Thermodynamics in combustion

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Kathleen Georgiana Newton
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Thermodynamics in combustion 2nd ste in toolbox Thermodynamics deals with a equilibrium state and how chemical comosition can be calculated for a system with

1 Thermodynamics in combustion 2nd ste in toolbox Thermodynamics deals with a equilibrium state and how chemical comosition can be calculated for a system with known atomic or molecular comosition if 2 indeendent thermodynamic roerties are known. Assumed no rate control, no gradients i.e. no fluid dynamics FPK1-29/MZ 1/5

2 Chemical equilibria and high temeratures Equilibrium constants Enthaly and entroy Free energy FPK1-29/MZ 2/5

3 Terms and Concets Equilibrium equation, law of mass action K H and S delta S and gas evolution delta H endothermic, exothermic delta G Energy/enthali/entroi of formation of comounds delta G for any reaction FPK1-29/MZ 3/5

4 Equilibrium A system is at chemical equilibrium when no aarent change in concentrations of roducts and reactants occur over time Forward and backward reaction rates are equal FPK1-29/MZ 4/5

5 a a a a l C j A m D k B Equilibrium constant K Law of mass action ja+kb lc+md K = equilibrium constant, a =activity Remember what was activity?? FPK1-29/MZ 5/5

6 Equilibrium constant K: activity Activity is a concentration-like measure of deviation from the standard state. i.e. it is a way to describe the non-ideality of a system For a gas, a=p/p where P =1 atm. (actually fugacity) For a solute, a=c/c where c = 1 mol/l For a solvent, a=x. For a solid, there is no sensible measure of concentration. A solid is either there or it isn t, so a=1 rovided some solid is resent. FPK1-29/MZ 6/5

7 FPK1-29/MZ 7/5 Equilibrium constant K k B j A A a a k RT E k r ex m D l C A a a k RT E k r ex At equilibrium k B j a A a k r r m D l a C a k 2 and k B j A m D l C a a a a k k K 2 1

8 Equilibrium constant K: examle H 2 (g) + Cl 2 (g) 2 HCl(g) a H a 2 2 HCl a Cl 2 K FPK1-29/MZ 8/5

9 FPK1-29/MZ 9/5 K vs K If the standard state for (G rxn) is 1 atm, we can use artial ressures as a good aroximation for activity: Law of mass action It is often convenient to use mole fractions: Since i = x i tot : tot i i n n x K k B j A m D l C k tot B j tot A m tot D l tot C K x x x x k j m l tot k B j A m D l C x x x x K

10 Generally: K x n tot K where n is the change in mole numbers in the reaction. FPK1-29/MZ 1/5

11 Thus for our examle: H 2 HCl 2 Cl2 K HCl tot x x H 2 x tot Cl 2 2 tot K x x 2 HCl x H 2 Cl2 K x (2 2) tot K FPK1-29/MZ 11/5

12 K vs K c Similarily, using the ideal gas relation: PV nrt P x n V x RT c x RT The equilibrium constant in terms of concentrations is found to be: K l C j A m D k B c c l m C D. RT j k A B c c lm jk n K c RT FPK1-29/MZ 12/5

13 K vs K c Thus for our examle the equilibrium constant in terms of concentrations is found to be: c RT c RT H 2 c HCl RT Cl 2 2 K 2 HCl c ch c 2 Cl2 K c 22 RT K And K c = K x if number of moles does not change FPK1-29/MZ 13/5

14 Equilibrium constant K. summary The two systems are not comletely unrelated. K is numerically equal to the true thermodynamic equilibrium constant rovided you always measure ressures in atmosheres. Kc is numerically equal to both K and K when the number of gas molecules on either side of the reaction is the same because the conversion factors then cancel. FPK1-29/MZ 14/5

15 Excersize Give K, K c and K x for the reaction of methane with oxygen FPK1-29/MZ 15/5

16 Entroy and enthaly FPK1-29/MZ 16/5

17 Entroy In the the diagram on the left, the area under the PV lot gives the work done by the system. W dv The diagram on the right is analogous to the one on the left. The area under the TS diagram gives the heat sulied to the system. Q TdS FPK1-29/MZ 17/5

18 Entroy Describes the number of arrangements that are available to a system existing in a given state The universe tries to increase its entroy Change in entroy of the system is determined by the changes in the system An increase in gasmolecules after reaction gives a increase in entroy S = n S(roduct) n r S(reactant) FPK1-29/MZ 18/5

19 Entroy: examle Does the entroy increase or decrease for the following reactions? FPK1-29/MZ 19/5

20 Entroy: ex FPK1-29/MZ 2/5

21 Entroy: examle standard state 2NO 2 (g) N 2 O 4 (g) S = (24.6) = J/K mol FPK1-29/MZ 21/5

22 Ex Calculate S for combustion of CH 4 FPK1-29/MZ 22/5

23 Entroy at other temeratures S T T c T dt Tables and databases give the integrated enthalies and entroies directly. f S T f S298 ST S298 FPK1-29/MZ 23/5

24 Examle: T=12K 2NO 2 (g) N 2 O 4 (g) For NO2: 267,76 J/molK f S T f S298 For N2O4: J/molK f S T S rxn = x267.76= J/molK ST S f S298 ST S FPK1-29/MZ 24/5

25 Ex Calculate the entroy of combustion of CH4 at 12K FPK1-29/MZ 25/5

26 Enthaly H = U+PV It is somewhat arallel to the first law of thermodynamics for a constant ressure system Q = U + P V since in this case Q= H At constant volume and temerature, the differential change in enthaly as function of ressure and entroy S is therefore dh = TdS+VdP where (all units given in SI) H is the enthaly (joules) U is the internal energy, (joules) is the ressure of the system, (ascals) V is the volume, (cubic meters) S is entroy H = n H(roduct) n r H(reactant) H < exothermic reaction H> endothermic reaction FPK1-29/MZ 26/5

27 Enthaly at other temeratures H f T H T H 298 f H 298 T 298 c dt HT H 298 FPK1-29/MZ 27/5

28 Enthaly: examle at T=12K 2NO 2 (g) N 2 O 4 (g) For NO2: f H T f H 298 HT H For N2O4: H rxn f H T f H 298 HT H FPK1-29/MZ 28/5

29 Ex Calculate reaction enthaly at 12 K for combustion of methane FPK1-29/MZ 29/5

30 Gibbs free energy The Gibbs free energy is defined as: G = U + V TS which is the same as: G = H TS where: U is the internal energy (SI unit: joule) is ressure (SI unit: ascal) V is volume (SI unit: m3) T is the temerature (SI unit: kelvin) S is the entroy (SI unit: joule er kelvin) H is the enthaly (SI unit: joule) Note: H and S are Thermodynamic values found at Standard Temerature and Pressure FPK1-29/MZ 3/5

31 Gibbs free energy: sontaneous rocesses FPK1-29/MZ 31/5

32 Gibbs free energy: sontaneous rocesses Max amount of work that can be done by a reaction FPK1-29/MZ 32/5

33 Free energy vs equilibrium constant K is usually not known for your reaction to be studied, but you may be able to deduce the K via G rxn. G = H-TS and when infinite small changes in the system occur dg= dh-d(ts) = dh-tds-sdt if T = Cst dg=dh-tds (Gibbs Helmholtz equation) Definition of enthaly: H=U+PV or dh=du+dv+vdp And du = q+w Thus dg = q+w+dv+vd-tds If T=Cst, and reversible reaction TdS=q dg=dv+vd+w Assuming the work is only done to exand V against a constant P surr :W=-dV And dg=vdp FPK1-29/MZ 33/5

34 Free energy vs equilibrium constant dg=vdp In case of an ideal gas V=nRT/P and thus: dg nrt dp G G dg nrt d G RT ln since 1atm FPK1-29/MZ 34/5

35 Free energy vs equilibrium constant G rxn n roduct. G ro duct nreac tan t. G reac tant In reaction of gases ja+kb? lc+md This means - G RT ln rxn K If T= cst, reversible reaction and work is only done to change volume against surrounding ressure at equilibrium A ositive logarithm of an equilibrium constant and a negative free energy of reaction both corresond to a sontaneous reaction FPK1-29/MZ 35/5

36 Free energy Ex. Given: (I) C(s) + O 2 CO 2, (II) CO + ½ O 2 CO 2, G I G II What is G rxn for: C(s) + ½ O 2 CO, G rxn Free energies are additive: G rxn = G I - G II FPK1-29/MZ 36/5

37 Generally for the reaction a A b B G rxn c C d D G rxn c G d G a G b G f C f D f A f B f G i refers to formation reaction of the secies i from its elements. Elements are assumed to be in their natural state : T > T B T B > T > T M T < T M gaseous liquid solid FPK1-29/MZ 37/5

38 Free Energy Temerature Diagrams FPK1-29/MZ 38/5

39 Free Energy - Temerature Diagram (Rosenquist) FPK1-29/MZ 39/5

40 According to equations above free energies are roughly linearly deendent on temerature: G k T (Cf. Rosenqvist Tables) For metal oxides the k is almost the same and indeendent of the comound! k = -S In reactions where gases evolve or disaear : S 25 cal K mol n g mol A(s) + B(g) AB(s) AB(s) A(s) + B(g) cal n g 1 S 25 K mol cal n g 1 S 25 K mol FPK1-29/MZ 4/5

41 Free Energy Temerature Diagrams FPK1-29/MZ 41/5

42 FPK1-29/MZ 42/5

43 Eq constant vs. temerature: ln ln ln K K K T T T G RT H H R T T T TST RT 1 T S R T If H T and S T indeendent of T: ln H R K T T S R ln K H R 1/T FPK1-29/MZ 43/5

44 K r increases with increasing temerature for endothermic reactions (H > ) K r decreases with increasing temerature for exothermic reactions (H < ) H < if roducts have stronger bonds than the reactants. FPK1-29/MZ 44/5

45 Terms and Concets Equilibrium equation, law of mass action K delta G H and S delta S and gas evolution delta H endothermic, exothermic Energy/enthali/entroi of formation of comounds -> delta G for any reaction FPK1-29/MZ 45/5

46 JANAF CH4 FPK1-29/MZ 46/5

47 JANAF O2 FPK1-29/MZ 47/5

48 JANAF H2O FPK1-29/MZ 48/5

49 JANAF CO2 FPK1-29/MZ 49/5

50 JANAF NO2

51 JANAF N2O4

Chapter 17.3 Entropy and Spontaneity Objectives Define entropy and examine its statistical nature Predict the sign of entropy changes for phase

Chapter 17.3 Entropy and Spontaneity Objectives Define entropy and examine its statistical nature Predict the sign of entropy changes for phase changes Apply the second law of thermodynamics to chemical