Calculating Constants & Bond Lengths for N2 and O2 - Prof. Stefan Franzen, Assignments of Physical Chemistry

Download Calculating Constants & Bond Lengths for N2 and O2 - Prof. Stefan Franzen and more Assignments Physical Chemistry in PDF only on Docsity! 1 Homework #3 Due: September 17 Name _________________________ Chemistry 331 1. The atmosphere is made up of 79% N2 and 20% O2. A. To apply rotational or vibrational spectroscopy formulae you will need to use the reduced mass for the diatomic. This is given by: µ = m1m2m1 + m2 Calculate the reduced mass for both N2 and O2 in kilograms. Solution: for oxygen. µ = mOmOmO + mO = mO 2 = 16 2 (1.672 x 10 – 27kg) = 1.337 x 10– 26kg and for nitrogen µ = mNmNmN + mN = mN 2 = 14 2 (1.672 x 10 – 27kg) = 1.170 x 10– 26kg B. Given the rotational constant B = 1.41cm-1 for N2 and 1.00 cm-1 for O2 determine the bond length of each molecule. Solution: If we solve for the rotational energy from the Schrodinger equation the energy levels are: E = h 2 8π2µR2 J(J + 1) or in cm– 1 ν = Ehc = h 8π2µcR2 J(J + 1) Note that we used h and not h_bar above so there is an extra factor of 4π2 in the denominator. Recall that the difference between any two levels is two 2J so that the rotational spectrum is a progression of lines with a spacing that is equal to twice the rotational constant B. If given B you can solve for the internuclear distance of a diatomic as follows. For nitrogen R= h8π2cµB = (6.626 x 10– 34Js) 8(3.14)2(2.99 x1010 cm/s)(1.337 x 10– 26 kg)(1.0 cm– 1) R = 1.1 x 10-10 m or 1.1 A for nitrogen For oxygen R= h8π2cµB = (6.626 x 10– 34Js) 8(3.14)2(2.99 x1010 cm/s)(1.170 x 10– 26 kg)(1.41 cm– 1) R = 1.22 x 10-10 m or 1.22 A for oxygen NOTE: In this problem all of the quantities are in MKS units except the speed of light. We cm/s because this way our answer is consistent with units of cm-1. 2 If you are given the bond length, then you can calculate the rotational constant. For oxygen the rotational constant is: B = h 8π2cµR2 = 6.626 x 10 – 34Js 8(3.14)2(2.99 x1010 cm/s)(1.337 x 10– 26)(1.22 x 10– 10)2 = 1.00 cm-1. For nitrogen the rotational constant is: B = h 8π2cµR2 = 6.626 x 10 – 34Js 8(3.14)2 (2.99 x1010 cm/s)(1.170 x 10– 26)(1.10 x 10– 10)2 = 1.41 cm-1. C. Describe the appearance the rotational spectra of N2 and O2. Which one has a greater spacing between the lines? Solution: Neither N2 nor O2 has a dipole moment. Therefore, neither has a pure rotational (microwave) absorption spectrum. D. Given the force constants for N2 and O2 are 4573 and 2267 N/m, respectively, calculate their vibrational frequencies. Solution: Again you will need to use the reduced mass that you calculated in part A. Recall that the classical relationship between the frequency and the force constant holds also in quantum mechanics. ω = kµ The quantity ω is the angular frequency, which is related to the frequency in Hz as ω = 2πν. Therefore, ν = 12π k µ To obtain the answer in wavenumbers (cm-1) we use the fact that ν = νc = 1 2πc k µ n_tilde is the answer in wavenumbers. For nitrogen we have ν = 12πc k µ = 1 2(3.141)(2.99 x 1010 cm/s) 4573 N/m 2.34 x 10– 26 kg = 2353 cm– 1 For oxygen we have: ν = 12πc k µ = 1 2(3.141)(2.99 x 1010 cm/s) 2267 N/m 2.67 x 10– 26 kg = 1551 cm– 1 If you are given the vibrational wavenumbers you can obtain the force constants as follows: For oxygen k = µω2 = 4π2cµν2 = 4(3.14159)2(2.99 x 1010 cm/s)2(2.67 x 10-26 kg)(1551 cm-1) 2 = 2267 N/m For nitrogen k = µω2 = 4π2cµν2 = 4(3.14159)2(2.99 x 1010 cm/s)2(2.34 x 10-26 kg)(2353 cm-1) 2 5 4. Water is an asymmetric top that has three rotational degrees of freedom. This means that water rotates around three axes. The rotational constants of water are 698.4 GHz, 436.3 GHz and 268.5 GHz. At what wavelengths does water absorb radiation? Please note: rotational transitions have multiple bands. Solution: The rotational spectrum is a series of lines spaced at 2B where B. There are three rotations possible for water. Their rotational constants are: 698.4 x 109 Hz = 6.984 x 1011 s-1 436.3 x 109 Hz = 4.363 x 1011 s-1 268.5 x 109 Hz = 2.685 x 1011 s-1 The rotational spectrum consists of a series of lines with a spacing equal to twice the rotational constant so: 1.397 x 1012 s-1, 2.794 x 1012 s-1, 4.190 x 1012 s-1, … 8.726 x 1011 s-1, 1.745 x 1012 s-1, 2.818 x 1012 s-1, … 5.370 x 1011 s-1, 1.074 x 1012 s-1, 2.148 x 1012 s-1, … However, of these one will not be observed since there is no change in dipole moment when water rotates about its axis of symmetry. Since the different axes of the moment of inertia were not given in the problem we will simply calculate all three. Since λν = c we have λ = c/ν. NOTE: we can use c in m/s and then we will multiply by 109 to obtain the value in nanometers. The wavelengths are: (109)2.99 x 108 m/s/1.397 x 1012 s-1 = 214030 nm ~ 214 microns (109)2.99 x 108 m/s/2.794 x 1012 s-1 = 107015 nm ~ 107 microns (109)2.99 x 108 m/s/4.190 x 1012 s-1 = 71369 nm ~ 73 microns … and (109)2.99 x 108 m/s/8.726 x 1011 s-1 = 342645 nm ~ 343 microns (109)2.99 x 108 m/s/ 1.745 x 1012 s-1= 171346 nm ~ 171 microns (109)2.99 x 108 m/s/2.818 x 1012 s-1 = 106103 nm ~ 106 microns … and (109)2.99 x 108 m/s/5.370 x 1011 s-1 = 556797 nm ~ 557 microns (109)2.99 x 108 m/s/1.074 x 1012 s-1 = 278394 nm ~ 278 microns (109)2.99 x 108 m/s/2.148 x 1012 s-1 = 139199 nm ~ 139 microns … This progression of lines does not increase without bound. In fact, the rotational levels are populated by thermal energy. What does this look like in practice? As an example I have calculated some typical rotational spectra at room temperature (298 K). I assumed different values of the rotational constant in cm-1. The formula I used is: E(J) = (2J + 1)e– 2BJ/kT J is the rotational quantum number. 2J+1 is the degeneracy and the exponential is like an equilibrium constant that tells you how many molecules are in the level with quantum number J. As might imagine that number gets smaller as J gets larger. This function will be peaked at around thermal energy kT = 200 cm-1. 6 This is a pretty large spacing (2B = 60 cm-1) that corresponds to the rotational constant for the H2 molecule. The value of B = 10-20 cm-1 is typical for molecule that have X-H bonds where X is any atom. 1 cm-1 ~ 30 GHz so you can about how big the spacing is for the water rotations above. Note that the maximum in the energy occurs at about the same place, but the spacing of the lines is closer together for smaller B. This trend continues. and 0 I I I I 1 ». Winn T T T T 200 400 0 1 600 , 80 1000 Wavenumber (cm ) T T T T 1 0 200 400 600 800 1000 Wavenumber (cm”)