Atkins' Physical Chemistry
Atkins' Physical Chemistry
11th Edition
ISBN: 9780198769866
Author: ATKINS, P. W. (peter William), De Paula, Julio, Keeler, JAMES
Publisher: Oxford University Press
Question
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Chapter 11, Problem 11C.6P
Interpretation Introduction

Interpretation:

The Morse potential energy curve from 50pm to 800pm around Re=236.7pm has to be plotted. The explanation for the way by which the rotation of a molecule may weaken its bond through the permission for the kinetic energy of rotation of a molecule has to be stated. The graph for J=40,80 and 100 is to be drawn. The affect on the dissociation energy through the rotation is to be stated.

Concept introduction:

The curve that represents the interatomic interaction model corresponding to the diatomic molecule’s potential energy is known as Morse potential curve. The curve gives the approximation for the vibrational structure of a diatomic molecule.

Expert Solution & Answer
Check Mark

Answer to Problem 11C.6P

The Morse potential energy curve from 50pm to 800pm around Re=236.7pm is shown below.

Atkins' Physical Chemistry, Chapter 11, Problem 11C.6P , additional homework tip  1

The rotation of a molecule may weaken its bond through the permission for the kinetic energy of rotation of a molecule as the value of V is attained after considering the kinetic energy. So, at high value of V, the increase in the value of R is observed.

The graph for J=40,80 and 100 is shown below.

Atkins' Physical Chemistry, Chapter 11, Problem 11C.6P , additional homework tip  2

The dissociation energy through the rotation is not affected by rotation.

Explanation of Solution

It is given that in the study of 85Rb1H, the value of wave number is v˜=936.8cm1 and xev˜=14.15cm1.

The value of Re is 236.7pm.

The value of rotational constant, B˜=3.020cm1.

The given mass of 85Rb is 84.911mu.

The value of 1mu is 1.66054×1027kg.

The value of Planck’s constant is 6.626×1034Js.

The value of speed of light is 2.998×1010cm/s.

The expression that to calculate the Morse potential energy is mentioned as follows.

    V(R)=hcDe˜{1ea(RRe)}2        (1)

Where,

  • De˜ is the molecular partition function.
  • h is the Planck’s constant.
  • c is the speed of light.
  • Re is the equilibrium bond length.
  • R is the bond length.

The expression for calculating the wave number is given below.

    v˜=ω2πc=936.8cm1        (2)

Where,

  • ω is the Wave number.

The expression for calculating the anharmonicity constant is given below.

    xe=a22meffωa22meffω=v˜4De˜=v˜4xe        (3)

Where,

  • meff is the effective mass.

The expression for calculating the effective mass of RbH is given below.

    meff(RbH)=mHmRbmH+mRb        (4)

Where,

  • mH is the molar mass of hydrogen.
  • mRb is the molar mass of rubidium.

Substitute mH as 1.008u, and mRb as 85.47u in the above equation.

    meff(RbH)=1.008u×85.47u1.008u+85.47u=86.15386.478u×1.66×1027kg=1.654×1027kg

Thus, the effective mass is 1.654×1027kg.

The expression for calculating the dissociation constant is given below.

    De˜=v¯4xev˜        (5)

Substitute the values of v2˜ as (936.8cm1)2, and xev˜=14.15cm1 in the above equation.

    De˜=(936.8cm1)24×14.15cm1=877594.256.6cm1=15505cm1

As v=cv¯, so, the value of a is calculated by the expression given below.

    a=2πcv¯×((meff)2hcDe˜)1/2        (6)

Substitute the value of h, c, meff and De˜ in the above equation.

    a=2×3.14×2.998×1010cms1×936.8cm1×((1.654×1027kg)2×15505cm1×6.626×1034Kgm2s2×2.998×1010cm/s)1/2=17637.54×1010s1×((1.654×1027kg)616005.83Kgm2s2)1/2=9.140×109m1

Thus, the value of a is 9.140×109m1. The reciprocal value of 1a=19.140×109m1=0.1094×109m1, In nm, it becomes a=10.1094nm.

Substitute the value of a in equation (1).

    V(R)=hcDe˜{1e(R236.7pm)/0.1094nm}2V(R)hcDe˜={1e(R236.7pm)/0.1094nm}2

The value of V(R)hcDe˜ at different values of R is given below in the table.

R/pm50100 200 300 400 500 600 700800
V(R)hcDe˜20.46.200.1590.1930.6010.8280.9290.9710.988

The graph between V(R)hcDe˜ and R is shown below.

Atkins' Physical Chemistry, Chapter 11, Problem 11C.6P , additional homework tip  3

Figure 1

The expression for  rotational constant is given below.

    B˜1R2

So, the given expression, V=V+hcBJ(J+1) can be written as follows.

    V=V+hcBeJ(J+1)(Re2R2)

At the value of Be equal to 3.020cm1, Re=236.4 and R=50pm to 800pm. The different values are given below.

Re/RVhcDe˜ (for J=40,80 and 100)(Re2/R2)hcBe40(40+1) hcBe80(80+1) hcBe100(100+1)
4.73420.3409314822.4107569.84518×10203.89005×10196.06319×1019
2.3676.193867115.6026899.84518×10203.89005×10196.06319×1019
1.18350.1588757641.400672259.84518×10203.89005×10196.06319×1019
0.7890.1930048490.6225219.84518×10203.89005×10196.06319×1019
0.591750.600984960.35016819.84518×10203.89005×10196.06319×1019
0.39450.929059950.155630259.84518×10203.89005×10196.06319×1019
0.2958750.9884232230.0875420169.84518×10203.89005×10196.06319×1019
0.23670.998134950.056026899.84518×10203.89005×10196.06319×1019

Thus, the graph for V=V+hcBJ(J+1) with B˜=h/4πcμR2 at J=40,80 and 100 is shown below.

Atkins' Physical Chemistry, Chapter 11, Problem 11C.6P , additional homework tip  4

Figure 2

The value of V is attained after considering the kinetic energy. Thus, at high value of V, the increase in the value of R is observed. In case of weak bond, the bond length increases. So, the increased bond length corresponds to the weaker bond. The weaker bond is the result of the involved kinetic energy.

In the above shown graph, the three lines coincides each other. The equation for rotational constant B is given below.

    B=h8π2cI=BRe2        (7)

Where,

  • I is the moment of inertia equals to μR2
  • R is the bond length.
  • μ is the reduced mass.

So, the above equation can be written as follows.

    h8π2cμ=BRe2

Substitute the value of B as 3.020cm1 and Re as 236.7×1010cm in the above equation.

    h8π2cμ=3.020cm1×(236.7×1010cm)2=1.692×1010cm

Therefore, the equation (1) can be written as follows.

    V(R)=hcDe{1ea(RRe)}2+hcR21.692×1010cm(J+1)

The above equation suggests that a non linear area will be obtained in the curve. Therefore, there will be no effect on the dissociation energy through rotation.

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Chapter 11 Solutions

Atkins' Physical Chemistry

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