Question

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

Short answer

The standard enthalpy of formation of acetone is equal to \(-247.9 \mathrm{~kJ/mol}\).

Step-by-step solution

Write down the enthalpy of formation equation for the reaction of interest.

Recall the general equation for standard enthalpy of formation: \[ \Delta H^{\circ} = \sum \nu n^{'} H_{f}^{\circ}(\text{products}) - \sum \nu n^{'} H_{f}^{\circ}(\text{reactants}) \] In this case, the reaction is: \[ \mathrm{C}_{3} \mathrm{H}_{6} \mathrm{O}(l)+4 \mathrm{O}_{2}(g) \longrightarrow 3 \mathrm{CO}_{2}(g) + 3 \mathrm{H}_{2} \mathrm{O}(l) \]

Write down the enthalpies of formation for the known substances

From Appendix C, we have the standard enthalpies of formation for \(\mathrm{O}_{2}(g)\), \(\mathrm{CO}_{2}(g)\), and \(\mathrm{H}_{2} \mathrm{O}(l)\): 1. \(\mathrm{O}_{2}(g): \, H_{f}^{\circ}[\mathrm{O}_{2}(g)] = 0 \mathrm{~kJ/mol}\) 2. \(\mathrm{CO}_{2}(g): \, H_{f}^{\circ}[\mathrm{CO}_{2}(g)] = -393.5 \mathrm{~kJ/mol}\) 3. \(\mathrm{H}_{2} \mathrm{O}(l): \, H_{f}^{\circ}[\mathrm{H}_{2} \mathrm{O}(l)] = -285.8 \mathrm{~kJ/mol}\)

Plug in the known enthalpies of formation into the enthalpy of formation equation

Using the general equation, plug in the known enthalpies of formation, and solve for the standard enthalpy of formation for acetone: \[ \begin{aligned} -1790 \mathrm{~kJ} &= \Big[3 \times H_{f}^{\circ}[\mathrm{CO}_{2}(g)] + 3 \times H_{f}^{\circ}[\mathrm{H}_{2} \mathrm{O}(l)] \Big] - \Big[H_{f}^{\circ}[\mathrm{C}_{3} \mathrm{H}_{6} \mathrm{O}(l)] + 4 \times H_{f}^{\circ}[\mathrm{O}_{2}(g)] \Big]\\ &= \Big[3 \times (-393.5) + 3 \times (-285.8) \Big] - \Big[H_{f}^{\circ}[\mathrm{C}_{3} \mathrm{H}_{6} \mathrm{O}(l)] + 4 \times 0 \Big] \end{aligned} \]

Solve for the standard enthalpy of formation of acetone

Now we solve the equation for \(H_{f}^{\circ}[\mathrm{C}_{3}\mathrm{H}_{6}\mathrm{O}(l)]\): \[ \begin{aligned} H_{f}^{\circ}[\mathrm{C}_{3} \mathrm{H}_{6} \mathrm{O}(l)] &= \Big[3 \times (-393.5) + 3 \times (-285.8) \Big] - (-1790) \\ &= (-1180.5 - 857.4) + 1790 \\ &= 1790 - 2037.9\\ &= -247.9 \mathrm{~kJ/mol} \end{aligned} \] So, the standard enthalpy of formation of acetone is equal to \(-247.9 \mathrm{~kJ/mol}\).

Key concepts

Standard Enthalpy of Formation

The standard enthalpy of formation is a key concept in thermochemistry, representing the heat absorbed or released when one mole of a compound is formed from its elements in their standard states. When we talk about standard states, we mean the most stable form of an element at 1 atm pressure and a specified temperature, generally 298 K (25°C).

Understanding the process involves knowing how to use a standard enthalpy of formation table, which lists values like those for \(\[ H_{f}^{\circ}[\mathrm{CO}_{2}(g)] = -393.5 \, \mathrm{~kJ/mol} \]\) and \(\[ H_{f}^{\circ}[\mathrm{H}_{2} \mathrm{O}(l)] = -285.8 \, \mathrm{~kJ/mol} \]\).- For compounds that are given, like acetone (\(\mathrm{C}_{3} \mathrm{H}_{6} \mathrm{O}\)), we often have to calculate this value using combustion reactions. - The values are crucial to determine reaction enthalpies by using Hess's Law where the sum of products’ enthalpies are subtracted by the sum of reactants’ enthalpies.Additionally, it's helpful to remember that the standard enthalpy of formation for any element in its elemental form, like \(\mathrm{O}_2(g)\), is zero.This intuitive approach allows for systematic calculations to unravel the energy changes in chemical reactions, such as in the combustion of acetone.

Combustion Reaction

A combustion reaction is a fundamental reaction type in chemistry where a substance combines with oxygen to release energy in the form of heat or light. This is commonly known as burning. In our example, acetone reacts with oxygen to form carbon dioxide and water.

This specific reaction can be written as: \(\[\mathrm{C}_{3} \mathrm{H}_{6} \mathrm{O}(l) + 4 \mathrm{O}_{2}(g) \longrightarrow 3 \mathrm{CO}_{2}(g) + 3 \mathrm{H}_{2} \mathrm{O}(l)\]\) with an enthalpy change, \(\Delta H^{\circ} = -1790 \, \mathrm{~kJ} \).- The negative value of \(\Delta H^{\circ}\) indicates the reaction is exothermic, meaning it releases energy. - These reactions are crucial for calculating the enthalpy of formation for various compounds, as shown in the exercise's step-by-step solution.Understanding these reactions lets us predict not only the energy output but also the necessary reactants’ amounts to achieve complete combustion. Combustion reactions are significantly important in various applications, such as energy production and material synthesis.

Chemical Thermodynamics

Chemical thermodynamics helps us understand energy transformations that occur during chemical reactions. It involves the study of enthalpy, entropy, and Gibbs free energy, which collectively describe the feasibility and energy change of chemical processes.

At the heart of chemical thermodynamics is the First Law of Thermodynamics, which states that energy cannot be created or destroyed, only converted. This principle guides us in calculating enthalpy changes in reactions like acetone combustion. - Using the enthalpy of formation values, we can apply Hess’s Law, which states that the total enthalpy change for a chemical reaction is the same regardless of the steps taken. - This is demonstrated in the exercise by comparing the energy absorbed and released to find the enthalpy of formation of acetone. Thermodynamics also dictates that spontaneous reactions typically result in a decrease in Gibbs free energy, highlighting the balance between enthalpy and entropy. By understanding these principles, we can predict whether reactions will occur and how much energy will be exchanged, as was determined for the combustion reaction in the solution.