The value of `DeltaH^(@)` for the reaction `Cu^(+)(g)+I^(-)(g)toCuI(g)` is -446kJ `mol^(-1)`. If the ionisation energy of `Cu(g)` is 745 kJ `mol^(-1)` and the electron affinity of I(g) is -295 kJ `mol^(-1)`, then the value of `DeltaH^(@)` for the formation of one mole of CuI(g) from Cu(g) and I(g) is:

Explanation

To find the value of ΔH° for the formation of one mole of CuI(g) from Cu(g) and I(g), we will use Hess's law, which states that the total enthalpy change for a reaction is the sum of the enthalpy changes for the individual steps of the reaction.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- ΔH° for the reaction Cu⁺(g) + I⁻(g) → CuI(g) = -446 kJ/mol
- Ionization energy of Cu(g) = 745 kJ/mol
- Electron affinity of I(g) = -295 kJ/mol

2. **Define the Steps of the Reaction:**
- Step 1: Ionization of Cu(g) to Cu⁺(g):
\[ \text{Cu(g)} \rightarrow \text{Cu}^+(g) + e^- \quad \Delta H_1 = +745 \text{ kJ/mol} \]
- Step 2: Electron affinity of I(g) to I⁻(g):
\[ \text{I(g)} + e^- \rightarrow \text{I}^-(g) \quad \Delta H_2 = -295 \text{ kJ/mol} \]
- Step 3: Formation of CuI(g) from Cu⁺(g) and I⁻(g):
\[ \text{Cu}^+(g) + \text{I}^-(g) \rightarrow \text{CuI(g)} \quad \Delta H_3 = -446 \text{ kJ/mol} \]

3. **Apply Hess's Law:**
According to Hess's law, the total enthalpy change (ΔH°) for the formation of CuI(g) from Cu(g) and I(g) can be expressed as:
\[ \Delta H° = \Delta H_1 + \Delta H_2 + \Delta H_3 \]

4. **Substitute the Values:**
\[ \Delta H° = 745 \text{ kJ/mol} + (-295 \text{ kJ/mol}) + (-446 \text{ kJ/mol}) \]

5. **Perform the Calculation:**
\[ \Delta H° = 745 - 295 - 446 \]
\[ \Delta H° = 745 - 741 = 4 \text{ kJ/mol} \]

6. **Conclusion:**
The value of ΔH° for the formation of one mole of CuI(g) from Cu(g) and I(g) is:
\[ \Delta H° = 4 \text{ kJ/mol} \]