Question

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

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!