Problem 111 - Chemistry The Central Science

Question

Consider benzene \(\left(\mathrm{C}_{6} \mathrm{H}_{6}\right)\) in the gas phase. (a) Write the reaction for breaking all the bonds in \(\mathrm{C}_{6} \mathrm{H}_{6}(g)\), and use data in Appendix \(\mathrm{C}\) to determine the enthalpy change for this reaction. (b) Write a reaction that corresponds to breaking all the carbon-carbon bonds in \(\mathrm{C}_{6} \mathrm{H}_{6}(g) .\) (c) By combining your answers to parts (a) and (b) and using the average bond enthalpy for \(\mathrm{C}-\mathrm{H}\) from Table \(8.4,\) calculate the average bond enthalpy for the carbon-carbon bonds in \(\mathrm{C}_{6} \mathrm{H}_{6}(g)\). (d) Comment on your answer from part (c) as compared to the values for \(\mathrm{C}-\mathrm{C}\) single bonds and \(\mathrm{C}=\mathrm{C}\) double bonds in Table 8.4

Step-by-Step Solution

Answer

The average bond enthalpy for the carbon-carbon bonds in benzene is approximately -400 kJ/mol, which is exactly halfway between the bond enthalpies for carbon-carbon single bonds (346 kJ/mol) and carbon-carbon double bonds (614 kJ/mol). This suggests that benzene's carbon-carbon bond enthalpy shares characteristics of both single and double bonds equally.

Step 1: (a) Reaction for breaking all bonds in benzene

The reaction to break all bonds in benzene (C6H6) results in separate carbon and hydrogen atoms in the gas phase, as shown below \[ \mathrm{C}_{6} \mathrm{H}_{6}(g) \rightarrow 6 \,\mathrm{C}(g) + 6 \,\mathrm{H}(g) \] Now let's determine the enthalpy change for this reaction using the data from Appendix C.

Step 2: (a) Determining enthalpy change for the given reaction

To determine the enthalpy change for the given reaction, we need to use the heat of formation of the compounds involved. The heat of formation of benzene, carbon, and hydrogen gas can be found in Appendix C. \(\Delta H_{f}^{\circ}(\mathrm{C}_{6} \mathrm{H}_{6})=-82.9 \,\mathrm{kJ \cdot mol^{-1}}\) \(\Delta H_{f}^{\circ}(\mathrm{C})=0 \,\mathrm{kJ \cdot mol^{-1}}\) (since it's the standard state) \(\Delta H_{f}^{\circ}(\mathrm{H})=0 \,\mathrm{kJ \cdot mol^{-1}}\) (since it's the standard state) Using the formula \(\Delta H_{rxn}=\sum \Delta H_{f(\mathrm{products})} - \sum \Delta H_{f(\mathrm{reactants})}\), we get: \[ \begin{aligned} \Delta H_{rxn} &=[6\Delta H_{f}^{\circ}(\mathrm{C}) + 6\Delta H_{f}^{\circ}(\mathrm{H})] - \Delta H_{f}^{\circ}(\mathrm{C}_{6}\mathrm{H}_{6}) \\ &=[6(0)+6(0)]-(-82.9\,\mathrm{kJ}) \\ &=82.9\,\mathrm{kJ} \end{aligned} \] The enthalpy change for breaking all bonds in benzene is 82.9 kJ.

Step 3: (b) Reaction for breaking all carbon-carbon bonds in benzene

The reaction to break all carbon-carbon bonds in benzene results in separate carbon and hydrogen atoms in the gas phase but the hydrogen atoms are still bonded to the carbon atoms: \[ C_6 H_6 (g) \rightarrow 6CH(g) \]

Step 4: (c) Calculate the average bond enthalpy of carbon-carbon bonds in benzene

To calculate the average bond enthalpy of carbon-carbon bonds in benzene, we can use our answers from parts (a) and (b) combined with the average bond enthalpy of the C-H bond from Table 8.4. The average bond enthalpy for the C-H bond is 414 kJ/mol. First, let's determine the energy required to break all the C-H bonds. \(6 \times 414 \,\mathrm{kJ/mol} = 2484 \,\mathrm{kJ}\) Now subtract this energy from the energy required to break all bonds in benzene calculated in part (a). \(82.9\,\mathrm{kJ} - 2484\,\mathrm{kJ}= -2401\,\mathrm{kJ}\) Since there are six carbon-carbon bonds in benzene, we can now calculate the average bond enthalpy for carbon-carbon bonds in benzene. \(\frac{-2401\,\mathrm{kJ}}{6} \approx -400\,\mathrm{kJ/mol}\) The average bond enthalpy for the carbon-carbon bonds in benzene is approximately -400 kJ/mol.

Step 5: (d) Comparing average bond enthalpy to carbon-carbon single and double bonds

From Table 8.4, the bond enthalpies for carbon-carbon single and double bonds are: C-C single bond: 346 kJ/mol C=C double bond: 614 kJ/mol Our calculated average bond enthalpy for carbon-carbon bonds in benzene is -400 kJ/mol, which is halfway between 346 kJ/mol (C-C single bond) and 614 kJ/mol (C=C double bond) is 480 kJ/mol. Since our calculated value is not exactly halfway between the two bond types, we can determine which type of bond is more similar to that of benzene by comparing the differences. Difference to C-C single bond: 480 - 346 = 134 kJ/mol Difference to C=C double bond: 614 - 480 = 134 kJ/mol Interestingly, the differences are equal, indicating that benzene's carbon-carbon bond enthalpy is actually exactly halfway between the enthalpies of carbon-carbon single and double bonds.

Key Concepts

Benzene

Benzene is a fascinating and unique aromatic compound with the chemical formula \( \mathrm{C}_{6} \mathrm{H}_{6} \). In its structure, it forms a hexagonal ring composed of six carbon atoms, each bonded to one hydrogen atom. The bonds within benzene are of special interest due to their resonance stability. Unlike typical alkanes or alkenes where bonds are distinct single or double bonds, benzene exhibits a blend of these characteristics. This phenomenon is known as resonance, where the actual bond order is somewhere between a single and double bond. This confers added stability to the molecule, making it less reactive than might be expected for an unsaturated hydrocarbon. In practical terms, the stability of benzene affects how we calculate and interpret the enthalpy changes associated with reactions involving its bonds.

Enthalpy Change

Enthalpy change, denoted as \( \Delta H \), is an important concept when discussing chemical reactions, particularly those involving bond breaking or formation. It represents the heat absorbed or released during a reaction at constant pressure. In the context of benzene, determining the enthalpy change when breaking all its bonds gives insight into the strength and stability of its internal structure. Using the heats of formation from standard tables, we can calculate the enthalpy change for breaking all bonds in benzene in the gas phase. This calculation involves considering each type of bond present and their respective bond enthalpies. By deducing the total energy required to completely disassemble benzene into its constituent atoms, we gain a clearer picture of the energy dynamics in this aromatic compound.

Carbon-Carbon Bonds

Within benzene, the carbon-carbon (C-C) bonds warrant particular attention due to the unique sp2 hybridization and resonance effects. Typically, carbon-carbon bonds can exist as single (sp3 hybridized) with a bond enthalpy around 346 kJ/mol, or as double bonds (sp2 hybridized) with a bond enthalpy approximately 614 kJ/mol. However, the C-C bonds in benzene are not purely single or double bonds but display partial characteristics of both. This results in an average enthalpy that interestingly sits between those of traditional single and double bonds. The calculated average bond enthalpy of -400 kJ/mol reflects this mixed character. The negative sign here indicates net release when forming the bond from separate atoms, implying that breaking them would require input of energy. This distinct nature is what gives benzene its chemical resonance stability.

C-H Bond

Though often overshadowed by discussions about carbon-carbon interactions in benzene, the carbon-hydrogen (C-H) bonds are equally essential. Each carbon in benzene is bonded to a hydrogen atom in a typical single bond pattern. In contrast to the C-C bonds, the C-H bonds in benzene display more regular behavior, with a consistent bond enthalpy of about 414 kJ/mol. They do not participate in resonance as the C-C bonds do. This regularity simplifies their inclusion in enthalpy calculations, where the energy required to break all such bonds can be directly calculated by multiplying the bond enthalpy by the number of C-H bonds present. Understanding the nature and enthalpy of C-H bonds is crucial since breaking these specific bonds often composes a significant part of the overall energy change when deconstructing benzene.