Question

sulfur reacts with oxygen to form sulfur dioxide ( \text{so}_2 (g) ), ( delta h_f = -296.8 , \text{kj/mol} ) according to the equation: ( \text{s}(s) + \text{o}_2(g)
ightarrow \text{so}_2(g) ) what is the enthalpy change for the reaction? use ( delta h_{\text{rxn}} = sum (delta h_{f, \text{products}}) - sum (delta h_{f, \text{reactants}}) ). options: 296.8 kj, 583.6 kj, -296.8 kj, -583.6 kj

Explanation:

Step1: Identify reactants and products

Reactants: \( S(s) \) and \( O_2(g) \); Product: \( SO_2(g) \).

Step2: Recall standard enthalpy of formation

Standard enthalpy of formation (\( \Delta H_f \)) for elements in their standard state (like \( S(s) \), \( O_2(g) \)) is \( 0 \, \text{kJ/mol} \). \( \Delta H_f(SO_2(g)) = -296.8 \, \text{kJ/mol} \).

Step3: Apply the formula \( \Delta H_{rxn} = \sum (\Delta H_{f, products}) - \sum (\Delta H_{f, reactants}) \)

\( \sum (\Delta H_{f, products}) = \Delta H_f(SO_2(g)) = -296.8 \, \text{kJ/mol} \) (since 1 mole of \( SO_2 \) is formed).
\( \sum (\Delta H_{f, reactants}) = \Delta H_f(S(s)) + \Delta H_f(O_2(g)) = 0 + 0 = 0 \, \text{kJ/mol} \).
So, \( \Delta H_{rxn} = (-296.8) - 0 = -296.8 \, \text{kJ} \).

Answer:

\(-296.8 \, \text{kJ}\) (the option with this value)