Given the following reactions and their enthalpy changes:

1. \( \text{S (s)} + \text{O}_2 \text{(g)} \rightarrow \text{SO}_2 \text{(g)} \)
\( \Delta H^\circ = -296.1 \, \text{kJ/mol} \)

2. \( 2 \text{SO}_3 \text{(g)} \rightarrow 2 \text{SO}_2 \text{(g)} + \text{O}_2 \text{(g)} \)
\( \Delta H^\circ = +198.2 \, \text{kJ/mol} \)

Calculate the enthalpy change \( \Delta H^\circ \) for the reaction:

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

Options:
A. \(-394.0 \, \text{kJ}\)
B. \(-790.4 \, \text{kJ}\)
C. \(-97.9 \, \text{kJ}\)
D. \(+97.9 \, \text{kJ}\)

To find the value of δh° for the reaction 2 S (s) + 3 O2 (g) → 2 SO3 (g), we can use Hess's law. This law states that the enthalpy change of a reaction is independent of the pathway taken to reach the products and depends only on the initial and final states of the reaction.
We can use the given values of δh° for the reactions involving S, O2, SO2, and SO3 to calculate the δh° for the desired reaction.
First, we need to reverse the reaction 2 SO3 (g) → 2 SO2 (g) + O2 (g) and change the sign of its δh° value to obtain the correct stoichiometry.
2 SO3 (g) → 2 SO2 (g) + O2 (g) δh° = -198.2 kJ/mol (reversed and sign changed)
Next, we need to multiply the reaction by 2 to obtain the desired stoichiometry.
2 (2 SO3 (g) → 2 SO2 (g) + O2 (g)) δh° = -396.4 kJ/mol
Finally, we need to add the δh° values for the reactions involving S and O2 to obtain the δh° for the desired reaction.
2 S (s) + 3 O2 (g) → 2 SO3 (g) δh° = (-296.1 kJ/mol) + (-396.4 kJ/mol) = -692.5 kJ/mol.

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