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Benzene is more stable than a ring of three single and three double bonds (one Kekule structure)…

Try it on paper first. A hint, the key idea and the worked solution are here when you need them.

Benzene is more stable than a ring of three single and three double bonds (one Kekule structure) would be. Estimate this resonance energy two ways, for gaseous benzene. Data, in $\text{kJ mol}^{-1}$ : $\Delta_f H^\circ$ of $\text{C}_6\text{H}_6(g)$ $+82.9$ and of cyclohexane $\text{C}_6\text{H}_{12}(g)$ $-123.1$ ; enthalpy of atomisation of graphite $716.7$ ; mean bond enthalpies $\text{H{-}H}$ $435.8$ , $\text{C{-}H}$ $414$ , $\text{C{-}C}$ $347$ and $\text{C{=}C}$ $611$ ; enthalpy of hydrogenation of cyclohexene to cyclohexane (one double bond, gases) $-119.7$ .

What is the enthalpy of atomisation of gaseous benzene, $\text{C}_6\text{H}_6(g) \rightarrow 6\,\text{C}(g) + 6\,\text{H}(g)$ ?
From the bond enthalpies, what would the enthalpy of formation of a Kekule benzene be?
What is the resonance energy by this route? (Give it as a positive number: how much more stable real benzene is.)
What is the enthalpy of hydrogenation of benzene to cyclohexane, $\text{C}_6\text{H}_6(g) + 3\,\text{H}_2(g) \rightarrow \text{C}_6\text{H}_{12}(g)$ ?
Taking a Kekule benzene to behave like three cyclohexene double bonds, what is the resonance energy by this route (again as a positive number)?

The worked solution

Have you tried it?

Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.

Plan. Compare the real benzene with an imagined Kekule benzene, once through free atoms and once through hydrogenation.

(a) Atomisation. The elements go to free atoms with $6 \times 716.7 + 3 \times 435.8 = 5607.6\ \text{kJ}$ . Benzene lies $82.9\ \text{kJ}$ above the elements, so it needs $5607.6 - 82.9 = 5524.7\ \text{kJ mol}^{-1}$ .

(b) Kekule benzene. Its bonds hold $3 \times 347 + 3 \times 611 + 6 \times 414 = 1041 + 1833 + 2484 = 5358\ \text{kJ}$ . So $\Delta_f H = 5607.6 - 5358 = +249.6\ \text{kJ mol}^{-1}$ .

(c) First route. Real benzene lies at $+82.9$ , the Kekule one at $+249.6$ : the resonance energy is $249.6 - 82.9 = 166.7\ \text{kJ mol}^{-1}$ . (The same as $5524.7 - 5358$ : real benzene needs that much more energy to pull apart.)

(d) Hydrogenation. $\Delta H = \Delta_f H(\text{C}_6\text{H}_{12}) - \Delta_f H(\text{C}_6\text{H}_6) = -123.1 - 82.9 = -206.0\ \text{kJ mol}^{-1}$ .

(e) Second route. Three isolated double bonds would give $3 \times (-119.7) = -359.1\ \text{kJ}$ . Benzene gives only $-206.0\ \text{kJ}$ , so it is $-206.0 - (-359.1) = 153.1\ \text{kJ mol}^{-1}$ more stable.

Check. The two routes agree roughly but not exactly, as expected: mean bond enthalpies are averages over many molecules, while hydrogenation compares real molecules. Both say benzene is $153\ \text{kJ mol}^{-1}$ to $167\ \text{kJ mol}^{-1}$ more stable than one Kekule structure.

Watch out. Bond enthalpies are for gases. That is why every species here is taken as a gas.

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