Chapter 4, Problem 4B.13BE

(i)

Interpretation Introduction

Interpretation: The enthalpy of vaporization of hexane has to be estimated.

Concept introduction: The Trouton’s rule states that all liquid shows same entropy of vaporization at their boiling points.  Mathematically, Trouton’s rule is stated as follows:

    $\Delta S_{\text{vaporization}}=10.5\times R$

Answer to Problem 4B.13BE

The enthalpy of vaporization of hexane is $\underset{\_}{29853.18Jmo{l}^{-1}}$.

Explanation of Solution

The change in entropy of hexane is calculated by the Trouton’s formula.

    $\Delta S_{\text{vaporization}}=10.5\times R$                                                                                     (1)

Where,

• • $\Delta S_{\text{vaporization}}$ is change in entropy of vaporization.
• • $R$ is gas constant.

The value of gas constant $R$ is $8.314\text{}\text{J}{\text{K}}^{\text{-1}}{\text{mol}}^{\text{-1}}$.

Substitute the value of $R$ in equation (1).

    $\begin{array}{c}\Delta S_{\text{vaporization}}=10.5\times 8.314\text{J}{\text{K}}^{\text{-1}}{\text{mol}}^{\text{-1}}\\ =87.29\text{J}{\text{K}}^{\text{-1}}{\text{mol}}^{\text{-1}}\end{array}$

Therefore, the change in entropy of vaporization of hexane is $87.29\text{J}{\text{K}}^{\text{-1}}{\text{mol}}^{\text{-1}}$.

The enthalpy of vaporization is calculated by the following formula.

    $\Delta H_{\text{vaporization}}=\Delta S_{\text{vaporization}}\times T$                                                                          (2)

It is given that the temperature $T$ is $69.0\text{}{}^{°}\text{C}$.

The conversion of ${}^{°}\text{C}$ in $\text{K}$ is done as,

    $1{}^{°}\text{C}=\text{273}\text{K}$

The conversion of $69.0{}^{°}\text{C}$ in $\text{K}$ is done as,

    $\begin{array}{c}69.0{}^{°}\text{C}=69.0+\text{273}\text{K}\\ =\text{342}\text{}\text{K}\end{array}$

Therefore, the temperature $\left(T\right)$ is $342\text{}\text{K}$.

Substitute the value of $\Delta S_{\text{vaporization}}$ and $T$ in equation (2).

    $\begin{array}{c}\Delta H_{\text{vaporization}}=87.29\text{J}{\text{K}}^{\text{-1}}{\text{mol}}^{\text{-1}}\times 342\text{K}\\ =\underset{\_}{29853.18Jmo{l}^{-1}}\end{array}$

Hence, the enthalpy of vaporization of hexane is $\underset{\_}{29853.18Jmo{l}^{-1}}$.

(ii)

Interpretation Introduction

Interpretation: The vapour pressure at ${25}^{\circ }\text{C}$ and at ${60}^{\circ }\text{C}$ has to be estimated.

Concept introduction: The Trouton’s rule states that all liquid shows same entropy of vaporization at their boiling points.  Mathematically Trouton’s rule is stated as follows.

    $\Delta S_{\text{vaporization}}=10.5\times R$

Answer to Problem 4B.13BE

The vapour pressure at $25{}^{\circ }\text{C}$ and at ${60}^{\circ }\text{C}$ are $\underset{\_}{0.184atm}$ and $\underset{\_}{0.729atm}$ respectively.

Explanation of Solution

The vapour pressure at $298\text{K}$ is calculated by the Clausius- Clapeyron  formula.

    $\ln \left(\frac{p_2}{p_1}\right)=\frac{\Delta H_{\text{vaporization}}}{R}\left(\frac{1}{T_1}-\frac{1}{T_2}\right)$                                                                (3)

Where,

• • $p_1$ and $p_2$ are the pressures.
• • $\Delta H_{\text{vaporization}}$ is the enthalpy of vaporization.
• • $T_1$ and $T_2$ are the temperatures.
• • $R$ is the gas constant.

For $25{}^{o}C$:

The boiling point of hexane is calculated at $1\text{atm}$ pressure.

Therefore, pressure $p_1=1\text{atm}$.

It is given that the temperature is $69.0{}^{\circ }\text{C}$.

The conversion of ${}^{°}\text{C}$ in $\text{K}$ is done as,

    $1{}^{°}\text{C}=\text{273}\text{K}$

The conversion of $69.0{}^{°}\text{C}$ in $\text{K}$ is done as,

    $\begin{array}{c}69.0{}^{°}\text{C}=69.0+\text{273}\text{K}\\ =\text{342}\text{}\text{K}\end{array}$

Therefore, the temperature $\left(T_1\right)$ is $342\text{}\text{K}$.

It is given that the temperature is $25{}^{\circ }\text{C}$.

The conversion of ${}^{\circ }\text{C}$ in $\text{K}$ is done as,

    $1{}^{\circ }\text{C}=\text{273}\text{K}$

The conversion of $25{}^{\circ }\text{C}$ in $\text{K}$ is done as,

    $\begin{array}{c}25{}^{\circ }\text{C}=25+\text{273}\text{K}\\ =\text{298}\text{K}\end{array}$

Therefore, the temperature $T_2$ is $298\text{K}$.

The enthalpy of vaporization for hexane is $29853.18\text{J}{\text{mol}}^{\text{-1}}$.

Substitute the values of $p_1,T_1,T_2,\Delta H_{\text{vaporization}}$ and $R$ in equation (3).

    $\begin{array}{c}\ln \left(\frac{p_2}{1\text{atm}}\right)=\frac{29853.18\text{J}{\text{mol}}^{\text{-1}}}{8.314\text{J}{\text{K}}^{\text{-1}}{\text{mol}}^{\text{-1}}}\left(\frac{1}{342\text{K}}-\frac{1}{298\text{K}}\right)\\ \ln \left(\frac{p_2}{1\text{atm}}\right)=3950.712{\text{mol}}^{\text{-1}}\text{K}\left(0.00292{\text{K}}^{-1}-0.00335{\text{K}}^{-1}\right)\\ \ln \left(\frac{p_2}{1\text{atm}}\right)=-1.69\\ p_2=e^{-1.69}\times 1\text{atm}\end{array}$

Or,

    $p_2=\underset{\_}{0.184atm}$

Hence, the vapour pressure at ${25}^{\circ }\text{C}$ is $\underset{\_}{0.184atm}$.

For $60{}^{o}C$:

The boiling point of hexane is calculated at $1\text{atm}$ pressure.

Therefore, pressure $p_1=1\text{atm}$.

It is given that the temperature is $69.0{}^{\circ }\text{C}$.

The conversion of ${}^{°}\text{C}$ in $\text{K}$ is done as,

    $1{}^{°}\text{C}=\text{273}\text{K}$

The conversion of $69.0{}^{°}\text{C}$ in $\text{K}$ is done as,

    $\begin{array}{c}69.0{}^{°}\text{C}=69.0+\text{273}\text{K}\\ =\text{342}\text{}\text{K}\end{array}$

Therefore, the temperature $\left(T_1\right)$ is $342\text{}\text{K}$.

It is given that the temperature $\left(T_2\right)$ is $60{}^{\circ }\text{C}$.

The conversion of ${}^{\circ }\text{C}$ in $\text{K}$ is done as,

    $1{}^{\circ }\text{C}=\text{273}\text{K}$

The conversion of $60{}^{\circ }\text{C}$ in $\text{K}$ is done as,

    $\begin{array}{c}60{}^{\circ }\text{C}=60+\text{273}\text{K}\\ =333\text{K}\end{array}$

Therefore, the temperature $T_2$ is $333\text{K}$.

The enthalpy of vaporization for hexane is $29853.18\text{J}{\text{mol}}^{\text{-1}}$.

Substitute the values of $p_1,T_1,T_2,\Delta H_{\text{vaporization}}$ and $R$ in equation (3).

    $\begin{array}{c}\ln \left(\frac{p_2}{1\text{atm}}\right)=\frac{29853.18\text{J}{\text{mol}}^{\text{-1}}}{8.314\text{J}{\text{K}}^{\text{-1}}{\text{mol}}^{\text{-1}}}\left(\frac{1}{342\text{K}}-\frac{1}{333\text{K}}\right)\\ \ln \left(\frac{p_2}{1\text{atm}}\right)=3950.712{\text{mol}}^{\text{-1}}\text{K}\left(0.00292{\text{K}}^{-1}-0.00300{\text{K}}^{-1}\right)\\ \ln \left(\frac{p_2}{1\text{atm}}\right)=-0.316\\ p_2=e^{-0.316}\times 1\text{atm}\end{array}$

Or,

    $p_2=\underset{\_}{0.729atm}$

Hence, the vapour pressure at ${60}^{\circ }\text{C}$ is $\underset{\_}{0.729atm}$.