PHYSICAL CHEMISTRY-WEBASSIGN
PHYSICAL CHEMISTRY-WEBASSIGN
11th Edition
ISBN: 9780357087411
Author: ATKINS
Publisher: CENGAGE L
Question
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Chapter 11, Problem 11B.4P
Interpretation Introduction

Interpretation:

The moment of inertia and bond length of H1C35l and positions of corresponding lines in H2C35l have to be calculated.

Concept introduction:

A rotational spectrum is a spectrum in which only the rotation of molecule changes.  For a molecule to exhibit pure rotational spectrum, the molecule must possess a permanent electric dipole moment.  The molecules possessing dipole moment is considered as a handle that stirs the electromagnetic field into oscillation.  The energy for a rotational spectrum relates the quantum number and rotational constant to give the expression, E=B˜J(J+1).  The rotational constant further gives a relation with the moment of inertia of the molecule.

Expert Solution & Answer
Check Mark

Answer to Problem 11B.4P

The moment of inertia and bond length of H1C35l is 2.73×10-47 kg m2_ and 1.30×1010 m_ respectively.

The separation of lines in H2C35l is 42.74 cm1_, 53.41 cm_1, 63.98 cm_1, 74.57 cm_1, 85.10 cm_1, 95.53 cm_1, 105.98 cm_1, 116.38 cm_1.

Explanation of Solution

From the wavenumbers at various transitions, the separation is calculated as follows.

  2B˜1(in cm1)=104.13 cm183.32 cm1=20.81 cm1

For the next transition,

  2B˜2(in cm1)=124.73 cm1104.13 cm1=20.6 cm1

For the next transition,

  2B˜3(in cm1)=145.37 cm1124.73 cm1=20.61 cm1

For the next transition,

2B˜4(in cm1)=165.89 cm1145.37 cm1=20.52 cm1

For the next transition,

  2B˜5(in cm1)=186.23 cm1165.89 cm1=20.34 cm1

For the next transition,

  2B˜6(in cm1)=206.60 cm1186.23 cm1=20.37 cm1

For the next transition,

  2B˜7(in cm1)=226.86 cm1206.60 cm1=20.26 cm1

The mean of the separation values is shown below.

  2B˜mean=20.81+20.6+20.61 +20.52+20.34+20.37+20.26 7cm1=20.50 cm1

The rotational constant is calculated as shown below.

  2B˜=20.50 cm1B˜=20.502 cm1=10.25 cm1

The value of rotational constant helps in determining the moment of inertia of a molecule as shown below.

  B˜=4πIc        (1)

Where,

  • is the reduced Planck’s constant.
  • I is the moment of inertia
  • c is the speed of light.

Substitute the value of B˜ as 10.25 cm1 and c as 2.998×1010 cm s1, as 1.05457×1034 J s in equation (1).

  10.25 cm1=1.05457×1034 Js4×3.14×I×2.998×1010 cm s1I=1.05457×1034 J s3.86×1012 s1I=2.73×10-47 kg m2_

The moment of inertia of H1C35l is 2.73×10-47 kg m2_

The moment of inertia is given by the equation as shown below.

  I=μR2        (2)

Where,

  • μ is the reduced mass
  • R is the bond length of the molecule.

From the masses of the molecules, the reduced mass is calculated as shown below.

  μ=mambma+mb        (3)

Where,

  • ma is the mass of first atom
  • mb is the mass of second atom

For H1C35l, substitute the value of ma as 1 g/mol and mb as 35 and in equation (3).

  μ=1 g/mol×35 g/mol1+35 g/mol=3536g/mol=0.9722 g/mol

Convert the reduced mass in kg as shown below.

  1 g/mol =103 kg6.022×1023/mol0.9722 g/mol=0.9722 g/mol×103 kg6.022×1023 /mol=1.61×1027 kg

Substitute μ as 1.61×1027 kg and I as 2.73×1047 kg m2 in equation (2) as shown below.

  2.73×1047 kg m2=1.61×1027 kg×R2R2=1.69×1020 m2R=1.30×1010 m_

Therefore, the value of bond length of H1C35l is 1.30×1010 m_.

On replacing the H1 of H1C35l with H2C35l, the separation of line differs as the rotational constant changes with the change is effective mass.

The change is rotational constant is inversely related to the effective mass as shown below.

  B˜1IImeff

Therefore, the relation between rotational constant and effective mass is shown below.

  B˜1meff

Therefore, the ratio rotational constant of H1C35l and H2C35l is calculated as shown below.

  B˜H2C35lBH1C35l=(meff)H1C35l(meff)H2C35l        (4)

The reduced mass of H2C35l is shown below.

For H2C35l, substitute the value of ma as 2 g/mol and mb as 35 and in equation (3).

  μ=2 g/mol×35 g/mol2+35 g/mol=9037g/mol=1.89 g/mol

Convert the reduced mass in kg as shown below.

  1 g/mol =103 kg6.022×1023/mol1.89 g/mol=1.89 g/mol×103 kg6.022×1023 /mol=3.138×1027 kg

Substitute the value of (meff)H2C35l as 3.138×1027 kg and (meff)H1C35l as 1.61×1027 kg in equation (4).

  B˜H2C35lBH1C35l=1.61×1027 kg3.138×1027 kg=0.5130

The separation of lines of H2C35l is calculated as shown below.

  SH2C35l=(meff)H1C35l(meff)H2C35l×SH1C35l cm1        (5)

Where,

  • SH2C35l is the separation of H2C35l
  • (meff)H1C35l(meff)H2C35l is the ratio of effective masses
  • SH1C35l is the separation of H1C35l.

Substitute SH1C35l as 83.32 cm1 and (meff)H1C35l(meff)H2C35l as 0.5130 in equation (5).

  (SH2C35l)1=0.5130×83.32 cm1=42.74 cm1_

Substitute SH1C35l as 104.13 cm1 and (meff)H1C35l(meff)H2C35l as 0.5130 in equation (5).

  (SH2C35l)2=0.5130×104.13 cm1=53.41 cm1_

Substitute SH1C35l as 124.73 cm1 and (meff)H1C35l(meff)H2C35l as 0.5130 in equation (5).

  (SH2C35l)3=0.5130×124.73 cm1=63.98 cm1_

Substitute SH1C35l as 145.37 cm1 and (meff)H1C35l(meff)H2C35l as 0.5130 in equation (5).

  (SH2C35l)4=0.5130×145.37 cm1=74.57 cm1_

Substitute SH1C35l as 165.89 cm1 and (meff)H1C35l(meff)H2C35l as 0.5130 in equation (5).

  (SH2C35l)5=0.5130×165.89 cm1=85.10 cm1_

Substitute SH1C35l as 186.23 cm1 and (meff)H1C35l(meff)H2C35l as 0.5130 in equation (5).

  (SH2C35l)6=0.5130×186.23 cm1=95.53 cm1_

Substitute SH1C35l as 206.60 cm1 and (meff)H1C35l(meff)H2C35l as 0.5130 in equation (5).

  (SH2C35l)7=0.5130×206.60 cm1=105.98 cm1_

Substitute the value of SH1C35l as 226.86 cm1 and (meff)H1C35l(meff)H2C35l as 0.5130 in equation (5).

  (SH2C35l)8=0.5130×226.86 cm1=116.38 cm1_

Therefore, the separation of lines in H2C35l are 42.74 cm1_, 53.41 cm_1, 63.98 cm_1, 74.57 cm_1, 85.10 cm_1, 95.53 cm_1, 105.98 cm_1, 116.38 cm_1.

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Chapter 11 Solutions

PHYSICAL CHEMISTRY-WEBASSIGN

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