Outline of the Course

Size: px

Start display at page:

Download "Outline of the Course"

Barbara Leonard
5 years ago
Views:

1 Outline of the Course 1) Review and Definitions 2) Molecules and their Energies 3) 1 st Law of Thermodynamics Conservation of Energy. 4) 2 nd Law of Thermodynamics Ever-Increasing Entropy. 5) Gibbs Free Energy 6) Phase Diagrams and REAL Phenomena 7) Non-Electrolyte Solutions 8) Chemical Equilibrium 9) Chemical Kinetics and Rates of Processes

2 Section 7.0. Non-Electrolyte Solutions & Simple Mixtures PART II (Chapter 7 in Chang Text Most of It!)

3 7.8. Real Solutions..most solutions do not behave ideally Remember.in a solution we have two components solvent and solute In a real (non-ideal) solution how do we write the chemical potentials of the solute and the solvent??

4 The Solvent Component the chemical potential of the solvent in an ideal solution is given by the following equation.. 1 (l) = 1 * (l) + RT ln x 1 where x 1 = P 1 /P 1 * and P 1 is the vapour pressure of 1 when it is a component of a solution while P 1 * is the equilibrium vapor pressure of pure component 1 at T. for a non-ideal solution we write.. 1 (l) = 1 * (l) + RT ln a 1 where a 1 is the activity of the solvent.

5 non-idealityis the consequence of unequal intermolecular forces between solvent solvent and solvent solute molecules the solvent s activity can be expressed in terms of vapour pressure as follows: a 1 = P 1 /P 1 * the solvent s activity is related to mole fraction (concentration) as follows: a 1 = 1 x 1 where 1 is the activity coefficient. 1 (l) = 1 * (l) + RT ln a 1 1 (l) = 1 * (l) + RT ln 1 + RT ln x 1

6 The Activity Coefficient ( 1 ) is a measure of the degree of deviation from ideality: a 1 = 1 x 1.solvents obey Raoult s Law as concentration of solute approaches zero (i.e. x 2 0 ). In this case (i.e. x 1 1 ) the activity of the solvent approaches the mole fraction.. so a 1 x 1 as x 1 1..in fact as x

8 Solute Component for ideal solutions the solute and the solvent obey Raoult s Law for ideal-dilute solutions (also referred to as non-ideal dilute solutions) the solute obeys Henry s Law and the solvent obeys Raoult s Law The chemical potential for the solute according to Raoult s Law (i.e. for solute in an ideal solution) is given by: 2 (l) = 2 * (l) + RT ln x 2 = 2 * (l) + RT ln (P 2 /P 2 *)

9 The chemical potential for the solute according to Henry s Law (i.e. for solute in an ideal-dilute solution) is given by: [ recall: P 2 = K x 2 ] 2 (l) = 2 * (l) + RT ln (P 2 /P 2 *) = 2 * (l) + RT ln [(K x 2 )/P 2 *] = 2 * (l) + RT ln [K/P 2 *] + RT ln x 2 = 2o (l) + RT ln x 2 where: = 2o (l) + RT ln (P 2 /K) 2o (l) = 2 * (l) + RT ln [K/P 2 *]

10 For non-ideal solutions (beyond the dilute solution limit) the chemical potential of the solute is commonly given as follows: 2 (l) = 2 (l) + RT ln a 2 where a 2 is the activity of the solute. As with the solvent for the solute: a 2 = 2 x 2 where 2 is the activity coefficient of the solute..solute obey Henry s Law as its concentration approaches zero (i.e. x 2 0 ). In this case (i.e. x 1 1 ) the activity of the solute approaches the mole fraction.. so a 2 x 2 as x 2 0..in fact as x

11 Example: Acetone (A) Chloroform (C) Mixture (propanone (H 3 C) 2 C=O ) (trichloromethane CHCl 3 ) X C P C /torr P A /torr Henry s Law Constants: K A K C = 175 torr = 165 torr

12 Solvent Convention (Raoult s Law): a C = P C / P C * γ C = a C / x C P C * = 273 torr Solute Convention (Henry s Law): a C = P C / K C γ C = a C / x C K C = 165 torr

13 Chloroform X C P C /torr Chloroform As Solvent (Raoult s Law): a C γ C Chloroform As Solute (Henry s Law): a C γ C γ 1 as x 1 (Raoult s Law) γ 1 as x 0 (Henry s Law)

14 Variation of Activity and Activity Coefficient with Mole Fraction of Chloroform according to.. Raoult s Law (chloroform considered solvent) Henry s Law (chloroform considered solute) As x As x Atkins 7 th Edn Phys. Chem

15 Activities In Terms Of Molalities (alternate form of Ideal-Dilute solution) Selection of a standard state is entirely arbitrary so choose one that best suits our purpose. In chemistry, compositions are usually expressed as molalities (m 2 ) rather than mole fractions (x 2 ). For 2 component system: 1 = solvent, 2 = solute x 2 = m 2 / [ m 2 + (1000/M 1 ) ] = m 2 M 1 / [ m 2 M ] If: m 2 M 1 << 1000 then: x 2 m 2 M 1 / 1000 = m 2 (M 1 /1000) Therefore: x 2 proportional to m 2 as: m 2 0

16 The alternate form of the chemical potential for the solute according to Henry s Law (for solute in an ideal-dilute solution) is given by: [ recall: P 2 = K m ] [choose: m o = 1 mol solute / kg solvent] 2 (l) = 2 * (l) + RT ln (P 2 /P 2 *) = 2 * (l) + RT ln [(K m )/P 2 *] = 2 * (l) + RT ln [(K m m o )/(P 2 *m o )] = 2 * (l) + RT ln [(K m o )/P 2 *] + RT ln [m/m o ] where: = 2o (l) + RT ln (m/m o ) 2o (l) = 2 * (l) + RT ln [(K m o )/P 2 *]

17 Check out dimensions of the two ln ratios: (K m o )/P 2 * = (atm mol -1 kg solvent)(mol kg -1 solvent)/(atm) = dimensionless!!! (m/m o ) = (mol solute kg -1 solvent)/(1 mol solute kg -1 solvent) = dimensionless!!! Correct!!! Can t take a logarithm of a dimensioned quantity!!!

18 For non-ideal solutions (beyond the dilute solution limit) the chemical potential of the solute is commonly given as follows: 2 (l) = 2 (l) + RT ln a 2 where a 2 is the activity of the solute. As with the solvent for the solute: a 2 = γ 2 (m 2 /m o ) where γ 2 is the activity coefficient of the solute solute obeys Henry s Law as its concentration approaches zero (i.e. m 2 0). In this case (i.e. x 1 1) the activity of the solute approaches the ratio (m 2 /m o )... so: a 2 (m 2 /m o ) and γ 2 1 as: m 2 0

19 Chang Figure 7.8 (a) Chemical potential of a solute plotted against the logarithm of molality for a non-ideal solution. (b) Activity of a solute as a function of molality for a nonideal solution. The standard state is at m 2 /m o =1

20 Phase Equilibria of Two Component Systems DISTILLATION: separation of two volatile liquid components may be achieved by fractional distillation.. to use distillation to separate two components we have to understand how pressure and temperature affect the vapor liquid equilibrium of binary liquid mixtures.need to be familiar with phase diagram for this two component system.

21 Phase Diagram for Mixture of Benzene and Toluene consists of A) Plot of vapor pressure vs. mole fraction of benzene in solution B) Plot of vapor pressure vs. mole fraction of benzene in vapor phase A) B)

22 Therefore, constructing the phase diagram involves two steps 1- construct plot of P versus mole fraction of benzene in solution (P vs x b ) 2- construct plot of P versus mole fraction of benzene in the vapor phase (P vs x b v ) Let s start with Step 1:..What is the relationship between P and x b? According to Raoult s Law the vapor pressure of the individual components may be described as follows: P b = x b P b * and P t = x t P t * where x b and x t are the mole fractions of benzene and toluene in solution. (* denotes pure component)

23 The total pressure may be given by: P = P b + P t = x b P b * + x t P t * = x b P b * + (1 x b )P t * = P t * + (P b * - P t * ) x b P = P t * + (P b * - P t * ) x b

24 STEP 2. Construct plot of P versus mole fraction of benzene in the vapor phase (P vs x b v ) According to Dalton s Law the mole fraction of benzene in the vapor phase is given by: x b v = (P b / P) = [ (x b P b *) / (P t * + (P b *- P t *) x b )] Solve for x b x b = [ (x bv P t *) / (P b * - (P b *- P t *) x bv )] By Dalton s and Raoult s Laws: P b = x bv P =x b P b *

25 P = (x b P b *) / x b v = (P b * P t *) / (P b * - (P b *- P t *) x bv )..equation may be used to construct plot of P vs. x b v

26 above straight linethe system is in liquid state (..as expected at high pressures system is in liquid state) below the curve the system exists as vapor (i.e. at low pressures system is a vapor) within enclosed area the system exists as vapor and liquid

27 At point a: we are in the liquid phase remember the phase rule f = c p + 2 At point a.f = 3 since f = = 3 at constant mole fraction and temperature we can lower pressure and reach point b. At point b: the liquid can exist in equilibrium with its vapor (at this point liquid begins to vaporize.there is virtually no vapor present but what is present has composition according to point ))

28 At point c: the composition of liquid is described by f while composition of vapor is defined by g horizontal tie - line Relative amounts of liquid and vapor present in equilibrium are given by the LEVER RULE: n L l L = n V l V n L and n V are the number of moles of liquid and vapor. At point c, we have two phases and f = 2 f= *remember: f, gives no. of variables (e.g. pressure, temperature and composition) that can be changed independently of one another without changing no. of phases in equilibrium.

29 The LEVER Rule n L l L = n V l V where l L and l V.as shown in Fig (Chang text) Use the Lever Rule to find relative amounts of two phases that are present in a quantitative manner.

30 Now with phase diagram in hand we can figure out a way to separate benzene and toluene (at a fixed temperature) Start at point a.and lower pressure until system begins to vaporize at this point x b = 0.2 and x t = = 0.8 The composition of vapor in equilibrium with the solution is x bv = 0.5 and x tv = 0.5.at this point the vapor phase is richer in benzene than liquid phase is

31 now condense vapor from b to c.now mole fraction of benzene in vapor phase is even higher.. x bv = 0.8 and x tv = 0.2 repeat process and eventually achieve complete separation of benzene and toluene.

32 Temperature - Composition Diagram -in practice we carry distillation out at constant P not constant T b.pt. Benzene = 80.1 C b.pt. Toluene = C -we vary T in order to separate components of solution At constant P - the liquid phase is most stable at low T - vapor most stable at high T To separate solution at point a is evaporated (a to b)..the vapor (richer in benzene) is then condensed (b to c) and evaporated (c to d)

33 -each vaporization-condensation step is called a theoretical plate - the efficiency of a fractionating column is expressed in terms of the no. of theoretical plates Atkins 7 th Edn Phys. Chem

34 Setup for Fractional Distillation used to separate volatile liquids flask contains benzene and toluene solution boils, vapor condenses on beads in column and falls back into flask beads gradually heat up allowing vapor to move up slowly vapor that migrates towards top of fractionating column becomes more and more rich in most volatile component ** vapor that reaches top of column is pure benzene Chang Text page

35 Azeotropes many liquids have temperature composition phase diagrams that resemble previous example..yet in some cases there are deviations from this including either a maximum or minimum in the phase diagram. Maximum in phase diagram occurs when there are favorable interactions between unlike (A and B) molecules. A-B interactions stabilize liquid. Negative deviation from Raoult s Law. Excess G E is negative more favorable to mixing than in ideal system. Atkins 7 th Edn Phys. Chem

36 Minimum in phase diagram occurs when the mixture is destabilized relative to ideal solution. A-B interactions are unfavorable. Positive deviation from Raoult s Law G E is positive (i.e. less favorable to mixing than in ideal solution) Atkins 7 th Edn Phys. Chem

37 once azeotrope distillate has been produced further distillation does not result in further separation Azeotrope : distillate (i.e. vapor) that has same composition as solution in flask)

Boiling point elevation and freezing point depression

Class 4.2 More Colligative Properties, Solutions & Distillation CHEM 102H T. Hughbanks Boiling point elevation and freezing point depression When nonvolatile solutes are present in a solvent, the solution