Chapter 14, Problem 14.18E

Interpretation Introduction

Interpretation:

The values of rotational constants $A,B,$ and $C$ for ${\text{H}}_{\text{2}}\text{O}$ in $\text{J}$ and ${\text{cm}}^{-1}$ are to be calculated when the moments of inertia for ${\text{H}}_{\text{2}}\text{O}$ are $1.09\times {10}^{-40}\text{ kg}\cdot {\text{m}}^2$, $1.91\times {10}^{-40}\text{ kg}\cdot {\text{m}}^2$, and $3.00\times {10}^{-40}\text{ kg}\cdot {\text{m}}^2$.

Concept introduction:

In a molecule, the rotational energy level corresponds to the different probable ways in which the portion of a molecule rotates around the chemical bond that binds it to the rest of the molecule. Every rotational energy level possesses degeneracy.

The asymmetric top refers to a rotor that has the different value for all moments of inertia. For asymmetric top molecule, $I_{\text{a}}<I_{\text{b}}<I_{\text{c}}$.

Answer to Problem 14.18E

The values of rotational constant $A$ in $\text{J}$ and ${\text{cm}}^{-1}$ are $5.110\times {10}^{-29}\text{J}$ and $2.572\times {10}^{-6}{\text{ cm}}^{-1}$.

The values of rotational constant $B$ in $\text{J}$ and ${\text{cm}}^{-1}$ are $2.915\times {10}^{-29}\text{J}$ and $1.467\times {10}^{-6}{\text{ cm}}^{-1}$.

The values of rotational constant $C$ in $\text{J}$ and ${\text{cm}}^{-1}$ are $1.855\times {10}^{-29}\text{J}$ and $9.339\times {10}^{-7}{\text{ cm}}^{-1}$.

Explanation of Solution

Water is an asymmetric top molecule. For asymmetric top molecule, $I_{\text{a}}<I_{\text{b}}<I_{\text{c}}$. The higher of the defined inertial moments is expected to be $I_{\text{c}}=3.00\times {10}^{-40}\text{ kg}\cdot {\text{m}}^2$. The inertial moments, $I_{\text{a}}$ and $I_{\text{b}}$ are lower of the inertial moments. So, $I_{\text{a}}=1.09\times {10}^{-40}\text{ kg}\cdot {\text{m}}^2$ and $I_{\text{b}}=1.91\times {10}^{-40}\text{ kg}\cdot {\text{m}}^2$.

The value of $A$ is calculated by the formula shown below.

$A=B=\frac{{\hslash }^2}{2I_{\text{a}}}$ …(1)

Where,

• $\hslash$ is the reduced Planck’s constant

• $I_{\text{a}}$ is the moment of inertia for $A$.

• $h$ is the Planck’s constant $\left(6.626\times {10}^{-34}\text{ J}\cdot \text{s}\right)$.

The relation between reduced Planck’s constant and Planck’s constant is shown below.

$\hslash =\frac{h}{2\pi }$. Therefore, the formula in equation (1) is represented as shown below.

$A=B=\frac{h^2}{{\left(2\pi \right)}^2\cdot 2I_{\text{a}}}$ …(2)

Substitute the values of $h$ and $I_{\text{a}}$ in the equation (2) to calculate the value of $A$.

$\begin{array}{rll}A& =& \frac{{\left(6.626\times {10}^{-34}\text{ J}\cdot \text{s}\right)}^2}{{\left(2\times 3.14\right)}^2\cdot 2\times 1.09\times {10}^{-40}\text{ kg}\cdot {\text{m}}^2}\\ & =& \frac{4.3903\times {10}^{-67}{\text{ J}}^2\cdot {\text{s}}^2}{39.438\times 2\times 1.09\times {10}^{-40}\text{ kg}\cdot {\text{m}}^2}\\ & =& \frac{4.3903\times {10}^{-67}{\text{ J}}^2\cdot {\text{s}}^2}{8.59\times {10}^{-39}\text{ kg}\cdot {\text{m}}^2}\text{, 1 J}=\text{kg}\cdot {\text{m}}^2\cdot {\text{s}}^{-2}\\ & =& 5.110\times {10}^{-29}\text{J}\end{array}$

Therefore, the rotational constant $A$ is $5.110\times {10}^{-29}\text{J}$.

The conversion of J into ${\text{cm}}^{-1}$ is done as shown below.

$\text{1 J}=5.035\times {10}^{\text{22}}{\text{ cm}}^{-1}$

Therefore, the conversion of $5.110\times {10}^{-29}\text{J}$ into ${\text{cm}}^{-1}$ is done as shown below.

$\begin{array}{rll}\text{1 J}& =& 5.110\times {10}^{-29}\times 5.035\times {10}^{\text{22}}{\text{ cm}}^{-1}\\ & =& 2.572\times {10}^{-6}{\text{ cm}}^{-1}\end{array}$

Therefore, the rotational constant $A$ is $2.572\times {10}^{-6}{\text{ cm}}^{-1}$.

The rotational constant $B$ for ${\text{H}}_{\text{2}}\text{O}$ is calculated by the formula shown below.

$A=B=\frac{h^2}{{\left(2\pi \right)}^2\cdot 2I_{\text{b}}}$ …(3)

Substitute the values of $h$ and $I_{\text{b}}$ in the equation (3) to calculate the value of $B$.

$\begin{array}{rll}A& =& \frac{{\left(6.626\times {10}^{-34}\text{ J}\cdot \text{s}\right)}^2}{{\left(2\times 3.14\right)}^2\cdot 2\times 1.91\times {10}^{-40}\text{ kg}\cdot {\text{m}}^2}\\ & =& \frac{4.3903\times {10}^{-67}{\text{ J}}^2\cdot {\text{s}}^2}{39.438\times 2\times 1.91\times {10}^{-40}\text{ kg}\cdot {\text{m}}^2}\\ & =& \frac{4.3903\times {10}^{-67}{\text{ J}}^2\cdot {\text{s}}^2}{1.506\times {10}^{-38}\text{ kg}\cdot {\text{m}}^2}\text{, 1 J}=\text{kg}\cdot {\text{m}}^2\cdot {\text{s}}^{-2}\\ & =& 2.915\times {10}^{-29}\text{J}\end{array}$

Therefore, the rotational constant $B$ is $2.915\times {10}^{-29}\text{J}$.

The conversion of J into ${\text{cm}}^{-1}$ is done as shown below.

$\text{1 J}=5.035\times {10}^{\text{22}}{\text{ cm}}^{-1}$

Therefore, the conversion of $2.915\times {10}^{-29}\text{J}$ into ${\text{cm}}^{-1}$ is done as shown below.

$\begin{array}{rll}\text{1 J}& =& 2.915\times {10}^{-29}\text{J}\times 5.035\times {10}^{\text{22}}{\text{ cm}}^{-1}\\ & =& 1.467\times {10}^{-6}{\text{ cm}}^{-1}\end{array}$

Therefore, the rotational constant $B$ is $1.467\times {10}^{-6}{\text{ cm}}^{-1}$.

The rotational constant $C$ for ${\text{H}}_{\text{2}}\text{O}$ is calculated by the formula shown below.

$A=B=\frac{h^2}{{\left(2\pi \right)}^2\cdot 2I_{\text{c}}}$ …(4)

Substitute the values of $h$ and $I_{\text{c}}$ in the equation (4) to calculate the value of $C$.

$\begin{array}{rll}A& =& \frac{{\left(6.626\times {10}^{-34}\text{ J}\cdot \text{s}\right)}^2}{{\left(2\times 3.14\right)}^2\cdot 2\times 3.00\times {10}^{-40}\text{ kg}\cdot {\text{m}}^2}\\ & =& \frac{4.3903\times {10}^{-67}{\text{ J}}^2\cdot {\text{s}}^2}{39.438\times 2\times 3.00\times {10}^{-40}\text{ kg}\cdot {\text{m}}^2}\\ & =& \frac{4.3903\times {10}^{-67}{\text{ J}}^2\cdot {\text{s}}^2}{2.366\times {10}^{-38}\text{ kg}\cdot {\text{m}}^2}\text{, 1 J}& =& \text{kg}\cdot {\text{m}}^2\cdot {\text{s}}^{-2}\\ & =& 1.855\times {10}^{-29}\text{J}\end{array}$

Therefore, the rotational constant $C$ is $1.855\times {10}^{-29}\text{J}$.

The conversion of J into ${\text{cm}}^{-1}$ is done as shown below.

$\text{1 J}=5.035\times {10}^{\text{22}}{\text{ cm}}^{-1}$

Therefore, the conversion of $1.855\times {10}^{-29}\text{J}$ into ${\text{cm}}^{-1}$ is done as shown below.

$\begin{array}{rll}\text{1 J}& =& 1.855\times {10}^{-29}\text{J}\times 5.035\times {10}^{\text{22}}{\text{ cm}}^{-1}\\ & =& 9.339\times {10}^{-7}{\text{ cm}}^{-1}\end{array}$

Therefore, the rotational constant $C$ is $9.339\times {10}^{-7}{\text{ cm}}^{-1}$.

Conclusion

The values of rotational constant $A$ in $\text{J}$ and ${\text{cm}}^{-1}$ are $5.110\times {10}^{-29}\text{J}$ and $2.572\times {10}^{-6}{\text{ cm}}^{-1}$.

The values of rotational constant $B$ in $\text{J}$ and ${\text{cm}}^{-1}$ are $2.915\times {10}^{-29}\text{J}$ and $1.467\times {10}^{-6}{\text{ cm}}^{-1}$.

The values of rotational constant $C$ in $\text{J}$ and ${\text{cm}}^{-1}$ are $1.855\times {10}^{-29}\text{J}$ and $9.339\times {10}^{-7}{\text{ cm}}^{-1}$.