   Chapter 8, Problem 51PS

Chapter
Section
Textbook Problem

Ethanol can be made by the reaction of ethylene and water:H2C=CH2(g) + H2O(g) → CH3CH2OH(g)Use bond dissociation enthalpies to estimate the enthalpy change in this reaction. Compare the value obtained to the value calculated from 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:

H2C=CH2(g)+H2O(g)CH3CH2OH(g)

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

C=Cis610kJ/mol

C-Cis346kJ/mol

C-Ois358kJ/mol

O-His463kJ/mol.

Examining the reaction shows that there are 1C=C and2O-H broken in reactant side and the bond formed in product sides are 1C-C ,1C-H,1O-Hand1C-O

1mol(610kJ/mol)+2mol(463kJ/mol)1mol(346kJ/mol)+1mol(413kJ/mol)+1mol(463kJ/mol)+1mol(358kJ/mol)1536 kJ1580kJ

ΔrH=ΔH(bondsbroken)

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