University Chemistry Quiz 4 2014/12/11 1. (5%) Wht is the dimensionlity of the three-phse coexistence region in mixture of Al, Ni, nd Cu? Wht type of geometricl region dose this define? Strtegy: Use the Gibbs phse rule (Eqution 5.3). How mny components re present? : The Gibbs phse rule (Eqution 5.3) is f c r + 2 There re three components present (Al. Ni nd Cu), so c 3. We re interested in three-phse coexistence, so r 3. The number of degrees of freedom (f) is then f 3 3 + 2 2 The number of degrees of freedom is equl to the dimensionlity of the phse-coexistence region, so the region of three-phse coexistence forms two-dimensionl surfce of the overll phse digrm. 2. (5%) Without referring to tble, select from the following list the gs tht hs the lrgest vlue of b in the vn der Wls eqution: CH 4, O 2, H 2 O, CCl 4, or Ne. CCl 4 will hve the lrgest effective moleculr size of the list becuse of the four Cl toms, which hve lrger tomic rdius thn ny of the elements mking up the other compounds. 3. (5%) The boiling point nd freezing point of sulfur dioxide t 1 br re -10 C nd -72.7 C, respectively. The triple point is t -75.5 C nd 1.67 x 10-3 br, nd its criticl point is t 157 C nd 79 br. On the bsis of this informtion, drw rough sketch of the phse digrm of SO 2.
79 br 79 br, 157 C P 1.00 br solid liquid 0.00167 br 72.7 C vpor 75.5 C 10 C 157 C T 4. (5%) To wht temperture must He toms be cooled so tht they hve the sme rms speed s O 2 t 25 C? (Hint: You do not hve to clculte the vlue of the rms speeds to solve this problem.) From Eqution 5.35 we hve u rms 3 M Equting u rms for He nd O 2, we hve u rms (He) u rms (O 2 ) 3 He M He 3 O2 M O2 Squring, cnceling like terms nd rerrnging gives T He T O2 M He M O2 So 4.003 g mol 1 T He (298.15 K) 37.3 K 1 32.00 g mol
5. (5%) Which of the following molecules hs the lrgest vlue: CH 4, F 2, C 6 H 6, or Ne? Explin your nswer in terms of intermoleculr forces. The vlue of indictes how strongly molecules of given type of gs ttrct one nother. C 6 H 6 hs the gretest intermoleculr ttrctions due to its lrger size compred to the other choices. Therefore, it hs the lrgest vlue. 6. (5%) Clculte the verge trnsltionl kinetic energy for n N 2 molecule nd for 1 mole of N 2 molecules t 20 C. From Eqution 5.34, the verge kinetic energy per molecule is kinetic energy per molecule 3k T B 2 For n N 2 molecule t 20ºC, we hve 23 3(1.38065 10 kinetic energy/molecule For mole of N 2 molecules: J K 1 )(273.15 + 20)K 2 6.07 10 21 J kinetic energy/mol (6.07 10 21 J/molecule)(6.022 10 23 molecules mol 1 ) 3.65 10 3 J mol 1 3.65 kj mol 1 7. (5%) Using the Lennrd-Jones model (Eqution 4.12), plot on the sme grph the potentil energy of interction between () two Ar toms nd (b) two Xe toms. If you knew nothing bout these two gses, wht could you conclude bout reltive physicl properties of Ar nd Xe bsed on this plot? ( Eqution 4.12: 44 σ r 11 σ r 6 )
The potentil for the xenon hs more negtive potentil well indicting tht xenon-xenon intermoleculr interctions re stronger thn rgon-rgon interctions nd tht xenon should hve higher boiling point. The xenon potentil become positive t lrge r, indicting tht xenon toms re lrger thn rgon toms. 8. (5%) Nme the kinds of intermoleculr interctions tht must be overcome in order to () boil liquid mmoni, (b) melt solid phosphorus (P 4 ), nd (c) dissolve CsI in liquid HF. () Boiling liquid mmoni requires breking hydrogen bonds between molecules. Dipole dipole nd dispersion forces must lso be overcome. (b) P 4 is nonpolr molecule, so the only intermoleculr forces re of the dispersion type. (c) CsI is n ionic solid. To dissolve in ny solvent ion ion interprticle forces must be overcome. To dissolve something in HF, some of the hydrogen bonds, dipole-dipole nd dispersion forces between HF molecules must lso be overcome. 9. (5%) Clculte the number of spheres tht would be found within smple cubic, body-centered cubic, nd fce-centered cubic cells. Assume tht the spheres re the sme. A corner sphere is shred eqully mong eight unit cells, so only one-eighth of ech corner sphere "belongs" to ny one unit cell. A fce-centered sphere is divided eqully between the two unit cells shring the fce. A body-centered sphere belongs entirely to its own unit cell. In simple cubic cell there re eight corner spheres. One-eighth of ech belongs to the individul cell giving totl of one whole sphere per cell. In body-centered cubic cell, there re eight corner spheres nd one body-center sphere giving totl of two spheres per unit cell (one from the corners nd one from the body-center). In fce-center sphere, there re eight corner spheres nd six fce-centered spheres (six fces). The totl number of spheres would be four: one from the corners nd three from the fces.
10. (10%) nd the curvture 2 P 2 long the criticl isotherm re zero. Use T these conditions to derive Eqution 6.1, 6.2 nd 6.3 for the criticl volume, pressure nd temperture of system obeying the vn der Wls eqution of stte (Eqution 5.42). Eqution 6.1: c 33 Eqution 6.2: P c Eqution 6.3: T c 22b 2 88 2222 Eqution 5.42: P nnn nn n 2 Strtegy We hve three unknowns (P c, T c nd c ) tht we need to determine. To do this we need to hve three simultneous equtions to solve. One of these is the vn der Wls eqution itself (Eqution 5.43), the other two re given in the problem; nmely, P ( ) 0 nd T 2 P 2 T 0 where P T nd 2 P 2 T re the first nd second prtil derivtives of the pressure with respect to the volume t constnt temperture (See Appendix 1 for the definition of the prtil derivtive.) : At the criticl point, the vn der Wls eqution (Eqution 5.43) is P c c c b 2 (1) c Setting the first derivtive to zero gives P T c ( c b) 2 + 2 c 3 0 (2) Setting the second derivtive to zero gives
2 P 2 T 2 c ( c b) 3 6 c 4 0 (3) This gives us three equtions in three unknowns, which we cn solve by the stndrd process of elimintion. If we multiply Eqution (2) by ( c b) 2 3 c to cler the denomintors, we get c 3 c + 2( c b) 2 0 (2 ) Similrly, multiplying Eqution (3) by ( c b)3 c 4 gives 2 c 4 c 6( c b) 3 0 (3 ) Solving Eqution (2 ) for gives 3 c 2( c b) 2 (4) Substituting Eqution (4) into Eqution (3 ) gives 3 2 4 c c 6 c 2( c b) 2 ( c b)3 0 Dividing both sides by 2 c 3 c nd simplifying gives the following expression for the criticl molr volume, which grees with Eqution 6.1: c 3b (5) Now, we cn go bck nd substitute this vlue of c into Eqution 2 to solve for the criticl temperture: (3b) 3 c + 2(3b b) 2 0 27b 3 c + 2(4b 2 ) 0
Rerrnging gives n expression for T c, which grees with Eqution 6.3: T c 8 27Rb (6) Finlly, we substitute T c nd c into the vn der Wls Eqution [Eqution (1) bove] to get n expression for the criticl pressure: P c 8 R 27Rb 3b b (3b) 2 which gives, fter simplifiction, the expression for the criticl pressure in terms of the vn der Wls prmeters nd b: P c 27b 2 (7) This expression grees with Eqution 6.2 in the text. 11. (5%) The compound dichlorodifluoromethne (CCl 2 F 2 ) hs norml boiling point of -30 C, criticl temperture of 112 C, nd corresponding criticl pressure of 40.5 br. If the gs is compressed to 18 br t 20 C, Will the gs condense? Your nswer should be bsed on grphicl interprettion. From the dt given, the liquid-vpor phse digrm on the right cn be constructed. From this, we see tht when gs t 1 tm (~1.01 br) nd 20ºC is compressed isothermlly to 18 br, it crosses the liquid-vpor coexistence line, indicting condenstion into liquid.
12. (5%) A quntittive mesure of how efficiently spheres pck into unit cells is clled pcking efficiency, Which is the percentge of the cell spce occupied by the spheres. Clculte the pcking efficiencies of simple cubic cell, body-centered cubic cell, nd fce-centered cubic cell. (Hint: Refer to the following figure nd use the reltionship tht the volume of sphere is 4πr 3 /3, where r is the rdius of the sphere.) The pcking efficiency is: volume of toms in unit cell volume of unit cell 100% An tom is ssumed to be sphericl, so the volume of n tom is (4/3)πr 3. The volume of cubic unit cell is 3 ( is the length of the cube edge). The pcking efficiencies re clculted below: () Simple cubic cell: cell edge () 2r
3 4πr 100% 3 3 4 100% Pcking efficiency π r π 100% 52.4% 3 3 (2 r) 24r 6 (b) Body-centered cubic cell: cell edge 4r 3 Pcking efficiency 2 4πr3 3 100% 4r 3 3 2 4πr3 3 100% 64r 3 3 3 2π 3 16 100% 68.0% Remember, there re two toms per body-centered cubic unit cell. (c) Fce-centered cubic cell : cell edge 8r 3 3 4πr 16πr 4 100% 100% 3 3 2π Pcking efficiency 100% 74.0% 3 3 8r 8 3 8 ( 8r) Remember, there re four toms per fce-centered cubic unit cell. 13. (5%) Clulte the vlues of u rrr nd u mm for N 2 t 298.15K. Strtegy The equtions for the two mesures of verge speed re Equtions 5.35 nd 5.38. In generl, the 5.35 nd 5.38 employing R nd M re most often used for this type of clcultion. The constnts nd prmeters needed re R 8.314 J mol -1 K -1, T 298.15 K nd M 28.02 g mol -1 0.02802 kg mol -1. The most probble speed is given by (Eqution 5.38) u mm 2RR M 2 8.314 J K 1 mmm 1 298.15K 0.02802 kk mmm 1 420.6 m s 1 For the rms speed, we hve (Eqution 5.35) u rrr 3RR M 3 8.314 J K 1 mmm 1 298.15K 0.02802 kk mmm 1 515 m s 1
14. (5%) Clculte the molr volume of crbon dioxide (CO 2 ) t 300K nd 20 br, nd compre your result with tht obtined using the idel gs eqution. The second viril coefficient for CO 2 t this temperture (B) is -0.123 L mol -1. ( Z PP nnn 1 + B + C 2 + D 3 + ) Strtegy Becuse we re only given the second viril coefficient, we must truncte the viril eqution (Eqution 5.44) fter the second term nd ssume tht the ffect of C, D,, re negligible. Neglecting terms contining viril coefficients higher thn B, the viril eqution (Eqution 5.44) become Z P RR 1 + B Multiplying by RR nd rerrnging gives qudrtic eqution for : P 2 RR P BBB P 0 Substituting in the given vlues (T 300K, P 100 br, R 0.08314 L br mol -1 k -1 ) gives 2 1.247 + 0.1534 0 Where is in L mol -1.Using the qudrtic eqution (see Appendix 1) gives 1.247 ± 1.2472 4(1)(0.1534) 2 1. 11 L mmm 1 oo 0. 111 L mmm 1 Becuse we hve two solution, we must choose the more physicl one. The idel gs eqution ( RR ) gives vlue of 1.247 L P mol-1, which is obtined from the coefficient of in the qudrtic. For the truncted viril eqution to be vlid, the molr volume must no be too smll, so we cn reject the smller vlue bove nd conclude tht 1.11 mol -1. 15. (5%) Gold (Au) crystllizes in cubic close-pcked structure (the fce-centered cubic unit cell) nd hs density of 19.3 g cm -3. Clculte the tomic rdius of gold in picometers.
Sttegy We wnt to clculte the rdius of gold tom. For fce-centered cubic unit cell, the reltionship between rdius (r) nd edge length (), ccording to Figure 6.21, is 8r. Therefore, to determine r of gold tom, we need to find. The volume of cube is 3 3 or. Thus, if we cn determine the volume of the unit cell, we cn clculte. We re given the density in the problem. density mmmm vvvvvv The sequence of step is summrized s follows: Density of volume of Edge length rdius of unit cell unit cell of unit cell Au tom Step 1: We know the density, so to determine the volume, we find the mss of the unit cell. Ech unit cell hs eight corners nd six fces. The totl number of toms within such cell, ccording to Figure 6.16 nd 6.18, is The mss of unit cell in grms is m 1 1 uuuu cccc 1 mmm 6.022 10 23 8 1 8 + 6 1 2 4 197.0 g AA 1mmm AA 1.31 10 21 g ppp uuuu cccc From the definition of density (d m/), we clculte the volume of the unit cell s follows: m d 1.31 10 21 g 19.3 g cc 3 6.79 10 23 cc 3 Step 2 Becuse volume is length cubed, we tke the cubic root of the volume of the unit cell to obtin the edge length () of the cell 3 3 6.79 10 23 cc 3 4.08 10 8 cc Step 3 From Figure 6.21 we see tht the rdius of n Au sphere (r) is relted to the edge length by Therefore r 8r 8 4.08 10 8 cc 1.44 10 8 cc 111 pp 8
16. (5%) The edge length of the NCl unit cell is 564 pm. Wht is the density of NCl in g cm -3? From Exmple 6.3 we see tht there re four NA+ ions nd four Cl- ions ech unit cell. So the totl mss (in u) of unit cell is mss 4(22.99 u + 35.45u) 233.8 u Converting u to grms, we write 1 g 233.8 u 6.022 10 23 u 3.882 10 22 g The volume of the unit cell is 3 (564 pp) 3 (564 10 10 ) 3 1.794 10 22 cc 3 Finlly, from the definition of density mmmm density vvvvvv 3.882 10 22 g 1.794 10 22 2. 11 g cc 3 cc3 17. (5%) () Express the vn der Wls eqution in the form of Eqution 5.44 (the viril eqution). Derive reltionships between the vn der Wls constnts ( nd b) nd the viril coefficients (B, C, nd D), given tht 1 1 x 1 + x + x2 + x 3 + fff x < 1 (Eqution 5.44 : Z PP nnn 1 + B + C 2 + D 3 + ) The vn der Wls eqution (Eqution 5.44) is P b 2 We re trying to put this into the form of viril eqution (Eqution 5.44): P 1+ B + C 2 + D 3 +... which is power series in 1. First, multiply the vn der Wls eqution by so tht the left hnd side is
the sme s in the viril expnsion: P b P 1 1 b / 1 2 The second term on the right hnd side is in the correct form for the viril eqution, using the expnsion for 1/(1 x) we cn write 1 1 b / 1+ b + b 2 + b 3 + b 4 +... Substituting into the bove gives P 1+ b + b 2 + b 3 + b 4 +... 1 Collecting like terms in 1 gives P b 1 + + b 2 + b 3 + b 4 +... so we cn identify the viril coefficients of the vn der Wls eqution s B b C b 2 D b 2, etc. (b) Using your results, determine the Boyle temperture for vn der Wls gs in term of nd b. Becuse idel gs: Z 1, then B 0 So b B 0 T B RR
18. (5%) The following digrm shows the Mxwell-Boltzmnn speed distributions for gs t two different temperture T 1 nd T 2. Clculte the vlue of T 2. The mximum of the Mxwell-Boltzmnn distribution is locted t the vlue of the most probble speed, given by Eqution 5.37. The rtio of the most probble speeds for given gs t different tempertures is given by u mp (1) u mp (2) 2 1 M 2 2 M 2 1 M 2 2 M u (1) mp u (2) T 1 T mp 2 Looking t the digrm, let's ssume tht the most probble speed t T 1 300 K is 500 m s 1, nd the most probble speed t T 2 is 1000 m s 1. Substitute into the bove eqution to solve for T 2. 500 m s 1 K 300 1000 m s 1 T 2 2 300 (0.5) T 2 T 2 1200 K 19. (10%) Given the phse digrm of crbon shown here, nswer the following questions: () How mny triple points re there nd wht re the phses tht cn coexist t ech triple point?
(b) Which hs higher density, grphite or dimond? (c) Synthetic dimond cn be mde from grphite. Bsed on the phse digrm, how would you do this? () Two triple points: Dimond/grphite/liquid nd grphite/liquid/vpor. (b) Dimond. (c) Apply high pressure t high temperture. 20. (5%) Wht type(s) of intermoleculr interctions exist between the following pirs: () HBr nd H 2 S, (b) Cl 2 nd CBr 4, (c) I 2 nd NO - 3, (d) NH 3 nd C 6 H 6? Exmple 4.10