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/mol"`
B. `-"14.6kJ/mol"`
C. `"16.8kJ/mol"`
D. `"33.5 kJ/mol"`

Answer

To find the standard enthalpy of formation of ICl(g), we can use the given bond dissociation energies and the enthalpy of sublimation of iodine. Here’s a step-by-step solution:

### Step 1: Write the relevant reactions
We need to consider the following reactions based on the bond dissociation energies provided:

1. Dissociation of Cl2:
\[ \text{Cl}_2(g) \rightarrow 2 \text{Cl}(g) \quad \Delta H_1 = 242.3 \, \text{kJ/mol} \]

2. Dissociation of I2:
\[ \text{I}_2(s) \rightarrow 2 \text{I}(g) \quad \Delta H_2 = 151.0 \, \text{kJ/mol} \]

3. Formation of ICl from its elements:
\[ \text{I}(g) + \text{Cl}(g) \rightarrow \text{ICl}(g) \quad \Delta H_3 = 211.3 \, \text{kJ/mol} \]

4. Sublimation of iodine (from solid to gas):
\[ \text{I}_2(s) \rightarrow \text{I}_2(g) \quad \Delta H_{\text{sublimation}} = 62.8 \, \text{kJ/mol} \]

### Step 2: Use Hess's Law
According to Hess's Law, the total enthalpy change for a reaction is the sum of the enthalpy changes for the individual steps. We need to manipulate the above reactions to find the enthalpy of formation of ICl.

### Step 3: Manipulate the reactions
We want to find the enthalpy change for the reaction:
\[ \frac{1}{2} \text{I}_2(g) + \frac{1}{2} \text{Cl}_2(g) \rightarrow \text{ICl}(g) \]

To do this, we will:
- Use half of the dissociation energy of Cl2:
\[ \frac{1}{2} \text{Cl}_2(g) \rightarrow \text{Cl}(g) \quad \Delta H = \frac{242.3}{2} = 121.15 \, \text{kJ/mol} \]

- Use half of the dissociation energy of I2:
\[ \frac{1}{2} \text{I}_2(g) \rightarrow \text{I}(g) \quad \Delta H = \frac{151.0}{2} = 75.5 \, \text{kJ/mol} \]

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

### Step 4: Calculate the total enthalpy change
Now, we can sum these enthalpy changes:
\[ \Delta H = \left( \frac{151.0}{2} \right) + \left( \frac{242.3}{2} \right) - 211.3 \]
\[ \Delta H = 75.5 + 121.15 - 211.3 \]
\[ \Delta H = 196.65 - 211.3 = -14.65 \, \text{kJ/mol} \]

### Step 5: Final calculation
Now, we need to add the sublimation energy of iodine:
\[ \Delta H = -14.65 + 62.8 = 48.15 \, \text{kJ/mol} \]

### Conclusion
Thus, the standard enthalpy of formation of ICl(g) is approximately:
\[ \Delta H_f^\circ \approx 48.15 \, \text{kJ/mol} \]