Chapter 6, Problem 6.33E

Four alcohols have the formula ${\text{C}}_{\text{4}}{\text{H}}_{\text{9}}\text{OH}$: $\text{1}$-butanol, $\text{2}$-butanol (or sec-butanol), isobutanol (or $\text{2}$-methyl- $\text{2}$-propanol). They are examples of isomers, or compounds that have the same molecular formula but different molecular structures. The following table gives data on the isomers:

Compound	${\Delta }_{vap}H\left(kJ/mol\right)$	Normal boiling point $\left(°C\right)$
$\text{1}$-Butanol	$45.90$	$117.2$
$\text{2}$-Butanol	$44.82$	$99.5$
Isobutanol	$45.76$	$108.1$
tert-Butanol	$43.57$	$82.3$

Using the Clausius-Clapeyron equation, rank the isomers of butanol in order of decreasing vapor pressure at $\text{25°C}$. Does the ranking agree with any conventional wisdom based on the ${\Delta }_{\text{vap}}H$ or the normal boiling points?

Interpretation Introduction

Interpretation:

The isomers of butanol are to be ranked in order of decreasing vapor pressure. Also the validation of the fact that the ranking agrees with the conventional wisdom based on ${\Delta }_{\text{vap}}H$ values or the normal boiling point is to be stated.

Concept introduction:

The Clausius-Clapeyron equation is obtained from the rearrangement and integration of Clapeyron equation. The Clausius-Clapeyron equation is generally used for gas-phase equilibria, to predict the equilibrium temperatures and pressures and also for the determination of enthalpy for phase transition.

Answer to Problem 6.33E

The decreasing order of vapor pressures of isomers of butanol is $\text{tert-butanol}<\text{2-butanol}<\text{isobutanol}<\text{1-butanol}$.

On comparison of the vapor pressure of isomers of butanol, the value of vapor pressure decreases with the increase in the values of enthalpy of vaporization and normal boiling point.

Explanation of Solution

The table containing the data for the isomers of butanol is shown below.

Compound	${\Delta }_{vap}H\left(kJ/mol\right)$	Normal boiling point $\left(°C\right)$
$\text{1}$-Butanol	$45.90$	$117.2$
$\text{2}$-Butanol	$44.82$	$99.5$
Isobutanol	$45.76$	$108.1$
tert-Butanol	$43.57$	$82.3$

Table 1

To calculate the vapor pressure of the given isomers of butanol the formula used is,

$\ln \frac{p_2}{p_1}=-\frac{\Delta \overline{H}}{R}\left(\frac{1}{T_2}-\frac{1}{T_1}\right)$

Where,

• $p$ represents the pressure.

• $T$ represents the temperature.

• $\Delta \overline{H}$ represents the enthalpy for phase transition.

Substitute the values of enthalpy of vaporization, vapor pressure at normal boiling point that is $1\text{atm}$ and temperature of $\text{1}$-butanolin the given formula.

$\begin{array}{rll}\ln \frac{p_2}{1}& =& -\frac{\text{45900J/mol}}{8.314\text{J/mol}.\text{K}}\left(\frac{1}{298.2\text{K}}-\frac{1}{390.4\text{K}}\right)\\ \ln \frac{p_2}{1}& =& -\left(5520.808\right)\left(7.92\times {10}^{-4}\right)\\ \ln p_2& =& -4.37\\ p_2& =& 0.01265\end{array}$

Thus, the vapor pressure of $\text{1}$-butanol at $\text{25°C}$ is $0.01265$.

Similarly, substitute the values of enthalpy of vaporization, vapor pressure at normal boiling point that is $1\text{atm}$ and temperature of $\text{2}$-butanol in the given formula.

$\begin{array}{rll}\ln \frac{p_2}{1}& =& -\frac{\text{44820 J/mol}}{8.314\text{J/mol}.\text{K}}\left(\frac{1}{298.2\text{K}}-\frac{1}{372.7\text{K}}\right)\\ \ln \frac{p_2}{1}& =& -\left(5390.91\right)\left(6.7033\times {10}^{-4}\right)\\ \ln p_2& =& -3.614\\ p_2& =& 0.027\end{array}$

Thus, the vapor pressure of $\text{2}$-butanol at $\text{25°C}$ is $0.027$.

Similarly, substitute the values of enthalpy of vaporization, vapor pressure at normal boiling point that is $1\text{atm}$ and temperature of isobutanol in the given formula.

$\begin{array}{rll}\ln \frac{p_2}{1}& =& -\frac{\text{45760J/mol}}{8.314\text{J/mol}.\text{K}}\left(\frac{1}{298.2\text{K}}-\frac{1}{381.3\text{K}}\right)\\ \ln \frac{p_2}{1}& =& -\left(5503.97\right)\left(7.3085\times {10}^{-4}\right)\\ \ln p_2& =& -4.0226\\ p_2& =& 0.018\end{array}$

Thus, the vapor pressure of isobutanol at $\text{25°C}$ is $0.018$.

Similarly, substitute the values of enthalpy of vaporization, vapor pressure at normal boiling point that is $1\text{atm}$ and temperature of tert-butanol in the given formula.

$\begin{array}{rll}\ln \frac{p_2}{1}& =& -\frac{\text{43570 J/mol}}{8.314\text{J/mol}.\text{K}}\left(\frac{1}{298.2\text{K}}-\frac{1}{355.5\text{K}}\right)\\ \ln \frac{p_2}{1}& =& -\left(5240.558\right)\left(5.4051\times {10}^{-4}\right)\\ \ln p_2& =& -2.83\\ p_2& =& 0.059\end{array}$

Thus, the vapor pressure of tert-butanol at $\text{25°C}$ is $0.059$.

The decreasing order of vapor pressures of isomers of butanol is $\text{tert-butanol}<\text{2-butanol}<\text{isobutanol}<\text{1-butanol}$.

On comparison of the vapor pressure of isomers of butanol, the value of vapor pressure decreases with the increase in the values of enthalpy of vaporization and normal boiling point.

Conclusion

The decreasing order of vapor pressures of isomers of butanol is $\text{tert-butanol}<\text{2-butanol}<\text{isobutanol}<\text{1-butanol}$.

On comparison of the vapor pressure of isomers of butanol, the value of vapor pressure decreases with the increase in the values of enthalpy of vaporization and normal boiling point.