If `DeltaH_(f)^(@)` of `Icl(g)`, `Cl(g)` and `I(g)` is `17.57`, `121.34` and `106.96 J mol^(-1)` respectively. Then bond dissociation energy of `I-Cl` bond is

Options

A. `35.15 J mol^(-1)`
B. `106.69 J mol^(-1)`
C. `210.73 J mol^(-1)`
D. `420.9 J mol^(-1)`

Explanation

To find the bond dissociation energy of the I-Cl bond using the given enthalpy of formation values, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the given values**:
- ΔH_f(ICl(g)) = 17.57 J/mol
- ΔH_f(Cl(g)) = 121.34 J/mol
- ΔH_f(I(g)) = 106.96 J/mol

2. **Write the reaction for the formation of ICl**:
The formation of iodine monochloride (ICl) from its elements can be represented as:
\[ I(g) + \frac{1}{2}Cl_2(g) \rightarrow ICl(g) \]

3. **Use Hess's law**:
According to Hess's law, the change in enthalpy (ΔH) for the reaction can be calculated using the enthalpy of formation of the products and reactants:
\[ ΔH = ΔH_f(\text{products}) - ΔH_f(\text{reactants}) \]

4. **Calculate the enthalpy change for the reaction**:
The enthalpy change for the formation of ICl can be expressed as:
\[ ΔH = ΔH_f(ICl) - [ΔH_f(I) + ΔH_f(Cl)] \]
Since we need to consider the enthalpy of formation of Cl2, we will use half of the ΔH_f(Cl(g)) because Cl2 is diatomic:
\[ ΔH = ΔH_f(ICl) - [ΔH_f(I) + ΔH_f(Cl2)] \]
Substituting the values:
\[ ΔH = 17.57 - [106.96 + 121.34/2] \]
\[ ΔH = 17.57 - [106.96 + 60.67] \]
\[ ΔH = 17.57 - 167.63 \]
\[ ΔH = -150.06 \text{ J/mol} \]

5. **Calculate the bond dissociation energy**:
The bond dissociation energy (BDE) of the I-Cl bond can be calculated as:
\[ BDE(I-Cl) = -ΔH \]
Therefore:
\[ BDE(I-Cl) = 150.06 \text{ J/mol} \]

### Final Answer:
The bond dissociation energy of the I-Cl bond is approximately **150.06 J/mol**.