Butane, \(\mathrm{C}_{4} \mathrm{H}_{10}\), is widely used as a fuel for disposable lighters. When one mole of butane is burned in oxygen, carbon dioxide and steam are formed and \(2658.3 \mathrm{~kJ}\) of heat is evolved. (a) Write a thermochemical equation for the reaction. (b) Using Table \(8.3\), calculate the standard heat of formation of butane.

A combustion reaction is a reaction between a hydrocarbon (such as butane) and oxygen, producing carbon dioxide and water. The balanced equation for the combustion of butane is: $$\mathrm{C}_{4}\mathrm{H}_{10} + \mathrm{O}_{2} \rightarrow \mathrm{CO}_{2} + \mathrm{H}_{2}\mathrm{O}$$ Now, we need to balance the equation, making sure all the atoms are equal on both sides of the equation: $$\mathrm{C}_{4}\mathrm{H}_{10} + \dfrac{13}{2}\mathrm{O}_{2} \rightarrow 4\,\mathrm{CO}_{2} + 5\,\mathrm{H}_{2}\mathrm{O}$$

Now that the equation is balanced, we can add the heat evolved in the reaction. The problem states that 1 mole of butane releases 2658.3 kJ of heat. Thus, the thermochemical equation is: $$\mathrm{C}_{4}\mathrm{H}_{10} + \dfrac{13}{2}\mathrm{O}_{2} \rightarrow 4\,\mathrm{CO}_{2} + 5\,\mathrm{H}_{2}\mathrm{O} \quad \Delta H=-2658.3\,\text{kJ}$$

We are given the standard heats of formation for several compounds in Table 8.3, which can be used to find the standard heat of formation for butane. The relationship between the standard heats of formation and the heat of reaction is given by: $$\Delta H_{reaction} = \sum \Delta H_{f}(\text{products}) - \sum \Delta H_{f}(\text{reactants})$$ Thus, the equation for calculating the heat of formation for butane is: $$\Delta H_{f\mathrm{(C_{4}H_{10})}} = \dfrac{\left[ 4\,\Delta H_{f\mathrm{(CO_{2})}} + 5\,\Delta H_{f\mathrm{(H_{2}O)}} - \Delta H_{reaction} \right] - \dfrac{13}{2}\Delta H_{f\mathrm{(O_{2})}}}{1}$$ Using the values from Table 8.3, we have: $$\Delta H_{f\mathrm{(C_{4}H_{10})}} = \dfrac{\left[ 4\,\times(-393.5\,\text{kJ/mol}) + 5\,\times(-285.8\,\text{kJ/mol}) - (-2658.3\,\text{kJ/mol}) \right] - \dfrac{13}{2}\times(0\,\text{kJ/mol})}{1}$$ Now, calculate the standard heat of formation for butane: $$\Delta H_{f\mathrm{(C_{4}H_{10})}} = \dfrac{-2445.7\,\text{kJ/mol}}{1} = -124.1\,\text{kJ/mol}$$ The standard heat of formation of butane is approximately \(-124.1\,\text{kJ/mol}\).