   # Calculate Δ S °(system), Δ S °(surroundings), and Δ S °(universe) for each of the following processes at 298 K, and comment on how these systems differ. (a) HNO 3 (g) → HNO 3 (aq) (b) NaOH(s) → NaOH(aq) ### Chemistry & Chemical Reactivity

9th Edition
John C. Kotz + 3 others
Publisher: Cengage Learning
ISBN: 9781133949640

#### Solutions

Chapter
Section ### Chemistry & Chemical Reactivity

9th Edition
John C. Kotz + 3 others
Publisher: Cengage Learning
ISBN: 9781133949640
Chapter 18, Problem 40GQ
Textbook Problem
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## Calculate ΔS°(system), ΔS°(surroundings), and ΔS°(universe) for each of the following processes at 298 K, and comment on how these systems differ. (a) HNO3(g) → HNO3(aq) (b) NaOH(s) → NaOH(aq)

(a)

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.

### 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

(b)

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.

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