Answer :
We begin with the two reference reactions and their reaction enthalpies:
[tex]$$
\begin{array}{rcl}
\text{(1)}\quad S_\omega + O_2 &\to& SO_2\quad\quad \Delta H_1 = -296.0\ \text{kJ/mol}, \\[1mm]
\text{(2)}\quad 2\,SO_2 + O_2 &\to& 2\,SO_3\quad \Delta H_2 = -198.2\ \text{kJ/mol}.
\end{array}
$$[/tex]
Our target reaction is:
[tex]$$
S_\omega + SO_2 + 2\,O_2 \to 2\,SO_3.
$$[/tex]
Step 1. Notice that reaction (1) produces [tex]$SO_2$[/tex] starting from sulfur, and reaction (2) consumes [tex]$2\,SO_2$[/tex] to yield [tex]$2\,SO_3$[/tex]. In order to combine them appropriately, we add reaction (1) and reaction (2):
[tex]\[
\begin{array}{rcl}
S_\omega + O_2 &\to& SO_2 \quad\quad\quad\quad\quad\quad (\Delta H_1=-296.0\ \text{kJ/mol})\\[1mm]
2\,SO_2 + O_2 &\to& 2\,SO_3 \quad\quad\quad\quad (\Delta H_2=-198.2\ \text{kJ/mol})
\end{array}
\][/tex]
Step 2. Write the combined reaction by adding the left-hand sides and the right-hand sides:
[tex]\[
\begin{array}{rcl}
\text{Left side:} && S_\omega + O_2 + 2\,SO_2 + O_2 = S_\omega + 2\,SO_2 + 2\,O_2, \\[1mm]
\text{Right side:} && SO_2 + 2\,SO_3.
\end{array}
\][/tex]
Step 3. Notice that [tex]$SO_2$[/tex] appears on both sides. Subtract one mole of [tex]$SO_2$[/tex] from both sides to simplify the overall reaction:
[tex]$$
S_\omega + SO_2 + 2\,O_2 \to 2\,SO_3.
$$[/tex]
Step 4. Since enthalpy is a state function, the enthalpy change for the overall reaction is the sum of the enthalpy changes for the individual steps:
[tex]$$
\Delta H = \Delta H_1 + \Delta H_2 = (-296.0) + (-198.2) = -494.2\ \text{kJ/mol}.
$$[/tex]
Thus, the enthalpy change for the reaction is
[tex]$$
\boxed{-494.2 \text{ kJ/mol}}
$$[/tex]
This is the final answer.
[tex]$$
\begin{array}{rcl}
\text{(1)}\quad S_\omega + O_2 &\to& SO_2\quad\quad \Delta H_1 = -296.0\ \text{kJ/mol}, \\[1mm]
\text{(2)}\quad 2\,SO_2 + O_2 &\to& 2\,SO_3\quad \Delta H_2 = -198.2\ \text{kJ/mol}.
\end{array}
$$[/tex]
Our target reaction is:
[tex]$$
S_\omega + SO_2 + 2\,O_2 \to 2\,SO_3.
$$[/tex]
Step 1. Notice that reaction (1) produces [tex]$SO_2$[/tex] starting from sulfur, and reaction (2) consumes [tex]$2\,SO_2$[/tex] to yield [tex]$2\,SO_3$[/tex]. In order to combine them appropriately, we add reaction (1) and reaction (2):
[tex]\[
\begin{array}{rcl}
S_\omega + O_2 &\to& SO_2 \quad\quad\quad\quad\quad\quad (\Delta H_1=-296.0\ \text{kJ/mol})\\[1mm]
2\,SO_2 + O_2 &\to& 2\,SO_3 \quad\quad\quad\quad (\Delta H_2=-198.2\ \text{kJ/mol})
\end{array}
\][/tex]
Step 2. Write the combined reaction by adding the left-hand sides and the right-hand sides:
[tex]\[
\begin{array}{rcl}
\text{Left side:} && S_\omega + O_2 + 2\,SO_2 + O_2 = S_\omega + 2\,SO_2 + 2\,O_2, \\[1mm]
\text{Right side:} && SO_2 + 2\,SO_3.
\end{array}
\][/tex]
Step 3. Notice that [tex]$SO_2$[/tex] appears on both sides. Subtract one mole of [tex]$SO_2$[/tex] from both sides to simplify the overall reaction:
[tex]$$
S_\omega + SO_2 + 2\,O_2 \to 2\,SO_3.
$$[/tex]
Step 4. Since enthalpy is a state function, the enthalpy change for the overall reaction is the sum of the enthalpy changes for the individual steps:
[tex]$$
\Delta H = \Delta H_1 + \Delta H_2 = (-296.0) + (-198.2) = -494.2\ \text{kJ/mol}.
$$[/tex]
Thus, the enthalpy change for the reaction is
[tex]$$
\boxed{-494.2 \text{ kJ/mol}}
$$[/tex]
This is the final answer.