Elements Of Physical Chemistry
Elements Of Physical Chemistry
7th Edition
ISBN: 9780198796701
Author: ATKINS, P. W. (peter William), De Paula, Julio
Publisher: Oxford University Press
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Chapter 12, Problem 12.1P

(a)

Interpretation Introduction

Interpretation:

With the given data, the ratio of the populations of CO2 molecules with J=4 and J=2 has to be calculated.

Concept Introduction:

The ratio of the population can be found using the Boltzmann distribution formula:

N2N1=Ng2e-2kTqNg1e-1kTqCancellingNandqN2N1=g2g1e(2-1)kT

N2isthefinalpopulationN1istheinitialpopulationg2degeneracyforthequantumlevelJ=2g1degeneracyforthequantumlevelJ=1

1=EnergyofthequantumlevelJ=12=EnergyofthequantumlevelJ=2k=Boltzmannconstant=1.381×10-23JK-1T=Temperature

(a)

Expert Solution
Check Mark

Answer to Problem 12.1P

With the given data, the ratio of the populations of CO2 molecules with J=4 and J=2 has been calculated as 1.7531.

Explanation of Solution

Calculating the degeneracy and energy for the given quantum states J=0andJ=5:

The given molecule is CO2. It is a linear molecule. So, it’s degeneracy for any of its quantum state J can be calculated as:

gJ=2J+1J=QuantumstategJ=DegeneracyofthequantumstateJ

For the quantum level or state J=2, the degeneracy is:

  g2=2(2)+1=4+1= 5

For the quantum level or state J=4, the degeneracy is:

  g4=2(4)+1= 8+1= 9

The energy for any of the quantum level J can be calculated as:

J=hBJ(J+1)

  B=Rotationalconstanth=Plank'sconstant=6.626×10-34JsJ=QuantumstategJ=DegeneracyofthequantumstateJ

  J=EnergyofthequantumstateJ

The energy for the quantum level J=2 can be calculated as:

  2=hB(2)(2+1)=2hB×3=6hB

The energy for the quantum level J=4 can be calculated as:

4=hB(4)(4+1)=4hB×5=20hB

The ratio of the population of the levels with J=2 and J=4 at the given temperature of 25oC can be determined as shown here:

Converting temperature from oC to kelvin scale:

25oC+273=298KT=298K

Given: The rotational constant is 11.70GHz.

B=11.70GHz=11.70×109s-1GHz=109s-1

N4N2=g4g2e(4-2)kT=95e(20hB-6hB)1.381×10-23JK-1×298K=95e(14hB)1.381×10-23JK-1×298K=95e(14×6.626×10-34Js×11.70×109s-1)1.381×10-23JK-1×298K

  =95×e-(0.0264)=95×0.9739=1.7531

Therefore, the ratio of the populations of CO2 molecules with J=4 and J=2 has been calculated as 1.7531.

(b)

Interpretation Introduction

Interpretation:

With the given data, the ratio of the populations of CH4 molecules with J=4 and J=2 has to be calculated.

Concept Introduction:

The ratio of the population can be found using the Boltzmann distribution formula:

  N2N1=Ng2e-2kTqNg1e-1kTqCancellingNandqN2N1=g2g1e(2-1)kT

N2isthefinalpopulationN1istheinitialpopulationg2degeneracyforthequantumlevelJ=2g1degeneracyforthequantumlevelJ=1

1=EnergyofthequantumlevelJ=12=EnergyofthequantumlevelJ=2k=Boltzmannconstant=1.381×10-23JK-1T=Temperature

(b)

Expert Solution
Check Mark

Answer to Problem 12.1P

With the given data, the ratio of the populations of CH4 molecules with J=4 and J=2 has been calculated as 2.2743

Explanation of Solution

The given molecule is CH4. It is a spherical molecule. So, it’s degeneracy for of its any quantum state J can be calculated as:

gJ=(2J+1)2J=QuantumstategJ=DegeneracyofthequantumstateJ.

For the quantum level or state J=2, the degeneracy is:

g2=(2(2)+1)2=(4+1)2= (5)2=25

For the quantum level or state J=4, the degeneracy is:

g4=(2(4)+1)2(8+1)2= (9)2=81

The energy for any of the quantum level J can be calculated as:

J=hBJ(J+1)

B=Rotationalconstanth=Plank'sconstant=6.626×10-34JsJ=QuantumstategJ=DegeneracyofthequantumstateJ

J=EnergyofthequantumstateJ

The energy for the quantum level J=2 can be calculated as:

2=hB(2)(2+1)=2hB×3=6hB

The energy for the quantum level J=4 can be calculated as:

4=hB(4)(4+1)=4hB×5=20hB

The ratio of the population of the levels with J=2 and J=4 at the given temperature of 25oC can be determined as shown here:

Converting temperature from oC to kelvin scale:

25oC+273=298KT=298K

Given: The rotational constant is 157GHz.

B=157GHz=157×109s-1GHz=109s-1

N4N2=g4g2e(4-2)kT=8125e(20hB-6hB)1.381×10-23JK-1×298K=8125e(14hB)1.381×10-23JK-1×298K=8125e(14×6.626×10-34Js×157×109s-1)1.381×10-23JK-1×298K

=8125×e-(0.3539)=8125×0.7019=2.2743

Therefore, the ratio of the populations of CH4 molecules with J=4 and J=2 has been calculated as 2.2743.

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