The values of heat of formation of `SO_2` and `SO_3` and `-298.2 kJ` and `-98.2`kJ. The heat of reaction of the following reaction will be: `SO_2 + 1/2 O_2 to SO_3`,___________.

Options

A. `-200`kJ
B. `-356.2`kJ
C. `+200`kJ
D. `-396.2`kJ

Answer

To find the heat of reaction for the given reaction \( \text{SO}_2 + \frac{1}{2} \text{O}_2 \rightarrow \text{SO}_3 \), we will use the standard enthalpy of formation values provided for \( \text{SO}_2 \) and \( \text{SO}_3 \).

### Step-by-Step Solution:

1. **Identify the given data:**
- Heat of formation of \( \text{SO}_2 \) (\( \Delta H_f \text{SO}_2 \)) = -298.2 kJ/mol
- Heat of formation of \( \text{SO}_3 \) (\( \Delta H_f \text{SO}_3 \)) = -98.2 kJ/mol
- Heat of formation of \( \text{O}_2 \) (\( \Delta H_f \text{O}_2 \)) = 0 kJ/mol (as it is in its standard state)

2. **Write the formula for the heat of reaction:**
The heat of reaction (\( \Delta H_r \)) can be calculated using the formula:
\[ \Delta H_r = \sum (\Delta H_f \text{ of products}) - \sum (\Delta H_f \text{ of reactants}) \]

3. **Apply the formula to the given reaction:**
- For the products, we have \( \text{SO}_3 \):
\[ \Delta H_f \text{ of products} = \Delta H_f \text{SO}_3 = -98.2 \text{ kJ/mol} \]
- For the reactants, we have \( \text{SO}_2 \) and \( \frac{1}{2} \text{O}_2 \):
\[ \Delta H_f \text{ of reactants} = \Delta H_f \text{SO}_2 + \Delta H_f \left(\frac{1}{2} \text{O}_2\right) = -298.2 \text{ kJ/mol} + 0 \text{ kJ/mol} = -298.2 \text{ kJ/mol} \]

4. **Calculate \( \Delta H_r \):**
\[ \Delta H_r = (-98.2 \text{ kJ/mol}) - (-298.2 \text{ kJ/mol}) \]
\[ \Delta H_r = -98.2 + 298.2 = 200 \text{ kJ/mol} \]

5. **Final answer:**
The heat of reaction for the reaction \( \text{SO}_2 + \frac{1}{2} \text{O}_2 \rightarrow \text{SO}_3 \) is \( 200 \text{ kJ} \).