Chapter Five: Simple Mixtures PDF

FOCUS 5 Simple mixtures Mixtures are an essential part of chemistry, either in their own right or as starting materials for chemical reactions. This 5C Phase diagrams of binary systems: group of Topics deals with the rich physical...

FOCUS 5 Simple mixtures Mixtures are an essential part of chemistry, either in their own right or as starting materials for chemical reactions. This 5C Phase diagrams of binary systems: group of Topics deals with the rich physical properties of mix- liquids tures and shows how to express them in terms of thermody- namic quantities. One widely employed device used to summarize the equilibrium properties of mixtures is the phase diagram. The Topic describes phase diagrams of systems of liquids with gradually increasing complexity. In each case the phase diagram for the system sum- 5A The thermodynamic description of marizes empirical observations on the conditions under which mixtures the liquid and vapour phases of the system are stable. 5C.1 Vapour pressure diagrams; 5C.2 Temperature–composition The first Topic in this Focus develops the concept of chemi- diagrams; 5C.3 Distillation; 5C.4 Liquid–liquid phase diagrams cal potential as an example of a partial molar quantity and explores how to use the chemical potential of a substance to describe the physical properties of mixtures. The under- lying principle to keep in mind is that at equilibrium the 5D Phase diagrams of binary systems: chemical potential of a species is the same in every phase. solids By making use of the experimental observations known as Raoult’s and Henry’s laws, it is possible to express the In this Topic it is seen how the phase diagrams of solid mix- chemical potential of a substance in terms of its mole frac- tures summarize experimental results on the conditions under tion in a mixture. which the liquid and solid phases of the system are stable. 5A.1 Partial molar quantities; 5A.2 The thermodynamics of mixing; 5D.1 Eutectics; 5D.2 Reacting systems; 5D.3 Incongruent melting 5A.3 The chemical potentials of liquids 5E Phase diagrams of ternary systems 5B The properties of solutions Many modern materials (and ancient ones too) have more In this Topic, the concept of chemical potential is applied than two components. This Topic shows how phase diagrams to the discussion of the effect of a solute on certain thermo- are extended to the description of systems of three compo- dynamic properties of a solution. These properties include nents and how to interpret triangular phase diagrams. the lowering of vapour pressure of the solvent, the elevation 5E.1 Triangular phase diagrams; 5E.2 Ternary systems of its boiling point, the depression of its freezing point, and the origin of osmotic pressure. It is possible to construct a model of a certain class of real solutions called ‘regular solutions’, which have properties that diverge from those of 5F Activities ideal solutions. 5B.1 Liquid mixtures; 5B.2 Colligative properties The extension of the concept of chemical potential to real solutions involves introducing an effective concentration called an ‘activity’. In certain cases, the activity may be Web resources What is an application interpreted in terms of intermolecular interactions. An important example is an electrolyte solution. Such solutions of this material? often deviate considerably from ideal behaviour on account of the strong, long-range interactions between ions. This Two applications of this material are discussed, one from Topic shows how a model can be used to estimate the devia- biology and the other from materials science, from among the tions from ideal behaviour when the solution is very dilute, huge number that could be chosen for this centrally impor- and how to extend the resulting expressions to more con- tant field. Impact 7 shows how the phenomenon of osmosis centrated solutions. contributes to the ability of biological cells to maintain their 5F.1 The solvent activity; 5F.2 The solute activity; 5F.3 The activities of shapes. In Impact 8, phase diagrams of the technologically regular solutions; 5F.4 The activities of ions important liquid crystals are discussed. TOPIC 5A The thermodynamic description of mixtures (a) Partial molar volume ➤ Why do you need to know this material? Imagine a huge volume of pure water at 25 °C. When a fur- Chemistry deals with a wide variety of mixtures, including ther 1 mol H2O is added, the volume increases by 18 cm3 and mixtures of substances that can react together. Therefore, it follows that the molar volume of pure water is 18 cm3 mol−1. it is important to generalize the concepts introduced in However, upon adding 1 mol H2O to a huge volume of Focus 4 to deal with substances that are mingled together. pure ethanol, the volume is found to increase by only 14 cm3. ➤ What is the key idea? The reason for the different increase in volume is that the vol- ume occupied by a given number of water molecules depends The chemical potential of a substance in a mixture is a on the identity of the molecules that surround them. In the logarithmic function of its concentration. latter case there is so much ethanol present that each H2O ➤ What do you need to know already? molecule is surrounded by ethanol molecules. The network of hydrogen bonds that normally hold H2O molecules at certain This Topic extends the concept of chemical potential distances from each other in pure water does not form; as a to substances in mixtures by building on the concept result the H2O molecules are packed more tightly and so in- introduced in the context of pure substances (Topic 4A). crease the volume by only 14 cm3. The quantity 14 cm3 mol−1 It makes use of the relation between the temperature is the ‘partial molar volume’ of water in pure ethanol. In gen- dependence of the Gibbs energy and entropy (Topic 3E), eral, the partial molar volume of a substance A in a mixture and the concept of partial pressure (Topic 1A). Throughout is the change in volume per mole of A added to a large volume this and related Topics various measures of concentration of the mixture. of a solute in a solution are used: they are summarized in The partial molar volumes of the components of a mix- The chemist’s toolkit 11. ture vary with composition because the environment of each type of molecule changes as the composition changes from pure A to pure B. This changing molecular environ- ment, and the consequential modification of the forces The consideration of mixtures of substances that do not react acting between molecules, results in the variation of the together is a first step towards dealing with chemical reactions thermodynamic properties of a mixture as its composition (which are treated in Topic 6A). At this stage the discussion is changed. The partial molar volumes of water and etha- centres on binary mixtures, which are mixtures of two compo- nol across the full composition range at 25 °C are shown in nents, A and B. In Topic 1A it is shown how the partial pressure, Fig. 5A.1. which is the contribution of one component to the total pres- The partial molar volume, VJ, of a substance J at some gen- sure, is used to discuss the properties of mixtures of gases. For eral composition is defined formally as follows: a more general description of the thermodynamics of mixtures other analogous ‘partial’ properties need to be introduced.  ∂V  Partial molar volume VJ =   (5A.1)  ∂nJ  p ,T ,n′ [definition] where the subscript n′ signifies that the amounts of all other 5A.1 Partial molar quantities substances present are constant. The partial molar volume is the slope of the plot of the total volume as the amount of J is The easiest partial molar property to visualize is the ‘partial changed, the pressure, temperature, and amount of the other molar volume’, the contribution that a component of a mix- components being constant (Fig. 5A.2). Its value depends on ture makes to the total volume of a sample. the composition, as seen for water and ethanol. 144 5 Simple mixtures 58 This equation can be integrated with respect to nA and nB pro- V(C2H5OH)/(cm3 mol–1) Partial molar volume of ethanol, Partial molar volume of water, vided that the amounts of A and B are both increased in such Water a way as to keep their ratio constant. This linkage ensures that 18 56 the partial molar volumes VA and VB are constant and so can be taken outside the integrals: V(H2O)/(cm3 mol–1) 16 nA nB nA nB 54 V = ∫ VA dnA + ∫ VB dnB = VA ∫ dnA + VB ∫ dnB (5A.3) 0 0 0 0 Ethanol = VAnA + VBnB 14 0 0.2 0.4 0.6 0.8 1 Although the two integrations are linked (in order to preserve Mole fraction of ethanol, x(C2H5OH) constant relative composition), because V is a state function the final result in eqn 5A.3 is valid however the solution is in Figure 5A.1 The partial molar volumes of water and ethanol at 25 °C. Note the different scales (water on the left, ethanol on fact prepared. the right). Partial molar volumes can be measured in several ways. One method is to measure the dependence of the volume on the composition and to fit the observed volume to a function of the amount of the substance. Once the function has been found, its slope can be determined at any composition of inter- est by differentiation. Total volume, V V(b) Example 5A.1 Determining a partial molar volume V(a) A polynomial fit to measurements of the total volume of a water/ethanol mixture at 25 °C that contains 1.000 kg of water is v = 1002.93 + 54.6664z − 0.363 94z2 + 0.028 256z3 a b Amount of J, nJ where v = V/cm3, z = nE/mol, and nE is the amount of CH3CH2OH present. Determine the partial molar volume of Figure 5A.2 The partial molar volume of a substance is the slope ethanol. of the variation of the total volume of the sample plotted against the amount of that substance. In general, partial molar quantities Collect your thoughts Apply the definition in eqn 5A.1, vary with the composition, as shown by the different slopes at taking care to convert the derivative with respect to n to a a and b. Note that the partial molar volume at b is negative: the derivative with respect to z and keeping the units intact. overall volume of the sample decreases as A is added. The solution The partial molar volume of ethanol, VE , is  ∂V   ∂(V /cm3 )  cm3 A note on good practice The IUPAC recommendation is to VE =   =   ∂nE  p ,T ,nW  ∂(nE /mol)  p ,T ,n mol denote a partial molar quantity by X , but only when there is the W possibility of confusion with the quantity X. For instance, to  ∂v  =  cm3 mol −1 avoid confusion, the partial molar volume of NaCl in water could  ∂ z  p ,T ,nW be written V(NaCl,aq) to distinguish it from the total volume of the solution, V. Then, because The definition in eqn 5A.1 implies that when the composi- dv tion of a binary mixture is changed by the addition of dnA of A = 54.6664 − 2(0.363 94)z + 3(0.028 256)z 2 dz and dnB of B, then the total volume of the mixture changes by it follows that  ∂V   ∂V  dV =  dn + dnB (5A.2) VE/(cm3 mol−1) = 54.6664 − 0.727 88z + 0.084 768z2  ∂nA  p ,T ,n A  ∂nB  p ,T ,n B A = VA dnA + VB dnB    Figure 5A.3 shows a graph of this function.  5A The thermodynamic description of mixtures 145 Partial molar volume, VE/(cm3 mol–1) 56 μ(b) Gibbs energy, G 55 μ(a) 54 a b 53 0 5 10 Amount of J, nJ z = nE/mol Figure 5A.4 The chemical potential of a substance is the slope of Figure 5A.3 The partial molar volume of ethanol as expressed the total Gibbs energy of a mixture with respect to the amount by the polynomial in Example 5A.1. of substance of interest. In general, the chemical potential varies with composition, as shown for the two values at a and b. In this Self-test 5A.1 At 25 °C, the mass density of a 50 per cent by case, both chemical potentials are positive. mass ethanol/water solution is 0.914 g cm−3. Given that the partial molar volume of water in the solution is 17.4 cm3 mol−1, what is the partial molar volume of the ethanol? 54.6 cm3 mol−1 by the formula above Answer: 56.4 cm3 mol−1 by using eqn 5A.3; where µA and µB are the chemical potentials at the composi- tion of the mixture. That is, the chemical potential of a sub- stance, multiplied by the amount of that substance present in Molar volumes are always positive, but partial molar quan- the mixture, is its contribution to the total Gibbs energy of the tities need not be. For example, the limiting partial molar vol- mixture. Because the chemical potentials depend on composi- ume of MgSO4 in water (its partial molar volume in the limit tion (and the pressure and temperature), the Gibbs energy of of zero concentration) is −1.4 cm3 mol−1, which means that the a mixture may change when these variables change, and for a addition of 1 mol MgSO4 to a large volume of water results in a system of components A, B, …, eqn 3E.7 (dG = Vdp − SdT) for decrease in volume of 1.4 cm3. The mixture contracts because a general change in G becomes the salt breaks up the open structure of water as the Mg2+ and SO42− ions become hydrated, so the structure collapses slightly. dG = Vdp − SdT + µAdnA + µBdnB + …  Fundamental equation of chemical thermodynamics (5A.6) (b) Partial molar Gibbs energies The concept of a partial molar quantity can be broadened to This expression is the fundamental equation of chemical any extensive state function. For a substance in a mixture, the thermodynamics. Its implications and consequences are ex- chemical potential is defined as the partial molar Gibbs energy: plored and developed in this and the next Focus. At constant pressure and temperature, eqn 5A.6 simplifies  ∂G  Chemical potential to µJ =   (5A.4)  ∂nJ  p ,T ,n′ [definition] dG = µAdnA + µBdnB + … (5A.7) where n′ is used to denote that the amounts of all other compo- nents of the mixture are held constant. That is, the chemical po- As established in Topic 3E, under the same conditions dG = tential is the slope of a plot of Gibbs energy against the amount dwadd,max. Therefore, at constant temperature and pressure, of the component J, with the pressure, temperature, and the amounts of the other substances held constant (Fig. 5A.4). For dwadd,max = µAdnA + µBdnB + … (5A.8) a pure substance G = nJGJ,m, and from eqn 5A.4 it follows that µJ = GJ,m: in this case, the chemical potential is simply the molar That is, additional (non-expansion) work can arise from the Gibbs energy of the substance, as is used in Topic 4A. changing composition of a system. For instance, in an electro- By the same argument that led to eqn 5A.3, it follows that chemical cell the chemical reaction is arranged to take place the total Gibbs energy of a binary mixture is in two distinct sites (at the two electrodes) and the electrical work the cell performs can be traced to its changing composi- G = nA µA + nB µB (5A.5) tion as products are formed from reactants. 146 5 Simple mixtures The wider significance of the chemical (c) This equation is a special case of the Gibbs–Duhem equation: potential ∑n dµ = 0 J J Gibbs–Duhem equation (5A.12b) The chemical potential does more than show how G varies J with composition. Because G = U + pV − TS, and therefore The significance of the Gibbs–Duhem equation is that U = −pV + TS + G, the general form of an infinitesimal change the chemical potential of one component of a mixture can- in U for a system of variable composition is not change independently of the chemical potentials of the other components. In a binary mixture, if one chemical dU = −pdV − Vdp + SdT + TdS + dG potential increases, then the other must decrease, with the two   = −pdV − Vdp + SdT + TdS changes related by eqn 5A.12a and therefore    + (Vdp − SdT + µAdnA + µBdnB + …) nA dµB = − dµ  (5A.13)   = −pdV + TdS + µAdnA + µBdnB + … nB A Brief illustration 5A.1 This expression is the generalization of eqn 3E.1 (that dU = TdS − pdV) to systems in which the composition may change. If the composition of a mixture is such that nA = 2nB, and a It follows that at constant volume and entropy, small change in composition results in µA changing by ∆µA = +1 J mol−1, µ B will change by dU = µAdnA + µBdnB + … (5A.9) ∆ µB = −2 × (1Jmol −1 ) = −2Jmol −1 and hence that  ∂U  The same line of reasoning applies to all partial molar quan- µJ =   (5A.10)  ∂nJ  S ,V ,n′ tities. For instance, changes in the partial molar volumes of the species in a mixture are related by Therefore, not only does the chemical potential show how G changes when the composition changes, it also shows how the ∑n dV = 0 J J (5A.14a) internal energy changes too (but under a different set of condi- J tions). In the same way it is possible to deduce that For a binary mixture, nA  ∂H   ∂A  dVB = − dV  (5A.14b) (a) µJ =  (b) µJ =   (5A.11) nB A  ∂nJ  S , p ,n′  ∂nJ  T ,V ,n′ As seen in Fig. 5A.1, where the partial molar volume of water Thus, µJ shows how all the extensive thermodynamic proper- increases, the partial molar volume of ethanol decreases. ties U, H, A, and G depend on the composition. This is why the Moreover, as eqn 5A.14b implies, and as seen from Fig. 5A.1, a chemical potential is so central to chemistry. small change in the partial molar volume of A corresponds to a large change in the partial molar volume of B if nA/nB is large, but the opposite is true when this ratio is small. In practice, the (d) The Gibbs–Duhem equation Gibbs–Duhem equation is used to determine the partial molar Because the total Gibbs energy of a binary mixture is given by volume of one component of a binary mixture from measure- eqn 5A.5 (G = nA µA + nB µB), and the chemical potentials de- ments of the partial molar volume of the second component. pend on the composition, when the compositions are changed infinitesimally the Gibbs energy of a binary system is expected Example 5A.2 Using the Gibbs–Duhem equation to change by The experimental values of the partial molar volume of dG = µAdnA + µBdnB + nAdµA + nBdµB K 2SO4(aq) at 298 K are found to fit the expression However, at constant pressure and temperature the change in vB = 32.280 +18.216z 1/2 Gibbs energy is given by eqn 5A.7. Because G is a state func- where v B = VK 2 SO4 /(cm3 mol−1) and z is the numerical value of tion, these two expressions for dG must be equal, which im- ⦵ the molality of K 2SO4 (z = b/b ; see The chemist’s toolkit 11). plies that at constant temperature and pressure Use the Gibbs–Duhem equation to derive an equation for the molar volume of water in the solution. The molar volume of nAdµA + nBd µB = 0 (5A.12a) pure water at 298 K is 18.079 cm3 mol−1.  5A The thermodynamic description of mixtures 147 Collect your thoughts Let A denote H2O, the solvent, and B Self-test 5A.2 Repeat the calculation for a salt B for which VB/(cm3 mol−1) = 6.218 + 5.146z − 7.147z2 with z = b/b. ⦵ denote K 2SO4, the solute. Because the Gibbs–Duhem equation for the partial molar volumes of two components implies that Answer: VA/(cm3 mol−1) = 18.079 − 0.0464z2 + 0.0859z3 dvA = −(nB/nA)dv B, vA can be found by integration: vB nB vA = v*A − ∫ dv 0 nA B 5A.2 The thermodynamics of mixing where vA* = V A*/(cm3 mol−1) is the numerical value of the molar volume of pure A. The first step is to change the variable of The dependence of the Gibbs energy of a mixture on its com- ⦵ integration from v B to z = b/b ; then integrate the right-hand position is given by eqn 5A.5, and, as established in Topic 3E, side between z = 0 (pure A) and the molality of interest. at constant temperature and pressure systems tend towards The solution It follows from the information in the question lower Gibbs energy. This is the link needed in order to apply that, with B = K 2SO4, dv B/dz = 9.108 z −1/2. Therefore, the inte- thermodynamics to the discussion of spontaneous changes of gration required is composition, as in the mixing of two substances. One simple example of a spontaneous mixing process is that of two gases b/b −○ − nB −1/2 vA = vA* − 9.108 ∫ z dz introduced into the same container. The mixing is spontane- 0 nA ous, so it must correspond to a decrease in G. The amount of A (H2O) is nA = (1 kg)/MA, where MA is the molar mass of water, and nB/(1 kg), which then occurs in the The Gibbs energy of mixing of perfect (a) ratio nB/nA, will be recognized as the molality b of B: gases nA = (1 kg)/MA nB /(1 kg) = b Let the amounts of two perfect gases in the two containers nB nB nM before mixing be nA and nB; both are at a temperature T and = = B A = bM A = zb M A ○  nA (1 kg)/M A 1 kg a pressure p (Fig. 5A.6). At this stage, the chemical potentials of the two gases have their ‘pure’ values, which are obtained by applying the definition µ = Gm to eqn 3E.15 (Gm(p) = G m + ⦵ Hence RT ln(p/p )): ⦵ ○  b/b vA = vA* − 9.108 M Ab ∫ z 1/2 d z − ○− 0 p Variation of chemical µ = µ +RT ln  −○ − = vA* − 23 (9.108 M Ab )(b / b )3/2 − ○− − ○− potential with pressure (5A.15a) p − − ○ [perfect gas] It then follows, by substituting the data (including MA = 1.802 × where µ is the standard chemical potential, the chemical po- ⦵ 10−2 kg mol−1, the molar mass of water), that tential of the pure gas at 1 bar. The notation is simplified by replacing p/p by p itself, for ⦵ VA/(cm3 mol−1) = 18.079 − 0.1094(b/b )3/2 ⦵ eqn 5A.15a then becomes The partial molar volumes are plotted in Fig. 5A.5. µ = µ +RT ln p  − ○− (5A.15b) 40 18.079 Initial nA, T, p V(K2SO4)/(cm3 mol–1) 38 18.078 nB, T, p (cm3 mol–1) V(H2O)/ 36 K2SO4 Water 34 18.076 Final T, pA, pB with pA + pB = p 32 18.075 0 0.05 0.1 b/(mol kg–1) Figure 5A.5 The partial molar volumes of the components of Figure 5A.6 The arrangement for calculating the thermodynamic an aqueous solution of potassium sulfate. functions of mixing of two perfect gases. 148 5 Simple mixtures The chemist’s toolkit 11 Measures of concentration Let A be the solvent and B the solute. The molar concentra- 2. The relation between molality and molar tion (informally: ‘molarity’), c B or [B], is the amount of solute concentration molecules (in moles) divided by the volume, V, of the solution: The total mass of a volume V of solution (not solvent) of mass density ρ is m = ρV. The amount of solute molecules in this nB cB = volume is nB = c BV. The mass of solute present is mB = nBMB = V c BVMB. The mass of solvent present is therefore mA = m – mB It is commonly reported in moles per cubic decimetre = ρV − c BVMB = (ρ − c BMB)V. The molality is therefore (mol dm−3) or, equivalently, in moles per litre (mol L −1). It is con- venient to define its ‘standard’ value as c = 1 mol dm−3. ⦵ nB cBV cB bB = = = The molality, bB, of a solute is the amount of solute species (in mA ( ρ − cB M B )V ρ − cB M B moles) in a solution divided by the total mass of the solvent (in The inverse of this relation, the molar concentration in terms kilograms), mA: of the molality, is nB bB = bB ρ mA cB = 1 + bB M B Both the molality and mole fraction are independent of tempera- ture; in contrast, the molar concentration is not. It is convenient 3. The relation between molar concentration and to define the ‘standard’ value of the molality as b = 1 mol kg−1. ⦵ mole fraction By inserting the expression for bB in terms of x B into the expres- 1. The relation between molality and mole fraction sion for c B, the molar concentration of B in terms of its mole Consider a solution with one solute and having a total amount fraction is n of molecules. If the mole fraction of the solute is x B, the amount of solute molecules is nB = x Bn. The mole fraction of xB ρ cB = solvent molecules is xA = 1 − x B, so the amount of solvent mol- x A M A + xB MB ecules is nA = xAn = (1 − x B)n. The mass of solvent, of molar mass MA, present is mA = nA M A = (1− x B )nM A. The molality of with xA = 1 − x B. For a dilute solution in the sense that the solute is therefore x B M B  x A M A, nB x Bn xB  ρ  bB = = = cB ≈  x mA (1− x B )nM A (1− x B )M A  x A M A  B The inverse of this relation, the mole fraction in terms of the If, moreover, x B 1, so x A ≈ 1, then molality, is bB M A  ρ  cB ≈  x xB = 1 + bB M A  M A  B In practice, the replacement of p/p by p means using the nu- ⦵ pA p ∆ mixG = nA RT ln + nB RT ln B  (5A.16c) merical value of p in bars. The total Gibbs energy of the sepa- p p rated gases is then given by eqn 5A.5 as At this point nJ can be replaced by x Jn, where n is the total amount of A and B, and the relation between partial pressure Gi = nAµA + nBµB = nA(µ A + RT ln p) + nB(µ B + RT ln p) ⦵ ⦵ and mole fraction (Topic 1A, pJ = x Jp) can be used to write  (5A.16a) pJ/p = x J for each component. The result is After mixing, the partial pressures of the gases are pA and pB, Gibbs energy with pA + pB = p. The total Gibbs energy changes to ∆mixG = nRT(xA ln xA + x B ln x B) of mixing (5A.17) [perfect gas] Gf = nA(µ A + RT ln pA) + nB(µ B + RT ln pB) ⦵ ⦵ (5A.16b) Because mole fractions are never greater than 1, the loga- rithms in this equation are negative, and ∆mixG < 0 (Fig. 5A.7). The difference Gf − Gi, the Gibbs energy of mixing, ΔmixG, is The conclusion that ∆mixG is negative for all compositions con- therefore firms that perfect gases mix spontaneously in all proportions.  5A The thermodynamic description of mixtures 149 0 The solution Given that the pressure of nitrogen is p, the pressure of hydrogen is 3p. Therefore, the initial Gibbs energy is –0.2 Gi = (3.0 mol){µ (H2) + RT ln 3p} ⦵ ΔmixG/nRT    + (1.0 mol){µ (N2) + RT ln p} ⦵ –0.4 When the partition is removed and each gas occupies twice –0.6 the original volume, the final total pressure is 2p. The partial pressure of nitrogen falls to 12 p and that of hydrogen falls to 2 p. Therefore, the Gibbs energy changes to 3 –0.8 0 0.5 1 Mole fraction of A, xA Gf = (3.0 mol){µ (H2) + RT ln 32 p} ⦵ Figure 5A.7 The Gibbs energy of mixing of two perfect gases   + (1.0 mol){µ (N2) + RT ln 12 p} ⦵ at constant temperature and pressure, and (as discussed later) of two liquids that form an ideal solution. The Gibbs energy of The Gibbs energy of mixing is the difference of these two mixing is negative for all compositions, so perfect gases mix quantities: spontaneously in all proportions. 2p p 3 1 ∆mixG = (3.0mol)RT ln + (1.0mol)RT ln 2 3p p      = − (3.0mol)RT ln2 − (1.0mol)RT ln2 Example 5A.3 Calculating a Gibbs energy of mixing      = − (4.0mol)RT ln2 = − 6.9kJ A container is divided into two equal compartments (Fig. 5A.8). One contains 3.0 mol H2(g) at 25 °C; the other Comment. In this example, the value of ∆mixG is the sum contains 1.0 mol N2(g) at 25 °C. Calculate the Gibbs energy of of two contributions: the mixing itself, and the changes in mixing when the partition is removed. Assume that the gases pressure of the two gases to their final total pressure, 2p. Do are perfect. not be misled into interpreting this negative change in Gibbs energy as a sign of spontaneity: in this case, the pressure changes, and ΔG < 0 is a signpost of spontaneous change only at constant temperature and pressure. When 3.0 mol H2 mixes 3.0 mol H2 with 1.0 mol N2 at the same pressure, with the volumes of the Initial 1.0 mol N2 vessels adjusted accordingly, the change of Gibbs energy is 3p −5.6 kJ. Because this value is for a change at constant pres- p sure and temperature, the fact that it is negative does imply spontaneity. 3.0 mol H2 Self-test 5A.3 Suppose that 2.0 mol H2 at 2.0 atm and 25 °C Final 2p 1.0 mol N2 and 4.0 mol N2 at 3.0 atm and 25 °C are mixed by removing p(H2) = 3 p the partition between them. Calculate ∆mixG. 2 1 Answer: −9.7 kJ p(N2) = 2 p Figure 5A.8 The initial and final states considered in the calculation of the Gibbs energy of mixing of gases at different (b) Other thermodynamic mixing functions initial pressures. In Topic 3E it is shown that (∂G/∂T)p = −S. It follows immedi- Collect your thoughts Equation 5A.17 cannot be used directly ately from eqn 5A.17 that, for a mixture of perfect gases ini- because the two gases are initially at different pressures, so tially at the same pressure, the entropy of mixing, ∆mixS, is proceed by calculating the initial Gibbs energy from the chemical potentials. To do so, calculate the pressure of each  ∂∆ G  ∆mix S = −  mix  = − nR( xA ln xA + xB ln xB ) gas: write the pressure of nitrogen as p, then the pressure of  ∂T  p hydrogen as a multiple of p can be found from the gas laws. Entropy of mixing  [perfect gases, constant T and p] (5A.18) Next, calculate the Gibbs energy for the system when the par- tition is removed. The volume occupied by each gas doubles, Because ln x < 0, it follows that ∆mixS > 0 for all compositions so its final partial pressure is half its initial pressure. (Fig. 5A.9). 150 5 Simple mixtures 0.8 5A.3 The chemical potentials of liquids 0.6 To discuss the equilibrium properties of liquid mixtures it is necessary to know how the Gibbs energy of a liquid varies ΔmixS/nR 0.4 with composition. The calculation of this dependence uses the fact that, as established in Topic 4A, at equilibrium the chemi- 0.2 cal potential of a substance present as a vapour must be equal to its chemical potential in the liquid. 0 0 0.5 1 (a) Ideal solutions Mole fraction of A, xA Quantities relating to pure substances are denoted by a super- Figure 5A.9 The entropy of mixing of two perfect gases at script *, so the chemical potential of pure A is written µA* and constant temperature and pressure, and (as discussed later) of as µA*(l) when it is necessary to emphasize that A is a liquid. two liquids that form an ideal solution. The entropy increases Because the vapour pressure of the pure liquid is pA* it fol- for all compositions, and because there is no transfer of heat to lows from eqn 5A.15b that the chemical potential of A in the the surroundings when perfect gases mix, the entropy of the vapour (treated as a perfect gas) is µ A + RT ln pA (with pA to be ⦵ surroundings is unchanged. Hence, the graph also shows the interpreted as pA/p ). These two chemical potentials are equal ⦵ total entropy of the system plus the surroundings; because the total entropy of mixing is positive at all compositions, at equilibrium (Fig. 5A.10), so perfect gases mix spontaneously in all proportions. liquid vapour     µA*(1) = µA (g) + RT ln pA*  ○  (5A.20a) If another substance, a solute, is also present in the liquid, the Brief illustration 5A.2 chemical potential of A in the liquid is changed to µA and its vapour pressure is changed to pA. The vapour and solvent are For equal amounts of perfect gas molecules that are mixed at still in equilibrium, so the same pressure, set xA = x B = 12 and obtain ∆mixS = −nR { 12 ln 12 + 12 ln 12 } = nR ln 2 µA (1) = µA (g) + RT ln pA (5A.20b) ○  with n the total amount of gas molecules. For 1 mol of each species, so n = 2 mol, ∆mixS = (2 mol) × R ln 2 = +11.5 J K−1 An increase in entropy is expected when one gas disperses into the other and the disorder increases. A(g) + B(g) μA(g, p) = Under conditions of constant pressure and temperature, the μA(l) enthalpy of mixing, ∆mixH, the enthalpy change accompany- A(l) + B(l) ing mixing, of two perfect gases can be calculated from ∆G = ∆H − T∆S. It follows from eqns 5A.17 and 5A.18 that Enthalpy of mixing ∆mixH = 0  [perfect gases, constant T and p] (5A.19) Figure 5A.10 At equilibrium, the chemical potential of the The enthalpy of mixing is zero, as expected for a system in gaseous form of a substance A is equal to the chemical potential which there are no interactions between the molecules form- of its condensed phase. The equality is preserved if a solute is ing the gaseous mixture. It follows that, because the entropy also present. Because the chemical potential of A in the vapour of the surroundings is unchanged, the whole of the driv- depends on its partial vapour pressure, it follows that the ing force for mixing comes from the increase in entropy of chemical potential of liquid A can be related to its partial vapour the system. pressure.  5A The thermodynamic description of mixtures 151 The next step is the combination of these two equations to p* Total pressure eliminate the standard chemical potential of the gas, µ A(g). To ⦵ B do so, write eqn 5A.20a as µ A(g) = µA*(1) − RT ln pA* and substi- ⦵ tute this expression into eqn 5A.20b to obtain Partial Pressure µA  ⦵(g) pressure of B pA*  p µA (1) = µA* (1) − RT ln pA* + RT ln pA* = µA* (1) + RT ln A  (5A.21) pA* Partial pressure of A The final step draws on additional experimental informa- tion about the relation between the ratio of vapour pressures Mole fraction of A, xA and the composition of the liquid. In a series of experiments on mixtures of closely related liquids (such as benzene and Figure 5A.11 The partial vapour pressures of the two methylbenzene), François Raoult found that the ratio of the components of an ideal binary mixture are proportional to the partial vapour pressure of each component to its vapour pres- mole fractions of the components, in accord with Raoult’s law. sure when present as the pure liquid, pA/pA*, is approximately The total pressure is also linear in the mole fraction of either equal to the mole fraction of A in the liquid mixture. That is, component. he established what is now called Raoult’s law: 80 Raoult’s law pA = xA pA* [ideal solution] (5A.22) 60 Benzene This law is illustrated in Fig. 5A.11. Some mixtures obey Pressure, p/Torr Total Raoult’s law very well, especially when the components are 40 structurally similar (Fig. 5A.12). Mixtures that obey the law throughout the composition range from pure A to pure B are 20 called ideal solutions. Methylbenzene 0 Mole fraction of methylbenzene, x(C6H5CH3) 0 1 Brief illustration 5A.3 Figure 5A.12 Two similar liquids, in this case benzene and The vapour pressure of pure benzene at 20 °C is 75 Torr and methylbenzene (toluene), behave almost ideally, and the variation that of pure methylbenzene is 25 Torr at the same tempera- of their vapour pressures with composition resembles that for an ture. In an equimolar mixture x benzene = xmethylbenzene = 12 so the ideal solution. partial vapour pressure of each one in the mixture is pbenzene = 1 × 80 Torr = 40 Torr For an ideal solution, it follows from eqns 5A.21 and 5A.22 2 that pmethylbenzene = 1 × 25 Torr = 12.5 Torr µA (1) = µA*(1) + RT ln x A  2 Chemical potential (5A.23) [ideal solution] The total vapour pressure of the mixture is 48 Torr. Given the This important equation can be used as the definition of an two partial vapour pressures, it follows from the definition ideal solution (so that it implies Raoult’s law rather than stem- of partial pressure (Topic 1A) that the mole fractions in the ming from it). It is in fact a better definition than eqn 5A.22 vapour are because it does not assume that the vapour is a perfect gas. The molecular origin of Raoult’s law is the effect of the sol- xvap,benzene = (40 Torr)/(48 Torr) = 0.83 ute on the entropy of the solution. In the pure solvent, the mol- ecules have a certain disorder and a corresponding entropy; and the vapour pressure then represents the tendency of the sys- tem and its surroundings to reach a higher entropy. When a xvap,methylbenzene = (12.5 Torr)/(48 Torr) = 0.26 solute is present, the solution has a greater disorder than the The vapour is richer in the more volatile component pure solvent because a molecule chosen at random might or (benzene). might not be a solvent molecule. Because the entropy of the solution is higher than that of the pure solvent, the solution 152 5 Simple mixtures 500 KB Total 400 Ideal–dilute solution Pressure, p/Torr Carbon disulfide Pressure, p (Henry) 300 pB* Real solution 200 Acetone 100 Ideal solution (Raoult) 0 0 Mole fraction of B, xB 1 0 Mole fraction of carbon disulfide, x(CS2) 1 Figure 5A.13 Strong deviations from ideality are shown by Figure 5A.14 When a component (the solvent) is nearly pure, it dissimilar liquids (in this case carbon disulfide and acetone has a vapour pressure that is proportional to the mole fraction (propanone)). The dotted lines show the values expected from with a slope pB* (Raoult’s law). When it is the minor component (the Raoult’s law. solute) its vapour pressure is still proportional to the mole fraction, but the constant of proportionality is now KB (Henry’s law). has a lower tendency to acquire an even higher entropy by the solvent vaporizing. In other words, the vapour pressure of the solvent in the solution is lower than that of the pure solvent. Some solutions depart significantly from Raoult’s law (Fig. 5A.13). Nevertheless, even in these cases the law is obeyed increasingly closely for the component in excess (the solvent) as it approaches purity. The law is another example of a lim- iting law (in this case, achieving reliability as xA → 1) and is a good approximation for the properties of the solvent if the solution is dilute. (b) Ideal–dilute solutions Figure 5A.15 In a dilute solution, the solvent molecules (the blue In ideal solutions the solute, as well as the solvent, obeys spheres) are in an environment that differs only slightly from Raoult’s law. However, William Henry found experimentally that of the pure solvent. The solute particles (the red spheres), however, are in an environment totally unlike that of the pure that, for real solutions at low concentrations, although the va- solute. pour pressure of the solute is proportional to its mole fraction, the constant of proportionality is not the vapour pressure of the pure substance (Fig. 5A.14). Henry’s law is: Henry’s law which is entirely different from their environment when it is in pB = x BKB (5A.24) [ideal–dilute solution] its pure form. Thus, the solvent behaves like a slightly modified In this expression x B is the mole fraction of the solute and KB pure liquid, but the solute behaves entirely differently from its is an empirical constant (with the dimensions of pressure) pure state unless the solvent and solute molecules happen to be chosen so that the plot of the vapour pressure of B against its very similar. In the latter case, the solute also obeys Raoult’s law. mole fraction is tangent to the experimental curve at x B = 0. Henry’s law is therefore also a limiting law, achieving reliabil- Example 5A.4 Investigating the validity of Raoult’s and ity as x B → 0. Henry’s laws Mixtures for which the solute B obeys Henry’s law and the solvent A obeys Raoult’s law are called ideal–dilute solutions. The vapour pressures of each component in a mixture of pro- The difference in behaviour of the solute and solvent at low panone (acetone, A) and trichloromethane (chloroform, C) concentrations (as expressed by Henry’s and Raoult’s laws, were measured at 35 °C with the following results: respectively) arises from the fact that in a dilute solution the xC 0 0.20 0.40 0.60 0.80 1 solvent molecules are in an environment very much like the one they have in the pure liquid (Fig. 5A.15). In contrast, pC/kPa 0 4.7 11 18.9 26.7 36.4 the solute molecules are surrounded by solvent molecules, pA/kPa 46.3 33.3 23.3 12.3 4.9 0  5A The thermodynamic description of mixtures 153 Confirm that the mixture conforms to Raoult’s law for the x 0.005 0.009 0.019 0.024 component in large excess and to Henry’s law for the minor p/kPa 27.3 48.4 101 126 component. Find the Henry’s law constants. Estimate the Henry’s law constant for chloromethane. Collect your thoughts Both Raoult’s and Henry’s laws are Answer: 5 MPa statements about the form of the graph of partial vapour pressure against mole fraction. Therefore, plot the partial vapour pressures against mole fraction. Raoult’s law is tested For practical applications, Henry’s law is expressed in terms by comparing the data with the straight line pJ = x Jp*J for each of the molality, b, of the solute, pB = bBKB. Some Henry’s law component in the region in which it is in excess (and acting data for this convention are listed in Table 5A.1. As well as as the solvent). Henry’s law is tested by finding a straight line providing a link between the mole fraction of the solute and pJ = x JKJ that is tangent to each partial vapour pressure curve its partial pressure, the data in the table may also be used to at low x, where the component can be treated as the solute. calculate gas solubilities. Knowledge of Henry’s law constants for gases in blood and fats is important for the discussion of The solution The data are plotted in Fig. 5A.16 together with the Raoult’s law lines. Henry’s law requires KA = 24.5 kPa for respiration, especially when the partial pressure of oxygen is acetone and KC = 23.5 kPa for chloroform. abnormal, as in diving and mountaineering, and for the dis- cussion of the action of gaseous anaesthetics. 50 Table 5A.1 Henry’s law constants for gases in water at 298 K* p*(acetone) 40 K/(kPa kg mol−1) p*(chloroform) Pressure, p/kPa CO2 3.01 × 103 30 Raoult’s law H2 1.28 × 105 K(acetone) N2 1.56 × 105 20 O2 7.92 × 104 K(chloroform) 10 * More values are given in the Resource section. Henry’s law 0 0 Mole fraction of chloroform, x(CHCl3) 1 Brief illustration 5A.4 Figure 5A.16 The experimental partial vapour pressures To estimate the molar solubility of oxygen in water at 25 °C of a mixture of chloroform (trichloromethane) and acetone and a partial pressure of 21 kPa, its partial pressure in the (propanone) based on the data in Example 5A.4. The values atmosphere at sea level, write of K are obtained by extrapolating the dilute solution vapour pO2 21kPa pressures, as explained in the Example. bO2 = = = 2.9 ×10−4 mol kg −1 K O2 7.9 ×104 kPa kg mol −1 The molality of the saturated solution is therefore 0.29 mmol Comment. Notice how the system deviates from both Raoult’s kg−1. To convert this quantity to a molar concentration, as- and Henry’s laws even for quite small departures from sume that the mass density of this dilute solution is essentially x = 1 and x = 0, respectively. These deviations are discussed that of pure water at 25 °C, or ρ = 0.997 kg dm−3. It follows that in Topic 5E. the molar concentration of oxygen is Self-test 5A.4 The vapour pressure of chloromethane at [O2 ] = bO2 ρ = (2.9 ×10−4 mol kg −1 ) × (0.997 kgdm −3 ) various mole fractions in a mixture at 25 °C was found to be = 0.29mmoldm −3 as follows: Checklist of concepts ☐ 1. The partial molar volume of a substance is the contri- ☐ 2. The chemical potential is the partial molar Gibbs bution to the volume that a substance makes when it is energy and is the contribution to the total Gibbs energy part of a mixture. that a substance makes when it is part of a mixture. 154 5 Simple mixtures ☐ 3. The chemical potential also expresses how, under a ☐ 7. Raoult’s law provides a relation between the vapour pres- variety of different conditions, the thermodynamic sure of a substance and its mole fraction in a mixture. functions vary with composition. ☐ 8. An ideal solution is a solution that obeys Raoult’s law ☐ 4. The Gibbs–Duhem equation shows how the changes in over its entire range of compositions; for real solutions chemical potentials (and, by extension, of other partial it is a limiting law valid as the mole fraction of the spe- molar quantities) of the components of a mixture are cies approaches 1. related. ☐ 9. Henry’s law provides a relation between the vapour ☐ 5. The Gibbs energy of mixing is negative for perfect pressure of a solute and its mole fraction in a mixture; it gases at the same pressure and temperature. is the basis of the definition of an ideal–dilute solution. ☐ 6. The entropy of mixing of perfect gases initially at the ☐ 10. An ideal–dilute solution is a solution that obeys same pressure is positive and the enthalpy of mixing is Henry’s law at low concentrations of the solute, and for zero. which the solvent obeys Raoult’s law. Checklist of equations Property Equation Comment Equation number Partial molar volume VJ = (∂V/∂nJ)p,T,n′ Definition 5A.1 Chemical potential µJ = (∂G/∂nJ)p,T,n′ Definition 5A.4 Total Gibbs energy G = nAµA + nBµB Binary mixture 5A.5 Fundamental equation of chemical dG = Vdp − SdT + µAdnA + µBdnB + … 5A.6 thermodynamics Gibbs–Duhem equation ∑JnJdµJ = 0 5A.12b ○  ○  Chemical potential of a gas µ = µ + RT ln( p / p ) Perfect gas 5A.15a Gibbs energy of mixing ΔmixG = nRT(xA ln xA + xB ln xB) Perfect gases and ideal solutions 5A.17 Entropy of mixing ΔmixS = −nR(xA ln xA + xB ln xB) Perfect gases and ideal solutions 5A.18 Enthalpy of mixing ΔmixH = 0 Perfect gases and ideal solutions 5A.19 Raoult’s law pA = xA pA* True for ideal solutions; limiting law as xA → 1 5A.22 Chemical potential of component µA (1) = µA* (1) + RT ln x A Ideal solution 5A.23 Henry’s law pB = xBKB True for ideal–dilute solutions; limiting law as xB → 0 5A.24 TOPIC 5B The properties of solutions gases (Topic 5A). The total Gibbs energy before the liquids are ➤ Why do you need to know this material? mixed is Mixtures and solutions play a central role in chemistry, Gi = nA µA* + nBµ *B  (5B.2a) and so it is important to understand how their composi- tions affect their thermodynamic properties, such as their where the * denotes the pure liquid. When they are mixed, the boiling and freezing points. One very important physi- individual chemical potentials are given by eqn 5B.1 and the cal property of a solution is its osmotic pressure, which total Gibbs energy is is used, for example, to determine the molar masses of macromolecules. Gf = nA(µA* + RT ln xA) + nB(µ *B + RT ln x B) (5B.2b) ➤ What is the key idea? Consequently, the Gibbs energy of mixing, the difference of The chemical potential of a substance in a mixture is the these two quantities, is same in every phase in which it occurs. ➤ What do you need to know already? ∆mixG = nRT(xA ln xA + x B ln x B) Gibbs energy of mixing This Topic is based on the expression derived from Raoult’s   [ideal solution] (5B.3) law (Topic 5A) in which chemical potential is related to where n = nA + nB. As for gases, it follows that the ideal entropy mole fraction. The derivations make use of the Gibbs– of mixing of two liquids is Helmholtz equation (Topic 3E) and the effect of pressure on chemical potential (Topic 5A). Some of the derivations Entropy of mixing are the same as those used in the discussion of the mixing ∆mixS = −nR(xA ln xA + x B ln x B) [ideal solution] (5B.4) of perfect gases (Topic 5A). Then from ΔmixG = Δmix H − TΔmixS it follows that the ideal enthalpy of mixing is zero, Δmix H = 0. The ideal volume of Thermodynamics can provide insight into the properties of mixing, the change in volume on mixing, is also zero. To see liquid mixtures, and a few simple ideas can unify the whole why, consider that, because (∂G/∂p)T = V (eqn 3E.8), ΔmixV = field of study. (∂ΔmixG/∂p)T. But ΔmixG in eqn 5B.3 is independent of pressure, so the derivative with respect to pressure is zero, and therefore ΔmixV = 0. 5B.1 Liquid mixtures Equations 5B.3 and 5B.4 are the same as those for the mixing of two perfect gases and all the conclusions drawn The development here is based on the relation derived in Topic there are valid here: because the enthalpy of mixing is zero 5A between the chemical potential of a component (which there is no change in the entropy of the surroundings so the here is called J, with J = A or B in a binary mixture) in an ideal driving force for mixing is the increasing entropy of the sys- mixture or solution, µJ, its value when pure, µ*, J and its mole tem as the molecules mingle. It should be noted, however, fraction in the mixture, x J: that solution ideality means something different from gas perfection. In a perfect gas there are no interactions between Chemical potential µJ = µJ* + RT ln x J  [ideal solution] (5B.1) the molecules. In 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 (a) Ideal solutions in the pure liquids. The variation of the Gibbs energy and entropy of mixing with composition is the same as that for The Gibbs energy of mixing of two liquids to form an ideal gases (Figs. 5A.7 and 5A.9); both graphs are repeated here solution is calculated in exactly the same way as for two (as Figs. 5B.1 and 5B.2). 156 5 Simple mixtures 0 The enthalpy of mixing is zero (presuming that the solution is ideal). –0.2 Real solutions are composed of molecules for which the ΔmixG/nRT –0.4 A–A, A–B, and B–B interactions are all different. Not only may there be enthalpy and volume changes when such liq- –0.6 uids 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 –0.8 with the others. If the enthalpy change is large and positive, or 0 0.5 1 Mole fraction of A, xA if the entropy change is negative (because of a reorganization Figure 5B.1 The Gibbs energy of mixing of two liquids that form of the molecules that results in an orderly mixture), the Gibbs an ideal solution. energy of mixing might be positive. In that case, separation is spontaneous and the liquids are immiscible. Alternatively, the liquids might be partially miscible, which means that they are miscible only over a certain range of compositions. 0.8 (b) Excess functions and regular solutions 0.6 The thermodynamic properties of real solutions are expressed in terms of the excess functions, X E, the difference between ΔmixS/nR 0.4 the observed thermodynamic function of mixing and the function for an ideal solution: 0.2 Excess function XE = ∆mix X − ∆mix Xideal [definition] (5B.5) 0