Chapter 17, Problem 17.101QE

(a)

Interpretation Introduction

Interpretation:

The vapor pressure of ${\text{CS}}_{\text{2}}\left(\text{l}\right)\text{at}{\text{5}}^{\text{o}}\text{C}$ has to be calculated.

Explanation of Solution

The given reaction as follows,

  ${\text{CS}}_{\text{2}}\left(\text{l}\right)\to {\text{CS}}_{\text{2}}\left(\text{g}\right)$

The value of $\Delta {\text{H}}_{rxn}^{\circ }$ is calculated by standard enthalpy change as follows,

  $\begin{array}{l}{\text{CS}}_{\text{2}}\left(\text{l}\right)\to {\text{CS}}_{\text{2}}\left(\text{g}\right)\\ {\text{ΔH}}_r^{\text{o}}\left(\text{kJ/mol}\text{.K}\right)89.70117.36\end{array}$

  $\begin{array}{c}{\text{ΔH}}_{\text{rxn}}^{\text{o}}\text{=}{\displaystyle \sum \left(\text{n×}{\text{ΔH}}_f^{\circ }\text{products}\right)}-{\displaystyle \sum \left(\text{n×}{\text{ΔH}}_f^{\circ }\text{ reactants}\right)}\\ \\ \text{=}\left(1\text{×}117.36\text{kJ/mol}\right)-\left(1\text{×}89.70\text{kJ/mol}\text{.K}\right)\\ \\ =27.66kJ/mol.\end{array}$

Hence, value of $\Delta {\text{H}}_{rxn}^{\circ }$ for the process is $27.66kJ/mol.$

The integrated Clausius-Clapeyron equation is used to determine the vapor pressure as follows,

  $\begin{array}{l}\text{ln}\left(\frac{{\text{P}}_{\text{2}}}{{\text{P}}_{\text{1}}}\right)\text{=}\frac{{\text{ΔH}}_{\text{v}}}{\text{R}}\left(\frac{\text{1}}{{\text{T}}_{\text{1}}}-\frac{\text{1}}{{\text{T}}_{\text{2}}}\right)\\ \\ \text{ln}\left(\frac{{\text{P}}_{\text{2}}}{{\text{P}}_{\text{1}}}\right)=\frac{{\text{ΔH}}_{\text{v}}}{\text{R}}\frac{\left({\text{T}}_{\text{2}}-{\text{T}}_{\text{1}}\right)}{{\text{T}}_{\text{1}}{\text{T}}_{\text{2}}}\end{array}$

As known, the normal boiling point is $\text{46}{\text{.3}}^{\text{o}}\text{C}$, the vapor pressure of ${\text{CS}}_{\text{2}}\left(\text{l}\right)$ is $\text{760}\text{torr}\left(\text{1}\text{atm}\right)$, the given data’s are,

  $\begin{array}{l}{\text{P}}_{\text{1}}\text{=}\text{760}\text{torr}\\ {\text{T}}_{\text{1}}\text{=}\text{46}\text{.3+273}\text{=}\text{319}\text{.3}\text{K}\\ {\text{T}}_{\text{2}}\text{=}\text{5}\text{+}\text{273}\text{=}\text{278}\text{K}\\ {\text{ΔH}}_{\text{v}}\text{=}\text{27}\text{.66}\text{kJ}\text{×}\frac{{\text{10}}^{\text{3}}\text{J}}{\text{1}\text{kJ}}\text{=}\text{27}\text{.66}{\text{×10}}^{\text{3}}\text{J}\\ \text{R}\text{=}\text{8}\text{.314}\text{J/K}\end{array}$

By substituting the values in the integrated Clausius-Clapeyron equation as shown below,

  $\begin{array}{c}\text{ln}\left(\frac{{\text{P}}_{\text{2}}}{{\text{P}}_{\text{1}}}\right)=\frac{\text{27}\text{.66}\times {\text{10}}^{3}J}{\text{8}\text{.314}\text{J/K}}\times \frac{\left(\text{278}K-\text{319}\text{.3}\text{K}\right)}{\text{319}\text{.3}\text{K}\times \text{278}\text{K}}\\ \\ =-1.55\\ \\ \left(\frac{{\text{P}}_{\text{2}}}{{\text{P}}_{\text{1}}}\right)=e^{-1.55}=0.207\\ \\ {\text{P}}_{\text{2}}=\left(0.207\right){\text{P}}_{\text{1}}=\left(0.207\right)\left(\text{1}\text{.0}\text{atm}\right)\\ \\ \text{=}0.207atm.\end{array}$

Hence, value of vapor pressure ${\text{CS}}_{\text{2}}\left(\text{l}\right)\text{at}{\text{5}}^{\text{o}}\text{C}$ is $0.207atm.$

(b)

Interpretation Introduction

Interpretation:

The vapor pressure of ${\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)\text{at}{\text{50}}^{\text{o}}\text{C}$ has to be calculated.

Explanation of Solution

The given reaction as follows,

  ${\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)\to {\text{H}}_{\text{2}}\text{O}\left(\text{g}\right)$

The value of $\Delta {\text{H}}_{rxn}^{\circ }$ is calculated by standard entropy change as follows,

  $\begin{array}{l}{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)\to {\text{H}}_{\text{2}}\text{O}\left(\text{g}\right)\\ {\text{ΔH}}_v^{\text{o}}\left(\text{kJ/mol}\text{.K}\right)-285.83-241.82\end{array}$

  $\begin{array}{c}{\text{ΔH}}_{\text{rxn}}^{\text{o}}\text{=}{\displaystyle \sum \left(\text{n×}{\text{ΔH}}_f^{\circ }\text{products}\right)}-{\displaystyle \sum \left(\text{n×}{\text{ΔH}}_f^{\circ }\text{ reactants}\right)}\\ \\ \text{=}\left(1\text{×}-241.82\text{kJ/mol}\right)-\left(1\text{×}-285.83\text{kJ/mol}\text{.K}\right)\\ \\ =44.01kJ/mol.\end{array}$

Hence, value of $\Delta {\text{H}}_{rxn}^{\circ }$ for the process is $44.01kJ/mol.$

The integrated Clausius-Clapeyron equation is used to determine the vapor pressure as follows,

  $\begin{array}{l}\text{ln}\left(\frac{{\text{P}}_{\text{2}}}{{\text{P}}_{\text{1}}}\right)\text{=}\frac{{\text{ΔH}}_{\text{v}}}{\text{R}}\left(\frac{\text{1}}{{\text{T}}_{\text{1}}}-\frac{\text{1}}{{\text{T}}_{\text{2}}}\right)\\ \\ \text{ln}\left(\frac{{\text{P}}_{\text{2}}}{{\text{P}}_{\text{1}}}\right)=\frac{{\text{ΔH}}_{\text{v}}}{\text{R}}\frac{\left({\text{T}}_{\text{2}}-{\text{T}}_{\text{1}}\right)}{{\text{T}}_{\text{1}}{\text{T}}_{\text{2}}}\end{array}$

As known, the normal boiling point is ${\text{100}}^{\text{o}}\text{C}$, the vapor pressure of water is $\text{760}\text{torr}\left(\text{1}\text{atm}\right)$, the given data’s are,

  $\begin{array}{l}{\text{P}}_{\text{1}}\text{=}\text{760}\text{torr}\\ {\text{T}}_{\text{1}}\text{=}\text{100+273}\text{=}\text{373}\text{K}\\ {\text{T}}_{\text{2}}\text{=}\text{50}\text{+}\text{273}\text{=}\text{323}\text{K}\\ {\text{ΔH}}_{\text{v}}\text{=}44.01\text{kJ}\text{×}\frac{{\text{10}}^{\text{3}}\text{J}}{\text{1}\text{kJ}}\text{=}\text{44}\text{.01}{\text{×10}}^{\text{3}}\text{J}\\ \text{R}\text{=}\text{8}\text{.314}\text{J/K}\end{array}$

By substituting the values in the integrated Clausius-Clapeyron equation as shown below,

  $\begin{array}{c}\text{ln}\left(\frac{{\text{P}}_{\text{2}}}{{\text{P}}_{\text{1}}}\right)=\frac{\text{44}\text{.01}\times {\text{10}}^{3}J}{\text{8}\text{.314}\text{J/K}}\times \frac{\left(\text{323}K-\text{373}\text{K}\right)}{\text{323}\text{K}\times 373\text{K}}\\ \\ =-2.196\\ \\ \left(\frac{{\text{P}}_{\text{2}}}{{\text{P}}_{\text{1}}}\right)=e^{-2.196}=0.111\\ \\ {\text{P}}_{\text{2}}=\left(0.111\right){\text{P}}_{\text{1}}=\left(0.111\right)\left(\text{1}\text{.0}\text{atm}\right)\\ \\ \text{=}0.111atm.\end{array}$

Hence, value of vapor pressure ${\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)\text{at}{\text{50}}^{\text{o}}\text{C}$ is $0.111atm.$

(c)

Interpretation Introduction

Interpretation:

The vapor pressure of ${\text{C}}_6{\text{H}}_6\left(\text{l}\right)\text{at}4{\text{5}}^{\text{o}}\text{C}$ has to be calculated.

Explanation of Solution

The given reaction as follows,

  ${\text{C}}_6{\text{H}}_6\left(\text{l}\right)\to {\text{C}}_6{\text{H}}_6\left(\text{g}\right)$

The value of $\Delta {\text{H}}_{rxn}^{\circ }$ is calculated by standard entropy change as follows,

  $\begin{array}{l}{\text{C}}_6{\text{H}}_6\left(\text{l}\right)\to {\text{C}}_6{\text{H}}_6\left(\text{g}\right)\\ {\text{ΔH}}_v^{\text{o}}\left(\text{kJ/mol}\text{.K}\right)49.0382.93\end{array}$

  $\begin{array}{c}{\text{ΔH}}_{\text{rxn}}^{\text{o}}\text{=}{\displaystyle \sum \left(\text{n×}{\text{ΔH}}_f^{\circ }\text{products}\right)}-{\displaystyle \sum \left(\text{n×}{\text{ΔH}}_f^{\circ }\text{ reactants}\right)}\\ \\ \text{=}\left(1\text{×}82.93\text{kJ/mol}\right)-\left(1\text{×}49.03\text{kJ/mol}\text{.K}\right)\\ \\ =33.9kJ/mol.\end{array}$

Hence, value of $\Delta {\text{H}}_{rxn}^{\circ }$ for the process is $33.9kJ/mol.$

The integrated Clausius-Clapeyron equation is used to determine the vapor pressure as follows,

  $\begin{array}{l}\text{ln}\left(\frac{{\text{P}}_{\text{2}}}{{\text{P}}_{\text{1}}}\right)\text{=}\frac{{\text{ΔH}}_{\text{v}}}{\text{R}}\left(\frac{\text{1}}{{\text{T}}_{\text{1}}}-\frac{\text{1}}{{\text{T}}_{\text{2}}}\right)\\ \\ \text{ln}\left(\frac{{\text{P}}_{\text{2}}}{{\text{P}}_{\text{1}}}\right)=\frac{{\text{ΔH}}_{\text{v}}}{\text{R}}\frac{\left({\text{T}}_{\text{2}}-{\text{T}}_{\text{1}}\right)}{{\text{T}}_{\text{1}}{\text{T}}_{\text{2}}}\end{array}$

As known, the normal boiling point is $\text{80}{\text{.1}}^{\text{o}}\text{C}$, the vapor pressure of ${\text{C}}_6{\text{H}}_6\left(\text{l}\right)$ is $\text{760}\text{torr}\left(\text{1}\text{atm}\right)$, the given data’s are,

  $\begin{array}{l}{\text{P}}_{\text{1}}\text{=}\text{760}\text{torr}\\ {\text{T}}_{\text{1}}\text{=}\text{80}\text{.1}\text{+273}\text{=}\text{353}\text{.1}\text{K}\\ {\text{T}}_{\text{2}}\text{=}4\text{5}\text{+}\text{273}\text{=}\text{318}\text{K}\\ {\text{ΔH}}_{\text{v}}\text{=}33.90\text{kJ}\text{×}\frac{{\text{10}}^{\text{3}}\text{J}}{\text{1}\text{kJ}}\text{=}\text{33}\text{.90}{\text{×10}}^{\text{3}}\text{J}\\ \text{R}\text{=}\text{8}\text{.314}\text{J/K}\end{array}$

By substituting the values in the integrated Clausius-Clapeyron equation as shown below,

  $\begin{array}{c}\text{ln}\left(\frac{{\text{P}}_{\text{2}}}{{\text{P}}_{\text{1}}}\right)=\frac{\text{33}\text{.90}\times {\text{10}}^{3}J}{\text{8}\text{.314}\text{J/K}}\times \frac{\left(\text{318}K-\text{353}\text{.1}\text{K}\right)}{\text{318}\text{K}\times \text{353}\text{.1}\text{K}}\\ \\ =-1.27\\ \\ \left(\frac{{\text{P}}_{\text{2}}}{{\text{P}}_{\text{1}}}\right)=e^{-1.27}=0.280\\ \\ {\text{P}}_{\text{2}}=\left(0.280\right){\text{P}}_{\text{1}}=\left(0.280\right)\left(\text{1}\text{.0}\text{atm}\right)\\ \\ \text{=}0.280atm.\end{array}$

Hence, value of vapor pressure ${\text{C}}_6{\text{H}}_6\left(\text{l}\right)\text{at}4{\text{5}}^{\text{o}}\text{C}$ is $0.280atm.$