## Ethanol Vs Gasoline

**Gasoline**– C₈H₁₈ + 12.5 O₂ → 8 CO₂ + 9 H₂O DH (heat released) = -5460 KJ/mol

**Ethanol**– C₂H₆O + 3 O₂ → 2 CO₂ + 3 H₂O DH = -1368 KJ/mol

Now when we apply this where oxygen is the limiting factor, things get interesting! When air is the limiting factor, you can burn 4.16 moles of ethanol vs one mole of gasoline. If we were to assume that the exhaust products were ideal gasses, the temperature of the burned fuel would be 4.4% higher. “Huh? But didn’t we say ethanol burns colder… and… now you are saying you get more heat from it?”

**Gasoline** – C₈H₁₈ + 12.5 O₂ → 8 CO₂ + 9 H₂O DH (heat released) = -5460 KJ/moL

**Ethanol** – 4.16 C₂H₆O + 12.5 O₂ → 8.3 CO₂ + 12.5 H₂O DH = -5700 KJ/mol

Simply by scaling down the ethanol content in the equation by 4.4%, we can reach the same thermal energy you get from octane gas.

**Ethanol** – 3.98 C₂H₆O + 11.97 O₂ → 7.95 CO₂ + 11.97 H₂O DH = -5700/1.044 = -5460 KJ/mole

Based off what you have read so far, mathematics and chemistry have shown that ethanol burns colder. But, the amount of molecules in 4.16 moles of ethanol are actually more than in a single mole of gasoline. Avogadros law (or as I like to call it Avocado’s Law) states that “equal volumes of all gases, at the same temperature and pressure, have the same number of molecules.” Based on this, we also know that the pressure is already going to be higher, and it turns out when we factor in the temperature and the fact that there are more combustion products, we actually see an increase in cylinder pressures of around 18%!

Now let’s raise our nerdy level even higher, and talk about the ideal gas law a little more! P=nRT/V. We know that pressure is proportional to temperature and the number of moles on the exhaust at the constant V. So if ethanol produces 18% more moles of exhaust gas at the same thermal energy as octane, then you could also drop the temperature by that 18% to produce the same cylinder pressure as gasoline. #MindBlown.