   Chapter 8, Problem 96SCQ

Chapter
Section
Textbook Problem

The answer in Study Question 95 does not exactly match the value for the bond dissociation enthalpy of P—Cl in Table 8.8. The reason is that the numbers in the table are averages, derived from data on a number of different compounds. Values calculated from different compounds do vary, sometimes widely. To illustrate this, let’s do another calculation. Next look at the following reaction:PCl3(g) + Cl2(g) → PCl5(g)The enthalpy of formation of PCl5(g) is −374.9 kJ/mol. Use this and Δ f H ∘ of PCl3(g) to determine the enthalpy change for this reaction. Then, use that value of Δ r H ∘ , along with the bond dissociation enthalpy of Cl2, to calculate the enthalpy change for the formation of a P—Cl bond in this reaction.

Interpretation Introduction

Interpretation:

The enthalpy change for the given reaction has to be determined and enthalpy change for the formation of PCl bond in the reaction also has to be determined.

Concept Introduction:

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:

PCl3(g)+Cl2(g)PCl5(g)

Enthalpy of formations values are given below:

PCl5ΔfHο=374.9kJ/molPCl3ΔfHο=287.0kJ/mol

ΔrH=ΔfH0(products)ΔfH0(reactants)

ΔrH=(374.9kJ/mol)(287.0kJ/mol)=87.9kJ/mol

Thus, the enthalpy of formation of the reaction is calculated as 87.9kJ/mol.

The enthalpy change for the formation of a PCl bond can be calculated using bond dissociation enthalpy of chlorine molecule and the calculated ΔrH

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