   Chapter 8, Problem 52PS

Chapter
Section
Textbook Problem

Methanol can be made by partial oxidation of methane using O2 in the presence of a catalyst:2 CH4(g) + O2(g) → 2 CH3OH(ℓ)Use bond dissociation enthalpies to estimate the enthalpy change for this reaction. Compare the value obtained to the value calculated using standard enthalpies of formation.

Interpretation Introduction

Interpretation: The enthalpy change for the given reaction has to be determined and has to be compared with the calculated value of enthalpies of formation.

Concept Introduction:

Bond dissociation enthalpy:

Bond energy or more correctly the bond dissociation enthalpy is the enthalpy change when breaking a bond in a molecule with the reactant and products in the gas phase.

ΔrH=ΔH(bondsbroken)ΔH(bondsformed)

Enthalpy of formation:

ΔrH=ΔfH0(products)ΔfH0(reactants)

Explanation

Given:

2CH4(g)+O2(g)2CH3OH(l)

Using the bond dissociation enthalpy, we can estimate the enthalpy change for this reaction.

Bond dissociation enthalpies for different bonds are given below:

C-His413kJ/mol

O=Ois498kJ/mol

C-Ois358kJ/mol

O-His463kJ/mol.

Examining the reaction shows that there are 2CH and1O=O broken in the reactant side and the bond formed in product sides are 2C-O and2O-H

2mol(413kJ/mol)+1mol(498kJ/mol)[2mol(358kJ/mol)+2mol(463kJ/mol)]1324 kJ1642kJ

ΔrH=ΔH(bondsbroken)ΔH(bondsformed)=1324 - 1642=-318kJ

Enthalpy change for the reaction is

ΔrH0=-318kJ

Now the value has to be compared with the calculated value from enthalpies of formation

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