Step 1: half reaction 1: H2 > 2H^+ + 2e^-

Question

half reaction 2: F2+ 2e^- > 2F^-

When H^+ and F^- sit in solution together, they form HF according to the reaction below

step 2: H^+ F^- > HF

The Bond energies for several of the reaction components are below

Bond: Bond dissociation energy (kJ)

H-F. 436

F-F. 158

H-F. 565

F=F. 45

Answer 1

Let's analyze the information you've provided.

You have two half-reactions:

1. \( \text{H}_2 \rightarrow 2\text{H}^+ + 2\text{e}^- \)
2. \( \text{F}_2 + 2\text{e}^- \rightarrow 2\text{F}^- \)

When hydrogen ions (\( \text{H}^+ \)) and flouride ions (\( \text{F}^- \)) are in solution together, they combine to form hydrogen fluoride (HF):

\[ \text{H}^+ + \text{F}^- \rightarrow \text{HF} \]

You've provided bond dissociation energies for some bonds here:

• \( \text{H-F} \): 436 kJ/mol (it looks like there are two values for \( \text{H-F} \) which could indicate a mistake; typically, we'd expect just one average bond dissociation energy).
• \( \text{F-F} \): 158 kJ/mol
• \( \text{F=F} \): 45 kJ/mol

To evaluate the overall reaction and its energetics, let's apply the bond dissociation method.

Steps to Calculate the Reaction Energy

Step 1: Bond Breakages and Formations in Reactions

When looking at the formation of HF from its ions (\( \text{H}^+ \) and \( \text{F}^- \)), one bond is formed. There are no bonds broken in this step, since \( \text{H}^+ \) does not have any bonds.

Step 2: Energy Considerations

Since the energy required to form a bond is negative (indicating exothermic reaction) and the energy associated with bond breaking is positive (indicating endothermic reaction), we can evaluate the process:

• Energy released when forming one \( \text{H-F} \) bond = -436 kJ/mol

According to Hess's law, the overall energy change for the formation of HF will be equal to the energy of bonds broken minus the energy of bonds formed.

Overall Energy Change Calculation

Since \( \text{H}^+ \) and \( \text{F}^- \) do not have bonds in them, there's no energy associated with their form (the energies given above for H-F will only apply when we consider them).

Thus, the reaction can be summarized as: \[ \Delta H = \text{Bonds broken (none)} - \text{Bonds formed (-436 kJ/mol)} \] \[ \Delta H = 0 - (-436 , \text{kJ/mol}) \] \[ \Delta H = -436 , \text{kJ/mol} \]

Conclusion

The formation of hydrogen fluoride from hydrogen and fluoride ions is an exothermic reaction with an enthalpy change of approximately -436 kJ/mol, indicating the process releases energy. If you have any further questions or need clarification on any part of this process, feel free to ask!