Answered

Discover answers to your most pressing questions at Westonci.ca, the ultimate Q&A platform that connects you with expert solutions. Our platform provides a seamless experience for finding reliable answers from a knowledgeable network of professionals. Get detailed and accurate answers to your questions from a dedicated community of experts on our Q&A platform.

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]

Sagot :

GinnyAnswer
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].
Thank you for your visit. We are dedicated to helping you find the information you need, whenever you need it. We hope you found what you were looking for. Feel free to revisit us for more answers and updated information. We're dedicated to helping you find the answers you need at Westonci.ca. Don't hesitate to return for more.