Calculate the bond energy of `Cl-Cl` bond from the following data:
`CH_(4)(g)+Cl_(2)(g)rarrCH_(3)Cl(g)+HCl(g), Delta H= -100.3 kJ`. Also the bond enthalpies of `C-H, C-Cl,H-Cl` bonds are `413,326` and `431 kJ mol^(-1)` respectively.

Explanation

To calculate the bond energy of the Cl-Cl bond from the given reaction and bond enthalpies, we can follow these steps:

### Step 1: Write down the given reaction and data
The reaction is:
\[ \text{CH}_4(g) + \text{Cl}_2(g) \rightarrow \text{CH}_3\text{Cl}(g) + \text{HCl}(g) \]
The change in enthalpy (\( \Delta H \)) for this reaction is given as:
\[ \Delta H = -100.3 \, \text{kJ} \]

The bond enthalpies provided are:
- C-H bond: \( 413 \, \text{kJ/mol} \)
- C-Cl bond: \( 326 \, \text{kJ/mol} \)
- H-Cl bond: \( 431 \, \text{kJ/mol} \)

### Step 2: Identify the bonds broken and formed
**Bonds broken (reactants):**
- In methane (CH₄), there are 4 C-H bonds.
- In Cl₂, there is 1 Cl-Cl bond.

**Bonds formed (products):**
- In CH₃Cl, there are 3 C-H bonds and 1 C-Cl bond.
- In HCl, there is 1 H-Cl bond.

### Step 3: Write the bond energy equation
Using the bond dissociation energy formula:
\[ \Delta H = \text{(Bonds broken)} - \text{(Bonds formed)} \]
We can express this as:
\[ -100.3 = [4 \times \text{(C-H)} + 1 \times \text{(Cl-Cl)}] - [3 \times \text{(C-H)} + 1 \times \text{(C-Cl)} + 1 \times \text{(H-Cl)}] \]

### Step 4: Substitute the known values
Substituting the known bond enthalpies:
\[ -100.3 = [4 \times 413 + 1 \times \text{(Cl-Cl)}] - [3 \times 413 + 1 \times 326 + 1 \times 431] \]

### Step 5: Simplify the equation
Calculating the left side:
- For bonds broken:
\[ 4 \times 413 = 1652 \, \text{kJ} \]
- For bonds formed:
\[ 3 \times 413 = 1239 \, \text{kJ} \]
\[ 326 + 431 = 757 \, \text{kJ} \]
Thus,
\[ \text{Total bonds formed} = 1239 + 757 = 1996 \, \text{kJ} \]

Now substituting these values into the equation:
\[ -100.3 = [1652 + \text{(Cl-Cl)}] - 1996 \]
This simplifies to:
\[ -100.3 = 1652 + \text{(Cl-Cl)} - 1996 \]
\[ -100.3 = \text{(Cl-Cl)} - 344 \]

### Step 6: Solve for Cl-Cl bond energy
Rearranging gives:
\[ \text{(Cl-Cl)} = -100.3 + 344 \]
\[ \text{(Cl-Cl)} = 243.7 \, \text{kJ/mol} \]

### Final Answer:
The bond energy of the Cl-Cl bond is:
\[ \text{Bond energy of Cl-Cl} = 243.7 \, \text{kJ/mol} \]