Chemistry: The Molecular Science
Chemistry: The Molecular Science
5th Edition
ISBN: 9781285199047
Author: John W. Moore, Conrad L. Stanitski
Publisher: Cengage Learning
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Chapter 9, Problem ISP

(a)

Interpretation Introduction

Interpretation:

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

(a)

Expert Solution
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Explanation of Solution

Chemistry: The Molecular Science, Chapter 9, Problem ISP , additional homework tip  1

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.

(b)

Expert Solution
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Answer to Problem ISP

The temperature of 1-butanol at 500mmHg of vapour pressure is approximately 104oC.

Explanation of Solution

Chemistry: The Molecular Science, Chapter 9, Problem ISP , additional homework tip  2

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 104oC.

(c)

Interpretation Introduction

Interpretation:

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

(c)

Expert Solution
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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.

(c)

Expert Solution
Check Mark

Explanation of Solution

Chemistry: The Molecular Science, Chapter 9, Problem ISP , additional homework tip  3

Figure 3

The boiling point of diethyl ether is 34.6C.  So it will evaporate first at body temperature which is C.  Draw a vertical line from 37C on the x-axis which will touch each curve.  It implies that none of the liquids has a vapour pressure close to 760mmHg at 37C, 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 ΔvapH of 1-butanol under given condition has to be determined.

Concept Introduction:

Clausius-Clapeyron equation:

  ln(P2P1)=ΔVapHR[1T21T1]

Where,

ΔVapH= 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.

(d)

Expert Solution
Check Mark

Answer to Problem ISP

The ΔvapH 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 117.3C.  So the vapour pressure at its normal boiling point is 760mmHg. Hence, P1=760mmHg.

  P2=389mmHgT1=117.3C+273.15=390.5KT2=100.0C+273.15=373.2KR=8.314JK1mol1

Clausius-Clapeyron equation:

  ln(P2P1)=ΔVapHR[1T21T1]

Substituting all the data in the above equation and solving for ΔvapH gives the following result.

  ln(389mmHg760mmHg)=ΔvapH8.314JK1mol1[1373.2K1390.5K]ln(0.512)=ΔvapH8.314JK1mol1(0.0026800.002561)ΔvapH=46.9kJ/mol.

Therefore, the ΔvapH of 1-butanol under given condition has been determined to be 46.9kJ/mol.

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Chapter 9 Solutions

Chemistry: The Molecular Science

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