The enthalpy of formation of hypothetical `CaCl(s)` is theoretically found to be `-188 k J mol^(-1)` and `Delta_(f) H^(@)` for `CaCl_(2) (s) is -795 k J mol^(-1)` . Calculate `Delta_(f) H^(@)` (in kJ `mol^(-1)`) for the disproportionation reaction .
`2 CaCl(s) to CaCl_(2) + Ca (s)`

Explanation

To calculate the enthalpy of formation for the disproportionation reaction \(2 \text{CaCl}(s) \rightarrow \text{CaCl}_2(s) + \text{Ca}(s)\), we will use the enthalpy of formation values provided for \( \text{CaCl}(s) \) and \( \text{CaCl}_2(s) \).

### Step-by-Step Solution:

1. **Identify the Reaction**:
The given reaction is:
\[ 2 \text{CaCl}(s) \rightarrow \text{CaCl}_2(s) + \text{Ca}(s) \]

2. **Write the Enthalpy of Formation Values**:
- For \( \text{CaCl}(s) \): \( \Delta_f H^\circ (\text{CaCl}) = -188 \, \text{kJ/mol} \)
- For \( \text{CaCl}_2(s) \): \( \Delta_f H^\circ (\text{CaCl}_2) = -795 \, \text{kJ/mol} \)
- For \( \text{Ca}(s) \): Since it is in its elemental form, \( \Delta_f H^\circ (\text{Ca}) = 0 \, \text{kJ/mol} \)

3. **Apply the Enthalpy of Formation Formula**:
The enthalpy change for the reaction can be calculated using the formula:
\[ \Delta H^\circ = \Delta_f H^\circ (\text{products}) - \Delta_f H^\circ (\text{reactants}) \]

4. **Calculate the Enthalpy Change**:
- Products:
- \( \Delta_f H^\circ (\text{CaCl}_2) = -795 \, \text{kJ/mol} \)
- \( \Delta_f H^\circ (\text{Ca}) = 0 \, \text{kJ/mol} \)
- Reactants:
- Since there are 2 moles of \( \text{CaCl} \), we have \( 2 \times \Delta_f H^\circ (\text{CaCl}) = 2 \times (-188 \, \text{kJ/mol}) = -376 \, \text{kJ/mol} \)

Now substituting into the formula:
\[ \Delta H^\circ = \left(-795 + 0\right) - \left(-376\right) \]
\[ \Delta H^\circ = -795 + 376 = -419 \, \text{kJ/mol} \]

5. **Final Answer**:
The enthalpy of formation for the disproportionation reaction is:
\[ \Delta_f H^\circ = -419 \, \text{kJ/mol} \]