Principles of General Chemistry
Principles of General Chemistry
3rd Edition
ISBN: 9780073402697
Author: SILBERBERG, Martin S.
Publisher: McGraw-Hill College
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Chapter 6, Problem 6.61P

(a)

Interpretation Introduction

Interpretation:The value of ΔH0rxn of fermentation and respiration needs to be determined.

Concept Introduction : The reactions of fermentation, respiration and combustion of glucose are as follows:

  1. Fermentation of glucose
  2.   C6H12O6(s)2C2H5OH(l)+2CO2(g)

  3. Respiration

  C6H12O6(s)+6O2(g)6CO2(g)+6H2O(g)

The amount of energy released on combustion of an organic compound is known as enthalpy of combustion. The formula of determination of ΔHrxn is as follows:

  ΔH0rxn=mΔH0f(products)nΔH0f(reactants)

Here, represents the sum of terms, m and n are number of moles of products and reactants respectively. The term H0f is standard heat of formation.

(a)

Expert Solution
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Answer to Problem 6.61P

  ΔH0rxn of fermentation reaction is 68.96kJ

  ΔH0rxn of respiration reaction is 2538.656kJ .

Explanation of Solution

The value of standard heat of formation of compounds is as follows:

  H0f(C6H12O6(s))=1273.3kJmol1H0f(CO2)=393.5kJmol1H0f(H2O(g))=241.826kJmol1H0f(O2)=0kJmol1

Now calculate ΔH0rxn of fermentation reaction.

Fermentation of glucose

  C6H12O6(s)2C2H5OH(l)+2CO2(g)

  ΔH0rxn=[2molΔH0f(C2H5OH(l))+2molΔH0f(CO2(g))][1molΔH0f(C6H12O6(s))]=[2mol×(277.63 kJ mol)+2mol×(393.5 kJ mol)][1mol×(1273.3 kJ mol)]=555.26kJ787kJ+1273.3kJ=68.96kJ

So, ΔH0rxn of fermentation reaction is 68.96kJ .

Now calculate ΔH0rxn of respiration reaction.

  C6H12O6(s)+6O2(g)6CO2(g)+6H2O(g)

  ΔH0rxn=[6molΔH0f(CO2(g))+6molΔH0f(H2O(g))][1molΔ H 0 f( C 6 H 12 O 6(s))+6molΔ H 0 f( O 2(g))]=[6mol×(393.5 kJ mol)+6mol×(241.826 kJ mol)][1mol×( 1273.3 kJ mol )+6mol×0]=2361kJ1450.956kJ+1273.3kJ=2538.656kJ

Hence, ΔH0rxn of respiration reaction is 2538.656kJ .

(b)

Interpretation Introduction

Interpretation:A combustion reaction for ethanol needs to be determined and whether ethanol or sugar has the higher ΔH0rxn for combustion per mole of C needs to be determined.

Concept Introduction : The amount of energy released on combustion of an organic compound is known as enthalpy of combustion. The formula of determination of ΔHrxn is as follows:

  ΔH0rxn=mΔH0f(products)nΔH0f(reactants)

Here, represents the sum of terms, m and n are number of moles of products and reactants respectively. The term H0f is standard heat of formation.

(b)

Expert Solution
Check Mark

Answer to Problem 6.61P

The balanced chemical reaction for combustion of ethanol is,

  CH3CH2OH(l)+3O2(g)2CO2(g)+3H2O

Ethanol has higher ΔH0rxn for the combustion of per mole of carbon than sugar.

Explanation of Solution

The balanced chemical reaction for combustion of ethanol is,

  CH3CH2OH(l)+3O2(g)2CO2(g)+3H2O

Use appendix B to get the value of standard heat of formation of compound.

  H0f(C2H5OH(l))=277.63kJmol1H0f(CO2)=393.5kJmol1H0f(H2O(g))=241.826kJmol1H0f(O2)=0kJmol1

Putting the values,

  ΔH0rxn=[2molΔH0f(CO2(g))+3molΔH0f(H2O(g))][1molΔ H 0 f( CH 3 CH 2OH(l))+3molΔ H 0 f( O 2(g))]=[2mol×(393.5 kJ mol)+3mol×(241.826 kJ mol)][1mol×( 277.63 kJ mol )+3mol×0]=787kJ725.478kJ+277.63kJ=1234.848kJ

Thus, standard heat of reaction for combustion of ethanol per mole will be 1234.848kJ

The molar relationship of carbon in sugar and ethanol are as follows:

  1molC6H12O6=6molC1molCH3CH2OH=2molC

Use the molar relationships and heats of reaction of sugar and ethanol to determine the standard heat of reaction for the combustion of per mole of carbon.

  ΔH0rxn=1.00molC×1molC6H 12O66mol C×2538.6kJ1molC6H 12O6=423.1kJ(FromC6H12O6)ΔH0rxn=1.00molC×1mol CH3 CH2OH2mol C×1234.8kJ1mol CH3 CH2OH=617.4kJ(FromCH3CH2OH)

Hence, ethanol has higher ΔH0rxn for the combustion of per mole of carbon than sugar.

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Chapter 6 Solutions

Principles of General Chemistry

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