LECTURE 6 NON ELECTROLYTE SOLUTION

Size: px

Start display at page:

Download "LECTURE 6 NON ELECTROLYTE SOLUTION"

Bertram Watkins
6 years ago
Views:

1 LECTURE 6 NON ELECTROLYTE SOLUTION Ch 45.5 pplied Phy Chem First Sem Ch 45.5 Exam II September 1/3 (Multiple Choice/Problem Solving) Coverage: Second/Third Laws of Thermodynamics Nonelectrolyte Solution Electrolyte Solutions Phase Equilibria EXM III October 6 and 7 Chemical Equilibria, Electrochemistry 1

2 NONELECTROLYTE SOLUTIONS SOLUTIONS single phase homogeneous mixture of two or more components NONELECTROLYTES do not contain ionic species. CONCENTRTION UNITS percent by weight = wt.of wt.of solute solvent + wt. of solute X100% wt of solute = wt. of solution X100% mole fraction (xi ) = no. of moles of component i no.of moles of all components 2

3 CONCENTRTION UNITS Molarity (M) = no.of moles of solute L of solution Molality (m) = no.of moles of solute wt. of solvent in kg PRTIL MOLR VOLUME Imagine a huge volume of pure water at 25 C. If we add 1 mol H 2 O, the volume increases 18 cm 3 (or 18 ml). So, 18 cm 3 mol -1 is the molar volume of pure water. 3

4 PRTIL MOLR VOLUME Now imagine a huge volume of pure ethanol and add 1 mol of pure H 2 O it. How much does the total volume increase by? PRTIL MOLR VOLUME When 1 mol H 2 O is added to a large volume of pure ethanol, the total volume only increases by ~ 14 ml. The packing of water in pure water ethanol (i.e. the result of H-bonding interactions), results in only an increase of 14 ml. 4

5 PRTIL MOLR VOLUME The quantity 14 cm 3 mol -1 is the partial molar volume of water in pure ethanol. The partial molar volumes of the components of a mixture varies with composition as the molecular interactions varies as the composition changes from pure to pure B. 5

6 V = ΔV Δ n J p,t,n' PRTIL MOLR VOLUME n When a mixture is changed by dn of and dn B of B, then the total volume changes by: dv = V n p,t,n B dn + V n B p,t,n dn B 6

7 PRTIL MOLR VOLUME for a two component system (,B), the volume of thesystem at a particular composition is given by : V = V n + V B n B PRTIL MOLR VOLUME How to measure partial molar volumes? Measure dependence of the volume on composition. Fit a function to data and determine the slope by differentiation. 7

8 PRTIL MOLR VOLUME Molar volumes are always positive, but partial molar quantities need not be. The limiting partial molar volume of MgSO 4 in water is -1.4 cm 3 mol -1, which means that the addition of 1 mol of MgSO 4 to a large volume of water results in a decrease in volume of 1.4 cm 3. 8

9 SMPLE PROBLEM 5.01 What is the total volume of a mixture of 50.0 g ethanol and 50.0 g of water at 25 o C? The partial molar volumes of the 2 substances in mixtures of this compositions are 56.0 ml/mole and 18.0 ml per mole respectively. (nswer: 110 ml) (0.84) nswer: 110 ml) SMPLE PROBLEM 5.02 commercial rubbing alcohol is 82.5% in ethyl alcohol and 17.5% water. Partial molar volumes of water at this composition is 16.8 ml/mole and 57.4 ml/mole for ethanol. What is the density of the rubbing alcohol? (What is the density0.84) 9

10 PRTIL MOLR GIBBS ENERGIES The concept of partial molar quantities can be extended to any extensive state function. For a substance in a mixture, the chemical potential, µ is defined as the partial molar Gibbs energy. G = µ i i ΔG = Δn T,P,n 10

11 PRTIL MOLR GIBBS ENERGIES Using the same arguments for the derivation of partial molar volumes, G = n + µ nbµ B ssumption: Constant pressure and temperature THERMODYNMICS OF MIXING So we ve seen how Gibbs energy of a mixture depends on composition. We know at constant temperature and pressure systems tend towards lower Gibbs energy. When we combine two ideal gases they mix spontaneously, so it must correspond to a decrease in G. 11

12 THERMODYNMICS OF MIXING DLTON S LW The total pressure is the sum of all the partial pressure. p j = x j p p + p B +L = (x + x B +L )p = p 12

13 THERMODYNMICS OF MIXING G m = G m θ + RT ln p p θ µ = µ θ + RT ln p p θ µ = µ θ + RT ln p Δ mix G = nrt ( x ln x + x B ln x B ) x, x B n = total number are mole fractions of of moles components, B, respectively THERMODYNMICS OF MIXING µ = µ θ + RT ln p G i = n ( µ θ + RT ln p)+ n B ( µ θ B + RT ln p) G f = n ( µ θ + RT ln p )+ n B µ θ B + RT ln p B Δ mix G = n RT ln p p + n RT ln p B B p ( ) 13

14 THERMODYNMICS OF MIXING Δ mix G = n RT ln p p + n BRT ln p B p p p = x p p = x B Δ mix G = n RT lnx + n B RT lnx B x n = n x B n = n B Δ mix G = nrt(x lnx + x B lnx B ) Δ mix G < 0 THERMODYNMICS OF MIXING 14

15 SMPLE PROBLEM 5.03 container is divided into two equal compartments. One contains 3.0 mol H 2 (g) at 25 C; the other contains 1.0 mol N 2 (g) at 25 C. Calculate the Gibbs energy of mixing when the partition is removed. GIBBS ENERGY OF MIXING Δ mix G = nrt(x lnx + x B lnx B ) Δ mix G = 3.0(RT ln 3 4 ) +1.0(RT ln 1 4 ) p Δ mix G = 2.14 kj 3.43 kj Δ mix G = 5.6 kj p 15

16 ENTHLPY, ENTROPY OF MIXING Δ mix G = ΔH mix TΔ mix S for ideal solutions, ΔH = 0 Δ mixg T = Δ mix S Δ mix S = nr( x ln x B + x B ln x B ) 16

17 OTHER MIXING FUNCTIONS Δ Δ mix mix G S = = nrt ( x nr( x ln x ln x + + x x B B ln x ln x B B ) ) ΔG = ΔH TΔS ΔH = 0 (constant p and T) SMPLE PROBLEM 5.03 Calculate the Gibbs Energy and entropy of mixing 1.6 moles of rgon at 1.0 atm with 2.6 moles of Nitrogen at 1.0 atm and 25 o C. ssume ideal behavior. 17

IDEL SOLUTIONS To discuss the equilibrium properties of liquid mixtures we need to know how the Gibbs energy of a liquid varies with composition.

18 IDEL SOLUTIONS To discuss the equilibrium properties of liquid mixtures we need to know how the Gibbs energy of a liquid varies with composition. We use the fact that, at equilibrium, the chemical potential of a substance present as a vapor must be equal to its chemical potential in the liquid. IDEL SOLUTIONS Chemical potential of vapor equals the chemical potential of the liquid at equilibrium. µ * = µ θ + RT ln p * * denotes pure substance n If another substance is added to the pure liquid, the chemical potential of will change. µ µ θ = µ θ + RT ln p is the standard chemical potential at p = 1 bar 18

19 IDEL SOLUTIONS µ * = µ θ + RT ln p * µ θ = µ * RT ln p * µ = µ θ + RT ln p µ = µ * RT ln p * + RT ln p µ = µ * + RT ln p p * ROULT S LW p = x p * The vapor pressure of a component of a solution is equal to the product of its mole fraction and the vapor pressure of the pure liquid. 19

20 SMPLE PROBLEM 5.04 Liquids and B form an ideal solution. t 45 o C, the vapor pressure of pure and pure B are 66 torr and 88 torr, respectively. Calculate the composition of the vapor in equilibrium with a solution containing 36 mole percent at this temperature. IDEL SOLUTIONS: µ = µ * + RT ln p p * p = x p * µ = µ * + RT lnx 20

21 NON-IDEL SOLUTIONS IDEL-DILUTE SOLUTIONS Even if there are strong deviations from ideal behaviour, Raoult s law is obeyed increasingly closely for the component in excess as it approaches purity. 21

22 HENRY S LW For real solutions at low concentrations, although the vapor pressure of the solute is proportional to its mole fraction, the constant of proportionality is not the vapor pressure of the pure substance. HENRY S LW p B = x B K B Even if there are strong deviations from ideal behaviour, Raoult s law is obeyed increasingly closely for the component in excess as it approaches purity. 22

23 IDEL-DILUTE SOLUTIONS Mixtures for which the solute obeys Henry s Law and the solvent obeys Raoult s Law are called ideal-dilute solutions. PROPERTIES OF SOLUTIONS We ve looked at the thermodynamics of mixing ideal gases, and properties of ideal and idealdilute solutions, now we shall consider mixing ideal solutions, and more importantly the deviations from ideal behavior. 23

24 IDEL SOLUTIONS G i = n µ * + n B µ B * µ = µ * + RT lnx G f = n {µ * + RT lnx } + n B {µ B * + RT lnx B } Δ mix G = n RT lnx + n B RT lnx B Δ mix G = nrt{x lnx + x B lnx B } IDEL SOLUTIONS Δ mix G = nrt{x lnx + x B lnx B } Δ mix S = nr{x lnx + x B lnx B } Δ mix H = 0 24

25 IDEL SOLUTIONS Δ mix G = nrt{x RT lnx + x B RT lnx B } G p T = V Δ mix G p T = Δ mix V = 0 REL SOLUTIONS Real solutions are composed of particles for which -, -B and B-B interactions are all different. There may be enthalpy and volume changes when liquids mix. ΔG=ΔH-TΔS So if ΔH is large and positive or ΔS is negative, then ΔG may be positive and the liquids may be immiscible. 25

26 EXCESS FUNCTIONS Thermodynamic properties of real solutions are expressed in terms of excess functions, X E. n excess function is the difference between the observed thermodynamic function of mixing and the function for an ideal solution. REL SOLUTIONS S E = Δ mix S Δ mix S ideal G E = Δ mix G Δ mix G ideal H E = Δ mix H Δ mix H ideal = Δ mix H V E = Δ mix V Δ mix V ideal = Δ mix V 26

27 Real Solutions Benzene/cyclohexane Tetrachloroethene/cyclopentane COLLIGTIVE PROPERTIES: colligative property is a property that depends only on the number of solute particles present, not their identity. The properties we will look at are: lowering of vapor pressure; the elevation of boiling point, the depression of freezing point, and the osmotic pressure arising from the presence of a solute. Only applicable to dilute solutions. 27

28 COLLIGTIVE PROPERTIES: ll the colligative properties stem from the reduction of the chemical potential of the liquid solvent as a result of the presence of solute. µ = µ * + RT ln x VPOR PRESSURE LOWERING: s solute molecules are added to a solution, the solvent become less volatile (=decreased vapor pressure). Solute-solvent interactions contribute to this effect. 28

29 VPOR PRESSURE LOWERING: Therefore, the vapor pressure of a solution is lower than that of the pure solvent. VPOR PRESSURE LOWERING: ROULT S LW where P = x P X is the mole fraction of compound P is the normal vapor pressure of at that temperature NOTE: This is one of those times when you want to make sure you have the vapor pressure of the solvent. 29

30 SMPLE PROBLEM: Glycerin (C 3 H 8 O 3 ) is a nonvolatile nonelectrolyte with a density of 1.26 g/ml at 25 o C. Calculate the vapor pressure at 25 o C of a solution made by adding 50.0 ml of glycerin to ml of water. The vapor pressure of pure water at 25 C is 23.8 torr. PRCTICE EXERCISE The vapor pressure of pure water at 110 o C is 1070 torr. solution of ethylene glycol and water has a vapor pressure of 1.00 atm at 110 C. ssuming that Raoult s law is obeyed, what is the mole fraction of ethylene glycol in the solution? 30

31 BOILING POINT ELEVTION: How do we figure out where the new boiling point is when a solute is present? Look for the temperature at which at 1 atm, the vapor of pure solvent vapor has the same chemical potential as the solvent in the solution. BOILING POINT ELEVTION: Let s denote solvent and solute B. Equilibrium is established when: µ (g) = µ * (g) + RT lnx = µ * (l) + RT lnx ΔT = Kx B K = RT *2 Δ vap H 31

$solute, only to its mole fraction. n So the elevation of boiling point is a colligative property.$

32 BOILING POINT ELEVTION: ΔT = Kx B K = RT *2 Δ vap H n ΔT makes no reference to the identity of the solute, only to its mole fraction. n So the elevation of boiling point is a colligative property. 32

33 BOILING POINT ELEVTION: ΔT = Kx B For practical purposes : ΔT = K b m K b = boiling point constant; m = molality 33

34 FREEZING POINT DEPRESSION: Let s denote solvent and solute B. Equilibrium is established when: µ (s) = µ * (l) + RT lnx n Same calculation as before (Justification 5.1) ΔT = ʹ K x B K ʹ = RT*2 Δ fus H FREEZING POINT DEPRESSION: ΔT = Kʹ x B For practical purposes : ΔT = K f m K f = freezing point constant; m = molality 34

35 SMPLE PROBLEM: What will be the freezing point and boiling point of an aqueous solution containing 55.0 g of glycerol, C 3 H 5 (OH) 3, and 250 g of water? K b (H 2 O) = 0.51 o C/m and K f = 1.86 o C/m. 35

SMPLE PROBLEM: solution is prepared from 25.0 g of benzene, C 6 H 6, and 2.50 g of an unknown compound. The freezing point of this solution is 4.3 o C. The normal freezing point of benzene is 5.

36 SMPLE PROBLEM: solution is prepared from 25.0 g of benzene, C 6 H 6, and 2.50 g of an unknown compound. The freezing point of this solution is 4.3 o C. The normal freezing point of benzene is 5.5 o C and the freezing point depression constant for benzene is 5.12 o C/m. Determine the molar mass of the compound. Cryoscopy ΔT = K f b b = K f = freezing point constant; b = molality n 1 kg solvent = m M 1 kg solvent 36

Solubility lthough solubility is not strictly a colligative property (because solubility varies with the identity of the solute), it may be estimated

37 Solubility lthough solubility is not strictly a colligative property (because solubility varies with the identity of the solute), it may be estimated using the same techniques. When a solid solute is left in contact with a solvent, it dissolves until the solution is saturated with the dissolved solute. 37

38 See Justification 5.2 Solubility µ B (s) = µ B * (l) + RT lnx B lnx B = Δ fush R 1 T f 1 T OSMOSIS: Osmosis refers to the spontaneous passage of a pure solvent into a solution separated from it by a semi-permeable membrane. In this case, the membrane is permeable to the solvent but not to the solute. 38

39 OSMOSIS: The osmotic pressure, π, is the pressure that must be applied to the solution to stop the influx of solvent. Examples of osmosis includes the transport of fluids across cell membranes and dialysis. 39

40 OSMOSIS: π is the osmotic c 2 is the concentration of M 2 π M is molarity of π c is the molar pressure of 2 = MRT or = the solution RT M 2 mass of the solution the solute in g/l the solute SMPLE PROBLEM: n aspartic acid solution will produce an osmotic pressure of 1.80 atm measured against pure water at 18.5 C. How many g of spartic cid(c 4 H 7 NO 4, M=133.1 g/mol) must be dissolved in a liter of water to produce an isotonic solution? 40

Lecture 6. NONELECTROLYTE SOLUTONS

Lecture 6. NONELECTROLYTE SOLUTONS NONELECTROLYTE SOLUTIONS SOLUTIONS single phase homogeneous mixture of two or more components NONELECTROLYTES do not contain ionic species. CONCENTRATION UNITS percent