Question

Complete combustion of $1\mathrm{mol}$ of acetone $\left({\mathrm{C}}_3{\mathrm{H}}_6\mathrm{O}\right)$ liberates $1790\mathrm{kJ}$ : ${\mathrm{C}}_3{\mathrm{H}}_6\mathrm{O}(l)+4{\mathrm{O}}_2(g)⟶3{\mathrm{CO}}_2(g)+3{\mathrm{H}}_2\mathrm{O}(l)$ $\mathrm{\Delta }{H}^{\circ }=-1790\mathrm{kJ}$ Using this information together with data from Appendix $C$, calculate the enthalpy of formation of acetone.

Short answer

The enthalpy of formation of acetone is calculated using Hess's Law and given standard enthalpies of formation for oxygen, carbon dioxide, and water. The enthalpy of formation of acetone is found to be 609.5 kJ/mol.

Step-by-step solution

Write the reaction for the enthalpy of formation of acetone

The enthalpy of formation of acetone can be expressed by the following reaction: $$\mathrm{C}(s)+3{\mathrm{H}}_2(g)+\frac{1}{2}{\mathrm{O}}_2(g)⟶{\mathrm{C}}_3{\mathrm{H}}_6\mathrm{O}(l)$$

Write the given reaction and its enthalpy change

We are given the complete combustion of acetone reaction and its enthalpy change: $${\mathrm{C}}_3{\mathrm{H}}_6\mathrm{O}(l)+4{\mathrm{O}}_2(g)⟶3{\mathrm{CO}}_2(g)+3{\mathrm{H}}_2\mathrm{O}(l)\mathrm{\Delta }{H}^{\circ }=-1790\mathrm{kJ}$$

Use data from Appendix C to find standard enthalpies of formation

Use standard enthalpies of formation for oxygen ($O_2$), carbon dioxide ($C{O}_2$), and water ($H_{2}O$) given in the appendix table: $\mathrm{\Delta }{H}_f^{\circ }[{\mathrm{O}}_2]$ = 0 kJ/mol (as it is an element in its standard state) $\mathrm{\Delta }{H}_f^{\circ }[{\mathrm{CO}}_2]$ = -393.5 kJ/mol $\mathrm{\Delta }{H}_f^{\circ }[{\mathrm{H}}_2\mathrm{O}]$ = -285.8 kJ/mol Comment: Oxygen is an element in its standard state, so the standard enthalpy of formation is zero.

Apply Hess's Law to find enthalpy of formation of acetone

Using Hess's Law, we can relate the enthalpy changes of the given reaction and the enthalpy of formation of acetone to the standard enthalpies of formation of the other substances: $\mathrm{\Delta }{H}_{rxn}^{\circ }$ = $\mathrm{\Sigma }(\mathrm{\Delta }{H}_f^{\circ }\times \mathrm{moles}\mathrm{of}\mathrm{products})$ - $\mathrm{\Sigma }(\mathrm{\Delta }{H}_f^{\circ }\times \mathrm{moles}\mathrm{of}\mathrm{reactants})$ Substituting the known values: -1790 kJ = $[3\times (-393.5)+3\times (-285.8)]-[\mathrm{\Delta }{H}_f^{\circ }[{\mathrm{C}}_3{\mathrm{H}}_6\mathrm{O}]+4\times 0]$

Solve for the unknown enthalpy of formation of acetone

Rearranging the equation and solving for $\mathrm{\Delta }{H}_f^{\circ }[{\mathrm{C}}_3{\mathrm{H}}_6\mathrm{O}]$: $\mathrm{\Delta }{H}_f^{\circ }[{\mathrm{C}}_3{\mathrm{H}}_6\mathrm{O}]$ = $[3\times (-393.5)+3\times (-285.8)]-(-1790)$ $\mathrm{\Delta }{H}_f^{\circ }[{\mathrm{C}}_3{\mathrm{H}}_6\mathrm{O}]$ =$-1180.5+1790$ $\mathrm{\Delta }{H}_f^{\circ }[{\mathrm{C}}_3{\mathrm{H}}_6\mathrm{O}]$ = $609.5$ kJ/mol The enthalpy of formation of acetone is 609.5 kJ/mol.

Key concepts

Hess's Law

Hess's Law is a critical principle in chemistry that allows us to calculate the enthalpy change for a given reaction. It states that the total enthalpy change for a chemical reaction is the same, no matter how it is carried out, as long as it goes from the same reactants to the same products. This means we can measure enthalpy changes in steps and add them together to find the overall enthalpy change.

This law is especially useful when the reaction occurs in multiple steps or when it's hard to directly measure the enthalpy change. By using known enthalpies of formation (like in the problem statement), we can find unknown enthalpies, such as that of acetone, by adding and subtracting these values appropriately.

Hess's Law makes extensive use of the concept of standard enthalpies of formation, which are based on the formation of 1 mole of a compound from its elements in their standard states. These values are tabulated and widely used because they simplify calculations. Thanks to Hess's Law, even complex multi-step reactions become manageable.

Combustion Reaction

A combustion reaction is a high-energy chemical reaction where a substance reacts rapidly with oxygen, releasing energy in the form of heat and light. The typical products of combustion reactions involving hydrocarbons, such as acetone, are carbon dioxide and water.

These reactions are exothermic, meaning they release energy. The example in the exercise involves the combustion of acetone where acetone (${\mathrm{C}}_3{\mathrm{H}}_6\mathrm{O}$) reacts with oxygen to produce carbon dioxide and water. This specific reaction has a known enthalpy change ($\mathrm{\Delta }{H}^{\circ }=-1790\mathrm{kJ}$), indicating the amount of energy released per mole of acetone burned.

Understanding combustion reactions helps in comprehending concepts of energy transfer, pollution, and even real-world applications such as fuel consumption in engines. The negative sign of the enthalpy change in combustion indicates the exothermic nature—energy is leaving the system.

Standard Enthalpy

Standard enthalpy refers to the heat content change under standard conditions, which include a pressure of 1 atmosphere and a temperature of 25°C (298 K). When we talk about the standard enthalpy of formation, $\mathrm{\Delta }{H}_f^{\circ }$, it's the enthalpy change when one mole of a compound is formed from its elements in their standard states.

These values are critical for calculating unknown enthalpy changes using Hess's Law. In the problem solution, standard enthalpies of formation were used for elements and compounds like ${\mathrm{CO}}_2$, ${\mathrm{H}}_2\mathrm{O}$, and ${\mathrm{O}}_2$, combining these allows the determination of acetone's enthalpy of formation.

The elements in their most stable state, such as ${\mathrm{O}}_2$ (gas), and $\mathrm{C}$ (graphite), have a standard enthalpy of formation of zero, serving as reference points. These enthalpies are often tabulated for convenience and are essential for solving thermodynamic problems across chemistry. Got a reaction? Just pick up the standard enthalpy values and apply the magic of Hess's Law for answers!