Question

In exercise 71 in chapter 13 the lewis structure for benzene was drawn. Using one of the lewis structures estimate $\mathbf{∆}{\mathit{H}}_{\mathbf{f}}^{\mathbf{\circ }}$for ${\mathit{C}}_{\mathbf{6}}{\mathit{H}}_{\mathbf{6}}\left(g\right)$. Using bond energies and given standard enthalpies of formation of C(g) is 717kJ/mol. The experimental _{$\mathbf{∆}{\mathit{H}}_{\mathbf{f}}^{\mathbf{\circ }}$}for ${\mathit{C}}_{\mathbf{6}}{\mathit{H}}_{\mathbf{6}}\mathbf{}$is 83 kJ/mol. Explain the discrepancy between the experimental value and the calculated $\mathbf{∆}{\mathit{H}}_{\mathbf{f}}^{\mathbf{\circ }}\mathbf{}$value of ${\mathit{C}}_{\mathbf{6}}{\mathit{H}}_{\mathbf{6}}\left(g\right)$.

Short answer

The value of $∆H_f^{\circ }$for $C_6{H}_6$is 5361KJ. Yes, there is somedifference between the experimental value and the calculated value.

Step-by-step solution

Step 1- Calculating the value for  ∆Hf∘

To determine the formation enthalpy, we must first count the broken bonds. The bond enthalpy is calculated by adding the bond energy of all the bonds that have been broken. The energy of broken bonds is

$6C-H\left(413kJ\right),3C=H\left(614kJ\right),3C-C\left(347kJ\right).$

$$\begin{gathered} ∆H_f^{\circ }=6\left(413\right)+3\left(614\right)+3\left(347\right) \\ =5361kJ \end{gathered}$$

Step 2- Difference in both the values.

Bond energies are used to ignore whether the molecule was gaseous, liquid, or solid, however, this is no longer the case in reaction enthalpies. Molecules also store heat as rotations and vibrations, which bond energy does not count properly. Also, bond energy changes due to environmental facts so we consider an average value, not an exact one. For different molecules bond energy is different for the same bond which causes variation in calculating the enthalpy of reaction.