Question

Complete combustion of 1 mol of acetone \(\left(\mathrm{C}_{3} \mathrm{H}_{6} \mathrm{O}\right)\) liberates \(1790 \mathrm{kJ} :\) $$\begin{aligned} \mathrm{C}_{3} \mathrm{H}_{6} \mathrm{O}(l)+4 \mathrm{O}_{2}(g) \longrightarrow & 3 \mathrm{CO}_{2}(g)+3 \mathrm{H}_{2} \mathrm{O}(l) \\ & \quad \quad \quad \quad \quad \quad \quad \Delta H^{\circ}=-1790 \mathrm{kJ} \end{aligned}$$ 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 -1380.9 kJ/mol.

Step-by-step solution

Write the equations for the standard enthalpies of formation

The standard enthalpies of formation are the enthalpies required to form 1 mol of a substance from its elements in their standard states. The equation for the formation of acetone is: \(C(s) + 3H_2(g) + \frac{1}{2}O_2(g) \to C_{3}H_{6}O(l)\) The equations for the formation of CO2 and H2O are: \(C(s) + O_2(g) \to CO_{2}(g)\) \(H_2(g) + \frac{1}{2}O_2(g) \to H_{2}O(l)\)

Determine the ∆H°f for CO2 and H2O

The standard enthalpies of formation for CO2(g) and H2O(l) can be found in Appendix C. Using this information, we have: ∆H°f(CO2(g))=-393.5 kJ/mol ∆H°f(H2O(l))=-285.8 kJ/mol

Set up the Hess's Law equation

Hess's Law states that the total enthalpy change for a reaction is the sum of the individual enthalpy changes for each step in the reaction. For the given problem, we can write the Hess's Law equation using the balanced equation of acetone combustion and the standard enthalpies of formation, as follows: \(\Delta H_{combustion}^{°} = 3\Delta H_{f(CO2(g))}^° + 3\Delta H_{f(H2O(l))^° − (\Delta H_{f(acetone)}^° + 4\Delta H_{f(O2(g))}^°)\) We know that, for O2(g), ∆H°f = 0 kJ/mol, since it is an element in its standard state. So we can simplify the equation: \(\Delta H_{combustion}^{°} = 3\Delta H_{f(CO2(g))}^° + 3\Delta H_{f(H2O(l))^° − \Delta H_{f(acetone)}^°\)

Calculate the ∆H°f of acetone

Now, plug in the known values and solve for the standard enthalpy of formation of acetone: \(-1790 kJ/mol = 3(-393.5 kJ/mol) + 3(-285.8 kJ/mol) – \Delta H_{f(acetone)}^°\) Solve the equation for ∆H°f(acetone): \(\Delta H_{f(acetone)}^° = 3(-393.5 kJ/mol) + 3(-285.8 kJ/mol) + 1790 kJ/mol\) \(\Delta H_{f(acetone)}^° = -1380.9 kJ/mol\) So, the standard enthalpy of formation of acetone is -1380.9 kJ/mol.

Key concepts

Hess's Law

Hess's Law is a key principle in thermochemistry. It states that the total enthalpy change for a chemical reaction is the same, no matter how it is carried out, in one step or multiple steps. This means that we can calculate the enthalpy change for a complex reaction by breaking it down into a series of simpler steps for which enthalpy changes are known.

Here's how it works in practice:

• Identify the initial and final states of the reaction.
• Break the reaction into multiple steps.
• Use known enthalpy changes for these steps to find the total enthalpy change.

In our exercise, we used Hess's Law by writing an equation that combined the enthalpy changes for the formation of carbon dioxide and water and related them to the combustion of acetone. This allowed us to find the enthalpy change for forming acetone from its elements.

Combustion Reaction

A combustion reaction occurs when a substance reacts quickly with oxygen, releasing energy in the form of heat or light. This energy release often makes combustion reactions very exothermic and noticeable.

In the context of our exercise:

• The combustion of acetone involves reacting with oxygen to produce carbon dioxide and water.
• The reaction is highly exothermic, releasing 1790 kJ of energy.
• Combustion reactions are important in energy production, like in engines and fires.

Understanding the energy changes in combustion reactions helps us calculate unknown values, like the standard enthalpy of formation, using known data.

Standard Enthalpies

Standard enthalpies of formation (\( \Delta H_{f}^{\circ} \)) are defined as the enthalpy change when one mole of a compound is formed from its elements in their standard states. These values are crucial in calculating the overall enthalpy change of reactions using known enthalpy changes.

Key points include:

• Elements in their standard states have a formation enthalpy of zero.
• Standard enthalpies are usually measured under conditions of 1 atm and 25°C (298 K).
• These values help us use Hess's Law effectively to calculate unknown enthalpies, like in the acetone example.

In our exercise, we used the standard enthalpies of CO2 and H2O to deduce the enthalpy of formation for acetone, tying all these concepts together into a clear calculation.