Diethyl ether, \(\mathrm{C}_{4} \mathrm{H}_{10} \mathrm{O}(l),\) a flammable compound that was once used as a surgical anesthetic, has the structure $$\mathrm{H}_{3} \mathrm{C}-\mathrm{CH}_{2}-\mathrm{O}-\mathrm{CH}_{2}-\mathrm{CH}_{3}$$ The complete combustion of 1 mol of \(\mathrm{C}_{4} \mathrm{H}_{10} \mathrm{O}(l)\) to \(\mathrm{CO}_{2}(g)\) and \(\mathrm{H}_{2} \mathrm{O}(l)\) yields \(\Delta H^{\circ}=-2723.7 \mathrm{kJ}\) . (a) Write a balanced equation for the combustion of 1 \(\mathrm{mol}\) of \(\mathrm{C}_{4} \mathrm{H}_{10} \mathrm{O}(l) .\) (b) By using the information in this problem and data in Table \(5.3,\) calculate \(\Delta H_{f}^{\circ}\) for diethyl ether.

To write a balanced chemical equation for the combustion of diethyl ether, we need to make sure that the number of each atom in the reactants is equal to the number in the products. The general equation for the combustion of a hydrocarbon is: \[ Hydrocarbon + O_{2}(g) \rightarrow CO_{2}(g) + H_{2}O(l) \] For diethyl ether, the hydrocarbon is \(\mathrm{C}_{4} \mathrm{H}_{10} \mathrm{O}\), so our balanced equation will be: \[ \mathrm{C}_{4} \mathrm{H}_{10} \mathrm{O}(l) + 6\mathrm{O}_{2}(g) \rightarrow 4\mathrm{CO}_{2}(g) + 5\mathrm{H}_{2} \mathrm{O}(l) \]

Now, we want to find the standard enthalpy of formation for diethyl ether. We will use the known values of ΔH° for the combustion reaction and the standard enthalpies of formation of the reaction products to calculate this value. According to Hess's Law: \[ ΔH_{f}^{\circ}(\mathrm{C}_{4}\mathrm{H}_{10}\mathrm{O}) = -ΔH^{\circ}_{combustion} + \sum_{i=1}^{4}\left(ΔH_{f}^{\circ}(\mathrm{CO}_{2})\right) + \sum_{i=1}^{5}\left(ΔH_{f}^{\circ}(\mathrm{H}_{2}\mathrm{O})\right) - 6\left(ΔH_{f}^{\circ}(\mathrm{O}_{2})\right) \] Note that the standard enthalpy of formation of an element in its standard state (oxygen gas in this case) is always zero, so we can simplify: \[ ΔH_{f}^{\circ}(\mathrm{C}_{4}\mathrm{H}_{10}\mathrm{O}) = -ΔH^{\circ}_{combustion} + 4\left(ΔH_{f}^{\circ}(\mathrm{CO}_{2})\right) + 5\left(ΔH_{f}^{\circ}(\mathrm{H}_{2}\mathrm{O})\right) \] We are given ΔH°\(_{combustion} = -2723.7\,\mathrm{kJ}\), and we can find the standard enthalpies of formation for carbon dioxide and liquid water in Table 5.3: ΔH°\(_{f}(\mathrm{CO}_{2}) = -393.5\,\mathrm{kJ/mol}\) ΔH°\(_{f}(\mathrm{H}_{2}\mathrm{O}) = -285.8\,\mathrm{kJ/mol}\) Now plug these values into the equation to get the standard enthalpy of formation for diethyl ether: \[ ΔH_{f}^{\circ}(\mathrm{C}_{4}\mathrm{H}_{10}\mathrm{O}) = -(-2723.7\,\mathrm{kJ}) + 4(-393.5\,\mathrm{kJ/mol}) + 5(-285.8\,\mathrm{kJ/mol}) \] \[ ΔH_{f}^{\circ}(\mathrm{C}_{4}\mathrm{H}_{10}\mathrm{O}) = 2723.7\,\mathrm{kJ} - 1574\,\mathrm{kJ} - 1429\,\mathrm{kJ} \] \[ ΔH_{f}^{\circ}(\mathrm{C}_{4}\mathrm{H}_{10}\mathrm{O}) = -279.3\,\mathrm{kJ/mol} \] So, the standard enthalpy of formation for diethyl ether is about -279.3 kJ/mol.