Chapter 4 phases and solutions
This presentation is the property of its rightful owner.
Sponsored Links
1 / 76

Chapter 4 Phases and Solutions PowerPoint PPT Presentation


  • 268 Views
  • Uploaded on
  • Presentation posted in: General

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

Download Presentation

Chapter 4 Phases and Solutions

An Image/Link below is provided (as is) to download presentation

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.


- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

Presentation Transcript


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 Raoults and Henrys Laws

  • 4.5 The chemical potentials of liquids

  • 4.6 The properties of solutions


New words and expressions

New Words and Expressions

  • Homogeneous

  • Heterogeneous

  • solution

  • solvent

  • solute

  • mole fraction

  • molality

  • molarity


New words and expressions1

New Words and Expressions

  • Mixture

  • Partial molar quantity

  • Chemical potential

  • Multi component

  • ideal solution

  • Ideal dilute solution

  • Activity and activity factor


New words and expressions2

New Words and Expressions

  • nonvolatile solute

  • positive (negative) deviation

  • dilute solution

  • colligative properties

  • freezing point

  • boiling point

  • osmotic pressure


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

  • 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,

    dGv=dGl


Chapter 4 phases and solutions

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/TVm (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

  • 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)


Chapter 4 phases and solutions

  • 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,


Chapter 4 phases and solutions

  • 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/T11/T2) (4.6)


Chapter 4 phases and solutions

  • 4.5 or 4.6

  • 1--

  • 2

  • (3) V(l) V(g)


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: Troutons 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 Troutons rule and was pointed out in 1884:

  • vapHm/Tb=vapSm88 JK-1mol-1 (4.7)


Chapter 4 phases and solutions

  • Example 4.2 Estimate the enthalpy of vaporization of CS2 if its boiling point is 319.40K.

  • Solution vapHm/Tb=vapSm88 JK-1mol-1

  • vapHm=88319.40=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)


Chapter 4 phases and solutions

,


Chapter 4 phases and solutions


Chapter 4 phases and solutions


Chapter 4 phases and solutions

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)


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

  • 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)


Chapter 4 phases and solutions

  • 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)


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 phases 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

  • 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)


Chapter 4 phases and solutions

  • 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.


Chapter 4 phases and solutions

  • 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 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.


Chapter 4 phases and solutions

  • mole fraction

molality


Chapter 4 phases and solutions

  • molarity

mass fraction


4 4 raoult s law and henry s law

4.4 Raoults law and Henrys law

  • Raoults 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 Raoults law .

  • PA=x1 P*A (4.22)


Raoult s and ideal solutions

Raoults and Ideal solutions

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

  • A number of pairs of liquids obey Raoults law over a wide range of compositions.


Raoult s and ideal solutions1

Raoults and Ideal solutions


Raoult s and ideal solutions2

Raoults 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.


Chapter 4 phases and solutions

  • Deviations from Raoults 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


Henry s law and ideal dilute solutions

Henrys 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

Henrys law and Ideal-dilute solutions

  • m2k2 =P2 or P2=kx2 =kc2 (4.27)

    Where k is the Henrys law constant, which is an empirical constant. Mixtures for which the solute obeys Henrys law and the solvent obeys Raoults law are called ideal-dilute solutions.


Henry s law and ideal dilute solutions2

Henrys law and Ideal-dilute solutions


Henry s law and ideal dilute solutions3

Henrys law and Ideal-dilute solutions

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

  • Thus Henrys 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 Raoults law but that under the same conditions the solute obeys Henrys 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


Chapter 4 phases and solutions

  • 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.


Chapter 4 phases and solutions

  • (b) Ideal-dilute solutions


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.


Chapter 4 phases and solutions

  • Vapor pressure lowering

    p=p*A pA

    =p*A xB (4.31)


Chapter 4 phases and solutions

  • 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)


Chapter 4 phases and solutions

  • 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


Chapter 4 phases and solutions

  • =0.14701*8.3145*326.282/17880

  • =7.28Kkgmol-1


Chapter 4 phases and solutions

  • 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.


Chapter 4 phases and solutions

  • 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.


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.

  • Vant Hoffs Equation

  • = n2RT/V or =cRT (4.37)


Chapter 4 phases and solutions

  • 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.


Chapter 4 phases and solutions

  • Answer: (1) mB = 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) and chloroform (trichloromethane) 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 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' CC14 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 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.


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 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. 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. 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.


  • Login