If `{:(H_(2)+1//2O_(2)rarrH_(2)O",",,,,DeltaH= -68 kcal),(K+H_(2)OrarrKOH(aq)+1//2H_(2)",",,,,DeltaH= -48 kcal),(KOH+"water"rarrKOH(aq)"," ,,,,DeltaH= -14 kcal):}`
Find the heat of formation of `KOH`.
Find the heat of formation of `KOH`.

Answer

To find the heat of formation of KOH, 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 of the reaction.

### Step 1: Write the reactions and their enthalpy changes
1. **Reaction 1**: \( H_2 + \frac{1}{2} O_2 \rightarrow H_2O \), \( \Delta H = -68 \, \text{kcal} \)
2. **Reaction 2**: \( K + H_2O \rightarrow KOH(aq) + \frac{1}{2} H_2 \), \( \Delta H = -48 \, \text{kcal} \)
3. **Reaction 3**: \( KOH + \text{water} \rightarrow KOH(aq) \), \( \Delta H = -14 \, \text{kcal} \)

### Step 2: Write the desired formation reaction for KOH
The formation reaction for KOH can be written as:
\[ K(s) + \frac{1}{2} O_2(g) + \frac{1}{2} H_2(g) \rightarrow KOH(s) \]

### Step 3: Combine the reactions
To find the heat of formation of KOH, we need to combine the first two reactions and subtract the third reaction.

1. **Add Reaction 1 and Reaction 2**:
\[ H_2 + \frac{1}{2} O_2 \rightarrow H_2O \quad (\Delta H = -68 \, \text{kcal}) \]
\[ K + H_2O \rightarrow KOH(aq) + \frac{1}{2} H_2 \quad (\Delta H = -48 \, \text{kcal}) \]
Adding these gives:
\[ K + \frac{1}{2} O_2 + H_2 \rightarrow KOH(aq) + H_2O \quad (\Delta H = -68 - 48 = -116 \, \text{kcal}) \]

2. **Subtract Reaction 3**:
\[ KOH + \text{water} \rightarrow KOH(aq) \quad (\Delta H = -14 \, \text{kcal}) \]
To subtract this, we reverse the reaction:
\[ KOH(aq) \rightarrow KOH + \text{water} \quad (\Delta H = +14 \, \text{kcal}) \]

### Step 4: Combine the results
Now we can combine the results:
\[ K + \frac{1}{2} O_2 + H_2 \rightarrow KOH(aq) + H_2O \quad (\Delta H = -116 \, \text{kcal}) \]
\[ KOH(aq) \rightarrow KOH + \text{water} \quad (\Delta H = +14 \, \text{kcal}) \]
Adding these gives:
\[ K + \frac{1}{2} O_2 + \frac{1}{2} H_2 \rightarrow KOH \quad (\Delta H = -116 + 14 = -102 \, \text{kcal}) \]

### Final Answer
Thus, the heat of formation of KOH is:
\[ \Delta H_f(KOH) = -102 \, \text{kcal} \]