Chapter 11, Problem 11A.7P

(a)

Interpretation Introduction

Interpretation:

The integrated absorption coefficient of ${\text{CH}}_{\text{3}}\text{I}$ has to be evaluated.

Concept introduction:

The maximum value of the molar absorption coefficient indicates the intensity of a transition.  However, as the absorption bands are spread over a range of wavenumbers, the absorption coefficient at a single wavenumber might not give a true indication of the intensity of transition.  The integrated absorption coefficient is the sum of the absorption coefficients over the entire band and corresponds to the area under the plot of the molar absorption coefficient against wavenumber.

Answer to Problem 11A.7P

The integrated absorption coefficient of ${\text{CH}}_{\text{3}}\text{I}$ is $\underset{\_}{2.42\times 10^{5}d{m}^{3}mo{l}^{-1}c{m}^{-2}}$.

Explanation of Solution

The integrated absorption coefficient is calculated by the formula given below.

    $A=\frac{1}{2}{\epsilon }_{\max }\Delta \tilde{\nu }$        (1)

Where,

• $A$ is the integrated absorption coefficient.
• ${\epsilon }_{\max }$ is the maximum molar extinction coefficient.
• $\Delta \tilde{\nu }$ is the wavenumber.

The $\Delta \tilde{\nu }$ is given by the expression below.

    $\Delta \tilde{\nu }={\tilde{\nu }}_{\text{f}}-{\tilde{\nu }}_{\text{i}}$

Where,

• ${\tilde{\nu }}_{\text{f}}$ is the final wavenumber.
• ${\tilde{\nu }}_{\text{i}}$ is the initial wavenumber.

Therefore, equation (1) can be written as shown below.

    $A=\frac{1}{2}{\epsilon }_{\max }\left({\tilde{\nu }}_{\text{f}}-{\tilde{\nu }}_{\text{i}}\right)$        (2)

The value of ${\epsilon }_{\max }$ is $150{\text{dm}}^3{\text{mol}}^{-1}{\text{cm}}^{-1}$.

The value of ${\tilde{\nu }}_{\text{f}}$ is $34483{\text{cm}}^{-1}$.

The value of ${\tilde{\nu }}_{\text{i}}$ is $31250{\text{cm}}^{-1}$.

Substitute the value of ${\epsilon }_{\max }$, ${\tilde{\nu }}_{\text{f}}$, and ${\tilde{\nu }}_{\text{i}}$ in equation (2).

    $\begin{array}{c}A=\frac{1}{2}\times 150{\text{dm}}^3{\text{mol}}^{-1}{\text{cm}}^{-1}\times \left(34483{\text{cm}}^{-1}-31250{\text{cm}}^{-1}\right)\\ =\frac{1}{2}\times 150{\text{dm}}^3{\text{mol}}^{-1}{\text{cm}}^{-1}\times 3233{\text{cm}}^{-1}\\ =\underset{\_}{2.42\times 10^{5}d{m}^{3}mo{l}^{-1}c{m}^{-2}}\end{array}$

Therefore, the integrated absorption coefficient is $\underset{\_}{2.42\times 10^{5}d{m}^{3}mo{l}^{-1}c{m}^{-2}}$.

(b)

Interpretation Introduction

Interpretation:

The absorbance expected at the mid-point of the absorption lineshape in a sample cell of length $12.0\text{cm}$ when $1\%$ of the ${\text{CH}}_{\text{3}}\text{I}$ units exist as dimers has to be calculated.has to be evaluated.

Concept introduction:

The Beer-lambert law states that the intensity of light absorbed by a solution is directly proportional to the concentration of the solution and the path length of that the light has to travel in the solution.  The mathematical expression for Beer-Lambert law is given below.

    $A=\epsilon cL$

Answer to Problem 11A.7P

The absorbance expected at the mid-point of the absorption lineshape in a sample cell of length $12.0\text{cm}$ is $\underset{\_}{0.1836}$.

Explanation of Solution

The concentration of ${\text{CH}}_{\text{3}}\text{I}$ is calculated by the formula given below.

    ${\left[{\text{CH}}_{\text{3}}\text{I}\right]}_{\text{total}}=\frac{p}{RT}$        (3)

Where,

• ${\left[{\text{CH}}_{\text{3}}\text{I}\right]}_{\text{total}}$ is the concentration of ${\text{CH}}_{\text{3}}\text{I}$.
• $p$ is the pressure.
• $R$ is the gas constant. $\left(62.364\text{Torr}{\text{dm}}^3{\text{mol}}^{-1}{\text{K}}^{-1}\right)$
• $T$ is the temperature.

The value of $p$ is $2.4\text{Torr}$.

The value of $T$ is $373\text{K}$.

Substitute the value of $p$, $R$, and $T$ in equation (3).

    $\begin{array}{c}{\left[{\text{CH}}_{\text{3}}\text{I}\right]}_{\text{total}}=\frac{2.4\text{Torr}}{62.364\text{Torr}{\text{dm}}^3{\text{mol}}^{-1}{\text{K}}^{-1}\times 373\text{K}}\\ =\frac{2.4\text{Torr}}{23261.772\text{Torr}{\text{dm}}^3{\text{mol}}^{-1}}\\ =1.03\times {10}^{-4}\text{mol}{\text{dm}}^{-3}\end{array}$

About $1\%$ of the ${\text{CH}}_{\text{3}}\text{I}$ units exist as dimmers.

Therefore, the percentage of monomers is $99\%$.

Therefore, the concentration of ${\text{CH}}_{\text{3}}\text{I}$ units that exist as monomers is calculated below.

    $\begin{array}{c}{\left[{\text{CH}}_{\text{3}}\text{I}\right]}_{\text{monomer}}=\frac{99}{100}\times 1.03\times {10}^{-4}\text{mol}{\text{dm}}^{-3}\\ =1.02\times {10}^{-4}\text{mol}{\text{dm}}^{-3}\end{array}$

The mathematical form of Beer-Lambert law is given below.

  $A=\epsilon cL$        (4)

Where,

• $A$ is the absorbance.
• $\epsilon$ is the molar extinction coefficient.
• $L$ is the length.
• $c$ is the concentration.

The value of $\epsilon$ is $150{\text{dm}}^3{\text{mol}}^{-1}{\text{cm}}^{-1}$.

The value of $L$ is $12.0\text{cm}$.

The value of $c$ is $1.016\times {10}^{-4}\text{mol}{\text{dm}}^{-3}$.

Substitute the value of $\epsilon$, $L$, and $c$ in equation (4).

    $\begin{array}{c}A=150{\text{dm}}^3{\text{mol}}^{-1}{\text{cm}}^{-1}\times 1.02\times {10}^{-4}\text{mol}{\text{dm}}^{-3}\times 12.0\text{cm}\\ \text{=1836}\times {10}^{-4}\\ =\underset{\_}{0.1836}\end{array}$

Therefore, the absorbance expected at the mid-point of the absorption lineshape in a sample cell of length $12.0\text{cm}$ is $\underset{\_}{0.1836}$.

(c)

Interpretation Introduction

Interpretation:.

The absorbance expected at the mid-point of the absorption lineshape in a sample cell of length $12.0\text{cm}$ when $18\%$ of the ${\text{CH}}_{\text{3}}\text{I}$ units exist as dimers has to be calculated.  The molar absorption coefficient that would be inferred if dimerization is not considered has to be computed.

Concept introduction:

As mentioned in concept introduction in part (b).

Answer to Problem 11A.7P

The absorbance expected at the mid-point of the absorption lineshape in a sample cell of length $12.0\text{cm}$ when $18\%$ of the ${\text{CH}}_{\text{3}}\text{I}$ units exist as dimmers is $\underset{\_}{6.3}$.  The molar absorption coefficient that would be inferred if dimerization is not considered is $\underset{\_}{0.122\times 10^{-3}d{m}^{3}mo{l}^{-1}c{m}^{-1}}$.

Explanation of Solution

The concentration of ${\text{CH}}_{\text{3}}\text{I}$ is calculated by the formula given below.

    ${\left[{\text{CH}}_{\text{3}}\text{I}\right]}_{\text{total}}=\frac{p}{RT}$        (3)

Where,

• ${\left[{\text{CH}}_{\text{3}}\text{I}\right]}_{\text{total}}$ is the concentration of ${\text{CH}}_{\text{3}}\text{I}$.
• $p$ is the pressure.
• $R$ is the gas constant. $\left(62.364\text{Torr}{\text{dm}}^3{\text{mol}}^{-1}{\text{K}}^{-1}\right)$
• $T$ is the temperature.

The value of $p$ is $100\text{Torr}$.

The value of $T$ is $373\text{K}$.

Substitute the value of $p$, $R$, and $T$ in equation (3).

    $\begin{array}{c}{\left[{\text{CH}}_{\text{3}}\text{I}\right]}_{\text{total}}=\frac{100\text{Torr}}{62.364\text{Torr}{\text{dm}}^3{\text{mol}}^{-1}{\text{K}}^{-1}\times 373\text{K}}\\ =\frac{100\text{Torr}}{23261.772\text{Torr}{\text{dm}}^3{\text{mol}}^{-1}}\\ =4.299\times {10}^{-3}\text{mol}{\text{dm}}^{-3}\end{array}$

About $18\%$ of the ${\text{CH}}_{\text{3}}\text{I}$ units exist as dimmers.

Therefore, the percentage of monomers is $82\%$.

Therefore, the concentration of ${\text{CH}}_{\text{3}}\text{I}$ units that exist as monomers is calculated below.

    $\begin{array}{c}{\left[{\text{CH}}_{\text{3}}\text{I}\right]}_{\text{monomer}}=\frac{82}{100}\times 4.299\times {10}^{-3}\text{mol}{\text{dm}}^{-3}\\ =3.5\times {10}^{-3}\text{mol}{\text{dm}}^{-3}\end{array}$

The mathematical form of Beer-Lambert law is given below.

  $A=\epsilon cL$        (4)

Where,

• $A$ is the absorbance.
• $\epsilon$ is the molar extinction coefficient.
• $L$ is the length.
• $c$ is the concentration.

The value of $\epsilon$ is $150{\text{dm}}^3{\text{mol}}^{-1}{\text{cm}}^{-1}$.

The value of $L$ is $12.0\text{cm}$.

The value of $c$ is $3.5\times {10}^{-3}\text{mol}{\text{dm}}^{-3}$.

Substitute the value of $\epsilon$, $L$, and $c$ in equation (4).

    $\begin{array}{c}A=150{\text{dm}}^3{\text{mol}}^{-1}{\text{cm}}^{-1}\times 3.5\times {10}^{-3}\text{mol}{\text{dm}}^{-3}\times 12.0\text{cm}\\ \text{=6300}\times {10}^{-3}\\ =\underset{\_}{6.3}\end{array}$

Therefore, the absorbance expected at the mid-point of the absorption lineshape in a sample cell of length $12.0\text{cm}$ is $\underset{\_}{6.3}$.

The concentration, $c$, if dimerization is not considered is $4.299\times {10}^{-3}\text{mol}{\text{dm}}^{-3}$.

Substitute the value of $A$, $L$ and $c$ in equation (4).

    $\begin{array}{c}\text{6}\text{.3}=\epsilon \times 4.299\times {10}^{-3}\text{mol}{\text{dm}}^{-3}\times 12.0\text{cm}\\ \epsilon =\frac{\text{6}\text{.3}}{4.299\times {10}^{-3}\text{mol}{\text{dm}}^{-3}\times 12.0\text{cm}}\\ =\underset{\_}{0.122\times 10^{-3}d{m}^{3}mo{l}^{-1}c{m}^{-1}}\end{array}$

Therefore, the molar extinction coefficient is $\underset{\_}{0.122\times 10^{-3}d{m}^{3}mo{l}^{-1}c{m}^{-1}}$.