Problem 75 - Chemistry: The Central Science in SI Units, Global Edition

Question

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

Step-by-Step Solution

Answer

The standard enthalpy of formation of acetone is 247.9 kJ/mol.

Step 1: Write the combustion equation with standard enthalpy changes

Write the given balanced combustion equation and the standard enthalpy change: \[C_3H_6O(l) + 4O_2(g) \rightarrow 3CO_2(g) + 3H_2O(l)\] \[\Delta H_1^\circ = -1790 kJ\]

Step 2: Write the formation equations for each substance involved in the combustion

Write the formation equations for acetone, oxygen, carbon dioxide, and water: 1. Acetone (C3H6O) formation: \[3C(s) + 3H_2(g) + \frac{1}{2}O_2(g) \longrightarrow C_3H_6O(l)\] \[\Delta H_2^\circ= ?\] 2. Oxygen (O2) formation: \[\frac{1}{2}O_2(g) \longrightarrow O_2(g)\] \[\Delta H_3^\circ= 0kJ\] (standard enthalpy of formation of an element in its standard state is zero) 3. Carbon dioxide (CO2) formation: \[C(s) + O_2(g) \longrightarrow CO_2(g)\] \[\Delta H_4^\circ= -393.5 kJ/mol\] (use value from Appendix C) 4. Water (H2O) formation: \[H_2(g) + \frac{1}{2}O_2(g) \longrightarrow H_2O(l)\] \[\Delta H_5^\circ= -285.8 kJ/mol\] (use value from Appendix C)

Step 3: Multiply each formation equation with an appropriate factor

Multiply the formation equations to match the coefficients of the substances in the combustion equation, keeping the enthalpy changes in mind: 1. Acetone formation: \[3C(s) + 3H_2(g) + \frac{1}{2}O_2(g) \longrightarrow C_3H_6O(l)\] \[\Delta H_2^\circ= ?\] 2. Oxygen formation: (multiplied by 4) \[2O_2(g) \longrightarrow 4O_2(g)\] \[\Delta H_3^\circ= 4 \times 0kJ = 0kJ\] 3. Carbon dioxide formation: (multiplied by 3) \[3C(s) + 3O_2(g) \longrightarrow 3CO_2(g)\] \[\Delta H_4^\circ= 3 \times -393.5 kJ/mol = -1180.5 kJ/mol\] 4. Water formation: (multiplied by 3) \[3H_2(g) + \frac{3}{2}O_2(g) \longrightarrow 3H_2O(l)\] \[\Delta H_5^\circ= 3 \times -285.8 kJ/mol = -857.4 kJ/mol\]

Step 4: Use Hess's Law to find the enthalpy of formation of acetone

Hess's Law states that the sum of the series of reactions is equal to the sum of the reaction enthalpies. So, \(\Delta H_1^\circ = \Delta H_2^\circ + \Delta H_3^\circ + \Delta H_4^\circ + \Delta H_5^\circ\). Plug in the known values and solve for the standard enthalpy of formation of acetone, \(\Delta H_2^\circ\): \[-1790 kJ = \Delta H_2^\circ + 0 kJ - 1180.5 kJ - 857.4 kJ\] Solve for \(\Delta H_2^\circ\): \[\Delta H_2^\circ = -1790kJ + 1180.5kJ + 857.4kJ\] \[\Delta H_2^\circ = -1790 + 2037.9\] \[\Delta H_2^\circ = 247.9 kJ/mol\] The standard enthalpy of formation of acetone is 247.9 kJ/mol.

Key Concepts

Hess's Law

Hess's Law is a foundational principle in chemistry that allows us to determine the enthalpy change of a reaction when it is not feasible to measure it directly. By leveraging the idea that the enthalpy change of an overall process is the sum of the enthalpy changes of its individual steps, we can calculate otherwise inaccessible values.
It operates under the premise that enthalpy is a state function. This means the total enthalpy change for a chemical transformation depends only on the initial and final states and not on the path taken.
This is advantageous when dealing with intricate reactions, as it lets us break them into more straightforward reactions, each with known enthalpy changes. Hence, the equation \(abla H_1 = abla H_2 + abla H_3 + \ldots \) sums these simpler reactions to get the overall enthalpy change.
For example, in the solution, we utilize Hess's Law to compute the standard enthalpy of formation of acetone. By knowing the combustion reaction's enthalpy, one can backtrack through the formation equations of the involved substances to find the desired value.

Combustion Reactions

Combustion reactions occur when a substance combines with oxygen to release energy in the form of heat. They're crucial in everyday applications, from engines to energy production.
A typical combustion reaction for organic compounds, like hydrocarbons, involves complete oxidation that produces carbon dioxide and water.

• The general format for combustion is: \ Hydrocarbon \(+(O_{2})\) → \((CO_{2}) + (H_{2}O)\).
• This transformation releases a significant amount of energy, as shown in our acetone example with the liberation of \(1790\) kJ/mol.

These reactions are not only vital for their energy yield but also role in calculating thermodynamic properties of substances like standard enthalpies. By assessing the products like \((CO_{2})\) and \((H_{2}O)\) and knowing their formation enthalpies, we can deduce the involved reactant's standard enthalpy, as we demonstrated with acetone.

Standard Enthalpies of Formation

Standard enthalpies of formation are the heat changes associated with the formation of one mole of a compound from its elements in their standard states. This is an essential concept for constructing complex chemical calculations like those involving Hess's Law.
The standard state refers to the most stable form of an element at \(1\) atm pressure and a specified temperature, typically \(25\)°C. This consistent baseline allows chemists to compile extensive datasets, often found in appendices, like Appendix C in textbooks.
In chemical reactions, especially when calculating unknowns like in our acetone problem, these values are indispensable. The known enthalpies of formation for reactants or products allow us to reverse-engineer enthalpies of less easily measured reactions by using the calculated sums.

• For example, \((O_{2})\) has a formation enthalpy of zero, as it's in its elemental form.
• \((CO_{2})\) and \((H_{2}O)\) have known enthalpies of \(-393.5\) kJ/mol and \(-285.8\) kJ/mol, respectively, values critical to our acetone enthalpy calculation.

Utilizing these standard values effectively facilitates finding unknown enthalpy changes in diverse chemical processes.