passage of a pure solvent into a solution separated from it by a semipermeable membrane, a membrane permeable to the solvent but not to the solute. The osmotic pressure,  , is the pressure that must be applied to the solution to stop the influx of solvent.  transport of fluid through cell membrane, dialysis (透析), osmometry (the determination of molar mass of macromolecules). Equilibrium is reached when the hydrostatic pressure of the column of solution match the osmotic pressure.

40. The chemical potential of the solvent is lowered by the solute, but it is restored to its “pure” value by the application of pressure. The van ’t Hoff equation for the osmotic pressure is Π = [B]RT [B] = the molar concentration of the solute. Justification 5.3 On the pure solvent side: A*(p) One the solution side: A(A, p + Π) = *A(p + Π) + RT lnA *A(p + Π) = *A(p) + p Vm dp  A(A, p + Π) = *A(p) + p Vm dp + RT lnA  – RT lnA = p Vm dp Π = [B]RT

41. Because the effect is so readily measurable and large, osometry is commonly used to measure molar mass of macromolecules (exp. Protein…). As these huge molecules dissolve to produce solutions that are far from ideal,  Π = [J]RT {1+ B[J] + …………} B: osmotic viral coefficient. Exp. 5.4 Osmotic pressure of PVC: c (g dm–3) vs. h/cm [J] = c/M; osmotic pressure = gh  h/c = RT/ gM (1 + Bc/M + …….) = RT/ gM + (RTB/ gM) c  To find M, h/c vs. c , and expect a straight line with intercept RT/ gM at c = 0.  1.2 x 102 kg mol-1 = 120 kDa (1 Da = 1g mol-1)

42. I5.2 Osmosis in physiology and Biochemistry: Cell membranes are semipermeable and allow water, small molecules, and hydrated ions to pass, while blocking the passage of biopolymers. The influx of water also keeps the cell swollen, whereas dehydration causes the cell to shrink. To maintain the integrity of the cells, solution injected into the blood must be isotonic with the blood.  the same osmotic pressure. Dialysis: a common technique for the removal of impurities from solution of biological macromolecules and for the study of binding of small molecules to macromolecules. In a purification experiments, a solution of macromolecules and impurities (ions and small proteins) is place in a bag made of semipermeable membrane, and the bag is immersed in a solvent.

43. In a binding experiment, a solution contains a macromolecules ([M]) and smaller molecules A ( the total concentration of A = [A]in) in a bag. The total concentration A is the sum of [A]in = [A]free and [A]bound. At equilibrium, the chemical potential of free A in macromolecule solution is equal to that in the solution on the other side.  [A]free = [A]out (the activity coefficient of A is the same in both solutions) The [A]bound = [A]in – [A]out The average number of A molecules bound to M molecules, v, is then the ratio v = [A]bound/[M] = ([A]in – [A]out)/M

44. If there are N identical and independent binding sites on each macromolecule, each macromolecule behaves like N smaller macromolecules, M’, with same value of K for each site. The bound and unbound A molecule to each binding site in the macromolecule are in equilibrium, M’ + A  M’A K = [M’A]/[M’]free[A]free [M’A]/[M’]free = (v/N) / (1 – v/N)  K = v/N/{(1 – v/N) [A]out}  v/[A]out = KN – Kv (Scatchard equation) If a straight line is not obtained we can conclude that the binding sites are not equivalent or independent.

45. Activities Account the deviations from ideal behavior. It is important to be aware of the different definitions of standard states and activities.

46. 5.6 The solvent activity The general form of the chemical potential of a real or ideal solvent is: A = A* + RT ln (pA/pA*) For the ideal solution: A = A* + RT ln A The solvent activity of the real solution is related to its chemical potential by A = A* + RT ln aA aA is the activity of A, a kind of ‘effective’ mole fraction.  The activity is defined as aA = pA/pA*. Illustration 5.3 The vapour pressure of 0.500 M KNO3 at 100oC is 99.95 kPa, so the activity of the solution at this temperature is aA = 99.95 kPa/ 101.325 kPa = 0.9864

47. Because all solvents obey the Raoult’s law at A 1 aA  A as A  1 A convenient way of expressing this convergence is to introduce the activity coefficient, aA = A A A  1 as A  1 A = A* + RT ln A + RT ln A 5.7 The solute activity: For solutes, they approach ideal-dilute (Henry’s law) behavior as B 0. (a) Ideal-dilute solutions: (Henry’s law: pB = KBB) The chemical potential of solute, B: B = B* + RT ln (pB/pB*) = B* + RT ln (KB/pB*) + RT ln B  a new standard chemical potential B0 = B* + RT ln (KB/pB*)  B = B0 + RT ln B For ideal solution: KB = pB* ;  B0 = B*

48. (b) Real solution: We now permit deviations from ideal-dilute solution: B = B0 + RT ln aB ; aB = pB/KB It is sensible to introduce an activity coefficient through aB = BB Because the solute obeys Henry’s law: [solute] 0 aB  B and B  1 as B 0. Deviations of the solute form idealitydisappear as zero concentration is approached. Example 5.5 Measuring activity c (chloroform); pc/kPa ; pA (acetone)/kPa When chloroform is solvent: aC = pC/pC* If it is solute: aC = pc/Kc ; C = aC/C

49. Raoult’s law Henry’s law

50. (c) Activities in terms of molarities: In chemistry, compositions are often expressed as molalities, b, in place of mole fractions. B = B0 + RT ln bB The chemical potential of the solute has its standard value of o when molality of B is equal to bo (1.0 mol kg-1). as bB 0 , B –   The solute becomes increasingly stabilized. In practical consequence, it is very difficult to remove the last trace of a solute from a solution. aB = B (bB/b0) where B 1 as bB  0 For the chemical potential of a real solution at any molality:  = 0 + RT ln a