Question

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 \(\Delta H_{f}^{\circ}=-239 \mathrm{~kJ} / \mathrm{mol}\).

Short answer

The enthalpy change for the combustion of 1 L of liquid butane is -29704 kJ, while the enthalpy change for the combustion of 1 L of liquid methanol is -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.

Step-by-step solution

Calculate the number of moles of butane and methanol

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 = \( \frac{792g}{32.04 \frac{g}{mol}} = 24.72 mol \)

Calculate the enthalpy change for the combustion of butane

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

Calculate the enthalpy change for the combustion of methanol

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

Compare the enthalpy changes for the combustions

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.

Key concepts

Combustion Reaction

In a combustion reaction, a substance combines with oxygen to produce oxides, along with the release of energy, usually in the form of heat and light. For hydrocarbons like butane - The general formula for a complete combustion reaction is: Hydrocarbon + O extsubscript{2} ightarrow CO extsubscript{2} + H extsubscript{2}O - For our butane example: \( C_4H_{10}(g) + \frac{13}{2} O_2(g) \rightarrow 4CO_{2}(g) + 5H_2O(g) \) This equation is already balanced, ensuring that we have the same number of each type of atom on both sides. - Combustion reactions are exothermic, meaning they release heat. - The energy released in these reactions primarily comes from breaking the chemical bonds in the reactants and forming new bonds in the products. Understanding how these reactions work is crucial for calculating the enthalpy change when a hydrocarbon like butane undergoes combustion.

Hess's Law

Hess's Law states that the total enthalpy change during a chemical reaction is the same, no matter how many steps the reaction takes. This principle allows us to calculate enthalpy change even if the reaction doesn't occur in a single step. Here's how it works: - When calculating the energy change in a reaction, you can add up the individual enthalpy changes from various sub-reactions. - For the combustion of butane, we used Hess’s Law to determine the enthalpy change by combining the following elements: - The enthalpy of vaporization, as we started with liquid butane (to convert it to vapor). - The formation enthalpies of the products (CO extsubscript{2} and H extsubscript{2}O). - Subtracting the formation enthalpy of butane. By summing these values, we determined the enthalpy change for the entire combustion reaction. Hess’s Law simplifies these complex calculations by utilizing known enthalpy values.

Molar Mass Calculation

Molar mass calculation is a fundamental concept in chemistry, allowing us to convert between the mass of a compound and the number of moles. This is key for stoichiometric calculations in chemical reactions. Here's how you calculate molar mass: - First, determine the atomic masses of each element in the compound from the periodic table. For example, in butane ( \(C_4H_{10}\), we have four carbon atoms (12.01 amu each) and ten hydrogen atoms (1.01 amu each). - Multiply the atomic mass of each element by the number of times it appears in the molecule. - Sum these values to get the molar mass. For butane: \(4(12.01) + 10(1.01) = 58.12 \, g/mol\) - This molar mass can then be used to convert between grams and moles, crucial for solving problems involving reaction stoichiometry and energy changes.

Enthalpy of Formation

The enthalpy of formation is a measure of the energy change when one mole of a compound is formed from its basic elements in their standard states. These values are critical for calculating the overall enthalpy change in chemical reactions. Here's what you need to know:- Standard enthalpy of formation is denoted as \( \Delta H_f^\circ \) and typically given in units of kJ/mol.- For example, the enthalpy of formation for \( C_4H_{10}(g) \) is \(-124.7 \, kJ/mol\). - These values are typically negative, indicating a release of energy when the compound forms. - In reaction calculations, you use the enthalpies of formation to determine the change in energy as you go from reactants to products. Knowing each substance's enthalpies of formation helps us employ Hess's Law to find the overall energy change for a reaction, such as the combustion of butane.