Let's break down the problem step by step to determine the enthalpy of combustion per mole of butane ([tex]$C_4 H_{10}$[/tex]):
### 1. Write the Balanced Chemical Equation:
[tex]\[
2 C_4 H_{10} (g) + 13 O_2 (g) \rightarrow 8 CO_2 (g) + 10 H_2O (g)
\][/tex]
### 2. Identify the Enthalpy of Formation ([tex]$\Delta H_f$[/tex]):
- [tex]$\Delta H_f$[/tex] of [tex]$CO_2 (g) = -393.5$[/tex] kJ/mol
- [tex]$\Delta H_f$[/tex] of [tex]$H_2O (g) = -241.82$[/tex] kJ/mol
- [tex]$\Delta H_f$[/tex] of [tex]$C_4 H_{10} (g) = -125.6$[/tex] kJ/mol
- [tex]$\Delta H_f$[/tex] for [tex]$O_2 (g)$[/tex] is 0 kJ/mol (since it's in its standard state)
### 3. Calculate the Total Enthalpy of Formation of Reactants:
[tex]\[
\text{Reactants} = (2 \times \Delta H_f \text{ of } C_4 H_{10}) + (13 \times \Delta H_f \text{ of } O_2)
\][/tex]
[tex]\[
\text{Reactants} = (2 \times -125.6) + (13 \times 0) = -251.2 \text{ kJ}
\][/tex]
### 4. Calculate the Total Enthalpy of Formation of Products:
[tex]\[
\text{Products} = (8 \times \Delta H_f \text{ of } CO_2) + (10 \times \Delta H_f \text{ of } H_2O)
\][/tex]
[tex]\[
\text{Products} = (8 \times -393.5) + (10 \times -241.82) = -3148 + (-2418.2) = -5566.2 \text{ kJ}
\][/tex]
### 5. Calculate the Enthalpy Change of the Reaction ([tex]$\Delta H_{\text{reaction}}$[/tex]):
[tex]\[
\Delta H_{\text{reaction}} = \sum (\Delta H_f \text{ of products}) - \sum (\Delta H_f \text{ of reactants})
\][/tex]
[tex]\[
\Delta H_{\text{reaction}} = -5566.2 - (-251.2) = -5566.2 + 251.2 = -5315.0 \text{ kJ}
\][/tex]
### 6. Calculate the Enthalpy of Combustion per Mole of Butane:
Since the given balanced equation represents the combustion of 2 moles of butane, we need to divide the total enthalpy change by 2 to find the enthalpy change per mole of butane.
[tex]\[
\Delta H_{\text{combustion per mole}} = \frac{-5315.0}{2} = -2657.5 \text{ kJ/mol}
\][/tex]
### Conclusion:
The enthalpy of combustion per mole of butane is \(-2657.5 \text{ kJ/mol}\).
Therefore, the correct answer is:
[tex]\[
\boxed{-2657.5 \text{ kJ/mol}}
\][/tex]
