Chemistry: Principles and Practice
Chemistry: Principles and Practice
3rd Edition
ISBN: 9780534420123
Author: Daniel L. Reger, Scott R. Goode, David W. Ball, Edward Mercer
Publisher: Cengage Learning
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Chapter 17, Problem 17.102QE

(a)

Interpretation Introduction

Interpretation:

The vapor pressure of CH3OH(l)at58oC has to be calculated.

(a)

Expert Solution
Check Mark

Explanation of Solution

The given reaction as follows,

  CH3OH(l)CH3OH(g)

  • The value of ΔHrxn is calculated by standard entropy change as follows,

  CH3OH(l)CH3OH(g)ΔHvo(kJ/mol.K)238.66200.66

  ΔHrxno=(ΔHfproducts)(ΔHf reactants)=(1×200.66kJ/mol)(1×238.66kJ/mol.K)=38kJ/mol.

Hence, value of ΔHrxn for the process is 38kJ/mol.

  • The integrated Clausius-Clapeyron equation is used to determine the vapor pressure as follows,

  ln(P2P1)=ΔHvR(1T11T2)ln(P2P1)=ΔHvR(T2T1)T1T2

As known, the normal boiling point of methanol is 64.7oC, the vapor pressure of CH3OH(l) is 760torr(1atm), the given data’s are,

  P1=760torrT1=64.7+273=338KT2=58+273=331KΔHv=38.0kJ×103J1kJ=38.0×103JR=8.314J/K

By substituting the values in the integrated Clausius-Clapeyron equation as shown below,

  ln(P2P1)=38.0×103J8.314J/K×(331K338K)338K×331K=0.286(P2P1)=e0.286=0.751P2=(0.751)P1=(0.751)(1.0atm)=0.751atm.

Hence, value of vapor pressure CH3OH(l)at58oC is 0.751atm.

(b)

Interpretation Introduction

Interpretation:

The vapor pressure of C2H5OH(l)at29oC has to be calculated.

(b)

Expert Solution
Check Mark

Explanation of Solution

The given reaction as follows,

  C2H5OH(l)C2H5OH(g)

  • The value of ΔHrxn is calculated by standard entropy change as follows,

  C2H5OH(l)C2H5OH(g)ΔHvo(kJ/mol.K)277.69235.10

  ΔHrxno=(ΔHfproducts)(ΔHf reactants)=(1×235.10kJ/mol)(1×277.69kJ/mol.K)=42.59kJ/mol.

Hence, value of ΔHrxn for the process is 42.59kJ/mol.

  • The integrated Clausius-Clapeyron equation is used to determine the vapor pressure as follows,

  ln(P2P1)=ΔHvR(1T11T2)ln(P2P1)=ΔHvR(T2T1)T1T2

As known, the normal boiling point is 79oC, the vapor pressure of C2H5OH(l) is 760torr(1atm), the given data’s are,

  P1=760torrT1=79+273=352KT2=29+273=302KΔHv=42.59kJ×103J1kJ=42.59×103JR=8.314J/K

By substituting the values in the integrated Clausius-Clapeyron equation as shown below,

  ln(P2P1)=42.59×103J8.314J/K×(303K352K)352K×302K=2.409(P2P1)=e2.409=0.0898P2=(0.0898)P1=(0.0898)(1.0atm)=0.0898atm.

Hence, value of vapor pressure C2H5OH(l)at29oC is 0.0898atm.

(c)

Interpretation Introduction

Interpretation:

The vapor pressure of Hg(l)at45oC has to be calculated.

(c)

Expert Solution
Check Mark

Explanation of Solution

The given reaction as follows,

  Hg(l)Hg(g)

  • The value of ΔHrxn is calculated by standard entropy change as follows,

  Hg(l)Hg(g)ΔHvo(kJ/mol.K)061.32

  ΔHrxno=(ΔHfproducts)(ΔHf reactants)=(1×61.32kJ/mol.K)(1×0kJ/mol)=61.32kJ/mol.

Hence, value of ΔHrxn for the process is 61.32kJ/mol.

  • The integrated Clausius-Clapeyron equation is used to determine the vapor pressure as follows,

  ln(P2P1)=ΔHvR(1T11T2)ln(P2P1)=ΔHvR(T2T1)T1T2

As known, the normal boiling point is 356.7oC, the vapor pressure of Hg(l) is 760torr(1atm), the given data’s are,

  P1=760torrT1=356.7+273=630KT2=45+273=318KΔHv=61.32kJ×103J1kJ=61.32×103JR=8.314J/K

By substituting the values in the integrated Clausius-Clapeyron equation as shown below,

  ln(P2P1)=61.32×103J8.314J/K×(318K630K)630K×318K=11.49(P2P1)=e11.49=0.0000123P2=(0.0000123)P1=(0.0000123)(1.0atm)=0.0000123atm.

Hence, value of vapor pressure Hg(l)at45oC is 0.0000123atm.

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Chapter 17 Solutions

Chemistry: Principles and Practice

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