To determine the enthalpy of combustion per mole of butane, we need to follow a few steps.
1. Identify the chemical reaction:
The balanced combustion reaction of butane is:
[tex]\[
2 \, \text{C}_4\text{H}_{10} (g) + 13 \, \text{O}_2 (g) \rightarrow 8 \, \text{CO}_2 (g) + 10 \, \text{H}_2\text{O} (g)
\][/tex]
2. Identify the given enthalpies of formation:
[tex]\[
\Delta H_{f, \text{C}_4\text{H}_{10}} = -125.6 \, \text{kJ/mol}
\][/tex]
[tex]\[
\Delta H_{f, \text{CO}_2} = -393.5 \, \text{kJ/mol}
\][/tex]
[tex]\[
\Delta H_{f, \text{H}_2\text{O}} = -241.82 \, \text{kJ/mol}
\][/tex]
3. Calculate the sum of enthalpies of formation for the products:
[tex]\[
\sum \Delta H_{f, \text{products}} = 8 \times \Delta H_{f, \text{CO}_2} + 10 \times \Delta H_{f, \text{H}_2\text{O}}
\][/tex]
Plugging in the values:
[tex]\[
\sum \Delta H_{f, \text{products}} = 8 \times (-393.5) + 10 \times (-241.82) = -3148 + (-2418.2) = -5566.2 \, \text{kJ}
\][/tex]
4. Calculate the sum of enthalpies of formation for the reactants:
[tex]\[
\sum \Delta H_{f, \text{reactants}} = 2 \times \Delta H_{f, \text{C}_4\text{H}_{10}}
\][/tex]
Plugging in the values:
[tex]\[
\sum \Delta H_{f, \text{reactants}} = 2 \times (-125.6) = -251.2 \, \text{kJ}
\][/tex]
5. Calculate the enthalpy change of the reaction:
[tex]\[
\Delta H_{\text{rxn}} = \sum \Delta H_{f, \text{products}} - \sum \Delta H_{f, \text{reactants}}
\][/tex]
Plugging in the values:
[tex]\[
\Delta H_{\text{rxn}} = -5566.2 - (-251.2) = -5566.2 + 251.2 = -5315.0 \, \text{kJ}
\][/tex]
6. Determine the enthalpy of combustion per mole of butane:
Since the reaction involves 2 moles of C[tex]\(_4\)[/tex]H[tex]\(_{10}\)[/tex], we need to divide the total enthalpy change by 2:
[tex]\[
\Delta H_{\text{combustion per mole}} = \frac{\Delta H_{\text{rxn}}}{2} = \frac{-5315.0}{2} = -2657.5 \, \text{kJ/mol}
\][/tex]
Thus, the enthalpy of combustion per mole of butane is [tex]\(\boxed{-2657.5 \, \text{kJ/mol}}\)[/tex].
