Calculate the enthalpy change for the process
`C Cl_(4)(g) rarr C(g)+4Cl(g)`
and calculate bond enthalpy of `C-Cl` in `C Cl_(4)(g)`.
`Delta_(vap)H^(Θ)(C Cl_(4))=30.5 kJ mol^(-1)`
`Delta_(f)H^(Θ)(C Cl_(4))=-135.5 kJ mol^(-1)`
`Delta_(a)H^(Θ)(C )=715.0 kJ mol^(-1)`, where `Delta_(a)H^(Θ)` is enthalpy of atomisation
`Delta_(a)H^(Θ)(Cl_(2))=242 kJ mol^(-1)`

Explanation

To calculate the enthalpy change for the process \( \text{CCl}_4(g) \rightarrow \text{C}(g) + 4\text{Cl}(g) \) and the bond enthalpy of \( \text{C-Cl} \) in \( \text{CCl}_4(g) \), we will follow these steps:

### Step 1: Write the reaction and identify the enthalpy changes involved

The reaction we are considering is:
\[ \text{CCl}_4(g) \rightarrow \text{C}(g) + 4\text{Cl}(g) \]

To find the enthalpy change for this reaction, we will use the following data:
- \( \Delta_{vap}H^{\Theta}(\text{CCl}_4) = 30.5 \, \text{kJ/mol} \)
- \( \Delta_{f}H^{\Theta}(\text{CCl}_4) = -135.5 \, \text{kJ/mol} \)
- \( \Delta_{a}H^{\Theta}(\text{C}) = 715.0 \, \text{kJ/mol} \)
- \( \Delta_{a}H^{\Theta}(\text{Cl}_2) = 242 \, \text{kJ/mol} \)

### Step 2: Use Hess's Law to calculate the enthalpy change

Using Hess's Law, we can express the enthalpy change for the reaction as follows:
\[ \Delta H = \Delta_{a}H^{\Theta}(\text{C}) + 2 \times \Delta_{a}H^{\Theta}(\text{Cl}_2) - \Delta_{vap}H^{\Theta}(\text{CCl}_4) - \Delta_{f}H^{\Theta}(\text{CCl}_4) \]

### Step 3: Substitute the values into the equation

Substituting the known values:
\[ \Delta H = 715.0 \, \text{kJ/mol} + 2 \times 242 \, \text{kJ/mol} - 30.5 \, \text{kJ/mol} - (-135.5 \, \text{kJ/mol}) \]

Calculating the individual terms:
\[ \Delta H = 715.0 + 484.0 - 30.5 + 135.5 \]

### Step 4: Perform the calculations

Calculating the total:
\[ \Delta H = 715.0 + 484.0 - 30.5 + 135.5 = 1304.0 \, \text{kJ/mol} \]

### Step 5: Calculate the bond enthalpy of C-Cl

The bond enthalpy of \( \text{C-Cl} \) in \( \text{CCl}_4 \) can be calculated by dividing the total enthalpy change by the number of \( \text{C-Cl} \) bonds broken in the reaction. Since there are 4 \( \text{C-Cl} \) bonds in \( \text{CCl}_4 \):
\[ \text{Bond enthalpy of C-Cl} = \frac{\Delta H}{4} = \frac{1304.0 \, \text{kJ/mol}}{4} = 326.0 \, \text{kJ/mol} \]

### Final Answers
- The enthalpy change for the process is \( 1304.0 \, \text{kJ/mol} \).
- The bond enthalpy of \( \text{C-Cl} \) in \( \text{CCl}_4(g) \) is \( 326.0 \, \text{kJ/mol} \).