Determine the enthalpy for the reaction represented by the equation:

[tex]\[ SO_{2(g)} + \frac{1}{2} O_{2(g)} \rightarrow SO_{3(g)} \][/tex]

Use Hess' law along with the following equations:

[tex]\[
\begin{array}{ll}
S_{(s)} + O_{2(g)} \rightarrow SO_{2(g)} & \Delta H = -296.8 \, \text{kJ} \\
2 S_{(s)} + 3 O_{2(g)} \rightarrow 2 SO_{3(g)} & \Delta H = -795.45 \, \text{kJ}
\end{array}
\][/tex]



Answer :

To determine the enthalpy change [tex]\((\Delta H)\)[/tex] for the reaction:

[tex]\[ \text{SO}_{2(g)} + \frac{1}{2} \text{O}_{2(g)} \rightarrow \text{SO}_{3(g)} \][/tex]

using Hess's law, we can utilize the given reactions:

1.
[tex]\[ \text{S}_{(s)} + \text{O}_{2(g)} \rightarrow \text{SO}_{2(g)} \quad \Delta H = -296.8 \ \text{kJ} \][/tex]

2.
[tex]\[ 2\text{S}_{(s)} + 3\text{O}_{2(g)} \rightarrow 2\text{SO}_{3(g)} \quad \Delta H = -795.45 \ \text{kJ} \][/tex]

Let's proceed step by step:

### Step 1: Understand the Goal

We need to find the [tex]\(\Delta H\)[/tex] for the reaction:

[tex]\[ \text{SO}_{2(g)} + \frac{1}{2} \text{O}_{2(g)} \rightarrow \text{SO}_{3(g)} \][/tex]

### Step 2: Manipulate the Given Equations

Equation 1:

[tex]\[ \text{S}_{(s)} + \text{O}_{2(g)} \rightarrow \text{SO}_{2(g)} \quad \Delta H = -296.8 \ \text{kJ} \][/tex]

Equation 2 (let's express it in a way that involves only one mole of [tex]\(\text{SO}_{3(g)}\)[/tex]):

[tex]\[ 2\text{S}_{(s)} + 3\text{O}_{2(g)} \rightarrow 2\text{SO}_{3(g)} \][/tex]

Divide Equation 2 by 2 to get it per mole of [tex]\(\text{SO}_{3(g)}\)[/tex]:

[tex]\[ \text{S}_{(s)} + \frac{3}{2}\text{O}_{2(g)} \rightarrow \text{SO}_{3(g)} \][/tex]

Since we divided the reaction by 2, we must also divide [tex]\(\Delta H\)[/tex] by 2:

[tex]\[ \Delta H = \frac{-795.45 \ \text{kJ}}{2} = -397.725 \ \text{kJ} \][/tex]

### Step 3: Construct the Target Equation

Now, we need to write the target reaction in terms of the modified equations:

[tex]\[ \text{SO}_{2(g)} + \frac{1}{2}\text{O}_{2(g)} \rightarrow \text{SO}_{3(g)} \][/tex]

Notice that:

- From the modified second equation, we have:
[tex]\[ \text{S}_{(s)} + \frac{3}{2}\text{O}_{2(g)} \rightarrow \text{SO}_{3(g)} \][/tex]

- And from the first equation, we know:
[tex]\[ \text{S}_{(s)} + \text{O}_{2(g)} \rightarrow \text{SO}_{2(g)} \][/tex]

By reversing and subtracting the first equation from the modified second equation, we get our target reaction:

[tex]\[ (\text{S}_{(s)} + \frac{3}{2} \text{O}_{2(g)} \rightarrow \text{SO}_{3(g)}) - (\text{S}_{(s)} + \text{O}_{2(g)} \rightarrow \text{SO}_{2(g)}) \][/tex]

This simplifies to:

[tex]\[ \text{SO}_{2(g)} + \frac{1}{2} \text{O}_{2(g)} \rightarrow \text{SO}_{3(g)} \][/tex]

### Step 4: Calculate the Enthalpy Change

The enthalpy change for the reaction will be:

[tex]\[ \Delta H_{\text{reaction}} = -397.725 \ \text{kJ} - (-296.8 \ \text{kJ}) \][/tex]

[tex]\[ \Delta H_{\text{reaction}} = -397.725 \ \text{kJ} + 296.8 \ \text{kJ} \][/tex]

[tex]\[ \Delta H_{\text{reaction}} = -100.925 \ \text{kJ} \][/tex]

### Conclusion

The enthalpy change for the reaction:

[tex]\[ \text{SO}_{2(g)} + \frac{1}{2}\text{O}_{2(g)} \rightarrow \text{SO}_{3(g)} \][/tex]

is [tex]\(\Delta H = -100.925 \ \text{kJ}\)[/tex].