substances (among other variables as well). ( ) Thus the change in volume of a mixture can be written as


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1 Mxtures and Solutons Partal Molar Quanttes Partal molar volume he total volume of a mxture of substances s a functon of the amounts of both V V n,n substances (among other varables as well). hus the change n volume of a mxture can be wrtten as V V dv = dn + dn n n n n he change n volume due to a change n the amount of a substance s known as ts partal molar volume. V V = dv= Vdn + Vdn n n he molar volume for a onecomponent system s the nverse of the densty and s ndependent of the amount of substance (t s an ntensve quantty). V V = = n ρ However, when consderng a mxture, the partal molar densty can be greater than, equal to or less than the molar densty. he partal molar densty can even be negatve as occurs when magnesum or alumnum salts are added to water. he partal molar volume s not the same as the molar volume: V V 3 Partal Molar Volume of EtOH (cm /mol) X EtOH Partal Molar Volume of EtOH (cm /mol)
2 Example: Calculate the total volume of soluton when mol of MgSO 4 s added to.00 kg of water. he partal molar volume of MgSO 4 depends on ts concentraton wth the followng expresson: V = cm 3 /mol V= V n + V n φ 3 VMgSO = m m cm mol 4 MgSO4 MgSO4 HO HO m m cm mol mol 8.063cm mol 000 g 8.0 g mol φ = + = cm cm = 00.3cm HO Other partal molar quanttes Other quanttes such as partal molar heat capactes are useful for thermodynamc work wth solutons. C p C p = n n,n 3,n3 Partal molar Gbbs free energy s seen n the last chapter, the partal molar Gbbs free energy at constant temperature and pressure s the chemcal potental = G n,p,n,n 3, So the total Gbbs free energy of a mxture can be found n terms of the chemcal potental. G = n+ n + 3n3+ 4n4 +
3 Gbbs Duhem equaton 3 he chemcal potentals of the components of a mxture are not ndependent of each other. Change n the chemcal potental depends on the changes of the other chemcal potentals. Let us examne the Gbbs free energy of a twocomponent mxture. G = n+ n he change n the Gbbs free energy (at constant temperature and pressure) can be wrtten n terms of changes n the chemcal potentals and changes n the amounts. dg = d n + n = n d + dn + n d + dn However recall that the Gbbs free energy s a functon of temperature, pressure and amount. G G p,,n,n dg = Vdp Sd + dn + dn dg = dn + dn,p Comparng the two expressons for dg leads to the Gbbs Duhem equaton. nd + dn+ nd + dn = dn+dn nd nd + nd = 0 d = n Smlar relatonshps hold for the partal molar volumes and partal molar entropes of a mxture. dv ndv nds = ds = n n Changes n one partal molar quantty depend on the changes of the other partal molar quanttes.
4 ctvtes and deal solutons 4 Consder a soluton of a volatle solvent wth a nonvolatle solute. It s an expermental fact that the vapor of the solvent above the soluton wll be reduced as more solute s added to the soluton. For the solvent at equlbrum, the chemcal potental of the vapor s equal to the chemcal potental of the lqud. ( s a customary label for the solvent when examnng solutons.) ( g) ( l) = he chemcal potental of the vapor s generally lower than ts standard state. p g g Rln p φ = + φ hus the chemcal potental of the solvent n soluton can be wrtten n terms of the chemcal potental of ts vapor. p l g Rln p φ () = + φ For a pure solvent, a smlar expresson can be wrtten for the solvent vapor n equlbrum wth the pure lqud. (he ndcates pure solvent.) φ p () l = ( g) + Rln p φ Examnng the dfference between the chemcal potental of the lqud n soluton and the chemcal potental of the pure lqud yelds φ p φ p () l () l = ( g) + Rln ( g) Rln φ + φ p p p p p = R ln R ln R ln φ φ = p p p he dfference n the chemcal potental of the solvent can be expressed n terms of actvty. p () l () l = R ln = R ln a p
5 hus actvty s ln a l l p = a = R p 5 If the actvty of the solvent n the soluton s equal to the mole fracton, the soluton s called an deal soluton as t follows Raoult s law. Propertes of deal soluton Volume a = x p = x p P P 0 Recall that G = V p 0 X G G hen = = = V= V = V n p p n p n p () () () l l = Rlnx l l = Rlnx p () l () l R ln x R ln x = = = p p p ln x = 0 V V = 0 V = V p V V he equalty of the partal molar volumes mples that n an deal soluton, all the molecules are the same sze.
6 Enthalpy 6 Recall the GbbsHelmholtz equaton: G H G H = = p n p n G H H = = p p l l H H Rlnx R lnx = = = ln x = 0 H H = 0 H = H he equalty of the partal molar enthalpes mples that n an deal soluton, all the nteractons between molecules are the same. hat s, Δ H =Δ H =Δ H solvent solvent solute solvent solute solute sde: Very often the solute solute nteractons are assumed to be zero. Δ H = 0 solute solute Compare the mcroscopc condtons of an deal soluton to that of an deal gas. deal soluton deal gas volume nonzero, all the same none ntermolecular forces nonzero, all the same none
7 hermodynamcs of Mxng 7 Gbbs free energy of mxng he chemcal potental of a substance changes when t s part of a soluton and the dfference between the two chemcal potentals s what s used to defne the actvty. Let G be the Gbbs free energy of a mxture G = n+ n+ n3+ 3 G be the Gbbs free energy of all of the components before mxng. G = n + n + n hen the Gbbs free energy of mxng can be defned as the dfference between two quanttes. Δ G = G G = n = n Rlna = nr x lna mx he Gbbs free energy can be rewrtten n terms of the mole fracton by recallng that n = nx, where n s the total number of moles of all substances n the soluton. Δ G = R n lna = nr x lna mx If the soluton s deal, then Δ Gmx = nr x lnx Entropy of mxng Recall that G = S p Let us takng the dervatve of the Gbbs free energy of mxng to fnd the entropy of mxng. Gmx nr x lnx Δ = ΔG mx = nr xln x Δ S = nr x lnx mx
8 Enthalpy of mxng 8 Recall the Gbbs Helmholtz equaton G H = p pply the equaton to the Gbbs free energy of mxng. ΔG mx nr x ln x = = nr xln x = ΔSmx ΔGmx ΔHmx = = nr x ln x = 0 ΔHmx = 0 Δ H mx = 0 herefore an deal soluton has no enthalpy of mxng. he mxng of an deal soluton s strctly an entropc process. Collgatve Propertes Collgatve propertes are propertes of solutons that depend on the amount of solute and not the dentty of the solute (at frst approxmaton). ddng solute decreases the chemcal potental of the solvent. l l = Rlna = Rlnx Lower the chemcal potental of a lqud component n a soluton, lowers the freezng pont and rases the bolng pont. sold lqud soluton gas m m b b
9 Freezng pont depresson 9 t equlbrum (at the meltng pont) ( l) ( s) = [Subscrpt labels the solvent.] he chemcal potental of the lqud s related to ts mole fracton n soluton. () l () s l = l + Rlnx s = l + Rlnx = R However, consder that Δ G = ( l) ( s) ln x ΔG R = ln x ake the temperature dervatve, consderng the Gbbs Helmholtz equaton. Δ G G ln x H ln x ln x Δ Δ = = = R R R ln x ΔH = R Integrate the equaton assumng ΔH s constant. (Excellent assumpton consderng the temperature range nvolved.) ΔH ΔH = = R R m m dlnx d ln lnx x m m ln x ΔH = R m m Change the emphasze from the amount of solvent to the amount of solute (labeled ), rememberng that x+ x = ΔH m m ln ( x ) = R mm ssume that m so that m = lso, from a aylor seres, ln ( x ) = x m m m ΔH m m x = R m
10 Rearrangng yelds 0 xr m m m = Δ H Consder a relatvely dlute soluton such that n << n. hen x n n he moles of solute can be calculated n terms of molalty (c ) and mass of the solvent, n = c m. he moles of solvent can be calculated n terms of mass of the solvent and ts molar mass, n = m M. hus xr nr c mr cm R Δ m = = = = ΔH nδh ( m M ) ΔH ΔH MRm Δ m = c = kfc ΔH m m m m he k f s called the cryoscopc constant and t s unque for each solvent. k f M R = Δ H m t equlbrum (at the bolng pont) ( l) = ( v) From here, a smlar analyss can be done to fnd the bolng pont elevaton as Δ = k c b b where k b, the ebulloscopc constant, s k b M R = Δ H b vap
11 Osmotc pressure Osmoss solutons n contact wth a sempermeable membrane that allows flow of only solvent through the membrane. Osmoss π t equlbrum, the chemcal potental of the soluton on the left of the sempermeable membrane s equal to the chemcal potental of the soluton on the rght of the membrane. ssume the soluton on the rght s actually the pure solvent.,p +π, x =,p atm atm he chemcal potental on the left has a concentraton dependence.,p +π, x =,p +π + Rln x atm atm hus,,p =,p +π + Rln x atm atm,p +π,p + Rln x = 0 atm atm he dfference n chemcal potentals can be cleverly rewrtten as an ntegral. However, snce patm +π atm atm patm (,p ) (,p ) d ( ) +π = G = V = V d= Vdp p p patm +π patm +π patm +π p atm patm patm d + Rlnx = 0 V dp+ Rlnx = 0 V p + Rlnx = 0 V p + π p + Rlnx =π V + Rlnx = 0 atm atm Recall our trck wth the freezng pont depresson, ln x = ln x x = x
12 n π V = Rlnx = Rx R n n n π V = R π V n = n R π V = n R π= R n V π= [ cr ] nother one of the van t Hoff equatons. he specal case of onc solutons Because collgatve propertes are ndependent of the specfc nature of the solute, no dstncton s made between a molecule n soluton and an on n soluton. hus, when takng nto account the collgatve propertes of onc solutons, the dssocaton of the onc compound must be consdered. 0. M = 0. M + 0. M = 0.4 M NaCl + Na Cl 0.45M = 0.90M M =.35M K SO + 4 K SO4 partcles partcles he proper way to wrte the van t Hoff equaton for osmotc pressure s π= cr [ ] where s the called the van t Hoff factor whch descrbes the amount of dssocaton of electrolytes (strong and weak). Example: 4.0 mg of human nsuln s dssolved n 0.7 ml of water. he osmotc pressure of the soluton s.5 torr at 37 C. Calculate the molar mass of the proten. Calculate concentraton of solute atm.5 torr π 760 torr 5 c= = = 6.47x0 M R L atm 30K mol K Calculate moles of solute mol = = = L 5 7 n c V 6.47 x L mol Calculate molar mass M 3 m 4.0x0 g = = 460g 7 n 6.9x0 mol = mol
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