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
Chemical equilibria and high temeratures Equilibrium constants Enthaly and entroy Free energy FPK1-29/MZ 2/5
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
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
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
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
FPK1-29/MZ 7/5 Equilibrium constant K k B j A A a a k RT E k r 1 1 1 ex m D l C A a a k RT E k r 2 2 2 ex At equilibrium k B j a A a k r r 1 2 1 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
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
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
Generally: K x n tot K where n is the change in mole numbers in the reaction. FPK1-29/MZ 1/5
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
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
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
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
Excersize Give K, K c and K x for the reaction of methane with oxygen FPK1-29/MZ 15/5
Entroy and enthaly FPK1-29/MZ 16/5
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
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
Entroy: examle Does the entroy increase or decrease for the following reactions? FPK1-29/MZ 19/5
Entroy: ex FPK1-29/MZ 2/5
Entroy: examle standard state 2NO 2 (g) N 2 O 4 (g) S = 34.29-2(24.6) = -175.83 J/K mol FPK1-29/MZ 21/5
Ex Calculate S for combustion of CH 4 FPK1-29/MZ 22/5
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
Examle: T=12K 2NO 2 (g) N 2 O 4 (g) For NO2: 267,76 J/molK f S T f S298 For N2O4: 366.18 J/molK f S T S rxn =366.18-2x267.76= -169.34 J/molK ST S298 24.3 267.76-24.3 f S298 ST S298 34.38 366.18-34.376 FPK1-29/MZ 24/5
Ex Calculate the entroy of combustion of CH4 at 12K FPK1-29/MZ 25/5
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
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
Enthaly: examle at T=12K 2NO 2 (g) N 2 O 4 (g) For NO2: f H T f H 298 HT H 298 33.9 42.95- For N2O4: H rxn f H T f H 298 HT H 298 9.8 18.66- FPK1-29/MZ 28/5
Ex Calculate reaction enthaly at 12 K for combustion of methane FPK1-29/MZ 29/5
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
Gibbs free energy: sontaneous rocesses FPK1-29/MZ 31/5
Gibbs free energy: sontaneous rocesses Max amount of work that can be done by a reaction FPK1-29/MZ 32/5
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
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
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
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
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
Free Energy Temerature Diagrams FPK1-29/MZ 38/5
Free Energy - Temerature Diagram (Rosenquist) FPK1-29/MZ 39/5
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
Free Energy Temerature Diagrams FPK1-29/MZ 41/5
FPK1-29/MZ 42/5
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 298 1 298 K T T S R ln K H R 1/T FPK1-29/MZ 43/5
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
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
JANAF CH4 FPK1-29/MZ 46/5
JANAF O2 FPK1-29/MZ 47/5
JANAF H2O FPK1-29/MZ 48/5
JANAF CO2 FPK1-29/MZ 49/5
JANAF NO2
JANAF N2O4