Chemistry 432 Problem Set 5 Spring 2018 Solutions −1 1

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Chemistry 432 Problem Set 5 Spring 2018 Solutions −1 1. The vibrational frequency of the Cl2 molecule in the gas phase is 559.7 cm . Calculate the ratio of the number of Cl2 molecules in the ﬁrst excited vibrational state to the number of molecules in the ground state at 298K. Answer: N hν ! 1 = exp − N0 kBT (6.626 × 10−34 Js)(559.7 cm−1)(2.998 × 1010 cm s−1)! = exp − = 0.067 (1.381 × 10−23 JK−1)(298 K)

2. Show that for a diatomic molecule interacting with a Lennard-Jones potential

"σ 12 σ 6# V (r) = 4 − r r

the minimum of the potential energy function occurs at r = 21/6σ with an energy of -. Answer: V 0(r) = 4(−12σ12r−13 + 6σ6r−7) = 0 at 12σ12 r13 = 6σ6 r7 2σ6 = r6 or r = 21/6σ " σ 12 σ 6# V (21/6σ) = 4 − 21/6σ 21/6σ 1 1 = 4 − = − 4 2

1 3. In the gas phase the IR spectrum of the HBr molecule consists of a single line at 2648.97 cm−1. Calculate the force constant, the vibrational frequency and the vibrational period of the HBr molecule in the gas phase. Answer:

ν = (2648.97 cm−1)(2.998 × 1010 cm s−1) = 7.942 × 1013 s−1 1 τ = = 1.259 × 10−14 s ν 79.9 k = 4π2µν2 = 4π2 kg (7.942 × 1013 s−1)2 = 408.39 N m−1 80.9 × 6.022 × 1026 4. In the gas phase the bond length of the IF molecule is 1.910 A.˚ Calculate the frequencies of the ﬁrst 2 lines in the microwave spectrum of IF. Answer: (126.9)(19.0) 1 ! I = µR2 = kg (1.910 × 10−10 m)2 = 1.001 × 10−45 kg m2 126.9 + 19.0 6.022 × 1026

h¯2 (1.055 × 10−34 Js)2 B = = = 5.56 × 10−24 J 2I 2(1.001 × 10−45 kg m2) 2B ν = = 1.68 × 1010 s−1 1 h 10 −1 ν2 = 2ν1 = 3.36 × 10 s

5. The second and third lines in the microwave spectrum of the gas-phase LiCl molecule occur at frequencies 4.248 cm−1 and 5.664 cm−1 respectively. Calculate the bond length of the gas phase LiCl molecule. Answer:

2B = h∆ν = (6.626×10−34 Js)(5.664 cm−1−4.248 cm−1)(2.998×1010 cm s−1) = 2.813×10−23 J

h¯2 (1.055 × 10−34 Js)2 I = = = 3.957 × 10−46 kg m2 2B 2.813 × 10−23 J v s u −46 2 I u 3.957 × 10 kg m −10 R = = u = 2.026 × 10 m = 2.026 A˚ µ u (6.941)(35.45) 1 t × kg (6.941 + 35.45) 6.022 × 1026

6. In the gas phase the rotational spectrum of the HS molecule consists of a series of lines separated by 19.36 cm−1. Calculate the bond length of HS in the gas phase. Answer:

2B = (19.36 cm−1)(2.998 × 1010 cm s−1)(6.626 × 10−34 Js) = 3.846 × 10−22 J

2 h¯2 (1.055 × 10−34 Js)2 I = = = 2.894 × 10−47 kg m2 2B 3.846 × 10−22 J I = µR2  1/2 −47 2  2.894 × 10 kg m  −10 R =   = 1.34 × 10 m = 1.34 A˚  32  kg 33 × 6.022 × 1026 7. The ﬁrst two lines in the gas-phase microwave spectrum of the OH molecule occur at 8.562 cm−1 and 17.124 cm−1. Calculate the bond length of the OH molecule in the gas phase. Answer 2B ∆ν = (17.124 cm−1 − 8.562 cm−1)(2.998 × 1010 cm s−1) = 2.567 × 1011 s−1 = h 2B = (2.567 × 1011 s−1)(6.626 × 10−34 Js) = 1.701 × 10−22 J h¯2 (1.055 × 10−34 Js)2 I = = = 6.544 × 10−47 kg m2 = µR2 2B 1.701 × 10−22 J  1/2 −47 2  6.544 × 10 kg m  −10 R =   = 2.046 × 10 m = 0.2046 nm = 2.046 A˚ 16 1  kg 17 6.022 × 1026 8. The rotational constant of the ICl molecule is .114 cm−1. What is the value of J in the most highly populated rotational level of ICl at 298K? Answer: −BJ(J+1)/kB T NJ = A(2J + 1)e

dNJ B = 2Ae−BJ(J+1)/kB T − A(2J + 1)2 e−BJ(J+1)/kB T dJ kBT = 0 at B 2 2 = (2Jmax + 1) kBT or  !1/2  1 2kBT Jmax =  − 1 2 B   1 2 × 1.38 × 10−23JK−1 × 298K !1/2 =  − 1 2 (.114cm−1)(3 × 1010cms−1)(6.626 × 10−34Js)

= 29.7 or Jmax = 30.

3 9. The IBr molecule has a bond length of 2.469 A.˚ Calculate the ratio of the number of molecules in the ground rotational state to the number of molecules in the most highly populated rotational state of a sample of IBr gas at 500K. Answer: (79.9)(126.9) kg I = µR2 = (2.469 × 10−10m)2 = 4.963 × 10−45kg m2 79.9 + 126.9 6.022 × 1026 h¯2 (1.055 × 10−34Js)2 B = = = 1.121 × 10−24J 2I 2(4.963 × 10−45kg m2) Using  !1/2  1 2kBT Jmax =  − 1 2 B   1 2(1.381 × 10−23JK−1)(500K)!1/2 =  − 1 = 55 2 1.121 × 10−24J Then N(0) 1 = N(55) [2(55) + 1] exp(−1.121 × 10−24J(55)(56)/(1.381 × 10−23JK−1 × 500K) = .015

10. The ﬁrst two lines in the gas-phase microwave spectrum of the BrF molecule occur at 0.3558 cm−1 and 0.7116 cm−1. Calculate the rotational energy level J that is most highly populated in a sample of BrF gas at 298.K. Remember that for rotational energy 2 EJ = BJ(J + 1), gJ = 2J + 1,B =h ¯ /2I. Answer: 2 −B(J +J)/kB T PJ = A(2J + 1)e

dPJ 2 B 2 = 2Ae−B(J +J)/kB T − A (2J + 1)2e−B(J +J)/kB T = 0 dJ kBT at 2k T B = (2J + 1)2 B max s  1 2kBT Jmax =  − 1 2 B

2B = (0.7116 cm−1−0.3558 cm−1)(2.998×1010 cm s−1)(6.626×10−34 J s) = 7.067×10−24 J B = 3.534 × 10−24 J   1 "2(1.381 × 10−23 JK−1)(298. K)#1/2 Jmax =  − 1 ≈ 24 2 3.534 × 10−24 J

4 11. In the gas phase the bond length of the MgS molecule is 2.1435 A.˚ Calculate the relative intensities of the ﬁrst two lines in the gas-phase microwave absorption spectrum of MgS at 298K. Answer: h¯2 (1.055 × 10−34 Js)2 2B = = = 1.055 × 10−23 J I "(24.3)(32.1) 1 # kg (2.1435 × 10−10 m)2 24.3 + 32.1 6.022 × 1026

J = BJ(J + 1)

∆ = 1 − 0 = B(1)(2) − 0 = 2B

gJ = 2J + 1 " −23 # I1 g1 3 1.055 × 10 J = e−(1−0)/kB T = exp − = 2.992 −23 −1 I0 g0 1 (1.381 × 10 JK )(298 K) 12. The gas-phase tellurium oxide (TeO) molecule has a natural vibrational frequency of 797.11 cm−1 and a bond length of 1.815 × 10−10 m. At a temperature of 300. K, calculate the ratio of the intensities of the ﬁrst two lines in the microwave spectrum of TeO. Answer: "(127.6)(16.0) kg # I = µR2 = (1.815 × 10−10 m)2 = 7.778 × 10−46 kg m2 127.6 + 16.0 6.022 × 1026

h¯2 (1.055 × 10−34 Js)2 B = = = 7.156 × 10−24 J 2I 2(7.778 × 10−46 kg m2)

Ii gi = e−(Ei−Ej )/kB T Ij gj

E1 − E0 = 2B g1 = 3 g0 = 1 I ( 2(7.156 × 10−24 J) ) 1→2 = 3 exp − = 2.99 −23 −1 I0→1 (1.381 × 10 JK )(300. K) 13. The gas-phase 39K127I molecule has a fundamental vibrational frequency of 5.592 ×1012 s−1 and a bond length of 3.048 ×10−10 m. Given the atomic mass of 39K is 38.96 g mol−1 and the atomic mass of 127I is 126.9 g mol−1, calculate the ratio of the frequency of the ﬁrst line in the microwave spectrum of 39K127I to the frequency of the most intense line in the infrared spectrum of 39K127I. Answer: (38.96)(126.9)! kg ! I = µR2 = (3.048×10−10 m)2 = 4.599×10−45 kg m2 38.96 + 126.9 6.022 × 1026

5 h¯2 (1.055 × 10−34 Js)2 2B = = = 2.420 × 10−24 J I (4.599 × 10−45 kg m2) 2B 2.420 × 10−24 J ν = = = 3.649 × 109 s−1 1←0 h 6.626 × 10−34 J s 9 ν1←0 3.649 × 10 −4 R = = 12 = 6.525 × 10 νIR 5.592 × 10 14. At 298K the relative intensity of the ﬁrst two lines in the microwave spectrum of the gas-phase hydrogen ﬂuoride molecule (HF) is I1→2/I0→1 = 2.451. Calculate the bond length of HF in the gas phase. Answer: I g m m −(m−n)/kB T = e gm = 2m + 1 1 − 0 = 2B In gn I 2B ! 1→2 = 3 exp − = 2.451 −23 −1 I0→1 (1.381 × 10 JK )(298 K) h¯2 (1.055 × 10−34 Js)2 2B = 8.318 × 10−22 J = = I I I = 1.338 × 10−47 kg m2 = µR2  1/2  1.338 × 10−47 kg m2    −11 R =   = 9.17 × 10 m  (1.01)(19.0) kg  1.01 + 19.0 6.022 × 1026 15. The infra-red spectrum of HCl consists of an intense band at 2991 cm−1. Assuming the potential energy curve for HCl and DCl to be identical, predict the frequency for the fundamental infra-red transition in DCl. Answer: 1 s k ν = 2π µ ν sµ (1)(35)/36!1/2 DCl = HCl = = .72 νHCl µDCl (2)(35)/37 −1 −1 νDCl = (2991cm )(.72) = 2154.cm