PS3_08-Solution - Chem 444B Homework Set #3 - Solutions...

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Problem Set #3 -1- Chem 444B Homework Set #3 - Solutions DUE: Friday, February 8, 2008 1) a) Calculate ε rot (in cm -1 ) for the J = 2 level of O 2 given r = 0.12074 nm () 2 2 12 2 9 2 9 23 46 2 2 34 46 1 , 2 16.0 / 16.0 / 0.12074 10 16.0 / 16.0 / 8.0 1 1 0.12074 10 1000 6.022 10 1.937 10 1.054 10 2 2 1 2 1.937 10 rot rot JJ mm Ir Im m gm o l o l g mol g mol gk g m o l mol g Ik g m Js εµ µ ε + == = + ⎡⎤ ⎢⎥ + ⎣⎦ × × × ×⋅ + = × = 22 2 1.72 10 J kg m 22 1 5.034 10 1 cm J × × 1 8.66 rot cm = b)Calculate ε vib for the v = 0 level of O 2 given i υ =1580cm -1 ( ) ( ) 34 10 1 20 11 vv 22 1 6.626 10 3 10 / 1580 0 2 1.57 10 vib vib vib hv hcv cms cm J −− ⎛⎞ =+ = + ⎜⎟ ⎝⎠ =×⋅ × + ± 22 1 5.034 10 1 cm J × × 1 790 vib cm =
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Problem Set #3 -2- 2) Calculate q trans for Ne(g) in a 200.0 mL container at 1200K () 3 2 2 3 2 23 3 23 3 2 34 3 2 29 3 29 2 2 11 2 20.18 1.38 10 1200 1000 6.022 10 1 200.0 100 6.626 10 1 1.415 10 1.415 10 B trans trans trans mk T qV h gk g m o l J K mol g K m q cm cm Js kg qm π ⎛⎞ = ⎜⎟ ⎝⎠ ⎡⎤ ×× × ⎢⎥ × ⎣⎦ ×⋅ 3 2 3 2 29 1.415 10 trans m m q 3) (McQ&S 18-2) 3 22 2 1 9 2 34 2 26 23 2 2 Eq. 18.4: ( , ) exp 8 Let's look at several terms in the summation near n=10 . .. 6.626 10 88 1 8 10 1 1.38 10 300 10 8 trans n B hn T ma hh ma ma k T mJ kg dm K dm K h ma β = −− =− == = ( ) ( ) ( ) ( ) 19 2 91 9 9 2 2 9 9 2 9 9 2 1.326 10 (Unitless!) 10 : exp exp 1.326 10 10 0.8758503709 8 10 1: exp exp 1.326 10 10 1 0.8758503707 8 10 2: exp exp 1.326 10 10 8 n ma n ma n ma × = × = =+ = × + = = × ( ) 2 -10 2 0.8758503705 These terms only differ by 2 10 , so the difference really is very small += ×
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Problem Set #3 -3- 4) a)McQ&S, 18-10, but use temperatures of 1000K, 2000K and 5000K, and plot from v=0 to v=5 () /v / v vib From Eq. 18.28, 1 Use 4227 from Table 18.3 vib vib TT fe e K −Θ − Θ =− Θ= vf v (1000) f v (2000) f v (5000) 0 0.9854 0.8792 0.5706 1 0.0144 0.1062 0.2450 2 2.099E-04 0.0128 0.1052 3 3.064E-06 0.0016 0.0452 4 4.473E-08 1.873E-04 0.0194 5 6.528E-10 2.263E-05 0.0083 Fraction of HCl(g) molecules in vibrational states 0.0 0.2 0.4 0.6 0.8 1.0 012345 v f v fv (1000) fv (2000) fv (5000) At low temperatures, the majority of the molecules are in the ground vibrational state. As we increase the temperature, we begin to populate excited vibrational states. b)Calculate the temperature at which 25% of the HCl molecules would be expected to be in states above the ground state ( ) 0 0.25 v f > = / 0 0 Using Eq. 18.29, 4227 3049 ln ln(0.25) vib T v vib v K TK f > > = == =
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Problem Set #3 -4- 5) (McQ&S 18-17) () (1 ) / J From Eq. 18.35, 2 1 T Use 2.39, and T=300K and 1000K and plot : rot JJ T rot J rot fJ e f −Θ + Θ ⎛⎞ =+ ⎜⎟ ⎝⎠ Θ= Fraction of NO(g) molecules in rotational states 0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0 5 10 15 20 25 30 35 40 45 50 J f J 300K 1000K The plot of the rotational states looks very different from the plot of the vibrational states in the previous problem. As the temperature increases, the number of available rotational states increases, and the distribution becomes broader.
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This homework help was uploaded on 04/13/2008 for the course CHEM 444 taught by Professor Jameslis during the Spring '08 term at University of Illinois at Urbana–Champaign.

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PS3_08-Solution - Chem 444B Homework Set #3 - Solutions...

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