Rotational and Vibrational Spectra of HCl and DCl

Goux Spring 2016

Physical Measurements Daniel Gonzalez

Lab Partner: Jenny Joiu

ABSTRACT:

Ft IR spectroscopy was performed on samples of HCl and DCl gas and measurements of

frequency (ṽ) in cm-1 were measured against % transmittance to produce a spectra for

each diatomic molecule. Using plots and regression analysis of m values (either –J or

J+1) assigned from the p branch or r branch peaks of the spectrum verses ṽ ,various

portions of the equation

T ( ν , J )=E (v , J )hc

=ṽe (v+ 12 )−ṽ eXe (v+ 1

2 )2

+Be J ( J+1 )−D eJ2 (J+1 )2−α e(v+1

2 )J (J+1) were

solved. The constants I, r, μ, ṽe αe ,De, Be , Xe ,k and ṽ0 were solved for and used to compare

the energy transition of HCl and DCl. It was found that for HCl k=1210 N/m

I=1.3427x10-20m, r=1.1708x10-10m , ṽo=2875.75cm-1 sd 1.5724 αe= .34823 sd ..0489 ,De=.01084 sd

,00288 Be=12.557 sd .2743 , Xe = .0178 ,and ṽe =2890.1 cm-1. It was found for DCl k= 1210 N/m ,

I=5.311x10-20m, r=1.67x10-10m , ṽo=2072.81 cm-1 αe= .0755 cm-1 sd .0157 ,De=.00099 sd .000522,

Be=6.15 cm-1 sd .0889 , Xe =.0200,and ṽe =2094.25 cm-1.

INTORDUCTION:

Infrared spectroscopy utilizes the principle of bond stretching and vibration to produce a

spectrum analyzed by chemists. Organic molecules produce complex patterns and

spectrums due to the large number of possible vibrations, rotations, and isotopes

present in the organic molecule. With this taken into consideration diatomic molecules

are the simplest molecules that could possibly experience stretching, vibration and

rotation. This intuitively makes sense as a diatomic molecule can be imagined as a two

balls connected by a spring. Now if this model is placed on a Cartesian coordinate

system with the length of the molecule in the positive and negative x direction, one can

easy imagine how the stretching of the bond can be modeled as a simple harmonic

oscillator. Namely by the equation ( ν )=hν (ν+ 12) , discrete allowed energy levels of ν

where ν=1,2,3.. and h being Plank’s constant can be used to solved for energy in the

harmonic oscillator model. Similarly, if one end of the molecule was held in place while

the other end was flicked in the positive y direction, the intuitive, and Newtonian

consequence, would be that the molecule will vibrate in the positive and negative y

directions until it reaches equilibrium. The third movement one could imagine would be

the entire molecule spinning as a simple rotor around the midpoint of the bond with

energy levels allowed by the equation E (J )=h2 J (J+1)8π2 I

with J being the rotational

equivalent of ν and I being the moment of inertia (or the point about where the rotor

turns). These 3 movements rotation, vibration, and stretching are all contributing

factors to spectra produced by IR and can be investigating using diatomic molecules

such as HCl and DCl. When taking into consideration rotation, it is important to

remember that all rotating things have an inherent moment of inertia about which the

object is rotating. This concept can easily be visualized classically by spinning a top and

observing how the top spins around its’ moment of inertia, or the central axis.

However, most molecules are not perfectly symmetrical like a top and will have a

shifted moment of inertia depending on the mass of the two atoms bonded to each

other. In the case of HCl and DCl, by gathering data of two identical molecules aside

from a proton, changes in the rotational behavior can be monitored.

As discussed earlier, IR spectroscopy works on the principles of vibration, rotation and

stretching ect. yet primarily give the observer useful information from inducing these

movements to occur. By blasting the sample with different wavelengths of energy in the

IR region the electrons in the molecule will become excited and different/ more

energetic movements can be recorded that are characteristic of the actual bonds that

are stuck together. Being able to mathematically model this energy transition then

becomes the true value in understanding how IR spectrums, the harmonic oscillator

model, and rigid rotor model work to explain the observed phenomena in an IR

spectrum. If the equation

T ( ν , J )=E (v , J )hc

=ṽe (v+ 12 )−ṽ eXe (v+ 1

2 )2

+Be J ( J+1 )−D eJ2 (J+1 )2−α e(v+1

2 )J (J+1) is

used, both the rotational energy and the vibrational energy are represented in a single

function with correction factors ṽe αe ,De, Be , and Xe. ṽe is the correction factor that takes into account

the molecules vibrating about its internuclear separation distance re which is assumed to be equal to r (just

the radius of the rotor). αe takes into account how the internuclear distance changes as the molecule

experiences rotational vibration. De takes into account stretching due to the rotation of the rotor., Be ,

When graphed, it is clear that a harmonic oscillator takes on a parabolic shape as the

spring constant is proportional to the square of the distance traveled. When these

parameters are fit to the Schrodinger equation, rather than a particle in a box, you have

a particle in a parabola in a box, when the wave function must be between the two arms

of the parabola and the wave’s amplitude is centered at discrete quantized energy

levels that are evenly space. Although this is fine and good, in reality, this is not how

bonds work. We can not continuously excite electrons to infinite excited states using the

same amount of energy per excitation indefinitely. Eventually, the bond will break

because too much energy has been applied. Also, the energy transitions will start to

converge meaning at some point it becomes harder and harder to excite already

excited electrons. With this in mind, it is important to consider the harmonic oscillator

as a model of what is really happening. It is simply just that, a model. In reality the

harmonic oscillator model is only true at lower energy levels, and if one wishes to model

all energy levels a new more complicated model, (the Morse potential) must be adopted

and Xe takes into account the effect of anharmonicity which is this deviation form the expected harmonic

oscillator..

As energy is applied to the simple harmonic oscillator model we will see transitions in a

predictable harmonic fashion. As energy is applied to the simple rotational model, we

will also see transition in energy denoted as J states. Although these are merely modes,

in reality both happen simultaneously and must be accounted for at the same time and

the only way to reconcile the interacting forces is to calculate ṽe αe ,De, Be , and Xe. ṽe from a IR

spectrum.

With this in mind, the purpose of this experiment are to characterize DCl and HCl using

the simple harmonic oscillator approach and the rigid rotor model to characterize what

electronic transitions are occurring, the distance between the two nuclei, the moment of

inertia between the two, and to investigate how these calculated values differ from

more inclusive approximations.

METHODS:

All methods were executed at The University of Texas at Dallas Berkner Teaching

Laboratory under Dr. Warren Goux. An FTIR instrument with at least 2 cm ! 1

resolution was used in this experiment.. CO and nitrogen were used for calibration. A

gas tube consisting of NaCl windows was used with clamps to prevent the leaking of gas

and was stored in a desiccator when not in use.

1ml Deuterated Sulfuric acid was mixed with 1.0120g KCl resulting in DCl and HCl as seen in equation 17.

17) HDSO4+2KCl 2 K+ SO4- +HCl +DCl

Figure 1HCl DCl collection apparatus

Figure 1 The above is an image of the apparatus used to collect the sample diatomic gasses. The tubes were clamed exceptionally tight to prevent any gas from escaping. All reactions were carried out under the fume hood.

RESULTS: Figure 2

Figure 3

Graph 1

-10 -8 -6 -4 -2 0 2 4 6 8 102500

2600

2700

2800

2900

3000

3100

f(x) = − 0.0433651530787755 x³ − 0.34823271413829 x² + 24.4178878568971 x + 2875.75758513932R² = 0.998844870689318

HCL m vs ṽ(m)

(m)

ṽ(cm

^-1)

Graph 2

-15 -10 -5 0 5 10 151800

1850

1900

1950

2000

2050

2100

2150

2200

2250

f(x) = − 0.00397860057280379 x³ − 0.0755236309584098 x² + 12.1566111129879 x + 2072.81266567571R² = 0.999095237758672

DCL m vs ṽ

m values

Cm^-

1

SUMMARY OUTPUT HCL

Regression StatisticsMultiple R 0.999422268R Square 0.998844871Adjusted R Square 0.998578302Standard Error 4.309105786Observations 17

ANOVA df SS MS F Significance F

Regression 3 208730.1109 69576.70363 3747.050424 2.45264E-19Residual 13 241.3891047 18.56839267Total 16 208971.5

CoefficientsStandard

Error t Stat P-value Lower 95%Upper 95%

Lower 95.0%

Upper 95.0%

Intercept 2875.757585 1.572244639 1829.077686 1.47309E-36 2872.360957 2879.154 2872.361 2879.154X Variable 1 12.55717664 0.274382433 45.76523544 9.41866E-16 11.96440943 13.14994 11.96441 13.14994X Variable 2 0.348232714 0.048941838 7.115235746 7.87151E-06 0.242500302 0.453965 0.2425 0.453965X Variable 3 0.010841288 0.002883925 3.759212382 0.002385297 0.004610946 0.017072 0.004611 0.017072

SUMMARY OUTPUT DCL

Regression StatisticsMultiple R 0.999547517R Square 0.999095238Adjusted R Square 0.998959523Standard Error 2.641571398

Observations 24

ANOVA

df SS MS FSignificance

FRegression 3 154108.6883 51369.56275 7361.751645 1.35951E-30Residual 20 139.557989 6.97789945

Total 23 154248.2463

CoefficientsStandard

Error t Stat P-value Lower 95% Upper 95% Lower 95.0%Upper 95.0%

Intercept 2072.812666 0.839766421 2468.320493 2.56008E-56 2071.060944 2074.564388 2071.060944 2074.564X Variable 1 6.153829187 0.088909982 69.21415391 2.72351E-25 5.968366215 6.33929216 5.968366215 6.339292X Variable 2 0.075523631 0.01573272 4.800417937 0.000108901 0.042705752 0.10834151 0.042705752 0.108342

X Variable 3 0.00099465 0.000522339 1.904224106 0.071367654-9.49295E-

05 0.00208423 -9.49295E-05 0.002084

Table 1

DCL Be Ae De Vocm-1 6.15 0.0755 0.00099 2072.81error 0.0889 0.0157 0.000522 0.8397

Table 2

HCl Be Ae De Vocm-1 12.557 0.34823 0.01084 2875.75error 0.2743 0.0489 0.00288 1.5724

Table 3

DCL Ie (m) R (m) K (N/m) μ (amu)5.311x10-20 1.67x10-10 518.76 1.9044

Table 4

HCL Ie (m) R (m) K (N/m) μ (amu)

1.3427x10-20 1.1708x10-10 518.76 .97959

Calculating μ

μ=m1m2

(m1+m2)

Now Substitute values for HCL mass and Chlorine

μ= 1.007825amu x 34.968853amu(1.007825amu+34.968853amu)

μHCl =.981077 amu

μHCl =.981077 amu(1.66054x10-27kg/1 amu)

μHCl =1.6291192x10-27kg

Calculating r

Be=h

8π2 I ec where I=μr2

r2= h8 π2 I Becμ

r=√ h8 π2Be cμ

Now substituting in values for HCl or DCl

r=√6.62607004 x 10−34m2kg s−¿

8 (3.14 )2¿¿¿¿

r = 1.708x10-10mTable 5

Isotopes Be¿ μ¿ re

Ve

.Xe VeXe

DCl37 6.1319 1.9003 1.1991x10-10

2148.7 .0178 38.24

HCl37 12.53796 .981077 1.1708x10-10

2995.9 .02 59.9

Calculating Isotope Effect for HCl

Be¿

Be=μμ¿

Be¿

12.557 cm−1=.981077amu

μ¿

μ¿=m1

m∗¿2

¿¿ ¿ with m2* being the mass of Cl 37

B e¿

1 =.981077amu.981077amu (12.557 cm−1)

Be*=12.53796 cm-1

Calculation of k for HCl

ṽo=√ kμṽo

2μ=k

(2875.75 cm-1)=6.2140x1013Hz

(2 π6.2140 x1013Hz )2(3.1623 x10−27 kg)=k

k=518.76 N/mCalculation of Ve for HCl

ṽe=√ kμ∗¿

¿

ṽe HCL=√ 518.76 kgmm−1 s−2

3.1623 x10−27 kg

ṽe HCL= 2995.9 cm-1

Calculation of Xe for HCl

V o−¿V e

−2V e=X e¿

2875.75−¿2995.9

−2995.9=Xe ¿

Xe =.02

ṽ e¿X e

¿

ṽ eX e= μμ¿

1.936≈1.5664

Table

Calculations of re using equation 5) Be=h

8π2 I ec I=μr2 and Bv=Be-αe(v+1/2)where

v=0,1,2

Bv HCl Bv DCl Re DCl Re HClBe 12.557 6.15 1.67E-10 1.99E-100 12.38289 6.11225 1.69E-10 2.01-101 11.86054 5.999 1.71E-10 2.23E-102 10.98997 5.81025 1.73E-10 2.34-10

To calculate overtones use equation 4) where v=+/- 2, +/-3vand J= +/-1

T ( ν , J )=E (v , J )hc

=ṽe (v+ 12 )−ṽ eX e (v+ 1

2 )2

+Be J ( J+1 )−D eJ2 (J+1 )2−α e(v+1

2 )J (J+1)

For first overtone of DCl use the following values form table 6 in equation 4 to calculate T(J,V)

T ( ν , J )=E (v , J )hc

=(2148.7 )(2+12 )−38.24 (2+1

2 )2

+12.557e1 (1+1 ) .01084 (1¿¿2) (1+1 )2−.34823(2+ 12 )1(1+1)¿

Table 6Calculations of predicted overtones T(J,V)

Ve cm-1Vexe cm-

1 Be cm-1 J v De cm-1 Ae cm-1T(J,V) cm-1

DCL 2148.7 38.24 12.557 1 2 0.01084 0.34823 5156.112DCL 2148.7 38.24 12.557 1 3 0.01084 0.34823 7074.676HCl 2995.9 59.9 6.15 1 2 0.00099 0.0755 7127.297HCL 2995.9 59.9 6.15 1 3 0.00099 0.0755 9763.646

Table 7Calculation of Vibrational Rotational and Translational Heat Capacities

Cv HCl 295.7 Cv HCl295.7 Cv DClCv 17.1517335 88.22511583

Cv Vib 0.00012309 0.000123086

Cv(trans)= 3/2 R Cv(rot)= R Cv(vib)=R ¿

R= 8.3144598 KJ-1mol-1

T=Kelvin 298.5

DISCUSSION:Ultimately the experiment was successful. HCl and Dcl were successfully isolated and IR

spectroscopy was used to generate spectra for each diatomic molecule. It was observed

that indeed the Values for De are very small and that the αe for HCl and DCl are very

different with HCl having an αe value nearly 3 times larger than that of DCl. The Be value

for DCl was nearly twice as large than the Be value for HCL. ṽo of DCl was smaller than

that of DCl. This Data makes sense as HCl is clearly lighter than DCl as deuterium has 2

neutrons rather than 1. This effect will change μ (the adjusted mass) of each molecule

and will directly result in a change in I by equation 3. Larger μ values will yield a larger I

which as depicted in equation 5 is inversely proportional to Be. In table the calculated

values are compared to literature values and come very close.

In table 23 re is calculated and rv values are calculated using different values of Be for

HCL and DCl. As v increases Bv decreases which causes rv to increase. This is clearly

demonstrated in the fact that HCl’s I value was 3 times smaller than that of DCl yet

HCl’s Be value was larger than DCl’s.

Intuitively these vales make sense if considering two masses connected by a spring.

When considering 2 of these systems: 2 masses of Cl but one connected to Deuterium

and one connected to Hydrogen, with all else assumed equal the classical Newtonian

physics principles should apply. Intuitively, if one was to spin objects that are heavier on

a spring the spring should stretch more and the moments of inertia should be different.

One fundamental difference is that the nucli are charged yet this is not that important

as Deuterium has the same charge ad Hydrogen. When analyzing the results it was

concluded in table 16 that the radius of DCl is greater than that of HCl, which intuitively

is what was predicted. (Heavier objects should stretch the spring more when rotating,

vibrating, and translating).

Xe is used to take into account anharmonicity which by definition should be the

deviation form the ideal harmonic oscillator. Intuitively, if something is oscillating back

and forth such as a mass on a spring, by increasing the mas you increase external

forces that can act upon the system. For example, more mass on a sliding mass on a

spring would introduce more friction and thus deviate from harmonicity. In the case of

molecules however, friction is not necessarily a factor, but gravitational, coulombic, and

torqe forces will all be introduced as mass or charge of particles change. I the case of

this experiment it was seen that HCL is more anharmonic than DCl. This could be

explained by the moment of inertia in HCL being less centered between the atoms due

to a larger size difference and thus less able to generate a gyroscopic effect to resist

anharmonicity. This can be confirmed by comparing the VeXe values of HCl to DCl which

clearly shows HCl to be more anharmonic.

REFFERENCES:1) Shoemaker, P David; Experiments in Physical Chemistry,8th ed.;New York, 2009;pp 492-4992)Doudy, Patrick,Inrared Spectroscopy of Protium Chloride35 and Deuterium Chloride 35http://www.slideshare.net/tiannadrew/vibrational-rotational-spectrum-of-hcl-and-dcl 2/24/16

APPENDIX

EQUATIONS:

1) E (ν )=hν (ν+ 12) where ν is the vibrational quantum number such that ν= 0,1,2,….

2) E (J )=h2 J (J+1)8π2 I

where h is planks J is the rotational quantum number, such that

J=0,1,2,… I is the moment of inertia dependent on the reduced mass μ and r the atomic radius, as shown in equation 3

3) I=μr2 where μ=m1m2

(m1+m2)

4)T ( ν , J )=E (v , J )hc

=ṽe (v+ 12 )−ṽ eXe (v+ 1

2 )2

+Be J ( J+1 )−D eJ2 (J+1 )2−α e(v+1

2 )J (J+1)

Where c is the speed of light, ṽe is the frequency in cm-1 of a molecule vibrating about its

internuclear separation distance re . Xe is a constant that will take into account

anharmonicity (as was discussed the system infact is not exactly a harmonic oscillator),

De takes into account centrifugal stretching, which would result as the rotor turned and

force is felt on either molecule, while the last term accounts for changes in r due to

vibrations that would throw off the rotational inertia.

5) Be=h

8π2 I ec I=μr2

6) ṽr=ṽ̥" +¿ ( 2Be−3α e)+(2Be−4α e)J ' '2¿ where J’’=0,1,2…

7)ṽ p=ṽ¿ (2 Be−2αe )J ' '−α eJ ' '2 where J’’=0,1,2..

8) ṽ0=ṽe−ṽe Xe this is the formula for forbidden transition states

9) ṽ (m )=ṽ o ( 2Be−2α e)m−αem2 this equation makes the substitution m=J+1 for the R

branch and m=-J for the P branch

10)ṽ (m )=ṽ o ( 2Be−2α e)m−αem2−4Dem

3 if the m takes an integral value and m=0 yields frequency ṽ of the forbidden purely vibrational transition. If D’’=D’=De

11)Be¿

Be= μμ¿ this is the isotope effect with the asterisk being the isotope

12) ṽe=1

2cπ( kμ¿ )

1/2

ṽe for a harmonic oscillator is gen by this equation

13) ṽ e¿

ṽ e=¿ the ratio ṽ e

¿

ṽ e differs slightly due to the fact that a harmonic oscillator is an

approximation

14) ṽ e¿Xe

¿

ṽ eXe= μμ¿

15) μμ¿= 1.946 for HCl to DCl

16) Cv(trans)= 3/2 R Cv(rot)= R Cv(vib)=R ¿

17) V o−¿V e

−V e=Xe¿