Chapter 9, Problem ISP

(a)

Interpretation Introduction

Interpretation:

The reason behind less volatility of hexane than diethyl ether has to be described.

Explanation of Solution

Figure 1

Vapour pressure of a liquid is defined as the pressure of the vapour when liquid and vapour are in dynamic equilibrium.  It increases with increase in temperature and a liquid with stronger intermolecular force of attractions has a lower vapour pressure at a given temperature.

If a liquid is less volatile, then it must have a lower vapour pressure which can be accounted for its stronger intermolecular force of attraction.  Examining the above figure, it can be concluded that hexane has a lower vapour pressure than diethyl ether at any temperature.  The non-covalent intermolecular forces of attraction acting between the molecules of hexane is London forces while in diethyl ether they are dipole-dipole forces and London forces.  Because liquid hexane has a lower vapour pressure, it can be predicted that the larger hexane molecules experience larger collective intermolecular London forces than diethyl ether.

So due to this reason, hexane is less volatile than diethyl ether.

(b)

Interpretation Introduction

Interpretation:

The temperature at which 1- butanol have a pressure of $500mmHg$ has to be determined.

Answer to Problem ISP

The temperature of 1-butanol at $500mmHg$ of vapour pressure is approximately $104^{o}C.$

Explanation of Solution

Figure 2

Examine the above figure carefully.  Draw a horizontal line from $500mmHg$ on the y-axis to the 1-butanol curve (blue color curve), then a vertical line down to the temperature axis.  The temperature of 1-butanol at $500mmHg$ of vapour pressure is approximately $104^{o}C.$

(c)

Interpretation Introduction

Interpretation:

The reason why the boiling point of 1-butanol is greater than that of water has to be described.

Explanation of Solution

The liquid having greater boiling point means it has stronger collective intermolecular forces.  Both water and 1-butanol can experience hydrogen bonding interactions.  Water is a small molecule with a total of $10$ electrons and 1-butanol is a larger molecule with $42$ electrons.  Therefore, it can be predicted that the larger 1-butanol molecule in the liquid state experience larger collective intermolecular London forces than liquid water making its boiling point larger.

(c)

Interpretation Introduction

Interpretation:

The substance which would evaporate immediately and which would remain as liquid has to be determined.

Explanation of Solution

Figure 3

The boiling point of diethyl ether is ${34.6}^{\circ }C.$  So it will evaporate first at body temperature which is C.  Draw a vertical line from ${37}^{\circ }C$ on the x-axis which will touch each curve.  It implies that none of the liquids has a vapour pressure close to $760mmHg$ at ${37}^{\circ }C$, which would  make it boil.  Hexane has a vapour pressure of around $280mmHg$ at this temperature.  Both water and 1-butanol have nearly zero vapour pressure at this temperature.  Hence, they will remain as liquid.

(d)

Interpretation Introduction

Interpretation:

The ${\Delta }_{vap}H$ of 1-butanol under given condition has to be determined.

Concept Introduction:

Clausius-Clapeyron equation:

  $\ln \left(\frac{P_2}{P_1}\right)=-\frac{{\Delta }_{Vap}H}{R}\left[\frac{1}{T_2}-\frac{1}{T_1}\right]$

Where,

${\Delta }_{Vap}H=$ Enthalpy of vaporization

P₁ and P₂ are two sets of pressures and T₁ and T₂ are two sets of absolute temperatures.

R is universal gas constant.

Answer to Problem ISP

The ${\Delta }_{vap}H$ of 1-butanol under given condition has been determined to be $46.9kJ/mol.$

Explanation of Solution

Given data:

The normal boiling point of 1-butanol is ${117.3}^{\circ }C.$  So the vapour pressure at its normal boiling point is $760mmHg$. Hence, $P_1=760mmHg.$

  $\begin{array}{l}{P}_2=389mmHg\\ \\ T_1={117.3}^{\circ }C+273.15=390.5K\\ \\ T_2={100.0}^{\circ }C+273.15=373.2K\\ \\ R=8.314J{K}^{-1}mo{l}^{-1}\end{array}$

Clausius-Clapeyron equation:

  $\ln \left(\frac{P_2}{P_1}\right)=-\frac{{\Delta }_{Vap}H}{R}\left[\frac{1}{T_2}-\frac{1}{T_1}\right]$

Substituting all the data in the above equation and solving for ${\Delta }_{vap}H$ gives the following result.

  $\begin{array}{c}\ln \left(\frac{389mmHg}{760mmHg}\right)=-\frac{{\Delta }_{vap}H}{8.314J{K}^{-1}mo{l}^{-1}}\left[\frac{1}{373.2K}-\frac{1}{390.5K}\right]\\ \\ \ln \left(0.512\right)=-\frac{{\Delta }_{vap}H}{8.314J{K}^{-1}mo{l}^{-1}}\left(0.002680-0.002561\right)\\ \\ {\Delta }_{vap}H=46.9kJ/mol.\end{array}$

Therefore, the ${\Delta }_{vap}H$ of 1-butanol under given condition has been determined to be $46.9kJ/mol.$