   Chapter 8.10, Problem 3.2ACP

Chapter
Section
Textbook Problem

Predict the bond dissociation enthalpy for a nitrogen-iodine bond in Nl3 using bond dissociation enthalpy values (Table 8.8) and electronegativity values (Figure 8.11).

Interpretation Introduction

Interpretation: The bond dissociation enthalpy for nitrogen and iodine in NI3 should be calculated using average bond dissociation enthalpy and electronegativity values.

Concept Introduction:

Electronegativity: It is defined as the capacity of the atom to abstract the pair of electrons towards itself results to have high negative charge.

Polar molecule: The molecule consists of atoms bonded with different electronegativity. Dipole moment is used to measure the polarity of the molecule.

Polarity of a molecule is measured in term of dipole moment.

Dipole moment for a polar molecule is non-zero and for a non-polar molecule dipole moment is zero.

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. The process of breaking bonds in a molecule is always endothermic, so ΔrH for bond breaking is always positive.

Explanation

The reaction enthalpy formula is ΔrH=ΔH(bondsbroken)ΔH(bondsformed)

Linus found that bond dissociation enthalpies for different atoms are larger than the average of the bond dissociation enthalpy between identical atoms.

This difference is mathematically in relation with the electronegativity’s difference by the following equation.

XA-XB=0.102[ΔdissH(AB)-ΔdissH(AA)dissH(BB)2]1/2

The electronegativity difference between hydrogen and chlorine in the hydrogen chloride using average bond dissociation enthalpies is as follows,

XN-XI=0.102[ΔdissH(NI)-ΔdissH(NN)dissH(II)2]1/2

The electronegativity difference between nitrogen and iodine in NI3 using electronegativity value is as follows,

Electronegativity of nitrogen is 3 and for iodine is 2

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