Vibrational Rotational Spectrum of HCl and DCl

24,137 views

Published on

Physical Chemistry, Semester 2.
The observation of the isotope effect on protium chloride and deuterium chloride

Published in: Education, Technology
0 Comments
2 Likes
Statistics
Notes
  • Be the first to comment

No Downloads
Views
Total views
24,137
On SlideShare
0
From Embeds
0
Number of Embeds
44
Actions
Shares
0
Downloads
101
Comments
0
Likes
2
Embeds 0
No embeds

No notes for slide

Vibrational Rotational Spectrum of HCl and DCl

  1. 1. A Presentation by Patrick Doudy & Tianna Drew
  2. 2. Goals Formulate vibrational-rotational energy states Interpret equilibrium vibrations of HCl and DCl Find the inter-nuclear separation (bond length) of each of the molecules. Calculate the harmonic oscillator force constant for HCl and DCl Calculate constant volume heat capacities Observe the isotope effect in diatomic molecules
  3. 3.  100mm glass sample cell placed in IR spectrometer, which was purged with N2 to produce a background scan. Sample cell was filled with HCl gas generated from the reaction of NaCl and D20 with Sulfuric Acid The spectrometer was purged again, and a reading of the sample was taken
  4. 4. Four types of peaks are seen in the spectra 1H35Cl Interested in 35Cl in this experiment 2H35Cl 1H37Cl 2H is Deuterium 2H37Cl
  5. 5. HClDCl
  6. 6.  The anharmonic oscillator and rigid rotor energy expressions are as follows: These can be combined to form the expression: T(v,J) = E(v,J)/hc = ṽe(v + ½ ) - ṽexe(v + ½ )2 + BeJ(J+1) – DeJ2(J+1)2 – αe(v + ½ )J(J+1) Where the final term accounts for vibrational-rotational interaction due to centrifugal stretching.
  7. 7.  Selection rules allow for ΔJ = ±1 Δv = ±1 Substitutinginto the previous energy equation lets us find expressions for the P and R branch energies ṽR = ṽ0 + (2Be - 2αe) + (2Be - 4αe)J’’ - αeJ’’2 ṽP = ṽ0 + (2Be - 2αe)J’’ - αeJ’’2
  8. 8.  Thesecan be combined to form the expression ṽ(m) = ṽ0 + (2Be - 2αe)m – αem2 – 4Dem3 This equation can be used to analyze the spectra
  9. 9. m = 0 gives the “forbidden” transition from ṽ(m). Directly assigning m=0 to the Q- branch and indirectly assigning values of m to all of the other peaks We can then produce a plot of (m) vs. wavenumber (ṽ) and use a third-order polynomial curve fit to determine the coefficients of ṽ(m) for both HCl and DCl
  10. 10. M=3 M=-2 M=4 M=-3 M=2 M=5 M=-4 M=-1 M=1 M=-5 M=6 M=-6 M=7 M=-7 M=8 M=-8M=9 M=-9 M=0
  11. 11. M=4 M=5 M=-4 M=3 M=-5 M=6 M=-3 M=-6 M=7 M=-9 M=2 M=-2 M=-7 M=8 M=-8 M=-10M=9 M=-1 M=1 M=0
  12. 12. y = -0.002x3 - 0.3035x2 + 20.572x + 2885.7 H 35 Cl 3100.00 3050.00 3000.00 H 35 Cl 2950.00 Poly. (H 35 Cl) 2900.00Wavenumber ṽ (1/cm) 2850.00 2800.00 y = -0.002x3 - 0.303x2 + 20.57x + 2885. R² = 1 2750.00 2700.00 2650.00 -10 -8 -6 -4 -2 0 2 4 6 8 10 m
  13. 13. y = -0.0102x3 - 0.0715x2 + 11.043x + 2090 D 35 Cl peak (cm-1) 2200.00 2150.00 2100.00Wavenumber ṽ (1/cm) D 35 Cl peak (cm-1) 2050.00 Poly. (D 35 Cl peak (cm-1)) y = -0.010x3 - 0.071x2 + 11.04x + 2090 R² = 0.999 2000.00 1950.00 -15 -10 -5 0 5 10 m
  14. 14. y = -0.0102x3 - 0.0715x2 + 11.043x + 2090 DCL y = -0.002x3 - 0.3035x2 + 20.572x + 2885.7 HCLṽ(m) = ṽ0 + (2Be - 2αe)m – αem2 – 4Dem3These equations give the following coefficients for ṽ(m) HCl DCl ṽ0 = 2885.7 cm-1 ṽ0 = 2090.0 cm-1 Be= 10.590 cm-1 Be= 5.593cm-1 αe= 0.3035 cm-1 αe= 0.0715 cm-1 De= 5 E-4 cm-1 De= 2.55 E-3 cm-1
  15. 15.  The isotope effect 4,7 • The difference in mass between atoms effects the vibrational and rotational energies (2 peaks) Relationsbetween two diatomic molecules with an isotopic substitution are HCland DCl can be related using these expressions.
  16. 16.  ṽ0 = ṽe - 2ṽexe Usingthe above with the isotope effect allows us to solve for ṽe and ṽexe for HCl. HCl DCl ṽe = 2986.9 cm-1 ṽe = 2142.2 cm-1 ṽexe= 50.602 cm-1 ṽexe= 26.029 cm-1
  17. 17.  Harmonic oscillator expression Allows us to find the “spring” force constant for HCl and DCl k = 514.96 N/m
  18. 18.  Using the rotational constants from the polynomial curve fit with the definition of B gives the moment of inertia
  19. 19. Finding r is trivial once the moment ofinertia is known HCl DCl R = 1.2699 E -10 meters R = 1.2580 E -10 meters
  20. 20.  Heatcapacity expression and vibronic terms for HCl and DCl give constant volume heat capacities at different temperatures.
  21. 21. H35Cl µ ῦe ῦ ex e Be αe De Re K x10-27kg cm-1 cm-1 cm-1 cm-1 cm-1 A N/mCalculated 1.6267 2986.9 50.602 10.590 0.3035 5. E -4 1.2699 514.96 Literature 1.6267 2990.95 52.8186 10.5934 0.30718 5.319E-4 1.2746 516.35 % Error 0 0.135 4.197 0.032 1.198 5.997 0.369 0.269 D35Cl µ ῦe ῦ ex e Be αe De Re K x10-27kg cm-1 cm-1 cm-1 cm-1 cm-1 A N/mCalculated 3.1624 2142.2 26.029 5.593 0.0715 2.55 E -3 1.2580 514.96Literature 3.1624 2145.16 27.1825 5.44879 0.11329 1.39 E-4 1.2746 516.35 % Error 0.137 4.243 2.646 36.887 1734.532 1.302 0.269
  22. 22. H35Cl Cv (vib) Cv Cv (vib) Cv 298K J/molK 298K 1000K J/molK 1000K Calc. 1.435E-3 20.787 3.327 23.114 Lit. 9.278 E -4 20.788 2.110 22.898% Error 0.546 4.811E-05 0.577 0.00943 D35Cl Cv (vib) Cv Cv (vib) Cv 298K J/molK 1000K J/molK 298K 1000K Calc. 0.0351 20.821 4.113 24.901 Lit. 0.0283 20.816 3.790 24.578 % Error 0.240 0.000240 0.0852 0.0131
  23. 23.  Experimental values closely follow literature values Isotope effect observed • K and R remain essentially unchanged HClhas greater anharmonicity values than DCl
  24. 24. 1. http://chemistry.washcoll.edu/facilities.php2. http://www.wolframalpha.com/entities/isotopes/chlorine_35/s8/c3/mp/ http://www.wolframalpha.com/input/?i=hydrogen+23. http://webbook.nist.gov/cgi/cbook.cgi?ID=C7698057&Units=SI&Mask=1000#Diatomic http://webbook.nist.gov/cgi/cbook.cgi?ID=C7647010&Units=SI&Mask=1000#Diatomic4. http://www.colby.edu/chemistry/PChem/lab/VibRotHClDCl.pdf http://www.ptable.com/5. http://www.chem.ufl.edu/~itl/4411L_f00/hcl/hcl_il.html6. http://www.phys.ufl.edu/courses/phy4803L/group_III/infra_red/irspec.pdf7. Garland, Carl. Experiments in Physical Chemistry 7th ed. 2002, McGraw-Hill Publishing Co.

×