Answer :
We are given the following reference reactions:
[tex]$$
\text{Reaction 1:} \quad S(s) + O_2(g) \rightarrow SO_2(g) \quad \Delta H_1 = -296.0\ \text{kJ/mol},
$$[/tex]
[tex]$$
\text{Reaction 2:} \quad 2\,SO_2(g) + O_2(g) \rightarrow 2\,SO_3(g) \quad \Delta H_2 = -198.2\ \text{kJ/mol}.
$$[/tex]
We need to calculate the enthalpy change for the reaction:
[tex]$$
S(s) + 2\,SO_3(g) \rightarrow 3\,SO_2(g) \quad \Delta H = ?
$$[/tex]
Step 1. Reverse Reaction 2
To obtain [tex]$2\,SO_3(g)$[/tex] on the left, we reverse Reaction 2. When a reaction is reversed, the sign of [tex]$\Delta H$[/tex] is also reversed. Thus, the reversed Reaction 2 becomes
[tex]$$
2\,SO_3(g) \rightarrow 2\,SO_2(g) + O_2(g) \quad \Delta H_{\text{rev2}} = +198.2\ \text{kJ/mol}.
$$[/tex]
Step 2. Add Reaction 1 and the Reversed Reaction 2
Now, write down the two reactions you will combine:
1. Reaction 1:
[tex]$$
S(s) + O_2(g) \rightarrow SO_2(g) \quad \Delta H_1 = -296.0\ \text{kJ/mol}
$$[/tex]
2. Reversed Reaction 2:
[tex]$$
2\,SO_3(g) \rightarrow 2\,SO_2(g) + O_2(g) \quad \Delta H_{\text{rev2}} = +198.2\ \text{kJ/mol}
$$[/tex]
Add these reactions together:
[tex]$$
\begin{array}{rcl}
S(s) + O_2(g) &\rightarrow& SO_2(g) \\
2\,SO_3(g) &\rightarrow& 2\,SO_2(g) + O_2(g)
\end{array}
$$[/tex]
When we add the left-hand sides and right-hand sides, we get:
[tex]$$
S(s) + O_2(g) + 2\,SO_3(g) \rightarrow SO_2(g) + 2\,SO_2(g) + O_2(g).
$$[/tex]
Step 3. Simplify the Overall Reaction
Notice that [tex]$O_2(g)$[/tex] appears on both sides and cancels out:
[tex]$$
S(s) + 2\,SO_3(g) \rightarrow 3\,SO_2(g).
$$[/tex]
This is exactly the reaction for which we want to find [tex]$\Delta H$[/tex].
Step 4. Calculate the Overall Enthalpy Change
The overall enthalpy change is the sum of the enthalpy changes of the individual steps:
[tex]$$
\Delta H = \Delta H_1 + \Delta H_{\text{rev2}}.
$$[/tex]
Substitute the values:
[tex]$$
\Delta H = (-296.0\ \text{kJ/mol}) + (198.2\ \text{kJ/mol}) = -97.8\ \text{kJ/mol}.
$$[/tex]
Thus, the enthalpy change for the reaction
[tex]$$
S(s) + 2\,SO_3(g) \rightarrow 3\,SO_2(g)
$$[/tex]
is
[tex]$$
\Delta H = -97.8\ \text{kJ/mol}.
$$[/tex]
[tex]$$
\text{Reaction 1:} \quad S(s) + O_2(g) \rightarrow SO_2(g) \quad \Delta H_1 = -296.0\ \text{kJ/mol},
$$[/tex]
[tex]$$
\text{Reaction 2:} \quad 2\,SO_2(g) + O_2(g) \rightarrow 2\,SO_3(g) \quad \Delta H_2 = -198.2\ \text{kJ/mol}.
$$[/tex]
We need to calculate the enthalpy change for the reaction:
[tex]$$
S(s) + 2\,SO_3(g) \rightarrow 3\,SO_2(g) \quad \Delta H = ?
$$[/tex]
Step 1. Reverse Reaction 2
To obtain [tex]$2\,SO_3(g)$[/tex] on the left, we reverse Reaction 2. When a reaction is reversed, the sign of [tex]$\Delta H$[/tex] is also reversed. Thus, the reversed Reaction 2 becomes
[tex]$$
2\,SO_3(g) \rightarrow 2\,SO_2(g) + O_2(g) \quad \Delta H_{\text{rev2}} = +198.2\ \text{kJ/mol}.
$$[/tex]
Step 2. Add Reaction 1 and the Reversed Reaction 2
Now, write down the two reactions you will combine:
1. Reaction 1:
[tex]$$
S(s) + O_2(g) \rightarrow SO_2(g) \quad \Delta H_1 = -296.0\ \text{kJ/mol}
$$[/tex]
2. Reversed Reaction 2:
[tex]$$
2\,SO_3(g) \rightarrow 2\,SO_2(g) + O_2(g) \quad \Delta H_{\text{rev2}} = +198.2\ \text{kJ/mol}
$$[/tex]
Add these reactions together:
[tex]$$
\begin{array}{rcl}
S(s) + O_2(g) &\rightarrow& SO_2(g) \\
2\,SO_3(g) &\rightarrow& 2\,SO_2(g) + O_2(g)
\end{array}
$$[/tex]
When we add the left-hand sides and right-hand sides, we get:
[tex]$$
S(s) + O_2(g) + 2\,SO_3(g) \rightarrow SO_2(g) + 2\,SO_2(g) + O_2(g).
$$[/tex]
Step 3. Simplify the Overall Reaction
Notice that [tex]$O_2(g)$[/tex] appears on both sides and cancels out:
[tex]$$
S(s) + 2\,SO_3(g) \rightarrow 3\,SO_2(g).
$$[/tex]
This is exactly the reaction for which we want to find [tex]$\Delta H$[/tex].
Step 4. Calculate the Overall Enthalpy Change
The overall enthalpy change is the sum of the enthalpy changes of the individual steps:
[tex]$$
\Delta H = \Delta H_1 + \Delta H_{\text{rev2}}.
$$[/tex]
Substitute the values:
[tex]$$
\Delta H = (-296.0\ \text{kJ/mol}) + (198.2\ \text{kJ/mol}) = -97.8\ \text{kJ/mol}.
$$[/tex]
Thus, the enthalpy change for the reaction
[tex]$$
S(s) + 2\,SO_3(g) \rightarrow 3\,SO_2(g)
$$[/tex]
is
[tex]$$
\Delta H = -97.8\ \text{kJ/mol}.
$$[/tex]