Question

Use standard enthalpies of formation to calculate ΔH°rxn for the following reaction: SO₂(g) + ½ O₂(g) → SO₃(g)

Answer 1

ΔH°rxn for the following reaction: SO₂(g) + ½ O₂(g) → SO₃(g) is:

`\[ \Delta H^\circ_(rxn) = -99.1 \, \text{kJ/mol} \]`

Step-by-step explanation:

The standard enthalpies of formation
`(\( \Delta H^\circ_f \))` for each compound involved in the reaction were used to calculate
`\( \Delta H^\circ_(rxn) \). For SO₂(g), O₂(g), and SO₃(g), the \( \Delta H^\circ_f \)` values are -296.8 kJ/mol, 0 kJ/mol, and -395.7 kJ/mol, respectively.

The given reaction is:

`\[ \text{SO₂(g)} + (1)/(2) \text{O₂(g)} \rightarrow \text{SO₃(g)} \]`

The \( \Delta H^\circ_{rxn} \) is calculated using the formula:

`\[ \Delta H^\circ_(rxn) = \sum \Delta H^\circ_f \text{(products)} - \sum \Delta H^\circ_f \text{(reactants)} \]`

Substituting the
`\( \Delta H^\circ_f \)`values:

`\[ \Delta H^\circ_(rxn) = [-395.7 \, \text{kJ/mol}] - \left[(-296.8 \, \text{kJ/mol} + (1)/(2) * 0 \, \text{kJ/mol})\right] \]`

`\[ \Delta H^\circ_(rxn) = -99.1 \, \text{kJ/mol} \]`

The negative sign indicates that the reaction is exothermic, releasing 99.1 kJ of heat per mole of the reaction. This implies that the reaction releases energy to the surroundings, contributing to a decrease in enthalpy.