Answer each questin in the space prvided; use back f page if extra space is needed. Answer questins s the grader can READILY understand yur wrk; nly wrk n the exam sheet will be cnsidered. Write answers, where apprpriate, with reasnable numbers f significant figures. Yu may use nly the "Student Handbk," a calculatr, and a straight edge. 1. (20 Pints) Cmplete the sentence in the left clumn using the answers prvided in the right clumn. Where requested, explicitly write ut True r False s as t avid ambiguity. DO NOT WRITE IN THIS SPACE 1. Fr a multi-cmpnent, multi-regin system that allws the transfer f the different cmpnents between regins f the system, the f f an individual species is i thrughut the entire system at equilibrium. 2. The varius Fundamental Equatins f Thermdynamics are unrelated and prvide different ways f cnsidering the same infrmatin. True r False FALSE 3. One measure f the nn-ideality f a pure vapr r a mixture f vaprs is the b. 4. Fr gas-phase reactins f species described by the fllwing equatin f state: PV = nrt, the equilibrium cnstant accunts fr the e cntributins t the Gibbs Free Energy f reactin arising frm g f the individual gases. a. equal t b. fugacity cefficient c. spatially-dependent d. Cefficient f nnideality, C NI p. 1 /20 p. 2 /20 p. 3 /20 p. 4 /20 p. 5 /20 ============= p. 6 /15 (Extra credit) ============= TOTAL PTS /100 5. Fr ideal, dilute slutins, the Henry s Law standard state represents a real, stable state f a pure fluid at the relevant system cnditins f temperature, pressure, and cmpsitin. True r False FALSE e. entrpic f. chemical ptentials g. mixing h. less than i. the same and timeindependent
NAME: 2 2. (20 Pints) Cnsider the fllwing reactin at T = 298.15 K and P = 1 bar: H 2 (gas) + CO 2 (gas) = CO (gas) + H 2 O (gas) Starting with a mles f H 2 (gas) and b mles f CO 2 (gas), hw many mles f each f the fur cmpnents are present at equilibrium. Assume ideal gas behavir fr each species and standard pressure f 1 bar. Slutin: H 2 CO 2 H 2 O CO Initial mles a b 0 0 Mles @ equilibrium Mle fractin @ equilibrium a-n b- n n n a n a + b b n a + b n a + b n a + b At equilibrium, ΔG rxn = RT ln K P The standard mlar Gibbs free energy f reactin is given by: ΔG rxn =ν H2 ΔG frmatin,h 2 = ΔG frmatin,h 2 ΔG frmatin,co2 +ν CO2 ΔG frmatin,co 2 +ν CO ΔG frmatin,co +ν H2O ΔG frmatin,h 2O + ΔG frmatin,co = 0 ( 94.26 kcal ml ) + ( 32.81kcal kcal ) + ( 54.64 ml ml ) = 6.81 kcal kj = 28.51 ml ml Thus, the dimensinless equilibrium cnstant is: 28510 J ml = RT ln K P + ΔG frmatin,h 2O K P = e 28510 J ml RT = e 11.50144 =1.0x10 5 At equilibrium (nte all expnents are unity, s they are nt explicitly included here), K P =1.0x10 5 = p CO p H 2 O p H 2 p CO2 = x x CO H 2O n 2 = x H 2 x CO2 (a n)(b n) Scre fr Page
NAME: 3 K P =1.0x10 5 = p CO p H 2 O p H 2 p CO2 n 2 = K P n 2 (a + b)n + ab 0 = (K P 1)n 2 K P (a + b)n + K P ab = x x CO H 2O n 2 = x H 2 x CO2 (a n)(b n) = K P n 2 K P (a + b)n + K P ab n = K P (a + b) ± K 2 p(a + b) 2 4abK P (K P 1) 2(K P 1) If a =1ml and b=2ml, n = 3K P ± 9K 2 p 8K P (K P 1) 2(K P 1) = 3(0.00001) ± 9(0.00001)2 8(0.00001)(0.00001 1) 2(0.00001 1) = 0.0045ml If yu frget the quadratic frmula, ne can cnsider that Kp is small and make sme apprximatins. With small Kp, n will be small. We can retain nly certain terms in the quadratic plynmial: K P =1.0x10 5 = p CO p H 2 O p H 2 p CO2 = x CO x H 2O n 2 = x H 2 x CO2 (a n)(b n) n 2 (a)(b) n 2 abk P n abk P = 2(0.00001) = 0.0045ml Thus, small Kp apprximatin leads t essentially the same final equilibrium mles and the number is quite small as expected fr self-cnsistency. Scre fr Page
NAME: 4 3. (20 Pints) Determine the enthalpy f vaprizatin f methanl using the fllwing experimental data fr temperature versus equilibrium saturatin pressure (vapr pressure) f pure methanl. State any assumptins yu make in yur slutin f this prblem. Pressure (mm Hg) T (Celsius) 40 100 400 760 1520 3800 7600 15200 30400 45600 5.0 21.2 49.9 64.7 84.0 112.5 138.0 167.8 203.5 224.0 Slutin: Use Clausius-Clapeyrn equatin with assumptin f ideal gas and cnstant vaprizatin enthalpy (ver the given temperature range). ln P = -ΔH vap R 1 T + c Plt ln Pressure versus inverse temperature and perfrm linear regressin. Regressin Results: Number f bservatins = 10 Mean f independent variable = 0.002725033 Mean f dependent variable = 7.610287 Standard dev. f ind. variable = 0.0005397251 Standard dev. f dep. variable = 2.389071 Crrelatin cefficient = -0.9997924 Regressin cefficient (SLOPE) = -4425.54 Standard errr f cefficient = 31.8908 t - value fr cefficient = -138.7717 Regressin cnstant (INTERCEPT) = 19.67003 Standard errr f cnstant = 0.08842427 t - value fr cnstant = 222.4506 y = 19.67-4425.5 * x H vap = (-R)(slpe) = 36.8 kj/ml (see figure; circles = data; squares = regressin) Scre fr Page
NAME: 5 4. (20 Pints) The binary system acetnitrile(1)/nitrmethane(2) is sufficiently well-described as an ideal slutin (bth vapr and cndensed phase are ideal) fr cnsidering vapr-liquid equilibrium (VLE) (see figure). Fr this prblem, species 1 is acetnitrile and species 2 is nitrmethane. One can thus use Rault s expressin relating the mle fractins f species in the vapr t the cndensed phase mle fractins and individual species saturatin vapr pressures. Using this infrmatin, alng with the fllwing Antine relatins fr the temperature dependence f the individual species saturatin pressures, determine the equilibrium liquid and vapr cmpsitins fr a temperature f 75 Celsius and X 1 =0.6. Data: Antine equatins fr the vapr pressures f pure species. Temperature in Celsius and pressure in kilpascal (kpa). ln P 1 saturatin =14.2724 2945.47 T +224.0 ln P 2 saturatin =14.2043 2972.64 T +209.0 Slutin: First determine the vapr pressures f the tw species at the given temperature, T=75 Celsius = 348.15 K. P saturatin 1 = e 14.2724 2945.47 T +224.0 14.2724 2945.47 = e 75+224.0 = 83.2069kPa 2972.64 2972.64 P saturatin 14.2043 14.2043 T +209.0 75+209.0 2 = e = e = 41.9828kPa P Ttal = x 1 P saturatin 1 + x 2 P saturatin 2 = (0.6)(83.2069) + (0.4)(41.9828) = 49.92414 +16.79312 = 66.71726 saturatin x y 1 = 1 P 1 x 1 P saturatin 1 + x 2 P = 49.92414 saturatin 2 66.71726 = 0.748 y 2 =1 y 1 = 0.252 x 1 = 0.6 x 2 = 0.4 Scre fr Page
NAME: 6 5. (20 Pints). State which thermdynamic ptential reaches an extremum under the fllwing external cnstraints n a system; cnsider n external wrk ther than pressure-vlume wrk is pssible fr this prblem: A. Temperature, pressure, and system cmpsitin Gibbs Free Energy. Temperature, vlume, and system cmpsitin Helmhltz Free Energy C. Entrpy, vlume, and system cmpsitin Internal Energy D. Internal energy, vlume, and system cmpsitin Entrpy
NAME: 7 6. (15 nus Pints). It can be shwn that the partitin functin f an ideal gas f N diatmic mlecules in an external electric field, ε, is: Q = [q]n k with q = C T sinh µε N! µε k T Here, T is temperature, k is ltzmann s cnstant, µ is the diple mment f a single mlecule, and C is a cnstant independent f ε. The partitin functin, Q, relates t the Helmhltz Free Energy thrugh the fllwing equatin: A = k T ln Q = k T ln [q]n N! Using this infrmatin alng with the Fundamental Thermdynamic Relatin fr the ttal derivative f the Helmhltz Free energy: da = SdT PdV (Nµ ) dε where µ is the average diple mment f a mlecule in the directin f the external field, ε, shw that at cnstant temperature and vlume: µ = µ cth µε k T k T µε Slutin: Frm the relatin fr the ttal differential f A at cnstant vlume and temperature: da = SdT PdV (Nµ) dε = (Nµ ) dε thus A ε = Nµ µ = -1 A N ε That s it; simply a matter f ding sme differentiatin nw: Scre fr Page
NAME: 8 µ = -1 A N ε = k T ln Q N ε = k T N ln q N ε = -1 k T ln Q N ε k T ln N! N ε secnd term is zer since N! des nt depend n electric field µ = k T N ln q N ε = Nk T ln q N ε = k T ln q ε = k T lnε ε = k T ε ln sinh µε k T + k T ε µ + k T cth µε k T k T = µ cth µε k T k T µε Scre fr Page