Butane \(\mathrm{C}_{4} \mathrm{H}_{10}(l)\) boils at $-0.5^{\circ} \mathrm{C} ;\( at this temperature it has a density of \)0.60 \mathrm{~g} / \mathrm{cm}^{3}\(. The enthalpy of formation of \)\mathrm{C}_{4} \mathrm{H}_{10}(g)\( is \)-124.7 \mathrm{~kJ} / \mathrm{mol},$ and the enthalpy of vaporiza- tion of \(\mathrm{C}_{4} \mathrm{H}_{10}(l)\) is $22.44 \mathrm{~kJ} / \mathrm{mol} .\( Calculate the enthalpy change when \)1 \mathrm{~L}$ of liquid \(\mathrm{C}_{4} \mathrm{H}_{10}(l)\) is burned in air to give \(\mathrm{CO}_{2}(g)\) and \(\mathrm{H}_{2} \mathrm{O}(g) .\) How does this compare with \(\Delta H\) for the complete combustion of \(1 \mathrm{~L}\) of liquid methanol, \(\mathrm{CH}_{3} \mathrm{OH}(l) ?\) For $\mathrm{CH}_{3} \mathrm{OH}(l),\( the density at \)25^{\circ} \mathrm{C}\( is \)0.792 \mathrm{~g} / \mathrm{cm}^{3},\( and \)\Delta H_{f}^{\circ}=-239 \mathrm{~kJ} / \mathrm{mol}$.

Since we are given the volumes and densities of both butane and methanol, we can use the formula mass = density * volume to find the mass of 1 L of each liquid. Then, we can find the number of moles by dividing the mass of each substance by its molar mass. For butane, \( C_4H_{10}(l) \), the density is 0.60 g/cm3, the volume is 1000 cm3 (since 1 L = 1000 cm3), and the molar mass is 58.12 g/mol (12.01*4 + 1.01*10). mass of butane = \( 0.60 \frac{g}{cm^3} * 1000 cm^3 = 600g \) moles of butane = \( \frac{600g}{58.12 \frac{g}{mol}} = 10.32 mol \) For methanol, \( CH_3OH(l) \), the density is 0.792 g/cm3, the volume is 1000 cm3, and the molar mass is 32.04 g/mol (12.01 + 1.01*4 + 16). mass of methanol = \( 0.792 \frac{g}{cm^3} * 1000 cm^3 = 792g \) moles of methanol = \( \frac{792g}{32.04 \frac{g}{mol}} = 24.72 mol \)

Combusting butane in the gas state would yield the following balanced chemical equation: \( C_4H_{10}(g) + \frac{13}{2} O_2(g) \rightarrow 4CO_{2}(g) + 5H_2O(g) \) Enthalpy changes for this reaction can be found using Hess's law. Since we are given the enthalpy of formation for gaseous butane and the enthalpy of vaporization for liquid butane, we can calculate the enthalpy change for the combustion of liquid butane as follows: ΔH_combustion (C4H10) = ΔH_vaporization (C4H10) + 4ΔH_formation (CO2) + 5ΔH_formation (H2O) - ΔH_formation (C4H10) The given values are: ΔH_vaporization (C4H10) = 22.44 kJ/mol ΔH_formation (C4H10) = -124.7 kJ/mol ΔH_formation (CO2) = -393.5 kJ/mol ΔH_formation (H2O) = -241.8 kJ/mol Plugging in all the values, we get: ΔH_combustion (C4H10) = 22.44 + 4(-393.5) + 5(-241.8) - (-124.7) = -2877.9 kJ/mol Now, we can calculate the enthalpy change for the combustion of 1 L of liquid butane: ΔH_combustion (1L C4H10) = ΔH_combustion (C4H10) * moles of butane = -2877.9 kJ/mol * 10.32 mol = -29704 kJ

Combusting methanol would yield the following balanced chemical equation: \( CH_3OH(l) + \frac{3}{2} O_2(g) \rightarrow CO_{2}(g) + 2H_2O(g) \) We can use Hess's law to find the enthalpy change for the combustion of methanol as well: ΔH_combustion (CH3OH) = ΔH_formation (CO2) + 2ΔH_formation (H2O) - ΔH_formation (CH3OH) The given values are: ΔH_formation (CH3OH) = -239 kJ/mol Plugging in all the values, we get: ΔH_combustion (CH3OH) = -393.5 + 2(-241.8) - (-239) = -637.5 kJ/mol Now, we can calculate the enthalpy change for the combustion of 1 L of liquid methanol: ΔH_combustion (1L CH3OH) = ΔH_combustion (CH3OH) * moles of methanol = -637.5 kJ/mol * 24.72 mol = -15752 kJ

Having now calculated the enthalpy changes for both substances, we can compare them: ΔH_combustion (1L C4H10) = -29704 kJ ΔH_combustion (1L CH3OH) = -15752 kJ This comparison shows that the combustion of 1 L of liquid butane is more energetic than the combustion of 1 L of liquid methanol, as the enthalpy change for butane is significantly larger in magnitude.