To find the enthalpy of combustion per mole of propane ([tex]\( C_3H_8 \)[/tex]), we need to use the enthalpy changes of formation ([tex]\( \Delta H_f \)[/tex]) for the reactants and products in the balanced chemical equation:
[tex]\[ C_3H_8(g) + 5 O_2(g) \rightarrow 3 CO_2(g) + 4 H_2O(g) \][/tex]
Given data:
- [tex]\(\Delta H_f \)[/tex] of [tex]\(C_3H_8 \)[/tex] = -103.8 kJ/mol
- [tex]\(\Delta H_f \)[/tex] of [tex]\(CO_2 \)[/tex] = -393.5 kJ/mol
- [tex]\(\Delta H_f \)[/tex] of [tex]\(H_2O \)[/tex] = -241.82 kJ/mol
To find the enthalpy of combustion, we need the total enthalpy change for the products and the reactants. The enthalpy change for the products is found by summing the enthalpies of formation of each product, each multiplied by their respective coefficients in the balanced equation.
### Calculating the enthalpy change for the products:
- For 3 moles of [tex]\( CO_2 \)[/tex]:
[tex]\[ 3 \times \Delta H_{f, CO_2} = 3 \times (-393.5\, \text{kJ/mol}) = -1180.5\, \text{kJ} \][/tex]
- For 4 moles of [tex]\( H_2O \)[/tex]:
[tex]\[ 4 \times \Delta H_{f, H_2O} = 4 \times (-241.82\, \text{kJ/mol}) = -967.28\, \text{kJ} \][/tex]
[tex]\[ \Delta H_{\text{products}} = -1180.5\, \text{kJ} + -967.28\, \text{kJ} = -2147.78\, \text{kJ} \][/tex]
### Calculating the enthalpy change for the reactants:
Since oxygen ([tex]\(O_2\)[/tex]) in its elemental form does not contribute to the enthalpy change, we only consider propane ([tex]\(C_3H_8\)[/tex]):
[tex]\[ \Delta H_{\text{reactants}} = \Delta H_{f, C_3H_8} = -103.8\, \text{kJ} \][/tex]
### Finding the enthalpy of combustion:
The enthalpy of combustion ([tex]\(\Delta H\)[/tex]) is given by:
[tex]\[ \Delta H = \Delta H_{\text{products}} - \Delta H_{\text{reactants}} \][/tex]
[tex]\[ \Delta H = (-2147.78\, \text{kJ}) - (-103.8\, \text{kJ}) \][/tex]
[tex]\[ \Delta H = -2043.98\, \text{kJ/mol} \][/tex]
### Conclusion:
The enthalpy of combustion per mole of [tex]\( C_3H_8 \)[/tex] is approximately [tex]\(-2044.0\, \text{kJ/mol} \)[/tex].
Thus, the correct choice among the given options is:
[tex]\[ -2,044.0\, \text{kJ/mol} \][/tex]
