CHM 3410 - Physical Chemistry 1
Second Hour Exam
October 22, 2010
There are five problems on the exam. Do all of the problems. Show your work.
______________________________________________________________________________________
R = 0.08206 L.atm/mole.K
NA = 6.022 x 1023
.
.
R = 0.08314 L bar/mole K
1 L.atm = 101.3 J
.
R = 8.314 J/mole K
1 atm = 1.013 bar
______________________________________________________________________________________
1. (18 points) Carbon tetrachloride (CCl4) is a common nonpolar solvent for organic reactions. Because it
is a carcinogen and a mutagen there is interest in its vapor pressure and thermodynamic properties.
The normal boiling point and enthalpy of vaporization for carbon tetrachloride are T b* = 349.9 K
and Hvap = 30.00 kJ/mol.
a) Estimate the vapor pressure of carbon tetrachloride at T = 298. K.
b) The free energy of formation of liquid carbon tetrachloride is Gf(CCl4()) = - 65.21 kJ/mol at
T = 298. K. Using this information and the results from part a of this problem find the value for
Gf(CCl4(g)), the free energy of formation of carbon tetrachloride vapor, at T = 298. K.
2. (24 points) Methanol (CH3OH) and carbon tetrachloride (CCl4) are miscible liquids. Data for solutions
of these two liquids at T = 20.0 C have been reported by J. Timmermans Physiochemical Constants of
Binary Systems in Concentrated Solutions, Volume 2 (1959). In the data given below M = methanol and C
= carbon tetrachloride.
For a liquid solution at T = 20.0 C with XM = 0.700, it is found that pM = 0.104 bar and pC =
0.109 bar. The vapor pressures of the pure liquids at this temperature are p M* = 0.128 bar and pC* = 0.120
bar.
a) What are aM, M, aC, and C, the activity and activity coefficients for methanol and carbon
tetrachloride, for the above solution using Raoult’s law to define ideal behavior.
b) What is Gmix(ideal) the free energy of mixing for the formation of 1.000 mole of a liquid
solution with XM = 0.700 from pure methanol and pure carbon tetrachloride and assuming ideal mixing?
c) What is Gmix(real) actual value for the free energy of mixing for the formation of 1.000 mole
of a liquid solution with XM = 0.700 from pure methanol and pure carbon tetrachloride, based on the above
data?
3. (16 points) The Henry’s law constant for methane (CH4) in water is KH = 755 L.bar/mol at T = 298. K.
a) What is the equilibrium concentration of dissolved methane in water when p(CH 4) = 1.000 bar
and T = 298. K (see figure below)?
b) Given that Gf(CH4(g)) = - 50.72 kJ/mol at T = 298. K, and the data above, find the value for
Gf(CH4(aq)), the free energy of formation for aqueous methane, at T = 298. K
4. (22 points) The phase diagram given below is for two partially miscible liquids A and B in equilibrium
with their vapor (at p = 1.000 atm), and may be of use in answering the following questions.
a) Describe what happens in a closed system with ZA = 0.120 when it is heated from an initial
temperature Ti = 55. C to a final temperature T f = 90. C (line 1 in the diagram).
b) Consider the points a, b, and c in the phase diagram. Which of these points (if any) represent an
azeotropic mixture? Briefly justify your answer.
c) What is the normal boiling point for liquid B?
d) Consider 1.000 mole of a system with Z A = 0.800 and a temperature T = 80.0 C. Indicate the
phases present and the number of moles of each phase for this value of Z A. temperature, and total number
of moles. Points d, e, and f represent mole fractions of A equal to 0.640, 0.800, and 0.978, respectively.
5. (20 points) In the gas phase, hydrogen (H 2) and iodine (I2) will chemically react to form hydrogen iodide
(HI). The equation corresponding to this reaction is
H2(g) + I2(g)  2 HI(g)
(5.1)
a) Give the expression for K for the above reaction in terms of activities of reactants and products.
b) What is the numerical value for K for the above reaction at T = 298. K?
c) What is the numerical value for K for the above reaction at T = 350. K? You may assume that
Hrxn and Srxn are constant over the temperature range 298. K to 350. K.
Solutions.
1)
a) One form of the Clausius-Clapeyron equation is
ln(p2/p1) = - (Hvap/R) [ (1/T2) – (1/T1) ]
= - [(30000. J/mol)/(8.314 J/mol.K)] [ (1/298. K) – (1/349.9 K)] = - 1.796
So
p2/p1 = e-1.796 = 0.1660
p2 = 0.1660 p1 = 0.1660 (1 atm) = 0.1660 atm (equal to 126. torr, or equal to 0.168 bar)
b) If we write the vaporization reaction as
CCl4()  CCl4(g)
K = (aCCl4(g))/a(CCl4())  pCCl4
Then, at T = 298. K
Grxn = - RT ln K = - RT ln p = - (8.314 J/mol.K) (298. K) ln(0.168) = 4.42 kJ/mol
But
Grxn = Gf(CCl4(g)) - Gf(CCl4())
And so Gf(CCl4(g)) = Grxn + Gf(CCl4()) = 4.42 kJ/mol + (- 65.21 kJ/mol) = - 60.79 kJ/mol
2)
a) Using Raoult’s law as the basis for ideal behavior means
ai = pi/p*i = iXi, and so i = ai/Xi
For methanol
aM = (0.104)/(0.128) = 0.812
M = 0.812/0.700 = 1.161
For carbon tetrachloride
aC = (0.109)/(0.120) = 0.908
C = 0.908/0.300 = 3.028
b) For ideal mixing
Gmix(ideal)= nRT { XM ln XM + XC ln XC }
= (1.000 mol) (8.314 J/mol.K) (293. K) {0.700 ln(0.700) + 0.300 ln(0.300)}
= - 1488. J
c) For real mixing (as derived in class, or as can quickly be derived by following the procedure
used to find Gmix for ideal mixing) we have
Gmix(real)= nRT { XM ln aM + XC ln aC }
= (1.000 mol) (8.314 J/mol.K) (293. K) {0.700 ln(0.812) + 0.300 ln(0.908)}
= - 426. J
3)
a) Henry’s law is (in terms of molarity)
p = KH [CH4]
So
[CH4] = p/KH = 1.000 bar/(755. L/mol.bar) = 0.001325 mol/L
b) The solubility of methane in water may be written as
CH4(g)  CH4(aq)
K = (aCH4(aq))/(aCH4(g))  [CH4]/pCH4
= 0.001325
So
Grxn = - RT ln K = - (8.314 J/mol.K) (298. K) ln(0.001325) = + 16.42 kJ/mol
But
Grxn = Gf(CH4(aq)) - Gf(CH4(g))
And so Gf(CH4(aq)) = Grxn + Gf(CH4(g)) = 16.42 kJ/mol + (- 50.72 kJ/mol) = - 34.30 kJ/mol
4)
a) At T = 55. C and ZA = 0.12 there are two distinct liquid phases present in the system. This
remains the case until T = 60. C, at which point the system form a single liquid phase. This liquid phase
begins to boil at T = 72. C. Boiling continues until T = 80.5 C, at which point all of the liquid has been
converted into vapor. Additional heating will raise the temperature of the vapor to T = 90. C.
b) Point b is an azeotrope, since it is a mixture which has a single boiling temperature – the
definition of azeotrope.
c) 82. C
d) The two phases present are liquid and vapor. We have the following two relationships
ng + n = n
(ZA – YA) ng = (XA – ZA) n
where the second equation is the lever rule, with X A = 0.978, ZA = 0.800, and YA = 0.640.
From the second equation we get
ng = [(XA – ZA)/(ZA – YA)] n = [(0.978 – 0.800)/(0.800 – 0.640)] n = 1.1125 n
Substituting this result into the first equation gives
1.1125 n + n = 2.1125 n = n
n = n/2.1125 = (1.000 mol)/(2.1125) = 0.473 mol
and so
ng = n - n = (1.000 – 0.473) = 0.527 mol
5)
a) K =
(aHI(g))2
(aH2(g)) (aI2(g))
b) We will calculate both Grxn and Hrxn.
Grxn = 2 Gf(HI(g)) – [ Gf(H2(g)) + Gf(I2(g)) ]
= 2 (1.70) – [ 0.00 + 19.33 ] = - 15.93 kJ/mol
Hrxn = 2 Hf(HI(g)) – [ Hf(H2(g)) + Hf(I2(g)) ]
= 2 (26.48) – [ 0.00 + 62.44 ] = - 9.48 kJ/mol
For the equilibrium constant at T = 298. K
ln K = - Grxn/RT = - ( - 15930. J/mol)/(8.314 J/mol.K) (298. K) = 6.430
K = e6.430 = 620.
c) For K at 350. K we use
ln(K2/K1) = - (Hrxn/R) [ (1/T2) – (1/T1) ]
= - [( - 9480. J/mol)/(8.314 J/mol.K)] [ (1/350. K) – (1/298. K) ] = = 0.568
So
K2/K1 = e-0.568 = 0.566
K2 = 0.566 K1 = 0.566 (620.) = 351.