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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.
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.

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.

for example , when equilibrium occurs between ice and water

G(s)=G (l) (4.1)

and when there is equilibrium between water and steam

G(l)=G (g) (4.2)

4 2 physical transformations of pure substances
4.2 Physical transformations of pure substances
  • 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,


Sm(v)dT+Vm(v)dP= Sm(l)dT+Vm(l)dP

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


=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.
It should be pointed out that the enthalpies of sublimation, fusion, and vaporization are related at constant temperature by the expression
  • subHm=fusHm+vapHm (4.6)
The Clausius-Claperon Equation

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,

dP/dT=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)
运用方程4.5 or 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。
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

Enthalpy and Entropy of Vaporization: Trouton’s Rule

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)
Example 4.2 Estimate the enthalpy of vaporization of CS2 if its boiling point is 319.40K.
  • Solution vapHm/Tb=vapSm88 JK-1mol-1
  • vapHm=88319.40=28.11(kJmol-1)
  • The experimental value is 28.40 kJmol-1.
4 3 simple mixtures
4.3 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
  • 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)





  • 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
dV=(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)

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)
Gibbs-Duhem Equation
  • 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
  • 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)
One of the most common uses of the chemical potential is as the criterion of equilibrium for a component distributed between two or more phase. Under conditions of constant temperature and pressure,
  • dG=idni (4.16)
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)
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)
Thus, for a one-component system, the requirement for equilibrium is that the chemical potential of the substance i is the same in the two phases.
  • i =i (4.20)
  • For a nonequilibrium situation, we write it in the form
  • dG=(i-i)dni (4.21)
if 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.

  • 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,
the chemical potentials ofCondensed 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

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.


mass fraction

4 4 raoult s law and henry s law
4.4 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
  • 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 solutions2
Raoult’s and Ideal solutions

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.

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.
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.
henry s law and ideal dilute solutions
Henry’s law and Ideal-dilute solutions
  • 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
  • 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 solutions3
Henry’s law and Ideal-dilute solutions
  • 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 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
Since 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.

4 6 the properties of solutions
4.6 The properties of solutions
  • (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
  • 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
(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.
Vapor pressure lowering

△p=p*A –pA

=p*A xB (4.31)

Freezing Point Depression

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)
Example 4.3 Find the value of Kf 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
Boiling Point Elevation
  • vapTb=Tb- Tb* =Kbm2 (4.35)
  • where Kb= M1RTb*2/vapsHm (4.36)
  • Kb is the empirical ebullioscopic constant od the solvent.
Osmotic Pressure
  • 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.
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)
Example 4.4
  • 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.

Answer: (1) mB = Tf /Kf



for a dilute solution, mB =cB

 =cRT =776kpa

(2)M C12H22O11 =342.99g/mol

WC12H22O11 =M*cV=103g

1.The partial molar volumes of acetone (propanone)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?
2. 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.
3.The addition of 100 g of a compound to 750 g of\' CC14 lowered the freezing point of the solvent by 10.5 K. Calculate the molar mass of the compound.
4.Substances A and B are both volatile liquids with pA*=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.
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?
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.
7. 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.
8. 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.