# Methanol can be made by partial oxidation of methane by O 2 (g). CH 4 (Mg) + ½O 2 (g) ⇄ CH 3 OH( ℓ ) (a) Determine Δ S °(system), Δ S °(surroundings), and Δ S °(universe) for this process. (b) Is this reaction product-favored at equilibrium at 25 °C?

### Chemistry & Chemical Reactivity

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

Chapter
Section

### Chemistry & Chemical Reactivity

9th Edition
John C. Kotz + 3 others
Publisher: Cengage Learning
ISBN: 9781133949640
Chapter 18, Problem 56GQ
Textbook Problem
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## Methanol can be made by partial oxidation of methane by O2(g).CH4(Mg) + ½O2(g) ⇄ CH3OH(ℓ) (a) Determine ΔS°(system), ΔS°(surroundings), and ΔS°(universe) for this process. (b) Is this reaction product-favored at equilibrium at 25 °C?

(a)

Interpretation Introduction

Interpretation:

The value of ΔSo(universe), ΔSo(system) and ΔSo(surroundings) for formation of methanol should be determined.

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 ΔSo(universe) for the formation of methanol is calculated below.

Given:

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

The standard entropy of CH4(g) is 186.26 J/Kmol.

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

The standard entropy of CH3OH(l) is 127.19 J/Kmol.

The standard enthalpy of CH4(g) is 74.87 kJ/mol.

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

The standard enthalpy of CH3OH(l) is 238.4 kJ/mol.

The balanced chemical equation is:

CH4(g) + 12O2(g)CH3OH(l)

Calculation of ΔSo(system):

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

ΔS(system)=ΔrS°nS°(products)-nS°(reactants)=[(1 mol CH3OH(l)/mol-rxn)S°[CH3OH(l)]-[(1 mol CH4(g)/mol-rxn)S°[CH4(g)]+(0.5 mol O2(g)/mol-rxn)S°[O2(g)]]]

Substituting the respective values

ΔSo(system)=[(1 mol CH3OH(l)/mol-rxn)(127.19 J/K×mol)-[(1 mol CH4(g)/mol-rxn)(186.26 J/K×mol)+(0.5 mol O2(g)/mol-rxn)(205.07 J/K×mol)]]=-161

(b)

Interpretation Introduction

Interpretation:

It should be identified that the given reaction is product favoured at equilibrium or not at 25oC.

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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