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双语课件第7章对应 Atkins《Physical Chemistry》Chapter 7 Simple mixtures，讲述混合物的化学势、拉乌尔定律与亨利定律、理想溶液与实际溶液的偏差及依数性。
Part1:Equilibrium7.Simple mixtures
The chapter begins by developing the concept of chemical potential to show that it is a particular of a class of partial molar quantities.Then it explore show the chemical potential of a substance is used tod escribe the physical properties of mixture.The underlying principle is that at equilibrium the chemical potential of a speciesist he same in every phase.27.Simple mixtures
The thermodynamic description of mixtures1.Partial molar quantities2.The thermodynamics of mixing3.The chemical potentials of liquid The properties of solutions4.Liquid mixtures5.Coll i gative properties Activities6.The solvent activity7.The solute activity7.Simple mixtures3
1)Partial molar volume When the composition of the mixture is changed by the addition ofd n A of A andd n B of B,the total volume of the mixture changes by7.1Partial molar quantities BA4A)p,T,n B)p,T,n(((∂V)(∂V)d V=(∂n)dn A+(∂n)dn Bd V=V A d n A+V B d n BV A and V Bare the partial molar volumes,they vary with composition.
Once the partial molar volume soft he two component sofa mixture at the composition(and T)of interest are known,we can state the total volume,V,of the mixture by using5V=n AVA+n B V B7.1Partial molar quantities
7.1Partial molar quantities•The determination of partial molar volume With particular value soft he parameters A,B,and C,then the partial molar volume of A at any composition could be obtained from V A B=B+2Cn A)A)p,T,n((∂V)=(∂n J6J)p,T,n'(∂n(∂V)V=
7.1Partial molar quantities7n Bn BV-n VA-(n2+1)C V B=AA=A The determination of partial molar volume The partial molar volume of the second component is obtained from V=n AVA+n B V B
7.1Partial molar quantities The determination of partial molar volume The molar volume:the volume that1mol substance occupies,which is always positive.The partial molar volume:the contribution of1mol of a component to the volume of the mixture at a specific Tand p,which maybe positiveor negative.8
7.1Partial molar quantities Partial molar quantities The concept of a partial molar quantity can be extended to any extensive state function.Let X d e note any extensive properties of a system that contains K components X=X(T,p,n1,n2···n K)The change in X Whend T,dp,d n i9((K∂n(∂X)(∂X)(∂X)j≠ii=1(i)T,p,n)T,n d nid X=(∂T)d T+(∂p)dp+j)p,nj
Definition Xj≠i((∂n)(∂X)i=i)T,p,n def X and n i are extensive properties,and they are independent of the total quantities in systems;however,the partial molar quantities,Xi,is intensive property.KX=n1X1+n2X2···+n K X K=ni Xii=17.1Partial molar quantities Partial molar quantities10
7.1Partial molar quantities2)Partial molar Gibbs energies For apure substance))p,T((∂n)(∂G)=(μ=Gm Fora substance ina mixture J)p,T,n'11((∂G)μJ=(∂n)
2)Partial molar Gibbs energies7.1Partial molar quantities The chemical potential of a substance is the slope of the total Gibbs energy of a mixture with respect to the amount of substance of interest.In general,the chemical potential varies with composition.12J)p,T,n'((∂G)μJ=(∂n)The total Gibbs energy of binary mixture is G=n AμA+n BμB
7.1Partial molar quantities132)Partial molar Gibbs energies In an open system of constant composition,the Gibbs energy depends on then i,p,and T.Thus,G may change when p,T,and then i.Andd G=V d p-Sd T becomes:d G=V d p-Sd T+μA d n A+μB d n B+···At constant pand T:d G=μA d n A+μB d n B+···
2)Partial molar Gibbs energies147.1Partial molar quantities Non-expansion work can arise from the changing composition of a system.+μB d n B+···=d G=μA d n A d we,max
7.1Partial molar quantities3)Thew ider significance of the chemical potential Since G=U+p V-TS,d U=-pd V-V d p+Sd T+Td S+d G=-pd V-V d p+Sd T+Td S+(V d p-Sd T+μA d n A+μB d n B+···)The generalization equation(thatd U=Td S–pd V)to systems in which the composition may change:d U=-pd V+Td S+μA d n A+μB d n B+···15
7.1Partial molar quantities3)Thew ider significance of the chemical potentiald U=-pd V+Td S+μA d n A+μB d n B+···It follows that,at constant volume and entropy:d U=μA d n A+μB d n B+···J16)J)S,V,n'(∂n(∂U)μ=(
7.1Partial molar quantities When the composition changes,not only doesμJ show how G changes,it also show show U,H,and A change too(but und era different set of conditions).3)Thew ider significance of the chemical potential There lations ofμJ with G,U,H,A areas followings:μJ)J)S,V,n'((∂U)=(∂nμJ)J)p,T,n'((∂G)=(∂nμJ)J)S,p,n'((∂H)=(∂nμJ17)J)T,V,n'((∂A)=(∂n
7.1Partial molar quantities184)The Gibbs-D u hem equation The total Gibbs energy of a mixture depends onμJand n,fora binary mixture:G=n AμA+n BμBd G=μA d n A+μB d n B+n AdμA+n BdμB At constant pand T:It implies thatd G=μA d n A+μB d n B+···n AdμA+n BdμB=0This equation is a case of the Gibbs-D u hem equation.
J197.1Partial molar quantities4)The Gibbs-D u hem equation n JdμJ=0The significance of the Gibbs-D u hem equation is that the chemical potential of one component of a mixture cannot change independently of the chemical potential soft he other components.n Jd XJ=0J In general
7.1Partial molar quantities A20A Bndμ=-n Adμ4)The Gibbs-D u hem equation Ina binary mixture,if one partial molar quantity in creases,the other mustd e crease:This is true for all partial molar quantities.
4)The Gibbs-D u hem equation7.1Partial molar quantities Where V water in creases,Vet h a nolde creases.Moreover,a small change in V A correspond stoa large change in V B if n A/n Bis large,but the opposite is true when this ratio is small.The partial molar volumes of water and ethanol at25°C.21
Example22Using the Gibbs-D u hem equation The experimental value of the partial molar volume of K2S04(aq)at298Kis given by the expression VK2S04/(cm3mol-1)= 32.280+18.216b1/2where bis the numerical value of the molality of K2S04.Use the Gibbs-D u hem equation to derive an equation for the molar volume of water in the solution.The molar volume of pure water at298Kis18.079cm3mol-1
Example Method:Let A d e note K2S04and Bden o t e H2O,the solvent.The Gibbs-D u hem equation for the partial molar volumes of two components is AB Bnd V=-n Ad VA Bd V n n AV=V*-BB Change the variable V A to the molality b,and integrate the right-hand side between b=0(pure B)and the molality of interest.23n Ad XA+n Bd XB=0J this relation implies that n Jd XJ=0
Example Answer:It follows from the information in the question,that,with A=K2S04,and VA=32.280+18.216b1/2=9.108b-1/2dbd V An nbd b V=V-9.108-1/2AA*BB,n BM Bn Ab=A Bd V n n AV=V*-B B where bis the molar mass of water24
Example By substituting the data B3b3/2)V=V*-2(9.108M BBB V/(cm3mol-1)=18.079-0.1094b3/2The partial molar volume soft he components of an aqueous solution of potassium sulfate.b1/2db25BV=V*-9.108M BBB23*b3/2)(9.108MB-=V
Thethermodynamicdescriptionofmixtures1.Partialmolarquantities2.Thethermodynamicsofmixing3.ThechemicalpotentialsofliquidThepropertiesofsolutions4.Liquidmixtures5.ColligativepropertiesActivities6.Thesolventactivity7.Thesoluteactivity7.Simplemixtures版版权权所所有：有华：东华理工东大理学工物理大化学学物教研理室化学教研室26
7.2The thermodynamics of mixing27The dependence of the Gibbs energy of a mixture on its composition is given and we know that at constant Tand p systems tend towards alower Gibbs energy.This is the link we need in order to apply thermodynamics to the discussion of spontaneous changes of composition,as in the mixing of two substances.
7.2The thermodynamics of mixing281)The Gibbs energy of mixing Let thea mounts of two perfect gases in the two container sben A and n B;botha reata temperature Tand a pressure p.At this stage,the Gibbs energy of the total system is Gi=n AμA+n BμB|||||||||||||。。(p。)|(p)||+n(p。)|(p)||=nμ+R T l n|μ+R T l n B BAA
After mixing,the partial pressure soft he gases a rep A and p B,with p A+p B=p.The total Gibbs energy changes to29BAA ABBμ+R T l n p Gf=n|μ+R T l n p}+n|oo7.2The thermodynamics of mixing1)The Gibbs energy of mixing Tore place p/pøby p,before mixing,the total Gibbs energy is A ABB Gi=n|μ+R T l n p}+n|μ+R T l n p}oo
30)pp(p B)(p A)A mix+n BR T l n(ΔG=n R T l n7.2The thermodynamics of mixing1)The Gibbs energy of mixing The Gibbs energy of mixing,Δmix G,isΔmix G=Gf-Gi}B BAA A Bμ+R T l n p=n|μ+R T l n p}+n|oo)))(((-n}+n|μ+R T l n p}|μ+R T l n p B B A A o o
7.2The thermodynamics of mixing1)The Gibbs energy of mixing At this point we may rep lacen J by x Jand use Dalton'slaw to write p J/p=x J for each component,which givesΔmix G=n RT(x A l n x A+x B l n x B)Because x J≯1,l n x J<0,andΔmix G<0.The negativeΔmix G confirms perfect gases mix spontaneously in all proportions.AndΔmix Gis directly proportion alto the T but is independent of the total pressure.31
Example32Calculating a Gibbs energy of mixing A contain eris divided into two equal compartments.One contains3.0mol H2at25℃;the other contains1.0mol N3at25℃.Calculate the Gibbs energy of mixing when the partition is removed.Assume perfect behavior.
Method:We proceed by calculating the initial Gibbs energy from the chemical potentials.To do so,we need the pressure of each gas.Write the pressure of nitrogen asp;then the pressure of hydrogen asa multiple o fpc an be found from the gas laws.Next,calculate the Gibbs energy for the system when the partition is removed.The volume of each gas doubles,so its partial pressure falls by a factor of2.33Example
Answer:Given that the pressure of N2isp,the pressure of H2is3p;therefore,the initial Gibbs energy is B2A2μ(N)+R T l n p+(1.00mol)|μ(H)+R T l n3p}=(3.00mol)|oo22to.Therefore,the Gibbs energy changes to2The partial pressure of N falls to1pand that of H falls p322341B223A22μ((N)+R T l n p+(1.00mol)|μ(H)+R T l n p}|G=(3.00mol)oo Example B A B1Aμ+R T l n p}+n|μ+R T l n p}G=n|oo
The Gibbs energy of mixing is the difference of these two quantities:In this example,the value ofΔmix Gist he sum of two contributions:the mixing itself,and the changes in pressure of the two gasest o their final pressure,2p.When two gases mix at the same pressure,the change of Gibbs energy i sonly from mixing,that is-5.6k J.p35pp31Δmix G=(3.00mol)R T l n2+(1.00mol)R T l n23p=-(4.00mol)R T l n2=-6.9k J Example
7.2The thermodynamics of mixing2)Other thermodynamic mixing functions The entropy of mixing fora mixture of perfect gases Because(∂G∂T)p,n=-SAB36mix mix))p,n,n)(((∂T∂ΔGΔS=-=-n R(x A l n x A+x B l n x B)Because l n x<0,it follows thatΔmix S>0for all compositions
7.2The thermodynamics of mixing2)Other thermodynamic mixing functions The enthalpy of mixing fora mixture of perfect gases The isothermal,isobaric enthalpy of mixing,Δmix H,of two perfect gases maybe found fromΔG=ΔH-TΔSΔmix H=0The enthalpy of mixing is zero,as we should expect fora system in which there a reno interactions between the molecules forming the gaseous mixture.It follows that the whole of the driving force for mixing comes from the in crease in entropy of the system,because the entropy of the surroundings i sun changed.37
The thermodynamic description of mixtures1.Partial molar quantities2.The thermodynamics of mixing3.The chemical potentials of liquid The properties of solutions4.Liquid mixtures5.Coll i gative properties Activities6.The solvent activity7.The solute activity7.Simple mixtures38
7.3The chemical potentials of liquids To discuss the equilibrium properties of liquid mixtures we need to knowhow the chemical potential of a liquid varies with its composition.To calculate its value,we use the fact that,at equilibrium,the chemical potential of a substance present a sava pour must be equal to its chemical potential in the liquid.39
7.3The chemical potentials of liquids1)Raoult's Law The French chemist Francois Raoult found that the ratio of the part i alva pour pressure of each component to its v a pour pressure as apure liquid,p A/p*A,is approximately equal to the mole fraction of Ain the liquid mixture.That is AA Ap*p=x This expression is called Raoult'slaw Quantities relating to pure substances a rede noted by a superscript*in the following discussions.40
7.3The chemical potentials of liquids1)Raoult's Law AA Ap*p=x A41p*-p=p*x A A B The tot alva pour pressure and the two part i alva pour pressure sofa nideal binary mixture are proportion alto the mole fraction soft he components.Fora binary mixture:x A+x B=1
7.3The chemical potentials of liquids1)Raoult's Law-Definition for Ideal solutions AA Ap*p=x Mixtures that obey the Raoult'slaw throughout the composition range from pure A to pure Bare called ideal solutions.42
7.3The chemical potentials of liquids Some mixtures obey Raoult'slaw very well,especially when the components are structurally similar.But some mixtures do not obey Raoult'slaw;and strong deviations from ideality are shown by dis similar liquids,e.g.carbon disulfide and acetone.43
7.3The chemical potentials of liquids44p B=x B K B2)Henry'slaw Henry found that for real solutions at low concentrations,although the v a pour pressure of the solute is proportion alto its mole fraction,the constant of proportionality i snot the v a pour pressure of the pure substance:where x Bis the mole fraction of the solute and K Bis an empirical constant,the tangent to the experimental curve of plot of the v a pour pressure of Bag a inst its mole fraction at x B=0.
7.3The chemical potentials of liquids452)Henry'slaw-ideal dilute solutions p B=x B K B Mixtures for which the solute obeys Henry'slaw and the solvent obeys Raoult'slaw are called ideal-dilute solutions.
7.3The chemical potentials of liquids3)Chemical potential for ideal mixtures For apure substance:)(∂n)T,p(∂G)μ=(o(po)(p)=μ+R T l n Because the v a pour pressure of the pure liquid isp*A,it follows that)(po)(p**o Aμ=μ+R T l n(A)Where p*A is the relative pressure,p*A/pθ*46A=μ+R T l n p o
7.3The chemical potentials of liquids1)ideal solutions–The chemical potential If another substance,a solute,is alsop resent in the liquid,the chemical potential of Ain the liquid isμA and its v a pour pressure isp A,in this case Where p A is the relative pressure,p A/pθo(po)(p)AμA=μ+R T l n(A47)=μo+R T l n p
7.3The chemical potentials of liquids1)ideal solutions–The chemical potential A48A A p*μ=μ*+R T l n p A*AA=μ+R T l n p*oμA=μo+R T l n p A At equilibriumμ=(μ*-R T l n p*)+R T l n p AAA Aμ
7.3The chemical potentials of liquids1)ideal solutions–The chemical potential AA Ap*According to Raoult’slaw p=x AA Ap*μ=μ*+R T l n p A Its hows a relation between the ratio of v a pour pressure sand the composition of the liquid.μ=μ*+R T l n x AAA This important equation can be used as the definition of an ideal solution.49
Example5010.800.600.400.200x C29321914282350p C/Torr037102185270347p A/Torr Confirm that the mixture con forms to Raoult'slaw for the component in large excess and to Henry'slaw fort heminor component.Find the Henry'slaw constants.Investigating the validity of Raoult'sand Henry'slaws The v a pour pressures of each component ina mixture of acetone(A)and chloroform(C)were measured at35℃with the following results:
Example51Method:Both Raoult'sand Henry'slaws are statements about the form of the graph of part i alva pour pressure against mole fraction.Therefore,plot the part i alva pour pressures against mole fraction.Raoult'slaw is tested by comparing the data with the straight line p J=x Jp*J for each component in the region in which it is in excess(and actin gas the solvent).Henry'slaw is tested by finding a straight line p J=x JK J that is tangent to each part i alva pour pressure at low x,where the component can be treated as the solute.
Example The experimental part i alva pour pressure sofa mixture of chloroform and acetone.52Answer:The data are plotted in the Figure together with the Raoult'slaw lines..Henry'slaw requires K=175Torr for acetone and K=165Torr for chloroform
Example53Using Henry'slaw Estimate the molar solubility(the solubility in mol esper liter)of oxygen in water at25℃and a partial pressure of160Torr,its partial pressure in the atmosphere at sea level.
Example54Method:The mole fraction of solute is given by Henry'slaw as x=p/K,where pis the partial pressure of the gaseous solute.All we need do is to calculate the mole fraction that corresponds to the stated partial pressure,and then interpret that mole fraction asa molar concentration.For the latter part of the calculation,we calculate the amount of O2dissolved in1.00k go f water(which correspond stoa bout1.00L water).The solution is dilute,so the expressions for the mole fraction can be simp l i fied.
Example Answer:Because the amount of O2dissolved is small,its mole fraction is2222n(O2)n(O2)≈n(O)+n(HO)n(HO)x(O)=Kp(O2)n(H2O)n(O2)≈x(O2)n(H2O)=since p B=x B K B≈(160Torr)×(55.5mol)=2.69×10-4mol3.33×107Torr The molality of the saturated solution is therefore2.69×10-4mol k g-1,corresponding to a molar concentration of approx i-mately2.7x10-4mol L-155
Thethermodynamicdescriptionofmixtures1.Partialmolarquantities2.Thethermodynamicsofmixing3.ThechemicalpotentialsofliquidThepropertiesofsolutions4.Liquidmixtures5.ColligativepropertiesActivities6.Thesolventactivity7.Thesoluteactivity7.Simplemixtures版版权权所所有：有华：东华理工东大理学工物理大化学学物教研理室化学教研室56
The properties of solutions57In this section we consider the thermodynamics of mixing of liquids.First,we consider the simple case of mixtures of liquids that mix to for mani deal solution.In this way,we identify the thermodynamic consequences of molecules of one species mingling randomly with molecule soft he second species.The calculation provides a background for discussing the deviations from ideal behavior exhibited byreal solutions
7.4Liquid mixtures58BBB A**f AAμ+R T l n x}+n|μ+R T l n x}G=n|The Gibbs energy of mixing of two liquids to for mani deal solution is calculated inexactly the same way as for two gases.The total Gibbs energy before liquids are mixed is G=nμ*+nμ*i A A B B when they are mixed,the individual chemical potential isμ=μ*+R T l n xiii and the total Gibbs energy is
7.4Liquid mixtures The Gibbs energy of mixing,Δmix G,isΔmix G=Gf-GiΔmix G=n RT|x A l n x A+x B l n x B}Because(∂G∂T)p,n=-SIt follows that the ideal entropy and enthalpy of mixing of two liquids isΔmix S=-n R(x A l n x A+x B l n x B)Δmix H=059
7.4Liquid mixtures The variation of the Gibbs energy of mixing with composition is the same as that already depicted for gases(above);the same is true of the entropy of mixing(below).and all the conclusions drawn there are valid here:the driving force for mixing is the increasing entropy of the system as the molecules mingle,and the enthalpy of mixing is zero.60
7.4Liquid mixtures61It should be noted that solution ideality means something different from gas perfection.Ina perfect gas there a reno interactions between molecules.In ideal solutions there are interactions,but the average A-B interactions in the mixture are the same as the average A-A and B-B interactions in the pure liquids.
7.4Liquid mixtures62□The regular solution Real solutions are composed of particles for which A-A,A-B,and B-B interactions are all different.Not only may there bean enthalpy change when liquids mix,but there may also bean additional contribution to the entropy arising from thew ayin which the molecules of one type might cluster together instead of mingling freely with the others.
7.4Liquid mixtures□The excess functions The difference between the observed thermodynamic function(extensive)of mixing and the function fora nideal solution is defined as the excess functions,X E X E=Δmix X-Δmix Xid X E X-Xid def63
7.4Liquid mixtures64□The excess functions The thermodynamic properties of real solutions maybe expressed in term soft he excess functions,XE.Deviation soft he excess functions from zero indicate the extent to which the solutions a reno nideal.S E G E=Δmix G-Δmix Gid=Δmix S-Δmix Sid
7.4Liquid mixtures65□The regular solution HE≠0SE=0A regular solution can bethought of a sone in which the two kinds of molecules are distributed randomly(as in an ideal solution)but have different energies of interactions with each other.
The thermodynamic description of mixtures1.Partial molar quantities2.The thermodynamics of mixing3.The chemical potentials of liquid The properties of solutions4.Liquid mixtures5.Coll i gative properties Activities6.The solvent activity7.The solute activity7.Simple mixtures66
7.5Coll i gative properties671)The common features of coll i gative properties In dilute solutions,some properties depend on lyon the number of solute particles present,not their identity.They are called coll i gative properties.Some of these properties areas followings:The elevation of boiling point The depression of freezing point Solubility and osmotic properties
2)The elevation of boiling point7.5Coll i gative propertiesμ*(g)=μ*(l)+R T l n x AAA Considering boiling is between the solvent v a pour and the solvent in solution at1atm.Weden o tet he solvent by A and the solute by B.The equilibrium is established at a temperature for which T68p Tb*Tb Ap*p A1atm
7.5Coll i gative properties2)The elevation of boiling point The presence of a solute at a mole fraction x B causes an in crease in normal boiling point from T*to T*+ΔTH69v a p RT*2K=ΔΔT=Kx B Because the equation make no reference to the identity of the solute,only to its mole fraction,we conclude that the elevation of boiling point is a coll i gative property.The value ofΔT does depend on the properties of the solvent.
7.5Coll i gative properties702)The elevation of boiling point For practical applications,we note that the mole fraction of Bis proportion alto its molality,b,in dilute solutions,thenΔT=K b b K bis the empirical constant of the solvent
7.5Coll i gative properties RT*2Δf us HK'=3)The depression of freezing point The only difference between this calculation and the above is the appearance of the solid's chemical potential in place of the v a pour's.we can write the result directlyΔT=K'x B whereΔTis the freezing point depression,Δf us His the enthalpy of fusion of the solvent.T p f T*Tf Ap*Lp A Lp A*S71
7.5Coll i gative properties4)Solubility Whenas olid solute is left in contact with a solvent,it dissolves until the solution is saturated.Saturation is a state of equilibrium,with the und is solved solute in equilibriumμB*(s),and the chemical potential of Bin solution,μB(l),are equal The heterogeneous equilibrium involved in the calculation of the solubility is between pure solid Band Bin the solution.μ*(s)=μ(l)=μ*(l)+R T l n x BBB B72
7.5Coll i gative properties4)Solubility To find the mole fraction of Bin solution at equilibrium when the temperature is T:R T R T=-Δf us H+Δf us Sμ*(s)-μ*(l)ΔG l n x B=BB=-f us R T R At the melting point of the solute,T*,we know thatΔf us G=0,con sequently,)73(*BT(T)R l n x=-Δf us H(1-1)
7.5Coll i gative properties The osmotic pressure,·,5)Osmosis The phenomenon of osmosis is the spontaneous passage of apure 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,П,is the pressure that must be app lied to the solution to stop the influx of solvent.74
7.5Coll i gative properties5)Osmosis For dilute solutionsΠ=[B]R T V[B]=n B[B]is the molar concentration of the solute.The above equation is called van t’t Hoff equation.75
The thermodynamic description of mixtures1.Partial molar quantities2.The thermodynamics of mixing3.The chemical potentials of liquid The properties of solutions4.Liquid mixtures5.Coll i gative properties Activities6.The solvent activity7.The solute activity7.Simple mixtures76
Activities77Now we see how to adjust the expressions developed earl i erin the chapter to take into account deviations from ideal be h avi our.In Section5.4we saw how the fugacity was introduced to take into account the effects of gas imperfections ina manner that resulted in the least upset of the form of equations.Here we see how the expressions encountered in the treatment of ideal solution scan also be preserved almost intact by introducing the concept of activity.
7.6The solvent activity The general form of the chemical potential of arealor ideal solvent A78A A p*μ=μ*+R T l n(p A)When the solution does not obey Raoult'slaw the equation can be preserved asμ=μ*+R T l n a AAAμ=μ*+R T l n x AAA Raoult'slaw
7.6The solvent activity is true for both real and ideal solutions,AAA p*Becauseμ=μ*+R T l n(p A)A79Ap*a=p A□The activity,a Aμ=μ*+R T l n a AAA The quantity a A is the activity of A,akind of'effective'mole fraction,just as the fugacity is an effective pressure.
7.6The solvent activity□The activity coefficient,γA Because all solvents obey Raoult'slaw increasingly closely as the concentration of solute approaches zero,the activity of the solvent approaches the mole fraction as x A→1as x A→1a A→x A as x A→1a A=γAx AγA→1γA is called the activity coefficient A80Axγ=a A
7.6The solvent activity81x A→1A tall temperature sand pressures,the chemical potential of the solvent is thenμ=μ*+R T l n x+R T l nγAAA A The standard state of the solvent,the pure liquid solvent at1bar,is established when x A=1□The activity coefficient,γAa A→x A as x A→1a A=γAx AγA→1as
The thermodynamic description of mixtures1.Partial molar quantities2.The thermodynamics of mixing3.The chemical potentials of liquid The properties of solutions4.Liquid mixtures5.Coll i gative properties Activities6.The solvent activity7.The solute activity7.Simple mixtures82
7.7The solute activity*(B)*B B p(p B)μ=μ+R T l n1).
Ideal-dilute solutions A solute B that satisfies Henry'slaw has ava pour pressure given by p B=K Bx B,where K Bis an empirical constant.In this case,the chemical potential of Bis BB Bp*=μ*+R T l n(KB)μ+B**Bp+R T l n x(B)(KB)=μ+R T l nμ=μ++R T l n x BBB It follows that The new standard chemical potential is83
7.7The solute activity2).Real solutes When permit deviations from ideal-dilute,Henry'slaw be h avi our.For the solute,introduce a Bin place of x Bμ=μ++R T l n x BBBμ=μ++R T l n a BBB The standard state remain sun changed,and all the deviations from ideality are capture dint he activity a B.The value of the activity at any concentration can be obtained in the same way as for the solvent,but here84B B K a=p B
7.7The solute activity852).Real solutes a B=γBx B All the deviations from ideality are capture dint he activity coefficientγB.Because the solute obeys Henry'slaw as x B→0,it follows that a B→x BandγB→1as x B→0A tall temperature sand pressures.Deviation soft he solute from idealityd is appear as zero concentration is approached.
7.7The solute activity3).Activities in terms of molalities In dilute solutions the amount of solute is much less than that of solvent n B<<n Ax B≈n B/n A Bbθ=κb B Be causen Bis proportion alto the molality b B,s ox where bθ=1mol k g-1andκis a dimensionless constant.Fora nideal-dilute solution it follows that BB Bbθμ=μ++R T l nκ+R T l n(b B)When re place b/bθby b Bμ=μ++R T l nκ+R T l n b BB86
7.7The solute activity thenμB=μBo+R T l n b Bis defined as the chemical potential of the soluteμθB when the molality of Bis equal tobθ(at1mol k g-1).The deviations from ideality by introducing a dimension-less activity,a B a dimensionless activity coefficientγBb BB Bbθa=γwhereγB→1as b B→0Let be anew standard chemical potential BBμ=μ+R T l nκo+Bμ=μ++R T l nκ+R T l n b BB87
Summary on activities Limits Component Solvent Basis Raoult Standard state Pure solventγ→1as x→1a=p/p*and a=γx Pure solvent Solute Henry1)A hypothetical state of pure solute a=p/Kand a=γxγ→1as x→02)A hypothetical state of solute at molality bθγ→1as b→0Chemical potentialμ=μ*+R T l n aμ=μ++R T l n aμ=μo+R T l n aμo=μ++R T l nκBp*μ+=μ*+R T l n(KB)*o pop*μ=μ+R T l n(A)B88bθa=γb/bθx=κb B
10.800.600.400.200x C29321914282350p C/Torr037102185270347p A/Torr89Use the information in Example7.3to calculate the activity and activity coefficient of chloroform in acetone at25℃,treating it first asa solvent and the nasa solute.The v a pour pressures of each component ina mixture of acetone(A)and chloroform(C)were measured at35℃with the following results:KB=165Torr Example Measuring activity
ExampleaB=pBKBγB=aBxBMethod:Fortheactivityofchloroformasasolvent10.800.600.400.200xC29321914282350pC/TorrAp*aA=pAγA=aAxAForitsactivityasasolute(theHenry'slawactivity):00.120.280.490.751.000.600.700.820.941.0000.210.500.861.331.7811.051.251.431.661.7890
Exampleγ→1as x→1the Raoult'slaw caseγ→1as x→0the Henry'slaw case91
1.The properties of gases2.The First Law:the concepts3.The First Law:the machinery4.The Second Law:the concepts5.The Second Law:the machinery6.Physical transformations of pure substances7.Simple mixtures8.Phase diagrams9.Chemical equilibrium10.Electrochemistry92Part1:Equilibrium
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