1. A Presentation by Patrick Doudy & Tianna Drew

2. Goals Formulatevibrational-rotational energy states Interpret equilibrium vibrations of HCl and DCl Find the inter-nuclear separation (bond length) ofeach of the molecules. Calculate the harmonic oscillator force constantfor HCl and DCl Calculate constant volume heat capacities Observe the isotope effect in diatomic molecules 3. 100mmglass sample cell placed in IRspectrometer, which was purged with N2 toproduce a background scan. Sample cell was filled with HCl gasgenerated from the reaction of NaCl andD20 with Sulfuric Acid The spectrometer waspurged again, and a readingof the sample was taken 4. Four types of peaks are seen in the spectra 1H35ClInterested in 35Cl inthis experiment 2H35Cl 1H37Cl2H is Deuterium 2H37Cl 5. HClDCl 6. The anharmonic oscillator and rigid rotorenergy expressions are as follows: These can be combined to form the expression: T(v,J) = E(v,J)/hc = e(v + ) - exe(v + )2 + BeJ(J+1) DeJ2(J+1)2 e(v + )J(J+1) Where the final term accounts for vibrational-rotationalinteraction due to centrifugal stretching. 7. Selection rules allow for J = 1 v = 1 Substitutinginto the previous energy equation lets us find expressions for the P and R branch energies R = 0 + (2Be - 2e) + (2Be - 4e)J - eJ2 P = 0 + (2Be - 2e)J - eJ2 8. Thesecan be combined to form the expression (m) = 0 + (2Be - 2e)m em2 4Dem3 This equation can be used to analyze the spectra 9. m = 0 gives the forbidden transition from (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 () and use a third-order polynomial curve fit to determine the coefficients of (m) for both HCl and DCl 10. M=3 M=-2M=4 M=-3M=2M=5M=-4M=-1M=1M=-5M=6 M=-6M=7M=-7M=8 M=-8M=9M=-9M=0 11. M=4M=5M=-4M=3 M=-5M=6 M=-3 M=-6M=7M=-9M=2M=-2 M=-7M=8M=-8 M=-10M=9 M=-1M=1M=0 12. y = -0.002x3 - 0.3035x2 + 20.572x + 2885.7 H 35 Cl 3100.00 3050.00 3000.00H 35 Cl 2950.00Poly. (H 35 Cl) 2900.00Wavenumber (1/cm) 2850.00 2800.00 y = -0.002x3 - 0.303x2 + 20.57x + 2885.R = 1 2750.00 2700.00 2650.00 -10 -8 -6 -4 -2 0 2 4 6810 m 13. y = -0.0102x3 - 0.0715x2 + 11.043x + 2090 D 35 Cl peak (cm-1) 2200.00 2150.00 2100.00Wavenumber (1/cm)D 35 Cl peak (cm-1) 2050.00Poly. (D 35 Cl peak (cm-1)) y = -0.010x3 - 0.071x2 + 11.04x + 2090R = 0.999 2000.00 1950.00 -15 -10 -5 05 10m 14. y = -0.0102x3 - 0.0715x2 + 11.043x + 2090 DCLy = -0.002x3 - 0.3035x2 + 20.572x + 2885.7 HCL(m) = 0 + (2Be - 2e)m em2 4Dem3These equations give the following coefficients for (m) HCl DCl0 = 2885.7 cm-10 = 2090.0 cm-1Be= 10.590 cm-1 Be= 5.593cm-1e= 0.3035 cm-1 e= 0.0715 cm-1De= 5 E-4 cm-1De= 2.55 E-3 cm-1 15. The isotope effect 4,7 The difference in mass between atoms effects thevibrational and rotational energies (2 peaks) Relationsbetween two diatomic molecules with an isotopic substitution are HCland DCl can be related using these expressions. 16. 0 = e - 2exe Usingthe above with the isotope effect allows us to solve for e and exe for HCl.HCl DCl e = 2986.9 cm-1e = 2142.2 cm-1 exe= 50.602 cm-1 exe= 26.029 cm-1 17. Harmonic oscillator expression Allows us to find the spring force constant for HCl and DCl k = 514.96 N/m 18. Using the rotational constants from the polynomial curve fit with the definition of B gives the moment of inertia 19. Finding r is trivial once the moment ofinertia is knownHCl DCl R = 1.2699 E -10 meters R = 1.2580 E -10 meters 20. Heatcapacity expression and vibronic terms for HCl and DCl give constant volume heat capacities at different temperatures. 21. H35Cl e ex eBeeDe Re Kx10-27kg cm-1 cm-1 cm-1cm-1cm-1A N/mCalculated 1.6267 2986.950.60210.590 0.3035 5. E -4 1.2699514.96 Literature1.62672990.9552.818610.59340.30718 5.319E-41.2746516.35% Error0 0.1354.1970.032 1.198 5.997 0.369 0.269D35Cl e ex eBeeDe ReKx10-27kgcm-1 cm-1 cm-1cm-1cm-1AN/mCalculated3.16242142.2 26.0295.5930.0715 2.55 E -31.2580 514.96Literature3.1624 2145.1627.18255.448790.113291.39 E-4 1.2746 516.35 % Error0.1374.2432.646 36.8871734.532 1.302 0.269 22. H35Cl Cv (vib)Cv Cv (vib)Cv298K J/molK 298K 1000K J/molK 1000K Calc.1.435E-320.787 3.32723.114Lit. 9.278 E -4 20.7882.110 22.898% Error0.546 4.811E-05 0.5770.00943 D35ClCv (vib)Cv Cv (vib)Cv 298K J/molK 1000K J/molK 298K1000K Calc. 0.035120.8214.11324.901Lit. 0.028320.8163.79024.578% Error 0.240 0.0002400.0852 0.0131 23. Experimental values closely followliterature values Isotope effect observed K and R remain essentially unchanged HClhas greater anharmonicity values than DCl 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.