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].
[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].
