Chapter 11, Problem 11C.5P

(a)

Interpretation Introduction

Interpretation:

The dissociation energy of ${}^1{\text{H}}^{35}\text{Cl}$ 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.

    $a^2=\left(\frac{m_{eff}\omega x_e}{2hc{\tilde{D}}_e}\right)$

Answer to Problem 11C.5P

The value of dissociation energy of ${}^1{\text{H}}^{35}\text{Cl}$ is $\underset{\_}{5.15eV}$.

Explanation of Solution

The formula to calculate dissociation energy is shown below.

    ${\tilde{D}}_0={\tilde{D}}_e-\tilde{\nu }\text{'}$        (1)

Where,

• ${\tilde{D}}_0$ is the dissociation energy of the molecule.
• ${\tilde{D}}_e$ is the dissociation energy of the potential well.
• $\tilde{\nu }\text{'}$ is the wavenumber of anharmonic oscillator.

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

    $\tilde{\nu }\text{'}=\frac{1}{2}\tilde{\nu }-\frac{1}{4}{x}_{\text{e}}\tilde{\nu }$        (2)

Where,

• $x_{\text{e}}$ is the anharmonicity constant.
• $\tilde{\nu }$ is the wavenumber

The value of $\tilde{\nu }$ is $\text{2989}\text{.7}{\text{cm}}^{-1}$ and the value of $x_{\text{e}}\tilde{\nu }$ is $52.05{\text{cm}}^{-1}$.

Substitute the values of $\tilde{\nu }$ and $x_{\text{e}}\tilde{\nu }$ in equation (2).

    $\begin{array}{c}\tilde{\nu }\text{'}=\frac{1}{2}\tilde{\nu }-\frac{1}{4}{x}_{\text{e}}\tilde{\nu }\\ =\frac{1}{2}\left(\text{2989}\text{.7}{\text{cm}}^{-1}\right)-\frac{1}{4}\left(52.05{\text{cm}}^{-1}\right)\\ =\left(1494.85-13.01\right){\text{cm}}^{-1}\\ \approx 1482{\text{cm}}^{-1}\end{array}$

The value of $\text{1}\text{eV}$ is equal to $8065.5{\text{cm}}^{-1}$.  The conversion of wavenumber of anharmonic oscillator into $\text{eV}$ is shown below.

    $\begin{array}{c}\tilde{\nu }\text{'}=1482{\text{cm}}^{-1}\times \frac{\text{1}\text{eV}}{8065.5{\text{cm}}^{-1}}\\ =0.1837\text{eV}\\ \approx 0.184\text{eV}\end{array}$

The value of $hc{\tilde{D}}_e$ is given as $\text{5}\text{.33}\text{eV}$.  Substitute the values of $\tilde{\nu }\text{'}$ and $hc{\tilde{D}}_e$ in equation (1).

    $\begin{array}{c}{\tilde{D}}_0={\tilde{D}}_e-\tilde{\nu }\text{'}\\ =\text{5}\text{.33}\text{eV}-0.184\text{eV}\\ =5.146\text{eV}\\ \approx \underset{\_}{5.15eV}\end{array}$

Therefore, the value of dissociation energy of ${}^1{\text{H}}^{35}\text{Cl}$ is $\underset{\_}{5.15eV}$.

(b)

Interpretation Introduction

Interpretation:

The dissociation energy of ${}^2{\text{H}}^{35}\text{Cl}$ 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).

Answer to Problem 11C.5P

The value of dissociation energy of ${}^2{\text{H}}^{35}\text{Cl}$ is $\underset{\_}{5.2eV}$.

Explanation of Solution

The expression for the Morse potential is shown below.

    $a^2=\left(\frac{m_{eff}\omega x_{\text{e}}}{2hc{\tilde{D}}_{\text{e}}}\right)$        (1)

Where,

• $m_{eff}$ is the effective mass.
• $\omega$ is the vibrational frequency.
• $h$ is the Planck’s constant.
• $c$ is the velocity of light.
• ${\tilde{D}}_{\text{e}}$ is the depth of potential minimum.
• $x_{\text{e}}$ is the anharmonicity constant.

The relation between vibrational frequency $\omega$ and $\tilde{\nu }$ is shown below.

    $\omega =2\pi hc\tilde{\nu }$        (2)

Where,

• $\omega$ is the vibrational frequency.
• $h$ is the Planck’s constant.
• $c$ is the velocity of light.
• $\tilde{\nu }$ is the wavenumber.

Comparing equation (1) and (2), the relation between $\tilde{\nu }{x}_{\text{e}}$ and $m_{eff}$ can be written as shown below.

    $\tilde{\nu }{x}_{\text{e}}\propto \frac{1}{m_{eff}}$        (3)

The value of $D_{\text{e}}$ can be written as shown below.

    $D_{\text{e}}=\frac{{\tilde{\nu }}^2}{4x_{\text{e}}\tilde{\nu }}$        (4)

Comparing equation (3) and (4), the relation between $\tilde{\nu }$ and $m_{eff}$ can be established as shown below.

    $\begin{array}{l}{\tilde{\nu }}^2\propto \frac{1}{m_{eff}}\\ \tilde{\nu }\propto {\left(\frac{1}{m_{eff}}\right)}^{\frac{1}{2}}\end{array}$        (5)

The mass of hydrogen atom is $1.007825\text{u}$.

The mass of hydrogen atom in $\text{kg}$ is converted as shown below.

    $\begin{array}{c}{m}_{\text{H}}=\left(1.007825\text{u}\right)\left(\frac{1.66054\times {10}^{-27}\text{kg}}{1\text{u}}\right)\\ =1.6735\times {10}^{-27}\text{kg}\end{array}$

The mass of chlorine atom is $34.96885\text{u}$.

The mass of chlorine atom in $\text{kg}$ is converted as shown below.

    $\begin{array}{c}{m}_{\text{Cl}}=\left(34.96885\text{u}\right)\left(\frac{1.66054\times {10}^{-27}\text{kg}}{1\text{u}}\right)\\ =5.8067\times {10}^{-26}\text{kg}\end{array}$

The effective mass of a molecule $\left(m_{eff}\right)$ is given by the expression as shown below.

  $m_{eff}=\frac{m_{\text{A}}{m}_{\text{B}}}{m_{\text{A}}+m_{\text{B}}}$        (6)

Where,

• $m_{\text{A}}$ is the mass of atom $\text{A}$.
• $m_{\text{B}}$ is the mass of atom $\text{B}$.

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

  $\begin{array}{c}{m}_{eff}\left(\text{HCl}\right)=\frac{\left(1.6735\times {10}^{-27}\text{kg}\right)\left(5.8067\times {10}^{-26}\text{kg}\right)}{\left(1.6735\times {10}^{-27}\text{kg}\right)+\left(5.8067\times {10}^{-26}\text{kg}\right)}\\ =1.6266\times {10}^{-27}\text{kg}\end{array}$

The mass of deuterium atom $\left(\text{D}\text{or}{}^2\text{H}\right)$ is $2.0140\text{u}$.

The mass of deuterium atom in $\text{kg}$ is converted as shown below.

    $\begin{array}{c}{m}_{\text{D}}=\left(2.0140\text{u}\right)\left(\frac{1.66054\times {10}^{-27}\text{kg}}{1\text{u}}\right)\\ =3.3443\times {10}^{-27}\text{kg}\end{array}$

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

$\begin{array}{c}{m}_{eff}\left(\text{DCl}\right)=\frac{\left(3.3443\times {10}^{-27}\text{kg}\right)\left(5.8067\times {10}^{-26}\text{kg}\right)}{\left(3.3443\times {10}^{-27}\text{kg}\right)+\left(5.8067\times {10}^{-26}\text{kg}\right)}\\ =3.1622\times {10}^{-27}\text{kg}\end{array}$

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

    $\tilde{\nu }\left(\text{DCl}\right)={\left(\frac{m_{eff}\left(\text{HCl}\right)}{m_{eff}\left(\text{DCl}\right)}\right)}^{\frac{1}{2}}\cdot \tilde{\nu }\left(\text{HCl}\right)$        (7)

Substitute the values of $m_{eff}\left(\text{HCl}\right)$, $m_{eff}\left(\text{DCl}\right)$ and $\tilde{\nu }\left(\text{HCl}\right)$ in the above expression.

    $\begin{array}{c}\tilde{\nu }\left(\text{DCl}\right)={\left(\frac{m_{eff}\left(\text{HCl}\right)}{m_{eff}\left(\text{DCl}\right)}\right)}^{\frac{1}{2}}\cdot \tilde{\nu }\left(\text{HCl}\right)\\ ={\left(\frac{1.6266\times {10}^{-27}\text{kg}}{3.1622\times {10}^{-27}\text{kg}}\right)}^{\frac{1}{2}}\times \text{2989}\text{.7}{\text{cm}}^{-1}\\ =2144.2{\text{cm}}^{-1}\end{array}$

The equation obtained by multiplying $x_{\text{e}}$ on both sides of equation (7) is shown below.

    $x_{\text{e}}\tilde{\nu }\left(\text{DCl}\right)={\left(\frac{m_{eff}\left(\text{HCl}\right)}{m_{eff}\left(\text{DCl}\right)}\right)}^{\frac{1}{2}}\cdot x_{\text{e}}\tilde{\nu }\left(\text{HCl}\right)$        (8)

Substitute the values of $m_{eff}\left(\text{HCl}\right)$, $m_{eff}\left(\text{DCl}\right)$ and $x_{\text{e}}\tilde{\nu }\left(\text{HCl}\right)$ in the above expression.

    $\begin{array}{c}{x}_{\text{e}}\tilde{\nu }\left(\text{DCl}\right)={\left(\frac{m_{eff}\left(\text{HCl}\right)}{m_{eff}\left(\text{DCl}\right)}\right)}^{\frac{1}{2}}\cdot x_{\text{e}}\tilde{\nu }\left(\text{HCl}\right)\\ ={\left(\frac{1.6266\times {10}^{-27}\text{kg}}{3.1622\times {10}^{-27}\text{kg}}\right)}^{\frac{1}{2}}\times \text{52}\text{.05}{\text{cm}}^{-1}\\ =37.33{\text{cm}}^{-1}\end{array}$

The formula to calculate dissociation energy is shown below.

    ${\tilde{D}}_0={\tilde{D}}_e-\tilde{\nu }\text{'}$        (9)

Where,

• ${\tilde{D}}_0$ is the dissociation energy of the molecule.
• ${\tilde{D}}_e$ is the dissociation energy of the potential well.
• $\tilde{\nu }\text{'}$ is the wavenumber of anharmonic oscillator.

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

    $\tilde{\nu }\text{'}=\frac{1}{2}\tilde{\nu }-\frac{1}{4}{x}_{\text{e}}\tilde{\nu }$        (10)

Where,

• $x_{\text{e}}$ is the anharmonicity constant.
• $\tilde{\nu }$ is the wavenumber

The value of $\tilde{\nu }$ for $\text{DCl}$ is $2144.2{\text{cm}}^{-1}$ and the value of $x_{\text{e}}\tilde{\nu }$ for $\text{DCl}$ is $37.33{\text{cm}}^{-1}$.

Substitute the values of $\tilde{\nu }$ and $x_{\text{e}}\tilde{\nu }$ in equation (10).

    $\begin{array}{c}\tilde{\nu }\text{'}\left(\text{DCl}\right)=\frac{1}{2}\tilde{\nu }\left(\text{DCl}\right)-\frac{1}{4}{x}_{\text{e}}\tilde{\nu }\left(\text{DCl}\right)\\ =\frac{1}{2}\left(2144.2{\text{cm}}^{-1}\right)-\frac{1}{4}\left(37.33{\text{cm}}^{-1}\right)\\ =\left(1072.1-9.3325\right){\text{cm}}^{-1}\\ =1062.7675{\text{cm}}^{-1}\end{array}$

The value of $\text{1}\text{eV}$ is equal to $8065.5{\text{cm}}^{-1}$.  The conversion of wavenumber of anharmonic oscillator into $\text{eV}$ is shown below.

    $\begin{array}{c}\tilde{\nu }\text{'}=1062.7675{\text{cm}}^{-1}\times \frac{\text{1}\text{eV}}{8065.5{\text{cm}}^{-1}}\\ =0.1317\text{eV}\\ \approx 0.132\text{eV}\end{array}$

The value of $hc{\tilde{D}}_e$ is given as $\text{5}\text{.33}\text{eV}$.  Substitute the values of $\tilde{\nu }\text{'}$ and $hc{\tilde{D}}_e$ in equation (9).

    $\begin{array}{c}{\tilde{D}}_0={\tilde{D}}_e-\tilde{\nu }\text{'}\\ =\text{5}\text{.33}\text{eV}-0.132\text{eV}\\ =5.198\text{eV}\\ \approx \underset{\_}{5.2eV}\end{array}$

Therefore, the value of dissociation energy of ${}^2{\text{H}}^{35}\text{Cl}\left(\text{DCl}\right)$ is $\underset{\_}{5.2eV}$.