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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Textbook Question
Chapter 8, Problem 73GQ

The equation for the combustion of gaseous methanol is

2 CH3OH(g) + 3 O2(g) → 2 CO2(g) + 4 H2O(g)

  1. (a) Using the bond dissociation enthalpies in Table 8.8, estimate the enthalpy change for this reaction. What is the enthalpy of combustion of one mole of gaseous methanol?
  2. (b) Compare your answer in part (a) with the value of Δ t H calculated using enthalpies of formation data.

(a)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation:

Enthalpy change for the given reaction and the enthalpy of combustion of one mole of gaseous methanol has to be determined.

Concept Introduction:

Bond dissociation enthalpy is the energy required to break one mole of gaseous bonds to form gaseous atoms. The process of breaking bonds in a molecule is always endothermic therefore ΔrH for bond breaking is always positive and ΔrH value for formation of bond is always negative.

ΔrH=ΣΔH(bondsbrocken)ΣΔH(bondsformed)

Explanation of Solution

Equation for the given reaction is,

2CH3OH(g)+3O2(g)2CO2(g)+4H2O(g)

Chemistry & Chemical Reactivity, Chapter 8, Problem 73GQ

The enthalpy change for this reaction can be determined as follows,

Bonds Brocken:

6molofCH2molofOH2molofCO3molofO=O

ΣΔH(bondsbrocken)=6×413kJforCHbonds+2×463kJforOH+2×358kJforCObonds+3×498kJforO=O=5614kJ/mol

Bonds formed:

4molC=O8molOH

ΣΔH(bondsformed)=4×745kJforC=Obonds+8×463kJforOH+=6684kJ/mol

ΔrH=ΣΔH(bondsbrocken)-ΣΔH(bondsformed)=5614kJ-6684kJ=-1070kJ/mol

Enthalpy of combustion of 2mole methanol is found to be -1070kJ/mol and so the enthalpy of combustion of 1mol methanol is -1070kJ/mol2=-535kJ/mol

(b)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation:

Standard enthalpy change for the given reaction and the standard enthalpy of combustion of one mole of gaseous methanol has to be determined.

Concept Introduction:

Bond dissociation enthalpy is the energy required to break one mole of gaseous bonds to form gaseous atoms. The process of breaking bonds in a molecule is always endothermic therefore ΔrH for bond breaking is always positive and ΔrH value for formation of bond is always negative.

ΔrH=ΣΔH(bondsbrocken)ΣΔH(bondsformed)

Enthalpy of formation:

ΔrH=ΔfH0(products)ΔfH0(reactants)

Explanation of Solution

Equation for the given reaction is,

2CH3OH(g)+3O2(g)2CO2(g)+4H2O(g)

The standard enthalpy change for this reaction can be determined as follows,

ΔfH0(reactants)=2×-201kJ/molforCH3OH+3×0kJ/molforO2=-402kJ/mole

ΔfH0(products)=2×-393.509kJ/molforCO2+4×241.83kJ/molforH2O=-1754.3kJ/mole

ΔrH=ΣΔfH0(products)-ΣΔfH0(reactants)=-1754.3kJ/mole-402kJ/mol=-1352.3kJ/mole

Enthalpy of combustion of 2mole methanol is found to be -1352.3kJ/mole and so the enthalpy of combustion of 1mol methanol is -1352.3kJ/mole2=-676kJ/mol

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Chapter 8 Solutions

Chemistry & Chemical Reactivity

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