Consider the reactions `:`
`C_((s))+2H_(2(g)) rarr CH_(4(g)), DeltaH=-X kcal`
`C_((g))+4H_((g)) rarr CH_(4(g)),DeltaH-X_(1)kcal`
`CH_(4(g))rarr CH_(3(g))+H_((g)), DeltaH=+Ykcal`
The average bond energy of `C-H` bond is `:`

Options

A. `(X)/(4) kcal mol^(-1)`
B. `Y kcal mol^(-1)`
C. `(X_(1))/(4)kcal mol^(-1)`
D. `X_(1)kcal mol^(-1)`

Explanation

To find the average bond energy of the C-H bond in methane (CH₄), we can analyze the given reactions and their enthalpy changes (ΔH).

### Step-by-Step Solution:

1. **Understanding the Reactions**:
- The first reaction is:
\[ C_{(s)} + 2H_{2(g)} \rightarrow CH_{4(g)}, \Delta H = -X \text{ kcal} \]
- The second reaction is:
\[ C_{(g)} + 4H_{(g)} \rightarrow CH_{4(g)}, \Delta H = -X_1 \text{ kcal} \]
- The third reaction is:
\[ CH_{4(g)} \rightarrow CH_{3(g)} + H_{(g)}, \Delta H = +Y \text{ kcal} \]

2. **Identifying the Bond Dissociation**:
- In the third reaction, methane (CH₄) is dissociating into CH₃ and a hydrogen atom (H). The ΔH for this reaction is +Y kcal, indicating that energy is absorbed to break the C-H bond.

3. **Relating the Reactions**:
- The enthalpy change for the dissociation of CH₄ into its constituent atoms can be related to the bond energies. When CH₄ dissociates, it breaks 4 C-H bonds:
\[ CH_{4(g)} \rightarrow C_{(g)} + 4H_{(g)} \quad \text{(reverse of the second reaction)} \]
- The enthalpy change for this reverse reaction will be +X₁ kcal.

4. **Setting Up the Equation**:
- The total energy required to break all 4 C-H bonds in CH₄ can be expressed as:
\[ 4 \times \text{(average bond energy of C-H)} = Y + X_1 \]
- Here, Y is the energy required to break one C-H bond, and X₁ is the energy required to form CH₄ from C and 4H.

5. **Calculating Average Bond Energy**:
- To find the average bond energy of one C-H bond, we rearrange the equation:
\[ \text{Average bond energy of C-H} = \frac{Y + X_1}{4} \]

### Final Answer:
The average bond energy of the C-H bond is:
\[ \text{Average bond energy of C-H} = \frac{Y + X_1}{4} \text{ kcal} \]