You are provided with two precise thermochemical equations: \ $2\mathrm{S}_{(s)} + 3\mathrm{O}_{2(g

Question

You are provided with two precise thermochemical equations: \ $2\mathrm{S}_{(s)} + 3\mathrm{O}_{2(g)} ightarrow 2\mathrm{SO}_{3(g)} \quad \Delta H = -A 	ext{ kJ}$ \ $2\mathrm{SO}_{2(g)} + \mathrm{O}_{2(g)} ightarrow 2\mathrm{SO}_{3(g)} \quad \Delta H = -B 	ext{ kJ}$ \ the standard enthalpy of formation for one mole of Sulfur Dioxide ($\mathrm{SO}_{2(g)}$) based on above data is

Professor PC Thomas Classes · Answer

(a) : The fundamental target equation defining the standard enthalpy of formation for one single mole of $\mathrm{SO}_{2(g)}$ is: $\mathrm{S}_{(s)} + \mathrm{O}_{2(g)} \rightarrow \mathrm{SO}_{2(g)}$. We must systematically apply Hess's Law to reach this. Step 1: Divide both initial equations by 2. This yields: (1') $\mathrm{S} + \frac{3}{2}\mathrm{O}_2 \rightarrow \mathrm{SO}_3 \quad \Delta H = -A/2$. (2') $\mathrm{SO}_2 + \frac{1}{2}\mathrm{O}_2 \rightarrow \mathrm{SO}_3 \quad \Delta H = -B/2$. Step 2: Subtract equation (2') from equation (1') to eliminate $\mathrm{SO}_3$ and position $\mathrm{SO}_2$ as a product. The final resulting $\Delta H$ is $(-A/2) - (-B/2) = \frac{B - A}{2}$.