Chem 75 February 16, 2017 Exam 2 Solutions

Transcription

1 1. (6 + 6 pints) Tw quick questins: (a) The Handbk f Chemistry and Physics tells us, crrectly, that CCl 4 bils nrmally at 76.7 C, but its mlar enthalpy f vaprizatin is listed in ne place as 34.6 kj ml 1 and in anther as 31.9 kj ml 1. This handbk als says the CCl 4 vapr pressure is atm at 15.8 C. Which enthalpy value des this vapr pressure supprt? Let s stick t atm pressure units here: bils nrmally means P = 1 atm. We cnvert the temperatures t K units, and the Clausius-Clapeyrn equatin gives us the answer: ln atm 1.0 atm = ΔH vap R K K r ΔH vap = kj ml 1 (b) A chemist was given a small cylinder f gas by her clleague. The label n the cylinder read: P = 23.5 atm, V = L, 0.81 g f CF, but at that pint, a drp f acetne had bscured the label. The chemist knew that the gas was either CF 4 r CF 3 H. She used the ideal gas relatinship (with T = 300 K, the lab s temperature) and calculates V m = RT/P = 1.05 L ml 1, which let her calculate the mlar mass as (0.81 g)(1.05 L ml 1 )/(0.010 L) = 85 g ml 1. This is clser t the mlar mass f CF 4 (88 g ml 1 ) than t that fr CF 3 H (70 g ml 1 ), and CF 4 is what she assumed she has. Was she justified? Use the van der Waals equatin f state s CF 4 and CF 3 H parameters belw t find ut. a/atm L 2 ml 2 b/l ml 1 CF CF 3 H We knw the mass f the gas, and thus we can calculate the number f mles, n, f gas. We calculate n = 0.81 g/ 88 g ml 1 = mml fr CF 4 r n = 0.81 g/70 g ml 1 = mml fr CF 3 H. Next, we calculate the pressure the van der Waals equatin predicts fr each f these pssibilities. We find P = atm fr CF 4 (nt very clse t the bserved 23.5 atm) and P = atm fr CF 3 H, which is in very gd agreement! The sample must have been CF 3 H. 2. (3 pints each) Fr each pair f quantities belw, the ne n the left is either greater than (>), less than (<), r equal t (=) the ne n the right. Enter the apprpriate symbl (>, <, =) in the blanks prvided in the middle. (a) μ(h 2 O(s), T = 270 K) < μ(h 2 O(l), T = 270 K) Slid water is in equilibrium with liquid water at K. Belw this temperature, slid water (ice) is the stable phase. That is the phase with the smaller chemical ptential. (b) K fr I 2 (g) 2I(g) at 400 K < K fr I 2 (g) 2I(g) at 600 K

2 This reactin must be endthermic; a bnd is brken. Fr endthermic reactins, K increases with T. (c) G T P (Ar, 100 K)) > G T P (Ar, 300 K) ( G/ T) P = S, and S increases with T (fr anything). (d) ΔG f (N2 (g), T = 298 K) = ΔG f (H2 (g), T = 298 K) These species are bth elements in their mst stable frm at 298 K. Thus, the bth have ΔG f = 0. (e) van der Waals b cnst. fr He < van der Waals b cnst. fr SF 6 The b cnstant represents the mlar vlume f a real gas ccupied by the mlecules themselves. SF 6 is much bigger than He. (f) ΔG f mixing (ideal slutin) < zer Ideal slutins spntaneusly mix in all cmpsitins. (It s an entrpy thing!) Spntaneus means negative free energy change. 3. (12 pints) Slid cpper(ii) brmide, CuBr 2 (s), vaprizes via the fllwing reactin: 2CuBr 2 (s) 2CuBr(s) + Br 2 (g) Studies f the vapr pressure P f Br 2 (g) in equilibrium with CuBr 2 (s) and CuBr(s) shw that it fllws the expressin P = ( bar) e (11,642 K)/T 320 K T 380 K Find the equilibrium cnstant K, the reactin free energy change ΔG R, the reactin enthalpy change ΔH R, and the reactin entrpy change ΔSR fr this reactin at T = 350 K. At this temperature, 350 K, the expressin fr P tells us that P = P Br 2 = bar. But the reactin has an equilibrium cnstant expressin K = P Br 2 because the ther tw species in the reactin are pure slids (with unit activity). Thus, K = Next, we use ΔG R = RT ln K t find ΔGR = 35.6 kj ml 1. We then use dlnk dt = dlnp Br 2 = dt dt d ln e (11,642 K)/T = 11,642 K d(1/t) dt = 11,642 K T 2 = ΔH R RT 2

3 r ΔH R = (11,642 K)R = 96.8 kj ml 1. Finally, we have ΔS R = (ΔHR ΔGR )/T = 175 J ml 1 K 1 (which we nte is psitive, as expected, fr a reactin that prduces ne mle f gaseus prduct frm n gaseus reactants). 4. ( pints) A few mre quick questins: (a) Fr He(g) at 25 C, ne f the fur values in the table belw is 126 J ml 1 K 1. Fill in the table with numerical values fr all fur quantities. Mlar enthalpy f frmatin Mlar standard entrpy Mlar free energy f frmatin Cnstant pressure mlar heat capacity J ml 1 K 1 0 C P = 2.5R = 20.8 J ml 1 K 1 He is an ideal gas element. Thus, its frmatin enthalpy and free energy are bth zer, and its C P is 5R/2 = 2.5 R = 20.8 J ml 1 K 1. All that s left is the entrpy! (Dimensinal analysis wuld have tld yu that the value yu were given had t be either C P r S.) (b) The gas-phase reactin 3 H 2 (g) + N 2 (g) 2 NH 3(g) is held at equilibrium at a cnstant temperature. He(g) is added t the mixture while keeping the ttal pressure cnstant. When equilibrium is attained again, the amunt f NH 3 (g) will have (circle ne) increased decreased stayed the same If the ttal pressure is held cnstant, the vlume f the system must increase as we add helium! This perturbatin will drive the reactin back t reactants (t make mre mles f gas), meaning sme NH 3 has been lst. (c) Explain the trend in entrpies f fusin (i.e., f melting), ΔS fus. fr the five linear hydrcarbns, methane thrugh n-ctane, listed belw in J ml 1 K 1 units. methane 10.4 ethane 31.8 n-butane 34.6 n-hexane 73.3 n-ctane 95.9 This increase in ΔS fus as the hydrcarbns get lnger and lnger reflects the strng rder in the slids (lng mlecules packed like drinking straws in a bx) versus the ever-increasing freedm affrded t the mlecules in the liquid phase (where the mlecules are free t bend and wiggle arund).

4 5. (20 pints) Sketch in the graphs belw, as accurately as yu can, the qualitative behavir f the chemical ptential, µ, the entrpy, S, the enthalpy, H, and the heat capacity, C P, fr a pure substance raised in temperature at 1 atm pressure thrugh the nrmal biling pint, T vap. liquid gas liquid gas gas liquid S T vap T vap liquid gas liquid gas H C P T vap T vap Plt by plt: the μ(t) plt has the tw curves fr the tw phases in questin crss at the vaprizatin temperature, and the gas curve has the mre negative slpe since ( μ/ T) P = S and S(g) > S(l). The S(T) plt reflects the facts that (1) S always increases with T, (2) S increases mre rapidly with T fr the liquid because ( S/ T) P = C P /T and C P (l) > C P (g), and (3) there is a discntinuus jump up in S at the vaprizatin temperature reflecting the (psitive) entrpy change n vaprizatin: ΔS vap = ΔH vap /T vap. The H(T) plt has similarities t the S(T) with a discntinuus

5 jump up due t the enthalpy change n vaprizatin. The C P (T) plt reflects (1) the generally larger liquid heat capacity (cmpared t the gas), (2) the spike t an infinite heat capacity at the phase transitin temperature, and (3) the fairly cnstant value f heat capacity as temperature changes. 6. (10 pints) Pyridine (py) and acetne (ac) are cmpletely miscible as liquids in all prprtins near rm temperature. Their vapr pressures when pure are (at 300 K, the temperature yu may assume thrughut this prblem) P* py = 27 trr P* ac = 281 trr (a) The partial pressure f acetne ver a py/ac slutin is 17 trr in a slutin fr which x ac, the mle fractin f acetne in the liquid phase, is x ac = Hw can yu tell that this slutin is nt ideal? Ideal slutins fllw Rault s law fr bth cmpnents. If acetne fllwed Rault s law, its partial pressure wuld be P ac = x ac P* ac = trr = 14 trr which is nt the bserved vapr pressure f 17 trr. (b) While nt ideal, this slutin fllws Henry s law at and belw this acetne cncentratin. One can shw that in a binary slutin in the ideal dilute regin (which is what we have here), if ne cmpnent fllws Henry s law, the ther cmpnent must fllw Rault s law. With this in mind, tell me tw things: (i) What is the Henry s law cnstant fr acetne in pyridine? (ii) What is the ttal equilibrium pressure f a py/ac slutin with x ac = 0.05? (i) Henry s law (with Henry s law cnstant k H ) tells us P ac = x ac k H = 0.05k H = 17 trr r k H = (17 trr)/0.05 = 340 trr (ii) If acetne fllws Henry s law, then pyridine fllws Rault s law, making the ttal pressure equal t P = P ac + P py = 17 trr + x py P* py = 17 trr + (1 0.05)(27 trr) = 42.7 trr 7. ((18 pints) At 100 C, the vapr pressures f hexane and ctane are atm and atm. Slutins f these are ideal, i.e., they fllw Rault s law. One particular slutin is mixed and held at 100 C and 5.00 atm, where it is all in the liquid phase. The pressure is slwly reduced t 1.50 atm, where the system is fund t exist as a tw-phase (gas-liquid) system. (a) What is the liquid-phase hexane mle fractin?

6 Let s call hexane cmpnent 1. Then, Rault s law tells us hw the ttal pressure P is related t the liquid phase mle fractin x 1 and the pure cmpnent vapr pressures: P = P 2 * + P1 * P2 * x1 = atm + (0.466 atm atm)x 1 = 1.50 atm and we find x 1 = (b) What is the gas-phase hexane mle fractin? Nw we slve fr the gas phase cmpsitin mle fractin y 1 as fllws: P = * P 1 P2 * (2.416 atm)(0.466 atm) = = 1.50 atm P * 1 + P2 * P1 * y atm + (0.466 atm atm)y 1 which yields y 1 = (b) It is further fund that the rati f the ttal number f mles in the liquid phase, n(l), t the ttal number f mles in the gas phase, n(g), is What is the ttal system cmpsitin, i.e., the ispleth cmpsitin? We use the lever rule expressin t find the system ispleth cmpsitin mle fractin x: n(l) x x 1 = n(g) y 1 x r n(l) n(g) = 1.10 = y 1 x = x x x 1 x frm which we find x = Here s the full phase diagram: x x 1 y 1 P ctane mle fractin hexane