Calculate the enthalpy of formation of KOH(s) from the following data :
`K(s) + H_(2)O(l) + (aq) to KOH(aq) + 1/2 H_(2)(g) Delta_(r)H^(@) = -200.8 kJ`
`H_(2)(g) + 1/2 O_(2)(g) to H_(2)O(l) Delta_(r)H^(@) = - 286.3 kJ`
`KOH(s) + (aq) to KOH(aq) Delta_(r)H^(@) = - 58.6 kJ`

Explanation

To calculate the enthalpy of formation of KOH(s) from the given reactions, 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-by-Step Solution:

1. **Identify the Reactions and Their Enthalpy Changes:**
- Reaction 1:
\[ K(s) + H_2O(l) + \frac{1}{2} H_2(g) \rightarrow KOH(aq) \quad \Delta_rH^(@) = -200.8 \, \text{kJ} \]
- Reaction 2:
\[ H_2(g) + \frac{1}{2} O_2(g) \rightarrow H_2O(l) \quad \Delta_rH^(@) = -286.3 \, \text{kJ} \]
- Reaction 3:
\[ KOH(s) + (aq) \rightarrow KOH(aq) \quad \Delta_rH^(@) = -58.6 \, \text{kJ} \]

2. **Write the Target Reaction:**
The target reaction for the formation of KOH(s) from its elements is:
\[ K(s) + \frac{1}{2} O_2(g) + H_2(g) \rightarrow KOH(s) \]

3. **Manipulate the Given Reactions:**
- Keep Reaction 1 as is.
- Reverse Reaction 3 to express KOH(aq) as KOH(s):
\[ KOH(aq) \rightarrow KOH(s) + (aq) \quad \Delta_rH^(@) = +58.6 \, \text{kJ} \]
- Reverse Reaction 2 to express H2O(l) as reactants:
\[ H_2O(l) \rightarrow H_2(g) + \frac{1}{2} O_2(g) \quad \Delta_rH^(@) = +286.3 \, \text{kJ} \]

4. **Combine the Reactions:**
Now, we will add the modified reactions:
- From Reaction 1:
\[ K(s) + H_2O(l) + \frac{1}{2} H_2(g) \rightarrow KOH(aq) \quad \Delta_rH^(@) = -200.8 \, \text{kJ} \]
- From the reversed Reaction 3:
\[ KOH(aq) \rightarrow KOH(s) + (aq) \quad \Delta_rH^(@) = +58.6 \, \text{kJ} \]
- From the reversed Reaction 2:
\[ H_2O(l) \rightarrow H_2(g) + \frac{1}{2} O_2(g) \quad \Delta_rH^(@) = +286.3 \, \text{kJ} \]

After adding, we get:
\[ K(s) + \frac{1}{2} O_2(g) + H_2(g) \rightarrow KOH(s) \]

5. **Calculate the Overall Enthalpy Change:**
Now we sum the enthalpy changes:
\[ \Delta H = (-200.8) + (58.6) + (286.3) \]
\[ \Delta H = -200.8 + 58.6 + 286.3 = -200.8 + 344.9 = -428.5 \, \text{kJ} \]

Thus, the enthalpy of formation of KOH(s) is:
\[ \Delta H_f(KOH(s)) = -428.5 \, \text{kJ/mol} \]