Physical Chemistry I CHEM 4641 Name: _______KEY__________________Fall 2019 Test 2chapters 4, 5, 613 points

Gas constant R = 8.314 J·mol-1·K-1

Boltzmann constant kB = 1.381× 10-23 J/K = 0.695 cm-1 K-1

Planck's constant h = 6.63×10-34 J·s

Avogadro's number NA = 6.02×1023 mol-1

Speed of light c = 3.00×108 m/s

Units: 1 bar = 105 Pa. 1 GPa = 109Pa 1 amu = 1.6605×10-27 kg

1 Joule = 1 Pa·m3 . 1 L·bar = 100 J. 1L=10-3m3 1Å = 10-10m

1H1.01

2He4.00

3Li6.94

4Be9.01

5B

10.81

6C

12.01

7N

14.01

8O

16.00

9F

19.00

10Ne20.18

11Na22.99

12Mg24.31

13Al

26.98

14Si

28.09

15P

30.97

16S

32.06

17Cl

35.45

18Ar

39.95

19K

39.10

20Ca40.08

21Sc44.96

22Ti

47.88

23V

50.94

24Cr

52.00

25Mn54.94

26Fe55.85

27Co58.93

28Ni

58.70

29Cu63.55

30Zn65.38

31Ga69.72

32Ge72.59

33As74.92

34Se78.96

35Br

79.90

36Kr

83.80

37Rb85.47

38Sr

87.62

39Y

88.91

40Zr

91.22

41Nb92.91

42Mo95.94

43Tc98

44Ru101.1

45Rh102.9

46Pd106.4

47Ag107.9

48Cd112.4

49In

114.8

50Sn118.7

51Sb121.8

52Te

127.6

53I

126.9

54Xe131.3

55Cs132.9

56Ba137.3

57La138.9

72Hf

178.5

73Ta

180.9

74W

183.8

75Re186.2

76Os190.2

77Ir

192.2

78Pt

195.1

79Au197.0

80Hg200.6

81Tl

204.4

82Pb207.2

83Bi

209.0

84Po209

85At210

86Rn222

87Fr223

88Ra226.0

89Ac227.0

104Rf267

105Db268

106Sg269

107Bh270

108Hs269

109Mt278

110Ds281

111Rg280

112Cn285

113Nh286

114Fl289

115Mc289

116Lv293

117Ts294

118Og294

58Ce140.1

59Pr

140.9

60Nd144.2

61Pm145

62Sm150.4

63Eu152.0

64Gd157.3

65Tb158.9

66Dy162.5

67Ho164.9

68Er

167.3

69Tm168.9

70Yb173.0

71Lu175.0

90Th232.0

91Pa231.0

92U

238.0

93Np237

94Pu244

95Am243

96Cm247

97Bk247

98Cf251

99Es252

100Fm257

101Md258

102No259

103Lr262

chapter 4 section A1. Recall the phase rule: F = C - P + 2. At right are densities ofliquid and gas 1-octene. The graph is Figure 1a from J. PabloPalofox-Herndandez, et al., J. Phys. Chem. B, 8(123), 2915-2924,2019. Consider a sample of pure 1-octene that contains bothliquid and gas.

a) (1 point) What are the values (i.e., numerical values) of C,P and F?

b) (1 point) Suppose a particular temperature is chosen: T =500 K. At that temperature, is pressure free to vary or ispressure fixed?

Answers:a) C=1, P=2, F = 1-2+2 = 1. P and T both can vary, but they do not vary independently. F=1 corresponds to the system being on the liquid-vapor coexistence line in (P,T) space.b) fixing T reduces F by 1. F=0 so P is fixed and is not free to vary.

chapter 52. (5 points) Mixtures of methanol and carbon dioxide were studied by Masahiro Kato, et al., Open Thermo. J., 2009. They reported the tabulated partial molar volumes at 313 K and 10.0 MPa. Consider a mixture of 0.80 mol CO2 and 0.20 mol CH3OH.

xCO2 VCO2(cm3/mol) VCH3OH (cm3/mol)

0.00 - 40

0.80 62 22

1.00 71 -

In case it helps: the entropy of a perfect mixture is Δmix S =− R(x1 ln x1 + x2 ln x2).

a) Calculate the volume of the mixture.b) Calculate the volume if the mixture was a perfect ("ideal") mixture.c) Calculate the excess volume.d) Calculate the Gibbs free energy of mixing, ΔmixG, assuming perfect mixing.

e) (∂ΔGmix /∂T )P=− ΔS mix . What analogous derivative equals ΔmixV ?

Answers:a) n1V1 + n2 V2 = 0.80 × 62 + 0.20 × 22 = 54 cm3

b) n1V1* + n2 V2* =0.80 × 71 + 0.20 × 40 = 65 cm3

c) 54 - 65 = -11 cm3

d) 1 mol ×RT (0.20 ln(0.20) + 0.80 ln(0.80) = 2602 J (-0.322 - 0.179) = -1300 Je) (∂ ΔGmix /∂ P )T

= ΔV mix because dG = -S dT + V dP

Score: / 13

chapter 5Question 3. (2 points) The phase diagram is for mixtures ofCaCl2 and CaBr2 at constant pressure. Thecompounds form complete solutions in thesolid phase below 970K and in the liquidphase at high temperature. The curveswere constructed by measuringtemperature over time while heatingsamples of fixed composition.

Suppose one begins with a solid mixture of20% CaBr2 and 80% CaCl2 at 960 K("start" on the figure). Here, "composition"means percent CaBr2 and it can be readfrom the graph.

a) The solid is heated slowly to 1005K, point a. What phases arepresent and what are their compositions?

b) Heating continues until point b is reached. Then the sample is then slowly cooled. At what temperature will crystals first form, and what will be the composition of those crystals?

Answers:a) (1 point)Solid and liquid are present. The solid is 16% CaBr2, the liquid is 23% CaBr2. b) (1 point) Crystals of composition 14% CaBr2 start forming at 1010 K.

source: Sajal Ghosh, et al, Thermochimca Acta 505, 69-72, 2010.

chapter 6

4. (4 points) Recall that ΔrG = ΔrGº + RT lnQ and that (∂(G /T )∂ T )P=−

H

T 2.

Reaction: 2 UH3 ⇌(s) 2 U(s) + 3 H2(g)

U(s) means solid uranium. T = 500 K. At equilibrium the pressure of H2 is 139 Pascals.

a) Write an expression for the equilibrium constant, K, in terms of PH2, the pressure of H2.

b) Calculate K. Calculate ΔrGº in kJ/mol.

c) Calculate the standard Gibbs free energy of formation of UH3(s), ΔfGº.

d) At 500K, ΔfHº(UH3,s) = -127.2 kJ/mol, ΔfHº(U,s)=0, and ΔfHº(H2,g)=0. Calculate ΔrHº.

e) Use ΔrHº to calculate K at 600 K.

Answers:a)

b) K = ( 139 Pa / 105 Pa )3 = 2.69× 10 -9

ΔrGº = - RT ln K = - (8.314×10 -3kJ/(mol K))(500 K) ln( 2.69× 10 -9) = +82. kJ/mol

c) (1 point) ΔrGº = 3 ΔfGº (H2,g) + 2 ΔfGº (U,s) - 2 ΔfGº (UH3,s) = 0 + 0 - 2 ΔfGº (UH3,s) +82.0 kJ/mol = - 2 ΔfGº (UH3,s)

ΔfGº (UH3,s) = -41.0 kJ/mol d)

e)

K =

a H 2

3 aU

aUH 3

2=

(P H 2/Po

)3×1

1=(

PH 2

Po )3

(∂(ΔGro/T )

∂T )P=−

Δro H

T 2

−Rd lnKd T

=−Δr

o H

T 2

d lnKd T

=254.4×103

8.314 T 2= 30.55×103T−2

Integrate: ln (K600) = ln (K500) + 30.55×103 ∫500

600d T

T 2

ln (K 600) = ln (K 500) + 30.55×103(− 1600

+1

500 )ln (K 600) =−19.66+30.55×103

(3.333×10−4)=−19.66+10.18=−9.48

K=7.7×10−5

Δr H o= 3Δ f (H 2 , g )+2Δ f (U , s) − 2Δ f (UH 3 , s)

Δr H o= − 2Δ f (UH 3 , s) =−2×(−127.2 kJ/mol) = 254.4kJ/mol