If `H_(2)+1//2O_(2)rarr H_(2)O , Delta =-68.39` Kcal
`K+H_(2)O+ "water" rarr KOH(aq)+1//2H_(2) , Delta H=-48.0` Kcal
`KOH+"water" rarr KOH(aq)Delta H=-14.0` Kcal the heat of formation of KOH is -

Options

A. `-68.39+48-14.0`
B. `-68.39-48.0+14.0`
C. `+68.39-48.0+14.0`
D. `+68.39+48.0-14.0`

Explanation

To find the heat of formation of KOH, 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. **Write the Given Reactions and Their Enthalpy Changes:**
- Reaction 1:
\[ H_2 + \frac{1}{2} O_2 \rightarrow H_2O \quad \Delta H_1 = -68.39 \text{ kcal} \]
- Reaction 2:
\[ K + H_2O \rightarrow KOH(aq) + \frac{1}{2} H_2 \quad \Delta H_2 = -48.0 \text{ kcal} \]
- Reaction 3:
\[ KOH(s) + H_2O \rightarrow KOH(aq) \quad \Delta H_3 = -14.0 \text{ kcal} \]

2. **Combine the Reactions:**
- We want to find the heat of formation of KOH(s) from its elements, which can be represented as:
\[ K(s) + \frac{1}{2} O_2(g) + \frac{1}{2} H_2(g) \rightarrow KOH(s) \]
- To achieve this, we will add Reaction 1 and Reaction 2, and then subtract Reaction 3.

3. **Add Reactions 1 and 2:**
- Adding Reaction 1 and Reaction 2 gives:
\[ H_2 + \frac{1}{2} O_2 + K + H_2O \rightarrow H_2O + KOH(aq) + \frac{1}{2} H_2 \]
- Simplifying this, we get:
\[ K + \frac{1}{2} O_2 + \frac{1}{2} H_2 \rightarrow KOH(aq) \]
- The enthalpy change for this combined reaction is:
\[ \Delta H = \Delta H_1 + \Delta H_2 = -68.39 - 48.0 = -116.39 \text{ kcal} \]

4. **Subtract Reaction 3:**
- Now, we subtract Reaction 3:
\[ KOH(aq) \rightarrow KOH(s) + H_2O \]
- The enthalpy change for this reaction is:
\[ \Delta H = -(-14.0) = +14.0 \text{ kcal} \]
- Therefore, the total enthalpy change becomes:
\[ \Delta H = -116.39 + 14.0 = -102.39 \text{ kcal} \]

5. **Final Result:**
- The heat of formation of KOH(s) is:
\[ \Delta H_f(KOH) = -102.39 \text{ kcal} \]

### Conclusion:
The heat of formation of KOH is \(-102.39\) kcal.