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Simple mixtures Before dealing with chemical reactions, here we consider mixtures of substances that do not react together. At this stage we deal mainly with binary mixtures (mixtures of two components, A and B). We therefore often be able to simplify equations using the relation xA + xB = 1. A restriction in this chapter – we consider mainly non-electrolyte solutions – the solute is not present as ions. The thermodynamic description of mixtures We already considered the partial pressure – the contribution of one component to the total pressure, when we were dealing with the properties of gas mixtures. Here we introduce other analogous partial properties. Partial molar quantities: describe the contribution (per mole) that a substance makes to an overall property of mixture. The partial molar volume, VJ – the contribution J makes to the total volume of a mixture. Although 1 mol of a substance has a characteristic volume when it is pure, 1 mol of a substance can make different contributions to the total volume of a mixture because molecules pack together in different ways in the pure substances and in mixtures. Imagine a huge volume of pure water. When a further 1 3 mol H2O is added, the volume increases by 18 cm . When we add 1 mol H2O to a huge volume of pure ethanol, the volume increases by only 14 cm3. 18 cm3 – the volume occupied per mole of water molecules in pure water. 14 cm3 – the volume occupied per mole of water molecules in virtually pure ethanol. The partial molar volume of water in pure water is 18 cm3 and the partial molar volume of water in pure ethanol is 14 cm3 – there is so much ethanol present that each H2O molecule is surrounded by ethanol molecules and the packing of the molecules results in the water molecules occupying only 14 cm3. The partial molar volumes of the components of a mixture vary with composition because the environment of each type of molecules changes as the composition changes from pure A to pure B. This changing molecular environment and the modification of the forces acting between molecules results in the variation of the thermodynamic properties of a mixture as its composition is changed. ∂V The formal definition of the partial molar volume: VJ = ∂nJ p,T ,n ′ n' subscript means that the amount of all other substances present is constant. The partial molar volume is the slope of the plot of the total volume as the amount of J is changed, the pressure,€ temperature, and amount of the other components being constant. When the composition of the mixture is changed by the addition of dnA of A and dnB of B, then the total volume of the mixture changes by ∂V ∂V dV = dnA + dnB = VAdnA + VBdnB ∂n ∂n A p,T ,nB B p,T ,nA Once we know the partial molar volumes VA and VB, we can state the total volume of the mixture: V = nAVA + nBVB Consider a very large sample of the mixture. When an amount n of A is added, the € A composition remains virtually unchanged but the volume increases by nAVA. Similarly, when amount nB of B is added, the volume increases by nBVB. The total increase in volume is nAVA + nBVB. The mixture now occupies a larger volume but the proportions of components are still the same. Next, scoop out of this enlarged volume a sample containing nA of A and nB of B. Its volume is nAVA + nBVB. Because the volume is a state function, we could prepare the same sample simply by mixing the appropriate amount of A and B. One method of measuring partial molar volumes is to measure the dependence of the volume on the composition and to fit the observed volume to a function of the substance. Once the function has been found, its slope can be determined at any composition by differentiation. The total volume of an ethanol solution at 25°C containing 1.000 kg of water is found to be given by the expression V/mL = 1002.93 + 54.6664b – 0.36394b2 + 0.028256b3 b – the numerical value of the molality. Because the amount of ethanol in moles here is equal to the numerical value of the molality in moles per kg, we can write the partial molar volume of ethanol as -1 2 VE/(mL mol ) = ∂ V /mL ∂b = 54.6664 – 2(0.36394)b + 3(0.028256)b ( ( ) ) p,T ,nW Molar volumes are always positive, but partial molar quantities need not be. Example: the € limiting partial molar volume of MgSO4 in water (its partial molar volume in the limit of zero 3 -1 concentration) is –1.4 cm mol , meaning that the addition of 1 mol MgSO4 to a large volume of water results in a decrease in volume of 1.4 cm3: the salt breaks the open structure of water as the ions become hydrated, and it collapses slightly. Partial molar Gibbs energies The concept of a partial molar quantity can be extended to any extensive state function. For a pure substance, the chemical potential is just another name for the molar Gibbs energy. For a ∂G substance in a mixture: µJ = ∂nJ p,T ,n ′ The chemical potential is the slope of a plot of Gibbs energy against the amount of the component J, with the pressure and temperature (and the amounts of other substances) held constant. Using the same argument as for€ partial molar volumes, the total Gibbs energy of a binary mixture can be expressed as G = nAµA + nBµB The Gibbs energy depends on the composition, pressure, and temperature: dG = Vdp – SdT + µAdnA + µBdnB + … The fundamental equation of chemical thermodynamics At constant pressure and temperature: dG = µAdnA + µBdnB + … Since dG = dwadd,max dwadd,max = µAdnA + µBdnB + … Additional (non-expansion) work can arise from the changing composition of a system. The wider significance of the chemical potential G = U +pV –TS U = G – pV + TS dU = –pdV – Vdp + SdT + TdS + dG = –pdV – Vdp + SdT + TdS + (Vdp – SdT + µAdnA + µBdnB + …) = –pdV + TdS + µAdnA + µBdnB + … At constant volume and entropy, ∂U dU = µAdnA + µBdnB + … ⇒ µJ = ∂nJ S,V ,n ′ Not only does the chemical potential show how G changes when the composition changes, it also shows how the internal energy changes too (but under a different set of conditions). Also, in the same way it is easy to show that ∂H € ∂A µJ = µJ = ∂nJ S,p,n ′ ∂nJ V ,T ,n ′ The µJ shows how all the extensive thermodynamic properties U, H, A, G depend on the composition. This is why the chemical potential is so central to chemistry. The Gibbs-Duhem equation € € G = nAµA + nBµB Therefore, dG = µAdnA + µBdnB + nAdµA + nBdµB On the other hand, at constant temperature and pressure dG = µAdnA + µBdnB Thus, nAdµA + nBdµB = 0 This equation is a special case of the Gibbs-Duhem equation: ∑nJ dµJ = 0 J € The significance of the Gibbs-Duhem equation – the chemical potential of one component of a mixture cannot change independently of the other components. In a binary mixture, if one nA partial molar quantity increases, then the other must decrease: dµB = − dµ A nB The same line of reasoning applies to all partial molar quantities. For example, where the partial molar volume of water increases, that of ethanol decreases. Moreover, a 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. In practice, the Gibbs-Duhem equation is used to determine the partial molar volume of one component of a binary mixture from measurements of the partial molar volume of the second component. Example 1. Using the Gibbs-Duhem equation. The experimental values of the partial molar volume of K2SO4(aq) at 298 K are given by 3 -1 1/2 the expression VK2SO4/(cm mol ) = 32.280 + 18.216b b – the numerical value of molality of K2SO4. Use the Gibbs-Duhem equation to derive an expression for the partial molar volume of water in solution. The molar volume of pure water is 18.079 cm3 mol-1. A: K2SO4 B: H2O nAdVA + nBdVB = 0 * nA dVA −1/2 dVB = -(nA/nB)dVA V = V − dV = 9.108b B B ∫ n A db b n B V = V * − 9.108 A b−1/2db B B ∫0 Molality: b = nA /(nBM B) nA nB = M Bb nB b V = V * − 9.108M b1/2db =V * − 2/3 9.108M b3/2 3/2 B B b €∫0 B ( )( € B ) VB = 18.079 – 0.1094b € € € € The thermodynamics of mixing The dependence of the Gibbs energy of a mixture is given by G = nAµA + nBµB. At constant T and p systems tend towards lower Gibbs energy. Example: all gases mix spontaneously because the molecules of one gas can mingle with the molecules of another gas - ΔG < 0. Let the amounts of two perfect gases in the two containers be nA and nB; both are at temperature T and pressure p. Ø Ø Ø Ø Gi = nAµA + nBµB = nA[µA + RTln (p/p )] + nB[µB + RTln (p/p )] µØ - the standard chemical potential, the chemical potential of the pure gas at 1 bar. Ø Ø Ø If pressure is measured in bar, p = 1 bar and Gi = nA[µA + RTln p] + nB[µB + RTln p] When the partition is removed, the total pressure remains the same, but, according to Dalton’s law, the partial pressures fall to xAp and xBp, xA = nA/n and xB = nB/n. Ø Ø Gf = nA[µA + RTln xAp] + nB[µB + RTln xBp] ΔmixG = Gf - Gi ln xp – ln p = ln(xp/p) = ln x ΔmixG = RT[nAln xA + nBln xB] = nRT[xAln xA + xBln xB] Because both xA and xB are less than 1, the two logarithms are negative – ΔmixG < 0.
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