The enthalpy of atomization of `CH_(4)` is 1665 `kJ mol^(-1)`. What is the bond enthalpy of `C - H` bond?

Answer

To find the bond enthalpy of the C-H bond in methane (CH₄) given the enthalpy of atomization, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Concept of Enthalpy of Atomization**:
- The enthalpy of atomization is the energy required to convert one mole of a compound into its gaseous atoms. For methane (CH₄), this means breaking it down into one carbon atom and four hydrogen atoms.

2. **Identify the Given Data**:
- The enthalpy of atomization of CH₄ is given as 1665 kJ/mol.

3. **Determine the Number of C-H Bonds**:
- In the methane molecule (CH₄), there are 4 C-H bonds.

4. **Calculate the Bond Enthalpy**:
- The bond enthalpy for one C-H bond can be calculated using the formula:
\[ \text{Bond Enthalpy} = \frac{\text{Enthalpy of Atomization}}{\text{Number of C-H Bonds}} \]
- Substituting the values:
\[ \text{Bond Enthalpy} = \frac{1665 \text{ kJ/mol}}{4} \]

5. **Perform the Calculation**:
- Calculating the above expression:
\[ \text{Bond Enthalpy} = 416.25 \text{ kJ/mol} \]

6. **Conclusion**:
- The bond enthalpy of the C-H bond in methane is 416.25 kJ/mol.

### Final Answer:
The bond enthalpy of the C-H bond in methane (CH₄) is **416.25 kJ/mol**.

---