Respuesta :
The addition of the enthalpies obtained gives the enthalpy of the target reaction which is;
ΔH°rxn = 1582.8 kJ
In this scenario, we need to manipulate the equations given;
The equations given are;
- SO2(g) → S(s) + O2(g) ΔH°rxn = +296.8 kJ
- 2SO2(g) + O2(g) → 2SO3(g) ΔH°rxn = -197.8 kJ
The target equation is;
- 4 SO3(g) → 4 S(s) + 6 O2(g) ΔH°rxn = ?
The first equation needs to be multiplied by 4;
so that the 4S(s) product in the target equation can be accounted for.
The second equation needs to be multiplied by 2 and multiplied by -1, so as to reverse the reactants and products.
Therefore, the reactions given become;
- 4SO2(g) → 4S(s) + 4O2(g) ΔH°rxn = +1187.2 kJ
- 4SO3(g) → 4SO2(g) + 2O2(g) ΔH°rxn = +395.6 kJ
Ultimately, the addition of both equations yields the target equation;
- 4 SO3(g) → 4 S(s) + 6 O2(g)
Therefore, the addition of the enthalpies obtained gives the enthalpy of the target reaction.
ΔH°rxn = 1582.8 kJ
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Enthalpy is defined as the sum of internal energy. The enthalpy of a system changes continuously and is measured by its heat content. The addition of enthalpies of the given reaction is [tex]\Delta \text{H}^0_{\text{rxn}}[/tex] = 1582.8 kJ.
The equations and their enthalpies are given as:
- [tex]\begin{aligned}\text{SO}_2\;\rightarrow\;\text{S + O}_2\;\;\;\;\;\Delta \text H^{0}_{\text{rxn}}&=+296.8kJ\\\text{SO}_2 + \text O_2\;\rightarrow\;\text{SO}_3\;\;\;\;\;\Delta \text H^{0}_{\text{rxn}}&=-197.8kJ\\\end{aligned}[/tex]
To find, change in enthalpy of the equation:
[tex]\text{4 SO}_3\;\rightarrow\;\text{4 S} + \text {6 O}_2 \;\;\;\;\;\;\Delta \text H ^0_{\text{rxn}}&=?[/tex]
Now, the equation will be multiplied by 4S to correct the reaction stoichiometrically. The second equation will be multiplied by 2 and -1.
Thus, the equation becomes:
- [tex]\begin{aligned}\text{4 SO}_2\;\rightarrow\;\text{4 S + 4 O}_2\;\;\;\;\;\Delta \text H^{0}_{\text{rxn}}&=+1187.2\text{kJ}\\\end{aligned}[/tex]
- [tex]\text{4 SO}_3\;\rightarrow\;\text{4 SO}_2 + \text {2 O}_2 \;\;\;\;\;\;\Delta \text H ^0_{\text{rxn}}&=+395.6\text{kJ}[/tex]
The addition of both equation will result in:
- [tex]\text{4 SO}_3\;\rightarrow\;\text{4 S} + \text {6 O}_2 \;\;\;\;\;\;\Delta \text H ^0_{\text{rxn}}&=1582.8 \text{kJ}[/tex]
Therefore, the enthalpy change for the given equation is 1582.8 kJ.
To know more about enthalpy change, refer to the following link:
https://brainly.com/question/15744112?referrer=searchResults
