   Chapter 18, Problem 27PS

Chapter
Section
Textbook Problem

Calculate ΔS°(universe) for the decomposition of 1 mol of liquid water to form gaseous hydrogen and oxygen. Is this reaction spontaneous under standard conditions at 25 °C? Explain your answer briefly.

Interpretation Introduction

Interpretation:

It should be identified that whether the given reaction is spontaneous at 298.15K by calculating  ΔS(system), ΔS(surroundings) and ΔS(universe) also the reactants and products should be defined as system.

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.

ΔS(universe)= ΔS(system)+ΔS(surroundings)

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

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

ΔS(system)=ΔrS°=nS°(products)nS°(reactants)

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

ΔS(surroundings)=ΔrHT

Here, ΔrHo is the enthalpy change for the reaction.

Explanation

Given: Decomposition of 1 mol liquid water to form gaseous hydrogen and oxygen.

The standard entropy of H2O(l) is 69.95 J/Kmol.

The standard entropy of H2(g) is 130.7 J/Kmol.

The standard entropy of O2(g) is 205.07 J/Kmol.

The standard enthalpy of H2O(l) is 285.83 kJ/mol.

The standard enthalpy of H2(g) is 0 kJ/mol.

The standard enthalpy of O2(g) is 0 kJ/mol.

The ΔS(universe) for the reaction H2O(l)H2(g) + 12O2(g) is calculated below.

The balanced chemical equation is:

H2O(l)H2(g) + 12O2(g)

Calculation for ΔSo(system):

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

ΔS(system)=ΔrS°=nS°(products)nS°(reactants)=[(1 mol H2(g)/mol-rxn)S°[H2(g)]+(0.5 mol O2(g)/mol-rxn)S°[O2(g)](1 mol H2O(l)/mol-rxn)S°[H2O(l)]]

Substitute the values,

ΔS(system)=[(1 mol H2(g)/mol-rxn)(130.7 J/Kmol)+(0.5 mol O2(g)/mol-rxn)(205.07 J/Kmol)(1 mol H2O(l)/mol-rxn)(69.95 J/Kmol)]=163

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