Answer :
To determine [tex]\(\Delta H^{\circ}_{rxn}\)[/tex] for the reaction:
[tex]\[ 3 \text{Fe}_2\text{O}_3(s) + \text{CO}(g) \rightarrow 2 \text{Fe}_3\text{O}_4(s) + \text{CO}_2(g) \][/tex]
we need to use the standard enthalpies of formation ([tex]\(\Delta H^{\circ}_f\)[/tex]) provided. The enthalpy change for the reaction ([tex]\(\Delta H^{\circ}_{rxn}\)[/tex]) can be found using the following relation:
[tex]\[ \Delta H^{\circ}_{rxn} = \sum \Delta H^{\circ}_f (\text{products}) - \sum \Delta H^{\circ}_f (\text{reactants}) \][/tex]
### Step-by-Step Solution
1. List the given enthalpies of formation:
[tex]\[ \begin{align*} \Delta H^{\circ}_f (\text{Fe}_2\text{O}_3(s)) & = -824 \, \text{kJ/mol} \\ \Delta H^{\circ}_f (\text{Fe}_3\text{O}_4(s)) & = -1118 \, \text{kJ/mol} \\ \Delta H^{\circ}_f (\text{CO}(g)) & = -111 \, \text{kJ/mol} \\ \Delta H^{\circ}_f (\text{CO}_2(g)) & = -394 \, \text{kJ/mol} \end{align*} \][/tex]
2. Write the balanced chemical equation:
[tex]\[ 3 \text{Fe}_2\text{O}_3(s) + \text{CO}(g) \rightarrow 2 \text{Fe}_3\text{O}_4(s) + \text{CO}_2(g) \][/tex]
3. Calculate the total enthalpy of the products:
[tex]\[ \Delta H^{\circ}_\text{products} = 2 \left(\Delta H^{\circ}_f (\text{Fe}_3\text{O}_4(s))\right) + 1 \left(\Delta H^{\circ}_f (\text{CO}_2(g))\right) \][/tex]
[tex]\[ \Delta H^{\circ}_\text{products} = 2(-1118) + (-394) \][/tex]
[tex]\[ \Delta H^{\circ}_\text{products} = -2236 - 394 \][/tex]
[tex]\[ \Delta H^{\circ}_\text{products} = -2630 \, \text{kJ} \][/tex]
4. Calculate the total enthalpy of the reactants:
[tex]\[ \Delta H^{\circ}_\text{reactants} = 3 \left(\Delta H^{\circ}_f (\text{Fe}_2\text{O}_3(s))\right) + 1 \left(\Delta H^{\circ}_f (\text{CO}(g))\right) \][/tex]
[tex]\[ \Delta H^{\circ}_\text{reactants} = 3(-824) + (-111) \][/tex]
[tex]\[ \Delta H^{\circ}_\text{reactants} = -2472 - 111 \][/tex]
[tex]\[ \Delta H^{\circ}_\text{reactants} = -2583 \, \text{kJ} \][/tex]
5. Calculate the enthalpy change of the reaction:
[tex]\[ \Delta H^{\circ}_{rxn} = \Delta H^{\circ}_\text{products} - \Delta H^{\circ}_\text{reactants} \][/tex]
[tex]\[ \Delta H^{\circ}_{rxn} = -2630 - (-2583) \][/tex]
[tex]\[ \Delta H^{\circ}_{rxn} = -2630 + 2583 \][/tex]
[tex]\[ \Delta H^{\circ}_{rxn} = -47 \, \text{kJ} \][/tex]
Thus, the enthalpy change for the reaction [tex]\(\Delta H^{\circ}_{rxn}\)[/tex] is [tex]\(-47 \, \text{kJ}\)[/tex].
[tex]\[ 3 \text{Fe}_2\text{O}_3(s) + \text{CO}(g) \rightarrow 2 \text{Fe}_3\text{O}_4(s) + \text{CO}_2(g) \][/tex]
we need to use the standard enthalpies of formation ([tex]\(\Delta H^{\circ}_f\)[/tex]) provided. The enthalpy change for the reaction ([tex]\(\Delta H^{\circ}_{rxn}\)[/tex]) can be found using the following relation:
[tex]\[ \Delta H^{\circ}_{rxn} = \sum \Delta H^{\circ}_f (\text{products}) - \sum \Delta H^{\circ}_f (\text{reactants}) \][/tex]
### Step-by-Step Solution
1. List the given enthalpies of formation:
[tex]\[ \begin{align*} \Delta H^{\circ}_f (\text{Fe}_2\text{O}_3(s)) & = -824 \, \text{kJ/mol} \\ \Delta H^{\circ}_f (\text{Fe}_3\text{O}_4(s)) & = -1118 \, \text{kJ/mol} \\ \Delta H^{\circ}_f (\text{CO}(g)) & = -111 \, \text{kJ/mol} \\ \Delta H^{\circ}_f (\text{CO}_2(g)) & = -394 \, \text{kJ/mol} \end{align*} \][/tex]
2. Write the balanced chemical equation:
[tex]\[ 3 \text{Fe}_2\text{O}_3(s) + \text{CO}(g) \rightarrow 2 \text{Fe}_3\text{O}_4(s) + \text{CO}_2(g) \][/tex]
3. Calculate the total enthalpy of the products:
[tex]\[ \Delta H^{\circ}_\text{products} = 2 \left(\Delta H^{\circ}_f (\text{Fe}_3\text{O}_4(s))\right) + 1 \left(\Delta H^{\circ}_f (\text{CO}_2(g))\right) \][/tex]
[tex]\[ \Delta H^{\circ}_\text{products} = 2(-1118) + (-394) \][/tex]
[tex]\[ \Delta H^{\circ}_\text{products} = -2236 - 394 \][/tex]
[tex]\[ \Delta H^{\circ}_\text{products} = -2630 \, \text{kJ} \][/tex]
4. Calculate the total enthalpy of the reactants:
[tex]\[ \Delta H^{\circ}_\text{reactants} = 3 \left(\Delta H^{\circ}_f (\text{Fe}_2\text{O}_3(s))\right) + 1 \left(\Delta H^{\circ}_f (\text{CO}(g))\right) \][/tex]
[tex]\[ \Delta H^{\circ}_\text{reactants} = 3(-824) + (-111) \][/tex]
[tex]\[ \Delta H^{\circ}_\text{reactants} = -2472 - 111 \][/tex]
[tex]\[ \Delta H^{\circ}_\text{reactants} = -2583 \, \text{kJ} \][/tex]
5. Calculate the enthalpy change of the reaction:
[tex]\[ \Delta H^{\circ}_{rxn} = \Delta H^{\circ}_\text{products} - \Delta H^{\circ}_\text{reactants} \][/tex]
[tex]\[ \Delta H^{\circ}_{rxn} = -2630 - (-2583) \][/tex]
[tex]\[ \Delta H^{\circ}_{rxn} = -2630 + 2583 \][/tex]
[tex]\[ \Delta H^{\circ}_{rxn} = -47 \, \text{kJ} \][/tex]
Thus, the enthalpy change for the reaction [tex]\(\Delta H^{\circ}_{rxn}\)[/tex] is [tex]\(-47 \, \text{kJ}\)[/tex].
