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 11, Problem 11B.2E

(a)

Interpretation Introduction

Interpretation:

Moment of inertia of (i) 1H2(ii) 2H2(iii)12C16O2(iv) 13CO2 has to be determined.

Concept Introduction:

Moment of inertia of diatomic molecules is given by,

  Moment of inertia, I = μR2

Effective mass of a homonuclear diatomic molecule is given by,

  μ = 12m

Moment of inertia of the triatomic linear rotors AB2 is given as,

  ABAMoment of inertia, I = 2mAR2

(a)

Expert Solution
Check Mark

Answer to Problem 11B.2E

  1.         i.            Moment of inertia of 1H2 is 4.58×1048 kg.m2.
  2.       ii.            Moment of inertia of 2H2 is 9.16×1048 kg.m2.
  3.     iii.            Moment of inertia of 12C16O2 is 7.22×1046 kg.m2.
  4.     iv.            Moment of inertia of 13CO2 is 7.22×1046 kg.m2.

Explanation of Solution

  1.         i.            Moment of inertia of 1H2

Given data,

  m(1H)= 1.008muR = 74×1012 m

Effective mass of 1H2 is given by,

  μ= 12m= 12×(1.008mu)= 0.504 mu= 0.504×1.66054×1027kg= 8.369×1028 kg

Moment of inertia is calculated

  I = μR2=  (8.369×1028 kg)(74×1012 m)2= 4.58×1048 kg.m2

Moment of inertia of 1H2 is 4.58×1048 kg.m2.

  1.       ii.            Moment of inertia of 2H2

Given data,

  m(2H)= 2.0141muR = 74×1012 m

Effective mass of 2H2 is given by,

  μ= 12m= 12×(2.0141mu)= 1.007 mu= 1.007×1.66054×1027kg= 1.672×1027 kg

Moment of inertia is calculated

  I = μR2=  (1.672×1027 kg)(74×1012 m)2= 9.16×1048 kg.m2

Moment of inertia of 2H2 is 9.16×1048 kg.m2.

  1.     iii.            Moment of inertia of 12C16O2

Given data,

  m(16O)= 16muR = 116.2×1012 m

Moment of inertia is calculated

Moment of inertia, I = 2mAR2= 2×(16×1.66054×1027 kg)×(116.2×1012 m)2= 7.17×1046 kg.m2

Moment of inertia of 12C16O2 is 7.17×1046 kg.m2.

  1.     iv.            Moment of inertia of 13CO2

Given data,

  m(16O)= 16muR = 116.2×1012 m

Moment of inertia is calculated

Moment of inertia, I = 2mAR2= 2×(16×1.66054×1027 kg)×(116.2×1012 m)2= 7.17×1046 kg.m2

Moment of inertia of 13CO2 is 7.17×1046 kg.m2.

(b)

Interpretation Introduction

Interpretation:

Rotational constant of (i) 1H2(ii) 2H2(iii)12C16O2(iv) 13CO2 in hertz and as a wavenumber in cm1 has to be determined.

Concept Introduction:

Rotational constant of a molecule can be determined using the given formula.

  Rotational constant, B = 4πIwhere,I =  moment of inertia 

Wavenumber is the total number of wavelengths observed at a unit distance. Equation for wavenumber can be given as,

  Wavenumber, ν¯ = 1λ = νcwhere, λ=wavelengthν=frequencyc = speed of light

(b)

Expert Solution
Check Mark

Answer to Problem 11B.2E

  1.         i.            Rotational constant of 1H2 is 1.83 THz.

  Rotational constant of 1H2 in cm1 is 61.04 cm1

  1.       ii.            Rotational constant of 2H2 is 0.917 THz.

  Rotational constant of 2H2 in cm1 is 30.59 cm1.

  1.     iii.            Rotational constant of 12C16O2 is 11.6 GHz.

  Rotational constant of 12C16O2 in cm1 is 0.387 cm1.

  1.     iv.            Rotational constant of 13CO2 is 11.6 GHz.

  Rotational constant of 13CO2 in cm1 is 0.387 cm1.

Explanation of Solution

  1.         i.            Rotational constant of 1H2 in hertz and as a wavenumber in cm1

Moment of inertia of 1H2 is 4.58×1048 kg.m2.

Rotational constant of 1H2 in Hz is determined as given,

  Rotational constant, B = 1.05457×1034Js4×3.14×(4.58×1048 kg.m2)= 1.83×1012 s1= 1.83 THz

Rotational constant of 1H2 is 1.83 THz.

Rotational constant of 1H2 in cm1 is determined as given,

  B = Bc= (1.83×1012 s1)(2.998×108ms1)= 6104 m1= 61.04 cm1

Rotational constant of 1H2 in cm1 is 61.04 cm1

  1.       ii.            Rotational constant of 2H2 in hertz and as a wavenumber in cm1

Moment of inertia of 2H2 is 9.16×1048 kg.m2.

Rotational constant of 2H2 in Hz is determined as given,

  Rotational constant, B = 1.05457×1034Js4×3.14×(9.16×1048 kg.m2)= 9.17×1011 s1= 0.917 THz

Rotational constant of 2H2 is 0.917 THz.

Rotational constant of 2H2 in cm1 is determined as given,

  B = Bc= (9.17×1011 s1)(2.998×108ms1)= 3059 m1= 30.59 cm1

Rotational constant of 2H2 in cm1 is 30.59 cm1.

  1.     iii.            Rotational constant of 12C16O2 in hertz and as a wavenumber in cm1

Moment of inertia of 12C16O2 is 9.16×1048 kg.m2.

Rotational constant of 12C16O2 in Hz is determined as given,

  Rotational constant, B = 1.05457×1034Js4×3.14×(7.17×1046 kg.m2)= 1.17×1010 s1= 11.7 GHz

Rotational constant of 12C16O2 is 11.7 GHz.

Rotational constant of 12C16O2 in cm1 is determined as given,

  B = Bc= (1.17×1010 s1)(2.998×108ms1)= 39.0 m1= 0.39 cm1

Rotational constant of 12C16O2 in cm1 is 0.39 cm1.

  1.     iv.            Rotational constant of 13CO2 in hertz and as a wavenumber in cm1

Moment of inertia of 13CO2 is 9.16×1048 kg.m2.

Rotational constant of 13CO2 in Hz is determined as given,

  Rotational constant, B = 1.05457×1034Js4×3.14×(7.17×1046 kg.m2)= 1.17×1010 s1= 11.7 GHz

Rotational constant of 13CO2 is 11.7 GHz.

Rotational constant of 13CO2 in cm1 is determined as given,

  B = Bc= (1.17×1010 s1)(2.998×108ms1)= 39.0 m1= 0.39 cm1

Rotational constant of 13CO2 in cm1 is 0.39 cm1.

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Chapter 11 Solutions

Elements Of Physical Chemistry

Ch. 11 - Prob. 11C.3STCh. 11 - Prob. 11C.4STCh. 11 - Prob. 11C.5STCh. 11 - Prob. 11C.6STCh. 11 - Prob. 11C.7STCh. 11 - Prob. 11D.1STCh. 11 - Prob. 11D.2STCh. 11 - Prob. 11D.3STCh. 11 - Prob. 11E.1STCh. 11 - Prob. 11E.2STCh. 11 - Prob. 11E.3STCh. 11 - Prob. 11E.4STCh. 11 - Prob. 11A.1ECh. 11 - Prob. 11A.2ECh. 11 - Prob. 11A.3ECh. 11 - Prob. 11A.4ECh. 11 - Prob. 11A.5ECh. 11 - Prob. 11A.6ECh. 11 - Prob. 11A.7ECh. 11 - Prob. 11A.8ECh. 11 - Prob. 11B.1ECh. 11 - Prob. 11B.2ECh. 11 - Prob. 11B.3ECh. 11 - Prob. 11B.4ECh. 11 - Prob. 11B.5ECh. 11 - Prob. 11B.6ECh. 11 - Prob. 11B.7ECh. 11 - Prob. 11B.8ECh. 11 - Prob. 11B.9ECh. 11 - Prob. 11B.10ECh. 11 - Prob. 11B.11ECh. 11 - Prob. 11B.12ECh. 11 - Prob. 11B.13ECh. 11 - Prob. 11B.14ECh. 11 - Prob. 11B.15ECh. 11 - Prob. 11B.16ECh. 11 - Prob. 11C.1ECh. 11 - Prob. 11C.2ECh. 11 - Prob. 11C.3ECh. 11 - Prob. 11C.4ECh. 11 - Prob. 11C.5ECh. 11 - Prob. 11C.6ECh. 11 - Prob. 11C.7ECh. 11 - Prob. 11C.8ECh. 11 - Prob. 11C.9ECh. 11 - Prob. 11D.1ECh. 11 - Prob. 11D.2ECh. 11 - Prob. 11D.3ECh. 11 - Prob. 11D.4ECh. 11 - Prob. 11D.5ECh. 11 - Prob. 11D.6ECh. 11 - Prob. 11E.1ECh. 11 - Prob. 11E.2ECh. 11 - Prob. 11E.3ECh. 11 - Prob. 11.1DQCh. 11 - Prob. 11.2DQCh. 11 - Prob. 11.3DQCh. 11 - Prob. 11.4DQCh. 11 - Prob. 11.5DQCh. 11 - Prob. 11.6DQCh. 11 - Prob. 11.7DQCh. 11 - Prob. 11.8DQCh. 11 - Prob. 11.9DQCh. 11 - Prob. 11.10DQCh. 11 - Prob. 11.11DQCh. 11 - Prob. 11.12DQCh. 11 - Prob. 11.13DQCh. 11 - Prob. 11.1PCh. 11 - Prob. 11.2PCh. 11 - Prob. 11.4PCh. 11 - Prob. 11.5PCh. 11 - Prob. 11.6PCh. 11 - Prob. 11.7PCh. 11 - Prob. 11.8PCh. 11 - Prob. 11.9PCh. 11 - Prob. 11.11PCh. 11 - Prob. 11.12PCh. 11 - Prob. 11.13PCh. 11 - Prob. 11.14PCh. 11 - Prob. 11.15PCh. 11 - Prob. 11.1PRCh. 11 - Prob. 11.2PRCh. 11 - Prob. 11.3PRCh. 11 - Prob. 11.5PR
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