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Chemistry 362Dr. Jean M. Standard

Problem Set 6 Solutions

l . Calculate the reduced mass in kg for the OH radical.

The reduced mass for OH is

= m

Om

H

mO+m

H

.

To properly calculate the reduced mass since it correlates to a single molecule, the ISOTOPIC masses for1H and

1 6O (the most abundant species) must be used in order to get accurate values. The isotopic masses are available

in the CRC. They are mH =1.0078 amuand mO =15.9949amu . Note that 1 amu = 1.660565"10-27 kg.

For OH,

= m

Om

H

mO +mH

= 15.9949 amu( ) 1.0078amu( )15.9949 amu+1.0078amu( )1.660565"10

#27

kg1amu

$

%&

'

()

= 1.5743"10#27 kg .

2 . The harmonic vibrational frequency of HCl in wavenumbers is 2989.6 cm1.

a. ) Calculate the energies of the first two vibrational levels in Joules.

The vibrational energy in the harmonic oscillator picture is

Ev = h"0 v+

1

2( ) .

The harmonic frequency "0

is defined as

"0 = #

ec

= 2989.6 cm$1( ) 2.99793%1010 cm/s( )

"0 = 8.9626%1013s$1.

The ground state vibrational energy is therefore

E0 =

1

2h"0

= 1

26.62618#10

$34Js( ) 8.9626#1013 s$1( )

E0 = 2.96939#10

$20J .

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2 a.) Continued

The first excited state vibrational energy is

E1 = 3

2h"

0

= 32 6.62618 #10$34

Js( ) 8.9626 #1013

s$1

( )E1 = 8.90817 #10

$20J .

b . ) Determine the wavelength for a transition from the v=0 to the v=1 level. Is thistransition in the infrared region of the electromagn etic spectrum?

The energy of a photon for a transition from the v=0 to v=1 level is

Ephoton = "E = E1 # E0

=

3

2h$

0 #

1

2h$

0

Ephoton = h$0 .

Since the energy of a photon is given by Ephoton = h", the frequency of the transition from the first to the

second vibrational level is

Ephoton = h"0

h" = h"0,

or " = "0.

Since the wavelength of a photon is related to the frequency through the relation "=

c

# , the wavelength is

given by

" = c

#0

= 2.99793$10

8m/s

8.9626$1013

s-1

" = 3.3449$10%6

m or 3345 nm.

This wavelength is in the infrared region of the spectrum.

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3 . The fundamental vibrational transition (

v =0" v =1) for CO is 2170 cm1. Treat CO as aharmonic oscillator, and determine the harmonic force constant k in g/s2.

In wavenumbers, the fundamental vibrational transition for a harmonic oscillator is given by

"

e =

#0

c

,

where the harmonic frequency "0is defined as

"0 =

1

2#

k

$

%&

'

()

1/ 2

.

Using these two equations, we can solve for the force constant k,

k = 4"2c

2#

e

2.

In order to use this equation, we need the reduced mass. For CO,

= mCmO

mC +mO

= 12.0000 amu( ) 15.9949 amu( )12.0000 amu+15.9949amu( )

1.660565"10#24 g

1amu

$

%&

'

()

= 1.1385"10#23 g .

Note that to properly calculate the reduced mass, the isotopic masses for 12C and 16O must be used in order to

get an accurate value.

Substituting the reduced mass into the expression for the force constant k,

k = 4" 2 2.997925#1010 cms $1( )2

2170 cm$1( )2

1.1385#10$23 g( )

k = 1.902"106 gs#2.

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4 . A diatomic molecule HX (X is an unknown atom) has a harmonic vibrational forceconstant k=9.6800"105 g/s2. The harmonic vibrational frequency in wavenumbers is4143.3 cm1.

a. ) What is the reduced mass of the molecule?

Using the relation for the harmonic frequency,

"0 = 1

2#

k

$

%&

'

()

1/ 2

,

we can solve for the reduced mass ,

= k

4"2#0

2,

or in terms of wavenumbers,

= k4"

2c

2#

e

2.

Substituting,

=9.6800"105 gs#2( )

4$2 2.997925"1010 cms #1( )24143.3cm#1( )

2

= 1.5892"10#24 g .

4. b.) What atom is X?

In units of amu,

= 1.5892"10#24 g1amu

1.660565"10#24 g

$

%&

'

()

= 0.9570 amu .

The definition of the reduced mass for the HX molecule is

= mHmX

mH

+mX

.

Solving for mX

,

mX

=

mH

mH

" .

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4 b.) Continued

Substituting the value calculated for and mH =1.0078 amu,

mX =

0.9570 amu( ) 1.0078 amu( )1.0078 amu" 0.9570 amu( )

,

mX = 18.99amu; therefore, X = F.

5 . Determine the ratio of populations of the v=1 and v=0 levels of the CO molecule at 300K. The harmonic vibrational frequency is 2170.2 cm1. Repeat your calculation for 1000K .

The population ratio is given by the Maxwell-Boltzmann equation,

n1

n0 =

g1

g0e" E1 "E0( ) / kBT ,

where n jis the population, g jis the degeneracy factor (=1 for harmonic oscillators), Ejis the energy, kB is

the Boltzmann constant, and Tis the temperature.

The energy of a harmonic oscillator is defined as

Ev = h"0 v+1/2( ) ,

so the energy difference is

E1" E

0 = h#

0.

Substituting, the population ratio at 300 K becomes

n1

n0

= exp "h#0 /kBT{ }

= exp "6.62618$10"34 Js( ) 6.50610$1013 s-1( )

1.38066$10"23 J/K( ) 300K( )

%

&'

('

)

*'

+'

n1

n0

= 0.00003.

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5 . Continued

The population of the v=1 level at 300 K is almost negligible (0.003% of the v=0 population). Also notice

that in the calculation, the harmonic frequency "0is obtained by taking the value in wavenumbers (2170.2

cm1) times the speed of light,

"0 = c#e

= 2.99793$1010

cm/s( ) 2170.2 cm%1( )"0

= 6.50610$1013s%1.

At 1000 K, the population ratio is

n1

n0

= exp "h#0 /kBT{ }

= exp "6.62618$10"34 Js( ) 6.50610$1013 s-1( )

1.38066$10"23 J/K( ) 1000K( )

%

&'

('

)

*'

+'

n1

n0

= 0.044 .

At 1000 K, the population of the v=1 level has increased to about 4.4% of the v=0 population.

6 . Refer to the handout on vibrations. Assume that the vibrational force constant does not

change upon isotopic substitution, kHCl =kDCl .

a. ) Using the values from the handout, calculate the ratio of the experimental harmonicfrequencies for HCl and DCl, "0 (HCl) /"0(DCl) .

From the handout, "e HCl( ) = 2989.6cm#1

and "e DCl( ) =2989.6cm#1

. We have from the definition that

"e= #

0 /c , so therefore "0 = c#e .

The ratio of harmonic frequencies is

"0 (HCl)

"

0 (DCl)

= c#

e(HCl)

c#e (DCl)

= #

e(HCl)

#e (DCl)

.

Substituting the experimental harmonic frequencies,

"0 (HCl)

"0 (DCl) =

2989.6 cm#1

2144.7 cm#1

"0 (HCl)

"0 (DCl) = 1.394

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6 . Continued

b.) Using the definition of the harmonic frequency in terms of the force constant andreduced mass predict the theoretical value of the frequency ratio. Does it agree withthe experimental result?

By definition, the harmonic frequency "0is given by

"0 =

1

2#

k

$

%&

'

()

1/ 2

.

Therefore, the ratio of frequencies can be written

"0 (HCl)

"0 (DCl) =

1

2#

kHCl

HCl

$

%&

'

()

1/ 2

1

2#

kDCl

DCl

$

%

&'

(

)

1/ 2 .

Assuming that kHCl =kDCl , this equation simplifies to

"0 (HCl)

"0 (DCl) =

DCl

HCl

#

$%

&

'(

1/ 2

.

Using the isotopic masses mH =1.0078 amu, mD = 2.0140amu, and m35Cl =34.9689 amu leads to the

reduced mass values of HCl = 0.97957amuand DCl =1.9043amu . Substituting into the harmonic

frequency ratio leads to

"0 (HCl)"0 (DCl)

= 1.9043amu0.97957 amu

#

$% &

'(

1/ 2

"0 (HCl)

"0 (DCl) = 1.394 .

Thus, the theoretical result agrees with the experimental result for the harmonic frequency ratio.

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7 . Consider the diatomic molecule LiH to be an anharmonic oscillator. Use the spectroscopicconstants given in the handout on vibrations and determine:

a. ) the fundamental vibrational transition.

From the handout, for LiH, "e =1405.7 cm#1

and xe =0.0165. For an anharmonic oscillator, the

vibrational levels in wavenumbers are given by

"v = "

ev+ 1

2( ) # "e xe v+ 12( )2.

The fundamental transition corresponds to v = 0" v =1. This transition is

"0#1

= "1$"

0 = "

e

32( ) $ "e xe 32( )

2[ ] $ "e 12( ) $ "e xe 12( )2[ ]"

0#1 = "

e$ 2"

ex

e

"0#1

= 1405.7cm$1$ 2 1405.7cm

$1( ) 0.0165( )"

0#1

= 1359.3cm$1

b . ) the first and second overtone transitions.

The first overtone transition is v = 0" v = 2 .

"0#2

= "2$"

0 = "

e

52( ) $ "e xe 52( )

2[ ] $ "e 12( ) $ "e xe 12( )2[ ]"

0#2 = 2"

e$ 6"

ex

e

"0#2

= 2672.2cm$1

The second overtone transition is v =0" v =3 .

"0#3

= "3$"

0 = "

e

72( ) $ "e xe 72( )

2[ ] $ "e 12( ) $ "e xe 12( )2[ ]"

0#3 = 3"

e$ 12"

ex

e

"0#3

= 3938.8cm$1

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8 . Calculate E in Joules, in s1, and in cm1 for the J = 4 ! 5 rotational transition ofCO. The rotational constant is Be = 1.931 cm1.

For a rotational transition with "J=1, we have from the handout that "E=2hcBeJ' . For the J =4 " 5

transition, J"=4 and J'= 5 . Substituting,

"E = 2hcBeJ'

= 2 6.62618#10$34

Js( ) 2.99793#1010 cms $1( ) 1.931cm$1( ) 5( )"E = 3.8359#10

$22J .

Note that the speed of light is expressed in cm/s in order for the units of cm to cancel with the units of therotational constant.

To get the frequency of the transition, we use

" = #E

h =

3.8359$10%22

J

6.62618$10%34

Js,

" = 5.7889$1011s%1.

And, to get the wavenumbers of the transition, we use

" = #

c

= 5.7889$10

11s%1

2.99793$1010

cms%1

,

" = 19.31cm%1.

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9 . For the rotational spectrum of H127I, starting with re = 1.63 , calculate:

a. ) the reduced mass, (kg)

The isotopic masses are mH =1.0078 amuand mI"127 =126.9004 amu(from the CRC). The reduced

mass is therefore

= m

HmI

mH

+mI

= 1.0078amu( ) 126.9004 amu( )1.0078amu +126.9004 amu( )

1.660565"10#27 kg/amu( )

= 1.6603"10#27 kg.

b. ) the moment of inertia, I (kg m2)

I = re

2 = 1.6603"10#27 kg( ) 1.63"10#10 m( )2

I = 4.41"10#47 kgm2 .

c . ) the rotational constant, Be (cm1)

Be = h

8" 2cI =

6.62618#10$34 Js( )8"2 2.99793#108 ms $1( ) 4.41#10$47 kgm2( )

%1m

100 cm

&

'(

)

*+

Be = 6.35 cm$1 .

d.) the rotational energies, EJ (Joules), for J = 0, 1, and 2

The rotational energy levels are EJ = hcBeJ(J+1) . The factor hcBe for the HI molecule is

hcBe = 6.62618"10

#34Js( ) 2.99793"108 ms #1( ) 635m#1( )

hcBe = 1.26"10

#22J .

Using this factor, the rotational energy levels may be calculated for J= 0, 1, and 2. These values are listed

in the table.

J

EJ

(J)

0 0

1

2.52"10#22

2

7.56"10#22

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9. Continued

e . ) the transition frequencies (s1) and wavenumbers (cm1) for transitions from J" !J' with J' = 1, 2, and 3 (and J = 1).

For transitions with "J=1and from J"" J' , the expression for the transition frequency is " = 2cBeJ' ,

and the transition in wavenumbers is " = 2BeJ' . Using these expressions, the results are listed in the

table below.

J' Transition (

J"" J' )

" (s#1

)

" (cm#1)

1 0" 1

3.81"1011 12.7

2 1" 2

7.61"1011 25.4

3 2" 3

1.14 "1012 38.1

f . ) the spacing in frequency ( ) and in wavenumbers ( ) between adjacent spectrall i ne s .

The adjacent rotational line spacing is "# = 2cBe in frequency and "#=2Be in wavenumbers. For HI,

"# = 2cBe = 2 2.99793$10

10cms

%1( ) 6.35cm%1( ) ,or "# = 3.81$10

11s%1.

and "# = 2Be = 2 6.35cm$1( ) = 12.7 cm$1 .

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1 0 . From experiment, the rotational constants for H35Cl and D35Cl are Be = 10.591 cm1 and5.448 cm1, respectively. What is the experimental ratio, B e(DCl) / B e(HCl)? Use thedefinition of B e to predict the theoretical ratio.

The experimental ratio of rotational constants is

Be DCl( )B

eHCl( )

= 5.448cm"1

10.591cm"1

= 0.5144 .

From the definition of the rotational constant, Be =

h

8"2cI

, the theoretical ratio of rotational constants should

be

BeDCl( )

Be HCl( ) =

h

8"2

c I DCl( )h

8"2

c I HCl( )

= I HCl( )I DCl( )

.

Since the moment of inertia Ifor a diatomic molecule is defined as I = re2, we have

BeDCl( )

BeHCl( )

= HCl re,HCl

2

DCl re,DCl2

= HCl

DCl

.

The equilibrium bond lengths cancel out of this expression because they are independent of isotopic

substitution; that is, re,HCl = re,DCl.

To calculate the reduced masses, we have from the CRC, mH =1.0078 amu , mD = 2.0140 amu , and

m35Cl = 34.9689 amu. The reduced masses are calculated as HCl =0.97957 amuand DCl =1.9043 amu .

Substituting, the theoretical ratio is

BeDCl( )

BeHCl( )

= HCl

DCl

= 0.97957amu

1.9043amu

orB

eDCl( )

BeHCl( )

= 0.5144 .

Therefore, the experimental and theoretical ratios are in agreement.

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1 1 . The first two rotational transitions (0 ! 1 and 1 ! 2) in wavenumbers for LiH are15.0262 and 30.0524 cm1, respectively. Determine the moment of inertia I (in kg m2)and the equilibrium bond length re (in ).

The spacing "# in wavenumbers between adjacent lines in a rotational spectrum is "#=2Be . Therefore,

Be =

"#

2 =

30.0524 cm$1 $15.0262cm$1

2,

or Be = 7.5131cm

$1.

Once the rotational constant is calculated, the moment of inertia can be obtained from

I = h

8"2cBe

= 6.62618#10$34 Js( )

8" 2 2.99793#1010 cms $1( ) 7.5131cm$1( )I = 3.7259#10

$47 kgm2.

Finally, from the definition of the moment of inertia, I = re2

, we can solve for the equilibrium bond length

re ,

re =

I

"

#$

%

&'

1/ 2

.

To calculate the reduced mass, we use the isotopic masses, mH =1.0078 amuand mLi =7.0160 amu , to get

LiH

=1.4633"10#27 kg (using the conversion factor 1 amu = 1.660565"10#27

kg). Substituting,

re

= I

"

#$

%

&'

1/ 2

= 3.7259(10)47 kg m2

1.4633(10)27 kg

"

#$

%

&'

1/ 2

re

= 1.596(10)10 m = 1.596 .