In Exercise 89 in Chapter 8, the Lewis structures for benzene \(\left(\mathrm{C}_{6} \mathrm{H}_{6}\right)\) were drawn. Using one of the Lewis structures, estimate \(\Delta H_{\mathrm{f}}^{\circ}\) for \(\mathrm{C}_{6} \mathrm{H}_{6}(g)\) using bond energies and given that the standard enthalpy of formation of \(\mathrm{C}(g)\) is \(717 \mathrm{~kJ} / \mathrm{mol}\). The experimental \(\Delta H_{\mathrm{f}}^{\circ}\) value of \(\mathrm{C}_{6} \mathrm{H}_{6}(g)\) is \(83 \mathrm{~kJ} / \mathrm{mol} .\) Explain the discrepancy between the experimental value and the calculated \(\Delta H_{\mathrm{f}}^{\circ}\) value for \(\mathrm{C}_{6} \mathrm{H}_{6}(g)\)

Short Answer

Expert verified
The calculated standard enthalpy of formation for benzene (C6H6) in the gas phase is -1068 kJ/mol, using the bond energies from one of the resonance structures. This value is significantly different from the experimental value of 83 kJ/mol. The discrepancy can be attributed to the fact that we only used one resonance structure for our calculations, and benzene's bond character is an intermediate between single and double bonds. This hybridization lowers the energy of the molecule, making it more stable than represented by a single Lewis structure, resulting in an underestimation of the actual enthalpy of formation.

Step by step solution

01

Lewis structure of benzene

First, let's recall the Lewis structure of benzene. Benzene has a hexagonal ring structure where each carbon atom is bonded to two other carbon atoms and one hydrogen atom. The carbon-carbon bonds alternate between single and double bonds. The resonance structure of benzene can be depicted as two alternating structures: Structure 1: C1=C2-C3=C4-C5=C6-C1 (single bond with C6) Structure 2: C1-C2=C3-C4=C5-C6=C1 (single bond with C1) In this solution, we will use Structure 1 for our calculations.
02

Calculating bond energies

Now, let's calculate the bond energy for each type of bond in the benzene molecule. We will multiply the bond energies by the number of each bond type in the molecule. We have: - 3 C=C (carbon-carbon double) bonds with bond energy of 614 kJ/mol - 3 C-C (carbon-carbon single) bonds with bond energy of 348 kJ/mol - 6 C-H (carbon-hydrogen) bonds with bond energy of 414 kJ/mol Calculating the total bond energy: Total bond energy = (3 x 614 kJ/mol) + (3 x 348 kJ/mol) + (6 x 414 kJ/mol) Total bond energy = 1842 + 1044 + 2484 kJ/mol = 5370 kJ/mol
03

Calculating ΔH𝑓° for C6H6(g)

We need to find the standard enthalpy of formation, ΔH𝑓°, for benzene (C6H6) in the gas phase. This can be calculated using bond energies: ΔH𝑓° = 6 x ΔH𝑓° of C(g) - Total bond energy Given that the standard enthalpy of formation of C(g) is 717 kJ/mol, we can calculate ΔH𝑓° for C6H6(g): ΔH𝑓° = (6 x 717 kJ/mol) - 5370 kJ/mol = 4302 kJ/mol - 5370 kJ/mol = -1068 kJ/mol
04

Comparing to the experimental value

Now that we have calculated the standard enthalpy of formation for benzene, we can compare it to the experimental value, which is given as 83 kJ/mol. Our calculated value is -1068 kJ/mol, which is significantly different from the experimental value of 83 kJ/mol. This discrepancy can be attributed to the fact that we used only one of the resonance structures of benzene for our calculations. In reality, the bond character of benzene is an intermediate between single and double bonds. This hybridization lowers the energy of the molecule, making it more stable than represented by a single Lewis structure. Therefore, using bond energies for only one of the resonance structures results in an underestimation of the actual enthalpy of formation.

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Most popular questions from this chapter

Consider the following electron configuration: $$ \left(\sigma_{3 s}\right)^{2}\left(\sigma_{3 s}^{*}\right)^{2}\left(\sigma_{3 p}\right)^{2}\left(\pi_{3 p}\right)^{4}\left(\pi_{3 p}^{*}\right)^{4} $$ Give four species that, in theory, would have this electron configuration.

Cyanamide \(\left(\mathrm{H}_{2} \mathrm{NCN}\right)\), an important industrial chemical, is produced by the following steps: $$ \begin{aligned} \mathrm{CaC}_{2}+\mathrm{N}_{2} & \longrightarrow \mathrm{CaNCN}+\mathrm{C} \\\ \mathrm{CaNCN} & \stackrel{\text { Acid }}{\longrightarrow} \mathrm{H}_{2} \mathrm{NCN} \end{aligned} $$ Cyanamid Calcium cyanamide (CaNCN) is used as a direct-application fertilizer, weed killer, and cotton defoliant. It is also used to make cyanamide, dicyandiamide, and melamine plastics: a. Write Lewis structures for \(\mathrm{NCN}^{2-}, \mathrm{H}_{2} \mathrm{NCN}\), dicyandiamide, and melamine, including resonance structures where appropriate. b. Give the hybridization of the \(\mathrm{C}\) and \(\mathrm{N}\) atoms in each species. c. How many \(\sigma\) bonds and how many \(\pi\) bonds are in each species? d. Is the ring in melamine planar? e. There are three different \(\mathrm{C}-\mathrm{N}\) bond distances in dicyandiamide, \(\mathrm{NCNC}\left(\mathrm{NH}_{2}\right)_{2}\), and the molecule is nonlinear. Of all the resonance structures you drew for this molecule, predict which should be the most important.

Explain the difference between the \(\sigma\) and \(\pi\) MOs for homonuclear diatomic molecules. How are bonding and antibonding orbitals different? Why are there two \(\pi\) MOs and one \(\sigma\) MO? Why are the \(\pi\) MOs degenerate?

Complete the Lewis structures of the following molecules. Predict the molecular structure, polarity, bond angles, and hybrid orbitals used by the atoms marked by asterisks for each molecule.

In the molecular orbital model, compare and contrast \(\sigma\) bonds with \(\pi\) bonds. What orbitals form the \(\sigma\) bonds and what orbitals form the \(\pi\) bonds? Assume the \(z\) -axis is the internuclear axis.

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