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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Chapter 18, Problem 40GQ

Calculate ΔS°(system), ΔS°(surroundings), and ΔS°(universe) for each of the following processes at 298 K, and comment on how these systems differ.

  1. (a) HNO3(g) → HNO3(aq)
  2. (b) NaOH(s) → NaOH(aq)

(a)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation:

The entropy change for the system, surroundings and universe for the given reaction should be calculated and commented how this system differs.

Concept introduction:

The universe consists of two parts, systems and surroundings. The entropy change for the universe is the sum of entropy change for the system and for surroundings.

  ΔSo(universe)= ΔSo(system)+ΔSo(surroundings)

The ΔSo(universe) should be greater than zero for a spontaneous process.

The  ΔSo(system) can be calculated by the following expression,

  ΔSo(system)rS°nS°(products)-nS°(reactants)

The ΔSo(surroundings) can be calculated by the following expression,

  ΔSo(surroundings)=rHoT

Here, ΔrHo is the enthalpy change for the reaction.

Answer to Problem 40GQ

The  ΔSo(system) for the formation of HNO3(aq) is 119.98 J/Kmol-rxn.

The ΔSo(surroundings) for the formation of HNO3(aq) is +242.49 J/Kmol-rxn.

The ΔSo(universe) for the formation of HNO3(aq) is +122.51 J/Kmol-rxn.

Explanation of Solution

The entropy change for the system, surroundings and universe for the given reaction is calculated below.

Given:

The Appendix L referred for the values of standard entropies and enthalpies.

The standard entropy of HNO3(g) is 266.38 J/Kmol.

The standard entropy of HNO3(aq) is 146.4 J/Kmol.

The standard enthalpy of HNO3(g) is 135.06 kJ/mol.

The standard enthalpy of HNO3(aq) is 207.36 kJ/mol.

The balanced chemical equation is:

  HNO3(g)HNO3(aq)

The  ΔSo(system) can be calculated by the following expression,

ΔSo(system)rS°nS°(products)-nS°(reactants)=[(1 mol HNO3(aq)/mol-rxn)S°[HNO3(aq)]-(1 mol HNO3(g)/mol-rxn)S°[HNO3(g)]]

Substituting the respective values

  ΔSo(system)=[(1 mol HNO3(aq)/mol-rxn)(146.4 J/K×mol)-(1 mol HNO3(g)/mol-rxn)(266.38 J/K×mol)]=-119.98 J/K×mol-rxn

The ΔrHo can be calculated by the following expression,

ΔrH°=fH°(products)fH°(reactants)=[(1 mol HNO3(aq)/mol-rxn)ΔfH°[HNO3(aq)]-(1 mol HNO3(g)/mol-rxn)ΔfH°[HNO3(g)]]

Substituting the respective values

ΔrH°=[(1 mol HNO3(aq)/mol-rxn)(-207.36 kJ/mol)-(1 mol HNO3(g)/mol-rxn)(-135.06 kJ/mol)]=-72.3 kJ/mol-rxn

The ΔSo(surroundings) can be calculated by the following expression,

  ΔSo(surroundings)rHoT= -[-72.3 kJ/mol-rxn298.15 K](1000JkJ)= +242.49 J/K×mol-rxn

Now,

  ΔSo(universe)= ΔSo(system)+ΔSo(surroundings)=(-119.98 J/K×mol-rxn)+(+242.49 J/K×mol-rxn)=+122.51 J/K×mol-rxn

The  ΔSo(system) for the formation of HNO3(aq) is 119.98 J/Kmol-rxn.

The ΔSo(surroundings) for the formation of HNO3(aq) is +242.49 J/Kmol-rxn.

The ΔSo(universe) for the formation of HNO3(aq) is +122.51 J/Kmol-rxn.

The given reaction is exothermic. There is more decrease in entropy in formation of HNO3(aq) from HNO3(g) due to decrease in number of moles of gases.

(b)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation:

The entropy change for the system, surroundings and universe for the given reaction should be calculated and commented how this system differs.

Concept introduction:

The universe consists of two parts, systems and surroundings. The entropy change for the universe is the sum of entropy change for the system and for surroundings.

  ΔSo(universe)= ΔSo(system)+ΔSo(surroundings)

The ΔSo(universe) should be greater than zero for a spontaneous process.

The  ΔSo(system) can be calculated by the following expression,

  ΔSo(system)rS°nS°(products)-nS°(reactants)

The ΔSo(surroundings) can be calculated by the following expression,

  ΔSo(surroundings)=rHoT

Here, ΔrHo is the enthalpy change for the reaction.

Answer to Problem 40GQ

The  ΔSo(system) for the formation of NaOH(aq) is 16.36 J/Kmol-rxn.

The ΔSo(surroundings) for the formation of NaOH(aq) is +144.96 J/Kmol-rxn.

The ΔSo(universe) for the formation of NaOH(aq) is +128.6 J/Kmol-rxn.

Explanation of Solution

The entropy change for the system, surroundings and universe for the given reaction is calculated below.

Given:

The Appendix L referred for the values of standard entropies and enthalpies.

The standard entropy of NaOH(s) is 64.46 J/Kmol.

The standard entropy of NaOH(aq) is 48.1 J/Kmol.

The standard enthalpy of NaOH(s) is 425.93 kJ/mol.

The standard enthalpy of NaOH(aq) is 469.15 kJ/mol.

The balanced chemical equation is:

  NaOH(s)NaOH(aq)

The  ΔSo(system) can be calculated by the following expression,

  ΔS(system)=ΔrS°nS°(products)-nS°(reactants)=[(1 mol NaOH(aq)/mol-rxn)S°[NaOH(aq)](1 mol NaOH(s)/mol-rxn)S°[NaOH(s)]]

Substituting the respective values

ΔSo(system)=[(1 mol NaOH(aq)/mol-rxn)(48.1 J/K×mol)-(1 mol NaOH(s)/mol-rxn)(64.46 J/K×mol)]=-16.36 J/K×mol-rxn

The ΔrHo can be calculated by the following expression,

ΔrH°=fH°(products)fH°(reactants)=[(1 mol NaOH(aq)/mol-rxn)ΔfH°[NaOH(aq)](1 mol NaOH(s)/mol-rxn)ΔfH°[NaOH(s)]]

Substituting the respective values

  ΔrH°=[(1 mol NaOH(aq)/mol-rxn)(-469.15 kJ/mol)-(1 mol NaOH(s)/mol-rxn)(-425.93 kJ/mol)]= -43.22 kJ/mol-rxn

The ΔSo(surroundings) can be calculated by the following expression,

  ΔSo(surroundings)=rHoT=-[-43.22 kJ/mol-rxn298.15 K](1000JkJ)=+144.96 J/K×mol-rxn

Now,

ΔSo(universe)= ΔSo(system)+ΔSo(surroundings)(-16.36 J/K×mol-rxn)+(+144.96 J/K×mol-rxn)= +128.6 J/K×mol-rxn

The  ΔSo(system) for the formation of NaOH(aq) is 16.36 J/Kmol-rxn.

The ΔSo(surroundings) for the formation of NaOH(aq) is +144.96 J/Kmol-rxn.

The ΔSo(universe) for the formation of NaOH(aq) is +128.6 J/Kmol-rxn.

The given reactions is exothermic. The entropy change for the formation of NaOH(aq) from NaOH(s) should be positive. However, the value is negative due to hydrogen bonding.

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Chapter 18 Solutions

Chemistry & Chemical Reactivity

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