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To calculate the enthalpy change ([tex]\(\Delta H_{rxn}^{\circ}\)[/tex]) for the combustion of propane ([tex]\(C_3H_8\)[/tex]), we need to apply the principle of using standard heats of formation ([tex]\(\Delta H_f^{\circ}\)[/tex]) of the reactants and products. Here is a step-by-step solution:
1. Write the balanced chemical equation for the combustion of propane:
[tex]\[ \mathrm{C_3H_8(g)} + 5 \mathrm{O_2(g)} \rightarrow 3 \mathrm{CO_2(g)} + 4 \mathrm{H_2O(g)} \][/tex]
2. List the given heat of formation values:
[tex]\[ \begin{array}{|c|c|} \hline \text{Substance} & \Delta H_f^{\circ} \, (\text{kJ/mol}) \\ \hline \mathrm{C_3H_8(g)} & -104.7 \\ \hline \mathrm{CO_2(g)} & -393.5 \\ \hline \mathrm{H_2O(g)} & -241.8 \\ \hline \mathrm{O_2(g)} & 0 \quad (\text{element in standard state}) \\ \hline \end{array} \][/tex]
3. Calculate the total enthalpy change for the products:
[tex]\[ \Delta H_{products}^{\circ} = (3 \times \Delta H_f^{\circ}(\mathrm{CO_2})) + (4 \times \Delta H_f^{\circ}(\mathrm{H_2O})) \][/tex]
Substituting the values:
[tex]\[ \Delta H_{products}^{\circ} = (3 \times -393.5) + (4 \times -241.8) \][/tex]
[tex]\[ \Delta H_{products}^{\circ} = -1180.5 + (-967.2) = -2147.7 \, \text{kJ} \][/tex]
4. Calculate the total enthalpy change for the reactants:
[tex]\[ \Delta H_{reactants}^{\circ} = (1 \times \Delta H_f^{\circ}(\mathrm{C_3H_8})) + (5 \times \Delta H_f^{\circ}(\mathrm{O_2})) \][/tex]
Substituting the values:
[tex]\[ \Delta H_{reactants}^{\circ} = (1 \times -104.7) + (5 \times 0) \][/tex]
[tex]\[ \Delta H_{reactants}^{\circ} = -104.7 \, \text{kJ} \][/tex]
5. Calculate the enthalpy change of the reaction ([tex]\(\Delta H_{rxn}^{\circ}\)[/tex]):
[tex]\[ \Delta H_{rxn}^{\circ} = \Delta H_{products}^{\circ} - \Delta H_{reactants}^{\circ} \][/tex]
Substituting the values:
[tex]\[ \Delta H_{rxn}^{\circ} = -2147.7 - (-104.7) \][/tex]
[tex]\[ \Delta H_{rxn}^{\circ} = -2147.7 + 104.7 = -2043.0 \, \text{kJ} \][/tex]
6. Express the final answer with appropriate units:
[tex]\[ \Delta H_{rxn}^{\circ} = -2043.0 \, \text{kJ} \][/tex]
Therefore, the enthalpy change for the combustion of 1 mole of propane is [tex]\(-2043.0 \, \text{kJ}\)[/tex].
1. Write the balanced chemical equation for the combustion of propane:
[tex]\[ \mathrm{C_3H_8(g)} + 5 \mathrm{O_2(g)} \rightarrow 3 \mathrm{CO_2(g)} + 4 \mathrm{H_2O(g)} \][/tex]
2. List the given heat of formation values:
[tex]\[ \begin{array}{|c|c|} \hline \text{Substance} & \Delta H_f^{\circ} \, (\text{kJ/mol}) \\ \hline \mathrm{C_3H_8(g)} & -104.7 \\ \hline \mathrm{CO_2(g)} & -393.5 \\ \hline \mathrm{H_2O(g)} & -241.8 \\ \hline \mathrm{O_2(g)} & 0 \quad (\text{element in standard state}) \\ \hline \end{array} \][/tex]
3. Calculate the total enthalpy change for the products:
[tex]\[ \Delta H_{products}^{\circ} = (3 \times \Delta H_f^{\circ}(\mathrm{CO_2})) + (4 \times \Delta H_f^{\circ}(\mathrm{H_2O})) \][/tex]
Substituting the values:
[tex]\[ \Delta H_{products}^{\circ} = (3 \times -393.5) + (4 \times -241.8) \][/tex]
[tex]\[ \Delta H_{products}^{\circ} = -1180.5 + (-967.2) = -2147.7 \, \text{kJ} \][/tex]
4. Calculate the total enthalpy change for the reactants:
[tex]\[ \Delta H_{reactants}^{\circ} = (1 \times \Delta H_f^{\circ}(\mathrm{C_3H_8})) + (5 \times \Delta H_f^{\circ}(\mathrm{O_2})) \][/tex]
Substituting the values:
[tex]\[ \Delta H_{reactants}^{\circ} = (1 \times -104.7) + (5 \times 0) \][/tex]
[tex]\[ \Delta H_{reactants}^{\circ} = -104.7 \, \text{kJ} \][/tex]
5. Calculate the enthalpy change of the reaction ([tex]\(\Delta H_{rxn}^{\circ}\)[/tex]):
[tex]\[ \Delta H_{rxn}^{\circ} = \Delta H_{products}^{\circ} - \Delta H_{reactants}^{\circ} \][/tex]
Substituting the values:
[tex]\[ \Delta H_{rxn}^{\circ} = -2147.7 - (-104.7) \][/tex]
[tex]\[ \Delta H_{rxn}^{\circ} = -2147.7 + 104.7 = -2043.0 \, \text{kJ} \][/tex]
6. Express the final answer with appropriate units:
[tex]\[ \Delta H_{rxn}^{\circ} = -2043.0 \, \text{kJ} \][/tex]
Therefore, the enthalpy change for the combustion of 1 mole of propane is [tex]\(-2043.0 \, \text{kJ}\)[/tex].
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