Alright, let's work through solving this problem step-by-step.
1. Initial Reaction:
[tex]\[
2 \text{Al(OH)}_3(s) \rightarrow \text{Al}_2\text{O}_3(s) + 3 \text{H}_2\text{O}(l) \quad \Delta H = 41 \text{kJ}
\][/tex]
2. First Reaction (Half of the initial reaction):
[tex]\[
\text{Al(OH)}_3(s) \rightarrow \frac{1}{2} \text{Al}_2\text{O}_3(s) + \frac{3}{2} \text{H}_2\text{O}(l)
\][/tex]
Since this reaction involves half of the amount of the initial reaction, the enthalpy change will also be half:
[tex]\[
\Delta H = \frac{41 \text{kJ}}{2} = 20.5 \text{kJ}
\][/tex]
3. Second Reaction (Reverse and doubled initial reaction):
[tex]\[
2 \text{Al}_2\text{O}_3(s) + 6 \text{H}_2\text{O}(l) \rightarrow 4 \text{Al(OH)}_3(s)
\][/tex]
This reaction is the reverse of the initial reaction and doubled. The enthalpy change for the reverse reaction is the negative of the initial reaction and, since it is doubled, we multiply by 2:
[tex]\[
\Delta H = -2 \times 41 \text{kJ} = -82 \text{kJ}
\][/tex]
4. Third Reaction (Reverse of the initial reaction):
[tex]\[
\text{Al}_2\text{O}_3(s) + 3 \text{H}_2\text{O}(l) \rightarrow 2 \text{Al(OH)}_3(s)
\][/tex]
This reaction is simply the reverse of the initial reaction, so the enthalpy change is the negative of the initial reaction:
[tex]\[
\Delta H = -41 \text{kJ}
\][/tex]
Thus, the completed table is:
[tex]\[
\begin{array}{|c|c|}
\hline \text{reaction} & \Delta H \\
\hline \text{Al(OH)}_3(s) \rightarrow \frac{1}{2} \text{Al}_2\text{O}_3(s) + \frac{3}{2} \text{H}_2\text{O}(l) & 20.5 \text{kJ} \\
\hline 2 \text{Al}_2\text{O}_3(s) + 6 \text{H}_2\text{O}(l) \rightarrow 4 \text{Al(OH)}_3(s) & -82 \text{kJ} \\
\hline \text{Al}_2\text{O}_3(s) + 3 \text{H}_2\text{O}(l) \rightarrow 2 \text{Al(OH)}_3(s) & -41 \text{kJ} \\
\hline
\end{array}
\][/tex]
