Thermodynamics and Kinetics. Lecture 14 Properties of Mixtures Raoult s Law Henry s Law Activity NC State University

Transcription

1 Thermodynamics and Kinetics Lecture 14 Properties of Mixtures Raoult s Law Henry s Law Activity NC State University

2 Measures of concentration There are three measures of concentration: molar concentration per unit volume (molarity) c = n/(liter of solution) molar concentration per unit mass (molality) m = n/(kg of solution) mole fraction x j = n j n i i

3 Raoult's law states P j = x j P j Ideal solutions where P j is the vapor pressure of pure component j. The vapor pressure of component j in an ideal solution is given by the product of its mole fraction and P j. The chemical potential can be expressed as: m j = m j + RTln(P j /P j ) where P j is vapor the pressure of the pure component j in the standard state.

4 Ideal solutions The significance of this expression is that we can now consider the equilibrium between vapor and solution to write: m j soln = m j vap = m j 0 + RTln(P j /P j0 ) but for the vapor P j 0 = 1 bar. In the limit that the vapor becomes the pure vapor we have: m j = m j vap = m j 0 + RTln(P j /P j0 ) Thus m j soln = m j + RTln(P j /P j ) keeping in mind the notation means the pure component.

5 Ideal solutions The equation below is central equation of binary solution mixtures. m j soln = m j + RTln(P j /P j ) Using Raoult's law x j = P j /P j we see that the chemical potential can be expressed as: m j soln = m j + RTln(x j ) This equation defines an ideal solution.

6 The free energy of mixing The free energy for formation of a solution from individual components is given by G mix = G soln G 1 G 2 Since G soln = n 1 m 1 + n 2 m 2, G 1 = n 1 m 1 and G 2 = n 2 m 2. we have that G mix = n 1 m 1 + n 2 m 2 n 1 m 1 n 2 m 2 = nrt x 1 ln x 1 + x 2 ln x 2 for an ideal solution there is no enthalpy of mixing. The volume of the mixture also does not change for an ideal solution. The entropy change can be obtained from in agreement with a previous derivation.

7 Two component phase diagrams The total vapor pressure over an ideal solution is given by P total = P 1 + P 2 = x 1 P 1 + x 2 P 2 = (1 - x 2 )P 1 + x 2 P 2 = P 1 + x 2 (P 2 - P 1 ) A plot of the total pressure has the form of a straight line.

8 Liquid composition: specific example Consider the example in the book of 1-propanol and 2-propanol, which have P 1 = 20.9 torr and P 2 = 45.2 torr, respectively. So in this example, the phase diagram has the appearance:

9 The vapor mole fraction The value of the mole fraction in the vapor is not necessarily equal to that of the liquid. In the vapor phase the relative numbers of moles is given by Dalton's law. Applying Dalton's law we find: y 1 = P 1 /P total = x 1 P 1 /P total or y 2 = P 2 /P total = x 2 P 2 /P total The vapor curve is not the same as the liquid line.

10 Deriving the vapor curve Using the equation of the liquid line we find: When substituted into y 2 = x 2 P 2 /P total we have Solving for P total we find: x 2 = P total P 1 P 2 P 1 y 2 = P 2 P total P 1 P total P 2 P 1 y 2 P total = P 2 P total P 1 P total P 1 = P 2 P 2 P 1 P 2 P P 2 1 P 2 P 1 P 2 y 2 P 2 P 1 P total = P 2 P 1 P total = P 2 P 1 P 2 y 2 P 2 P 1

11 The vapor curve The is shown in the Figure below. The purple line was calculated using the Dalton's law expression. What lies between the blue and purple lines? This is the two phase region. Two Phase Region

12 The two phase region If we pick a composition and pressure that is inside this region then we can use a tie line to indicate the composition of each phase.

13 Boiling in a two component system If we reduce the pressure above a two component mixture of 1-propanol and 2-propanol with a mole fraction of propanol what is the composition of the vapor?

14 What is the composition of the vapor? a. y 2 = 0.60 b. y 2 = 0.48 c. y 2 = 0.78 d. y 2 = 0.98

15 What is the composition of the vapor? a. y 2 = 0.60 b. y 2 = 0.48 c. y 2 = 0.78 d. y 2 = 0.98

16 Boiling in a two component system If we continue to reduce the pressure the system moves into the two-phase region. In the two-phase region the composition of the liquid and the vapor is not the same.

17 What is the composition? a. x 2 = 0.40, y 2 = 0.60 b. x 2 = 0.50, y 2 = 0.70 c. x 2 = 0.50, y 2 = 0.50 d. x 2 = 0.40, y 2 = 0.70

18 What is the composition? a. x 2 = 0.40, y 2 = 0.60 b. x 2 = 0.50, y 2 = 0.70 c. x 2 = 0.50, y 2 = 0.50 d. x 2 = 0.40, y 2 = 0.70

19 What is the composition? a. x 2 = 0.40, y 2 = 0.60 b. x 2 = 0.50, y 2 = 0.70 c. x 2 = 0.50, y 2 = 0.50 d. x 2 = 0.40, y 2 = 0.70

20 Boiling in a two component system If we continue to reduce the pressure the system reaches the boundary with pure vapor.

21 What is the composition? a. x 2 = 0.40, y 2 = 0.60 b. x 2 = 0.50, y 2 = 0.70 c. x 2 = 0.50, y 2 = 0.50 d. x 2 = 0.40, y 2 = 0.70

22 What is the composition? a. x 2 = 0.40, y 2 = 0.60 b. x 2 = 0.50, y 2 = 0.70 c. x 2 = 0.50, y 2 = 0.50 d. x 2 = 0.40, y 2 = 0.70

23 The tie line The tie line shown in Figure (red line) is at a total pressure of 30 torr. You can read the x 2 and y 2 values from the plot (or calculate them using the equations above used to generate the blue and purple curves in the composition-pressure plot.

24 The lever rule The tie line can be used to define the quantity of each phase present in the two phase region. The total composition x a can be used together with x 2 and y 2. z a y 2 - z a z a

25 The lever rule The lever rule states that the amount of each phase present is inversely proportional to the length of distance along the tie line from the phase boundary to the total composition, z a. z a y 2 - z a z a

26 The lever rule states: The lever rule n liquid n vapor = y 2 z a z a x 2 The total composition is z a. z a y 2 - z a z a

27 Non-ideal solutions Many solutions are not ideal. For ideal solutions the role of intermolecular interactions can be ignored. This may be because they are small or because two components have the same interaction with each other that they have with themselves. In other words similar solvents will form ideal solutions. However, in many cases, intermolecular interactions cause deviations from Raoult's law. We can consider the "like" interactions between molecules of same species and "unlike" interactions between molecules of different species. If the unlike-molecule interactions are more attractive than the like molecule interactions, the vapor pressure above a solution will be smaller than we would calculate using Raoult's law. If the unlike-molecule interactions are more repulsive, then the vapor pressure is greater than for the ideal solution.

28 Henry s law The statement of Henry s law is: P 1 = x 1 k H,1 It looks like Raoult s law except that you have this funny constant k H,1 instead of P 1 (the vapor pressure of component 1). This law is only valid for dilute solutions, i.e. when component 1 is the solute. Under these conditions the vapor pressure of component 1 really does not matter, because component 1 is mostly surrounded by component 2 and so its properties really quite different from the properties of pure 1. Henry s law can be applied to mixtures of solvents And also to solutions of gases in liquids. For example, see the problems on the concentration of O2 and N2 in water at the end of the lecture.

29 Henry s law Attractive interactions between unlike molecules leads to negative deviations from Raoult's law (lower vapor pressure than ideal) and repulsive interactions lead to positive deviations (higher vapor pressure than ideal). As any solution approaches a mole fraction of one (i.e. approaches a pure solution of one component) it becomes an ideal solution. In other words, P 1 x 1 P 1 as x 1 1. However, as x 1 0 the component is surrounded by unlike molecules and the solution has the maximum deviation from ideal behavior. For this case we define Henry's law, P 1 x 1 k H,1 as x 1 0. In this expression k H,1 is the Henry's law constant. Although we have focused on component 1 the same holds true for component 2.

30 Non-ideal solutions For the example shown in the plot above we have assumed that P 1 = 100 torr. Note that as x 1 1 the slope approaches the ideal slope obtained from Raoult's law. However, as x 1 0 (and therefore x 2 1) the slope is quite different from ideal behavior. Note that the slope has the value of the Henry's law constant k H,1. This is depicted in the Figure below (purple line). The Henry's law value can be quite different from the ideal value.

31 Activity The activity in non-ideal solutions corresponds to mole fraction in ideal solutions. a j = P j P j The activity replaces mole fraction in the expression for the chemical potential. When considering a non-ideal solution the above expressions hold and thus the mole fraction x j is no longer equal to P j /P j. However, as the mole fraction approaches unity (a pure substance) the solution becomes ideal. Thus, as x j 1, a j x j. m j soln = m j + RT ln a j

32 The activity coefficient We can define an activity coefficient g 1 such that g 1 = a 1 /x 1. The property of the activity coefficient is that it approaches a value of 1 (ideal behavior) as the composition approaches the pure solvent: a 1 x 1 and g 1 1 as x 1 1. Thus, the solution This definition is based on a Raoult's law standard state, which is also known as a solvent standard state. The activities or chemical potentials are meaningless unless we know the standard state

33 Henry s law constant and the solubility of gases Henry s law constants in H 2 O (atm x 10 3 ) He 131 N 2 86 CO 57 O 2 43 Ar 40 CO Problem: Divers get the bends if bubbles of N 2 form in their blood because they rise too rapidly. Calculate the molarity of N 2 in water (i.e. blood) at sea level and 100 m below sea level. At sea level: a N2 = P N2 /k H,N2 = 0.8 atm/86 x 10 3 atm = 9.3 x 10-6 c N2 = 55.6 a N2 = 5 x 10-4 mol/l At 100 m: a N2 = P N2 /k H,N2 = 9.8 atm/86 x 10 3 atm = 1.1 x 10-4 c N2 = 55.6 a N2 = 6 x 10-3 mol/l

34 A note on conversion from mole fraction to molarity The conversion from mole fraction to molarity can be solved analytically. n x 1 = 1 c = 1 n 1 + n 2 c 1 + c 2 since n 1 = c 1 V and V cancels x 1 (c 1 + c 2 ) = c 1 x 1 c 2 = c 1 (1 x 1 ) x 1 c 2 = c 1 x 2 c 2 = c 1 x 2 x 1 Note that for water as solvent (component 1) x 1 ~ 1 and the Concentration of water is c 1 ~ 55.6 so that the conversion For a dilute solute such as a gas is c 2 ~ 55.6 x 2

35 Question Henry s law constants in H 2 O (atm x 10 3 ) He 131 N 2 86 CO 57 O 2 43 Ar 40 CO A species of fish requires a concentration of O 2 > 100 mm. A marine biologist is trying to determine the depth profile for O 2 in sea water. The first step is to calculate the concentration of O 2 in sea water at sea level. As an assistant you perform the calculation and find that the O 2 concentration is: A. 250 mm B. 430 mm C. 760 mm D mm

36 Question Henry s law constants in H 2 O (atm x 10 3 ) He 131 N 2 86 CO 57 O 2 43 Ar 40 CO Most fish require a concentration of O 2 that is greater than 100 mm. A marine biologist is trying to determine the depth profile for O 2 in sea water. The first step is to calculate the concentration of O 2 in sea water at sea level. As an assistant you perform the calculation and find that the O 2 concentration is: A. 250 mm B. 430 mm C. 760 mm D mm a O2 = P O2 /k H,O2 = 0.2 atm/43 x 10 3 atm = 4.7 x 10-6 c O2 = 55.6 a O2 = 2.5 x 10-4 mol/l

37 The Henry s law constant is an equilibrium constant Gases dissolve in liquids to form solutions. This dissolution is an equilibrium process for which an equilibrium constant can be written. For example, the equilibrium between oxygen gas and dissolved oxygen in water is O 2 (aq) <--> O 2 (g). The equilibrium constant for this equilibrium is: K H = p(o 2 )/c(o 2 ). The form of the equilibrium constant shows that the concentration of a solute gas in a solution is directly proportional to the partial pressure of that gas above the solution. The form of the equation can be Rearranged to give: p(o 2 ) = c(o 2 )K H = x(o 2 ) c(h 2 O)K H

38 The Henry s law constant is tabulated for c and x The Henry s law constant can be tabulated for mole fraction (as seen in the previous problem) or for molarity. If the units of the Henry s law constant are atm then it is valid for mole fraction. If the units are atm/(mol/l) then K H is tabulated for molarity. The relationship between the two is: p(o 2 ) = c(o 2 )K H = x(o 2 ) c(h 2 O)K H So that K H = 55.6 (mol/l)k H since the concentration of water is c(h 2 O) = molar. For example, K H is 757 atm/(mol/l). K H = K H 55.6 = = 13.6