Chapter 17, Problem 17.102QE

(a)

Interpretation Introduction

Interpretation:

The vapor pressure of ${\text{CH}}_{\text{3}}\text{OH}\left(\text{l}\right)\text{at}{\text{58}}^{\text{o}}\text{C}$ has to be calculated.

Explanation of Solution

The given reaction as follows,

  ${\text{CH}}_{\text{3}}\text{OH}\left(\text{l}\right)\to {\text{CH}}_{\text{3}}\text{OH}\left(\text{g}\right)$

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

  $\begin{array}{l}{\text{CH}}_{\text{3}}\text{OH}\left(\text{l}\right)\to {\text{CH}}_{\text{3}}\text{OH}\left(\text{g}\right)\\ {\text{ΔH}}_v^{\text{o}}\left(\text{kJ/mol}\text{.K}\right)-238.66-200.66\end{array}$

  $\begin{array}{c}{\text{ΔH}}_{\text{rxn}}^{\text{o}}\text{=}{\displaystyle \sum \left(\text{n×}{\text{ΔH}}_f^{\circ }\text{products}\right)}-{\displaystyle \sum \left(\text{n×}{\text{ΔH}}_f^{\circ }\text{ reactants}\right)}\\ \\ \text{=}\left(1\text{×}-200.66\text{kJ/mol}\right)-\left(1\text{×}-238.66\text{kJ/mol}\text{.K}\right)\\ \\ =38kJ/mol.\end{array}$

Hence, value of $\Delta {\text{H}}_{rxn}^{\circ }$ for the process is $38kJ/mol.$

• 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 of methanol is $\text{64}{\text{.7}}^{\text{o}}\text{C}$, the vapor pressure of ${\text{CH}}_{\text{3}}\text{OH}\left(\text{l}\right)$ is $\text{760}\text{torr}\left(\text{1}\text{atm}\right)$, the given data’s are,

  $\begin{array}{l}{\text{P}}_{\text{1}}\text{=}\text{760}\text{torr}\\ {\text{T}}_{\text{1}}\text{=}\text{64}\text{.7+273}\text{=}\text{338}\text{K}\\ {\text{T}}_{\text{2}}\text{=}\text{58}\text{+}\text{273}\text{=}\text{331}\text{K}\\ {\text{ΔH}}_{\text{v}}\text{=}\text{38}\text{.0kJ}\text{×}\frac{{\text{10}}^{\text{3}}\text{J}}{\text{1}\text{kJ}}\text{=}\text{38}\text{.0}{\text{×10}}^{\text{3}}\text{J}\\ \text{R}\text{=}\text{8}\text{.314}\text{J/K}\end{array}$

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

  $\begin{array}{c}\text{ln}\left(\frac{{\text{P}}_{\text{2}}}{{\text{P}}_{\text{1}}}\right)=\frac{\text{38}\text{.0}\times {\text{10}}^{3}J}{\text{8}\text{.314}\text{J/K}}\times \frac{\left(331K-\text{338}\text{K}\right)}{\text{338}\text{K}\times 331\text{K}}\\ \\ =-0.286\\ \\ \left(\frac{{\text{P}}_{\text{2}}}{{\text{P}}_{\text{1}}}\right)=e^{-0.286}=0.751\\ \\ {\text{P}}_{\text{2}}=\left(0.751\right){\text{P}}_{\text{1}}=\left(0.751\right)\left(\text{1}\text{.0}\text{atm}\right)\\ \\ \text{=}0.751atm.\end{array}$

Hence, value of vapor pressure ${\text{CH}}_{\text{3}}\text{OH}\left(\text{l}\right)\text{at}{\text{58}}^{\text{o}}\text{C}$ is $0.751atm.$

(b)

Interpretation Introduction

Interpretation:

The vapor pressure of ${\text{C}}_2{\text{H}}_5\text{OH}\left(\text{l}\right)\text{at}{\text{29}}^{\text{o}}\text{C}$ has to be calculated.

Explanation of Solution

The given reaction as follows,

  ${\text{C}}_2{\text{H}}_5\text{OH}\left(\text{l}\right)\to {\text{C}}_2{\text{H}}_5\text{OH}\left(\text{g}\right)$

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

  $\begin{array}{l}{\text{C}}_2{\text{H}}_5\text{OH}\left(\text{l}\right)\to {\text{C}}_2{\text{H}}_5\text{OH}\left(\text{g}\right)\\ {\text{ΔH}}_v^{\text{o}}\left(\text{kJ/mol}\text{.K}\right)-277.69-235.10\end{array}$

  $\begin{array}{c}{\text{ΔH}}_{\text{rxn}}^{\text{o}}\text{=}{\displaystyle \sum \left(\text{n×}{\text{ΔH}}_f^{\circ }\text{products}\right)}-{\displaystyle \sum \left(\text{n×}{\text{ΔH}}_f^{\circ }\text{ reactants}\right)}\\ \\ \text{=}\left(1\text{×}-235.10\text{kJ/mol}\right)-\left(1\text{×}-277.69\text{kJ/mol}\text{.K}\right)\\ \\ =42.59kJ/mol.\end{array}$

Hence, value of $\Delta {\text{H}}_{rxn}^{\circ }$ for the process is $42.59kJ/mol.$

• 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 ${\text{79}}^{\text{o}}\text{C}$, the vapor pressure of ${\text{C}}_2{\text{H}}_5\text{OH}\left(\text{l}\right)$ is $\text{760}\text{torr}\left(\text{1}\text{atm}\right)$, the given data’s are,

  $\begin{array}{l}{\text{P}}_{\text{1}}\text{=}\text{760}\text{torr}\\ {\text{T}}_{\text{1}}\text{=}\text{79}\text{+}\text{273}\text{=}\text{352}\text{K}\\ {\text{T}}_{\text{2}}\text{=}\text{29}\text{+}\text{273}\text{=}\text{302}\text{K}\\ {\text{ΔH}}_{\text{v}}\text{=}42.59\text{kJ}\text{×}\frac{{\text{10}}^{\text{3}}\text{J}}{\text{1}\text{kJ}}\text{=}42.59{\text{×10}}^{\text{3}}\text{J}\\ \text{R}\text{=}\text{8}\text{.314}\text{J/K}\end{array}$

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

  $\begin{array}{c}\text{ln}\left(\frac{{\text{P}}_{\text{2}}}{{\text{P}}_{\text{1}}}\right)=\frac{42.59\times {\text{10}}^{3}J}{\text{8}\text{.314}\text{J/K}}\times \frac{\left(\text{303}K-\text{352}\text{K}\right)}{\text{352}\text{K}\times 302\text{K}}\\ \\ =-2.409\\ \\ \left(\frac{{\text{P}}_{\text{2}}}{{\text{P}}_{\text{1}}}\right)=e^{-2.409}=0.0898\\ \\ {\text{P}}_{\text{2}}=\left(0.0898\right){\text{P}}_{\text{1}}=\left(0.0898\right)\left(\text{1}\text{.0}\text{atm}\right)\\ \\ \text{=}0.0898atm.\end{array}$

Hence, value of vapor pressure ${\text{C}}_2{\text{H}}_5\text{OH}\left(\text{l}\right)\text{at}{\text{29}}^{\text{o}}\text{C}$ is $0.0898atm.$

(c)

Interpretation Introduction

Interpretation:

The vapor pressure of $\text{Hg}\left(\text{l}\right)\text{at}{\text{45}}^{\text{o}}\text{C}$ has to be calculated.

Explanation of Solution

The given reaction as follows,

  $\text{Hg}\left(\text{l}\right)\to \text{Hg}\left(\text{g}\right)$

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

  $\begin{array}{l}\text{Hg}\left(\text{l}\right)\to \text{Hg}\left(\text{g}\right)\\ {\text{ΔH}}_v^{\text{o}}\left(\text{kJ/mol}\text{.K}\right)061.32\end{array}$

  $\begin{array}{c}{\text{ΔH}}_{\text{rxn}}^{\text{o}}\text{=}{\displaystyle \sum \left(\text{n×}{\text{ΔH}}_f^{\circ }\text{products}\right)}-{\displaystyle \sum \left(\text{n×}{\text{ΔH}}_f^{\circ }\text{ reactants}\right)}\\ \\ \text{=}\left(1\text{×}61.32\text{kJ/mol}\text{.K}\right)-\left(1\text{×}0\text{kJ/mol}\right)\\ \\ =61.32kJ/mol.\end{array}$

Hence, value of $\Delta {\text{H}}_{rxn}^{\circ }$ for the process is $61.32kJ/mol.$

• 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 ${356.7}^{\text{o}}\text{C}$, the vapor pressure of $\text{Hg}\left(\text{l}\right)$ is $\text{760}\text{torr}\left(\text{1}\text{atm}\right)$, the given data’s are,

  $\begin{array}{l}{\text{P}}_{\text{1}}\text{=}\text{760}\text{torr}\\ {\text{T}}_{\text{1}}\text{=}\text{356}\text{.7}\text{+273}\text{=}\text{630}\text{K}\\ {\text{T}}_{\text{2}}\text{=}4\text{5}\text{+}\text{273}\text{=}\text{318}\text{K}\\ {\text{ΔH}}_{\text{v}}\text{=}61.32\text{kJ}\text{×}\frac{{\text{10}}^{\text{3}}\text{J}}{\text{1}\text{kJ}}\text{=}\text{61}\text{.32}{\text{×10}}^{\text{3}}\text{J}\\ \text{R}\text{=}\text{8}\text{.314}\text{J/K}\end{array}$

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

  $\begin{array}{c}\text{ln}\left(\frac{{\text{P}}_{\text{2}}}{{\text{P}}_{\text{1}}}\right)=\frac{\text{61}\text{.32}\times {\text{10}}^{3}J}{\text{8}\text{.314}\text{J/K}}\times \frac{\left(\text{318}K-\text{630K}\right)}{\text{630}\text{K}\times 318\text{K}}\\ \\ =-11.49\\ \\ \left(\frac{{\text{P}}_{\text{2}}}{{\text{P}}_{\text{1}}}\right)=e^{-11.49}=0.0000123\\ \\ {\text{P}}_{\text{2}}=\left(0.0000123\right){\text{P}}_{\text{1}}=\left(0.0000123\right)\left(\text{1}\text{.0}\text{atm}\right)\\ \\ \text{=}0.0000123atm.\end{array}$

Hence, value of vapor pressure $\text{Hg}\left(\text{l}\right)\text{at}{\text{45}}^{\text{o}}\text{C}$ is $0.0000123atm.$