If `{:(SO_(2)+1/2O_(2) rarr SO_(3),DeltaH = -98.7 kJ),(SO_(3)+H_2O rarr H_(2)SO_(4),DeltaH = -130.2 kJ),(H_(2)+1/2O_(2) rarr H_2O , Delta H = -287.3 kJ),(S + H_2 + 2O_(2) rarr H_2SO_4, DeltaH = -814.4 kJ):}`
Then enthaply of `SO_(2)` at 298 K is:

Options

A. `-298.2 kJ`
B. `-650 .3 kJ`
C. `-320 kJ`
D. `-233.5 kJ`

Explanation

To find the enthalpy of formation of SO₂ at 298 K, we will use Hess's law and the given reactions. We need to manipulate the reactions to derive the formation reaction of SO₂ from its constituent elements (sulfur and oxygen).

### Step-by-Step Solution:

1. **List the Given Reactions:**
- Reaction 1: \( SO_2 + \frac{1}{2} O_2 \rightarrow SO_3, \Delta H_1 = -98.7 \, \text{kJ} \)
- Reaction 2: \( SO_3 + H_2O \rightarrow H_2SO_4, \Delta H_2 = -130.2 \, \text{kJ} \)
- Reaction 3: \( H_2 + \frac{1}{2} O_2 \rightarrow H_2O, \Delta H_3 = -287.3 \, \text{kJ} \)
- Reaction 4: \( S + H_2 + 2O_2 \rightarrow H_2SO_4, \Delta H_4 = -814.4 \, \text{kJ} \)

2. **Reverse Reaction 1:**
- We need \( SO_2 \) on the product side.
- Reversed Reaction 1: \( SO_3 \rightarrow SO_2 + \frac{1}{2} O_2, \Delta H = +98.7 \, \text{kJ} \)

3. **Use Reaction 4 as it is:**
- \( S + H_2 + 2O_2 \rightarrow H_2SO_4, \Delta H_4 = -814.4 \, \text{kJ} \)

4. **Reverse Reaction 2:**
- We want to cancel out \( H_2SO_4 \).
- Reversed Reaction 2: \( H_2SO_4 \rightarrow SO_3 + H_2O, \Delta H = +130.2 \, \text{kJ} \)

5. **Reverse Reaction 3:**
- We want to cancel out \( H_2O \).
- Reversed Reaction 3: \( H_2O \rightarrow H_2 + \frac{1}{2} O_2, \Delta H = +287.3 \, \text{kJ} \)

6. **Add the Reactions:**
- Now, we add the modified reactions:
- \( SO_3 \rightarrow SO_2 + \frac{1}{2} O_2 \) (from step 2)
- \( S + H_2 + 2O_2 \rightarrow H_2SO_4 \) (from step 3)
- \( H_2SO_4 \rightarrow SO_3 + H_2O \) (from step 4)
- \( H_2O \rightarrow H_2 + \frac{1}{2} O_2 \) (from step 5)

- After cancellation:
- The overall reaction simplifies to:
\[ S + \frac{1}{2} O_2 \rightarrow SO_2 \]

7. **Calculate the Enthalpy Change:**
- The total enthalpy change is:
\[ \Delta H = \Delta H_1 + \Delta H_2 + \Delta H_3 + \Delta H_4 \]
- Substituting the values:
\[ \Delta H = (-814.4) + (130.2) + (98.7) + (287.3) \]
- Calculating:
\[ \Delta H = -814.4 + 130.2 + 98.7 + 287.3 = -298.2 \, \text{kJ} \]

### Final Answer:
The enthalpy of formation of \( SO_2 \) at 298 K is \( -298.2 \, \text{kJ} \).