What is the bond enthalpy of `Xe-F` bond ?
`XeF_(4)(g)rarrXe^(+)(g)+F^(-)(g)+F_(2)(g)+F(g)," "Delta_(r)H=292"kcal/mol"`
Given : Ionization energy of `Xe=279"kcal/mol"`
`B.E.(F-F) = 30"kcal/mol"`, Electron affinity of F = 85 kcal/mol

Options

A. 24 kcal/mol
B. 34 kcal/mol
C. `8.5` kcal/mol
D. None of these

Explanation

To find the bond enthalpy of the `Xe-F` bond in the reaction given, we will use Hess's law and the provided thermodynamic data. Let's break down the steps:

### Step 1: Write the reaction and identify the components
The reaction given is:
\[ \text{XeF}_4(g) \rightarrow \text{Xe}^+(g) + \text{F}^-(g) + \text{F}_2(g) + \text{F}(g) \]
The enthalpy change for this reaction is given as \( \Delta_rH = 292 \, \text{kcal/mol} \).

### Step 2: Identify the energies involved
1. **Ionization energy of Xe**: The energy required to remove an electron from Xe to form `Xe^+` is given as \( 279 \, \text{kcal/mol} \).
2. **Electron affinity of F**: The energy released when F gains an electron to form `F^-` is given as \( 85 \, \text{kcal/mol} \). However, since this is energy released, we will consider it as negative: \( -85 \, \text{kcal/mol} \).
3. **Bond energy of F-F**: The bond energy of the F-F bond is given as \( 30 \, \text{kcal/mol} \). When breaking the F-F bond to form two F atoms, we will consider this as energy required, thus it will be \( +30 \, \text{kcal/mol} \).

### Step 3: Set up the equation using Hess's law
According to Hess's law, the total enthalpy change for the reaction can be expressed as:
\[ \Delta_rH = \text{Energy required to break bonds} + \text{Ionization energy} + \text{Electron affinity} - \text{Bond energy of F-F} \]
Let \( x \) be the bond enthalpy of the `Xe-F` bond. Since there are 4 `Xe-F` bonds in `XeF_4`, the energy required to break these bonds is \( 4x \).

So, we can write:
\[ 292 = 4x + 279 - 85 + 30 \]

### Step 4: Simplify and solve for x
Now simplify the equation:
\[ 292 = 4x + 279 - 85 + 30 \]
\[ 292 = 4x + 224 \]
Subtract 224 from both sides:
\[ 292 - 224 = 4x \]
\[ 68 = 4x \]
Now, divide by 4:
\[ x = \frac{68}{4} = 17 \, \text{kcal/mol} \]

### Step 5: Conclusion
Thus, the bond enthalpy of the `Xe-F` bond is:
\[ \boxed{17 \, \text{kcal/mol}} \]