The provided reactions are:1. \( \text{S(s)} + \frac{3}{2} \text{O}_2 \text{(g)} \to \text{SO}_3 \text{(g)} + 2x \text{ K} \)2. \( \text{SO}_2 \text{(g)} + \frac{1}{2} \text{O}_2 \text{(g)} \to \text{SO}_3 \text{(g)} + y \text{ K} \)We aim to calculate \( \Delta H \) for the reaction:\[\text{S(s)} + \text{O}_2 \text{(g)} \to \text{SO}_2 \text{(g)}\]Hess's Law, which states that the total enthalpy change of a reaction is the sum of the enthalpy changes of its constituent steps, will be applied. The second reaction is reversed to align with the target reaction, and its enthalpy change is adjusted accordingly.- Reversing the second reaction changes the sign of \( \Delta H \) and scales it: \[ \text{SO}_3 \text{(g)} \to \text{SO}_2 \text{(g)} + \frac{1}{2} \text{O}_2 \text{(g)}, \quad \Delta H = -y \text{ K} \]This reversed reaction is then added to the first reaction:\[\text{S(s)} + \frac{3}{2} \text{O}_2 \text{(g)} \to \text{SO}_3 \text{(g)} + 2x \text{ K}\]\[\text{SO}_3 \text{(g)} \to \text{SO}_2 \text{(g)} + \frac{1}{2} \text{O}_2 \text{(g)}, \quad \Delta H = -y \text{ K}\]The summation yields the following reaction:\[\text{S(s)} + \text{O}_2 \text{(g)} \to \text{SO}_2 \text{(g)}\]Consequently, the enthalpy change for this reaction is:\[\Delta H = 2x - (x + y) = - (x + y)\]
