The enthalpy change for the following reaction is 368 kJ. Calculate the average O-F bond energy.
`OF_(2)(g)rarrO(g)+2F(g)`

Options

A. 184 kJ/mol
B. 368 kJ/mol
C. 536 kJ/mol
D. 736 kJ/mol

Explanation

To calculate the average O-F bond energy from the given reaction, we can follow these steps:

### Step 1: Understand the Reaction
The reaction given is:
\[ \text{OF}_2(g) \rightarrow \text{O}(g) + 2\text{F}(g) \]
This indicates that one molecule of dioxygen difluoride (OF₂) is decomposing into one atom of oxygen and two atoms of fluorine.

### Step 2: Identify Bonds Broken
In the reaction, we are breaking two O-F bonds since there are two fluorine atoms in the OF₂ molecule.

### Step 3: Relate Enthalpy Change to Bond Energies
The enthalpy change (ΔH) for this reaction is given as 368 kJ. This value represents the energy required to break the bonds in the OF₂ molecule. Since two O-F bonds are being broken, we can express this as:
\[ \Delta H = 2 \times \text{(Average O-F bond energy)} \]

### Step 4: Set Up the Equation
Let \( x \) be the average O-F bond energy. Then we can write:
\[ 368 \text{ kJ} = 2x \]

### Step 5: Solve for x
To find the average O-F bond energy, we rearrange the equation:
\[ x = \frac{368 \text{ kJ}}{2} \]
\[ x = 184 \text{ kJ/mol} \]

### Conclusion
The average O-F bond energy is:
\[ \text{Average O-F bond energy} = 184 \text{ kJ/mol} \]

---