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Chapter 4 Phases and Solutions. 4.1 Phase Recognition 4.2 Physical transformations of pure substances 4.3 Simple mixtures 4.4 Raoult’s and Henry’s Laws 4.5 The chemical potentials of liquids 4.6 The properties of solutions. New Words and Expressions. Homogeneous 均相,同相

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Chapter 4 phases and solutions
Chapter 4 Phases and Solutions

  • 4.1 Phase Recognition

  • 4.2 Physical transformations of pure substances

  • 4.3 Simple mixtures

  • 4.4 Raoult’s and Henry’s Laws

  • 4.5 The chemical potentials of liquids

  • 4.6 The properties of solutions


New words and expressions
New Words and Expressions

  • Homogeneous 均相,同相

  • Heterogeneous 异相,不同相

  • solution 溶液

  • solvent 溶剂

  • solute 溶质

  • mole fraction 摩尔分数

  • molality 质量摩尔分数

  • molarity 物质的量浓度,容模


New words and expressions1
New Words and Expressions

  • Mixture 混合物

  • Partial molar quantity 偏摩尔量

  • Chemical potential 化学势

  • Multi component 多组分

  • ideal solution 理想溶液

  • Ideal dilute solution 理想稀溶液

  • Activity and activity factor 活度和活度系数


New words and expressions2
New Words and Expressions

  • nonvolatile solute 不挥发性溶质

  • positive (negative) deviation 正(负)偏离

  • dilute solution 稀溶液

  • colligative properties 依数性质

  • freezing point 凝固点

  • boiling point 沸点

  • osmotic pressure 渗透压


4 1 phase recognition
4.1 Phase Recognition

  • Homogeneous and Heterogeneous

    A single phase is uniform throughout both in chemical composition and physical state, and it is said to be homogeneous. In contrast to this, a heterogeneous system consists of more than one phase.

  • The phases are distinguished from each other through separation by distinct boundaries.


Chapter 4 phases and solutions

  • Solid and liquid mixtures may consist of a number of phases; gases exist in only one phase at normal pressures since gases mix in all proportions to give a uniform mixture.

    A phase transition, the spontaneous conversion of one phase into another phase, occurs at a characteristic temperature for a given pressure.


Chapter 4 phases and solutions

The transition temperature is the temperature at which the two phases are in equilibrium and the Gibbs energy is minimized at the given pressure.

Phase Equilibrium in a one-Component System: The change in Gibbs energy between the two equilibrium phase is zero under given conditions of temperature and pressure.


Chapter 4 phases and solutions


4 2 physical transformations of pure substances
4.2 Physical transformations of pure substances the two phases are in equilibrium and the Gibbs energy is minimized at the given pressure.

  • Thermodynamics of Vapor Pressure: The Clapeyron Equation

    We begin with the statement of phase equilibrium, written for a pure substance in the liquid and vapor states. If the pressure and temperature are changed infinitesimally in such a way that equilibrium is maintained,

    dGv=dGl


Chapter 4 phases and solutions

the two phases are in equilibrium and the Gibbs energy is minimized at the given pressure.Sm(v)dT+Vm(v)dP= Sm(l)dT+Vm(l)dP

or dP/dT= [Sm(v)  Sm(l)]/ [Vm(v) Vm(l)]

=Sm/Vm

=Hm/TVm (4.5)

  • This is known as the Clapeyron Equation and may be applied to vaporization, sublimation, fusion, or solid phase transitions of a pure substance.


Chapter 4 phases and solutions


Chapter 4 phases and solutions

  • The Clausius-Claperon Equation fusion, and vaporization are related at constant temperature by the expression

    When one of the phases in equilibrium is a vapor phase, we assume that Vm(v) is so much larger than Vm(l) that we may neglect Vm(l) in comparison to Vm(v) when the pressure is near 1 bar. The second assumption is to replace Vm(v) by its equivalent from the ideal gas law RT/P. then,


Chapter 4 phases and solutions

  • dP/dT= fusion, and vaporization are related at constant temperature by the expressionvapHmP/RT2 (4.3)

  • or dlnP/dT=vapHm/RT2 (4.4)

  • This expression is known as the Clausius-Claperon Equation.

  • Assuming vapHm to be independent of temperature and pressure. we thus obtain

  • lnP=vapHm/RT+C (4.5)

  • or lnP2/P1=vapHm/R(1/T11/T2) (4.6)


Chapter 4 phases and solutions

  • 运用方程4.5 fusion, and vaporization are related at constant temperature by the expressionor 4.6 ,必须满足以下三个条件:

  • (1)气-液或气-固两相平衡,气体可视为理想气体。

  • (2)在温度变化范围内摩尔蒸发焓可视为常数。

  • (3) V(l) 与V(g)相比可忽略不计。


Chapter 4 phases and solutions

Example 4.1 Benzene has a normal boiling point at 760 Torr of 353.25K and vapH=30.76kJmol-1, if benzene is to be boiled at 30.000C in a vacuum distillation, to what value of P must the pressure be lowered?

Solution: Using Eq.4.10 , we have

lnP2/760.0=30760/8.3145(1/353.25 - 1/303.15)

P2 =134.6Torr


Chapter 4 phases and solutions

  • Enthalpy and Entropy of Vaporization: Trouton’s Rule of 353.25K and

    The entropies of vaporization vapSm of most non-hydrogen-bonded compounds have values of vapSm in the neighborhood of 88 JK-1mol-1. This generalization is known as Trouton’s rule and was pointed out in 1884:

  • vapHm/Tb=vapSm88 JK-1mol-1 (4.7)


Chapter 4 phases and solutions


4 3 simple mixtures
4.3 of 353.25K and Simple mixtures

  • At 298k, 1atm,

    100cm3 water + 100cm3 ethanol 192cm3

    150cm3 water + 50cm3 ethanol 195cm3

    50cm3 water + 150cm3 ethanol 193cm3


Partial molar quantities
Partial Molar Quantities of 353.25K and

  • Partial Molar Quantities

  • Any extensive thermodynamic quantity such as the enthalpy, internal energy, or the Gibbs energy, each of these extensive functions depends on the amount variables for the particular function, Z=Z(T,P,n1,n2,,ni)


Chapter 4 phases and solutions

按偏摩尔量定义, of 353.25K and

在保持偏摩尔量不变的情况下,对上式积分


Chapter 4 phases and solutions

这就是偏摩尔量的集合公式,说明体系的总容量性质等于各组分偏摩尔量的加和。这就是偏摩尔量的集合公式,说明体系的总容量性质等于各组分偏摩尔量的加和。


Chapter 4 phases and solutions

  • 只有容量性质才有偏摩尔量这就是偏摩尔量的集合公式,说明体系的总容量性质等于各组分偏摩尔量的加和。

  • 偏摩尔量与系统的浓度有关

  • 纯物质的偏摩尔量就是其摩尔量


Chapter 4 phases and solutions

dV=(这就是偏摩尔量的集合公式,说明体系的总容量性质等于各组分偏摩尔量的加和。V/n1)T,P,n,n,dn1+(V/n2)T,P,n,n,dn2+ (4.8)

The increase in volume per mole of component 1 is known as the partial molar volume of component 1. It is given the symbol V1 and is written as

V1 (V/n1)T,P,n,n, (4.9)


Chapter 4 phases and solutions

In either case, the definition for the partial molar volume, may be used to rewrite as

dV=Vdn1 + Vdn2 +  (4.10)

  • once the partial molar volumes of the two components of a mixture at the composition(and temperature) of interest are known, we can state the total volume, V of the mixture by using

  • V=n1V1+n2V2+ (4.11)


Chapter 4 phases and solutions

  • Gibbs-Duhem Equation may be used to rewrite as

  • x1dV1+x2dV2+=0 (4.12)

  • the significance of the Gibbs- Duhem Equation is that the chemical potential of one component of a mixture cannot change independently of the chemical potentials of the other components.


The chemical potential
The Chemical Potential may be used to rewrite as

  • The partial molar Gibbs energy is called the chemical potential i, for the its component. Therefore,

  • dG =SdT + VdP +idni (4.13)

  • G=n1G1+n2G2+ (4.14)

  • A similar treatment of the other thermodynamic functions shows that

  • I=(G/ni)T,P,n=(U/ni)S,Vn=(H/ni)S,P,n

  • =(A/ni)T,Vn=T(S/ni)U,Vn (4.15)


Chapter 4 phases and solutions


Chapter 4 phases and solutions

  • This expression allows the calculation of the Gibbs energy for the change in both the amount of substance present in a phase and also the number of the phase’s components.

  • If a single phase is closed, and no matter is transferred across its boundary (dG=0 for a closed system),

  • idni =0 (4.17)


Chapter 4 phases and solutions

  • If a system consisting of several phases in contact is closed but matter is transferred between phases, the condition for equilibrium at constant T and P becomes

  • dG= dG +dG+ dG+ =0 (4.18)

  • We may write

  • dG=idni+idni +idni + =0 (4.19)


Chapter 4 phases and solutions


Chapter 4 phases and solutions

  • if equilibrium is that the chemical potential of the substance i is the same in the two phases.i < i , dG<0, the transfer of matter occurs i from  to ;

  • if i >i , dG>0, the transfer of matter occurs i from  to ;

  • if i =i , dG=0, no matter cross its boundary.


Chapter 4 phases and solutions

equilibrium is that the chemical potential of the substance i is the same in the two phases.

  • the chemical potential of a perfect gas

  • 0i is the standard chemical potential, the chemical potential of the pure gas at 1 bar.

  • for a real gas,


Chapter 4 phases and solutions

  • the chemical potentials of equilibrium is that the chemical potential of the substance i is the same in the two phases.Condensed phases

  • pure liquid

    l=g =0(g) +RTlnPl*/p0

    pure solid

    s=g =0(g) +RTlnPs*/p0

    P* is Saturated vapor pressureof pure liquid or pure solid.


Simple mixture
Simple mixture equilibrium is that the chemical potential of the substance i is the same in the two phases.

A solution is any homogeneous phase that contains more than one component. We call the component that constitutes the larger proportion of the solution the solvent; the component in lesser proportion is called the solute.


Chapter 4 phases and solutions

  • mole fraction equilibrium is that the chemical potential of the substance i is the same in the two phases.

molality


Chapter 4 phases and solutions

  • molarity equilibrium is that the chemical potential of the substance i is the same in the two phases.

mass fraction


4 4 raoult s law and henry s law
4.4 equilibrium is that the chemical potential of the substance i is the same in the two phases.Raoult’s law and Henry’s law

  • Raoult’s law: According to the French chemist Francois Raoult, the ratio of the partial vapor pressure of each component to its vapour pressure as a pure liquild PA / P*A , is approximately equal to the mole fraction of 1 in the liquild mixture, that is what we now call Raoult’s law .

  • PA=x1 P*A (4.22)


Raoult s and ideal solutions
Raoult’s and Ideal solutions equilibrium is that the chemical potential of the substance i is the same in the two phases.

  • Some mixtures obey Raoult’s law very well, especially when the components are structurally similae.

  • A number of pairs of liquids obey Raoult’s law over a wide range of compositions.


Raoult s and ideal solutions1
Raoult’s and Ideal solutions equilibrium is that the chemical potential of the substance i is the same in the two phases.


Raoult s and ideal solutions2
Raoult’s and Ideal solutions equilibrium is that the chemical potential of the substance i is the same in the two phases.

Mixtures that obey the law throughout the composition range from pure A to pure B are called ideal solutions.

The solution is considered to be ideal when there is a complete uniformity of intermolecular forces, arising from similarity in molecular size and structure.


Chapter 4 phases and solutions

  • Deviations from Raoult’s law do occur and may be explained if we consider again the interaction between molecules A and B. If the strength of the interaction between like molecules, A-A or B-B, is greater than that between A and B the tendency will be to force both components into the vapor phase.


Chapter 4 phases and solutions

  • This increases the pressure is known as a positive deviation. In addition to the aforementioned, negative deviations occur when the attractions between components A and B are strong. This may be visualized as a holding back of molecules that would otherwise go into the vapor state.


Positive deviation
positive deviation deviation. In addition to the aforementioned, negative deviations occur when the attractions between components A and B are strong. This may be visualized as a holding back of molecules that would otherwise go into the vapor state.


Henry s law and ideal dilute solutions
Henry’s law and Ideal-dilute solutions deviation. In addition to the aforementioned, negative deviations occur when the attractions between components A and B are strong. This may be visualized as a holding back of molecules that would otherwise go into the vapor state.

  • Another property of binary systems was discovered by the English physical chemist William Henry. He found experimentally that, for real solutions at low concentrations, the constant of proportionality is not the vapor pressure of the pure substance.


Henry s law and ideal dilute solutions1
Henry’s law and Ideal-dilute solutions deviation. In addition to the aforementioned, negative deviations occur when the attractions between components A and B are strong. This may be visualized as a holding back of molecules that would otherwise go into the vapor state.

  • m2k2 =P2 or P2=k’x2 =k”c2 (4.27)

    Where k is the Henry’s law constant, which is an empirical constant. Mixtures for which the solute obeys Henry’s law and the solvent obeys Raoult’s law are called ideal-dilute solutions.


Henry s law and ideal dilute solutions2
Henry’s law and Ideal-dilute solutions deviation. In addition to the aforementioned, negative deviations occur when the attractions between components A and B are strong. This may be visualized as a holding back of molecules that would otherwise go into the vapor state.


Henry s law and ideal dilute solutions3
Henry’s law and Ideal-dilute solutions deviation. In addition to the aforementioned, negative deviations occur when the attractions between components A and B are strong. This may be visualized as a holding back of molecules that would otherwise go into the vapor state.

  • If several gases from a mixture of gases dissolve in a solution, Henry’s law applies to each gas independently, regardless of the pressure of the other gases present in the mixture.

  • Thus Henry’s law may also be applied to dilute solutions of a binary liquid system. It is found that in the limit of infinite dilution most liquid solvents obey Raoult’s law but that under the same conditions the solute obeys Henry’s law.


4 5 the chemical potentials of liquilds
4.5 deviation. In addition to the aforementioned, negative deviations occur when the attractions between components A and B are strong. This may be visualized as a holding back of molecules that would otherwise go into the vapor state. The chemical potentials of liquilds

  • (a) Ideal solutions

  • For any component i of a solution in equilibrium with its vapor, we may write

  • i,sol = i,vap

  • If the vapor behaves ideally, the Gibbs energy for each component is given

  • Gi =G0i +niRTlnPi /P0


Chapter 4 phases and solutions

  • Since deviation. In addition to the aforementioned, negative deviations occur when the attractions between components A and B are strong. This may be visualized as a holding back of molecules that would otherwise go into the vapor state. i = Gi/ni , we may write

  • i,vap= 0i,vap + RTlnPi / P0

  • therefore, i,sol=i,vap = 0i,vap + RTlnPi /P0

  • Pi=xi P*i (4.23)

  • i,sol=i,vap = 0i,vap + RTlnPi /P0

  • = 0i,vap + RTln xi P*i /P0

  • = i*+ RTlnxi

    where the superscript * represents the value for the pure material.


Chapter 4 phases and solutions

  • ( deviation. In addition to the aforementioned, negative deviations occur when the attractions between components A and B are strong. This may be visualized as a holding back of molecules that would otherwise go into the vapor state. b) Ideal-dilute solutions


4 6 the properties of solutions
4.6 The properties of solutions deviation. In addition to the aforementioned, negative deviations occur when the attractions between components A and B are strong. This may be visualized as a holding back of molecules that would otherwise go into the vapor state.

  • (a) Liquid mixtures

  • Ideal Solutions

  • mixVid =0 or Vi= Vi* (4.24)

  • mixHid =0 or Hi= Hi* (4.25)

  • mixG id =RT(n1lnx1 +n2lnx2 +) (4.26)

  • mixS id =- R(n1lnx1 +n2lnx2 +) (4.27)

  • i,,id =i*+ RTlnxi (4.28)


Excess functions and regular solutions
Excess functions and regular solutions deviation. In addition to the aforementioned, negative deviations occur when the attractions between components A and B are strong. This may be visualized as a holding back of molecules that would otherwise go into the vapor state.

  • real Solutions: Activity and Activity Coefficient

  • Pi=ai P* i (4.29)

  • ai , Activity and ai /xi Activity Coefficient

  • i, =i*+ RTlnai (4.30)

  • The thermodynamic properties of real solutions are expressed in terms of the excess functions, the excess entropy, for example, is defined as

  • SE = mixS _mixS id


B the colligative properties
( deviation. In addition to the aforementioned, negative deviations occur when the attractions between components A and B are strong. This may be visualized as a holding back of molecules that would otherwise go into the vapor state. b) The Colligative Properties

  • The properties of dilute solutions that depend on only the number of solute molecules and not on the type of species present are called colligative properties.

  • All colligative properties (such as boiling point elevation, freezing point depression, and osmotic pressure) ultimately can be related to a lowering of the vapor pressure P*- P for dilute solutions of nonvolatile solutes.


Chapter 4 phases and solutions

  • Vapor pressure lowering deviation. In addition to the aforementioned, negative deviations occur when the attractions between components A and B are strong. This may be visualized as a holding back of molecules that would otherwise go into the vapor state.

    △p=p*A –pA

    =p*A xB (4.31)


Chapter 4 phases and solutions

  • Freezing Point Depression deviation. In addition to the aforementioned, negative deviations occur when the attractions between components A and B are strong. This may be visualized as a holding back of molecules that would otherwise go into the vapor state.

    fusT=Tf* - Tf M1RTf*2/fusHm m2 (4.32)

  • The freezing point depression or cryoscopic constant Kf , defined as

  • Kf= M1RTf*2/fusHm (4.33)

  • With the definition of Kf , Eq4.44. may be expressed as

  • fusT= Kfm2 (4.34)


Chapter 4 phases and solutions

  • Example 4.3 Find the value of K deviation. In addition to the aforementioned, negative deviations occur when the attractions between components A and B are strong. This may be visualized as a holding back of molecules that would otherwise go into the vapor state. f for the solvent 1,4-dichlorobenzene from the following data:

  • M=147.01gmol-1; Tf* =326.28K; fusHm =17.88kJmol-1

  • Solution The value of Kf depends only on the properties of the pure solvent.

  • Kf= M1RTf*2/fusHm


Chapter 4 phases and solutions

  • =0.14701*8.3145*326.28 deviation. In addition to the aforementioned, negative deviations occur when the attractions between components A and B are strong. This may be visualized as a holding back of molecules that would otherwise go into the vapor state. 2/17880

  • =7.28Kkgmol-1


Chapter 4 phases and solutions

  • Boiling Point Elevation deviation. In addition to the aforementioned, negative deviations occur when the attractions between components A and B are strong. This may be visualized as a holding back of molecules that would otherwise go into the vapor state.

  • vapTb=Tb- Tb* =Kbm2 (4.35)

  • where Kb= M1RTb*2/vapsHm (4.36)

  • Kb is the empirical ebullioscopic constant od the solvent.


Chapter 4 phases and solutions

  • Osmotic Pressure deviation. In addition to the aforementioned, negative deviations occur when the attractions between components A and B are strong. This may be visualized as a holding back of molecules that would otherwise go into the vapor state.

  • The solution and the pure solvent are separated from each other by a semipermeable membrane. The natural flow of solvent molecules can be stopped by applying a hydrostatic pressure to the solution side.


Chapter 4 phases and solutions

  • The effect of the pressure is to increase the tendency of the solvent molecules to flow from solution into pure solvent. The particular pressure that causes the net flow to be reduced to zero is known as the ostmotic pressure  of the solution.

  • Van’t Hoff’s Equation

  •  = n2RT/V or  =cRT (4.37)


Chapter 4 phases and solutions

  • Example 4.4 the solvent molecules to flow from solution into pure solvent. The particular pressure that causes the net flow to be reduced to zero is known as the ostmotic pressure

  • Blood can be regarded as aqueous solution, solidified at -0.560C and 101.325kpa. It is known that the freezing point lowering cofficients of water Kf = 1.86kg/mol. Please find

  • (1) the osmotic pressure of blood at 370C.

  • (2)at the same temperature, how much C12H22O11

    is needed to be put into 1 dm3 aqueous solution then the osmotic pressure be the same as blood.


Chapter 4 phases and solutions

  • Answer: (1) m the solvent molecules to flow from solution into pure solvent. The particular pressure that causes the net flow to be reduced to zero is known as the ostmotic pressure B = Tf /Kf

    =0-(-0.56)/1.86

    =0.301mol/kg

    for a dilute solution, mB =cB

     =cRT =776kpa

    (2)M C12H22O11 =342.99g/mol

    WC12H22O11 =M*cV=103g


Chapter 4 phases and solutions

  • 1.The partial molar volumes of acetone (propanone) the solvent molecules to flow from solution into pure solvent. The particular pressure that causes the net flow to be reduced to zero is known as the ostmotic pressure 丙酮 and chloroform (trichloromethane)氯仿 in a mixture in which the mo1e fraction of CHCl3 is 0.4693 are 74.166 cm3 mol -1 and 80.235 cm3 mol-1, respectively. What is the volume of a solution of mass 1.000 kg?


Chapter 4 phases and solutions

  • 2. the solvent molecules to flow from solution into pure solvent. The particular pressure that causes the net flow to be reduced to zero is known as the ostmotic pressure The vapour pressure of benzene is 400 Torr at 60.60C, but it fell to 386 Torr when 19.0 g of an involatile organic compound was dissolved in 500 g of benzene. Calculate the molar mass of the compound.


Chapter 4 phases and solutions

  • 3.The addition of 100 g of a compound to 750 g of' CC1 the solvent molecules to flow from solution into pure solvent. The particular pressure that causes the net flow to be reduced to zero is known as the ostmotic pressure 4 lowered the freezing point of the solvent by 10.5 K. Calculate the molar mass of the compound.


Chapter 4 phases and solutions

  • 4.Substances A and B are both volatile liquids with p the solvent molecules to flow from solution into pure solvent. The particular pressure that causes the net flow to be reduced to zero is known as the ostmotic pressure A*=300 Torr, pB*=250 Torr, and KB= 200 Torr (concentration expressed in mole fraction). When x A=0.9, bB= 2.22 mol kg -1, pA = 250 Torr and PB =25 Torr. Calculate the activities and activity coefifcients of A and B. Use the mole fraction, Raoult's law basis system, for A and the Henry's law basis system (both mole fractions and molalities) for B.


Chapter 4 phases and solutions

  • 5.Benzene and toluene form nearly ideal solutions. The boiling point of pure benzene is 80.10C. Calculate the chemical potential of benzene relative to that of pure benzene when Xbenzene = 0.30 at its boiling point. If the activity coefficient of benzene in this solution were actually 0.93 rather than 1.00, what would be its vapour pressure?


Chapter 4 phases and solutions

  • 6.The variation of the equilibrium vapor pressure with temperature for liquid and solid chlorine(氯) in the vicinity of the triple point is given by

  • In P1 = -2661/T +22.76

  • In P2 =- 3755/T+ 26.8

  • Use P/pascal in the equations. Calculate the triple point pressure and temperature.


Chapter 4 phases and solutions

  • 7. temperature for liquid and solid chlorine(At 250C, 0.5mol A and 0.5mol B can be regarded as ideal liquid mixture. Please calculate mixV , mixH , mixS , mixG , mixU, mixA of the mixing process.


Chapter 4 phases and solutions

  • 8. temperature for liquid and solid chlorine(At 1000C, the vapor pressure of CCl4 and SnCl4 is respectively 1.933*105 Pa and 0.666*105Pa. CCl4 and SnCl4 can form mixture of ideal liquid, when external pressure p =1.013* 105Pa, this mixture of ideal liquid is heated to boiling, please calculate:

  • (1) composition of this mixture of ideal liquid;

  • (2)composition of the first air bubble while the mixture of ideal liquid is boiling.