To find the enthalpy of formation of SO₂ at 298 K, we will rearrange the given reactions and their enthalpy changes. The goal is to derive the formation reaction of SO₂ from its elements in their standard states.
### Step-by-Step Solution:
1. **Identify the Target Reaction**:
The target reaction for the formation of SO₂ from its elements is:
\[
S + O_2 \rightarrow SO_2
\]
2. **List the Given Reactions**:
- Reaction 1: \( SO_2 + \frac{1}{2} O_2 \rightarrow SO_3, \Delta H = -98.7 \, \text{kJ} \)
- Reaction 2: \( SO_3 + H_2O \rightarrow H_2SO_4, \Delta H = -130.2 \, \text{kJ} \)
- Reaction 3: \( H_2 + \frac{1}{2} O_2 \rightarrow H_2O, \Delta H = -287.3 \, \text{kJ} \)
- Reaction 4: \( S + H_2 + 2O_2 \rightarrow H_2SO_4, \Delta H = -814.4 \, \text{kJ} \)
3. **Reverse the Necessary Reactions**:
- Reverse Reaction 1 to get SO₂ as a product:
\[
SO_3 \rightarrow SO_2 + \frac{1}{2} O_2, \Delta H = +98.7 \, \text{kJ}
\]
- Reverse Reaction 2 to remove H₂SO₄:
\[
H_2SO_4 \rightarrow SO_3 + H_2O, \Delta H = +130.2 \, \text{kJ}
\]
- Reverse Reaction 3 to remove H₂O:
\[
H_2O \rightarrow H_2 + \frac{1}{2} O_2, \Delta H = +287.3 \, \text{kJ}
\]
4. **Keep Reaction 4 as is**:
\[
S + H_2 + 2O_2 \rightarrow H_2SO_4, \Delta H = -814.4 \, \text{kJ}
\]
5. **Combine the Reactions**:
Now, we will add the modified reactions:
- From Reaction 4: \( S + H_2 + 2O_2 \rightarrow H_2SO_4 \)
- From the reversed Reaction 2: \( H_2SO_4 \rightarrow SO_3 + H_2O \)
- From the reversed Reaction 1: \( SO_3 \rightarrow SO_2 + \frac{1}{2} O_2 \)
- From the reversed Reaction 3: \( H_2O \rightarrow H_2 + \frac{1}{2} O_2 \)
When combined, the H₂SO₄ cancels out, and we are left with:
\[
S + O_2 \rightarrow SO_2
\]
6. **Calculate the Total Enthalpy Change**:
Now, we will sum the enthalpy changes:
\[
\Delta H = (-814.4) + 130.2 + 98.7 + 287.3
\]
\[
\Delta H = -298.2 \, \text{kJ}
\]
### Final Answer:
The enthalpy of formation of SO₂ at 298 K is:
\[
\Delta H_f(SO_2) = -298.2 \, \text{kJ}
\]
