Chem 116 POGIL Wrksheet - Week 3 - Slutins Intermlecular Frces, Liquids, Slids, and Slutins Key Questins 1. Is the average kinetic energy f mlecules greater r lesser than the energy f intermlecular frces f attractin in (a) slids, (b) liquids, and (c) gases? (a) In slids, kinetic energy is less than intermlecular energy. (b) In liquids, kinetic energy is less than intermlecular energy, but the disparity is less than in slids. (c) In gases, the kinetic energy is much greater than the energy f intermlecular frces f attractin. 2. Why des increasing the temperature cause a substance t change in successin frm a slid t a liquid t a gas? Kinetic energy increases with temperature. As the kinetic energy rises with temperature, the intermlecular frces f attractin are vercme by increasing mlecular mtin. 3. Why d substances with high surface tensin als tend t have high viscsities? Liquids with strnger intermlecular frces f attractin hld the mlecules clser tgether, which causes strnger surface tensin and greater resistance t flwing (viscsity). 4. Why d surface tensin and viscsity decrease with increasing temperature? The increased kinetic energy with rising temperature vercmes the chesive intermlecular frces f attractin. 5. Name the kind r kinds f intermlecular frces that must be vercme t cnvert the fllwing frm liquid r slid t gas: (a) Br, (b) CH OH, (c) CO, (d) HCN, (e) NH 2 3 2 3 (a) Br 2 Lndn dispersin (b) CH3OH Lndn dispersin, diple-diple, hydrgen bnding (c) CO 2 Lndn dispersin (CO 2 is linear and therefre nnplar.) (d) HCN Lndn dispersin, diple-diple (HCN is linear but plar.) (e) NH Lndn dispersin, diple-diple, hydrgen bnding 3 6. Nrmal alkanes are hydrcarbns with unbranched carbn chains, having a general frmula CnH 2n+2. At rm temperature, ethane, C2H 6, is a gas; hexane, C6H 14, is a liquid; and ctadecane, C18H 38, is a slid. Describe the intermlecular frces present in each substance and explain the differences in their rm-temperature phases.
All are nnplar and therefre nly have Lndn dispersin frces. Hwever, Lndn dispersin frces rise with mlecular weight, as the numbers f electrns increase, which in turn cause the plarizabilities t increase. Thus, the rder f increasing intermlecular frces is C2H 6 < C6H 14 < C18H 38. As the Lndn dispersin frces increase the tendency t be in a cndensed phase increases. 7. Arrange the fllwing in rder f increasing biling pint: The cmpunds that have -OH grups will have hydrgen bnding and therefre much higher biling pints than thse that d nt. Within the tw grups (hydrgen bnding r n hydrgen bnding), the rdering fllws increasing Lndn dispersin frces (increasing mlecular weight) and additinal hydrgen bnding. Nte that the last cmpund, ethylene glycl, has twice as many sites fr hydrgen bnding per mlecule as the tw alchls, CH OH and C H OH. The bserved biling pint ( C) is shwn fr each cmpund. 3 2 5 8. Hw much heat is required t heat 10.0 g f ice at -5.00 C t becme liquid water at +7.00 C? In this temperature range, the heat capacity f H2O(s) is 37.7 J/ml K, and the heat capacity f H O(l) is 75.8 J/ml K. The mlar heat f fusin f ice is 6.01 kj/ml. 2 Calculate the heat t warm the ice, then the heat t melt the ice, and finally the heat t warm the liquid water. The sum is the ttal heat required. -1 ml H2O = 10.0 g/18.02 g ml = 0.555 ml Heat ice t 0 C: q 1 = CpÄT = (37.7 J/ml K)(0.555 ml)(5.00 K) = 105 J Melt ice at 0 C: q 2 = ÄH melt mles = (6.01 kj/ml)(0.555 ml) = 3.34 kj = 3340 J Heat liquid t 7 C: q = C ÄT = (75.8 J/ml K)(0.555 ml)(7.00 K) = 294 J 3 p Ttal heat: q = q + q + q = 105 J + 3340 J + 294 J = 3739 J = 3.74 kj t 1 2 3
9. Explain hw each f the fllwing affects the vapr pressure f a liquid: (a) the vlume f the liquid, (b) the vlume f the cntainer, (c) the surface area f the liquid, (d) the temperature, (e) intermlecular frces f attractin, (f) the density f the liquid. (a) N effect (b) N effect (c) N effect (d) Higher vapr pressure at higher temperature (e) Lwer vapr pressure with strnger intermlecular frces f attractin (f) Density is mass per unit vlume. It tends t increase as mlecular weight increases. Lndn dispersin frces als increase with mlecular weight, and this wuld cause a decrease in vlatility, resulting in lwer vapr pressure. Thus, denser liquids tend t have lwer vapr pressures. 10. Describe the phases and/r phase transitins experienced by CO 2 under the fllwing cnditins: (a) Heating frm 100 C t 30 C at 1.0 atm Slid t gas, with a nrmal sublimatin pint at 78 C. (b) Heating frm 100 C t 50 C at 70 atm Slid t liquid t gas (c) A sample at 35 C and 100 atm A pint abve the critical pint, s a supercritical fluid (d) A sample at 50 C and 6.0 atm A pint in the liquid range, just abve the triple pint 11. Des carbn dixide have a nrmal biling pint? Explain. The nrmal biling pint is the temperature at which the vapr pressure abve a liquid reaches exactly 1 atm. At 1 atm there is n liquid-vapr equilibrium fr CO 2, s it des nt have a nrmal biling pint. 12. Describe the cnditins under which liquid carbn dixide bils. At a pressure and temperature just abve the triple pint ( 57 C, 5.1 atm), liquid CO 2 wuld exist. This culd be biled by raising the temperature t achieve a pint n the liquid-vapr line. 13. Identify the principal type f slute-slvent interactin that is respnsible fr frming the fllwing slutins: (a) KNO 3 in water; (b) Br 2 in benzene, C6H 6; (c) glycerl, CH 2(OH)CH(OH)CH2OH, in water; (d) HCl in acetnitrile, CH3CN [HCl des nt frm ins in CH CN]. 3 (a) in-diple (b) Lndn dispersin (c) hydrgen bnding (d) diple-diple
14. Fr the fllwing carbxylic acids, predict whether slubility will be greater in water r carbn tetrachlride, and give yur reasning: (a) acetic acid, CH3CO2H, (b) stearic acid, CH (CH ) CO H. 3 2 16 2 (a) Acetic acid s -OH grups make hydrgen bnding pssible, which is cmpatible with slvent water. Carbn tetrachlride has nly Lndn dispersin frces, which are less cmpatible. Therefre, acetic acid is mre sluble in water than carbn tetrachlride. (b) Stearic acid has a very lng chain and much higher Lndn dispersin frces than acetic acid. This makes it mre cmpatible with carbn tetrachlride, despite the ptential fr hydrgen bnding (which is largely mitigated by the lng chain getting in the way). Therefre, stearic acid is mre sluble in carbn tetrachlride than water. 15. Hexane (C6H 14) and heptane (C7H 16) are miscible in all prprtins with ÄH sln 0. (a) Why are these tw liquids miscible with each ther? They have very similar Lndn dispersin frces, wing t their similar mlar masses. (b) Why is ÄH sln 0 fr this pair f liquids? The intermlecular frces f attractin in the neat liquids are s similar t each ther that little change ccurs n mixing. It is the change in intermlecular attractin strength that is principally respnsible fr the sense and magnitude f ÄH. (b) Why d they spntaneusly frm slutins, given that ÄH sln 0? Mixing is a mre disrdered state than exists in the separate neat liquids. The increase in entrpy is the driving factr in making slutin frmatin spntaneus in this case. sln -3 16. The slubility f N 2 at p(n 2) = 1 atm is 1.75 x 10 g/100 ml f water. What is the slubility in water at an air pressure f 2.51 atm, the pressure at 50 ft belw the surface f the water? Air is 78.1 vl-% N 2. [Hint: What is the partial pressure f N 2(g) when the air pressure is 2.51 atm?] Frm Daltn's Law f Partial Pressures, at air pressure f 2.51 atm, the partial pressure f N 2 is = (0.781)(2.51 atm) = 1.96 atm Frm Henry's Law, the slubility is
17. Calculate the mlality f ethanl, C2H5OH (m.w. = 46.06) in a slutin prepared by disslving 5.00 g f ethanl in 25.00 g f water. 18. Calculate the ttal mlality f all ins in a slutin prepared by disslving 20.0 g f (NH ) SO in 95.0 g f water. [f.w. (NH ) SO = 132 u] 4 2 4 4 2 4 (NH ) SO (s) 2NH (aq) + SO + 2 4 2 4 4 4 (aq) 19. Cnsider a 2.00 m slutin f sugar in water at 25.00 C. (a) What is the value f the mle fractin f water in this slutin? [Hint: Imagine that the slutin was made up with exactly 1 Kg f water.] (m.w. H O = 18.02 u) 2 T calculate mle fractin, we need the numbers f mles f water and f sugar. If we assume that exactly 1000 g f water were used, then we already knw that the slutin cntains 2.00 mles f sugar. All we need is the number f mles in 1000 g f water, and then we can calculate (H O). ml sugar = 2.00 ml sugar 2 ml H2O = (1000 g H2O)(1 mle H2O/18.02 g H2O) = 55.49 ml H2O (H2O) = 55.49 ml/(55.49 + 2.00) ml = 55.49/57.49 = 0.9652
(b) Calculate the vapr pressure abve a 2.00 m slutin f sugar in water at 25.00 C, given that the vapr pressure f pure water at this temperature is 23.76 mm Hg. P sln = (H2O) P (H2O) = 0.9652 23.76 mm Hg = 22.93 mm Hg 20. Calculate the expected vapr pressure abve a 2.00 m slutin f Na2SO 4 in water at 25.00 C. Cmpare this result t what yu fund in part a f the preceding Key Questn. We must take accunt f the dissciatin f Na SO (s): Na SO (s) 2Na (aq) + SO 2 4 + 2 2 4 4 (aq) Again, assume a slutin prepared with exactly 1000 g f water. ml ins = 3 ml Na2SO 4 = 3 2.00 ml = 6.00 ml (H2O) = 55.49 ml/(55.49 + 6.00) ml = 55.49/61.49 = 0.9024 P sln = (H2O) P (H2O) = 0.9024 23.76 mm Hg = 21.44 mm Hg The amunt by which the vapr pressure f water has been lwered is almst three times greater in the Na SO slutin, cmpared t the sugar slutin. 2 4 21. What are the partial pressures and ttal vapr pressure abve a slutin at 20.0 C made by mixing 12.5 g benzene (C6H 6) with 44.2 g tluene (C6H5CH 3). At 20.0 C, P (C6H 6) = 74.7 trr and P (C H CH ) = 22.3 trr. [m.w. C H = 78.11; m.w. C H CH = 92.14] 6 5 3 6 6 6 5 3 ml C6H 6 = (12.5 g C6H 6)(ml C6H 6/78.11 g C6H 6) = 0.160 ml C6H6 ml C6H5CH 3 = (44.2 g C6H5CH 3)(ml C6H5CH 3/92.14 g C6H5CH 3) = 0.480 ml C6H5CH3 ttal mles = 0.160 ml + 0.480 ml = 0.640 ml (C6H 6, sln) = 0.160 ml/0.640 ml = 0.250 (C6H5CH 3, sln) = 1 0.250 = 0.750 P(C6H 6) = (0.250)(74.5 trr) = 18.7 trr P(C6H5CH 3) = (0.750)(22.3 trr) = 16.7 trr P t = (18.7 + 16.7) trr = 35.4 trr
22. In terms f mle fractins, what is the cmpsitin f the vapr abve the previusly described benzene-tluene mixture? We previusly fund P(C6H 6) = (0.250)(74.5 tr) = 18.7 trr P(C6H5CH 3) = (0.750)(22.3 trr) = 16.7 trr P t = (18.7 + 16.7) trr = 35.4 trr Using these values, we can calculate the mle fractins in the vapr as fllws: (C6H 6, vap) = 18.7 trr/35.4 trr = 0.528 (C6H5CH 3, vap) = 16.7 trr/35.4 trr = 1 0.528 = 0.473 Ntice that mre vlatile benzene has an increased mle fractin in the vapr. 23. Pure benzene has a freezing pint f 5.5 C and a biling pint f 80.1 C. What are the expected freezing pint and biling pint fr a 0.15 m slutin f a nnvlatile slute in benzene? Fr benzene, K = 5.12 C/m and K = 2.53 C/m. f sln ÄT f = (5.12 C/m)(0.15 m) = 0.77 C T f = T f ÄT f = (5.5 0.77) C = 4.7 C sln ÄT b = (2.53 C/m)(0.15 m) = 0.38 C T b = T b + ÄT b = (80.1 + 0.38) C = 80.5 C b 24. When 45.0 g f an unknwn nnvlatile nnelectrlyte is disslved in 500.0 g f water, the resulting slutin freezes at -0.930 C. What is the mlar mass f the unknwn substance? K = 1.86 C/m fr water. f ÄT f = 0.930 C m = ÄT f/k f = 0.930 C/1.86 C/m = 0.500 m = 0.500 ml X/kg H2O 25. What is the smtic pressure f a 0.100 M glucse slutin in trr at 25.0 C? ð = MRT = (0.100 ml/l)(0.0821 L atm/k ml)(298 K) = 2.45 atm (760 trr atm ) -1 = 1860 trr
26. Sea water is apprximately 0.60 M NaCl. What is the minimum applied pressure that must be exceeded t achieve water purificatin by reverse smsis at 25 C? We must use the mlarity f ins, nt the stated mlarity f NaCl (analytical cncentratin f NaCl). + NaCl(s) Na (aq) + Cl (aq) mlarity f ins = 2 C NaCl = 2 0.60 M = 1.20 M ð = MRT = (1.20 ml/l)(0.0821 L atm/k ml)(298 K) = 29.36 atm = 29.4 atm