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 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) Compare your answer from part (c) to the values for \(\mathrm{C}-\mathrm{C}\) single bonds and \(\mathrm{C}=\mathrm{C}\) double bonds in Table 8.4. Is benzene's \(\mathrm{C}-\mathrm{C}\) bond enthalpy exactly halfway between them? If not, which bond type is more similar to that of benzene, CC single or CC double bonds?

Step by step solution

01

(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.

02

(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.

03

(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: \[ C_6 H_6 (g) \rightarrow 6CH(g) \]

04

(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.

05

(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.

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