Three common hydrocarbons that contain four carbons are listed here, along with their standard enthalpies of formation: (a) For each of these substances, calculate the molar enthalpy of combustion to \(\mathrm{CO}_{2}(g)\) and \(\mathrm{H}_{2} \mathrm{O}(l) .\) (b) Calculate the fuel value in \(\mathrm{kJ} / \mathrm{g}\) for each of these compounds. \((\mathrm{c})\) For each hydrocarbon, determine the percentage of hydrogen by mass. (d) By comparing your answers for parts (b) and (c), propose a relationship between hydrogen content and fuel value in hydrocarbons.

To find the molar enthalpy of combustion for these hydrocarbons to \(\mathrm{CO_2}\) and \(\mathrm{H_2O}\), we will need the balanced chemical equations for their combustion. 1. Butane: \(\mathrm{C_4H_{10}(g) + \frac{13}{2}O_{2}(g) \rightarrow 4CO_{2}(g) + 5H_2O(l)}\) Using Hess's Law, the enthalpy change of the reaction is: \(\Delta H^\circ_\mathrm{comb} = 4\Delta H^\circ_\mathrm{f}[\mathrm{CO_2}] + 5\Delta H^\circ_\mathrm{f}[\mathrm{H_2O}] - \Delta H^\circ_\mathrm{f}[\mathrm{C_4H_{10}}]\) By plugging in the values for the standard enthalpies of formation, we get \(\Delta H^\circ_\mathrm{comb}= -2880\,\mathrm{kJ/mol}\) for Butane 2. Butene: \(\mathrm{C_4H_8(g) + 6O_2(g) \rightarrow 4CO_2(g) + 4H_2O(l)}\) Using Hess's Law, the enthalpy change of the reaction is: \(\Delta H^\circ_\mathrm{comb} = 4\Delta H^\circ_\mathrm{f}[\mathrm{CO_2}] + 4\Delta H^\circ_\mathrm{f}[\mathrm{H_2O}] - \Delta H^\circ_\mathrm{f}[\mathrm{C_4H_8}]\) By plugging in the values for the standard enthalpies of formation we get \(\Delta H^\circ_\mathrm{comb}=-2823\,\mathrm{kJ/mol}\) for Butene 3. Butyne: \(\mathrm{C_4H_6(g) + \frac{11}{2}O_2(g) \rightarrow 4CO_2(g) + 3H_2O(l)}\) Using Hess's Law, the enthalpy change of the reaction is: \(\Delta H^\circ_\mathrm{comb} = 4\Delta H^\circ_\mathrm{f}[\mathrm{CO_2}] + 3\Delta H^\circ_\mathrm{f}[\mathrm{H_2O}] - \Delta H^\circ_\mathrm{f}[\mathrm{C_4H_6}]\) By plugging in the values for the standard enthalpies of formation, we get \(\Delta H^\circ_\mathrm{comb}=-2776\,\mathrm{kJ/mol}\) for Butyne

To find the fuel value in kJ/g, we need to divide the molar enthalpy of combustion (in kJ/mol) by the molar mass of the hydrocarbon (in g/mol). 1. Butane: Fuel value = \(\frac{-2880\,\mathrm{kJ/mol}}{58.12\,\mathrm{g/mol}} = -49.5\,\mathrm{kJ/g}\) 2. Butene: Fuel value = \(\frac{-2823\,\mathrm{kJ/mol}}{56.11\,\mathrm{g/mol}} = -50.3\,\mathrm{kJ/g}\) 3. Butyne: Fuel value = \(\frac{-2776\,\mathrm{kJ/mol}}{54.10\,\mathrm{g/mol}} = -51.3\,\mathrm{kJ/g}\)

To find the percentage of hydrogen by mass, we need to divide the mass of hydrogen in the molecule by the total mass of the molecule. 1. Butane: Percentage of hydrogen = \(\frac{10 \times 1.01\,\mathrm{g/mol}}{58.12\,\mathrm{g/mol}} \times 100 = 17.4\%\) 2. Butene: Percentage of hydrogen = \(\frac{8 \times 1.01\,\mathrm{g/mol}}{56.11\,\mathrm{g/mol}} \times 100 = 14.4\%\) 3. Butyne: Percentage of hydrogen = \(\frac{6 \times 1.01\,\mathrm{g/mol}}{54.10\,\mathrm{g/mol}} \times 100 = 11.2\%\)

Comparing the results of parts (b) and (c), we can see that as the percentage of hydrogen by mass in a hydrocarbon increases, the fuel value (kJ/g) decreases. Therefore, we can propose that there is an inverse relationship between hydrogen content and fuel value in hydrocarbons.