Properties of Mixtures in Chemistry: Ideal and Non-Ideal Solutions - Prof. Stefan Franzen, Study notes of Physical Chemistry

Download Properties of Mixtures in Chemistry: Ideal and Non-Ideal Solutions - Prof. Stefan Franzen and more Study notes Physical Chemistry in PDF only on Docsity! 1 Chemistry 331 Lecture 21 Properties of Mixtures NC State University 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 niΣi Partial molar volume We have seen molar values. In a mixture or a solution individual components can have partial molar values. The easiest to see physically is the partial molar volume Vj,m = (∂V/∂n) j. The total volume is then V = V1.mn1 + V2,mn2 For example, when 1-propanol and water are mixed, the final volume, V of the solution is not equal to the volumes of pure 1-propanol and water. The mixture of two components that can interact in a non-ideal fashion leads to a solution volume that is greater or less than that of the pure components. The partial molar volumes allow this to be quantified. Ideal solutions Raoult's law states Pj = xjPj* where Pj* 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: µj = µj* + RTln(Pj/Pj*) where Pj* is vapor the pressure of the pure component j in the standard state. Ideal solutions The significance of this expression is that we can now consider the equilibrium between vapor and solution to write: µjsoln = µjvap = µj0 + RTln(Pj/Pj0) but for the vapor Pj0 = 1 bar and so µjsoln = µjvap = µj0 + RTln(Pj) In the limit that the vapor becomes the pure vapor we have: µjsoln = µj* + RTln(Pj/Pj*) keeping in mind the notation * means the pure component. Ideal solutions The equation below is central equation of binary solution mixtures. µjsoln = µj* + RTln(Pj/Pj*) Using Raoult's law xj = Pj/Pj* we see that the chemical potential can be expressed as: µjsoln = µj* + RTln(xj) This equation defines an ideal solution. 2 Question Which statement describes ideal solutions? A. The vapor pressure of component j in an ideal solution is given by the product of its mole fraction and P j*. B. P j* represents the vapor pressure of pure j. C. The chemical potential of pure j is µj*. D. All of the above. Question Which statement describes ideal solutions? A. The vapor pressure of component j in an ideal solution is given by the product of its mole fraction and P j*. B. P j* represents the vapor pressure of pure j. C. The chemical potential of pure j is µj*. D. All of the above. The free energy of mixing The free energy for formation of a solution from individual components is given by Since Gsoln = n1µ1 + n2µ2 , G1* = n1µ1* and G2* = n2µ2*. we have that 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. ∆Gmix = G soln – G1 * – G2 * ∆Gmix = n1µ1 + n2µ2 – n1µ1* – n2µ2* = nRT x1ln x1 + x2ln x2 Which is true? a. ∆Gmix = - T∆Smix b. ∆Gmix = T∆Smix c. ∆Gmix = - ∆Smix/T d. ∆Gmix = ∆Smix/T Which is true? a. ∆Gmix = - T∆Smix b. ∆Gmix = T∆Smix c. ∆Gmix = - ∆Smix/T d. ∆Gmix = ∆Smix/T Two component phase diagrams The total vapor pressure over an ideal solution is given by Ptotal = P1 + P2 = x1P1* + x2P2* = (1 - x2)P1* + x2P2* = P1* + x2(P2* - P1*) A plot of the total pressure has the form of a straight line. 5 What is the composition? a. x2 = 0.40 , y2 = 0.60 b. x2 = 0.50 , y2 = 0.70 c. x2 = 0.50 , y2 = 0.50 d. x2 = 0.40 , y2 = 0.70 What is the composition? a. x2 = 0.40 , y2 = 0.60 b. x2 = 0.50 , y2 = 0.70 c. x2 = 0.50 , y2 = 0.50 d. x2 = 0.40 , y2 = 0.70 The tie line The tie line shown in Figure (red line) is at a total pressure of 30 torr. You can read the x2 and y2 values from the plot (or calculate them using the equations above used to generate the blue and purple curves in the composition-pressure plot. 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 xa can be used together with x2 and y2. za y2 - za za The lever rule za y2 - za za 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, za. The lever rule za y2 - za za The lever rule states: nliquid nvapor = y2 – za za – x2 6 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. Henry’s law The statement of Henry’s law is: P1 = x1kH,1 It looks like Raoult’s law except that you have this funny constant kH,1 instead of P1* (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. 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, P1 x1P1* as x1 1. However, as x1 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, P1 x1kH,1 as x1 0. In this expression kH,1 is the Henry's law constant. Although we have focused on component 1 the same holds true for component 2. Non-ideal solutions A general type of expression for non-ideal behavior is shown below. Note that the mole fraction of component 2 appears in the exponent. A plot of this function for α = 1 and β = 1 (blue) compared to Raoult's law (black) is shown in the Figure below. P1 = x1P1 *eαx22 + βx23 Non-ideal solutions For the example shown in the plot above we have assumed that P1* = 100 torr. Note that as x1 1 the slope approaches the ideal slope obtained from Raoult's law. However, as x1 0 (and therefore x2 1) the slope is quite different from ideal behavior. We can see that as x2 1 the slope becomes P1*eα+β. Note that this is value of the Henry's law constant kH,1. This is depicted in the Figure below (purple line). The Henry's law value can be quite different from the ideal value. Activity The activity in non-ideal solutions corresponds to mole fraction in ideal solutions. 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 xj is no longer equal to Pj/Pj*. However, as the mole fraction approaches unity (a pure substance) the solution becomes ideal. Thus, as xj 1, aj xj. a j = Pj Pj * µ jsoln = µ j* + RT ln aj 7 Activity coefficient The ratio of the activity to the mole fraction is called the activity coefficient γj. So as a solution becomes ideal γj 1, as well. If we examine the expressions above for the Henry's law behavior of a solvent we see that the exponential terms that were used are related to the activity coefficient. Starting with the example from above: we see that γ j = a j x j P1 = x1P1 * exp αx22 + βx23 a1 = x1exp αx22 + βx23 Activity coefficient The activity coefficient can be defined in terms of the mole fraction as follows: Although this expressions is generally true we will be more often concerned with the limiting cases which are implicit in these expressions. These are summarized below. Raoult’s law: As xi 1, γi 1, and ai xi Henry’s law: As xi 0, γi kH,i/Pi*, and ai xikH,i/Pi* γ1 = exp αx22 + βx23 Problem Use the data from table below at 35 oC determine the activity coefficient for acetone (A) and chloroform (C). The Henry's law constants are kH,A = 175 torr and kH,C = 165 torr. What is activity coefficient for chloroform (C) at xC = 0.8? A. 0.747 B. 0.847 C. 0.934 D. 1.070 xC PC (torr) PA (torr) 0 0 347 0.2 35 270 0.4 82 142 0.6 142 102 0.8 219 37 1.0 293 0 Problem Use the data from table below at 35 oC determine the activity and activity coefficient for acetone (A) and chloroform (C). The Henry's law constants are kH,A = 175 torr and kH,C = 165 torr. What is activity coefficient for chloroform (C) at xC = 0.8? A. 0.747 B. 0.847 C. 0.934 D. 1.070 xC PC (torr) PA (torr) 0 0 347 0.2 35 270 0.4 82 142 0.6 142 102 0.8 219 37 1.0 293 0 aC = PC PC = 219293 = 0.747 γC = aC xC = 0.7470.800 = 0.934 Problem Use the data from table below at 35 oC determine the activity and activity coefficient for acetone (A) and chloroform (C). The Henry's law constants are kH,A = 175 torr and kH,C = 165 torr. What is activity coefficient for acetone (A) at xC = 0.4? A. 0.681 B. 0.747 C. 0.934 D. 1.466 xC PC (torr) PA (torr) 0 0 347 0.2 35 270 0.4 82 142 0.6 142 102 0.8 219 37 1.0 293 0 Problem Use the data from table below at 35 oC determine the activity and activity coefficient for acetone (A) and chloroform (C). The Henry's law constants are kH,A = 175 torr and kH,C = 165 torr. What is activity coefficient for acetone (A) at xC = 0.4? A. 0.681 B. 0.747 C. 0.934 D. 1.466 xC PC (torr) PA (torr) 0 0 347 0.2 35 270 0.4 82 142 0.6 142 102 0.8 219 37 1.0 293 0 aA = PA PA * = 142 347 = 0.409 γA = aA xA = 0.4090.600 = 0.681 Note: xA = 1 - xC