There are five problems on the exam. Do all of the problems. Show your work.

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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 N A = 6.022 x 10 23 R = 0.

1 CHM 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 = L. atm/mole. K N A = x R = L. bar/mole. K 1 L. atm = J R = J/mole. K 1 atm = bar 1. (18 points) Carbon tetrachloride (CCl 4) 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* = K and H vap = 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(ccl 4( )) = kj/mol at T = 298. K. Using this information and the results from part a of this problem find the value for G f(ccl 4(g)), the free energy of formation of carbon tetrachloride vapor, at T = 298. K. 2. (24 points) Methanol (CH 3OH) and carbon tetrachloride (CCl 4) 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 X M = 0.700, it is found that p M = bar and p C = bar. The vapor pressures of the pure liquids at this temperature are p M* = bar and p C* = bar. a) What are a M, M, a C, 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 G mix(ideal) the free energy of mixing for the formation of mole of a liquid solution with X M = from pure methanol and pure carbon tetrachloride and assuming ideal mixing? c) What is G mix(real) actual value for the free energy of mixing for the formation of mole of a liquid solution with X M = from pure methanol and pure carbon tetrachloride, based on the above data? 3. (16 points) The Henry s law constant for methane (CH 4) in water is K H = 755 L. bar/mol at T = 298. K. a) What is the equilibrium concentration of dissolved methane in water when p(ch 4) = bar and T = 298. K (see figure below)? b) Given that G f(ch 4(g)) = kj/mol at T = 298. K, and the data above, find the value for G f(ch 4(aq)), the free energy of formation for aqueous methane, at T = 298. K

2 4. (22 points) The phase diagram given below is for two partially miscible liquids A and B in equilibrium with their vapor (at p = atm), and may be of use in answering the following questions. a) Describe what happens in a closed system with Z A = when it is heated from an initial temperature T i = 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 mole of a system with Z A = 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 (I 2) will chemically react to form hydrogen iodide (HI). The equation corresponding to this reaction is H 2(g) + I 2(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.

3 Solutions. 1) a) One form of the Clausius-Clapeyron equation is ln(p 2/p 1) = - ( H vap/r) [ (1/T 2) (1/T 1) ] = - [( J/mol)/(8.314 J/mol. K)] [ (1/298. K) (1/349.9 K)] = So p 2/p 1 = e = p 2 = p 1 = (1 atm) = atm (equal to 126. torr, or equal to bar) b) If we write the vaporization reaction as CCl 4( ) CCl 4(g) K = (a CCl4(g))/a( CCl4( )) p CCl4 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(ccl 4(g)) - G f(ccl 4( )) And so G f(ccl 4(g)) = G rxn + G f(ccl 4( )) = 4.42 kj/mol + ( kj/mol) = kj/mol 2) a) Using Raoult s law as the basis for ideal behavior means For methanol a i = p i/p* i = ix i, and so i = a i/x i a M = (0.104)/(0.128) = M = 0.812/0.700 = For carbon tetrachloride a C = (0.109)/(0.120) = C = 0.908/0.300 = b) For ideal mixing G mix(ideal)= nrt { X M ln X M + X C ln X C } = (1.000 mol) (8.314 J/mol. K) (293. K) {0.700 ln(0.700) ln(0.300)} = J c) For real mixing (as derived in class, or as can quickly be derived by following the procedure used to find G mix for ideal mixing) we have G mix(real)= nrt { X M ln a M + X C ln a C } = (1.000 mol) (8.314 J/mol. K) (293. K) {0.700 ln(0.812) ln(0.908)} = J

4 3) a) Henry s law is (in terms of molarity) p = K H [CH 4] So [CH 4] = p/k H = bar/(755. L/mol. bar) = mol/l b) The solubility of methane in water may be written as CH 4(g) CH 4(aq) K = (a CH4(aq))/(a CH4(g)) [CH 4]/p CH4 = So But G rxn = - RT ln K = - (8.314 J/mol. K) (298. K) ln( ) = kj/mol G rxn = G f(ch 4(aq)) - G f(ch 4(g)) And so G f(ch 4(aq)) = G rxn + G f(ch 4(g)) = kj/mol + ( kj/mol) = kj/mol 4) a) At T = 55. C and Z A = 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 n g + n = n (Z A Y A) n g = (X A Z A) n where the second equation is the lever rule, with X A = 0.978, Z A = 0.800, and Y A = From the second equation we get n g = [(X A Z A)/(Z A Y A)] n = [( )/( )] n = n Substituting this result into the first equation gives n + n = n = n n = n/ = (1.000 mol)/(2.1125) = mol and so n g = n - n = ( ) = mol

5 5) a) K = (a HI(g)) 2 (a H2(g)) (a I2(g)) b) We will calculate both G rxn and H rxn. G rxn = 2 G f(hi(g)) [ G f(h 2(g)) + G f(i 2(g)) ] = 2 (1.70) [ ] = kj/mol H rxn = 2 H f(hi(g)) [ H f(h 2(g)) + H f(i 2(g)) ] = 2 (26.48) [ ] = kj/mol For the equilibrium constant at T = 298. K ln K = - G rxn/rt = - ( J/mol)/(8.314 J/mol. K) (298. K) = K = e = 620. c) For K at 350. K we use ln(k 2/K 1) = - ( H rxn/r) [ (1/T 2) (1/T 1) ] So K 2/K 1 = e = = - [( J/mol)/(8.314 J/mol. K)] [ (1/350. K) (1/298. K) ] = = K 2 = K 1 = (620.) = 351.

There are five problems on the exam. Do all of the problems. Show your work

CHM 3400 Fundamentals of Physical Chemistry Second Hour Exam March 8, 2017 There are five problems on the exam. Do all of the problems. Show your work R = 0.08206 L atm/mole K N A = 6.022 x 10 23 R = 0.08314