Chemical Principles: The Quest for Insight
Chemical Principles: The Quest for Insight
7th Edition
ISBN: 9781464183959
Author: Peter Atkins, Loretta Jones, Leroy Laverman
Publisher: W. H. Freeman
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Chapter 4, Problem 4J.8E

(a)

Interpretation Introduction

Interpretation:

The standard change in the enthalpy, entropy and Gibbs free energy for the given reaction has to be determined.

Concept Introduction:

The importance of Gibbs energy is that it gives information about the spontaneity of the reaction or a process at constant temperature and pressure.  A compound is considered thermodynamically stable when it has negative Gibbs free energy of formation.  The relation to determine Gibbs free energy of a reaction is shown below.

ΔG°=nΔGf(products)nΔGf(reactants)

(a)

Expert Solution
Check Mark

Answer to Problem 4J.8E

The standard change in the enthalpy, entropy and Gibbs free energy for the given reaction +206.1kJ_, +214.632JK-1_ and +142.12kJ_ respectively.

Explanation of Solution

The given chemical equation for the production of synthesis gas is shown below.

    CH4(g)+H2O(g)CO(g)+3H2(g)

The relation for the calculation of standard change in the enthalpy is shown below.

    ΔH°=nΔHf(products)nΔHf(reactants)

The standard change in the enthalpy of the given reaction is calculated by the expression is shown below.

    ΔH°={(1mol)×ΔHf(CO,g)+(3mol)×ΔHf(H2,g)}{(1mol)×ΔHf(CH4,g)+(1mol)×ΔHf(H2O,g)}        (1)

Where,

  • ΔHf(CO,g) is the standard enthalpy of formation of CO(g).
  • ΔHf(H2,g) is the standard enthalpy of formation of H2(g).
  • ΔHf(CH4,g) is the standard enthalpy of formation of CH4(g).
  • ΔHf(H2O,g) is the standard enthalpy of formation of H2O(g).

The value of ΔHf(CO,g) is 110.53kJmol1.

The value of ΔHf(H2,g) is 0.0kJmol1.

The value of ΔHf(CH4,g) is 74.81kJmol1.

The value of ΔHf(H2O,g) is 241.82kJmol1.

Substitute the value of ΔHf(CO,g), ΔHf(H2,g), ΔHf(CH4,g) and ΔHf(H2O,g) in equation (1).

    ΔH°={(1mol)×(110.53kJmol1)+(3mol)×(0.0kJmol1)}{(1mol)×(74.81kJmol1)+(1mol)×(241.82kJmol1)}={110.53kJ}{74.81kJ241.82kJ}=+206.1kJ

The relation for the calculation of standard change in the entropy is shown below.

    ΔS°=nΔSf(products)nΔSf(reactants)

The standard change in the entropy of the given reaction is calculated by the expression is shown below.

    ΔS°={(1mol)×ΔSf(CO,g)+(3mol)×ΔSf(H2,g)}{(1mol)×ΔSf(CH4,g)+(1mol)×ΔSf(H2O,g)}        (2)

Where,

  • ΔSf(CO,g) is the standard entropy of formation of CO(g).
  • ΔSf(H2,g) is the standard entropy of formation of H2(g).
  • ΔSf(CH4,g) is the standard entropy of formation of CH4(g).
  • ΔSf(H2O,g) is the standard entropy of formation of H2O(g).

The value of ΔSf(CO,g) is 197.67JK1mol1.

The value of ΔSf(H2,g) is 130.684JK1mol1.

The value of ΔSf(CH4,g) is 186.26JK1mol1.

The value of ΔSf(H2O,g) is 188.83JK1mol1.

Substitute the value of ΔSf(CO,g), ΔSf(H2,g), ΔSf(CH4,g) and ΔSf(H2O,g) in equation (2).

    ΔS°={(1mol)×(197.67JK1mol1)+(3mol)×(130.684JK1mol1)}{(1mol)×(186.26JK1mol1)+(1mol)×(188.83JK1mol1)}={197.67JK1+392.052JK1}{186.26JK1+188.83JK1}=+214.632JK1

The relation for the calculation of standard Gibbs free energy for the given reaction is shown below.

    ΔG°={(1mol)×ΔGf(CO,g)+(3mol)×ΔGf(H2,g)}{(1mol)×ΔGf(CH4,g)+(1mol)×ΔGf(H2O,g)}        (3)

Where,

  • ΔGf(CO,g) is the standard Gibbs free energyentropy of formation of CO(g).
  • ΔGf(H2,g) is the standard Gibbs free energy of formation of H2(g).
  • ΔGf(CH4,g) is the standard Gibbs free energy of formation of CH4(g).
  • ΔGf(H2O,g) is the standard Gibbs free energy of formation of H2O(g).

The value of ΔGf(CO,g) is 137.17kJmol1.

The value of ΔGf(H2,g) is 0.0kJmol1.

The value of ΔGf(CH4,g) is 50.72kJmol1.

The value of ΔGf(H2O,g) is 228.57kJmol1.

Substitute the value of ΔGf(CO,g), ΔGf(H2,g), ΔGf(CH4,g) and ΔGf(H2O,g) in equation (3).

    ΔG°={(1mol)×(137.17kJmol1)+(3mol)×(0.0kJmol1)}{(1mol)×(50.72kJmol1)+(1mol)×(228.57kJmol1)}={137.17kJ}{50.72kJ228.57kJ}=+142.12kJ

Thus, the standard change in the enthalpy, entropy and Gibbs free energy for the given reaction +206.1kJ_, +214.632JK-1_ and +142.12kJ_ respectively.

(b)

Interpretation Introduction

Interpretation:

The standard change in the enthalpy, entropy and Gibbs free energy for the given reaction has to be determined.

Concept Introduction:

Same as part (a).

(b)

Expert Solution
Check Mark

Answer to Problem 4J.8E

The standard change in the enthalpy, entropy and Gibbs free energy for the given reaction +205.79kJ_, +257.6JK-1_ and +59.5kJ_ respectively.

Explanation of Solution

The given chemical equation for the thermal decomposition of ammonium nitrate is shown below.

    NH4NO3(s)N2O(g)+H2O(g)

The standard change in the enthalpy of the given reaction is calculated by the expression is shown below.

    ΔH°={(1mol)×ΔHf(H2O,g)+(1mol)×ΔHf(N2O,g)}{(1mol)×ΔHf(NH4NO3,s)}        (4)

Where,

  • ΔHf(N2O,g) is the standard enthalpy of formation of N2O(g).
  • ΔHf(H2O,g) is the standard enthalpy of formation of H2O(g).
  • ΔHf(NH4NO3,s) is the standard enthalpy of formation of NH4NO3(s).

The value of ΔHf(N2O,g) is 82.05kJmol1.

The value of ΔHf(H2O,g) is 241.82kJmol1.

The value of ΔHf(NH4NO3,s) is 365.56kJmol1.

Substitute the value of ΔHf(N2O,g), ΔHf(H2O,g) and ΔHf(NH4NO3,s) in equation (4).

    ΔH°={(1mol)×(82.05kJmol1)+(1mol)×(241.82kJmol1)}{(1mol)×(365.56kJmol1)}={82.05kJ241.82kJ}{365.56kJ}=+205.79kJ

The standard change in the entropy of the given reaction is calculated by the expression is shown below.

    ΔS°={(1mol)×ΔSf(H2O,g)+(1mol)×ΔSf(N2O,g)}{(1mol)×ΔSf(NH4NO3,s)}        (5)

Where,

  • ΔSf(N2O,g) is the standard entropy of formation of N2O(g).
  • ΔSf(H2O,g) is the standard entropy of formation of H2O(g).
  • ΔSf(NH4NO3,s) is the standard entropy of formation of NH4NO3(s).

The value of ΔSf(N2O,g) is 219.85JK1mol1.

The value of ΔSf(H2O,g) is 188.83JK1mol1.

The value of ΔSf(NH4NO3,s) is 151.08JK1mol1.

Substitute the value of ΔSf(N2O,g), ΔSf(H2O,g) and ΔSf(NH4NO3,s) in equation (5).

    ΔS°={(1mol)×(219.85JK1mol1)+(1mol)×(188.83JK1mol1)}{(1mol)×(151.08JK1mol1)}={219.85JK1+188.83JK1}{151.08JK1}=+257.6JK1

The relation for the calculation of standard Gibbs free energy for the given reaction is shown below.

    ΔG°={(1mol)×ΔGf(H2O,g)+(1mol)×ΔGf(N2O,g)}{(1mol)×ΔGf(NH4NO3,s)}        (6)

Where,

  • ΔGf(N2O,g) is the standard Gibbs free energy of formation of N2O(g).
  • ΔGf(H2O,g) is the standard Gibbs free energy of formation of H2O(g).
  • ΔGf(NH4NO3,s) is the standard Gibbs free energy of formation of NH4NO3(s)

The value of ΔGf(N2O,g) is 104.20kJmol1.

The value of ΔGf(H2O,g) is 228.57kJmol1.

The value of ΔGf(NH4NO3,s) is 183.87kJmol1.

Substitute the value of ΔGf(N2O,g), ΔGf(H2O,g) and ΔGf(NH4NO3,s) in equation (6).

    ΔG°={(1mol)×(104.20kJmol1)+(1mol)×(228.57kJmol1)}{(1mol)×(183.87kJmol1)}={104.20kJ228.57kJ}{183.87kJ}=+59.5kJ

Thus, the standard change in the enthalpy, entropy and Gibbs free energy for the given reaction are +205.79kJ_, +257.6JK-1_ and +59.5kJ_ respectively.

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Chapter 4 Solutions

Chemical Principles: The Quest for Insight

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