Question

Stearic acid \(\left(\mathrm{C}_{18} \mathrm{H}_{36} \mathrm{O}_{2}\right)\) is a fatty acid, a molecule with a long hydrocarbon chain and an organic acid group (COOH) at the end. It is used to make cosmetics, ointments, soaps, and candles and is found in animal tissue as part of many saturated fats. In fact, when you eat meat, you are ingesting some fats containing stearic acid. (a) Write a balanced equation for the combustion of stearic acid to gaseous products. (b) Calculate \(\Delta H_{\mathrm{rm}}^{\circ}\) for this combustion \(\left(\Delta H_{\mathrm{i}}^{\circ}\right.\) of \(\mathrm{C}_{18} \mathrm{H}_{36} \mathrm{O}_{2}=\) \(-948 \mathrm{~kJ} / \mathrm{mol})\) (c) Calculate the heat \((q)\) released in \(\mathrm{kJ}\) and kcal when \(1.00 \mathrm{~g}\) of stearic acid is burned completely. (d) A candy bar contains \(11.0 \mathrm{~g}\) of fat and provides \(100 .\) Cal from fat; is this consistent with your answer for part (c)?

Short answer

The balanced equation is \[ \text{C}_{18}\text{H}_{36}\text{O}_{2} + 26 \text{O}_{2} \rightarrow 18 \text{CO}_{2} + 18 \text{H}_{2}\text{O} \]. \(abla H_r^{\text{o}} = -11021.4 \text{kJ/mol}\). Burning 1.00g releases -9.27 kcal. The candy bar data is consistent.

Step-by-step solution

Write the balanced combustion equation

To write the balanced equation, note that the combustion of stearic acid \(\text{C}_{18}\text{H}_{36}\text{O}_{2}\) reacts with oxygen \(\text{O}_{2}\) to produce carbon dioxide \(\text{CO}_{2}\) and water \(\text{H}_{2}\text{O}\). The balanced equation is: \[ \text{C}_{18}\text{H}_{36}\text{O}_{2} + 26 \text{O}_{2} \rightarrow 18 \text{CO}_{2} + 18 \text{H}_{2}\text{O} \]

Calculate \(abla H_r^{\text{o}}\) for the combustion

To find \( \text{abla H}_r^{\text{o}} \), use the enthalpy of formation of the products and reactants. The standard enthalpy of formation for \( \text{CO}_{2} \) is \( -393.5 \text{abla kJ/mol} \), and for \(\text{H}_{2}\text{O}\text{(g)} \) is \( -241.8 \text{abla kJ/mol} \). The equation is: \[ \text{abla H}_r^{\text{o}} = [18\text{abla fH}^{\text{o}}(\text{CO}_{2}) + 18\text{abla fH}^{\text{o}}(\text{H}_{2}\text{O})] - [\text{abla fH}^{\text{o}}(\text{C}_{18}\text{H}_{36}\text{O}_{2})] \] Substitute the values: \[ \text{abla H}_r^{\text{o}} = [18(-393.5) + 18(-241.8)] - (-948) = -11021.4 \text{kJ/mol} \]

Calculate the heat released (q) when 1.00g of stearic acid is burned

Find the molar mass of stearic acid: \[ \text{M}(\text{C}_{18}\text{H}_{36}\text{O}_{2}) = 18(12) + 36(1) + 2(16) = 284 \text{g/mol} \] Calculate moles of 1.00 g stearic acid: \[ \text{moles} = \frac{1.00 \text{g}}{284 \text{g/mol}} = 0.00352 \text{mol} \] Use the \(abla H_r^{\text{o}} \) value: \[ q = 0.00352 \text{mol} \times -11021.4 \text{kJ/mol} = -38.78 \text{kJ} \] Convert to kcal: \[ -38.78 \text{kJ} \times \frac{1 \text{kcal}}{4.184 \text{kJ}} = -9.27 \text{kcal} \]

Compare with candy bar data

Given: 11.0g of fat provides 100 Cal (or 100 kcal). Calculate kcal/gram: \[ \frac{100 \text{kcal}}{11.0 \text{g}} = 9.09 \text{kcal/g} \] The heat released per gram in part (c) is approximately 9.27 kcal/g. Compare: 9.27 kcal/g is consistent with 9.09 kcal/g considering significant figures.

Key concepts

enthalpy of formation

When discussing chemical reactions, we often need to know how much energy is needed to form molecules from their elements. This is known as the *enthalpy of formation*. The enthalpy of formation, symbolized as \(\text{ΔH}_f^{\text{o}}\), is the change in enthalpy when one mole of a compound is formed from its elements under standard conditions. It's a measure of energy change, typically in kJ/mol.
In the combustion of stearic acid, we use the enthalpies of formation for the reactants and products. The enthalpy of formation for carbon dioxide \(\text{CO}_2\) and water \(\text{H}_2\text{O}\) are known values, -393.5 kJ/mol and -241.8 kJ/mol, respectively. These values indicate that energy is released when these compounds form.
By using these known \(\text{ΔH}_f^{\text{o}}\) values, we can calculate the total enthalpy change for the combustion reaction. This total change is helpful for understanding the energy impact of burning stearic acid.

balanced combustion equation

A balanced combustion equation shows how reactants combine to form products in precise amounts. For the combustion of stearic acid, we start with the molecule \(\text{C}_{18}\text{H}_{36}\text{O}_2\), which reacts with oxygen \(\text{O}_2\). The products of this reaction are carbon dioxide \(\text{CO}_2\) and water \(\text{H}_2\text{O}\).
To balance the equation, we ensure that the number of atoms for each element is equal on both sides of the equation. The balanced combustion equation for stearic acid is:
\[ \text{C}_{18}\text{H}_{36}\text{O}_2 + 26 \text{O}_2 \rightarrow 18 \text{CO}_2 + 18 \text{H}_2\text{O} \]
This tells us that one molecule of stearic acid needs 26 molecules of oxygen to fully react, producing 18 molecules of carbon dioxide and 18 molecules of water. Balancing is crucial because it follows the Law of Conservation of Mass, ensuring that the atoms we start with are accounted for in the products.

heat calculation

Heat calculation in chemical reactions involves determining the energy changes during the reaction. For stearic acid combustion, we need to find out how much energy is released. We previously calculated enthalpy change \(\text{ΔH}_r^\text{o}\), which was -11021.4 kJ/mol.
First, determine the molar mass of stearic acid:
\[ \text{M}(\text{C}_{18}\text{H}_{36}\text{O}_2) = 18(12) + 36(1) + 2(16) = 284 \text{g/mol} \]
Then, convert the mass of stearic acid to moles. For instance, 1.00g of stearic acid is equivalent to:
\[ \frac{1.00 \text{g}}{284 \text{g/mol}} = 0.00352 \text{mol} \]
Using the enthalpy change value:
\[ q = 0.00352 \text{mol} \times -11021.4 \text{kJ/mol} = -38.78 \text{kJ} \]
Finally, converting kJ to kcal:
\[ -38.78 \text{kJ} \times \frac{1 \text{kcal}}{4.184 \text{kJ}} = -9.27 \text{kcal} \]
These calculations show the energy released when burning a small sample of stearic acid and are consistent with the caloric values found in foods containing fats.