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

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1 CHM 3400 Fundamentals o Physical Chemistry Final Exam April 23, 2012 There are eight problems on the exam. Do all o 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 = x 10 5 N/m 2 F = C/mol h = x J. s 1 ev = x J c = x 10 8 m/s 1 cm -1 = x m e = 9.11 x kg Thermodynamic data or several substances are given below, at T = 298. K and p = bar. The data may be useul in doing problems 1, 2, and 3. Substance M (g/mol) H (kj/mol) G (kj/mol) S (J/mol. K) C p,m (J/mol. K) Cl 2 (g) P(s) P 4 (g) PCl 3 (g) PCl 5 (g) (24 points) The temperature o mol o chlorine gas (Cl 2 (g)) is changed rom an initial value T i = K to a inal temperature T = K. The process is carried out reversibly at a constant pressure p = bar. For the conditions o the problem you may assume that chlorine gas behaves ideally, and that the value or the constant pressure molar heat capacity is independent o temperature. Find q, w, U, H, S syst, S surr, and S univ or the above process, or briely explain why there is insuicient inormation to ind the quantity requested. 2. (32 points) The element phosphorus is unusual in that the most stable orm o the element in the gas phase is P 4, a molecule with tetrahedral geometry. a) What is the rms average speed o a P 4 molecule in the gas phase at T = 298. K? Give your inal answer in units o m/s. b) What is the vapor pressure o solid phosphorus at T = 298. K? Give your inal answer in units o torr. c) The heat capacity o solid phosphorus in the vicinity o absolute zero is given by the expression C p,m = at 3 + bt 4 (2.1) where a = 1.04 x 10-3 J/mol. K 4 and b = x 10-6 J/mol. K 5. Find the value or H and S when the temperature o 1.00 mol o solid phosphorus is changed rom an initial value T = 0.0 K to a inal value T = 20.0 K reversibly and at constant pressure. 3. (30 points) In a system containing chlorine gas (Cl 2 ) an equilibrium is established between phosphorus trichloride (PCl 3 ) and phosphorus pentachloride (PCl 5 ) by the process PCl 3 (g) + Cl 2 (g) PCl 5 (g) (3.1) a) Give the expression or K or reaction 3.1 in terms o reactant and product activities. b) Give the expression or K or reaction 3.1 assuming ideal behavior or all reactants and products. c) Give the numerical value or K or reaction 3.1 at T = 298. K d) Give the numerical value or K or reaction 3.1 at T = 373. K. State any assumptions made in inding the value or K at this temperature.

2 4. (24 points) The density o a mixture o carbon monoxide (CO, M = g/mol) and propane (C 3 H 8, M = g/mol) was measured at T = 29.3 C and p = atm, and ound to be = g/l. a) Based on this above inormation, and assuming the gases obey the ideal gas law, ind X CO, the mole raction o CO in the gas mixture. b) How many rotational motions does a molecule o C 3 H 8 possess? c) How many vibrational motions does a molecule o C 3 H 8 posses? 5. (30 points) Hypochlorous acid (HOCl) plays an important role in stratospheric ozone chemistry. The ollowing question explores some o the properties o this molecule. a) Write the correctly balanced ormation reaction or HOCl(g). b) Using the table o bond enthalpies (Table 3.5, p 70 o Atkins) estimate the value or H, the enthalpy o ormation, or HOCl(g). c) The enthalpy o reaction or the process HOCl(g) + HCl(g) H 2 O(g) + Cl 2 (g) (5.1) is H rxn = kj/mol at T = 25.0 C. Based on this result and the data in the Appendix o Atkins, ind the value or H, the enthalpy o ormation, or HOCl(g). 6. (16 points) In a disproportionation reaction the same substance is simultaneously oxidized and reduced. The disproportionation reaction or Cu + ion is 2 Cu + (aq) Cu(s) + Cu 2+ (aq) (6.1) Using the table o standard cell potentials in the Appendix o Atkins, determine E rxn and K, the numerical value or the equilibrium constant or the above reaction, at T = 25.0 C. 7. (24 points) The initial concentration o a substance X is mol/l. Ater s, the concentration o X has decreased to mol/l. Calculate the value or concentration or X ater 3600 s (1 hour) assuming a) That the disappearance o X ollows irst order homogeneous kinetics, that is d[x]/dt = - k [X] (7.1) b) That the disappearance o X ollows second order homogeneous kinetics, that is d[x]/dt = - k [X] 2 (7.2) 8. (20 points) An argon ion laser provides monochromatic light at a wavelength = nm. a) How many photons are produced by the above laer when it emits a pulse o monochromatic light with total energy E = 4.45 x 10-3 J? b) A low energy pulse o light rom the above laser is used to illuminate a metal surace in a photoelectric eect experiment. The work unction or the metal is = 3.18 ev. Will electrons be produced? Justiy your answer. I your answer is yes, also give the maximum kinetic energy and speed o electrons that are produced.

3 Solutions. 1) For an ideal gas H = i n C p,m dt U = i n C V,m dt C p,m C V,m = R, so C V,m = C p,m R We are told that C p,m is constant over the temperature range o the problem, so H = i n C p,m dt = n C p,m (T T i ) U = i n C V,m dt = n C V,m (T T i ) C V,m = (33.91 J/mol. K) (8.314 J/mol. K) = J/mol. K So H = (1.00 mol) (33.91 J/mol. K) (350.0 K K) = J U = (1.00 mol) (25.60 J/mol. K) (350.0 K K) = J The process is carried out under conditions o constant pressure, and so q = H = J From the irst law U = q + w and so w = U q = (2560. J) (3391. J) = J Since the process is reversible, S univ = 0. Since S univ = S syst + S surr = 0, it ollows that S surr = - S syst S syst = i (dq) rev /T = i (nc p,m /T) dt = nc p,m ln(t /T i ) So S syst = (1.00 mol) (33.91 J/mol. K) ln(350.k/250.k) = J/K S surr = J/K 2) a) c rms = (3RT/M) 1/2 = [3(8.314 J/mol. K)(298. K)/( x 10-3 kg/mol)] 1/2 = 245. m/s b) We can consider the vaporization reaction as the process 4 P(s) P 4 (g) K = (a P4(g) ) p(p 4 (g)) (a P(s) ) 4 So the value o the equilibrium constant is equal to the vapor pressure o solid phosphorus (in bar). To ind K we need to ind G rxn or the above process. But G rxn = G (P 4 (g)) 4 G (P(s)) = kj/mol 4 (0) = kj/mol

4 Since ln K = - G rxn = - ( J/mol) = RT (8.314 J/mol. K) (298. K) K = e = 5.28 x 10-5 p vap = (5.28 x 10-5 bar) 1 atm 760. torr = torr bar 1 atm c) For a constant pressure reversible heating H = i (n C p,m ) dt = n i (at 3 + bt 4 ) dt = n {(at 4 /4) + (bt 5 /5)} Ti T S = i (n C p,m /T) dt = n i (at 2 + bt 3 ) dt = n {(at 3 /3) + (bt 4 /4)} Ti T Since T i = 0. K, we may write the above as H = n {(at 4 /4) + (bt 5 /5)} = (1.00 mol) {[(1.04 x 10-3 J/mol. K 4 )(20. K) 4 /4) + [(- 8.0 x 10-6 J/mol. K 5 )(20. K) 5 /5} = J + ( J) = 36.5 J S = n {(at 3 /3) + (bt 4 /4)} = (1.00 mol) {[(1.04 x 10-3 J/mol. K 4 )(20. K) 3 /3) + [(- 8.0 x 10-6 J/mol. K 5 )(20. K) 4 /4} = 2.77 J/K + ( J/K) = 2.45 J/K 3) a+b) K = (a PCl5 ) = (p PCl5 ) (a PCl3 )(a Cl2 ) (p PCl3 )(p Cl2 ) We calculate the ollowing to help in doing parts c and d G rxn = [ G (PCl 5 (g)) [ G (PCl 3 (g))+ G (Cl 2 (g))] = [( )] [( ) + (0.0)] = kj/mol H rxn = [ H (PCl 5 (g)) [ H (PCl 3 (g))+ H (Cl 2 (g))] = [( )] [( ) + (0.0)] = kj/mol c) At T = 298. K S rxn = [S (PCl 5 (g)) [S (PCl 3 (g))+ S (Cl 2 (g))] = [(364.6)] [(311.78) + (223.07)] = J/mol. K ln K = - G rxn = - ( J/mol) = RT (8.314 J/mol. K) (298. K) K = e = 3.3 x 10 6

5 d) At T = 373. K, and assuming H rxn and S rxn are independent o temperature G rxn = H rxn - T S rxn = ( kj/mol) ( x 10-3 kj/mol. K)(373. K) = kj/mol ln K = - G rxn = - ( J/mol) = 7.87 RT (8.314 J/mol. K) (373. K) K = e 7.87 = 2.6 x ) a) pv = nrt, and so V = p = (1.045 atm) = mol/l n RT ( L. atm/mol. K)(302.4) Since we know the density o the gas mixture, we can now ind the average molecular mass M ave = = (1.324 g/l) = g/mol (V/n) ( mol/l) But M ave = X CO M CO = X C3H8 M C3H8 Since X CO + X C3H8 = 1, X C3H8 = 1 X CO Mave = X CO M CO + (1 X CO ) M C3H8 = X CO M CO + M C3H8 X CO M C3H8 Solving or X CO gives X CO = (M C3H8 M ave ) = ( ) = (M C3H8 M CO ) ( ) b) Molecule is non-linear, so there are 3 rotations. c) Molecule is non-linear, and so there are 3(11) 6 = 27 bibrations. 5) a) Formation reaction is ½ H 2 (g) + ½ O 2 (g) + ½ Cl 2 (g) HOCl(g) b) We may estimate H rxn by keeping track o the bonds broken and ormed bonds broken bonds ormed ½ (H-H) = ½ (436) 1 (O-H) = 1 (463) ½ (O=O) = ½ (497) 1 (O-Cl) = 1 (203) ½ (Cl-Cl) = ½ (242) kj/mol kj/mol H rxn (bonds broken) (bonds ormed) = (587.5 kj/mol) (666.0 kj/mol) =- 78. kj/mol

6 c) H rxn = [ H (H 2 O(g)) + H (Cl 2 (g))] [ H (HOCl(g)) + H (HCl(g))] So H (HOCl(g))= [ H (H 2 O(g)) + H (Cl 2 (g))] [ H rxn + H (HCl(g))] = [( ) + (0.)] [(- 75.0) + ( )] = kj/mol 6) ox Cu + (aq) Cu 2+ (aq) + e - E = v red Cu + (aq) + e - Cu(s) E = 0.52 v net 2 Cu + (aq) Cu 2+ (aq) + Cu(s) E rxn = 0.36 v From reworking the Nernst equation, we may say ln K = FE rxn = (1) ( C/mol) (0.36 v) = RT (8.314 J/mol. K) (298. K) K = e = 1.23 x ) a) For a irst order reaction ln([a] 0 /[A] t ) = kt k = (1/t) ln([a] 0 /[A] t ) = (1/1000.) ln(0.287/0.185) = 4.39 x 10-4 s -1 Ater s, the concentration will be [A] t = [A] 0 e -kt = (0.287 mol/l) exp[-(4.39 x 10-4 s -1 )/(3600. s)] = mol/l b) For a irst order reaction [ (1/[A] t ) - (1/[A] 0 ) ] = kt k = (1/t) [ (1/[A] t ) - (1/[A] 0 ) ] Ater s, the concentration will be = (1/1000. s) [ (1/0.185 mol/l) - (1/0.287 mol/l) ] 1.92 x 10-3 L/mol. s [A] t = [A] 0 = (0.287 mol/l) (1 + kt[a] 0 ) [ 1 + (1.92 x 10-3 L/mol.s)(3600. s)(0.287 mol/l)] = mol/l 8) a) The energy o one photon rom the laser is E = hc = (6.626 x J s)(2.998 x 10 8 m/s) = x J (488.0 x 10-9 m) Since the energy o the laser pulse is 4.45 x 10-3 J, the number o photons produced is # photons = 4.45 x 10-3 J = 1.09 x photons x J/photon

7 b) For the photoelectric eect E = E K,max + or E K,max = E - where a negative value or E K means no electrons are produced. The work unction or the metal is = 3.18 ev (1.602 x J/eV) = x J The energy o one photon o light is E = x J I we insert these values into the equation or E K,max, we get a negative number, which is not physically possible. What this means is that electrons will not be produced by light at this wavelength.

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

CHM 3400 Fundamentals o Physical Chemistry First Hour Exam There are six problems on the exam. Do all o the problems. Show your work R = 0.08206 L. atm/mole. K N A = 6.022 x 10 23 R = 0.08314 L. bar/mole.