FLORIDA INTERNATIONAL UNIVERSITY



CHM 3410 – Problem Set 5

Due date: Wednesday, October 13th

Do all of the following problems. Show your work.

1) Consider the system below

The left side contains 1.000 mol of N2 gas, while the right side contains 2.000 mol of O2 gas. Both sides have a volume V = 10.00 L and are at a temperature T = 400.0 K.

The barrier between the two sides is removed and the two gases are allowed to mix until equilibrium is achieved. Find (Gmix, (Smix, and (Hmix for the process. (Hint: You may assume that N2 and O2 behave ideally, and that T remains constant at 400. K. Note that you cannot simply use the equations derived in class for (Gmix, (Smix, and (Hmix, since those equations were derived for the case of equal initial pressures of gas. However, you can use the same general method to obtain the appropriate equations for the change in the thermodynamic functions for mixing).

2) There are two ways in which the equation for boiling point elevation is expressed

(T = Kb XB (2.1)

(T = K(b bB (2.2)

As shown in class, the value for Kb (the constant in equn 2.1) is given by the expression

Kb = RTb*2/(H(vap (2.3)

a) Using equn 2.1 and 2.3, find the value for Kb( (the constant in equn 2.2). Note that your equation for Kb( will involve MA (molecular mass of the solvent). Also note that in deriving your equation for Kb( you will need to make the approximation that XB = nB/(nA + nB) ( nB/nA.

b) Using the data in Table 2.3 of the appendix of Atkins find the value for Kb( for benzene (C6H6) and carbon disulfide (CS2). Compare your calculated values to the experimental values given in Table 5.2 of the appendix of Atkins.

3) Data on the partial pressures of iodoethane (I = CH3CH2I) and ethyl acetate (A = CH3COOC2H5) in equilibrium with a liquid solution of iodoethane and ethyl acetate, at T = 50. (C, are given below.

XI 0.0000 0.0579 0.1095 0.1918 0.2353 0.3718 0.5478 0.6349 0.8253 0.9093 1.0000

pI(kPa) 0.00 3.73 7.03 11.70 14.05 20.72 28.44 31.88 39.58 43.00 47.12

pA(kPa) 37.38 35.48 33.64 30.85 29.44 25.05 19.23 16.39 8.88 5.09 0.00

a) Plot the above data. Plot pI, pA, and ptotal vs XI.

b) Find the numerical value for the Henry's law constant for ethyl acetate in iodoethane at T = 50. (C.

c) Find the values for activity (a) and activity coefficient (() for iodoethane and ethyl acetate for the data at XI = 0.2353 and XI = 0.6349. In both cases use the solvent standard state for both iodoethane and ethyl acetate.

Exercises

5.3b 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.

5.6b 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.

5.11b The mole fraction of N2 and O2 in air at sea level are approximately 0.78 and 0.21. Calculate the molalities of the solution formed in an open flask of water at T = 25. (C. (Hint: The appropriate Henry's law constants are given in Table 5.1 of the appendix of Atkins. Assume a total pressure of 1.000 bar. Note that since the solvent is water the molality and molarity of dissolved gas have approximately the same numerical value.)

Problems

5.15 The excess Gibbs free energy of solution of methylcyclohexane (MCH) and tetrahydrofuran (THF) at 303.15 K was found to fit the expression

GE = RTx(1 - x) {0.4857 - 0.1077 (2x - 1) + 0.0191 (2x - 1)2 }

where x is the mole fraction of the methylcyclohexane. Calculate the Gibbs free energy of mixing when a mixture of 1.00 mol of MCH and 3.00 mol of THF is prepared. (Hint: The equation above is for GE. Recall that GE = (Gmix - (Gmix,ideal. You are being asked for the value for (Gmix in this problem.)

Solutions.

1) Following the procedure we used in class, we may say

(Gmix = Gf – Gi

Assuming ideal gas behavior, we may say the chemical potential of the gas is given by the expression

(A = (A( + RT ln(pA)

Let us call the initial pressures of N2 and O2 pN2( and pO2( (we have sufficient information to calculate the initial pressures, but will not need to do so).

Then the partial pressures of N2 and O2 after mixing are (because Vf = 2 Vi)

pN2,f = (Vi/Vf) PN2( = pN2(/2

pO2,f = (Vi/Vf) PO2( = pO2(/2

So

Gf = nN2 [(N2( + RT ln(pN2(/2)] + nO2 [(O2( + RT ln(pO2(/2)]

Gi = nN2 [(N2( + RT ln(pN2()] + nO2 [(O2( + RT ln(pO2()]

(Gmix = Gf – Gi = { nN2 [(N2( + RT ln(pN2(/2)] + nO2 [(O2( + RT ln(pO2(/2)] }

- { nN2 [(N2( + RT ln(pN2()] + nO2 [(O2( + RT ln(pO2()] }

= nN2 RT ln(1/2) + nO2 RT ln(1/2)

= - (nN2 + nO2) RT ln(2)

= - (1.000 mol + 2.000 mol) (8.314 J/mol.K) (400. K) ln(2) = - 6915. J

Since the gases are ideal, (Hmix = 0.

Since G – H – TS, (Gmix = (Hmix - T(Smix , and so

(Smix = - (Gmix/T = - (- 6915. J)/(400 K) = + 17.29 J/K.

2) a) (T = Kb XB

But XB = nB/(nA + nB) ( nB/nA (when nB ................
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