Welcome to Westonci.ca, where curiosity meets expertise. Ask any question and receive fast, accurate answers from our knowledgeable community. Our Q&A platform offers a seamless experience for finding reliable answers from experts in various disciplines. Experience the convenience of finding accurate answers to your questions from knowledgeable experts on our platform.
Sagot :
To find the standard enthalpy change [tex]\(\Delta H^{\circ}_{\text{rxn}}\)[/tex] for the given reaction:
[tex]\[ 6 \text{Cl}_2(g) + 2 \text{Fe}_2\text{O}_3(s) \rightarrow 4 \text{FeCl}_3(s) + 3 \text{O}_2(g), \][/tex]
we need to use the provided standard enthalpy of formation ([tex]\(\Delta H_{f}^{\circ}\)[/tex]) values for each substance involved.
### Step-by-Step Solution:
1. Identify the enthalpy of formation ([tex]\(\Delta H_{f}^{\circ}\)[/tex]) for each substance:
[tex]\[ \begin{aligned} &\text{Cl}_2(g) & : &\quad 0.0 \, \text{kJ/mol} \\ &\text{Fe}_2\text{O}_3(s) & : &\quad -824.2 \, \text{kJ/mol} \\ &\text{FeCl}_3(s) & : &\quad -399.5 \, \text{kJ/mol} \\ &\text{O}_2(g) & : &\quad 0.0 \, \text{kJ/mol} \\ \end{aligned} \][/tex]
2. Use the stoichiometric coefficients from the balanced chemical equation:
[tex]\[ \begin{aligned} &\text{Cl}_2(g) & : &\quad 6 \\ &\text{Fe}_2\text{O}_3(s) & : &\quad 2 \\ &\text{FeCl}_3(s) & : &\quad 4 \\ &\text{O}_2(g) & : &\quad 3 \\ \end{aligned} \][/tex]
3. Calculate the total enthalpy of the reactants:
[tex]\[ \Delta H_{\text{reactants}} = (6 \times 0.0) + (2 \times -824.2) \][/tex]
[tex]\[ \Delta H_{\text{reactants}} = 0 + (-1648.4) = -1648.4 \, \text{kJ} \][/tex]
4. Calculate the total enthalpy of the products:
[tex]\[ \Delta H_{\text{products}} = (4 \times -399.5) + (3 \times 0.0) \][/tex]
[tex]\[ \Delta H_{\text{products}} = -1598.0 + 0 = -1598.0 \, \text{kJ} \][/tex]
5. Calculate the standard enthalpy change of the reaction ([tex]\(\Delta H^{\circ}_{\text{rxn}}\)[/tex]):
[tex]\[ \Delta H^{\circ}_{\text{rxn}} = \Delta H_{\text{products}} - \Delta H_{\text{reactants}} \][/tex]
[tex]\[ \Delta H^{\circ}_{\text{rxn}} = -1598.0 - (-1648.4) \][/tex]
[tex]\[ \Delta H^{\circ}_{\text{rxn}} = -1598.0 + 1648.4 = 50.4 \, \text{kJ} \][/tex]
Thus, the standard enthalpy change for the reaction, [tex]\(\Delta H^{\circ}_{\text{rxn}}\)[/tex], is [tex]\(50.4 \, \text{kJ}\)[/tex].
[tex]\[ 6 \text{Cl}_2(g) + 2 \text{Fe}_2\text{O}_3(s) \rightarrow 4 \text{FeCl}_3(s) + 3 \text{O}_2(g), \][/tex]
we need to use the provided standard enthalpy of formation ([tex]\(\Delta H_{f}^{\circ}\)[/tex]) values for each substance involved.
### Step-by-Step Solution:
1. Identify the enthalpy of formation ([tex]\(\Delta H_{f}^{\circ}\)[/tex]) for each substance:
[tex]\[ \begin{aligned} &\text{Cl}_2(g) & : &\quad 0.0 \, \text{kJ/mol} \\ &\text{Fe}_2\text{O}_3(s) & : &\quad -824.2 \, \text{kJ/mol} \\ &\text{FeCl}_3(s) & : &\quad -399.5 \, \text{kJ/mol} \\ &\text{O}_2(g) & : &\quad 0.0 \, \text{kJ/mol} \\ \end{aligned} \][/tex]
2. Use the stoichiometric coefficients from the balanced chemical equation:
[tex]\[ \begin{aligned} &\text{Cl}_2(g) & : &\quad 6 \\ &\text{Fe}_2\text{O}_3(s) & : &\quad 2 \\ &\text{FeCl}_3(s) & : &\quad 4 \\ &\text{O}_2(g) & : &\quad 3 \\ \end{aligned} \][/tex]
3. Calculate the total enthalpy of the reactants:
[tex]\[ \Delta H_{\text{reactants}} = (6 \times 0.0) + (2 \times -824.2) \][/tex]
[tex]\[ \Delta H_{\text{reactants}} = 0 + (-1648.4) = -1648.4 \, \text{kJ} \][/tex]
4. Calculate the total enthalpy of the products:
[tex]\[ \Delta H_{\text{products}} = (4 \times -399.5) + (3 \times 0.0) \][/tex]
[tex]\[ \Delta H_{\text{products}} = -1598.0 + 0 = -1598.0 \, \text{kJ} \][/tex]
5. Calculate the standard enthalpy change of the reaction ([tex]\(\Delta H^{\circ}_{\text{rxn}}\)[/tex]):
[tex]\[ \Delta H^{\circ}_{\text{rxn}} = \Delta H_{\text{products}} - \Delta H_{\text{reactants}} \][/tex]
[tex]\[ \Delta H^{\circ}_{\text{rxn}} = -1598.0 - (-1648.4) \][/tex]
[tex]\[ \Delta H^{\circ}_{\text{rxn}} = -1598.0 + 1648.4 = 50.4 \, \text{kJ} \][/tex]
Thus, the standard enthalpy change for the reaction, [tex]\(\Delta H^{\circ}_{\text{rxn}}\)[/tex], is [tex]\(50.4 \, \text{kJ}\)[/tex].
Visit us again for up-to-date and reliable answers. We're always ready to assist you with your informational needs. We hope you found what you were looking for. Feel free to revisit us for more answers and updated information. Get the answers you need at Westonci.ca. Stay informed with our latest expert advice.
