CHEM 301: Homework assignment #12

Transcription

1 CHEM 301: Homework assignment #12 Solutions 1. Let s practice converting between wavelengths, frequencies, and wavenumbers. (10%) Express a wavelength of 442 nm as a frequency and as a wavenumber. What is the photon energy for this wavelength? The frequency ν of a photon is related to its wavelength λ in vacuum as follows: ν c λ m/s m s, where c is the speed of light in vacuum. The energy of the photon is: E hν J s s J 2.79 ev. What are the wavenumber and the wavelength of the radiation used by an FM radio transmitter broadcasting at 88.0 MHz? The wavelength is related to the frequency as follows: The wavenumber is ν ν c λ c ν m/s Hz 3.41 m Hz m/s 2.94 m cm Suppose you were seeking the presence of (planar) SO 3 molecules in the microwave spectra of interstellar gas clouds. Calculate the rotational constants A and B for 32 S 16 O 3. Assume m ( 32 S ) a.m.u., m ( 16 O ) a.m.u., and an SO bond length of 143 pm. (15%) SO 3 is a trigonal planar molecule. I is perpendicular to the plane of the molecule and along the vertical axis passes through the S atom. I is shown in the figure. The bond angles are 120. I 2m 0 R 2 (1 cos θ), 1

2 where θ 120, I kg ( m ) 2 (1 cos 120 ) kg m 2. The second component of the moment of inertia I m 0 R 2 (1 cos θ) 1 2 I kg m kg m 2. The rotational constants are and A B 4πI kg m Hz 4πI kg m Hz. Could you use microwave spectroscopy to distinguish the relative abundances of 32 S 16 O 3 and 33 S 16 O 3? (10%) The S atom is located at the center of mass of the molecule and does not affect the moments of inertia, and hense the rotational constants. Consequently, microwave spectroscopy cannot be used to distinguish between 32 S 16 O 3 and 33 S 16 O How many rotational states have an energy equal to hbj (J + 1) with J 8 for (a) a methane molecule and (b) a chloromethane molecule described by the quantum numbers J, M J, and K? (10%) A methane molecule is a spherical rotor; hence I I and A B and the rotational energy levels only depend on J, not on K or M J : E J hbj (J + 1). There are 2J + 1 possible values of K and 2J + 1 possible values of M J for each value of J. Thus, for J 8 there are 17 possible K values and 17 possible M J values. All levels with the same J and any K and M J values have the same energy, so the level is fold degenerate. If methane is replaced by chloromethane (CH 3 Cl) then the molecule is no longer a spherical rotor (it is a symmetric rotor). Thus, the rotational energy levels depend both on J and K: E J,K hbj (J + 1) + h (A B) K 2. The energy is equal to hbj (J + 1) only for K 0, and for J 8 are 17 possible M J values, so this energy level is 17-fold degenerate. 4. Microwave spectroscopy can be used to determine the bond length of diatomic molecules. 2

3 The rotational constant of 127 I 35 Cl is cm 1. I-Cl bond length. The rotational constant is thus the moment of inertia B I 4πI, 4πB. On the other hand, for a diatomic molecule: Calculate the where I µr 2, µ m Im Cl a.m.u. m I + m Cl is the effective mass. Comparing the two expressions for the moment of inertia, we find µr 2 4πB, and we can express the interatomic distance (bond length) as R 4πBµ. To convert the rotational constant from cm 1 to Hz, multiply it by c (in cm/s): B cm cm/s Hz. Plugging this value and the value of µ that we calculated (in kg) into the expression for R, we find: R 34 J s Hz Å. kg The microwave spectrum of 1 H 127 I consists of a series of lines separated by 384 GHz. Compute its bond length. What would the separation of the lines be in the spectrum of 2 H 127 I? The selection rule for rotational spectra is J ±1. The transition energy for levels J and J + 1 is E J J+1 hb (J + 1) (J + 2) hbj (J + 1) hb ( J 2 + 2J + J + 2 J 2 J ) 2hB (J + 1). 3

4 Lines in the spectrum will appear at energies 2hB, 4hB, etc., with energy spacing 2hB between each pair of lines or frequency spacing 2B between each pair of lines. Thus, and Because we can express R as where 2B Hz, B Hz 2 B Hz. 4πI 4πµR 2, R 4πµB, µ m Im H m I + m H kg kg. Plugging in the values of B and µ, we obtain R kg Å. Hz Replacing 1 H by 2 H changes the effective mass, µ m Im H m I + m H kg kg. However, it almost does not effect the bond length, which we have calculated to be 1.62 Å. Thus, the rotational constant for 2 H 127 I can be calculated as B 4πµR J s kg (1.62 ) 96.9 GHz Å and the separation between spectral lines of 2 H 127 I is 2B 194 GHz. Assume m (1 H ) a.m.u., m (2 H ) a.m.u., m ( 35 Cl ) a.m.u., m ( 127 I ) a.m.u. (20%) 5. Suppose the CO group in a peptide bond can be regarded as isolated from the rest of the molecule. Given the force constant of the bond in a carbonyl group is 908 N/m, calculate the vibrational frequency of (a) 12 C 16 O and (b) 13 C 16 O. (15%) 4

5 The vibrational frequency is ν 1 k 2π µ, where for 12 C 16 O µ m Cm O m C + m O kg kg. Plugging in the values of k and µ into the expression for ν, we find ν Hz. Similarly, for 13 C 16 O the effective mass is calculated to be µ kg and ν Hz. Thus, vibrational spectroscopy is able to distinguish between different isotopes in a molecular group. 6. The wavenumber of the fundamental vibrational transition of Cl 2 is 565 cm 1. Calculate the force constant of the bond. (15%) The wavenumber is defined in spectroscopy as ν ν c 1 k 2πc µ. For a homonuclear molecule like Cl 2, the effective mass is simply one half of the atomic mass: µ 1 2 m (3 5Cl ). Expressing the force constant, we obtain k (2πc ν) 2 µ 1 2 (2πc ν)2 m (3 5Cl ) ( m/s m 1) kg 329 N/m