The enthaply of atomisation for the reaction : `CH_(4)(g) to C(g) + 4 H(g)` is `1665 kJ "mol"^(-1)`.What is the bond enthalpy of C-H bond? (Keep your answer in KJ/mol and three digits only)

Explanation

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

### Step-by-Step Solution:

1. **Understand the Reaction**: The reaction given is:
\[ \text{CH}_4(g) \rightarrow \text{C}(g) + 4 \text{H}(g) \]
The enthalpy of atomization for this reaction is provided as 1665 kJ/mol. This means that 1665 kJ of energy is required to break all the bonds in one mole of methane to form one mole of carbon atoms and four moles of hydrogen atoms.

2. **Identify the Number of Bonds**: In methane (CH₄), there are four C-H bonds. Therefore, when we atomize one mole of CH₄, we are breaking four C-H bonds.

3. **Calculate the Bond Enthalpy**: The bond enthalpy can be calculated by dividing the total enthalpy of atomization by the number of bonds broken:
\[ \text{Bond Enthalpy} = \frac{\text{Enthalpy of Atomization}}{\text{Number of C-H Bonds}} \]
Substituting the values:
\[ \text{Bond Enthalpy} = \frac{1665 \text{ kJ/mol}}{4} \]

4. **Perform the Calculation**:
\[ \text{Bond Enthalpy} = \frac{1665}{4} = 416.25 \text{ kJ/mol} \]

5. **Round the Answer**: Since the question asks for the answer in three digits, we round 416.25 to 416 kJ/mol.

### Final Answer:
The bond enthalpy of the C-H bond is **416 kJ/mol**.