From the following data, the heat of formation of `Ca(OH)_(2)(s)` at `18^(@)C` is …. Kcal.
(i) `CaO(s)+H_(2)O(l)=Ca(OH)_(2)(s) , Delta H18^(@)C=-15.26` Kcal…..
(ii) `H_(2)O(l)=H_(2)(g)+1//2O_(2)(g) , Delta H18^(@)C=68.37` Kcal….
(iii) `Ca(s)+1//2O_(2)(g)=CaO(s) , Delta H 18^(@)C=-151.80` Kcal....

Options

A. `-98.69`
B. `-235.43`
C. `194.91`
D. `98.69`

Explanation

To find the heat of formation of \( \text{Ca(OH)}_2(s) \) at \( 18^\circ C \), we can 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, regardless of the pathway taken.

### Step-by-Step Solution:

1. **Identify the target reaction**:
We want to find the heat of formation for the reaction:
\[ \text{Ca}(s) + \frac{1}{2} \text{O}_2(g) + \text{H}_2(g) \rightarrow \text{Ca(OH)}_2(s) \]
This is our Equation 1.

2. **List the given reactions and their enthalpy changes**:
- Reaction 2:
\[ \text{CaO}(s) + \text{H}_2O(l) \rightarrow \text{Ca(OH)}_2(s) \quad \Delta H = -15.26 \text{ Kcal} \]
- Reaction 3:
\[ \text{H}_2O(l) \rightarrow \text{H}_2(g) + \frac{1}{2} \text{O}_2(g) \quad \Delta H = 68.37 \text{ Kcal} \]
- Reaction 4:
\[ \text{Ca}(s) + \frac{1}{2} \text{O}_2(g) \rightarrow \text{CaO}(s) \quad \Delta H = -151.80 \text{ Kcal} \]

3. **Manipulate the given reactions to derive Equation 1**:
To obtain Equation 1, we can rearrange and combine the other reactions:
- Start with Reaction 4 (to form \( \text{CaO}(s) \)):
\[ \text{Ca}(s) + \frac{1}{2} \text{O}_2(g) \rightarrow \text{CaO}(s) \quad \Delta H = -151.80 \text{ Kcal} \]
- Then use Reaction 2 (to form \( \text{Ca(OH)}_2(s) \) from \( \text{CaO}(s) \)):
\[ \text{CaO}(s) + \text{H}_2O(l) \rightarrow \text{Ca(OH)}_2(s) \quad \Delta H = -15.26 \text{ Kcal} \]
- Finally, we need to reverse Reaction 3 (to convert \( \text{H}_2O(l) \) into \( \text{H}_2(g) \) and \( \frac{1}{2} \text{O}_2(g) \)):
\[ \text{H}_2(g) + \frac{1}{2} \text{O}_2(g) \rightarrow \text{H}_2O(l) \quad \Delta H = -68.37 \text{ Kcal} \]

4. **Combine the enthalpy changes**:
Now we can sum the enthalpy changes:
\[ \Delta H = (-151.80) + (-15.26) + (-68.37) \]

5. **Calculate the total enthalpy change**:
\[ \Delta H = -151.80 - 15.26 - 68.37 = -235.43 \text{ Kcal} \]

### Final Answer:
The heat of formation of \( \text{Ca(OH)}_2(s) \) at \( 18^\circ C \) is \( -235.43 \text{ Kcal} \).

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