The enthalpy of formation of `UF(g)` is `22kcal mol^(-1)` and that of `U(g)` is `128kcal mol^(-1)`.
The bond energy of the `F-F` bond is `37kcal mol^(-1)` . The bond dissociation energy of `UF(g)` is `(are):`

Options

A. `124.5 kcal mol^(-1)`
B. `131.1 kcal mol^(-1)`
C. `521 kJ mol^(-1)`
D. `623 kJ mol^(-1)`

Explanation

To solve the problem, we need to determine the bond dissociation energy of UF(g) using the given enthalpy values and bond energy. Let's break this down step by step.

### Step 1: Write the reaction for the formation of UF(g)
The formation of UF(g) from U(g) and F2(g) can be represented as:
\[ U(g) + \frac{1}{2} F_2(g) \rightarrow UF(g) \]

### Step 2: Identify the given data
- Enthalpy of formation of UF(g), \( \Delta H_f (UF) = 22 \, \text{kcal/mol} \)
- Enthalpy of formation of U(g), \( \Delta H_f (U) = 128 \, \text{kcal/mol} \)
- Bond energy of F-F bond, \( BE(F-F) = 37 \, \text{kcal/mol} \)

### Step 3: Write the equation for the enthalpy of formation
The enthalpy of formation can be expressed in terms of bond energies:
\[ \Delta H_f = BE(\text{reactants}) - BE(\text{products}) \]

For our reaction:
\[ \Delta H_f (UF) = BE(U) + \frac{1}{2} BE(F_2) - BE(UF) \]

### Step 4: Substitute the known values
We know that:
- \( BE(U) = 128 \, \text{kcal/mol} \)
- \( BE(F_2) = 37 \, \text{kcal/mol} \), but since we have \(\frac{1}{2} F_2\), we take half of that:
\[ \frac{1}{2} BE(F_2) = \frac{37}{2} = 18.5 \, \text{kcal/mol} \]

Now substituting these values into the equation:
\[ 22 = 128 + 18.5 - BE(UF) \]

### Step 5: Solve for BE(UF)
Rearranging the equation to solve for \( BE(UF) \):
\[ BE(UF) = 128 + 18.5 - 22 \]
\[ BE(UF) = 146.5 - 22 \]
\[ BE(UF) = 124.5 \, \text{kcal/mol} \]

### Conclusion
The bond dissociation energy of UF(g) is \( 124.5 \, \text{kcal/mol} \).

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