Question

Liquid butane \(\left(\mathrm{C}_{4} \mathrm{H}_{10}\right)\), the fuel used in many disposable lighters, has \(\Delta H_{\mathrm{f}}^{\circ}=-147.5 \mathrm{~kJ} / \mathrm{mol}\) and a density of \(0.579 \mathrm{~g} / \mathrm{mL}\). Write a balanced equation for the combustion of butane, and use Hess's law to calculate the enthalpy of combustion in \(\mathrm{kJ} / \mathrm{mol}, \mathrm{kJ} / \mathrm{g}\), and \(\mathrm{kJ} / \mathrm{mL}\)

Short answer

The enthalpy of combustion is \(-2877\ \mathrm{kJ/mol}\), \(-49.5\ \mathrm{kJ/g}\), and \(-28.7\ \mathrm{kJ/mL}\).

Step-by-step solution

Write the Balanced Chemical Equation for the Combustion

The balanced chemical equation for the combustion of butane (\(\mathrm{C}_4\mathrm{H}_{10}\)) is:\[2\mathrm{C}_4\mathrm{H}_{10}(l) + 13\mathrm{O}_2(g) \rightarrow 8\mathrm{CO}_2(g) + 10\mathrm{H}_2\mathrm{O}(l) \]This demonstrates that two moles of butane react with thirteen moles of oxygen to produce eight moles of carbon dioxide and ten moles of water.

Identify Enthalpies of Formation

Use standard enthalpies of formation for the reactants and products involved:- \(\Delta H_{\mathrm{f}}^{\circ}(\mathrm{CO}_2(g)) = -393.5\ \mathrm{kJ/mol}\)- \(\Delta H_{\mathrm{f}}^{\circ}(\mathrm{H}_2\mathrm{O}(l)) = -285.8\ \mathrm{kJ/mol}\)- \(\Delta H_{\mathrm{f}}^{\circ}(\mathrm{O}_2(g)) = 0\ \mathrm{kJ/mol}\)- \(\Delta H_{\mathrm{f}}^{\circ}(\mathrm{C}_4\mathrm{H}_{10}(l)) = -147.5\ \mathrm{kJ/mol}\) (provided)

Apply Hess's Law for Enthalpy Change Calculation

Using Hess's Law and the enthalpies of formation, calculate the enthalpy change (\(\Delta H_{\text{comb}}\)) for the combustion:\[\Delta H_{\text{comb}} = [8 \times \Delta H_{\mathrm{f}}^{\circ}(\mathrm{CO}_2) + 10 \times \Delta H_{\mathrm{f}}^{\circ}(\mathrm{H}_2\mathrm{O})] - [2 \times \Delta H_{\mathrm{f}}^{\circ}(\mathrm{C}_4\mathrm{H}_{10})]\]\[\Delta H_{\text{comb}} = [8 \times (-393.5) + 10 \times (-285.8)] - [2 \times (-147.5)] = -5754.0\ \mathrm{kJ}\]Thus, the enthalpy of combustion per mole of butane is \(-2877.0\ \mathrm{kJ/mol}\).

Calculate Enthalpy per Gram and per Milliliter

To find the enthalpy per gram, first determine the molar mass of butane: \(\mathrm{C}_4\mathrm{H}_{10}\) has a molar mass of 58.12 g/mol.\[\text{Enthalpy per gram} = \frac{-2877.0\ \mathrm{kJ/mol}}{58.12\ \mathrm{g/mol}} = -49.5\ \mathrm{kJ/g}\]For the enthalpy per milliliter, using the provided density of \(0.579\ \mathrm{g/mL}\):\[\text{Enthalpy per } \mathrm{mL} = -49.5\ \mathrm{kJ/g} \times 0.579\ \mathrm{g/mL} = -28.7\ \mathrm{kJ/mL}\]

Summarize the Calculations

The enthalpy of combustion of butane is \(-2877\ \mathrm{kJ/mol}\), \(-49.5\ \mathrm{kJ/g}\), and \(-28.7\ \mathrm{kJ/mL}\). These values show the energy released per mole, per gram, and per milliliter of butane, respectively.

Key concepts

Hess's Law

Hess's Law is a concept in thermochemistry which states that the total enthalpy change in a chemical reaction is the same, regardless of the pathway taken. This means you can calculate the enthalpy change (\(\Delta H\)) for a reaction by using known enthalpy values of each individual step rather than measuring it directly. It relies on the fact that enthalpy is a state function, which depends only on the initial and final states of a system, not the path taken.

In our exercise on butane combustion, Hess’s Law is used to calculate the enthalpy of combustion through the enthalpies of formation of the reactants and products. By knowing how much energy is required to form the compounds from their elements, we can rearrange reactions and their corresponding enthalpy changes to find the overall energy released when butane combusts. This method allows for precise calculations when direct measurement may be challenging.

Enthalpy of Formation

The enthalpy of formation is the change in energy when one mole of a compound is formed from its elements in their standard states. These values, typically measured in \(\mathrm{kJ/mol}\), are crucial for calculating reaction enthalpies using Hess's Law.

For example, the enthalpy of formation of carbon dioxide, water, and butane were utilized in the butane combustion exercise. These values are:

• \(\Delta H_{\mathrm{f}}^{\circ}(\mathrm{CO}_2(g)) = -393.5\ \mathrm{kJ/mol}\)
• \(\Delta H_{\mathrm{f}}^{\circ}(\mathrm{H}_2\mathrm{O}(l)) = -285.8\ \mathrm{kJ/mol}\)
• \(\Delta H_{\mathrm{f}}^{\circ}(\mathrm{C}_4\mathrm{H}_{10}(l)) = -147.5\ \mathrm{kJ/mol}\)

When we calculate the enthalpy of combustion for butane, we use these formation enthalpies in the equation, substituting them into Hess’s Law to get the total change in energy for the entire process.

Balanced Chemical Equation

In chemical reactions, a balanced chemical equation is essential. It represents the exact quantities of reactants and products involved, ensuring the conservation of mass and charge. For combustion reactions like that of butane, balancing the equation allows us to correctly calculate the enthalpy changes.

The balanced equation for butane combustion is:\[2\mathrm{C}_4\mathrm{H}_{10}(l) + 13\mathrm{O}_2(g) \rightarrow 8\mathrm{CO}_2(g) + 10\mathrm{H}_2\mathrm{O}(l) \]This equation shows that two moles of butane require thirteen moles of oxygen to produce eight moles of carbon dioxide and ten moles of water. Ensuring that this equation is balanced allows us to systematically apply concepts like Hess’s Law and calculate energy changes accurately. It is the starting point for any thermochemical calculation.

Density of Butane

Understanding the density of a substance like butane is essential for converting between different units of measurements, such as from grams to milliliters. Density is defined as mass per unit volume and is typically expressed in \(\mathrm{g/mL}\).

In our exercise, the density of butane is given as \(0.579\ \mathrm{g/mL}\). Knowing this allows us to determine the enthalpy of combustion per milliliter, providing a more practical understanding of the energy content in real-world scenarios.

For instance, after calculating the enthalpy change in \(\mathrm{kJ/g},\) you can multiply it by the density to get the energy change in \(\mathrm{kJ/mL}.\) This conversion is vital for applications where butane is stored in liquid form and consumed in volumetric units, like in lighters and other fuel-efficient devices.