To calculate the enthalpy change ([tex]\(\Delta H^\circ_\text{rxn}\)[/tex]) for the given combustion reaction:
[tex]\[ \text{C}_3\text{H}_8(g) + 5 \text{O}_2(g) \rightarrow 3 \text{CO}_2(g) + 4 \text{H}_2\text{O}(g) \][/tex]
we need to use the standard enthalpies of formation ([tex]\(\Delta H^\circ_f\)[/tex]) of the reactants and products involved in the reaction. The standard enthalpies of formation are provided as follows:
- [tex]\(\Delta H^\circ_f(\text{C}_3\text{H}_8(g)) = -103.9 \text{ kJ/mol}\)[/tex]
- [tex]\(\Delta H^\circ_f(\text{CO}_2(g)) = -393.5 \text{ kJ/mol}\)[/tex]
- [tex]\(\Delta H^\circ_f(\text{H}_2\text{O}(g)) = -241.8 \text{ kJ/mol}\)[/tex]
Step-by-step solution:
1. Identify the Balanced Chemical Equation:
[tex]\[ \text{C}_3\text{H}_8(g) + 5 \text{O}_2(g) \rightarrow 3 \text{CO}_2(g) + 4 \text{H}_2\text{O}(g) \][/tex]
2. Sum the Standard Enthalpies of Formation for the Products:
[tex]\[
\Delta H^\circ_\text{products} = (3 \cdot \Delta H^\circ_f(\text{CO}_2)) + (4 \cdot \Delta H^\circ_f(\text{H}_2\text{O}))
\][/tex]
[tex]\[
\Delta H^\circ_\text{products} = (3 \cdot -393.5 \text{ kJ/mol}) + (4 \cdot -241.8 \text{ kJ/mol})
\][/tex]
[tex]\[
\Delta H^\circ_\text{products} = -1180.5 \text{ kJ/mol} + (-967.2 \text{ kJ/mol})
\][/tex]
[tex]\[
\Delta H^\circ_\text{products} = -2147.7 \text{ kJ}
\][/tex]
3. Sum the Standard Enthalpies of Formation for the Reactants:
Since oxygen ([tex]\(\text{O}_2\)[/tex]) is in its elemental form, its enthalpy of formation is zero. Thus, we only need to consider propane ([tex]\(\text{C}_3\text{H}_8\)[/tex]).
[tex]\[
\Delta H^\circ_\text{reactants} = \Delta H^\circ_f(\text{C}_3\text{H}_8)
\][/tex]
[tex]\[
\Delta H^\circ_\text{reactants} = -103.9 \text{ kJ}
\][/tex]
4. Calculate the Standard Enthalpy Change of the Reaction:
[tex]\[
\Delta H^\circ_\text{rxn} = \Delta H^\circ_\text{products} - \Delta H^\circ_\text{reactants}
\][/tex]
[tex]\[
\Delta H^\circ_\text{rxn} = -2147.7 \text{ kJ} - (-103.9 \text{ kJ})
\][/tex]
[tex]\[
\Delta H^\circ_\text{rxn} = -2043.8 \text{ kJ}
\][/tex]
Therefore, the enthalpy change for the combustion of propane is:
[tex]\[
\Delta H^\circ_\text{rxn} = -2043.8 \text{ kJ}
\][/tex]
Expressed to four significant figures, the enthalpy change for this reaction is [tex]\(-2043.8 \text{ kJ}\)[/tex].
