OWLv2 for Moore/Stanitski's Chemistry: The Molecular Science, 5th Edition, [Instant Access], 1 term (6 months)
OWLv2 for Moore/Stanitski's Chemistry: The Molecular Science, 5th Edition, [Instant Access], 1 term (6 months)
5th Edition
ISBN: 9781285460420
Author: John W. Moore; Conrad L. Stanitski
Publisher: Cengage Learning US
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Chapter 16, Problem 120QRT

(a)

Interpretation Introduction

Interpretation:

Three balanced equations for the reactions of coal with steam to form ethane gas, propane gas and liquid methanol with carbon dioxide as a byproduct have to be stated.

Concept Introduction:

Reaction of carbon with steam results in the formation of various gases along with carbon dioxide as by product.  The materials whose reaction releases large quantities of Gibbs free energy are useful to society because this Gibbs free energy can be utilized to do some other fruitful works that may not occur by itself.  The burning of coal releases a large amount of Gibbs free energy.  This energy is utilized in doing many useful works.

(a)

Expert Solution
Check Mark

Answer to Problem 120QRT

Three balanced equations for the reactions of coal with steam to form ethane gas, propane gas and liquid methanol with carbon dioxide as a byproduct are shown below.

    7C(carbon)+6H2O(g)2C2H6(g)+3CO2(g)5C(carbon)+4H2O(g)C3H8(g)+2CO2(g)3C(carbon)+4H2O(g)2CH3OH(l)+CO2(g)

Explanation of Solution

Reaction of carbon with steam results in the formation of various gases along with carbon dioxide as by product.  Three balanced equations for the reactions of coal with steam to form ethane gas, propane gas and liquid methanol with carbon dioxide as a byproduct are shown below.

    7C(carbon)+6H2O(g)2C2H6(g)+3CO2(g)5C(carbon)+4H2O(g)C3H8(g)+2CO2(g)3C(carbon)+4H2O(g)2CH3OH(l)+CO2(g)

(b)

Interpretation Introduction

Interpretation:

The value of ΔrH°,ΔrS°,ΔrG° for each of the reaction has to be calculated.  It has to be commented that whether any of them would be a feasible way to produce the mentioned products.

Concept Introduction:

Refer to part (a).

(b)

Expert Solution
Check Mark

Answer to Problem 120QRT

For the first reaction, the value of ΔrH° is 101.02kJ/mol_, the value of ΔrG° is 122.71kJ/mol_ and the value of ΔrS° is -72.71J/molK_.

For the second reaction, the value of ΔrH° is 76.5kJ/mol_, the value of ΔrG° is 102.08kJ/mol_ and the value of ΔrS° is -86.6J/molK_.

For the third reaction, the value of ΔrH° is 96.44kJ/mol_, the value of ΔrG° is 187.39kJ/mol_ and the value of ΔrS° is -305.2J/molK_.

All these reactions has a positive value of ΔrG°.  Therefore, none of them is feasible.  The value of ΔrH° is positive and the value of ΔrS° is negative.  It means that there is no temperature at which the formation of products is favored.

Explanation of Solution

The given reaction is shown below.

  7C(g)+6H2O(g)2C2H6(g)+3CO2(g)

The value of ΔrH° is calculated by the formula shown below.

    ΔrH°=nProductsΔfH°(Products)nReactantsΔfH°(Reactants)        (1)

Where,

  • ΔfH° is the change in standard enthalpy of formation.
  • nReactants is the number of moles of reactants.
  • nProducts is the number of moles of products.

The value of ΔfH° for C2H6(g), CO2(g), C(g) and H2O(g) is 84.68kJ/mol, 393.509kJ/mol, 0kJ/mol and 241.818kJ/mol respectively.

Substitute the values in equation (1) as shown below.

    ΔrH°=(((nC2H6(g)×ΔfH°(C2H6(g)))+(nCO2(g)×ΔfH°(CO2(g))))(nC(g)×ΔfH°(C(g))+(nH2O(g)×ΔfH°(H2O(g)))))=(((2×84.68kJ/mol)+(3×393.509kJ/mol))(7×0kJ/mol+(6×241.818kJ/mol)))=101.02kJ/mol_

The value of ΔrS° is calculated by the formula shown below.

    ΔrS°=nProductsΔfS°(Products)nReactantsΔfS°(Reactants)        (2)

Where,

  • ΔfS° is the change in standard entropy of formation.
  • nReactants is the number of moles of reactants.
  • nProducts is the number of moles of products.

The value of ΔfS° for C2H6(g), CO2(g), C(g) and H2O(g) is 229.6J/molK, 213.74J/molK, 5.74J/molK and 188.825J/molK respectively.

Substitute the values in equation (2) as shown below.

    ΔrS°=(((nC2H6(g)×ΔfS°(C2H6(g)))+(nCO2(g)×ΔfS°(CO2(g))))(nC(g)×ΔfS°(C(g))+(nH2O(g)×ΔfS°(H2O(g)))))=(((2×229.6J/molK)+(3×213.74J/molK))(7×5.74J/molK+(6×188.825J/molK)))=-72.71J/molK_

The value of temperature is given as 25.0°C.

Convert the temperature of 25.0°C into Kelvin as follows.

T(K)=T(°C)+273.15=(25+273.15)K=298.15K

The value of ΔrG° is calculated by the formula shown below.

    ΔrG°=nProductsΔfG°(Products)nReactantsΔfG°(Reactants)        (3)

Where,

  • ΔfG° is the change in standard Gibbs free energy of formation.
  • nReactants is the number of moles of reactants.
  • nProducts is the number of moles of products.

The value of ΔfG° for C2H6(g), CO2(g), C(g) and H2O(g) is 32.82kJ/mol, 394.359kJ/mol, 0kJ/mol and 228.572kJ/mol respectively.

Substitute the values in equation (3) as shown below.

    ΔrG°=(((nC2H6(g)×ΔfG°(C2H6(g)))+(nCO2(g)×ΔfG°(CO2(g))))(nC(g)×ΔfG°(C(g))+(nH2O(g)×ΔfG°(H2O(g)))))=(((2×32.82kJ/mol)+(3×394.359kJ/mol))(7×0kJ/mol+(6×228.572kJ/mol)))=122.71kJ/mol_

The given reaction is shown below.

  5C(carbon)+4H2O(g)C3H8(g)+2CO2(g)

The value of ΔrH° is calculated by the formula shown below.

    ΔrH°=nProductsΔfH°(Products)nReactantsΔfH°(Reactants)        (1)

Where,

  • ΔfH° is the change in standard enthalpy of formation.
  • nReactants is the number of moles of reactants.
  • nProducts is the number of moles of products.

The value of ΔfH° for C3H8(g), CO2(g), C(g) and H2O(g) is 103.8kJ/mol, 393.509kJ/mol, 0kJ/mol and 241.818kJ/mol respectively.

Substitute the values in equation (1) as shown below.

    ΔrH°=(((nC2H6(g)×ΔfH°(C3H8(g)))+(nCO2(g)×ΔfH°(CO2(g))))(nC(g)×ΔfH°(C(g))+(nH2O(g)×ΔfH°(H2O(g)))))=(((1×103.8kJ/mol)+(2×393.509kJ/mol))(5×0kJ/mol+(4×241.818kJ/mol)))=76.5kJ/mol_

The value of ΔrS° is calculated by the formula shown below.

    ΔrS°=nProductsΔfS°(Products)nReactantsΔfS°(Reactants)        (2)

Where,

  • ΔfS° is the change in standard entropy of formation.
  • nReactants is the number of moles of reactants.
  • nProducts is the number of moles of products.

The value of ΔfS° for C3H8(g), CO2(g), C(g) and H2O(g) is 269.9J/molK, 213.74J/molK, 5.74J/molK and 188.825J/molK respectively.

Substitute the values in equation (2) as shown below.

    ΔrS°=(((nC2H6(g)×ΔfS°(C3H8(g)))+(nCO2(g)×ΔfS°(CO2(g))))(nC(g)×ΔfS°(C(g))+(nH2O(g)×ΔfS°(H2O(g)))))=(((1×269.9J/molK)+(2×213.74J/molK))(5×5.74J/molK+(4×188.825J/molK)))=-86.6J/molK_

The value of temperature is given as 25.0°C.

Convert the temperature of 25.0°C into Kelvin as follows.

T(K)=T(°C)+273.15=(25+273.15)K=298.15K

The value of ΔrG° is calculated by the formula shown below.

    ΔrG°=nProductsΔfG°(Products)nReactantsΔfG°(Reactants)        (3)

Where,

  • ΔfG° is the change in standard Gibbs free energy of formation.
  • nReactants is the number of moles of reactants.
  • nProducts is the number of moles of products.

The value of ΔfG° for C3H8(g), CO2(g), C(g) and H2O(g) is 23.49kJ/mol, 394.359kJ/mol, 0kJ/mol and 228.572kJ/mol respectively.

Substitute the values in equation (3) as shown below.

    ΔrG°=(((nC2H6(g)×ΔfG°(C3H8(g)))+(nCO2(g)×ΔfG°(CO2(g))))(nC(g)×ΔfG°(C(g))+(nH2O(g)×ΔfG°(H2O(g)))))=(((1×23.49kJ/mol)+(2×394.359kJ/mol))(5×0kJ/mol+(4×228.572kJ/mol)))=102.08kJ/mol_

The given reaction is shown below.

  3C(carbon)+4H2O(g)2CH3OH(l)+CO2(g)

The value of ΔrH° is calculated by the formula shown below.

    ΔrH°=nProductsΔfH°(Products)nReactantsΔfH°(Reactants)        (1)

Where,

  • ΔfH° is the change in standard enthalpy of formation.
  • nReactants is the number of moles of reactants.
  • nProducts is the number of moles of products.

The value of ΔfH° for CH3OH(l), CO2(g), C(g) and H2O(g) is 238.66kJ/mol, 393.509kJ/mol, 0kJ/mol and 241.818kJ/mol respectively.

Substitute the values in equation (1) as shown below.

    ΔrH°=(((nC2H6(g)×ΔfH°(CH3OH(l)))+(nCO2(g)×ΔfH°(CO2(g))))(nC(g)×ΔfH°(C(g))+(nH2O(g)×ΔfH°(H2O(g)))))=(((2×238.66kJ/mol)+(1×393.509kJ/mol))(3×0kJ/mol+(4×241.818kJ/mol)))=96.44kJ/mol_

The value of ΔrS° is calculated by the formula shown below.

    ΔrS°=nProductsΔfS°(Products)nReactantsΔfS°(Reactants)        (2)

Where,

  • ΔfS° is the change in standard entropy of formation.
  • nReactants is the number of moles of reactants.
  • nProducts is the number of moles of products.

The value of ΔfS° for CH3OH(l), CO2(g), C(g) and H2O(g) is 126.8J/molK, 213.74J/molK, 5.74J/molK and 188.825J/molK respectively.

Substitute the values in equation (2) as shown below.

    ΔrS°=(((nC2H6(g)×ΔfS°(CH3OH(l)))+(nCO2(g)×ΔfS°(CO2(g))))(nC(g)×ΔfS°(C(g))+(nH2O(g)×ΔfS°(H2O(g)))))=(((2×126.8J/molK)+(1×213.74J/molK))(3×5.74J/molK+(4×188.825J/molK)))=-305.2J/molK_

The value of temperature is given as 25.0°C.

Convert the temperature of 25.0°C into Kelvin as follows.

T(K)=T(°C)+273.15=(25+273.15)K=298.15K

The value of ΔrG° is calculated by the formula shown below.

    ΔrG°=nProductsΔfG°(Products)nReactantsΔfG°(Reactants)        (3)

Where,

  • ΔfG° is the change in standard Gibbs free energy of formation.
  • nReactants is the number of moles of reactants.
  • nProducts is the number of moles of products.

The value of ΔfG° for CH3OH(l), CO2(g), C(g) and H2O(g) is 166.27kJ/mol, 394.359kJ/mol, 0kJ/mol and 228.572kJ/mol respectively.

Substitute the values in equation (3) as shown below.

    ΔrG°=(((nC2H6(g)×ΔfG°(CH3OH(l)))+(nCO2(g)×ΔfG°(CO2(g))))(nC(g)×ΔfG°(C(g))+(nH2O(g)×ΔfG°(H2O(g)))))=(((2×166.27kJ/mol)+(1×394.359kJ/mol))(3×0kJ/mol+(4×228.572kJ/mol)))=187.39kJ/mol_

All these reactions has a positive value of ΔrG°.  Therefore, none of them is feasible.  The value of ΔrH° is positive and the value of ΔrS° is negative.  It means that there is no temperature at which the formation of products is favored.

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Chapter 16 Solutions

OWLv2 for Moore/Stanitski's Chemistry: The Molecular Science, 5th Edition, [Instant Access], 1 term (6 months)

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