Calculate `DeltaH^(@)` `("in" kJmol^(-1))` for the reaction
`CH_(2)Cl_(2)(g)rarrC(g)+2H(g)+2Cl(g).`
The average bond enthalpie of `C-H` and `C-C1` bonds are `414 kJ mol^(-1)`.

Answer

To calculate the standard enthalpy change (ΔH°) for the reaction

\[ \text{CH}_2\text{Cl}_2(g) \rightarrow \text{C}(g) + 2\text{H}(g) + 2\text{Cl}(g) \]

we will use the average bond enthalpies of the bonds broken and formed during the reaction.

### Step-by-Step Solution:

1. **Identify the Bonds in the Reactants:**
In the molecule CH₂Cl₂, there are:
- 2 C-H bonds
- 2 C-Cl bonds

2. **Calculate the Total Bond Enthalpy for the Reactants:**
We will use the average bond enthalpies provided:
- Average bond enthalpy of C-H = 414 kJ/mol
- Average bond enthalpy of C-Cl = 330 kJ/mol

The total bond enthalpy for the reactants (CH₂Cl₂) is calculated as follows:
\[ \text{Total bond enthalpy} = 2 \times \text{(C-H bond)} + 2 \times \text{(C-Cl bond)} \]
\[ = 2 \times 414 \, \text{kJ/mol} + 2 \times 330 \, \text{kJ/mol} \]
\[ = 828 \, \text{kJ/mol} + 660 \, \text{kJ/mol} \]
\[ = 1488 \, \text{kJ/mol} \]

3. **Identify the Bonds in the Products:**
In the products (C, H, and Cl in their gaseous forms), the bonds are considered to be in their elemental forms, which means their bond enthalpies are zero:
- C(g) has no bonds to break.
- H(g) has no bonds to break.
- Cl(g) has no bonds to break.

4. **Calculate ΔH° for the Reaction:**
Since the products are in their elemental forms, the total bond enthalpy for the products is 0 kJ/mol. Therefore, the ΔH° for the reaction can be calculated as:
\[ ΔH° = \text{Total bond enthalpy of reactants} - \text{Total bond enthalpy of products} \]
\[ ΔH° = 1488 \, \text{kJ/mol} - 0 \, \text{kJ/mol} \]
\[ ΔH° = 1488 \, \text{kJ/mol} \]

### Final Answer:
The standard enthalpy change (ΔH°) for the reaction is **1488 kJ/mol**.