Question

Complete combustion of \(1 \mathrm{~mol}\) of acetone \(\left(\mathrm{C}_{3} \mathrm{H}_{6} \mathrm{O}\right)\) liberates $$ \begin{array}{l} 1790 \mathrm{~kJ}: \\ \mathrm{C}_{3} \mathrm{H}_{6} \mathrm{O}(l)+4 \mathrm{O}_{2}(g) \longrightarrow 3 \mathrm{CO}_{2}(g)+3 \mathrm{H}_{2} \mathrm{O}(l) \\ \Delta H^{\circ}=-1790 \mathrm{~kJ} \end{array} $$ Using this information together with data from Appendix C, calculate the enthalpy of formation of acetone.

Short answer

The enthalpy of formation of acetone, \( \Delta H_{f, \mathrm{C}_{3} \mathrm{H}_{6} \mathrm{O}} \), is approximately \( -247.9\, \mathrm{kJ\, mol^{-1}} \).

Step-by-step solution

Identify given information

We are given the following information: 1. The combustion reaction of 1 mol of acetone (C3H6O): \( \mathrm{C}_{3} \mathrm{H}_{6} \mathrm{O}(l)+4 \mathrm{O}_{2}(g) \longrightarrow 3 \mathrm{CO}_{2}(g)+3 \mathrm{H}_{2} \mathrm{O}(l) \) 2. The enthalpy change of this reaction: \( \Delta H^{\circ} = -1790 \, \mathrm{kJ} \) Step 2: Recall the definition of enthalpy of formation

Understand the concept of enthalpy of formation

Enthalpy of formation (\( \Delta H_f \)) is the change in enthalpy when 1 mol of a substance is formed from its elements in their standard states. Step 3: Write down the equation relating enthalpy of combustion and enthalpy of formation

Equation for enthalpy of combustion and enthalpy of formation

The equation that relates enthalpy of combustion (\( \Delta H_c \)) and enthalpy of formation of reactants and products is: \[ \Delta H_c = \sum n_p \Delta H_{f,p} - \sum n_r \Delta H_{f,r} \] Where: - \( \Delta H_{c} \) is the enthalpy of combustion, - \( \Delta H_{f,p} \) is the enthalpy of formation of products, - \( \Delta H_{f,r} \) is the enthalpy of formation of reactants, - \( n_p \) is the moles of each product, - \( n_r \) is the moles of each reactant. Step 4: Compute the enthalpy of combustion using given data

Calculating the enthalpy of combustion

Using the given data from Appendix C, lookup the enthalpy of formation values for CO2 and H2O. We obtain the following values: - \( \Delta H_{f, \mathrm{CO}_{2}} = -393.5\, \mathrm{kJ}\, \mathrm{mol}^{-1} \) - \( \Delta H_{f, \mathrm{H}_{2} \mathrm{O}} = -285.8\, \mathrm{kJ}\, \mathrm{mol}^{-1} \) Now, substitute the above values and the given enthalpy change into Equation 1: \[ -1790 = \left[ 3 \times \left( -393.5 \right) + 3 \times \left( -285.8 \right) \right] - \left[ \Delta H_{f, \mathrm{C}_{3} \mathrm{H}_{6} \mathrm{O}} + 4 \times \left( 0 \right) \right] \] Since the enthalpy of formation of O2 in its standard state (O2 gas) is zero, we can simplify the equation as: \[ -1790 = \left[ -1180.5 - 857.4 \right] - \Delta H_{f, \mathrm{C}_{3} \mathrm{H}_{6} \mathrm{O}} \] Step 5: Solve for the enthalpy of formation of acetone

Calculate the enthalpy of formation of acetone

Rearrange the equation in Step 4 to find \( \Delta H_{f, \mathrm{C}_{3} \mathrm{H}_{6} \mathrm{O}} \): \[ \Delta H_{f, \mathrm{C}_{3} \mathrm{H}_{6} \mathrm{O}} = -1180.5 - 857.4 - \left(-1790\right) \] Now, calculate the final value: \[ \Delta H_{f, \mathrm{C}_{3} \mathrm{H}_{6} \mathrm{O}} = -2037.9 + 1790 = -247.9\, \mathrm{kJ\, mol^{-1}} \] The enthalpy of formation of acetone, \( \Delta H_{f, \mathrm{C}_{3} \mathrm{H}_{6} \mathrm{O}} \), is approximately \( -247.9\, \mathrm{kJ\, mol^{-1}} \).

Key concepts

Combustion Reaction

A combustion reaction is a fundamental chemical process where a substance, usually a hydrocarbon, reacts with oxygen to produce carbon dioxide, water, and energy. Acetone, with the chemical formula \( \text{C}_3\text{H}_6\text{O} \), undergoes complete combustion in the given scenario, meaning it burns entirely with \( \text{O}_2 \) to yield \( \text{CO}_2 \) and \( \text{H}_2\text{O} \). These reactions are exothermic, meaning they release a significant amount of energy as heat. This release of energy is crucial in everyday applications, from powering vehicles to heating homes.

Key features to remember about combustion reactions include:

• They require oxygen (\( \text{O}_2 \)).
• They produce energy, typically in the form of heat.
• They are generally characterized by the formation of \( \text{CO}_2 \) and \( \text{H}_2\text{O} \).

Understanding this concept is crucial when analyzing thermochemistry problems, as it helps predict and calculate the energy exchanges involved.

Enthalpy of Combustion

The enthalpy of combustion, represented as \( \Delta H_c \), is the heat change when one mole of a substance combusts completely with oxygen under standard conditions. In other words, it measures the energy released as a substance reacts with oxygen. Enthalpy values are negative for exothermic reactions, which include most combustion reactions, because they release heat. In the exercise provided, the enthalpy of combustion for acetone is \(-1790 \text{ kJ/mol}\), indicating a significant release of energy.

To understand the enthalpy of combustion, remember the following:

• It is a measure of energy change during a reaction.
• It has units of kilojoules per mole (\( \text{kJ/mol} \)).
• It is negative for exothermic reactions.

Combustion reactions are vital in energy production industries, where understanding and controlling the enthalpy changes allow for efficient energy utilization.

Thermochemistry

Thermochemistry is the branch of chemistry that studies the heat involved during chemical reactions. It focuses on the energy changes that accompany chemical transformations. In thermochemistry, we often use terms like enthalpy and exothermic or endothermic reactions to describe how energy is absorbed or released during a reaction.

Here's what you need to know to grasp thermochemistry better:

• **Enthalpy (\(H\))** is the total heat content of a system.
• **Exothermic reactions** release heat, characterized by negative \(\Delta H\).
• **Endothermic reactions** absorb heat, with positive \(\Delta H\).
• Thermochemistry provides insights into reaction spontaneity and energy conservation.

A solid understanding of thermochemistry principles allows chemists to design processes that are energy-efficient and safe, balancing the release and absorption of energy to optimize reactions.