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1. ME 525: CombustionLecture 4: Second Law, Products of Combustion, Chemical Equilibrium, Applications Second law of thermodynamics applied to reacting systems Product composition governed by second law Chemical equilibrium Equilibrium constant of formation Examples
2. Equilibrium Criteria: Simple Compressible Systems Consider a simple compressible system with fixed mass and uniform temperature and pressure: If volume change is the only work mode then Second Law of thermodynamics (Clausius inequality): Combining the first and second law of thermodynamics:
3. Equilibrium Criteria: Simple Compressible Systems Introducing enthalpy: Introducing Gibbs Free Energy (Gibbs Function): Introducing Helmholtz Free Energy (Helmholtz Function):
4. Summary of Equilibrium Criteria for Simple Compressible Systems Constant Volume and Internal Energy: Constant Temperature and Pressure Constant Temperature and Volume Depending on the constraints on the system the first law and second law combine to provide three different criteria for the equilibrium condition. Each requires evaluation of entropy “s”
5. Entropy of Reacting Mixtures Entropy of a species “i ” in a reacting mixture is: Entropy of the reacting mixture is summation of entropies 1 kmol of CO and 0.5 kmol of O2 initially at T1 = 298K and P1 = 1 atm. react in a constant volume, insulated reactor. As the reaction proceeds the temperature and pressure increase. If all the CO is converted to CO2, the temperature will be maximum but if equilibrium prevents all CO from burning to CO2, the temperature will be less and is defined as equilibrium temperature.
6. Constant Volume, Internal Energy Equillibrium First Law: Second Law:
7. Entropy of Reacting Mixtures Entropy of a species “i ” in a reacting mixture is: Entropy of the reacting mixture is summation of entropies
8. Plotting the results as a function of the parameter 1-aresults in the graph shown in Fig. 2.11 in Turns, Note that the reaction proceeds from a condition of 1-a = 0 (cold reactants) to a condition of 1-a = 0.52 with a system temperature of about 3500 K. At this point the entropy is a maximum. Any further creation of CO2 would require a decrease of entropy which cannot occur for an isolated system. Thus the system reaches chemical equilibrium at the vertical dashed line.
9. Gibbs Free Energy Change in a Reaction Gibbs Free Energy of a reacting mixture is a summation: First term is the molar Gibbs Free Energy at the standard pressure and the second term is the change in molar Gibbs Free Energy with pressure. Equating change in G to zero yields the equilibrium criterion for constant P constant T: Note: The constant P can be higher than P o
10. Equilibrium Composition and Constant Consider a general reaction system The Gibbs Free Energy Change can be written as: Regrouping yields:
11. Gibbs Free Energy and Kp of Formation Gibbs function of formation of a compound is defined using the formation reaction from elemental species. Gibbs function of formation of naturally occurring elements are assigned a reference value of 0 similar to the enthalpy of formation of naturally occurring elements. This allows the definition of an equilibrium constant of formation on the basis of Gibbs free energy of formation for each compound.
12. Gibbs Free Energy and Kp of Formation Gibbs function of formation of a compound is defined using the formation reaction from elemental species. Gibbs function of formation of naturally occurring elements are assigned a reference value of 0 similar to the enthalpy of formation of naturally occurring elements. This allows the definition of an equilibrium constant of formation on the basis of Gibbs free energy of formation for each compound.
13. Example 1: Equilibrium Composition Consider combustion of methanol with air in stoichiometrically correct amount at standard pressure and temperature and estimate equilibrium composition assuming P=1 atm., T = 2151 K and T= 2194 K and P=2 atm., T=2194 K. Find the concentrations of CO2, H2O, N2, O2, H2, and CO in the product species.
14. Example 2: Equilibrium Composition Consider combustion of methanol with air in stoichiometrically correct amount at standard pressure and temperature and estimate equilibrium composition assuming P=1 atm., T = 2151 K and T= 2194 K and P=2 atm., T=2194 K. Find the concentrations of CO2, H2O, N2, O2, H2, CO, H, O, OH, NO, and N in the product species.
15. Examples 1 & 2: Solution Consider combustion of methanol with air in stoichiometrically correct amount at standard pressure and temperature and estimate equilibrium composition assuming P=1 atm., T = 2151 K and T= 2194 K and P=2 atm., T=2194 K. Find the concentrations of CO2, H2O, N2, O2, H2, CO, H, O, OH, NO, and N in the product species. (1) (2) (3) 4 atom (CHON) balances provide the solutions to (1): a1=1, b1=2, c1=3.76(1.5), d1=0 (2) has 2 additional unknowns e2, f2. Equilibrium reactions of formation for CO, CO2 & H2O provide 3 additional equations. Two unknowns, three additional equations automatically indicate the presence of at least one additional species. Ex. C atoms (3) has 4 additional species H, O, OH, and NO and these bring 4 equilibrium equations but still not C atoms. C atoms exist in both cases (2) and (3) but in very small amounts!
16. Example 2: Equilibrium Composition Consider combustion of methanol with air in stoichiometrically correct amount at standard pressure and temperature and estimate equilibrium composition assuming P=1 atm., T = 2151 K and T= 2194 K and P=2 atm., T=2194 K. Find the concentrations of CO2, H2O, N2, O2, H2, CO, H, O, OH, NO, and N in the product species. (T, P given or else energy and mass equations would have to be solved) (1) (2) (3) (4) (5) (6) (7)
17. Example 2: Equilibrium Composition Consider combustion of methanol with air in stoichiometrically correct amount at standard pressure and temperature and estimate equilibrium composition assuming P=1 atm., T = 2151 K and T= 2194 K and P=2 atm., T=2194 K. Find the concentrations of CO2, H2O, N2, O2, H2, CO, H, O, OH, NO, and N in the product species. (T, P given or else energy and mass equations would have to be solved) (8) (1) (2) (9) (3) (4) (10) (5) (11) (6) (12) (7)
18. Example 2: Equilibrium Composition Consider combustion of methanol with air in stoichiometrically correct amount at standard pressure and temperature and estimate equilibrium composition assuming P=1 atm., T = 2151 K and T= 2194 K and P=2 atm., T=2194 K. Find the concentrations of CO2, H2O, N2, O2, H2, CO, H, O, OH, NO, and N in the product species. (T, P given or else energy and mass equations would have to be solved) High temperature leads to decomposition. High pressure leads to recombination. O, H are much smaller than OH and NO. CO, H2 and OH should be considered first before considering O, H. NO2, N and C concentrations can be checked by adding three more equations but are small.