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 down the reactions and their enthalpy changes:**
- Reaction 1: \( H_2 + \frac{1}{2} O_2 \rightarrow H_2O \)
\( \Delta H_1 = -68.39 \, \text{kcal} \)
- Reaction 2: \( K + H_2O \rightarrow KOH + \frac{1}{2} H_2 \)
\( \Delta H_2 = -48.0 \, \text{kcal} \)
- Reaction 3: \( KOH + H_2O \rightarrow KOH(aq) \)
\( \Delta H_3 = -14.0 \, \text{kcal} \)

2. **Identify the target reaction for the formation of KOH:**
- Target Reaction: \( K + \frac{1}{2} H_2 + \frac{1}{2} O_2 \rightarrow KOH \)

3. **Combine the reactions to derive the target reaction:**
- We need to manipulate the given reactions to arrive at the target reaction.
- From Reaction 2, we can rearrange it to:
\[ K + H_2O \rightarrow KOH + \frac{1}{2} H_2 \]
- From Reaction 1, we have:
\[ H_2 + \frac{1}{2} O_2 \rightarrow H_2O \]
- Reaction 3 does not need to be altered since it involves KOH in aqueous form.

4. **Add the enthalpy changes:**
- To find the enthalpy change for the target reaction, we add the enthalpy changes of the relevant reactions:
\[ \Delta H = \Delta H_1 + \Delta H_2 + \Delta H_3 \]
\[ \Delta H = (-68.39) + (-48.0) + 14.0 \]

5. **Calculate the total enthalpy change:**
\[ \Delta H = -68.39 - 48.0 + 14.0 = -102.39 \, \text{kcal} \]

6. **Conclusion:**
The heat of formation of KOH is \( -102.39 \, \text{kcal} \).