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1 Chemistry 362 Dr Jean M Standard Homework Problem Set 6 Solutions l Calculate the reduced mass in kg for the OH radical The reduced mass for OH is m O m H m O + m H To properly calculate the reduced mass since it correlates to a single molecule, the ISOTOPIC masses for 1 H and 16 O (the most abundant species must be used in order to get accurate values The isotopic masses are available in the CRC They are m H amu and m O amu Note that 1 amu kg For OH, m O m H m O + m H ( amu ( amu $ kg ' ( amu amu % 1amu ( kg 2 The harmonic vibrational frequency of HCl in wavenumbers is cm 1 a Calculate the energies of the first two vibrational levels in Joules The vibrational energy in the harmonic oscillator picture is The harmonic frequency ν 0 is defined as ν 0 ω e c E v hν 0 ( v ( ( cm/s cm 1 ν s 1 The ground state vibrational energy is therefore E hν 0 ( ( s Js E J

2 2 2 a Continued The first excited state vibrational energy is E hν ( Js ( s 1 E J b Determine the wavelength for a transition from the v0 to the v1 level Is this transition in the infrared region of the electromagn etic spectrum? The energy of a photon for a transition from the v0 to v1 level is E photon ΔE E 1 E 0 E photon hν hν hν 0 Since the energy of a photon is given by E photon hν, the frequency of the transition from the first to the second vibrational level is E photon hν 0 hν hν 0, or ν ν 0 Since the wavelength of a photon is related to the frequency through the relation λ c, the wavelength is ν given by λ c ν m/s s -1 λ m or 3345 nm This wavelength is in the infrared region of the spectrum

3 3 3 The fundamental vibrational transition ( v 0 v 1 for CO is cm 1 Treat CO as a harmonic oscillator, and determine the harmonic force constant k in g/s 2 In wavenumbers, the fundamental vibrational transition for a harmonic oscillator is given by where the harmonic frequency ν 0 is defined as ω e ν 0 c, ν 0 1 $ k ' 2π % µ ( Using these two equations, we can solve for the force constant k, k 4π 2 c 2 ω e 2 µ In order to use this equation, we need the reduced mass For CO, m C m O m C + m O ( amu ( amu $ g ' ( amu amu % 1amu ( g Note that to properly calculate the reduced mass, the isotopic masses for 12 C and 16 O must be used in order to get an accurate value Substituting the reduced mass into the expression for the force constant k, k 4π 2 ( cms cm 1 k gs 2 ( 2 ( g

4 4 4 A diatomic molecule HX (X is an unknown atom has a harmonic vibrational force constant k g/s 2 The harmonic vibrational frequency in wavenumbers is cm 1 a What is the reduced mass of the molecule? Using the relation for the harmonic frequency, ν 0 1 $ k ' 2π % µ (, we can solve for the reduced mass µ, k 4π 2 ν 0 2, or in terms of wavenumbers, k 4π 2 c 2 ω e 2 Substituting, ( gs 2 ( 2 ( 41433cm 1 2 4π cms g 4 b Which atom corresponds to X? In units of amu, $ amu ' g % g ( amu The definition of the reduced mass for the HX molecule is m H m X m H + m X Solving for m X, m X µ m H m H µ

5 5 4 b Continued Substituting the value calculated for µ and m H amu, m X ( amu ( amu ( amu amu, m X 1899 amu; therefore, X F 5 Determine the ratio of populations of the v1 and v0 levels of the CO molecule at 300 K The harmonic vibrational frequency is cm 1 Repeat your calculation for 1000 K The population ratio is given by the Maxwell-Boltzmann equation, n 1 g 1 e ( E 1 E 0 / k T B, n 0 g 0 where n j is the population, g j is the degeneracy factor (1 for harmonic oscillators, E j is the energy, k B is the Boltzmann constant, and T is the temperature The energy of a harmonic oscillator is defined as so the energy difference is E v hν 0 ( v +1/2, E 1 E 0 hν 0 Substituting, the population ratio at 300 K becomes n 1 n 0 exp{ hν 0 /k B T} ( ( s -1 ( ( 300 K % ' exp Js (' J/K ' * + ' n 1 n

6 6 5 Continued The population of the v1 level at 300 K is almost negligible (0003% of the v0 population Also notice that in the calculation, the harmonic frequency ν 0 is obtained by taking the value in wavenumbers (21702 cm 1 times the speed of light, ν 0 cω e ( ( cm cm/s ν s 1 At 1000 K, the population ratio is n 1 n 0 exp{ hν 0 /k B T} ( ( s -1 ( ( 1000 K % ' exp Js (' J/K ' * + ' n 1 n At 1000 K, the population of the v1 level has increased to about 44% of the v0 population 6 While the reduced mass changes upon isotopic substitution (for example, substitution of D for H in DCl vs HCl, the vibrational force constant does not change; therefore, k HCl k DCl a Using the values from the handout on vibrational spectroscopy, calculate the ratio of the experimental harmonic frequencies for HCl and DCl, ν 0 (HCl /ν 0 (DCl From the handout, ω e HCl ω e ν 0 /c, so therefore ν 0 cω e The ratio of harmonic frequencies is ( cm 1 and ω e ( DCl 21447cm 1 We have from the definition that ν 0 (HCl ν 0 (DCl cω e (HCl cω e (DCl ω e (HCl ω e (DCl Substituting the experimental harmonic frequencies, ν 0 (HCl ν 0 (DCl ν 0 (HCl ν 0 (DCl cm cm

7 7 6 Continued b Using the definition of the harmonic frequency in terms of the force constant and reduced mass predict the theoretical value of the frequency ratio Does it agree with the experimental result? By definition, the harmonic frequency ν 0 is given by ν 0 1 $ k ' 2π % µ ( Therefore, the ratio of frequencies can be written ν 0 (HCl ν 0 (DCl 1 $ 2π % 1 $ 2π % k ' HCl µ HCl ( k ' DCl µ DCl ( Assuming that k HCl k DCl, this equation simplifies to ν 0 (HCl ν 0 (DCl µ DCl ( µ HCl ' # % $ Using the isotopic masses m H amu, m D amu, and m 35Cl amu leads to the reduced mass values of µ HCl amu and µ DCl 19043amu Substituting into the harmonic frequency ratio leads to ν 0 (HCl ν 0 (DCl ν 0 (HCl ν 0 (DCl # 19043amu % ( $ amu ' 1394 Thus, the theoretical result agrees with the experimental result for the harmonic frequency ratio

8 7 Use the spectroscopic constants given in the handout on vibrational spectroscopy and determine for NaF: a the ground state energy in cm 1 assuming a harmonic oscillator and then assuming an anharmonic oscillator 8 From the handout, for NaF, ω e 5361cm 1 and x e For a harmonic oscillator, the vibrational energy levels in wavenumbers are given by ω v ω e ( v For an anharmonic oscillator, the vibrational energy levels in wavenumbers are given by ω v ω e ( v ω e x e ( v The ground state energy corresponds to v0 Substituting, we have for the harmonic oscillator model, ω 0 ω e ( ( cm 1 ω cm 1 And for the anharmonic oscillator model, we have ω 0 ω e ( ω e x e ( ( 5361cm 1 ( 5361cm 1 ( ( cm 1 096cm 1 ω cm 1 b the v6 energy in cm 1 assuming a harmonic oscillator and then assuming an anharmonic oscillator Substituting, for v6 we have for the harmonic oscillator model, ω 6 ω e ( ( cm 1 ω cm 1 And for the anharmonic oscillator model for v6, we have ω 6 ω e ( ω e x e ( ( 5361cm 1 ( 5361cm 1 ( ( cm cm 1 ω cm 1

9 9 7 Continued c Compare and contrast the harmonic and anharmonic oscillator results from parts (a and (b The table below provides a comparison of the NaF molecule treated as a harmonic oscillator and anharmonic oscillator for v0 and v6 v ω v (cm 1 Harmonic Osc ω v (cm 1 Anharmonic Osc For v0, the harmonic and anharmonic oscillator models predict energies that are very close, within 1 cm 1 (04% of each other This is in accord with the idea that real molecules have potential energy curves that are very close to harmonic at low energies, such as near the v0 ground state For v6, the energies predicted by the harmonic and anharmonic oscillator models deviate by more than 160 cm 1, or almost 5% For higher energies, the harmonic and anharmonic oscillator models are expected to show greater deviation because the anharmonic model takes into account the dissociative nature of a real chemical bond, while the harmonic oscillator model does not As a result, the anharmonic oscillator energies are always lower than those of the harmonic oscillator, and the differences increase with increasing quantum number

10 10 8 Consider the diatomic molecule LiH to be an anharmonic oscillator Use the spectroscopic constants given in the handout on vibrational spectroscopy and determine: a the fundamental vibrational transition From the handout, for LiH, ω e cm 1 and x e For an anharmonic oscillator, the vibrational energy levels in wavenumbers are given by ω v ω e ( v ω e x e ( v The fundamental transition corresponds to v 0 v 1 This transition is ω 0 1 ω 1 ω 0 ω e 3 2 ω 0 1 ω e 2ω e x e ω cm cm 1 ω cm 1 ( ω e x 3 [ e ( 2 2 ] ω e( 1 2 ω e x 1 e ( 2 2 ( ( [ ] b the first and second overtone transitions Express your answers in wavenumbers The first overtone transition is v 0 v 2 ω 0 2 ω 2 ω 0 ω e 5 2 ω 0 2 2ω e 6ω e x e ω cm 1 ( ω e x 5 [ e ( 2 2 ] ω e( 1 2 ω e x 1 e ( 2 2 [ ] The second overtone transition is v 0 v 3 ω 0 3 ω 3 ω 0 ω e 7 2 ω 0 3 3ω e 12ω e x e ω cm 1 ( ω e x 7 [ e ( 2 2 ] ω e( 1 2 ω e x 1 e ( 2 2 [ ]