# PS4+Solutions - Chemistry 440 Fall 2011 Problem Set {1 Due...

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Unformatted text preview: Chemistry 440 Fall 2011 Problem Set {1 Due Thursday October 13 Problem 1 . Calculate the lattice energy of calcium chloride, given that the heat of sublimation of Ca is 121 kJ mol" and a fﬁo(CaC12) = 495 to mol". (See Tables 14.4 and 14.5 for other data.) ____._._.....—_—»-1 Problem 2 The potential energy of the helium dimer (Hez) is given by ‘1 a: where a = 9.29 x 104 km” (mol dimer)" and C = 97.7 1d )1“ (mol dimer)". (:1) Calculate the equilibrium distance between the He atoms. (b) Calculate the binding energy of the dimer. ((2) Would you expect the dimer to be stable at room temperature (300 K)? Problem 3 Differentiate Equation 16.21 with respect to r to obtain an expression for c and s. Express the equilibrium distance. re. in terrns of or and show that V = —e. n- Pmblem 4 Calculate the induced dipole moment of I2 due to a Na: ion that is 5.0 A away from the center of the].2 molecule. The polarizability of I2 is 12.5 x 111‘3 m . -- man-m. lrp-n—mm—q-H— Problem 5 Two water molecules are separated by 2.76 A in air. Use Equation 16.9 to calculate the dipole-dipoie interaction. The dipole moment of water is 1.82 D. 490 'h N H n E. n a 5?. n H 5' {I a PP Chapter 20 The individual enthalpy changes are AH,“ =.." 155.21ch151-l -1 AH; = (150.51311101 ) err; = 520 kJrnol" AH: = —3ss 1:1 mor' -—e 15,}! = —5514.11<er51—t [enthalpy of sublimation of Li(.5)] = 75.3 M rnol'l [bond enthalpy of 13(3), given] [ﬁrst ioniZation energy of Li(g), Table 14.4] [electron afﬁnity of Hg). Table 14.5] [given] Thus, the lattice energy is I [#541552 + 75.5 + 520 — 323 + 594.1) kJmol" = 1017 ktmer' “"5 energy of calcium chloride, given that the heat of sublimation of Ca is 121 1:1 mer‘ and (1:41:12) = 495 1:1 merl. (See Tables 14.4 and 14.5 for other data.) [an —-—~—_WL_.___.__ Th 4 " aber cycle is Can) + 2Cl'(g) M033 T I A1304 Ca(g) + 2CI(g) AH”; I I AH“; Afar) CHIS) + (313(3) —"'—'—r—II- C3C12(S) The lattice energy is —AH§’. Since .91).?“ = AHf+AH§+AH§+AHf+AH§ The lattice energyls hen; = AH? + AH—f + AH; + 1515:— Ari?“ The various enthalpy changes are AH," =4" 12] ltJrnol‘l AH; = 242.7 kJmol" AH; *—'-' (539.5 + 1145) kJ mol" = 1734.5 kJ mol" [enthalpy of sublimation of Ca(s)] [bend enthalpy of (312(8). Table 4.4] [1" and 2"“ ionization energies of Ca(g), Table 14.4] AH: =4 2 (-4549 Id mer') = —698 kJ mol" [electron afﬁnity of (21(3), Table 14.5] AI? 2 —795 k] rnol'l [given] Thua, the lattice energy is (121 + 242.7 +1734.5 — 693 + 795) kJ I'nol'l = 2195 kJ mol"' —~—~———~—u_._________ where B = 9.29 x 10“ MA” (mol dimer)“ and C m 97.7 It] 1316 (mol dimer)“. (1:) Calculate the equilibrium distance between the He atoms. (b) Calculate the binding energy of the dimer. (1:) Would you expect the dimer to be stable at room temperature (300 K)? (a) The equilibrium distance, r“. can be calculated by setting ﬂ = 0. dr 41V 133 E 37=—js'+ ,1 -123 69- rd,” I”: _ 1/7 13 .29 104141181” 4' "‘ m r =(iééfg) =l [9 x (“‘01 "1"”) =2.975A=2.9sA " 6 [97.7 H A6 (mol dimer)“] (h) The binding energy, Wig), is B C Vr em”— ‘9 r.” 4 g 9.29 x 10" 141.81” (moi dimer)“ _ 97.7 1111365 (mol dimer)“ (2.975 A)” (2.975 A)“ = 4.150 x 10-7! kJ(mol dimer)“ (c) The thermal energy at 300 K is RT = (3.314 JK“ moi“) (300 K) = 2.49 x 103 Jmol“ = 2.49 kJrnol“ which is much larger than 7.60 x 10"2 kJ (mol dimer)“. Thus the dimer would not be able to form at room temperature. This species has been observed and studied at low temperature. Winn—m— 16.31 The internuclear distance between two closest Ar atoms in solid argon is about 3.8 A. The polarizability of argon is 1.66 x 10“'0 m3, and the ﬁrst ionization energy is 1521 Id moi“. Estimate the boiling point of argon. [Hint Calculate the potential energy due to dispersion interaction for solid argon. and equate this quantity to the average kinetic energy of 1 mole of argon gas, which is (3 /2)RT.] The potential energy due to the dispersion interaction for solid argon is 31142! “‘27? ﬂ 3 (1.515 x 10—30 1111*)2 (1521 kJmol“) 4 (3.3 x 10"" m)6 e 4.11414111141-l ..;;3~_fg»‘:;_-._L_ A _ L 420 Chapter 16 16.10 Direntiate Equation 16.21 with respect to r to obtain an expression for e and e. Express the ium distance. r”, in terms etc and show that V = —e. Starting with Equation 16.21, and differentiating gives The minimum of the potential energy occurs when r = re and ﬂ 2 0. air 1 12’ 6 4e|:— 2: +91]=0 re r 12a” and To calculate the potential energy at the equilibrium distance, substitute the expression for re, into that for the potential energy. 12 6 v :4. .T‘I... _ .2. 2 No 21/60' 1 l :4E[E"§] =r—e 16.11 Calculate the bond enthalpy of LiF using the Bum—Haber eyele. The bond length of LiF is 1.51 A. See Tables 14.4 and 14.5 for other information. Use it a 10in Equation 16.7. The Born~Haber cycle is Li+(s) + F'ta) It a w Li (31+ F (QT Li+F'(s) Intarmoletular Forces 16.8 Coulombie forces are usually referred to as longsrange forces (they depend on 1 /r2) whereas van der Waals forces are called short—range forces (they depend on III”). (a) Assuming that the forces (F) depend only on distances. plot F as a function of r at r = 1 A, 2 A, 3 A, 4 A, and 5 A. Cb) Based on your results, explain the fact that although a 0.2 M nonelectrolyte solution usually behaves ideally. nonideal behavior is quite noticeable in a 0.02 M electrolyte solution. (a) A plot with graphs of 1 H vs 1' and us" vs r is presented below. The forces will be proportional to these functions. ll’r2 W l 2 3 4 5 ﬁsh (13) In a nonelectrolyte solution, the attractive forces have a 1 #7 dependence, and as the graph shows, they fall off very rapidly with distance. In an electrolyte solution, the ionic (Coulombic) forces have a 1/r3 dependence that extends to large distances. These “long-range" forces are ‘ :. nsible for nonideal behavior, even at low concentrations. culate the induced dipole moment of I2 due to a Na“" ion that is 5.0 A away from the center 3 a 6 I2 molecule. The polarizability of I2 is 12.5 x 1040 m3. Using Equation 16.13 and the discussion following it in the text. “induced = “IE ii (5.0 x 10"“) us)2 to = 8.01x10‘3‘3C ( ) ( m) 3.336XIO'33Cm =2.4o 418 Chapter 16 (a) (351-16: Dispersion forces (b) CH3C1: Dipole—dipole and dispersion forces (c) P133: Dipole—dipole and dispersion forces (d) NaCl: Ionic and dispersion forces (e) C32: Dispersion forces —-—H——-uﬂﬂ—l—HH—n—_—Iu__—_— 16.6 The boiling points of the three different structural isomers of pentane ((25le) are 95°C, 279°C. and 361°C. Draw their structures, and arrange them in order of decreasing boiling points. Justify your arrangement. T 1" H3 H H3 ch— —-([2—t|3—CH3 H3C—T—-+—CH3 ch—C'—‘CH3 H H H H CH3 n-pentane 2-metl‘tylbutane 2.2-dimethylpropane 36.1 “C 27.9 "C 9.5 °C The boiling points depend on the ease of packing the molecules together. The n-pentane packs together most easily, and it has the highest boiling point. The packing is least favorable for 2,2-dimethylpropane. which has the lowest boiling point. 0 water molecules are separated by 2.76 A in air. Use Equation 16.9 to calculate the ole—dipole interaction. The dipole moment of water is 1.82 D. The dipole moment of H20 is -30 MHZO = (1.32 D) (W) = 6.072 x 10—30 Cm resulting in a dipole-dipole interaction of 2 V 2 _ #Hzof‘Hzo 4HEDF3 m 2 (6.072 x 10-30 c m) (6.072 x 10*“30 c m) 4” (3-8542 x 10*12 (sin-l m'z) (2.75 x 10-10 m)3 = —3.152 x 10-201 = 4.15 x 10-201 Expressing the energy on a per mole basis, . 1013 V = c (3.152 x .o-w.) (\$7) = “MW-n ...
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## This note was uploaded on 11/07/2011 for the course CHEM 440 taught by Professor Gennis during the Fall '08 term at University of Illinois, Urbana Champaign.

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PS4+Solutions - Chemistry 440 Fall 2011 Problem Set {1 Due...

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