To find the enthalpy change for the reaction, we shall use the given enthalpy of formation values and apply the formula:
[tex]\[
\Delta H_{rxn} = \sum \Delta H_{f,\text{products}} - \sum \Delta H_{f,\text{reactants}}
\][/tex]
Let's identify the relevant enthalpy of formation values given in the problem:
- The enthalpy of formation for [tex]\( \text{SO}_2(g) \)[/tex] is [tex]\( \Delta H_f = -296.8 \ \text{kJ/mol} \)[/tex].
- The enthalpy of formation for [tex]\( \text{S(s)} \)[/tex] and [tex]\( \text{O}_2(g) \)[/tex] are both [tex]\( 0 \ \text{kJ/mol} \)[/tex] because they are elements in their standard states.
Now, let's write the reaction:
[tex]\[
\text{S(s)} + \text{O}_2(g) \rightarrow \text{SO}_2(g)
\][/tex]
We calculate the sum of the enthalpies of formation for the reactants and products:
- For the products:
[tex]\[
\sum \Delta H_{f,\text{products}} = \Delta H_f(\text{SO}_2(g)) = -296.8 \ \text{kJ/mol}
\][/tex]
- For the reactants:
[tex]\[
\sum \Delta H_{f,\text{reactants}} = \Delta H_f(\text{S(s)}) + \Delta H_f(\text{O}_2(g)) = 0 \ \text{kJ/mol} + 0 \ \text{kJ/mol} = 0 \ \text{kJ/mol}
\][/tex]
Now, subtract the sum of the reactants' enthalpies from the sum of the products' enthalpies:
[tex]\[
\Delta H_{rxn} = \sum \Delta H_{f,\text{products}} - \sum \Delta H_{f,\text{reactants}} = -296.8 \ \text{kJ/mol} - 0 \ \text{kJ/mol} = -296.8 \ \text{kJ/mol}
\][/tex]
So, the enthalpy change for the reaction is:
[tex]\[
\Delta H_{rxn} = -296.8 \ \text{kJ/mol}
\][/tex]
Thus, the correct answer is [tex]\( \boxed{-296.8 \ \text{kJ}} \)[/tex].
