The bond dissociation energies for `Cl_(2)`, `I_(2)` and `IC l` are `242.3`, `151.0` and `211.3 kJ//"mole"` respectively. The enthalpy of sublimation of iodine is `62.8 kJ //"mole"`. What is the standard enthalpy of formation of `ICI(g)` nearly equal to

Options

A. `-211.3 kJ//"mole"`
B. `-14.6 kJ//"mole"`
C. `16.8 kJ//"mole"`
D. `33.5 kJ//"mole"`

Explanation

To find the standard enthalpy of formation of ICl(g), we can use the bond dissociation energies and the enthalpy of sublimation provided in the question. The process involves combining the relevant reactions and their associated enthalpy changes.

### Step-by-Step Solution:

1. **Identify the reactions and their enthalpies:**
- The bond dissociation energy for \( Cl_2 \) is \( 242.3 \, \text{kJ/mol} \).
- The bond dissociation energy for \( I_2 \) is \( 151.0 \, \text{kJ/mol} \).
- The bond dissociation energy for \( ICl \) is \( 211.3 \, \text{kJ/mol} \).
- The enthalpy of sublimation of iodine \( I \) is \( 62.8 \, \text{kJ/mol} \).

2. **Write the formation reaction for ICl:**
\[ \frac{1}{2} I_2 (s) + \frac{1}{2} Cl_2 (g) \rightarrow ICl (g) \]

3. **Set up the equations based on the given data:**
- For \( Cl_2 \) dissociation:
\[ Cl_2 (g) \rightarrow 2 Cl (g) \quad \Delta H = 242.3 \, \text{kJ} \]
Therefore, for \( \frac{1}{2} Cl_2 \):
\[ \frac{1}{2} Cl_2 (g) \rightarrow Cl (g) \quad \Delta H = \frac{242.3}{2} = 121.15 \, \text{kJ} \]

- For \( I_2 \) dissociation:
\[ I_2 (s) \rightarrow 2 I (g) \quad \Delta H = 151.0 \, \text{kJ} \]
Therefore, for \( \frac{1}{2} I_2 \):
\[ \frac{1}{2} I_2 (s) \rightarrow I (g) \quad \Delta H = \frac{151.0}{2} = 75.5 \, \text{kJ} \]

- For the formation of \( ICl \):
\[ I (g) + Cl (g) \rightarrow ICl (g) \quad \Delta H = -211.3 \, \text{kJ} \]

4. **Combine the equations:**
To find the standard enthalpy of formation of \( ICl(g) \), we combine the enthalpy changes:
\[ \Delta H_{f} = \left( \Delta H_{I} + \Delta H_{Cl} - \Delta H_{ICl} \right) \]
Substituting the values:
\[ \Delta H_{f} = \left( 75.5 + 121.15 - 211.3 \right) \, \text{kJ} \]

5. **Calculate the final value:**
\[ \Delta H_{f} = 75.5 + 121.15 - 211.3 = -14.65 \, \text{kJ/mol} \]

6. **Final adjustment with sublimation of iodine:**
Since we need to account for the sublimation of iodine:
\[ \Delta H_{f} = -14.65 + 62.8 = 48.15 \, \text{kJ/mol} \]

7. **Conclusion:**
The standard enthalpy of formation of \( ICl(g) \) is approximately \( 48.15 \, \text{kJ/mol} \).