Question

Consider the following reaction:
\[ \text{S(s)} + \frac{3}{2} \text{O}_2 \text{(g)} \to \text{SO}_3 \text{(g)} + 2x \text{ K} \] \[ \text{SO}_2 \text{(g)} + \frac{1}{2} \text{O}_2 \text{(g)} \to \text{SO}_3 \text{(g)} + y \text{ K} \] Calculate \( \Delta H \) for the following reaction (K):
\[ \text{S(s)} + \text{O}_2 \text{(g)} \to \text{SO}_2 \text{(g)} \]

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

Hint:

When using Hess's Law, carefully reverse reactions and adjust their enthalpy changes to match the desired overall reaction.

Answer 1

The Correct Option is A

The given 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 need to calculate \( \Delta H \) for the reaction: \[ \text{S(s)} + \text{O}_2 \text{(g)} \to \text{SO}_2 \text{(g)} \] To do this, we use Hess's Law, which states that the enthalpy change for the overall reaction is the sum of the enthalpy changes for the individual reactions. We reverse the second reaction to match the target reaction and adjust the coefficients accordingly. - Reverse the second reaction and change the sign of \( \Delta H \) and scale it: \[ \text{SO}_3 \text{(g)} \to \text{SO}_2 \text{(g)} + \frac{1}{2} \text{O}_2 \text{(g)}, \quad \Delta H = -y \text{ K} \] Now, add this reversed reaction 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 resulting reaction is: \[ \text{S(s)} + \text{O}_2 \text{(g)} \to \text{SO}_2 \text{(g)} \] Thus, the enthalpy change for the reaction is: \[ \Delta H = 2x - (x + y) = - (x + y) \]