(a) Use Trouton's rule, $\mathbf{∆}{\mathit{H}}_{\mathbf{v}\mathbf{a}\mathbf{p}}^{\mathbf{°}}\mathbf{\approx }\mathbf{(}\mathbf{88}\mathbf{}\mathbf{J}\mathbf{}{\mathbf{mol}}^{\mathbf{-}\mathbf{1}}\mathbf{}{\mathbf{K}}^{\mathbf{-}\mathbf{1}}\mathbf{)}{\mathit{T}}_{\mathbf{b}\mathbf{p}}$, to estimate the enthalpy of vaporization of octane (b.p. ${\mathbf{126}}^{\mathbf{°}}$).

(b) Use the form of the Clausius-Clapeyron equation below to estimate the vapor pressure of octane at the column temperature in Figure 24-9$\mathbf{\left(}{\mathbf{70}}^{\mathbf{°}}\mathbf{C}\mathbf{\right)}$

$\mathbf{In}\left(\frac{P_1}{P_2}\right)\mathbf{=}\mathbf{-}\left(\frac{∆H_{\mathrm{vap}}}{R}\right)\left(\frac{1}{T_1}-\frac{1}{T_2}\right)$

(c) Calculate the vapor pressure for hexane (b.p. ${\mathbf{69}}^{\mathbf{°}}\mathbf{}\mathbf{C}$) at${\mathbf{70}}^{\mathbf{°}}\mathbf{}\mathbf{C}$

(d) What is the relationship between solute vapor pressure and retention?

(e) Why is the technique called "gas chromatography” if retained analytes are only partially vaporized?

Given:${\mathrm{T}}_{\mathrm{bp}}={126}^{°}\mathrm{C}$

Find: enthalpy of vaporization role="math" localid="1657772978189" $(∆{\mathrm{H}}_{\mathrm{vap}})$of octane using the Trouton's rule

$∆{\mathrm{H}}_{\mathrm{vap}}\approx \left(88\frac{\mathrm{J}}{\mathrm{mol}.\mathrm{K}}\right).{\mathrm{T}}_{bp}$

Convert the given temperature to Kelvin (K):

${\mathrm{T}}_{\mathrm{bp}}={126}^{°}\mathrm{C}+273.15=399.15\mathrm{K}$

Compute the enthalpy of vaporization $(∆{\mathrm{H}}_{\mathrm{vap}})$of octane using the Trouton's rule:

role="math" localid="1657773979789" $$\begin{gathered} ∆{\mathrm{H}}_{\mathrm{vap}}\approx \left(88\frac{\mathrm{J}}{\mathrm{mol}.\mathrm{K}}\right).399.15\mathrm{K} \\ \approx 35125.2\frac{\mathrm{J}}{\mathrm{mol}} \end{gathered}$$

The Clausius-Clapeyron equation to be used in the problem is shown below:

$\mathrm{In}\left(\frac{{\mathrm{P}}_1}{{\mathrm{P}}_2}\right)=-\left(\frac{∆{\mathrm{H}}_{\mathrm{vap}}}{\mathrm{R}}\right)\left(\frac{1}{{\mathrm{T}}_1}-\frac{1}{{\mathrm{T}}_2}\right)$

where:

-${\mathrm{P}}_1=\mathrm{vapor}\mathrm{pressure}\mathrm{at}{\mathrm{T}}_1\left(\mathrm{K}\right)$

-${\mathrm{P}}_2=\mathrm{vapor}\mathrm{pressure}\mathrm{at}{\mathrm{T}}_2\left(\mathrm{K}\right)$

- $∆{\mathrm{H}}_{\mathrm{vap}}$= enthalpy of vaporization of the substance

-$\mathrm{R}=\mathrm{gas}\mathrm{constant}\left(8.314\frac{\mathrm{J}}{\mathrm{mol}\times \mathrm{K}}\right)$

Given:

- ${\mathrm{T}}_1={\mathrm{T}}_{\mathrm{bp}}=399.15\mathrm{K}$

- ${\mathrm{P}}_1=1\mathrm{atm}=760\mathrm{mmHg}$(the vapor pressure is equal to atmospheric pressure the boiling temperature)

-${\mathrm{T}}_2={70}^{°}\mathrm{C}+273.15=343.15\mathrm{K}$

- $∆{\mathrm{H}}_{\mathrm{vap}}=35125.2\frac{\mathrm{J}}{\mathrm{mol}}$

Find: ${\mathrm{P}}_2$= vapor pressure of octane at the column temperature${70}^{°}\mathrm{C}$

Substitute the given values to the Clausius-Clapeyron equation:

role="math" localid="1657774107541" $$\begin{gathered} \mathrm{In}\left(\frac{760\mathrm{mmHg}}{{\mathrm{P}}_2}\right)=\left(\frac{35125.2{\displaystyle \frac{\mathrm{J}}{\mathrm{mol}\times \mathrm{K}}}}{8.314{\displaystyle \frac{\mathrm{J}}{\mathrm{mol}}}}\right)\left(\frac{1}{399.15\mathrm{K}}-\frac{1}{343.15\mathrm{K}}\right) \\ \mathrm{In}\left(\frac{760\mathrm{mmHg}}{{\mathrm{P}}_2}\right)=1.7273 \\ \frac{760\mathrm{mmHg}}{{\mathrm{P}}_2}=5.6256 \\ {\mathrm{P}}_2=135.10\mathrm{mmHg} \end{gathered}$$

Given:

-${\mathrm{T}}_1={70}^{°}\mathrm{C}+273.15=343.15\mathrm{K}$

-${\mathrm{T}}_2={\mathrm{T}}_{\mathrm{bp}}={69}^{°}\mathrm{C}+273.15=342.15\mathrm{K}$

-${\mathrm{P}}_2\mathrm{atm}=760\mathrm{mmHg}$

Find: ${\mathrm{P}}_1=$vapor pressure of hexane at${70}^{°}\mathrm{C}$

Use Trouton's rule to compute for the heat of vaporization:

$$\begin{gathered} ∆{\mathrm{H}}_{\mathrm{vap}}\approx \left(88\frac{\mathrm{J}}{\mathrm{mol}.\mathrm{K}}\right).T_{bp} \\ \approx \left(88\frac{\mathrm{J}}{\mathrm{mol}.\mathrm{K}}\right).342.15K \\ \approx 30109.2\frac{\mathrm{J}}{\mathrm{mol}} \end{gathered}$$

Substitute the given values to the Clausius-Clapeyron equation to compute for the vapor pressure at${70}^{°}\mathrm{C}$

$$\begin{gathered} \mathrm{In}\left(\frac{P_2}{760\mathrm{mmHg}}\right)=\left(\frac{30109.2{\displaystyle \frac{\mathrm{J}}{\mathrm{mol}\times \mathrm{K}}}}{8.314{\displaystyle \frac{\mathrm{J}}{\mathrm{mol}}}}\right)\left(\frac{1}{343.15\mathrm{K}}-\frac{1}{343.15\mathrm{K}}\right) \\ \mathrm{In}\left(\frac{{\mathrm{P}}_2}{760\mathrm{mmHg}}\right)=0.0308 \\ \frac{P_2}{760\mathrm{mmHg}}=1.0313 \\ {\mathrm{P}}_2=783.81\mathrm{mmHg} \end{gathered}$$

d. The relationship between solute vapor pressure and retention is that as vapor pressure decreases, retention increases. In Figure 24-9 in the textbook, retention increases with boiling temperature. However, boiling temperature has an indirect relationship with vapor pressure (i.e., higher boiling point = lower vapor pressure).

e. The technique is called gas chromatography because it is a method used to analyze volatile compounds in the gas phase, although retained analytes are only partially vaporized. In gas chromatography, the sample components are dissolved in a solvent and the analytes are separated by vaporizing the mixture. The sample is then separated into two phases: mobile phase and stationary phase.

$$\begin{gathered} \mathrm{a})∆{\mathrm{H}}_{\mathrm{vap}}\approx 35125.2\frac{\mathrm{J}}{\mathrm{mol}} \\ \mathrm{b}){\mathrm{P}}_2=135.10\mathrm{mmHg} \\ \mathrm{c}){\mathrm{P}}_2=783.81\mathrm{mmHg} \end{gathered}$$