1. *The density of aqueous solutions of acetic acid varies with the mass fraction w 2 of acetic acid at 298 K as follows:

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1 Chem 42/523 Chemical Thermodynamics Homework Assignment # 4 1. *The density of aqueous solutions of acetic acid varies with the mass fraction w 2 of acetic acid at 298 K as follows: ρ (g cm 3 ) w 2 ρ (g cm 3 ) w 2 ρ (g cm 3 ) w Plot an appropriate volume unit vs. x 2, the mole fraction of acetic acid. Determine partial molar volumes of water and acetic acid at x 2 =,.25,.5,.75, 1.. PlotV 1 and V 2 as functions of x 2. Answer We first calculate the volume of acetic acid solutions containing 1 g of water. The mass fraction of acetic acid is then given by (using the letter µ for mass because we want to reserve m for molality) w 2 = µ 2 µ 1 + µ 2 µ 2 = 1 + µ 2 The mass of acetic acid in a solution with a weight fraction w 2 of acetic acid is given by µ 2 = 1w 2 1 w 2 and the total mass of the solution is, therefore, (µ 2 + 1) g. From this mass and the given density, we can calculate the molality of acetic acid m 2. ρ w 2 µ 2 µ sol x 2 m = n 2 V (g cm 3 ) g g mol kg 1 (cm 3 )

2 Plotting the volume as a function of molefraction, we get the figure below. 1 Volume vs. mole fraction of acetic acid 8 V (cm 3 ) x 2 To evaluate the partial molar volume, it is better to plot volume vs. molality: 1 Volume vs. molality of acetic acid 8 V (cm 3 ) m 2 (mol kg 1 ) We fit this data to a second order polynomial (even though the line is almost linear, we will allow for small variations of slope with molality). This results in V = m m 2 2. From this relationship, we get µ µ V V V 2 = = n 2 T,p,n 1 m 2 T,p,n 1 = m 2. (1) 2

3 Since the molality relationship does not apply when w 2 =1, this result is not valid for x 2 =1. However, since x 2 =1corresponds to pure acetic acid, the partial molar volume at this mole fraction is simply the molar volume of pure acetic acid, given by M gmol 1 = ρ(w 2 =1) gcm 3 = cm3 mol 1 To find the partial molar volume of water, we calculate the volume of solutions containing 1. kg of acetic acid (i.e., we treat the acetic acid as solvent and water as the solute). ρ w 1 µ 1 µ sol x 2 m = n 1 V (g cm 3 ) g g mol kg 1 (cm 3 ) We plot volume vs. molality of water to evaluate the partial molar volume of water: 1 Volume vs. molality of water 8 V(cm 3 ) m 1 (mol kg 1 ) Analysis similar to that above yields V = m 1 +.8m 2 µ µ 1. V V V 1 = = n 1 T,p,n 2 m 1 T,p,n 2 = m. (2) 3

4 Again, for reasons mentioned above, this expression is not valid at x 2 =. However, since this corresponds to pure water, V 1 = V m,1 =18.16 gmol 1 /.9982 gcm 3 =18.49 cm 3 mol 1. Therefore, we tabulate the partial molar volumes calculated from Eqs. (1) and (2) below. Note that m 2 is calculated for solutions containing 1. kg of water (55.51 mol H 2 O) and m 1 is calculated for solutions containing 1. kg of acetic acid (16.65 mol HAc). x 2 m 1 m 2 V 1 (cm 3 mol 1 ) V 2 (cm 3 mol 1 ) A plot of the partial molar volumes vs. mole fractions of acetic acid is given below. 6 Partial molar volumes vs. mole fraction of HAc Partial molar volumes x 2 Note: It should be clear to you by now that partial molar volumes are extremely sensitive to the method of analysis used. You could easily justify using a linear fit for the V vs. m i curves in which case the partial molar volumes would have no concentration dependence at all. Alternately, one could have used amodellike V = a + bm 1/2 + cm + dm 3/2 + em 2 and obtained somewhat different results from those reported above. The main reason I avoided the model with the m 1/2 term was to avoid the singularity in the derivative when m =. It must also be mentioned that one of the authors of the text book, Dr. Boerio-Goates, suggests that we start with Eqs. (5.33) and (5.34), use the Redlich-Kister equation given in Eq. (5.4), and find the partial molar volumes. We should expect those results to be qualitatively similar, but quantitatively different, from those obtained above. I will try to solve this problem using this method and provide you with the results. 4

5 2. At K, the total volume of a solution formed from 1. kg of water mixed with m moles of MgSO 4 is given by V/(cm 3 ) = (m.7) 2. The expression applies up to a molality of approximately m =1.1 mol kg 1. molar volumes of (a) the MgSO 4, (b) the water when m =.5 mol kg 1? Answer (a) The partial molar volume of MgSO 4 is given by (.5.7) = V 2 = µ V = (m.7) m T,p,n 1 = cm 3 mol 1 What are the partial (b) The partial molar volume of water is determined from the relationship (with m =.5 mol kg 1 and n 1 =1g/18.16 gmol 1 ) V =n 1 V 1 + n 2 V 2 V 1 =(V n 2 V 2 )/n 1 = 1 i h (m.7) (m.7) n 1 = 1 h i (.5.7) (.5.7) =18.62 cm 3 mol The apparent heat capacity of sucrose (2) in water (1) is given as a function of molality, m by the expression φc p /(J K 1 mol 1 ) = m.1948m 2 with C p,1 =75.4 JK 1 mol 1. (a) Derive expressions for C p,1 and C p,2. (b) Calculate the heat capacity of a solution consisting of.3 moles of sucrose and 6 g of water. Answer (a) Using Eq. (5.44), we get µ Cp m µ φcp m T,p,n 1 C p,2 = = φc p + m T,p,n 1 = m.1948m 2 + m ( m) = m.5844m 2 From the Gibbs-Duhem equation we have n 1 dc p,1 + n 2 dc p,2 =, dc p,1 = n 2 dc p,2 n 1 = m n 1 ( m) dm. 5

6 Simplifying and integrating between the limits of molality of solute (pure solvent) and the molality m, we get Z Cp,1 C p,1 dc p,1 = 1 Z m (9.456m m 2 )dm n 1 C p,1 Cp,1 = 1 Ã! m2 n m3 3 from which we get C p,1 = C p,1 1 n 1 ³ 4.728m m 3 (b) The molality of the solution is given by.3 mol m= 1 gkg 1 6 g =.5 mol kg 1 The heat capacity of the solution is calculated as (with n 1 =55.51 mol) C p =mφc p + n 1 C p,1 = = JK 1 4. The following data are given for aqueous solutions of calcium chloride solutions at 2 C: CaCl 2 Density CaCl 2 Density %byweight (gcm 3 ) % by weight (g cm 3 ) (a) Calculate the volume of the solutions containing 1. kg of water. (b) Use a graphical procedure to determine partial molar volume of CaCl 2 at molalities corresponding to 12, 2, 25, and 35 percent weight. (c) * Using a curve-fitting program (Mathcad or the freely available CurveExpert), fit the volume vs. molality data found in part (b) to an expression like V = a + bm 1/2 + cm + dm 3/2 + em 2 and report the values and units of the constants a,b,c,d,ande. (d) * Calculate the partial molar volumes at the same compositions as in part (b) and compare them. (e) * Derive an expression for the partial molar volume of water from the results of part (c). Answer 6

7 (a) The volume and molality data for solutions containing 1. kg of water are tabulated below. Molar mass of CaCl 2 is g mol 1. CaCl 2 Density Mass of Solute Mass of solution m V % by weight (g cm 3 ) (g) (g) (mol kg 1 ) (cm 3 ) (b) A plot of the volume vs. molality is given below. 12 Question 4 (b), (c) 115 V (cm 3 ) m (mol kg 1 ) The line is almost perfectly linear beyond the third data point, which corresponds to a molality of mol kg 1,or12%byweightCaCl 2. Therefore, we estimate the derivatives by central difference: µ V V 2 = = 1 µ Vi V i 1 + V i+1 V i. m i T,p,n 1 2 m i m i 1 m i+1 m i The partial molar volume of water is found from V = n 1 V 1 + n 2 V 2. The results are tabulated below. The number of moles of solute, n 1, is in all cases, corresponding to 1. g of water. Weight % m V 2 V

8 (c) * The curve shown in the figure above is the result of fitting the volume-molality data tabulated in part (a) to the model V = a + bm 1/2 + cm + dm 3/2 + em 2. The result of the fit are a = ,b = ,c = ,d = ,e = (d) * The partial molar volumes of the solute, V 2, are found by differentiation of the model above. The partial molar volume of the solvent is found as in part (b). The results are tabulated below. Weight % m V 2 V Except for the case of the 12 % by weight solution, the numerical derivatives found in part (b) and the analytical derivatives above are in good agreement. The model is clearly deviating from the data in the vicinity of the first concentration considered and, therefore, we would expect the model s prediction to be inaccurate at that value. (e) The expression for the partial molar volume of the solute is obtained directly from differentiating the model with respect to molality: V 2 = µ V m = T,p,n 1 2 m 1/ m1/ m =43.647m 1/ m 1/ m. From the Gibbs-Duhem equation, n 1 dv 1 + n 2 dv 2 =,andn 1 =55.51,we get dv 1 = n 2 dv 2 n 1 = m µ 1 n m 3/ m 1/ =.39315m 1/ m 1/2.5956m Integrating this expression from molality =(V 1 = V m,1) tom, andn 1 =55.51, weget Z V 1 V m,1 dv 1 = 1 n 1 Z m ³.39315m 1/ m 1/2.5956m dm V 1 = Vm, /2 m1/ /2 m3/ m 2 = m 1/ m 3/ m 2. 8

CHEM N-7 November 2005

CHEM N-7 November 2005 CHEM1909 2005-N-7 November 2005 Calcium chloride (3.42 g) is completely dissolved in 200 ml of water at 25.00 ºC in a coffee cup calorimeter. The temperature of the water after dissolution is 27.95 ºC.