Question

The following bond energies are given for \(298 \mathrm{K}: \mathrm{O}_{2}\) \(498 ; \mathrm{N}_{2}, 946 ; \mathrm{F}_{2}, 159 ; \mathrm{Cl}_{2}, 243 ;\) ClF, \(251 ; \mathrm{OF}\left(\text { in } \mathrm{OF}_{2}\right)\) \(213 ; \mathrm{ClO}\left(\operatorname{in} \mathrm{Cl}_{2} \mathrm{O}\right), 205 ;\) and \(\mathrm{NF}\left(\mathrm{in} \mathrm{NF}_{3}\right), 280 \mathrm{kJmol}^{-14}\) Calculate \(\Delta H_{f}\) at \(298 \mathrm{K}\) for \(1 \mathrm{mol}\) of \((\mathrm{a}) \mathrm{ClF}(\mathrm{g})\) (b) \(\mathrm{OF}_{2}(\mathrm{g}) ;(\mathrm{c}) \mathrm{Cl}_{2} \mathrm{O}(\mathrm{g}) ;(\mathrm{d}) \mathrm{NF}_{3}(\mathrm{g})\).

Short answer

The formation enthalpies of ClF(g), OF2(g), Cl2O(g) and NF3(g) at 298K are -151 kJmol-1, -231 kJmol-1, -331 kJmol-1 and -283 kJmol-1, respectively.

Step-by-step solution

Analyze the Provided Data

Evaluate the data given. The provided energies are bond dissociation energies, in kJ/mol which represent the energy required to break a specific type of bond in a molecule. The substances we have to find the formation enthalpies for are ClF(g), OF2(g), Cl2O(g), NF3(g). The equation for finding the enthalpy change is: \( \Delta H_f = \sum E_{\text{end bond}} - \sum E_{\text{start bond}} \).

Calculate \(\Delta H_f\) for ClF(g)

To form ClF(g), a Cl2 molecule and a F2 molecule should be broken first to get 2 Cl atoms and 2 F atoms, then 1 Cl atom combines with 1 F atom, while other atoms remain unchanged. So we use the formula: \( \Delta H_f = E_{\text{Cl - F}} - (E_{\text{Cl - Cl}} + E_{\text{F - F}}) = 251 - (243 + 159) = -151 \, \text{kJmol}^{-1} \).

Calculate \(\Delta H_f\) for OF2(g)

For the formation of OF2(g), an O2 molecule and an F2 molecule should be broken first to get 2 O atoms and 2 F atoms, then 1 O atom combines with 2 F atoms to form OF2(g), and the other O atom remains unchanged. \( \Delta H_f = 2 \times E_{\text{O - F}} - (E_{\text{O - O}} + E_{\text{F - F}}) = 2 \times 213 - (498 + 159) = -231 \, \text{kJmol}^{-1} \).

Calculate \(\Delta H_f\) for Cl2O(g)

For the formation of Cl2O(g), a molecule each of Cl2 and O2 should be broken first to get 2 Cl atoms and 2 O atoms, then 2 Cl atoms combine with 1 O atom, and the other O atom remains unchanged. \( \Delta H_f = 2 \times E_{\text{Cl - O}} - (E_{\text{Cl - Cl}} + E_{\text{O - O}}) = 2 \times 205 - (243 + 498) = -331 \, \text{kJmol}^{-1}\).

Calculate \(\Delta H_f\) for NF3(g)

To form NF3(g), an N2 molecule and three F2 molecules are broken first to get 2 N atoms and 6 F atoms, then 1 N atom combines with 3 F atoms to form NF3(g), and the other atoms remain unchanged. \( \Delta H_f = 3 \times E_{\text{N - F}} - (E_{\text{N - N}} + 3 \times E_{\text{F - F}}) = 3 \times 280 - (946 + 3 \times 159) = -283 \, \text{kJmol}^{-1}\).

Key concepts

Bond Energy

Bond energy refers to the amount of energy required to break one mole of a chemical bond in a molecule and separate the atoms into individual gaseous atoms.

This is a critical concept in thermochemistry because it helps us understand how much energy is involved in both breaking chemical bonds during the reactant phase and forming new bonds during the product phase of a chemical reaction.

For instance, when we look at the bond energies provided in the exercise, they indicate how much energy would be needed to break down one mole of a diatomic molecule, like \(\mathrm{O}_{2}\), into individual oxygen atoms in the gas phase. We use these bond energies to calculate the overall enthalpy change in a reaction.

Thermochemistry

Thermochemistry deals with the heat involved in chemical processes. Understanding how heat transfer is associated with chemical reactions is fundamental for predicting reaction behavior and determining stability.

Enthalpy, denoted by \( H \), is the measure of the total heat content of a system at constant pressure. The enthalpy of formation, or \( \Delta H_{f} \), is a specific kind of enthalpy change that measures the heat change when one mole of a substance is formed from its elements in their standard states.

The challenges faced in this exercise stem from thermochemistry concepts, as we're asked to calculate the enthalpy of formation for various compounds using the bond energies given, reflecting the concept's central role in understanding chemical processes.

Chemical Bonds

Chemical bonds are the attractive forces that hold atoms together in compounds. They result from the sharing or transfer of electrons between atoms. The primary types of chemical bonds are ionic, covalent, and metallic.

In the context of our exercise, covalent bonds are the focus. Covalent bonding involves the sharing of electron pairs between atoms. The bonds within \(\mathrm{O}_{2}\), \(\mathrm{N}_{2}\), and other molecules listed are examples of covalent bonds.

When forming a compound like \(\mathrm{ClF}(g)\), a covalent bond is formed between chlorine and fluorine atoms, which requires energy changes that are crucial for calculating the enthalpy of formation.

Enthalpy Change Calculation

The calculation of enthalpy change, or \( \Delta H \), in a chemical reaction involves the difference between the sum of bond energies for the bonds broken in reactants and the sum for the bonds formed in products.

An example from the exercise is the calculation of \( \Delta H_{f} \) for \(\mathrm{ClF}(g)\). We first determine the energy needed to break the bonds in the reactants (\(\mathrm{Cl}_{2}\) and \(\mathrm{F}_{2}\)) and then subtract the energy released when the new \(\mathrm{Cl-F}\) bond forms.

To summarize, the formula \( \Delta H_{f} = \sum E_{\text{end bond}} - \sum E_{\text{start bond}} \) represents this relationship. It underpins the step-by-step solution provided for the exercise, demonstrating how enthalpy of formation is inherently related to bond energies.