$$S(g) + \frac{3}{2}O_2(g) \rightarrow SO_3(g) + 2x\,\text{kcal} \\SO_2(g) + \frac{1}{2}O_2(g) \rightarrow SO_3(g) + y\,\text{kcal}\\\text{The heat of formation of } SO_2(g) \text{ is given by:}$$

We are given two thermochemical equations and need to find the heat of formation of $$SO_2(g)$$.

Reaction 1: $$S(g) + \frac{3}{2}O_2(g) \rightarrow SO_3(g) + 2x \text{ kcal}$$ ... (i)

Reaction 2: $$SO_2(g) + \frac{1}{2}O_2(g) \rightarrow SO_3(g) + y \text{ kcal}$$ ... (ii)

The heat of formation of $$SO_2(g)$$ corresponds to:

$$S(g) + O_2(g) \rightarrow SO_2(g) + \Delta H_f$$

To obtain the target reaction, subtract Reaction (ii) from Reaction (i):

$$[S(g) + \frac{3}{2}O_2(g) \rightarrow SO_3(g)] - [SO_2(g) + \frac{1}{2}O_2(g) \rightarrow SO_3(g)]$$

This gives:

$$S(g) + \frac{3}{2}O_2(g) - SO_2(g) - \frac{1}{2}O_2(g) \rightarrow SO_3(g) - SO_3(g)$$

$$S(g) + O_2(g) \rightarrow SO_2(g)$$

The enthalpy change for the target reaction is:

$$\Delta H_f = (-2x) - (-y) = -2x + y = y - 2x \text{ kcal}$$

(Note: the heat released in Reaction 1 is $$2x$$ kcal, meaning $$\Delta H_1 = -2x$$ kcal, and similarly $$\Delta H_2 = -y$$ kcal.)

The heat of formation of $$SO_2(g)$$ is $$(y - 2x)$$ kcal.

The correct answer is Option 2: $$y - 2x$$ kcal.