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
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
Four types of peaks are seen in the spectra 1H35Cl Interested in 35Cl in this experiment 2H35Cl 1H37Cl 2H is Deuterium 2H37Cl
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.
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
Thesecan be combined to form the expression ṽ(m) = ṽ0 + (2Be - 2αe)m – αem2 – 4Dem3 This equation can be used to analyze the spectra
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
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.
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
Harmonic oscillator expression Allows us to find the “spring” force constant for HCl and DCl k = 514.96 N/m
Using the rotational constants from the polynomial curve fit with the definition of B gives the moment of inertia
Finding r is trivial once the moment ofinertia is known HCl DCl R = 1.2699 E -10 meters R = 1.2580 E -10 meters
Heatcapacity expression and vibronic terms for HCl and DCl give constant volume heat capacities at different temperatures.
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