Discover the best answers to your questions with the help of IDNLearn.com. Find the answers you need quickly and accurately with help from our knowledgeable and experienced experts.
Sagot :
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].
Thank you for joining our conversation. Don't hesitate to return anytime to find answers to your questions. Let's continue sharing knowledge and experiences! Find clear and concise answers at IDNLearn.com. Thanks for stopping by, and come back for more dependable solutions.
