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Here we estimate the heat released in a chemical reaction by the difference in the total energy of stable bonds in the reactants and the products. This requires us to balance two reactions, the combustion of methane, and the combustion of octane.

Let's inspect the unbalanced combustion reaction for methane $\ce{CH4 + O2 -> CO2 + H2O}$ We can see that for each molecule of methane that is combusted, we'll form one molecule of carbon dioxide (since methane has one carbon) and two molecules of water (since methane has four molecules of hydrogen, and water has two). Producing one carbon dioxide and two water molecules requires four oxygen atoms overall, or two molecules of $\ce{O2}$ , therefore, we have for the balanced combustion of methane $\ce{CH4 + 2O2 -> CO2 + 2H2O}$ For the combustion of octane, we have the unbalanced reaction $\ce{C8H18 + O2 -> CO2 + H2O}$ By the same argument as above, for each molecule of octane that we combust, we produce eight molecules of $\ce{CO2}$ . Similarly, we create nine molecules of $\ce{H2O}$ . Altogether, this requires twenty five atoms of oxygen. In order to get a whole number of $\ce{O2}$ molecules, we double the amount of everything, to arrive at the balanced reaction $\ce{2C8H18 + 25O2 -> 16CO2 + 18H2O}$ Now, we have to translate these reactions into bonds. Let's start with the molecules that appear in both reactions.

Each molecule of $\ce{O2}$ contribute one oxygen-oxygen double bond, or, $\ce{O=O}$ .

Each molecule of $\ce{H2O}$ contribute two hydrogen-oxygen bonds $\ce{O-H}$ .

Each molecule of $\ce{CO2}$ contributes two carbon-oxygen double bonds $\ce{C=O}$ .

Now, carbon has a valence of four, so methane $\ce{CH4}$ consists of four carbon-hydrogen bonds $\ce{C-H}$ .

Octane has a total carbon valence of $4\times8=32$ , of which eighteen are accounted for by carbon-hydrogen bonds, leaving a valence of $32-18=14$ to be satisfied by carbon-carbon bonds of some kind. Each carbon-carbon single bond contributes one valence electron to each carbon, so all of the carbon-carbon bonds are single bonds. Thus, a molecule of octane is equivalent to seven carbon-carbon singles bonds and eighteen carbon-hydrogen bonds, or $\ce{C8H18 \leftrightarrow 7C-C + 18C-H}$ .

Now, the heat released in an energy is the sum of the bond energies of the reactants minus the bond energies of the products. The lower the energy of the formed products is relative to the combusted oxygen and hydrocarbon, the more energy that is produced. We have $\Delta E_\textrm{combustion} = \sum_\textrm{reactants} E_\textrm{bond} - \sum_\textrm{products} E_\textrm{bond}$ .

For methane, we have $\begin{aligned} \Delta E_\textrm{combustion}(\textrm{methane}) &= 4E(\ce{C-H}) + 2E(\ce{O=O}) - 2E(\ce{C=O}) - 4E(\ce{O-H}) \\ &= \SI{810}{\kilo\joule\per\mole} \end{aligned}$

For octane, we have $\begin{aligned} \Delta E_\textrm{combustion}(\textrm{octane}) &= 2\times18E(\ce{C-H}) + 2\times7E(\ce{C-C}) + 25E(\ce{O=O}) - 32E(\ce{C=O}) - 36E(\ce{O-H}) \\ &= \SI{10188}{\kilo\joule\per\mole} \end{aligned}$

Now, we are ultimately interested in the amount of energy released per moles of carbon dioxide produced. Each molecule of octane has eight carbon atoms, so we need to divide by 16 to get the amount of energy released per mole of carbon dioxide. Thus, we find that the combustion of conventional fuel (represented by octane) produces about $\SI{636.75}{\kilo\joule}$ per mole of carbon dioxide while the "clean" methane produces $\SI{810}{\kilo\joule}$ . The increase in efficiency is then

$\begin{aligned} \Delta f &= \frac{E_C(\textrm{methane}) - E_C(\textrm{octane})}{E_C(\textrm{octane})} \\ &= \frac{E_C(\textrm{methane})}{E_C(\textrm{octane})} - 1 \end{aligned}$

which is an improvement of just 27.2%.