Question

Calculate heat of formation of \(\text{SO}_2\) from following equations.
\( \text{S} + \frac{3}{2}\text{O}_2 \longrightarrow \text{SO}_3, \Delta \text{H} = -2\text{x kJ} \)
\( \text{SO}_2 + \frac{1}{2}\text{O}_2 \longrightarrow \text{SO}_3, \Delta \text{H} = -\text{y kJ} \)

• A. \(\text{y} - 2\text{x}\)
• B. \(2\text{x} - \text{y}\)
• C. \(2\text{x} + \text{y}\)
• D. \(\frac{2\text{x}}{\text{y}}\)

Hint:

Hess law: Reverse → sign changes, then add equations.

Answer 1

The Correct Option is B

Concept:
Apply Hess’s law.

Step 1: Write required reaction. \[ \text{S} + \text{O}_2 \rightarrow \text{SO}_2 \]

Step 2: Manipulate equations. Given: \[ (1)\quad \text{S} + \frac{3}{2}O_2 \rightarrow SO_3 \quad (\Delta H = -2x) \] \[ (2)\quad SO_2 + \frac{1}{2}O_2 \rightarrow SO_3 \quad (\Delta H = -y) \] Reverse (2): \[ SO_3 \rightarrow SO_2 + \frac{1}{2}O_2 \quad (\Delta H = +y) \]

Step 3: Add equations. \[ \text{S} + \frac{3}{2}O_2 \rightarrow SO_3 \] \[ SO_3 \rightarrow SO_2 + \frac{1}{2}O_2 \] Cancel \(SO_3\): \[ \text{S} + O_2 \rightarrow SO_2 \]

Step 4: Add enthalpies. \[ \Delta H = -2x + y = -(2x - y) \] Thus heat of formation: \[ = 2x - y \]

Step 5: Conclusion. \[ \text{Correct answer = } 2x - y \]