At 298 K, the following heats of reaction are known:

1. $ \text{S (s) + O}_2 \text{(g)} \rightarrow \text{SO}_2 \text{(g)} $
2. $ 2 \text{SO}_3 \text{(g)} \rightarrow 2 \text{SO}_2 \text{(g)} + \text{O}_2 \text{(g)} $

Calculate ΔH for the reaction:

$ 2 \text{S (s)} + 3 \text{O}_2 \text{(g)} \rightarrow 2 \text{SO}_3 \text{(g)} $

Given options for ΔH:
A. $ -97.9 \, \text{kJ} $
B. $ -394.0 \, \text{kJ} $
C. $ +97.9 \, \text{kJ} $
D. $ -790.4 \, \text{kJ} $
E. $ \Delta H = -296.1 \, \text{kJ} $
F. $ \Delta H = 198.2 \, \text{kJ} $

The enthalpy change (ΔH) for the reaction [tex]2 S (s) + 3 O_2 (g) → 2 SO_3[/tex] (g) is calculated as -792.2 kJ using Hess's Law which states that enthalpy change is same irrespective of the path.

In this problem, we are using the concept of Hess's Law, which is a significant concept in thermochemical calculations in chemistry. According to Hess's Law, the enthalpy change (ΔH) for a reaction is the same whether it takes place in one step or several steps. Therefore, we can calculate ΔH for the reaction [tex]2 S (s) + 3 O_2 (g) → 2 SO_3[/tex] (g) by adding the ΔH of the given reactions that add up to this reaction.

The first given reaction is: [tex]S (s) + O_2 (g) → SO_2[/tex](g) ΔH = -297.0 kJ. But we require 2 S (s), hence we multiply this equation by 2: [tex]2 S (s) + 2 O_2 (g) → 2 SO_2[/tex](g) ΔH = -2*(-297.0) = -594.0 kJ

The second given reaction is: [tex]2 SO_2(g) + O_2 (g) → 2 SO_3[/tex](g) ΔH = -198.2 kJ. These two equations on adding give us: [tex]2 S (s) + 3 O_2 (g) → 2 SO_3(g)[/tex], and adding the ΔH gives: ΔH = -594.0 kJ -198.2 kJ = -792.2 kJ

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