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

1. A Presentation by
Patrick Doudy &
Tianna Drew

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.  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. Four types of peaks are seen in the spectra
 1H35Cl
Interested in 35Cl in
this experiment
 2H35Cl
 1H37Cl
2H is Deuterium
 2H37Cl

5. HCl
DCl

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.  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.  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. 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. 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=-8
M=9
M=-9
M=0

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=-10
M=9 M=-1
M=1
M=0

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)
2900.00
Wavenumber
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. y = -0.0102x3 - 0.0715x2 + 11.043x + 2090
D 35 Cl peak (cm-1)
2200.00
2150.00
2100.00
Wavenumber
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. y = -0.0102x3 - 0.0715x2 + 11.043x + 2090 DCL
y = -0.002x3 - 0.3035x2 + 20.572x + 2885.7 HCL
ṽ(m) = ṽ0 + (2Be - 2αe)m – αem2 – 4Dem3
These 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.  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.  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.  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 of
inertia is known
HCl 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 µ ῦe ῦ ex e Be αe De Re K
x10-27kg cm-1 cm-1 cm-1 cm-1 cm-1 A N/m
Calculated 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/m
Calculated 3.1624 2142.2 26.029 5.593 0.0715 2.55 E -3 1.2580 514.96
Literature 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. 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.  Experimental values closely follow
literature values
 Isotope effect observed
• K and R remain essentially unchanged
 HClhas greater anharmonicity values
than DCl

24. 1. http://chemistry.washcoll.edu/facilities.php
2. http://www.wolframalpha.com/entities/isotopes/chlorine_35/s8/c3/mp/
http://www.wolframalpha.com/input/?i=hydrogen+2
3. 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#Diatomic
4. 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.html
6. http://www.phys.ufl.edu/courses/phy4803L/group_III/infra_red/irspec.pdf
7. Garland, Carl. Experiments in Physical Chemistry 7th ed. 2002, McGraw-Hill
Publishing Co.