Chapter 18, Problem 18.15P

(a)

Interpretation Introduction

Interpretation:

The conversion of transmittance to absorbance has to be given for the given value.

Concept introduction:

Beer-lambert law:

$\begin{array}{l}\text{A}\text{=}\text{-}\text{log}\frac{\text{P}}{{\text{P}}_{\text{o}}}\text{=}\text{εcb}\\ \text{where,}\\ \text{A}\text{-}\text{absorbance}\\ {\text{P}}_{\text{o}}\text{-}\text{incident}\text{irradiance}\text{and}\text{P}\text{-}\text{transmitted}\text{irradiance}\\ \text{ε}\text{-}\text{molar}\text{absorptivity}\\ \text{c-}\text{concentration}\text{and}\text{b}\text{-}\text{pathlength}\text{.}\end{array}$

Answer to Problem 18.15P

• The value of absorbance is $\text{0}\text{.613}\text{.}$

Explanation of Solution

Given information,

$\begin{array}{l}\text{Vapor}\text{pressure}\text{is}\text{30}\text{.3}\text{µbar}\text{.}\\ \text{transmittance}\text{is}\text{24}\text{.4}\text{\%}\text{at}\text{266}\text{nm}\text{.}\\ \text{pathlength}\text{is}\text{3}\text{.00}\text{cm}\text{.}\text{temperature}\text{is}\text{298}\text{K}\text{.}\end{array}$

Calculate the value of absorbance

$\begin{array}{l}\text{A}\text{=}\text{-}\text{log}\text{T}\\ \text{=}\text{-}\text{log}\left(\text{0}\text{.244}\right)\\ \text{=}\text{0}\text{.613}\text{.}\end{array}$

The value of absorbance is $\text{0}\text{.613}\text{.}$

(b)

Interpretation Introduction

Interpretation:

The conversion of pressure to concentration has to be given.

Concept introduction:

Ideal gas equation:

The relationship between $\text{P,}\text{R}\text{and}\text{T}$ is given by ideal gas equation

$\begin{array}{l}\text{PV=}\text{nRT}\\ \text{where,}\\ \text{P}\text{-}\text{pressure}\\ \text{V}\text{-}\text{volume}\\ \text{n-}\text{number}\text{of}\text{moles}\\ \text{R}\text{-}\text{gas}\text{constant}\text{and}\text{T}\text{-}\text{temperature}\text{.}\end{array}$

Answer to Problem 18.15P

• The concentration is $\text{1}\text{.223}\text{×}{\text{10}}^{\text{-6}}\text{M}\text{.}$

Explanation of Solution

Given information,

$\begin{array}{l}\text{Vapor}\text{pressure}\text{is}\text{30}\text{.3}\text{µbar}\text{.}\\ \text{transmittance}\text{is}\text{24}\text{.4}\text{\%}\text{at}\text{266}\text{nm}\text{.}\\ \text{pathlength}\text{is}\text{3}\text{.00}\text{cm}\text{.}\text{temperature}\text{is}\text{298}\text{K}\text{.}\\ \text{Gas}\text{constant}\left(\text{R}\right)\text{=}\text{0}\text{.08314}\frac{\text{L}\text{bar}}{\text{mol}\text{K}}\end{array}$

Apply ideal gas equation to convert pressure into concentration

$\begin{array}{l}\text{Concentration}\left(\frac{\text{mol}}{\text{L}}\right)\text{=}\frac{\text{n}}{\text{V}}\text{=}\frac{\text{P}}{\text{RT}}\\ \text{=}\frac{\text{30}\text{.3}\text{µbar}}{\left(\text{0}\text{.08314}\frac{\text{L}\cdot \text{bar}}{\text{mol}\cdot \text{K}}\right)\left(\text{298}\text{K}\right)}\\ \text{=}\frac{\text{30}\text{.3}\text{×}{\text{10}}^{\text{-6}}}{\left(\text{0}\text{.08314}\right)\left(\text{298}\right)}\left(\text{mol/L}\right)\\ \text{=}\text{1}\text{.223}\text{×}{\text{10}}^{\text{-6}}\text{M}\text{.}\end{array}$

The concentration is $\text{1}\text{.223}\text{×}{\text{10}}^{\text{-6}}\text{M}\text{.}$

(c)

Interpretation Introduction

Interpretation:

The value of molar absorptivity of gaseous pyrazine has to be calculated at $\text{266}\text{nm}$.

Concept introduction:

Beer-lambert law:

$\begin{array}{l}\text{A}\text{=}\text{-}\text{log}\frac{\text{P}}{{\text{P}}_{\text{o}}}\text{=}\text{εcb}\\ \text{where,}\\ \text{A}\text{-}\text{absorbance}\\ {\text{P}}_{\text{o}}\text{-}\text{incident}\text{irradiance}\text{and}\text{P}\text{-}\text{transmitted}\text{irradiance}\\ \text{ε}\text{-}\text{molar}\text{absorptivity}\\ \text{c-}\text{concentration}\text{and}\text{b}\text{-}\text{pathlength}\text{.}\end{array}$

Answer to Problem 18.15P

• The value of molar absorptivity is $\text{1}\text{.67}\text{×}{\text{10}}^{\text{5}}{\text{M}}^{\text{-1}}{\text{cm}}^{\text{-1}}$.

Explanation of Solution

Given information,

$\begin{array}{l}\text{Vapor}\text{pressure}\text{is}\text{30}\text{.3}\text{µbar}\text{.}\\ \text{transmittance}\text{is}\text{24}\text{.4}\text{\%}\text{at}\text{266}\text{nm}\text{.}\\ \text{pathlength}\text{is}\text{3}\text{.00}\text{cm}\text{.}\text{temperature}\text{is}\text{298}\text{K}\text{.}\\ \text{Concentration}\text{is}\text{1}\text{.223}\text{×}{\text{10}}^{\text{-6}}\text{M}\text{.}\end{array}$

Apply Beer’s law to calculate the value of molar absorptivity

$\begin{array}{l}\text{A}\text{=}\text{εcb}\\ \text{ε}\text{=}\text{A/bc}\\ \text{=}\frac{\text{0}\text{.613}}{\left(\text{1}\text{.223}\text{×}{\text{10}}^{\text{-6}}\text{M}\right)\left(\text{3}\text{.00}\text{cm}\right)}\\ \text{=}\text{167075}{\text{M}}^{\text{-1}}{\text{cm}}^{\text{-1}}\text{=}\text{1}\text{.67}\text{×}{\text{10}}^{\text{5}}{\text{M}}^{\text{-1}}{\text{cm}}^{\text{-1}}\text{.}\end{array}$

The value of molar absorptivity is $\text{1}\text{.67}\text{×}{\text{10}}^{\text{5}}{\text{M}}^{\text{-1}}{\text{cm}}^{\text{-1}}$.