Question

(a) From these data: $$ \begin{aligned} \mathrm{F}_{2}(g) \longrightarrow 2 \mathrm{~F}(g) & \Delta H_{\mathrm{rxn}}^{\circ}=156.9 \mathrm{~kJ} / \mathrm{mol} \\ \mathrm{F}^{-}(g) \longrightarrow \mathrm{F}(g)+e^{-} & \Delta H_{\mathrm{rxn}}^{\circ}=333 \mathrm{~kJ} / \mathrm{mol} \\ \mathrm{F}_{2}^{-}(g) \longrightarrow \mathrm{F}_{2}(g)+e^{-} & \Delta H_{\mathrm{rxn}}^{\circ}=290 \mathrm{~kJ} / \mathrm{mol} \end{aligned} $$ calculate the bond enthalpy of the \(\mathrm{F}_{2}^{-}\) ion. (b) Explain the difference between the bond enthalpies of \(\mathrm{F}_{2}\) and \(\mathrm{F}_{2}^{-}\).

Short answer

The bond enthalpy of the \( \mathrm{F}_{2}^{-} \) ion is -35.45 kJ/mol. The bond enthalpy of the neutral \( \mathrm{F}_{2} \) molecule is higher because it does not possess an extra electron, meaning it needs more energy to break.

Step-by-step solution

Title

Determine the enthalpy of formation for one F- ion from half of a F2 molecule which is calculated as \[ \Delta H_{\mathrm{rxn1}} = \Delta H_{\mathrm{rxn2}} - \frac{1}{2}\Delta H_{\mathrm{rxn0}} = 333 \, \mathrm{kJ/mol} - \frac{1}{2}(156.9 \, \mathrm{kJ/mol}) = 254.55 \, \mathrm{kJ/mol} . \]

Title

Next find the bond enthalpy of the \( \mathrm{F}_{2}^{-} \) ion. We can treat the breaking of an \( \mathrm{F}_{2}^{-} \) ion into a F- ion and an F atom as a reaction and calculate its enthalpy change as follows: \[ \Delta H_{\mathrm{rxn3}} = \Delta H_{\mathrm{rxn1}} - \Delta H_{\mathrm{rxn2}} = 254.55 \, \mathrm{kJ/mol} - 290 \, \mathrm{kJ/mol} = -35.45 \, \mathrm{kJ/mol} . \]

Title

Lastly, provide an explanation for the difference in the bond enthalpies of \( \mathrm{F}_{2} \) and \( \mathrm{F}_{2}^{-} \). The bond enthalpy of a neutrally charged \( \mathrm{F}_{2} \) molecule can be seen as the amount of energy needed to break that bond. It is higher because there is no extra electron unlike in \( \mathrm{F}_{2}^{-} \), thus it needs less energy to break. Hence, the negative value of the enthalpy change in \( \mathrm{F}_{2}^{-} \) illustrates that it's more stable than the \( \mathrm{F}_{2} \) molecule.

Key concepts

Enthalpy of Formation

The enthalpy of formation refers to the change in enthalpy when one mole of a substance is formed from its constituent elements in their standard states. It's a key concept in assessing how much energy is involved in creating compounds and is represented by the symbol \( \Delta H_f^\circ \). Consider our example involving fluorine: The provided data allows us to derive the enthalpy of formation for the \( \mathrm{F}^{-} \) ion.

Through the step by step solutions, we've understood that calculating the enthalpy of formation can be nuanced. To ensure better understanding, it’s crucial to grasp that these values are intrinsic to a compound and are part of larger chemical equations. For instance, the formation of a \( \mathrm{F}^{-} \) ion from elemental fluorine involves bond breaking and electron addition, each of which has an associated energy change. By using half the bond enthalpy of \( \mathrm{F}_2 \) and subtracting the electron affinity, we obtained the enthalpy of formation of a single \( \mathrm{F}^{-} \) ion.

Chemical Bonds

Chemical bonds are the glue that holds atoms together in molecules. They are formed when atoms share or transfer electrons, and they can be classified into several types, including ionic, covalent, and metallic bonds. The bond enthalpy, or bond energy, is a measure of the strength of a chemical bond. It is defined as the amount of energy required to break one mole of bonds in a gaseous substance under standard conditions (usually represented as \( \Delta H_{bond} \)).

In the case of the \( \mathrm{F}_2^{-} \) ion, the bond enthalpy calculation provides insight into the stability of the ion compared to its neutral counterpart \( \mathrm{F}_2 \). The exercise demonstrates that the presence of an extra electron influences the bond energy. It's essential for students to comprehend that chemical bonds vary widely in their strengths, and this variability plays a critical role in chemical reactivity, stability, and other properties of substances.

Thermochemical Equations

Thermochemical equations provide a bridge between chemical reactions and the energy changes that accompany them. These balanced equations not only show the reactants converting into products but also the heat gain or loss as \( \Delta H \), known as the enthalpy change. Enthalpy changes can be exothermic (releasing heat) or endothermic (absorbing heat), and they are crucial for predicting reaction behaviors.

Interpreting the given data in terms of thermochemical equations allows us to understand the enthalpy changes of \( \mathrm{F}_2 \) and its ions. By framing the bond dissociation and electron affinities as separate reactions, we can calculate specific enthalpy changes. It is vital to recognize that these enthalpy values in thermochemical equations are not just abstract numbers but quantifications of the energy transformations occurring during chemical reactions, which govern the feasibility and directionality of the reactions.