Chemistry & Chemical Reactivity
Chemistry & Chemical Reactivity
9th Edition
ISBN: 9781133949640
Author: John C. Kotz, Paul M. Treichel, John Townsend, David Treichel
Publisher: Cengage Learning
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Chapter 11, Problem 38GQ

The following data are the equilibrium vapor pressure of limonene, C10H16, at various temperatures. (Limonene is used as a scent in commercial products.)

Chapter 11, Problem 38GQ, The following data are the equilibrium vapor pressure of limonene, C10H16, at various temperatures.

  1. (a) Plot these data as ln P versus 1/T so that you have a graph resembling the one in Figure 11.13.
  2. (b) At what temperature does the liquid have an equilibrium vapor pressure of 250 mm Hg? At what temperature is it 650 mm Hg?
  3. (c) What is the normal boiling point of limonene?
  4. (d) Calculate the molar enthalpy of vaporization for limonene using the Clausius-Clapeyron equation.

(a)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation:

The graph of lnPversus1/T using the given data has to be plotted.

Concept Introduction:

Clausius-Clapeyron equation:

  lnP=(ΔvapH0RT)+C

From this relationship we can calculate the molar enthalpy of vaporization by knowing the corresponding temperature and pressure values.

If we have pressures at two different temperatures, then enthalpy of vaporization can be calculated by

  lnP2p1=-ΔvapH0R[1T2-1T1]

Explanation of Solution

The graph of lnP versus 1/T is plotted using given data.

Given:

  Temperature(K)VapourPressure(mmHg)2871326.810357.340381.3100424.4400

The values of lnP and 1/T is determined

  lnP1/T03.48×10-32.30253.059×10-33.68882.798×10-34.60512.622×10-35.99142.356×10-3

The graph of lnP versus 1/T is plotted

Chemistry & Chemical Reactivity, Chapter 11, Problem 38GQ , additional homework tip  1

(b)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation:

The temperature at which the liquid has vapour pressures of 250mmHg and 650mmHg has to be determined.

Concept Introduction:

Clausius-Clapeyron equation:

  lnP=(ΔvapH0RT)+C

From this relationship we can calculate the molar enthalpy of vaporization by knowing the corresponding temperature and pressure values.

If we have pressures at two different temperatures, then enthalpy of vaporization can be calculated by

  lnP2p1=-ΔvapH0R[1T2-1T1]

Answer to Problem 38GQ

The temperature at which liquid has a vapour pressure of 250mmHg is 396.96K.

The temperature at which liquid has a vapour pressure of 650mmHg is 434.50K.

Explanation of Solution

The temperature at which the liquid has a pressure of 250mmHgand650mmHg is determined.

Given:

  Temperature(K)VapourPressure(mmHg)2871326.810357.340381.3100424.4400

The values of lnP and 1/T is determined

  lnP1/T03.48×10-32.30253.059×10-33.68882.798×10-34.60512.622×10-35.99142.356×10-3

The graph of lnP versus 1/T is plotted

Chemistry & Chemical Reactivity, Chapter 11, Problem 38GQ , additional homework tip  2

From the slope of the graph we can calculate the molar enthalpy of vaporization.

Slope=ΔvapH0R

From the graph drawn,

  Slope=y2y1x2x1=(4.6051)(2.3025)(0.00279)(0.00305)=8856.1538

8856.1538=ΔvapH00.008314kJ/K.molΔvapH0=73.63kJ/mol.K

  • The temperature at which liquid has a vapour pressure of 250mmHg_ is calculated

We know the equation

lnP2p1=ΔvapH0R[1T21T1]

P1=100mmHg,P2=250mmHgT1=381.3KΔvapH0=73.63kJ/mol.K

Substituting the values

ln[250100]=73.63kJ/mol.K0.008314kJ/K.mol[1T21381.3K]0.9612=73.63kJ/mol.K0.008314kJ/K.mol[1T21381.3K]T2=396.96K

Thus the temperature at which liquid has a vapour pressure of 250mmHg is 396.96K

  • The temperature at which liquid has a vapour pressure of 650mmHg_ is calculated

P1=400mmHg,P2=650mmHgT1=424.4KΔvapH0=73.63kJ/mol.K

Substituting the values

ln[650400]=73.63kJ/mol.K0.008314kJ/K.mol[1T2K1424.4K]0.4855=73.63kJ/mol.K0.008314kJ/K.mol[1T21424.4K]T2=434.51K

Thus the temperature at which liquid has a vapour pressure of 650mmHg is 434.51K

(c)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation:

The normal boiling point of limonene has to be determined.

Concept Introduction:

Clausius-Clapeyron equation:

  lnP=(ΔvapH0RT)+C

From this relationship we can calculate the molar enthalpy of vaporization by knowing the corresponding temperature and pressure values.

If we have pressures at two different temperatures, then enthalpy of vaporization can be calculated by

  lnP2p1=-ΔvapH0R[1T2-1T1]

Boiling point of a liquid: The temperature at which external pressure and vapour pressure of the liquid become same.

Normal boiling point: When the external pressure is 760mmHg we can call it as normal boiling point.

Answer to Problem 38GQ

The normal boiling point of limonene is 437.86K.

The molar enthalpy of vaporization of limonene is 45.104kJ/mol.

Explanation of Solution

The normal boiling point of limonene is determined.

Normal boiling point is the temperature when the external pressure is 760mmHg

P1=400mmHg,P2=760mmHgT1=424.4KΔvapH0=73.63kJ/mol.K

Substituting the values

ln[760400]=73.63kJ/mol.K0.008314kJ/K.mol[1T21424.4K]0.6418=73.63kJ/mol.K0.008314kJ/K.mol[1T21424.4K]T2=437.86K

The normal boiling point of benzene is 437.86K

(d)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation:

The molar enthalpy of vaporization has to be determined

Concept Introduction:

Clausius-Clapeyron equation:

  lnP=(ΔvapH0RT)+C

From this relationship we can calculate the molar enthalpy of vaporization by knowing the corresponding temperature and pressure values.

If we have pressures at two different temperatures, then enthalpy of vaporization can be calculated by

  lnP2p1=-ΔvapH0R[1T2-1T1]

Molar enthalpy of vaporization: The energy required to convert liquid to gas of 1mol of a substance is called molar enthalpy of vaporization

Answer to Problem 38GQ

The molar enthalpy of vaporization of limonene is 45.11kJ/mol.

Explanation of Solution

The molar enthalpy of vaporization using Clausius-Clapeyron equation is determined

If we have pressures at two different temperatures, then enthalpy of vaporization can be calculated by

  lnP2p1=-ΔvapH0R[1T2-1T1]

Given:

  Temperature(K)VapourPressure(mmHg)2871326.810357.340381.3100424.4400

P1=1mmHg,P2=10mmHgT1=287K,T2=326.8KΔvapH0=?

Substituting the values

ln[101]=ΔvapH00.008314kJ/K.mol[1326.8K1287K]2.3025=ΔvapH00.008314kJ/K.mol[1326.8K1287K]ΔvapH0=45.11kJ/mol

The molar enthalpy of vaporization of limonene is 45.11kJ/mol.

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