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\[
\begin{array}{|l|l|}
\hline
\text{Substance} & \Delta H_f^{\circ} (\text{kJ/mol}) \\
\hline
SO_2(g) & -296.1 \\
\hline
SO_3(g) & -395.2 \\
\hline
O_2(g) & 0.0 \\
\hline
\end{array}
\]

Based on the information above, what is [tex]\(\Delta H^{\circ}\)[/tex] for the reaction [tex]\(SO_2(g) + \frac{1}{2} O_2(g) \rightarrow SO_3(g)\)[/tex]?

A. [tex]\(-591.3 \, \text{kJ/mol}\)[/tex]

B. [tex]\(-99.1 \, \text{kJ/mol}\)[/tex]

C. [tex]\(+99.1 \, \text{kJ/mol}\)[/tex]

D. [tex]\(+591.3 \, \text{kJ/mol}\)[/tex]

Answer :

Sure! Let's find the enthalpy change, [tex]\(\Delta H^{\circ}\)[/tex], for the given reaction:

[tex]\[ SO_2(g) + \frac{1}{2} O_2(g) \rightarrow SO_3(g) \][/tex]

We can calculate the enthalpy change for the reaction using the standard enthalpies of formation ([tex]\(\Delta H_f^{\circ}\)[/tex]) for the reactants and products. The formula to use is:

[tex]\[
\Delta H^{\circ}_{\text{reaction}} = \sum \Delta H_f^{\circ} (\text{products}) - \sum \Delta H_f^{\circ} (\text{reactants})
\][/tex]

From the table, we have:

- Enthalpy of formation for [tex]\(SO_2(g)\)[/tex], [tex]\(\Delta H_f^{\circ}(SO_2)\)[/tex] = -296.1 kJ/mol
- Enthalpy of formation for [tex]\(SO_3(g)\)[/tex], [tex]\(\Delta H_f^{\circ}(SO_3)\)[/tex] = -395.2 kJ/mol
- Enthalpy of formation for [tex]\(O_2(g)\)[/tex], [tex]\(\Delta H_f^{\circ}(O_2)\)[/tex] = 0.0 kJ/mol (since it is a pure element in its standard state)

Now let's calculate [tex]\(\Delta H^{\circ}_{\text{reaction}}\)[/tex]:

1. Calculate the sum of the enthalpy of formation for the products:
- Product: [tex]\(SO_3(g)\)[/tex]
- [tex]\(\Delta H_f^{\circ}(SO_3) = -395.2\)[/tex] kJ/mol

2. Calculate the sum of the enthalpy of formation for the reactants:
- Reactants: [tex]\(SO_2(g)\)[/tex] and [tex]\(\frac{1}{2} O_2(g)\)[/tex]
- [tex]\(\Delta H_f^{\circ}(SO_2) + 0.5 \times \Delta H_f^{\circ}(O_2) = -296.1 + 0.5 \times 0.0 = -296.1\)[/tex] kJ/mol

3. Apply the formula:

[tex]\[
\Delta H^{\circ}_{\text{reaction}} = (-395.2) - (-296.1) = -395.2 + 296.1 = -99.1 \, \text{kJ/mol}
\][/tex]

So, the enthalpy change for the reaction is [tex]\(-99.1 \, \text{kJ/mol}\)[/tex].

Therefore, the correct answer is (B) [tex]\(-99.1 \, \text{kJ/mol}_{\text{r n}}\)[/tex].

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