Question

Use bond energies (Table 8.4) to show that the preferred products for the decomposition of \(\mathrm{N}_{2} \mathrm{O}_{3}\) are \(\mathrm{NO}_{2}\) and \(\mathrm{NO}\) rather than \(\mathrm{O}_{2}\) and \(\mathrm{N}_{2} \mathrm{O}\). (The \(\mathrm{N}-\mathrm{O}\) single bond energy is \(201 \mathrm{~kJ} / \mathrm{mol} .\) ) Hint: Consider the reaction kinetics.

Short answer

The energy changes for the two possible decomposition reactions of \(\mathrm{N}_{2}\mathrm{O}_{3}\) are calculated as: 1. \(\mathrm{N}_{2}\mathrm{O}_{3} \to \mathrm{NO}_{2} + \mathrm{NO}\): energy change = \(-406\ \mathrm{kJ/mol}\) 2. \(\mathrm{N}_{2}\mathrm{O}_{3} \to \mathrm{O}_{2} + \mathrm{N}_{2}\mathrm{O}\): energy change = \(310\ \mathrm{kJ/mol}\) Since the first reaction has a lower energy change, the preferred decomposition products are \(\mathrm{NO}_{2}\) and \(\mathrm{NO}\).

Step-by-step solution

Gather bond energies from Table 8.4 and given information

We know the \(\mathrm{N}-\mathrm{O}\) single bond energy is \(201 \mathrm{~kJ}/\mathrm{mol}\). According to Table 8.4, we have the following bond energies: \(\mathrm{N-N}\) (Triple bond): \(941\ \mathrm{kJ/mol}\) \(\mathrm{N-O}\) (Double bond): \(607\ \mathrm{kJ/mol}\) \(\mathrm{O=O}\) bond: \(498\ \mathrm{kJ/mol}\)

Calculate the energy changes for the first reaction

For the reaction \(\mathrm{N}_{2}\mathrm{O}_{3} \to \mathrm{NO}_{2} + \mathrm{NO}\), we must break one \(\mathrm{N-O}\) single bond and one \(\mathrm{N-O}\) double bond, and then form two \(\mathrm{N-O}\) double bonds. Breaking bonds: 1 \(\times\) \(\mathrm{N-O}\) single bond = \(1 \times 201 = 201\ \mathrm{kJ/mol}\) 1 \(\times\) \(\mathrm{N-O}\) double bond = \(1 \times 607 = 607\ \mathrm{kJ/mol}\) Forming bonds: 2 \(\times\) \(\mathrm{N-O}\) double bond = \(2 \times 607 = 1214\ \mathrm{kJ/mol}\) Energy change for the first reaction = (bond energies of broken bonds) - (bond energies of formed bonds) = \((201 + 607) - 1214 = -406\ \mathrm{kJ/mol}\)

Calculate the energy changes for the second reaction

For the reaction \(\mathrm{N}_{2}\mathrm{O}_{3} \to \mathrm{O}_{2} + \mathrm{N}_{2}\mathrm{O}\), we must break one \(\mathrm{N-O}\) single bond and one \(\mathrm{N-O}\) double bond, and then form one \(\mathrm{N-N}\) triple bond and one \(\mathrm{O=O}\) double bond. Breaking bonds: 1 \(\times\) \(\mathrm{N-O}\) single bond = \(1 \times 201 = 201\ \mathrm{kJ/mol}\) 1 \(\times\) \(\mathrm{N-O}\) double bond = \(1 \times 607 = 607\ \mathrm{kJ/mol}\) Forming bonds: 1 \(\times\) \(\mathrm{N-N}\) triple bond = \(1 \times 941 = 941\ \mathrm{kJ/mol}\) 1 \(\times\) \(\mathrm{O=O}\) double bond = \(1 \times 498 = 498\ \mathrm{kJ/mol}\) Energy change for the second reaction = (bond energies of broken bonds) - (bond energies of formed bonds) = \((201 + 607) - (941 + 498) = 310\ \mathrm{kJ/mol}\)

Compare the energy changes and determine the preferred reaction

Now, let's compare the energy changes for both reactions: Energy change for the first reaction = \(-406\ \mathrm{kJ/mol}\) Energy change for the second reaction = \(310\ \mathrm{kJ/mol}\) Since the energy change for the first reaction is lower than the energy change for the second reaction, the decomposition of \(\mathrm{N}_{2}\mathrm{O}_{3}\) into \(\mathrm{NO}_{2}\) and \(\mathrm{NO}\) is the preferred reaction.

Key concepts

Chemical Bonds

Understanding chemical bonds is central to comprehending how molecules interact and react with each other. A chemical bond is an attraction between atoms that allows the formation of chemical substances containing two or more atoms. The bond energy is a measure of the strength of a chemical bond and is defined as the energy required to break one mole of the bond in a gaseous state.

Bonds can be classified into three main types: ionic, covalent, and metallic. In covalent bonds, which are of interest in the given exercise, atoms share pairs of electrons. There are also different types of covalent bonds, like single, double, and triple bonds, which involve the sharing of one, two, and three pairs of electrons, respectively. The more pairs of electrons shared, the stronger the bond tends to be, and thus the higher the bond energy.

Reaction Kinetics

Reaction kinetics studies the rates of chemical reactions and the factors affecting them. It's crucial for predicting how fast a reaction will proceed under certain conditions. Several factors can influence reaction rates, including concentration, temperature, and the presence of a catalyst.

In relation to the textbook exercise, reaction kinetics can also involve the energies required to break and form bonds during a reaction. Reactions typically occur in steps, each with its own transition state and energy profile. Understanding the kinetic requirements for a reaction can help explain why certain reaction pathways are preferred over others when multiple possibilities exist.

Enthalpy Change

Enthalpy change, denoted as \(\Delta H\), is a measure of the total heat content in a thermodynamic system, and its change is an indicator of the heat absorbed or released during a reaction at constant pressure. It's an essential concept in thermochemistry because it helps predict whether a reaction is endothermic (absorbs heat) or exothermic (releases heat).

In the context of the given exercise, calculating the enthalpy change involves subtracting the bond energies of the products from the reactants. A negative \(\Delta H\) value indicates that the reaction releases energy to the surroundings, often signifying a more spontaneous and preferred reaction pathway, as seen in the decomposition of \(\mathrm{N}_{2}\mathrm{O}_{3}\) into \(\mathrm{NO}_{2}\) and \(\mathrm{NO}\).

Thermochemistry

Thermochemistry is the branch of chemistry that deals with the relationship between chemical reactions and energy changes involving heat. It's based on the first law of thermodynamics which states that energy cannot be created or destroyed, only transformed.

This concept applies to bond energies as shown in our textbook exercise, where the total energy of a system is conserved when bonds are broken and formed. Thermochemical calculations allow us to understand the flow of energy within a reaction and determine whether it will occur spontaneously. Reactions with a negative enthalpy change, such as the preferred decomposition of \(\mathrm{N}_{2}\mathrm{O}_{3}\), are usually favorable because they release energy, which is a driving force for the reaction to happen.