Question

Sulfur reacts with oxygen to form sulfur dioxide (SO2(g), Delta.Hf = –296.8 kJ/mol) according to the equation below.

Upper S (s) plus upper O subscript 2 (g) right arrow upper S upper o subscript 2 (g).

What is the enthalpy change for the reaction?
Use Delta H r x n equals the sum of delta H f of all the products minus the sum of delta H f of all the reactants..
–593.6 kJ
–296.8 kJ
296.8 kJ
593.6 kJ

Answer 1

To calculate the enthalpy change (\( \Delta H_r \)) for the reaction, we can use the given formula:

\[
\Delta H_r = \sum \Delta H_f \text{(products)} - \sum \Delta H_f \text{(reactants)}
\]

From the reaction provided:

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

We know:

- The standard enthalpy of formation of sulfur dioxide (\( \Delta H_f \) for \( \text{SO}_2 \)) = –296.8 kJ/mol.
- The standard enthalpy of formation of elemental sulfur (\( \text{S (s)} \)) is 0 kJ/mol (as it is in its standard state).
- The standard enthalpy of formation of elemental oxygen (\( \text{O}_2 \text{(g)} \)) is also 0 kJ/mol (as it is in its standard state).

Now we can substitute these values into the equation:

\[
\Delta H_r = [\Delta H_f \text{(SO}_2\text{)}] - [\Delta H_f \text{(S)} + \Delta H_f \text{(O}_2\text{)}]
\]

Substituting the values:

\[
\Delta H_r = [-296.8 \text{ kJ/mol}] - [0 + 0]
\]

\[
\Delta H_r = -296.8 \text{ kJ/mol}
\]

Thus, the enthalpy change for the reaction is:

\(-296.8\) kJ.

The correct answer from the provided options is:

**–296.8 kJ**.