Problem 13 - Chemistry: Introducing Inorganic, Organic and Physical Chemistry

Question

Calculate the enthalpy change when gaseous benzene \(\left(\mathrm{C}_{6} \mathrm{H}_{6}\right)\) dissociates into gaseous atoms at \(298 \mathrm{K}\). Carry out the calculation by two different methods using the data in (a) and (b) below. Comment on the difference in the values you obtain by the two methods. (Section 13.3 ) (a) Assume benzene molecules contain three single and three double carbon- carbon bonds, and use mean bond enthalpy data. (Mean bond enthalpies/kJmol- \(^{-1}:\) C-C \(, 347\); \(\mathrm{C}=\mathrm{C}, 612 ; \mathrm{C}-\mathrm{H}, 412 .\) (b) The enthalpy change of combustion of liquid benzene at \(298 \mathrm{K}\) is \(-3267.4 \mathrm{kJmol}^{-1}\). The enthalpy change of vaporization of benzene at \(298 \mathrm{K}\) is \(+33.9 \mathrm{kJ} \mathrm{mol}^{-1}\) \\[ \begin{array}{l} \left(\mathrm{A}_{1} \mathrm{H}_{298}^{\circ} / \mathrm{kJmol}^{-1}: \mathrm{CO}_{2}(\mathrm{g}),-393.5 ; \mathrm{H}_{2} \mathrm{O}(0,-285.8 ; \mathrm{C}(\mathrm{g})\right. \\ 716.7 ; \mathrm{H}(\mathrm{g}), 218 .) \end{array} \\]

Step-by-Step Solution

Answer

Using mean bond enthalpies gives 5530 kJ/mol, while enthalpy changes yield 9266.9 kJ/mol due to different assumptions.

Step 1: Understanding Benzene Dissociation

In this problem, you need to calculate the enthalpy change when one mole of gaseous benzene dissociates into gaseous atoms. The chemical reaction can be represented as \( \mathrm{C}_6\mathrm{H}_6 (g) \rightarrow 6\mathrm{C} (g) + 6 \mathrm{H} (g) \).

Step 2: Using Mean Bond Enthalpies (Method a)

For this calculation, assume benzene contains three C-C single bonds, three C=C double bonds, and six C-H bonds. Calculate the total bond enthalpy of all bonds broken: \[\Delta H = [3 \times 347 + 3 \times 612 + 6 \times 412] = 5530 \text{ kJ/mol} \]Since dissociation requires breaking all the bonds, this value is the enthalpy change of the dissociation process.

Step 3: Using Enthalpy Changes & Atomization (Method b)

Find the standard enthalpy of atomization of benzene in this method. Use the given values: combustion enthalpy of benzene is \(-3267.4 \text{ kJ/mol}\), and vaporization is \(33.9 \text{ kJ/mol}\).1. Combustion reaction: \( \mathrm{C}_6\mathrm{H}_6(l) + \frac{15}{2} \mathrm{O}_2(g) \rightarrow 6\mathrm{CO}_2(g) + 3\mathrm{H}_2\mathrm{O}(l) \)2. Calculate enthalpy for gases formed: \[\Delta H = [6 \times (-393.5) + 3 \times (-285.8)] = -3267.4 \text{ kJ/mol} \]3. Adjust for vapor to atomize gasses: Add vaporization enthalpy to combustion: \[\Delta H = -3267.4 + 33.9 + 6(716.7) + 6(218) = 9266.9 \text{ kJ/mol} \]

Step 4: Compare the Two Methods

The results from the two methods are different: - Method (a) using mean bond enthalpies results in \(5530 \text{ kJ/mol}\). - Method (b) using enthalpy changes yields \(9266.9 \text{ kJ/mol}\).The difference arises due to the assumptions made for bond strengths using average values in method (a) versus precise enthalpy process measurements in method (b). Method (b) is generally more accurate due to incorporation of all specific energy changes.

Key Concepts

Mean Bond Enthalpy

Mean bond enthalpy is the average energy needed to break one mole of a particular type of bond in a gaseous molecule. It's an important concept in chemistry because it provides a way to estimate the energy change during a reaction. When calculating the enthalpy change for a reaction using mean bond enthalpies, one must take into account all the bonds that are broken and the ones that are formed during the reaction.
In the scenario of benzene dissociation, we assume each type of bond in benzene (C-C single, C=C double, and C-H) contributes equally to the bond breaking process. This is a simplification that helps us estimate the dissociation energy but might not always reflect the exact energy changes because the context of each bond, such as surrounding atoms or molecular structure, can affect its strength.

Combustion Enthalpy

Combustion enthalpy refers to the energy released when one mole of a substance fully reacts with oxygen under standard conditions. This process usually forms carbon dioxide and water if the substance is organic, like benzene.
For benzene, the combustion reaction involves a large energy release, which we can use to determine how benzene's structure contributes to its stability. The large negative value of the combustion enthalpy indicates that benzene releases a significant amount of energy, making it a powerful source of heat. This value, combined with other enthalpic data, can help to calculate the energy required for vaporization and subsequently for atomization.

Benzene Atomization

Atomization is the process where molecules are completely broken down into individual atoms. In the context of benzene, atomization involves breaking it into carbon and hydrogen atoms, which requires a significant energy input.
Method (b) to calculate enthalpy change involves indirect calculation from combustion and vaporization data. We use the known values of combustion, then adjust for the vaporization of benzene, and further calculate the energy necessary to convert the molecular substances into atoms. The comprehensive energy changes that include vaporization and standard states of elemental gases make this method accurate, though more complex than using mean bond enthalpies.

Enthalpy of Vaporization

Enthalpy of vaporization is the energy required to change a substance from liquid to gas at its boiling point under standard pressure. This energy is necessary to overcome the intermolecular forces in the liquid phase.
In the analysis of benzene, the enthalpy of vaporization plays a crucial role in moving from liquid benzene (in the combustion data) to its gaseous state, which is necessary before full atomization.
The relatively lower value of vaporization enthalpy compared to combustion or atomization reflects that intermolecular forces in the liquid phase are rather moderate. This concept shows the importance of phase change energies in calculated enthalpy changes, demonstrating how transitions between states (liquid to gas) affect the overall enthalpy calculation.