CHM 3410 – Problem Set 7

Due date: Monday, October 19th (The second hour exam is Friday, October 23rd. It will cover the following material - Chapter 3, section 3D; Chapter 4, all; Chapter 5, sections 5A, 5B. 5C, and 5E; Chapter 6, sections 6A and 6B; handouts).

Do all of the following problems. Show your work.

"Politics is like football. You have to be smart enough to understand the game and dumb enough to think it's important." - Eugene McCarthy

1) At 310. K the partial vapor pressures of a substance B dissolved in a liquid A are as follows:

XB 0.010 0.015 0.020

PB(kPa) 82.0 122.0 166.1

Show that this solution obeys Henry's law in this range of mole fractions and calculate the Henry's law constant at T = 310. K.

2) The addition of 5.000 g of a compound to 250.0 g of naphthalene lowered the freezing point of the solvent by 0.780 K. Calculate the molar mass of the compound. Note that for naphthalene Kf¢= 6.94 kg K/mol.

3) Calcium carbonate (CaCO3) will decompose into calcium oxide (CaO) and carbon dioxide (CO2) by the process

CaCO3(s) D CaO(s) + CO2(g) K = (aCaO)(aCO2) = (aCO2) (3.1)

(aCaCO3)

Thermodynamic data for CaO, CaCO3, and CO2 are given below (at T = 298.0 K) and will be of use in doing the following problems.

a) What is the numerical value for K for reaction 3.1 at T = 298.0 K?

b) If we assume that the value for DH°rxn for reaction 3.1 is independent of temperature. the value for K for the reaction at temperatures different than T = 298.0 K can be found from the expression

ln(K2/K1) = - (DH°rxn/R) [ (1/T2) - (1/T1) ] (3.2)

Using eq 3.2, find the value for K for reaction 3.1 at T = 500.0 K.

c) Since both DH°rxn and DS°rxn vary with temperature, a better way to find the value for K at temperatures different than T = 298.0 K is to calculate values for DH°rxn and DS°rxn at the new temperature, using the relationships

DH°rxn(T2) = DH°rxn(T1) + òT1T2 DCp,m dT (3.3)

DS°rxn(T2) = DS°rxn(T1) + òT1T2 (DCp,m/T) dT (3.4)

find the value for DG°rxn at the new temperature (using DG°rxn = DH°rxn - TDS°rxn), and then finding K from the expression ln K = - DG°rxn/RT. Use this procedure to find the numerical value for K for reaction 3.1 at T = 500.0 K. Compare your result to that found in part b of this problem.

Substance DH°f (kJ/mol) DG°f (kJ/mol) S° (J/mol K) Cp (J/mol K)

CaO(s) - 635.09 - 604.03 39.75 42.80

CaCO3(s) - 1206.9 - 1128.8 92.9 81.88

CO2(g) - 393.51 - 394.36 213.74 37.11

4) The equilibrium constant for the reaction

2 C3H6(g) D C2H4(g) + C4H8(g) (4.1)

is found, between 300.0 K and 600.0 K, to fit the expression

lnK = A + (B/T) + (C/T2) (4.2)

where A = - 1.02, B = - 1088. K, and C = 1.51 x 105 K2. Using this information. find DG°rxn, DH°rxn, and DS°rxn for reaction 4.1 at T = 400.0 K

5) A phase diagram for a solution of A and B (at a fixed pressure of 1.00 atm) is given below. Based on the phase diagram, answer the following questions.

a) Describe what happens when a closed system at an initial temperature Ti = 60. °C and an initial mole fraction ZA = 0.400 is heated at a constant pressure p = 1.00 atm.

b) What is the normal boiling point for A and the normal boiling point for B?

c) Do A and B form an azeotrope? If so, give the location (temperature and mole fraction) for the azeotropic mixture. Is the azeotrope a high boiling or a low boiling azeotrope?

d) Consider a system with 1.000 moles of substance, and a mole fraction of A in the system ZA = 0.400. At T = 80.0 °C, points a, b, and c are located at 0.284, 0.400, and 0.656, respectively. Find the total number of moles in the liquid phase and the total number of moles in the vapor phase.

6) A phase diagram for a solution of A and B (at a fixed pressure of 1.00 atm) is given below. Based on the phase diagram, answer the following questions.

a) Describe what happens when a closed system at an initial temperature Ti = 80. °C and an initial mole fraction ZA = 0.400 is heated at a constant pressure p = 1.00 atm.

b) What is the normal boiling point for A and the normal boiling point for B?

c) Do A and B form an azeotrope? If so, give the location (temperature and mole fraction) for the azeotropic mixture. Is the azeotrope a high boiling or a low boiling azeotrope.

d) Consider distillation of a mixture with a mole fraction of ZA = 0.400. Vapor is continuously removed by condensation during the distillation. When does the mixture first boil? What is the initial value for the mole fraction of A in the liquid phase and the vapor phase when the mixture first boils? How do the boiling temperature, the mole fraction of A in the liquid phase, and the mole fraction of A in the vapor phase, change as the distillation proceeds? Is it possible to obtain pure A by distillation from a mixture with an initial mole fraction ZA = 0.400? Why or why not?


Solutions.

1) There are several ways of doing this problem. One method is to calculate the Henry's law constant for each data point, using

pB = KB XB KB = pB/XB

where KB is the Henry's law constant.

XB KB

0.010 8200. kPa

0.015 8133. kPa

0.020 8305. kPa

The data are plotted below. The best fitting line has slope = 10500. kPa, intercept = 8055. kPa. The value for the intercept represents the value for the Henry's law constant in the limit XB ® 0, and so represents the best estimate for KB. Based on this, I get KB = 8.06 x 103 kPa (KB = 8060. kPa).


2) We can find the molality of the solution (assuming ideal behavior) using

DTf = Kf¢ bB

bB = DTf/Kf¢ = (0.780 K)/(6.94 kg K/mol) = 0.1124 mol/kg

The number of moles of solute is

nB = (0.1124 mol/kg)(0.2500 kg) = 0.0281 mol

and so M = mB/nB= (5.000 g)/(0.0281 mol) = 178. g/mol

3) I am going to find DG°rxn at T = 298. K in two ways.

Direct method. DG°rxn = [DG°f(CaO(s)) + DG°f(CO2(g))] - [DG°f(CaCO3(s))]

= [(- 604.03) + (- 394.36)] - [(- 1128.8)] = + 130.41 kJ/mol

Indirect method. First find DH°rxn and DS°rxn, then find DG°rxn using

DG°rxn = DH°rxn - T DS°rxn

DH°rxn = [DH°f(CaO(s)) + DH°f(CO2(g))] - [DH°f(CaCO3(s))]

= [(- 635.09) + (- 393.51)] - [(- 1206.9)] = + 178.30 kJ/mol

DS°rxn = [S°(CaO(s)) + S°(CO2(g))] - [S°(CaCO3(s))]

= [(39.75) + (213.74)] - [(92.9)] = + 160.59 J/mol K = 0.16059 kJ/mol K

Then DG°rxn = (178.30 kJ/mol) - (298. K)(0.16059 kJ/mol K) = + 130.44 kJ/mol

The two methods give essentially the same result for DG°rxn. I am going to use the value from the indirect method (DG°rxn = 130.44 kJ/mol) for consistency with the method used in part c of the problem.

lnK = - DG°rxn = - (130440. J/mol) = - 52.645

RT (8.3145 J/mol K)(298. K)

K = e-52.645 = 1.37 x 10-23

b) ln(K2/K1) = - (DH°rxn/R) [ (1/T2) - (1/T1) ]

or ln(K2) = ln(K1) - (DH°rxn/R) [ (1/T2) - (1/T1) ]

= (- 52.645) - [(178300. J/mol)/(8.3145 J/mol K)] [(1./500.0 K) - (1.298.0 K)]

= (- 52.645) + (29.07) = - 23.57

and so K(T = 500. K) = 5.80 x 10-11

c) To use this method we need to first find DCp,m, the change in heat capacity per mole of reaction. Note we are assuming that DCp,m is temperature independent.

DCp,m = [(42.80) + 37.11)] - [(81.88)] = - 1.97 J/mol K

Then DH°rxn(T2) = DH°rxn(T1) + òT1T2 DCp,m dT = DH°rxn(T1) + DCp,m (T2 - T1)

= 178.30 kJ/mol + (- 1.97 x 10-3 kJ/mol K)(500.0 K - 298.0 K)

= 177.90 kJ/mol

DS°rxn(T2) = DS°rxn(T1) + òT1T2 (DCp,m/T) dT = DS°rxn(T1) + DCp,m ln(T2/T1)

= 160.59 J/mol K + (- 1.97 J/mol K) ln(500/298)

= 159.57 J/mol K

Then DG°rxn = DH°rxn - T DS°rxn = (177.90 kJ/mol) - (500.0 K)(0.15957 kJ/mol K) = 98.115 kJ/mol

lnK = - DG°rxn = - (98115. J/mol) = - 23.601

RT (8.3145 J/mol K)(500. K)

K = e-23.601 = 5.63 x 10-11

That is a trivial difference! The small difference is due to the fact that DCp,m is so small. For a reaction where the difference in heat capacities between the products and reactants was larger, there would be a large rdifference in the results obtained by these two methods, with the result from the second method the more accurate result.

4) DG°rxn = - RT lnK

lnK = A + (B/T) + (C/T2) = - 1.02 + ( - 1088. K)/(400. K) + (1.51 x 105 K2)/(400. K)2

= - 2.796

so DG°rxn = - (8.3145 J/mol K)(400.0 K)(- 2.796) = 9.299 kJ/mol

DH°rxn = - R (d lnK/d(1/T)) = - R [ B + 2C/T ]

= - (8.3145 J/mol K)[ - 1088. K + 2(1.51 x 105 K2)/(400.0 K) ]

= 2.769 kJ/mol

Finally DS°rxn = DH°rxn - DG°rxn = [ (2769. J/mol) - (9229. J/mol) ] = - 16.15 J/mol K

T 400.0 K

The value for DS°rxn seems reasonable, since for this reaction there is no change in the number of moles of gas, and so we expect DS°rxn @ 0.

5) a) The system initially forms a single liquid phase. When the temperature reaches 73 °C, the liquid begins to boil, with the vapor phase having a smaller mole fraction of A than the liquid phase. Boiling occurs between 73 °C and 82 °C, at which point all of the liquid has been converted into vapor. Above 82 °C the system consists of a single vapor phase.

b) Normal boiling point for A = 90 °C

Normal boiling point for B = 67 °C

c) There is no azeotrope. The only liquids with a single boiling point temperature in this system are pure B (XA = 0.00) and pure A (XA = 1.00), and pure liquids do not count as azeotropes.

d) We have two equations we can use here

n + ng = n = 1.000 mol

(ZA – YA) ng = (XA – ZA) n

where n = total number of moles in the liquid phase, ng = total number of moles in the gas phase, n = total number of moles in the system. The second equation above is from the lever rule. Note that the mole fractions correspond to the points given on the phase diagram, so YA = 0.284, ZA = 0.400, and XA = 0.656.

Substituting into the second equation gives

(0.400 – 0.284) ng = (0.656 – 0.400) n

0.116 ng = 0.256 n

ng = (0.256/0.116) n = 2.21 n

If we substitute this result into our first equation, we get

n + (2.21 n) = 3.21 n = 1.000 mol

n = 1.000 mol/3.21 = 0.312 mol

Notice that this answer is qualitatively consistent with the phase diagram. The “lever” to the liquid is longer than the lever to the vapor, so the number of moles of liquid should be smaller than the number of moles of vapor.

6) a) The system initially forms a single liquid phase. When the temperature reaches 89 °C, the liquid begins to boil, with the vapor phase having a smaller mole fraction of A that the liquid phase. Boiling occurs between 89 °C and 96 °C, at which point all of the liquid has been converted into vapor. Above 96 °C the system consists of a single vapor phase.

b) Normal boiling point for A = 94 °C

Normal boiling point for B = 102 °C

c) Yes, there is an azeotrope at mole ZA = 0.71, T = 83 °C. It has a lower boiling point than the pure liquids, and so is a low boiling azeotrope.

d) The mixture first boils at T = 89 °C. Initially XA = 0.40, YA = 0.60. As distillation proceeds the mole fraction of A in the system (ZA) decreases, since vapor enriched in A is being removed by the distillation. Both XA and YA decrease as distillation proceeds. If we continue to remove vapor as we proceed (not what we would do in a real distillation) then at the end the system will be pure B and boiling at T = 102 °C (note that this is different behavior from that observed in a closed system – distillation represents an open system, as vapor is being removed from the system, changing the value of ZA for the system.

If the vapor is condensed and redistilled, the composition of the vapor in successive distillations will converge on the composition of the azeotrope (ZA = 0.71), at which point no further purification can occur by distillation. Therefore it is not possible to obtain pure A by distillation of a mixture with ZA = 0.40.