The enthaplpy changes state for the following processes are listed below:
`Cl_(2)(g)=2Cl(g)` : `242.3KJmol^(-1)`
`I_(2)(g)=2I(g)` , `151.0KJ mol^(-1)`
`ICl(g)=I(g)+Cl(g)` : `211.3KJ mol^(-1)`
`I_(2)(s)=l_(2)(g)` , `62.76KJ mol^(-1)`
Given that the standard states for iodine chlorine are `I_(2)(s)` and `Cl_(2)(g)` , the standard enthalpy of formation for `ICl(g)` is:

Options

A. ` + 2448.8 KJ mol^(-1)`
B. `-14.6 K J mol^(-1)`
C. `-16.8 K J mol ^(-1)`
D. `+16.8 KJ mol^(-1)`

Explanation

To find the standard enthalpy of formation for ICl(g), we will use the given enthalpy changes for the reactions provided. The goal is to manipulate these reactions to derive the formation reaction for ICl(g) from its standard states, which are I2(s) and Cl2(g).

### Step-by-Step Solution:

1. **Identify the target reaction**:
We need to find the standard enthalpy of formation for the reaction:
\[ \frac{1}{2} I_2(s) + \frac{1}{2} Cl_2(g) \rightarrow ICl(g) \]

2. **List the given reactions and their enthalpy changes**:
- \( Cl_2(g) \rightarrow 2Cl(g) \) : \( \Delta H = 242.3 \, \text{kJ} \)
- \( I_2(g) \rightarrow 2I(g) \) : \( \Delta H = 151.0 \, \text{kJ} \)
- \( ICl(g) \rightarrow I(g) + Cl(g) \) : \( \Delta H = 211.3 \, \text{kJ} \)
- \( I_2(s) \rightarrow I_2(g) \) : \( \Delta H = 62.76 \, \text{kJ} \)

3. **Manipulate the reactions**:
- For \( I_2(s) \rightarrow I_2(g) \):
\[ \Delta H = 62.76 \, \text{kJ} \]
- We need half of this reaction:
\[ \frac{1}{2} I_2(s) \rightarrow \frac{1}{2} I_2(g) \quad \Delta H = \frac{62.76}{2} = 31.38 \, \text{kJ} \]

- For \( Cl_2(g) \rightarrow 2Cl(g) \):
\[ \Delta H = 242.3 \, \text{kJ} \]
- We need half of this reaction:
\[ \frac{1}{2} Cl_2(g) \rightarrow Cl(g) \quad \Delta H = \frac{242.3}{2} = 121.15 \, \text{kJ} \]

- For \( ICl(g) \rightarrow I(g) + Cl(g) \):
\[ \Delta H = 211.3 \, \text{kJ} \]
- We need to reverse this reaction:
\[ I(g) + Cl(g) \rightarrow ICl(g) \quad \Delta H = -211.3 \, \text{kJ} \]

- For \( I_2(g) \rightarrow 2I(g) \):
\[ \Delta H = 151.0 \, \text{kJ} \]
- We need half of this reaction:
\[ I_2(g) \rightarrow 2I(g) \quad \Delta H = 75.5 \, \text{kJ} \]

4. **Combine the reactions**:
Now we add the manipulated reactions:
\[ \frac{1}{2} I_2(s) \rightarrow \frac{1}{2} I_2(g) \quad \Delta H = 31.38 \, \text{kJ} \]
\[ \frac{1}{2} Cl_2(g) \rightarrow Cl(g) \quad \Delta H = 121.15 \, \text{kJ} \]
\[ I(g) + Cl(g) \rightarrow ICl(g) \quad \Delta H = -211.3 \, \text{kJ} \]

5. **Cancel out terms**:
When we add these reactions, we will cancel out the I(g) and Cl(g) on both sides:
\[ \frac{1}{2} I_2(s) + \frac{1}{2} Cl_2(g) + I(g) + Cl(g) \rightarrow \frac{1}{2} I_2(g) + ICl(g) \]

6. **Calculate the total enthalpy change**:
\[ \Delta H_{total} = 31.38 + 121.15 - 211.3 + 75.5 \]
\[ \Delta H_{total} = 16.78 \, \text{kJ} \]

### Final Answer:
The standard enthalpy of formation for ICl(g) is:
\[ \Delta H_f^{\circ}(ICl(g)) = 16.78 \, \text{kJ/mol} \]