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.5P

(a)

Interpretation Introduction

Interpretation:

The dissociation energy of 1H35Cl has to be calculated assuming that the potential remains unchanged on deuteration.

Concept introduction:

The Morse potential is an interatomic interaction model that gives the expression for the potential energy of a molecule.  It also deals with interactions that exist between an atom and another surface.  The Morse potential can be expressed as shown below.

    a2=(meffωxe2hcD˜e)

(a)

Expert Solution
Check Mark

Answer to Problem 11C.5P

The value of dissociation energy of 1H35Cl is 5.15eV_.

Explanation of Solution

The formula to calculate dissociation energy is shown below.

    D˜0=D˜eν˜'        (1)

Where,

  • D˜0 is the dissociation energy of the molecule.
  • D˜e is the dissociation energy of the potential well.
  • ν˜' is the wavenumber of anharmonic oscillator.

The formula to calculate the wavenumber of anharmonic oscillator is shown below.

    ν˜'=12ν˜14xeν˜        (2)

Where,

  • xe is the anharmonicity constant.
  • ν˜ is the wavenumber

The value of ν˜ is 2989.7cm1 and the value of xeν˜ is 52.05cm1.

Substitute the values of ν˜ and xeν˜ in equation (2).

    ν˜'=12ν˜14xeν˜=12(2989.7cm1)14(52.05cm1)=(1494.8513.01)cm11482cm1

The value of 1eV is equal to 8065.5cm1.  The conversion of wavenumber of anharmonic oscillator into eV is shown below.

    ν˜'=1482cm1×1eV8065.5cm1=0.1837eV0.184eV

The value of hcD˜e is given as 5.33eV.  Substitute the values of ν˜' and hcD˜e in equation (1).

    D˜0=D˜eν˜'=5.33eV0.184eV=5.146eV5.15eV_

Therefore, the value of dissociation energy of 1H35Cl is 5.15eV_.

(b)

Interpretation Introduction

Interpretation:

The dissociation energy of 2H35Cl has to be calculated assuming that the potential remains unchanged on deuteration.

Concept introduction:

Concept introduction is the same as that discussed in part (a).

(b)

Expert Solution
Check Mark

Answer to Problem 11C.5P

The value of dissociation energy of 2H35Cl is 5.2eV_.

Explanation of Solution

The expression for the Morse potential is shown below.

    a2=(meffωxe2hcD˜e)        (1)

Where,

  • meff is the effective mass.
  • ω is the vibrational frequency.
  • h is the Planck’s constant.
  • c is the velocity of light.
  • D˜e is the depth of potential minimum.
  • xe is the anharmonicity constant.

The relation between vibrational frequency ω and ν˜ is shown below.

    ω=2πhcν˜        (2)

Where,

  • ω is the vibrational frequency.
  • h is the Planck’s constant.
  • c is the velocity of light.
  • ν˜ is the wavenumber.

Comparing equation (1) and (2), the relation between ν˜xe and meff can be written as shown below.

    ν˜xe1meff        (3)

The value of De can be written as shown below.

    De=ν˜24xeν˜        (4)

Comparing equation (3) and (4), the relation between ν˜ and meff can be established as shown below.

    ν˜21meffν˜(1meff)12        (5)

The mass of hydrogen atom is 1.007825u.

The mass of hydrogen atom in kg is converted as shown below.

    mH=(1.007825u)(1.66054×1027kg1u)=1.6735×1027kg

The mass of chlorine atom is 34.96885u.

The mass of chlorine atom in kg is converted as shown below.

    mCl=(34.96885u)(1.66054×1027kg1u)=5.8067×1026kg

The effective mass of a molecule (meff) is given by the expression as shown below.

  meff=mAmBmA+mB        (6)

Where,

  • mA is the mass of atom A.
  • mB is the mass of atom B.

Substitute the values of the mass of hydrogen atom and chlorine atom in the equation (6).

  meff(HCl)=(1.6735×1027kg)(5.8067×1026kg)(1.6735×1027kg)+(5.8067×1026kg)=1.6266×1027kg

The mass of deuterium atom (Dor2H) is 2.0140u.

The mass of deuterium atom in kg is converted as shown below.

    mD=(2.0140u)(1.66054×1027kg1u)=3.3443×1027kg

Substitute the values of the mass of deuterium atom and chlorine atom in the equation (6).

meff(DCl)=(3.3443×1027kg)(5.8067×1026kg)(3.3443×1027kg)+(5.8067×1026kg)=3.1622×1027kg

From the relation given in equation (5), the ratio of wavenumbers of HCl and DCl is shown below.

    ν˜(DCl)=(meff(HCl)meff(DCl))12ν˜(HCl)        (7)

Substitute the values of meff(HCl), meff(DCl) and ν˜(HCl) in the above expression.

    ν˜(DCl)=(meff(HCl)meff(DCl))12ν˜(HCl)=(1.6266×1027kg3.1622×1027kg)12×2989.7cm1=2144.2cm1

The equation obtained by multiplying xe on both sides of equation (7) is shown below.

    xeν˜(DCl)=(meff(HCl)meff(DCl))12xeν˜(HCl)        (8)

Substitute the values of meff(HCl), meff(DCl) and xeν˜(HCl) in the above expression.

    xeν˜(DCl)=(meff(HCl)meff(DCl))12xeν˜(HCl)=(1.6266×1027kg3.1622×1027kg)12×52.05cm1=37.33cm1

The formula to calculate dissociation energy is shown below.

    D˜0=D˜eν˜'        (9)

Where,

  • D˜0 is the dissociation energy of the molecule.
  • D˜e is the dissociation energy of the potential well.
  • ν˜' is the wavenumber of anharmonic oscillator.

The formula to calculate the wavenumber of anharmonic oscillator is shown below.

    ν˜'=12ν˜14xeν˜        (10)

Where,

  • xe is the anharmonicity constant.
  • ν˜ is the wavenumber

The value of ν˜ for DCl is 2144.2cm1 and the value of xeν˜ for DCl is 37.33cm1.

Substitute the values of ν˜ and xeν˜ in equation (10).

    ν˜'(DCl)=12ν˜(DCl)14xeν˜(DCl)=12(2144.2cm1)14(37.33cm1)=(1072.19.3325)cm1=1062.7675cm1

The value of 1eV is equal to 8065.5cm1.  The conversion of wavenumber of anharmonic oscillator into eV is shown below.

    ν˜'=1062.7675cm1×1eV8065.5cm1=0.1317eV0.132eV

The value of hcD˜e is given as 5.33eV.  Substitute the values of ν˜' and hcD˜e in equation (9).

    D˜0=D˜eν˜'=5.33eV0.132eV=5.198eV5.2eV_

Therefore, the value of dissociation energy of 2H35Cl(DCl) is 5.2eV_.

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Chapter 11 Solutions

Atkins' Physical Chemistry

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