Chapter 11, Problem 11C.17P

(a)

Interpretation Introduction

Interpretation:

The vibrational wavenumber of $\text{CO}$ has to be stated.

Concept introduction:

Each vibrational spectrum at high resolution exhibits closely spaced components.  There is very less spacing between these components signifying that these components are due to rotational transitions.  During simultaneous vibrational and rotational changes, the rotational quantum number changes by a factor of one during vibrational transition of a molecule.  On combining the vibration and rotation terms, the vibration-rotation spectra gives the equation as shown below.

    $\tilde{S}\left(\nu ,J\right)=\left(\nu +\frac{1}{2}\right)\tilde{\nu }+\tilde{B}J\left(J+1\right)$

Answer to Problem 11C.17P

The vibrational wavenumber of $\text{CO}$ is at $\underset{\_}{2150c{m}^{-1}}$.

Explanation of Solution

The vibrational wavenumber is given by the position of band centre. In case of carbon monoxide, a low resolution spectrum shows that the band centre occurs at $2150{\text{cm}}^{-1}$. Therefore, the vibrational wavenumber of Carbon monoxide occurs at $\underset{\_}{2150c{m}^{-1}}$.

(b)

Interpretation Introduction

Interpretation:

The molar zero-point vibrational energy of $\text{CO}$ has to be stated.

Concept introduction:

The same concept introduction as discussed in part (a).

Answer to Problem 11C.17P

The molar zero-point energy is $\underset{\_}{12869.014Jmo{l}^{-1}}$.

Explanation of Solution

The zero point energy is calculated by the formula shown below.

  $E=\left(v+\frac{1}{2}\right)hc\tilde{\nu }+\tilde{B}J(J+1)$        (1)

Where,

• $\nu$ is vibrational quantum number
• $h$ is the reduced Planck’s constant
• $c$ is the speed of light
• $\tilde{\nu }$ is the wavenumber
• $\tilde{B}$ is the rotational constant.
• $J$ is the rotational quantum number,.

Substitute $\tilde{\nu }$ as $2150{\text{cm}}^{-1}$, $h$ as $6.626\times {10}^{-34}\text{J}\cdot \text{s}$ , $c$ as $2.998\times {10}^{10}\text{cm}{\text{s}}^{-1}$ $\nu$ as $0$, $J$ as $0$ in equation (1).

  $\begin{array}{c}E=\left(0+\frac{1}{2}\right)\left(6.626\times {10}^{-34}\text{J}\cdot \text{s}\right)\left(2.998\times {10}^{10}\text{cm}{\text{s}}^{-1}\right)\left(2150{\text{cm}}^{-1}\right)+\tilde{B}\times 0(0+1)\\ =\left(\frac{1}{2}\right)\left(6.626\times {10}^{-34}\text{J}\cdot \text{s}\right)\left(2.998\times {10}^{10}\text{cm}{\text{s}}^{-1}\right)\left(2150{\text{cm}}^{-1}\right)\\ =\frac{4.274\times {10}^{-20}\text{J}}{2}\\ =2.137\times {10}^{-20}\text{J}\end{array}$

For molar zero energy, multiplying the above value by Avogadro’s number as shown below.

  $E\left(\text{J/mol}\right)=E\left(\text{J}\right)\times N_A$        (2)

Substitute $E\left(\text{J}\right)$        (2)

$\begin{array}{c}E\left(\text{J/mol}\right)=2.137\times {10}^{-20}\text{J}\times 6.022\times {10}^{23}{\text{mol}}^{-1}\\ =\underset{\_}{12869.014Jmo{l}^{-1}}\end{array}$

Therefore, the molar zero-point energy is $\underset{\_}{12869.014Jmo{l}^{-1}}$.

(c)

Interpretation Introduction

Interpretation:

The force constant of $\text{CO}$ has to be stated.

Concept introduction:

The strength of a bond in a bond is measured by force constant. The force constant of a molecule is related to the effective mass and vibrational frequency.  The vibrational energy of a molecule is given by the equation as shown below.

$E_{\nu }=\left(\nu +\frac{1}{2}\right)\hslash \omega$.

Answer to Problem 11C.17P

The force constant for $\text{CO}$ is $\underset{\_}{1863.01N{m}^{-1}}$.

Explanation of Solution

The wavenumber of a molecule is given by the equation as shown below.

  $\tilde{\nu }=\frac{1}{2\text{π}c}{\left(\frac{k_{\text{f}}}{m_{\text{eff}}}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$        (3)

Where,

• $k_{\text{f}}$ is the force constant.
• $m_{\text{eff}}$ is the effective mass.
• $c$ is the speed of light

Rearrange the above equation for force constant as shown below.

  $\begin{array}{c}\tilde{\nu }=\frac{1}{2\text{π}c}{\left(\frac{k_{\text{f}}}{m_{\text{eff}}}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\\ 4{\tilde{\nu }}^2{\text{π}}^2{c}^2=\frac{k_{\text{f}}}{m_{\text{eff}}}\end{array}$

Thus the equation reduces to

  $k_{\text{f}}=4{\text{π}}^2{c}^2{\tilde{\nu }}^2{m}_{\text{eff}}$        (4)

The effective mass of $\text{CO}$ is calculated as shown below.

  $m_{\text{eff}}=\frac{m_a{m}_b}{m_a+m_b}$        (5)

Where,

• $m_a$ is the mass of first atom
• $m_b$ is the mass of second atom

For $\text{CO}$, substitute the value of $m_a$ as $\text{12 g/mol}$ and $m_b$ as $16\text{ g/mol}$ and in equation (5).

  $\begin{array}{c}{m}_{\text{eff}}=\frac{\text{12 g/mol}\times 16\text{ g/mol}}{12+16\text{ g/mol}}\\ =\frac{192}{28}\text{ g/mol}\\ =\text{6}\text{.85}\text{g/mol}\end{array}$

Convert the effective mass in $\text{kg}$ as shown below.

  $\begin{array}{c}1\text{ g/mol }=\frac{{10}^{-3}\text{ kg}}{6.022\times {10}^{23}\text{/mol}}\\ 6.85\text{ g/mol}=\frac{\text{6}\text{.85 g/mol}\times {\text{10}}^{-3}\text{ kg}}{6.022\times {10}^{23}\text{ /mol}}\\ =1.138\times {10}^{-26}\text{ kg}\end{array}$

Substitute $\tilde{\nu }$ as $2150{\text{ cm}}^{-1}$, $m_{\text{eff}}$ as $\text{1}\text{.138}\times {\text{10}}^{-26}\text{ kg}$ and $c$ as $2.998\times {10}^{10}{\text{ cm s}}^{-1}$ in equation (4).

  $\begin{array}{c}{k}_{\text{f}}=4\times {\left(3.14\right)}^2\times {\left(2.998\times {10}^{10}\text{ cm}{\text{s}}^{-1}\right)}^2\times {\left(2150{\text{ cm}}^{-1}\right)}^2\times \left(1.138\times {10}^{-26}\text{kg}\right)\\ =4\times 9.8596\times 8.98\times {10}^{20}{\text{cm}}^2{\text{s}}^{-2}\times 4622500{\text{cm}}^{-\text{2}}\times 1.138\times {10}^{-26}\text{kg}\\ =\underset{\_}{1863.01N{m}^{-1}}\end{array}$

Therefore, the force constant for $\text{CO}$ is $\underset{\_}{1863.01N{m}^{-1}}$.

(d)

Interpretation Introduction

Interpretation:

The rotational constant of $\text{CO}$ has to be stated.

Concept introduction:

The energy for a rotational spectrum relates the quantum number and rotational constant to give the expression, $E=\tilde{B}J(J+1)$.  The rotational constant further gives a relation with the moment of inertia of the molecule.

Answer to Problem 11C.17P

The rotational constant is $\underset{\_}{3.8275c{m}^{-1}}$.

Explanation of Solution

In a high resolution spectrum, it is observed that the band centre is split into two peaks. As the vibrational spectrum splits to give two peaks with a separation of $7.655{\text{cm}}^{-1}$.  This separation between two peaks is given by $2\tilde{B}$.  Therefore, rotational constant is calculated as shown below.

  $\begin{array}{c}2\tilde{B}=7.655{\text{cm}}^{-1}\\ \tilde{B}=\frac{7.655}{2}{\text{cm}}^{-1}\\ =\underset{\_}{3.8275c{m}^{-1}}\end{array}$

Therefore, the rotational constant is $\underset{\_}{3.8275c{m}^{-1}}$.

(e)

Interpretation Introduction

Interpretation:

The bond length of $\text{CO}$ has to be stated.

Concept introduction:

The same concept introduction as mentioned in part (d)

Answer to Problem 11C.17P

The value of bond length of $\text{CO}$ is $\underset{\_}{8.02\times 10^{-11}m}$.

Explanation of Solution

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

  $\tilde{B}=\frac{\hslash }{4\text{π}Ic}$        (6)

Where,

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

Substitute the value of $\tilde{B}$        (6)

  $\begin{array}{c}3.8275{\text{ cm}}^{-1}=\frac{1.05457\times {10}^{-34}\text{ J}\cdot \text{s}}{4\times 3.14\times I\times 2.998\times {10}^{10}{\text{ cm s}}^{-1}}\\ I=\frac{1.05457\times {10}^{-34}\text{ J}\cdot \text{s}}{4\times 3.14\times 3.8275{\text{ cm}}^{-1}\times 2.998\times {10}^{10}{\text{ cm s}}^{-1}}\\ I=\frac{1.05457\times {10}^{-34}\text{ J}\cdot \text{s}}{1.44\times {10}^{12}}\\ I=7.32\times {10}^{-47}{\text{ kg m}}^2\end{array}$

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

  $I=\mu R^2$        (7)

Where,

• $\mu$ is the reduced mass
• $R$ is the bond length of the molecule.

Substitute $\mu$ as $1.138\times {10}^{-26}\text{ kg}$ and $I$ as $7.32\times {10}^{-47}{\text{ kg m}}^2$ in equation (7) as shown below.

  $\begin{array}{c}7.32\times {10}^{-47}{\text{ kg m}}^2=1.138\times {10}^{-26}\text{ kg}\times R^2\\ R^2=6.43\times {10}^{-21}{\text{ m}}^2\\ R=\underset{\_}{8.02\times 10^{-11}m}\end{array}$

Therefore, the value of bond length of $\text{CO}$ is $\underset{\_}{8.02\times 10^{-11}m}$.