Lecture 5 Thermodynamics of Combustion AOSC 434 AIR POLLUTION RUSSELL R. DICKERSON

Wark and Warner Chapter 7,8 Finlayson Pitts & Pitts, Appendix 1 II Wark and Warner Chapter 7,8 Finlayson Pitts & Pitts, Appendix 1 II. Thermodynamics (continued) a) Enthalpy, Bond Energy, and Entropy From the last lecture, you will remember the concept of enthalpy or heat. There is the heat of formation of substances from their elemental parts, and the heat of specific reactions leading to complete oxidation of the reactants is called the heat of combustion. EXAMPLE Remember the combustion of isooctane (C₈H₁₈), an approximation of gasoline C₈H₁₈ + 12.5 O₂ (+47N₂) → 8CO₂ + 9H₂O (+47N₂) The heat of combustion, is about 1240 kcal/mole.

Remember also the Air-Fuel-Ratio Remember also the Air-Fuel-Ratio. If the AFR is higher than stoichiometry dictates (here 15) than the mixture is lean; if it is less than 15, then the mixture is said to be rich and pollutants such as CO or H₂CO must be formed. For example in the combustion of natural gas, methane primarily, if insufficient oxygen is present the following reaction can occur: But there is more energy to be had from this reaction: This shows that burning rich not only produces pollutants, but cheats you out of some of the available heat (and energy).

Students: calculate the heat of combustion of methane Students: calculate the heat of combustion of methane. How does it compare the sun of the heats of the two above reactions? How is this related to ΔH being an equation of state? How do we find ΔHf° for new compounds? BOND ENERGIES Example: water, H-O-H Look at the table and see an O-H bound requires about 462 kJ/mole ( 110.6 kcal/mole) to break. Two such bonds must be broken to turns H₂O into 2H + O. ΔHf° H₂O = ? H - O - H → H + H + O H + H → H₂ O → ½ O₂ NET: H₂O → H₂ + ½ O₂ Check against the table for the reverse reaction ΔHf° (H₂O) = -242, pretty close

Enthalpy is a powerful tool, but is it the “criterion of feasibility” Enthalpy is a powerful tool, but is it the “criterion of feasibility”? There are reactions that proceed in spite of the ΔH being positive, such as air expanding to fill a vacuum. These processes are driven by another force and the Second Law of Thermodynamics defines that force. Entropy is the degree of mixing or disorder in a system. Systems tend toward the highest state of entropy. The products of the combustion of isooctane are favored by entropy because there is much more order to isooctane and oxygen than to eight CO₂’s and H₂O’s. Entropy is defined as: dS = DQ/T The total entropy of a system plus surroundings always increases, that is ΔS ≥ 0.0. Entropy has units of kcal/moleK of kJ/moleK.

By combining the First and Second Laws of Thermodynamics, we can derive the criterion of feasibility – Gibbs Free Energy. ΔG = ΔH – TΔS Units are kcal/mole or kJ/mole. If ΔG ≥0.0 reaction will not proceed spontaneously – without the input of energy. This does not mean that all reactions with ΔG ≥0.0 proceed immediately, otherwise the gas tank on your car would explode, of course the potential is there.

Calculating the ΔG of a reaction is the same as calculating the ΔH of a reaction. Consider the following particularly important reaction. This reaction can go “spontaneously,” that is without the input of energy. If n=1 we have formaldehyde combustion. But consider n=6 then (H₂CO)₆ is glucose and when glucose is oxidized 2870 kJ/mole (686 kcal/mole) of free energy are released. This process is respiration. But we know that this reaction proceeds in the opposite direction as well. This process is photosynthesis. Does this violate the second law of thermodynamics? No, because six photons are involved in the synthesis of glucose molecule. Thermodynamics tells us that respiration can provide energy to an organism, but that photosynthesis requires energy from outside the system; it even tells how much energy. The problem is that these processes are much more complicated than the reaction written above. Both respiration and photosynthesis require about 100 steps. Fortunately, ΔH and ΔG are “equations of state” and therefore are path independent.

Problem for students: Calculate the ΔGº for the complete combustion of isooctane. (-5229 kJ/mole)