CHM 3410 – Problem Set 5Due date: Friday, October 16th Do all of the following problems. Show your work.

"If things were simple word would have gotten around." - Jacques Derrida

1) If we assume Hvap is independent of temperature the following equation can be derived from the Clausius-Clapeyron equation

ln(p2/p1) = - (Hvap/R) [ (1/T2) - (1/T1) ] (1.1)

where p1 is the vapor pressure of the liquid at T1 and p2 is the vapor pressure of the liquid at T2.

a) Use equn 1.1 and the data given below to estimate the value for the vapor pressure of water (H2O) at T = 50.0 C. Give your final answer in units of torr.

b) Equation 1.1 is derived assuming that the value for Hvap is independent of temperature. A better equation may be obtained by including a term to account for the dependence of Hvap on temperature. Let

Hvap = a + b (T - T) (1.2)

where a = Hvap, the enthalpy of vaporization at T, the normal boiling point of the liquid, and b = Cp,m = Cp,m(g) - Cp,m() is the difference between the constant pressure molar heat capacities of the gas and liquid.

Using equn 1.2 and the Clausius-Clapeyron equation

d ln (p) = H vap (1.3) dT RT2

find an improved expression for ln(p2/p1) that takes into account the temperature dependence of Hvap.

c) Use your answer in b and the data below to estimate the value for the vapor pressure of water at T = 50.0 C. Give your final answer in units of torr.

d) Compare your estimated values for the vapor pressure of water found in part a and part c of this problem to the experimental value at T = 50.0 C, pvap = 92.6 torr.

Hvap(H2O) = 40656. J/mol T(H2O) = 100.00 C Cp,m(H2O) = - 41.71 J/mol.K

2) "Synthetic air" is a mixture of nitrogen (N2) and oxygen (O2) with XN2 = 0.80. Find Gmix, Hmix, and Smix

when 1.000 mol of synthetic air is produced by mixing an appropriate amount of pure nitrogen and pure oxygen at p = 1.000 atm and T = 25.0 C. Assume ideal behavior.

3) When ethylene bromide (EtBr) and propylene bromide (PrBr) are mixed they form, to a first approximation, an ideal solution.

Consider a solution of EtBr and PrBr at T = 80.0 C. The vapor pressure above the solution is p = 139. torr. The vapor pressures of the pure liquids at this temperature are p*(EtBr) = 172. torr and p*(PrBr) = 127. torr. Find the mole fraction of EtBr in the solution and the partial prssure of EtBr in the vapor above the solution.

4) In class we derived the following expression for the free energy of mixing of two ideal gases initially at the same temperature and pressure

Gmix = nRT [ XA lnXA + XB lnXB ] (4.1)

The same procedure can be used to find Gmix when the two gases are originally at different pressures.Consider the system below. 1.000 mol of argon (Ar) is initially confined to the left side of the container,

which has a volume V = 30.00 L, and 1.00 mol of helium (He) is initially confined to the right side of the container, which has a volume V = 10.00 L. The barrier between the two sides of the container is removed and the gases are allowed to mix until equilibrium is reached. Temperature remains constant at T = 300.0 K throughout the process. Find Gmix. You may assume that argon and helium behave ideally. HINT: You cannot simply use equn 4.1 to answer this question since it was derived assuming the initial pressures of the two gases were equal to each other.

5) A solution is formed by dissolving 66.83 g of vanadium tetrachloride (VCl 4, M = 192.75 g/mol), a nonvolatile solute, in 1000.0 g of carbon tetrachloride (CCl4, M = 153.82 g/mol). The normal freezing point and freezing point depression constant for carbon tetrachloride are Tfus = - 22.9 C and Kf = 30.0 kg.K/mol.

a) Based on the above data predict Tf, the depression of the freezing point for the above solution.b) The observed depression of the freezing point for the above solution is Tf = 5.97 C, which is smaller

than that predicted in part a. This is due to the fact that vanadium tetrachloride dimerizes in carbon tetrachloride. The reaction may be written as

2 VCl4(soln) V2Cl8(soln) Km = b(V2Cl8)/[b(VCl4)]2 (5.1)

where Km is the equilibrium constant for the reaction in terms of molality (b).Using the information given above find Km for the dimerization reaction at the freezing point of the

solution.

Also do the following from Atkins:

Exercises 5.1 a Note that M(CH3COCH3) = 58.08 g/mol; M(CHCl3) = 119.38 g/mol.

Problems 4.8

4.10 Give the phase diagram for the pressure range from 0 to 1.5 atm and the temperature range from - 40.0 C to 100.0 C.

5.4

EXTRA CREDIT - Is there evidence in the data in Problem 4.8 for temperature dependence in the value for Hvap for nitric acid? Justify your answer - and if your answer is yes, find an expression for Hvap(T), the enthalpy of vaporization as a function of T.

Solutions

1) a) If we assume Hvap is independent of temperature, then

ln(p2/p1) = - (Hvap/R) [ (1/T2) - (1/T1) ]

Let T1 = 100.0 C = 373.15 K T2 = 50.0 C = 323.15 Kp1 = 760. torr Hvap = 40656. J/mol R = 8.3145 J/mol.K

Substitution into the above equation gives

ln(p2/p1) = - [(40656. J/mol)/(8.3145 J/mol.K)] [ (1/323.15 K) - (1/373.15 K) ] = - 2.0275

p2/p1 = exp(-2.0275) = 0.13166

p2 = (0.13166) p1 = (0.13166) (760 torr) = 100.1 torr

b) To take the temperature dependence of Hvap into account we say

Hvap = a + b (T - T) where T = 100.0 C = 373.15 K

Substitution into the Clausius-Clapeyron equatin gives

d ln (p) = H vap = [ a + b (T - T ) ] multiply both sides by dT and rearrange dT RT2 RT2

d ln (p) = [ (a - bT ) + bT ] dT integrate from p1, T1 to p2, T2 RT2

p1p2 d ln (p) = T1

T2 [ (a - bT)/RT2 + b/RT ] dT

ln (p2/p1) = - [ (a - bT)/R] [ (1/T2) - (1/T1) ] + (b/R) ln(T2/T1)

c) Substituting the information from part a of the problem, and a = 40656. J/mol ; b = - 41.71 J/mol .K

ln(p2/p1) = - [ (40656. - (- 41.71) (373.15) ] J/mol [ (1/323.15 K) - (1/373.15 K) ] (8.314 J/mol .K)

+ ( - 41.71 J/mol . K) ln(323.15/373/15) ( 8.3145 J/mol .K)

= - 2.8039 + 0.7217 = - 2.0822

p2/p1 = exp(- 2.0822) = 0.12465

p2 = (0.12465) p1 = (0.12465) (760 torr) = 94.7 torr

d) The experimental value for the vapor pressure at T = 50.0 C is p = 92.6 torr. The result in part c (2.3 % higher) is a bit better than the result from part a (5.7 % higher). I'm not sure what is responsible for the remaining difference, but I believe the biggest remaining error is in the assumption that the vapor phase behaves like an ideal gas.

2) For ideal mixing

Gmix = nRT [XN2 lnXN2 + XO2 lnXO2] = (1.00 mol) (8.3145 J/mol.K) (298.15 K) [0.8 ln(0.8) + 0.2 ln(0.2)]

= - 1240. J

Smix = - nR [XN2 lnXN2 + XO2 lnXO2] = - (1.00 mol) (8.3145 J/mol .K) [0.8 ln(0.8) + 0.2 ln(0.2)] = 4.161 J/K

Hmix = 0.

3) Let XE = mole fraction of ethylene bromideXP = 1 - XE = mole fraction of propylene bromide

Then XE pE* + XP pP* = XE pE* + (1 - XE) pP* = p

pP* + XE (pE* - pP*) = p

XE = (p - pP*) = (139. - 127.) = 0.267 (and so XP = 0.733) (pE* - pP*) (172. - 127.)

Then pE = XE pE* = (0.267) (172. torr) = 45.9 torr (and so pP = 93.1 torr)

4) We use the same procedure as used in class to find Gmix when the pressures were initially the same.

Gmix = Gf - Gi

Gi = nAr Ar,i + nHe He,i

Gf = nAr Ar,f + nHe He,f

where j = j + RT ln pj

Let pAr,i = initial pressure of Ar

pHe,i = initial pressure of He

Then, using Boyle's law

pAr,f = final pressure of argon = (30./40.) pAr,i = (0.75) pAr,i

pHe,f = final pressure of helium = (10.40.) pHe,i = (0.25) pHe,i

So Gi = nAr (Ar + RT ln pAr,i) + nAr (He + RT ln pHe,i)

Gf = nAr (Ar + RT ln 0.75 pAr,i) + nAr (He + RT ln 0.25 pHe,i)

= nAr (Ar + RT ln pAr,i) + nAr (He + RT ln pHe,i) + nArRT ln(0.75) + nHeRT ln(0.25)

Gmix = Gf - Gi

= nAr (Ar + RT ln pAr,i) + nAr (He + RT ln pHe,i) + nArRT ln(0.75) + nHeRT ln(0.25)

- [ nAr (Ar + RT ln pAr,i) + nAr (He + RT ln pHe,i) ]

= nArRT ln(0.75) + nHeRT ln(0.25)

= (1.00 mol) (8.3145 J/mol.K) (300.0 K) ln(0.75) + (1.00 mol) (8.3145 J/mol .K) (300.0 K) ln(0.25)

= - 4175. J

Notice that we never had to find values for any of the pressures in the problem.

5) If all of the solute was in the orm of the monomer, VCl4, then

b = (66.83 g) (1. mol/192.75 g) = 0.3467 mol/kg 1.000 kg

The actual molality of solute particles can be found from the measured depression of the freezing point

Tf = Kf bsol

So bsol = T f = 5.97 K = 0.1990 mol/kg Kf (30.0 kg.K/mol)

This lower value occurs because of the dimerization reaction. To find the value for Km we need the equilibrium molalities of the monomer and dimer.

Compound initial change equilibrium

VCl4 0.3467 - 2x 0.3467 - 2x

V2Cl8 0.0 + x + x

Since bsol = b(VCl4) + b(V2Cl8)

0.1990 = (0.3467 - 2x) + x = 0.3467 - x

x = 0.3467 - 0.1990 = 0.1477 mol/kg

So b(VCl4) = 0.3467 - 2 (0.1477) = 0.0513 mol/kg

b(V2Cl8) = 0.1477 mol/kg

Km = b(V2Cl8) = (0.1477) = 56.1 [b(VCl4)]2 (0.0513)2

Exercise 5.1 a

Start by assuming 1.000 mol of the mixture. Let XA = mole fraction acetone

XC = mole frac tion chloroform

m = XA MA + XC MC = (0.5307) (58.08 g/mol) + (0.4693) (119.38 g/mol) = 86.85 g/mol soln

So the total number of moles of solution is

n = 1000.0 g 1 mol = 11.514 mol soln 86.85 g

The volume occupied by 1.00 mol of solution is

Vm = (0.5307) (74.166 cm3/mol) + (0.4693) (80.235 cm3/mol) = 77.014 cm3/mol

So V = n Vm = (11.514 mol) (77.014 cm3/mol) = 886.7 cm3

Problem 4.8

The data are tabulated below

T (C) T (K) p (kPa) 1/T (K-1) ln p

0.0 273.15 1.92 0.0036610 0.65233 20.0 293.15 6.38 0.0034112 1.85317

40.0 313.15 17.7 0.0031934 2.87356 50.0 323.15 27.7 0.0030945 3.32143

70.0 343.15 62.3 0.0029142 4.13196 80.0 353.15 89.3 0.0028317 4.49200

90.0 363.15 124.9 0.0027537 4.82751100.0 373.15 170.9 0.0026799 5.14108

The data are plotted below.

The best fitting line is

ln p = 17.429 - (4570. K)/T

a) At the normal boiling point p = 1.000 atm = 101.325 kPa.

ln p = 17.429 - (4570. K)/T

ln p - 17.429 = (- 4570. K)/T

1/T = ln p - 17.429 = ln (101.325) - 17.429 = 0.0028032 K-1 (- 4570. K) (- 4570. K)

and so T = 1 = 356.7 K 0.0028032 K-1

b) Hvap = - m R = (4570. K) (8.3145 J/mol.K) = 38.00 J/mol

Problem 4.10

Start at the triple point p = 36. torr = 0.047 atm T = 5.50 C = 278.65 K

The solid-liquid boundary is, over the pressure range of the phase diagram, essentially a vertical line.

For the solid-gas and liquid-gas boundaries we use the general relationship

ln(p2/p1) = - (Hpt/R) [ (1/T2) - (1/T1) ]

So, for the solid-gas boundary

ln(p/0.047) = - 1275. [ (1/T2) - 0.003589 ]

and for the liquid-gas boundary

ln(p/0.047) = - 3705. [ (1/T2) - 0.003589 ]

Points for the phase diagram are tabulated below

T(C) p(atm)

solid-gas - 40.0 0.019- 30.0 0.024- 20.0 0.030- 10.0 0.036 0.0 0.043

triple point 5.5 0.047liquid-gas 10.0 0.058

20.0 0.091 40.0 0.203 60.0 0.414 80.0 0.777100.0 1.363

The phase diagram is given below.

Problem 5.4

Let W = water, nW = moles of water, VW = partial molar volume of water

C = copper II sulfate, nC = moles of copper II sulfate, VC = partial molar volume of copper II sulfate

MW = 18.02 g/mol MC = 159.61 g/mol Then

V = nW VW + nC VC

dV = nW dVW + VW dnW + nC dVC + VC dnC

However, if we keep the mass (and therefore the number of moles) of water constant at 100.0 g, then dnW = 0.

From the Gibbs-Duhem equation

dVW = - (nC/nW) dVC

Substituting into our equation for V gives

dV = nW [ - (nC/nW) dVC ] + nC dVC + VC dnC = VC dnC

Solving for VC, we get

VC = dV/dnC

The data in the problem are tabulated below. Note we adjust the total mass of solution so that m(H 2O) remains constant at 100.0 g. This has a small effect on the mass of CuSO4, as indicated below.

m(soln) = 100.0 g [100.0/(100.0 - m(CuSO4))]

m(CuSO4) = m(soln) - 100.0 g

CuSO4 (g) adjusted adjusted adjusted (g/cm3) V (cm3)mass soln (g) mass CuSO4 (g) mol CuSO4

5.0 105.26 5.26 0.0330 1.051 100.1510.0 111.11 11.11 0.0696 1.107 100.3715.0 117.65 17.65 0.1106 1.167 100.8120.0 125.00 25.00 0.1566 1.230 101.63

If we assume

VC = dV/dnC V/nC

then we may use the above data to estimate the value for VC at each intermediate value of added CuSO4.

CuSO4 (g) VC (cm3) nC (mol) VC (cm3/mol) ave mass (g) mol fraction

8.18 0.0092 0.22 0.0366 6.014.38 0.0160 0.44 0.0410 10.721.32 0.0235 0.82 0.0460 17.8

The data are plotted below

The value for VC at XC = 0 (from extrapolation of the data) is VC = - 1.9 cm3/mol. This is close to the value VC = - 1.4 cm3/mol for MgSO4 (Atkins, page 138).

EXTRA CREDIT

To look for evidence of curvature we can fit the data in problem 4.8 to a second order polynomial

ln p = a + b/T + c/T2

and see if the value of c (coefficient of the 1/T2 term) is significantly different from zero.

Using the POLY program the best fitting second order polynomial to the data has

c = ( - 3.35 0.25) x 105 K2 (at 95% confidence limits)

Since c is significantly different from zero (at 95% confidence limits) there is evidence for curvature in the data.