AME 513 Priciples of Combustio " Lecture 3 Chemical thermodyamics I 2 d Law Outlie" Why do we eed to ivoke chemical equilibrium? Degrees Of Reactio Freedom (DORFs) Coservatio of atoms Secod Law of Thermodyamics for reactive systems Etropy of a ideal gas mixture Equilibrium costats Applicatio of chemical equilibrium to hydrocarbo-air combustio Applicatio of chemical equilibrium to compressio/ expasio AME 513 - Fall 2012 - Lecture 3 - Chemical Thermodyamics 2 2 1
Why do we eed chemical equilibrium?" (From lecture 2) What if we assume more products, e.g. 1CH 4 + 2(O 2 + 3.77N 2 )? 2 +? H 2 O +? N 2 +? I this case how do we kow the amout of vs. 2? Ad if if we assume oly 3 products, how do we kow that the preferred products are 2, H 2 O ad N 2 rather tha (for example), H 2 O 2 ad N 2 O? Need chemical equilibrium to decide - use 2d Law to determie the worst possible ed state of the mixture AME 513 - Fall 2012 - Lecture 3 - Chemical Thermodyamics 2 3 Degrees of reactio freedom (DoRFs)" If we have a reactig soup of, O 2, 2, H 2 O, H 2 ad OH, ca we specify chages i the amout of each molecule idepedetly? No, we must coserve each type of atom (3 i this case) Coservatio of C atoms: + 2 = costat O atoms: + 2 2 + 2 O2 + H2O + OH = costat H atoms: 2 H2O + 2 H2 + OH = costat 3 equatios, 6 ukow i s 3 degrees of reactio freedom (DoRFs) of DoRFs = of differet molecules () - of differet elemets Each DoRF will result i the requiremet for oe equilibrium costrait AME 513 - Fall 2012 - Lecture 3 - Chemical Thermodyamics 2 4 2
Coservatio of atoms" Typically we apply coservatio of atoms by requirig that the ratios of atoms are costat, i.e. the same i the reactats as the products C atoms: + 2 = costat O atoms: + 2 2 + 2 O2 + H2O + OH = costat H atoms: 2 H2O + 2 H2 + OH = costat C O = X + X 2 X + 2X 2 + 2X O2 + X H 2 O + X OH = costat C X + X 2 = H 2X + 2X = costat H 2 O H 2 + X OH Specifyig O / H also would be redudat, so the umber of atom ratio costraits = of atoms - 1 What are these costats? Depeds o iitial mixture, e.g. for stoichiometric CH 4 i O 2, C / O = 1/4, C / H = 1/4 AME 513 - Fall 2012 - Lecture 3 - Chemical Thermodyamics 2 5 2d law of thermo for reactig systems" Costraits for reactig system First law: de = δq - δw = δq - PdV Secod law: ds δq/t Combie: TdS - du - PdV 0 for ay allowable chage i the state of the system For a system at fixed T ad P (e.g. material i a pisto-cylider with fixed weight o top of pisto, all placed i isothermal bath: d(ts-u-pv) 0, or per uit mass d(ts-u-pv) 0 Defie Gibbs fuctio g h - Ts = u + Pv - Ts Thus for system at fixed T ad P: d(-g) 0 or dg 0 Thus at equilibrium, dg = 0 or dg = 0 (g or G is miimum) ad for each species i!g " µ i = 0 µ i = chemical potetial of species i! i Similarly, for system at fixed U ad V (e.g. isulated chamber of fixed volume): ds 0, at equilibrium ds = 0 or ds = 0 (s is maximum) AME 513 - Fall 2012 - Lecture 3 - Chemical Thermodyamics 2 6 3
Etropy of a ideal gas mixture" Depeds o P AND T (ulike h ad u, which deped ONLY o T) A B C A B C P = P ref P = P ref P = P ref P P ref P P ref P P ref A, B, C P P ref (1) (2) (3) (1) Start with gases A, B ad C at temperature T ad P o = 1 atm S = S A + S B + S C = A!!s A o (T )+ B!!s B o (T )+ C!!s C o (T ) = i!!s i o (T ) S = total etropy of all gases; i = moles of species i; = umber of species!!s i o (T ) = etropy per mole of i at ay temperature T ref. pressure = 1 atm (2) Raise/lower each gas to pressure P 1 atm (same P for all)!s A =! A CP l T 2 T 1 ( ) " l( P 2 P 1 ) S = ( i!!s o i (T )"( i l P P ref ' = A "!C P l( T T ) " l( P P ref ) ( ) = i!!s o i (T )" l( P P ref ) ( ) ' = " l P P A ref ( ' = ( i!!s o i (T )" T l P P ref ( ) AME 513 - Fall 2012 - Lecture 3 - Chemical Thermodyamics 2 7 Etropy of a ideal gas mixture" (3) Now remove dividers, V A V A + V B + V C (V = volume)!s A =! A CP l T 2 T 1 ( ) + "l( V 2 V 1 ) ( ) = ' A "l A T = A "l ( A + B + C ) A S = ) i!!s o i (T )' T "l P P ref ( ) ' " i ( ) = A!C P l( T T ) + "l ( V A +V B +V C ) V A ( );recall X A ( A T = mole fractio of A ) l i T ) ' T "l P P ref ( ) = i!!s o i (T )' "l X i Ca also combie X i ad P/P ref terms usig partial pressures P i ; for a ideal gas mixture X i = P i /P, where (as above) P without subscripts is the total pressure = ΣP i S = ( i!!s o i (T )" l X i '" T l P P ref ( ( ) = i!!s o i (T )" l P i P ref ' ( ) = ( i!!s o i (T )" l X i " l( P P ref ) ( ) ' AME 513 - Fall 2012 - Lecture 3 - Chemical Thermodyamics 2 8 4
Etropy of a ideal gas mixture" Most useful form - etropy per uit mass (m), sice mass is costat: ( s = S i!!s o i (T )" l X i '" T l( P P ref ) m = = ( i M i X i ( )!!s o i (T )" l X i ' " l P P ref Rl(P/P ref ) - etropy associated with pressure differet from 1 atm - P > P ref leads to decrease i s RX i l(x i ) - etropy associated with mixig (X i < 1 meas more tha 1 specie is preset - always leads to icrease i etropy sice -X i l(x i ) > 0) Deomiator is just the average molecular weight Uits of s are J/kgK ( X i ( M i AME 513 - Fall 2012 - Lecture 3 - Chemical Thermodyamics 2 9 Equilibrium of a ideal gas mixture" How to use this result? dg = 0 meas there are equatios of the form G/ i = 0 for ukow species cocetratios! Break system ito multiple sub-systems, each with 1 DoRF, e.g.. for a chemical reactio of the form ν A A + ν B B ν C C + ν D D e.g. 1 H 2 + 1 I 2 2 HI A = H 2, ν A = -1, B = I 2, ν B = -1, C = HI, ν C = 2, D = othig, ν D = -0 coservatio of atoms requires that d A! A =! d B! B = d C! C = d D! D = k (k = ay costat, + or -) G = H TS, where H =!! i hi (T ); h! i (T ) = [ h(t! )" h! 298 ] i + h! o ; S = f,i!!s o (T )" l ( P P )' i i i ref ( G = H "TS =!! i h i (T )"T!s o i (T )+ T l( P i P ref )' ( =! i!g o i (T )+ T l( P i P ref )' ( where!g o i (T ) ) h! i (T )"T!s o i (T ) is the Gibbs fuctio for pure i at temp. T P = P ref AME 513 - Fall 2012 - Lecture 3 - Chemical Thermodyamics 2 10 5
Equilibrium of a ideal gas mixture" dg =!!g o i (T )d i + "T l( P i P ref )d i + i "T dp i )' + i d!!g o i (T ) ( = 0 1st term: )!!g o i (T )d i = k )! i!!g o i (T ), e.g. 2!!g o HI (T )*1!!g o H2 (T )*1!!g o I2 (T ), 2d term: )"T l( P i P ref )d i = k"t l. + P i P ref - AME 513 - Fall 2012 - Lecture 3 - Chemical Thermodyamics 2 P i ( )! i /, 1, e.g. k"t l. 0. - 3rd term: i "T dp i dp ) = "T) i i = "T ) dp i = 0 sice P = costat i P P P i 4th term: ) i d!!g o i (T ) = 0 sice T = costat, 2 k )! i!!g o i (T ) + k"t l. + P i P ref - or i terms of mole fractios' ( )! i + X i! i ( P HI P ref ) 2 ( P H2 P ref ) 1 P I2 P ref ( ) 1 / 1 = 0 2 ( P i P ref )! i + = exp '*)! i!!g o i (T ) "T( 0 ( P P ref ( ) )! i = exp *!i!!g o i (T ) ' ) "T( / 1 1 0 11 Equilibrium of a ideal gas mixture"! For example " 2 X HI X 1 1 H2 X I2! P " P ref 2'1'1 ( ) )T = exp * + ' 2(!g o HI (T )'1(!g o H2 (T )'1(!g o I2 (T ) It is fairly icoveiet to work with the exp[-g/rt] terms, but we ca tabulate for every species its equilibrium costat K i, i.e. the value of this term for the equilibrium of the species with its formig elemets i their stadard states, e.g. ( ) T K HI = exp! "!g o HI (T )! 0.5"!g o H2 (T )! 0.5"!g o I2 (T ) ' So i geeral for a reactio of the form ν A A + ν B B ν C C + ν D D X C! C X D! D X A! A X B! B!! P C +! D '! A '! B " P = K! C! C K D D! ref K A! A K B B X i = mole fractio of i; ote( X i =1 P = total pressure, P ref = referece pressure = 1 atm K i = equilibrium costat of i (from tables) = fuctio of T oly (ot P) AME 513 - Fall 2012 - Lecture 3 - Chemical Thermodyamics 2, - 12 6
Equilibrium costats" Examples of tabulated data o K - (double-click table to ope Excel spreadsheet with all data for, O, 2, C, O 2, H, OH, H 2 O, H 2, N 2, NO at 200K - 6000K) Note K = 1 at all T for elemets i their stadard state (e.g. O 2 ) Molecular weight = 28.01054 g/mole!hf o (kj/mole) -110.541 T K 200 298 300 400 500 600 700 800 900 1000 1100 1200 1300 1400 1500 1600 1700 1800 1900 2000 2100 2200 2300 2400 2500 h-h_298 Keq kj/mole (o uits) -2.858 3.567E+25 0.000 1.190E+16 0.054 9.085E+15 2.975 1.360E+11 5.929 1.664E+08 8.941 1.854E+06 12.021 7.334E+04 15.175 6.416E+03 18.397 9.701E+02 21.686 2.067E+02 25.033 5.870E+01 28.426 2.050E+01 31.865 8.382E+00 35.338 3.886E+00 38.848 1.991E+00 42.384 1.106E+00 45.940 6.577E-01 49.522 4.136E-01 53.124 2.728E-01 56.739 1.876E-01 60.375 1.333E-01 64.019 9.767E-02 67.676 7.350E-02 71.346 5.656E-02 75.023 4.443E-02 O 2 Molecular weight = 31.99879 g/mole!h f o (kj/mole) 0.000 T h-h_298 Keq K kj/mole (o uits) 200-2.866 1.000E+00 298 0.000 1.000E+00 300 0.054 1.000E+00 400 3.029 1.000E+00 500 6.088 1.000E+00 600 9.247 1.000E+00 700 12.502 1.000E+00 800 15.841 1.000E+00 900 19.246 1.000E+00 1000 22.707 1.000E+00 1100 26.217 1.000E+00 1200 29.765 1.000E+00 1300 33.351 1.000E+00 1400 36.966 1.000E+00 1500 40.610 1.000E+00 1600 44.279 1.000E+00 1700 47.970 1.000E+00 1800 51.689 1.000E+00 1900 55.434 1.000E+00 2000 59.199 1.000E+00 2100 62.986 1.000E+00 2200 66.802 1.000E+00 2300 70.634 1.000E+00 2400 74.492 1.000E+00 2500 78.375 1.000E+00 2 Molecular weight = 44.00995 g/mole!h f o (kj/mole) -393.522 T h-h_298 Keq K kj/mole (o uits) 200-3.414 8.148E+94 298 0.000 1.395E+61 300 0.067 5.248E+60 400 4.008 3.678E+43 500 8.314 1.764E+33 600 12.916 2.265E+26 700 17.761 2.662E+21 800 22.815 5.295E+17 900 28.041 7.065E+14 1000 33.405 3.439E+12 1100 38.894 4.454E+10 1200 44.484 1.190E+09 1300 50.158 5.548E+07 1400 55.907 4.013E+06 1500 61.714 4.119E+05 1600 67.580 5.622E+04 1700 73.492 9.700E+03 1800 79.442 2.037E+03 1900 85.429 5.043E+02 2000 91.450 1.437E+02 2100 97.500 4.612E+01 2200 103.575 1.644E+01 2300 109.671 6.406E+00 2400 115.788 2.702E+00 2500 121.926 1.222E+00 AME 513 - Fall 2012 - Lecture 3 - Chemical Thermodyamics 2 13 Chemical equlibrium - example" For a mixture of, O 2 ad 2 (ad othig else, ote 1 DoRF for this case) at 10 atm ad 2500K with C:O = 1:3, what are the mole fractios of, O 2 ad 2? 1 2 1 +.5 O 2 X 1.5 1+.5(1 X O2 " 10 atm 1 ' = K 1.5 K O2 1 = 0.044431 1.5 = 0.0364 (chem. equilibrium) X 2 1 atm K 2 1.222 1 X + X O2 + X 2 =1 (sum of mole fractios = 1) C X + X 2 = = 1 O X + 2X O2 + 2X 2 3 (coservatio of atoms) 3 equatios, 3 ukows: X 2 = 0.6495, X O2 = 0.3376, X = 0.0129 At 10 atm, 1000K: X 2 =.6667, X O2 = 0.3333, X = 2.19 x 10-11 At 1 atm, 2500K: X 2 = 0.6160, X O2 = 0.3460, X = 0.0380 With N 2 additio, C:O:N = 1:3:6, 10 atm, 2500K: X 2 = 0.2141, X O2 = 0.1143, X = 0.0073, X N2 = 0.6643 (X 2 /X = 29.3 vs. 50.3 without N 2 dilutio) (ote still 1 DoRF i this case) Note high T, low P ad dilutio favor dissociatio AME 513 - Fall 2012 - Lecture 3 - Chemical Thermodyamics 2 14 7
Chemical equlibrium" How do I kow to write the equilibrium as 1 2 1 +.5 O 2 ad ot (for example) 2 + 1 O 2 2 2 For the first form X 1.5 X O2 " P 1 X 2 P ' ref 1+.5(1 ) X 2 " 2 P X 2 1 X O2 P ' ref = K 1.5 K O2 1 K 2 2(2(1 2 = K 2 K 2 1 K O2 ) X 2 X O2 1 X 2 2 which is the appropriate expressio of equilibrium for the 2d form - so the aswer is, it does t matter as log as you re cosistet " P P ref ' 2+1(2 = K 2 1 K O2 2 K 2 AME 513 - Fall 2012 - Lecture 3 - Chemical Thermodyamics 2 15 Chemical equlibrium - hydrocarbos" Reactats: C x H y + ro 2 + sn 2 (ot ecessarily stoichiometric) Assumed products: 2,, O 2, O, H 2 O, OH, H, H 2, N 2, NO How may equatios? 10 species, 4 elemets 6 DoRFs 6 equil. costraits 4 types of atoms 3 atom ratio costraits Coservatio of eergy (h reactats = h products ) (costat pressure reactio) or u reactats = u products (costat volume reactio) (products) " X i =1 Pressure = costat or (for cost. vol.) 6 + 3 + 1 + 1 + 1 = 12 equatios How may ukows? 10 species 10 mole fractios (X i ) Temperature Pressure 10 + 1 + 1 = 12 equatios P products P reactats = " i T products (reactats) T reactats i " AME 513 - Fall 2012 - Lecture 3 - Chemical Thermodyamics 2 16 8
Chemical equlibrium - hydrocarbos" Equilibrium costraits - ot a uique set, but for ay set Each species appear i at least oe costrait Each costrait must have exactly 1 DoRF Note that the iitial fuel molecule does ot ecessarily appear i the set of products! If the fuel is a large molecule, e.g. C 8 H 18, its etropy is so low compared to other species that the probability of fidig it i the equilibrium products is egligible! Example set (ot uique) 2 " + 0.5O 2 ; H 2 O " H 2 + 0.5O 2 ; 0.5H 2 " H 0.5O 2 " O H 2 O " 0.5H 2 + OH 0.5N 2 + 0.5O 2 " NO X 1.5 X O2 P 1 X 2 P ( ref ' etc. 1+.5)1 { } 0.5 { K O2 (T)} 1 { = K (T) } 1 K O2 (T) AME 513 - Fall 2012 - Lecture 3 - Chemical Thermodyamics 2 17 Chemical equlibrium - hydrocarbos" Atom ratios C = x O 2r = X + X 2 X + 2X 2 + 2X O2 + X H 2 O + X OH + X O + X NO C = x H y = X + X 2 2X H 2 O + 2X H 2 + X OH + X H = x N 2s = X + X 2 2X N2 + X NO Sum of all mole fractios = 1 " X i = X + X 2 + X O2 + X O + X H 2 O + X H 2 + X OH + X H + X N2 + X NO =1 C Coservatio of eergy (costat P show) h reactats = (reactats) i [ h (T) " h 298 ] i + h o ( ) f,i (reactats) i M i = h products = (products) i [ h (T) " h 298 ] i + h o ( ) f,i (products) i M i AME 513 - Fall 2012 - Lecture 3 - Chemical Thermodyamics 2 18 9
Chemical equlibrium - hydrocarbos" This set of 12 simultaeous oliear algebraic equatios looks hopeless, but computer programs (usig slightly differet methods more ameable to automatio) (e.g. GASEQ) exist Typical result, for stoichiometric CH 4 -air, 1 atm, costat P AME 513 - Fall 2012 - Lecture 3 - Chemical Thermodyamics 2 19 Chemical equlibrium - hydrocarbos" Most of products are 2, H 2 O ad N 2 but some dissociatio (ito the other 7 species) occurs Product γ is much lower tha reactats - affects estimatio of compressio / expasio processes usig Pv γ relatios Bad thigs like NO ad appear i relatively high cocetratios i products, but will recombie to some extet durig expasio By the time the expasio is completed, accordig to equilibrium calculatios, practically all of the NO ad should disappear, but i reality they do t - as T ad P decrease durig expasio, reactio rates decrease, thus at some poit the reactio becomes froze, leavig NO ad stuck at cocetratios MUCH higher tha equilibrium AME 513 - Fall 2012 - Lecture 3 - Chemical Thermodyamics 2 20 10
Adiabatic flame temp. - hydrocarbos" Adiabatic Flame Temp. (K) 3500 3000 2500 2000 1500 1000 Iitial temperature T! = 300K Iitial pressure = 1 atm CH4-air (cost. P) C3H8-air (cost. P) C8H18-air (cost. P) H2-air (cost. P) CH4-O2 (cost. P) CH4-air (cost. V) 500 0 0 0.5 1 1.5 2 2.5 3 Equivalece ratio AME 513 - Fall 2012 - Lecture 3 - Chemical Thermodyamics 2 21 Adiabatic flame temp - hydrocarbos" Adiabatic flame temperature (T ad ) peaks slightly rich of stoichiometric - sice O 2 is highly diluted with N 2, burig slightly rich esures all of O 2 is cosumed without addig a lot of extra uburable molecules T ad peaks at 2200K for CH 4, slightly higher for C 3 H 8, isooctae (C 8 H 18 ) practically idistiguishable from C 3 H 8 H 2 has far heatig value per uit fuel mass, but oly slightly higher per uit total mass (due to heavy air), so T ad ot that much higher Also - massive dissociatio as T icreases above 2400K, keeps peak temperature dow ear stoichiometric Also - sice stoichiometric is already 29.6 H 2 i air (vs. 9.52 for CH 4, 4.03 for C 3 H 8 ), so goig rich does ot add as may excess fuel molecules CH 4 - O 2 MUCH higher - o N 2 to soak up thermal eergy without cotributig ethalpy release Costat volume - same treds but higher T ad AME 513 - Fall 2012 - Lecture 3 - Chemical Thermodyamics 2 22 11
Compressio / expasio" Compressio / expasio processes are typically assumed to occur at costat etropy (ot costat h or u) Use s reactats = s products costat for compressio / expasio istead of h reactats = h products or u reactats = u products All other relatios (atom ratios, ΣX i = 1, equilibrium costraits) still apply AME 513 - Fall 2012 - Lecture 3 - Chemical Thermodyamics 2 23 Compressio / expasio" Three levels of approximatio Froze compositio (o chage i X i s), correspods to ifiitely slow reactio Equilibrium compositio (X i s chage to ew equilibrium), correspods to ifiitely fast reactio (sice oce we get to equilibrium, o further chage i compositio ca occur) Reactig compositio (fiite reactio rate, ot ifiitely fast or slow) - more like reality but MUCH more difficult to aalyze sice rate equatios for 100 s or 1000 s of reactios are ivolved Which gives most work output? Equilibrium - you re gettig everythig the gas has to offer; recombiatio (e.g. H + OH H 2 O) gives extra ethalpy release, thus more push o pisto or more kietic eergy of exhaust Froze - o recombiatio, o extra heat release Reactig - somewhere betwee, most realistic AME 513 - Fall 2012 - Lecture 3 - Chemical Thermodyamics 2 24 12
Compressio / expasio" Example - expasio of 2 -O 2 - mixture from 10 atm, 2500K to 1 atm i steady-flow cotrol volume (e.g. ozzle) or cotrol mass (e.g pisto/cylider) Iitial state (mixture from lecture 2 where h was calculated): X = 0.0129, X O2 = 0.3376, X 2 = 0.6495, T = 2500, P = 10 atm h = -3784 kj/kg, u = -4397 kj/kg " s o (T) = 266.755 J /molek ;" s O2 s = +X " s = 7382 J /kgk ( s o l X ) + X O2 " s o O2 o (T) = 277.207 J /molek;" o s 2 ( l X O2 ) + X 2 " s o 2 l X 2 X M + X O2 M O2 + X 2 M 2 (T) = 322.808 J /molek ( ) l(p /P ref ) (.0129)( 266.755 " 8.314 l(.0602) ) + (0.3376)( 277.207 " 8.314l(0.3376) ) + (0.6495) 322.808 " 8.314l(0.6495) s = ( ) " 8.314l(10/1) (.0129)(0.028) + (0.3376)(0.032) + (0.6495)(0.044) AME 513 - Fall 2012 - Lecture 3 - Chemical Thermodyamics 2 25 Compressio / expasio" Expad at costat etropy to 1 atm, froze compositio: T = 1738K, X = 0.0129, X O2 = 0.3376, X 2 = 0.6495,, h = -4795 kj/kg, u = -5159 kj/kg, s = 7381J/kgK Work doe (cotrol volume, steady flow) = h before - h after = +1011 kj/kg Work doe (cotrol mass) = u before - u after = +762 kj/kg Expad at costat etropy to 1 atm, equilibrium compositio: T = 1794K, X = 0.00022, X O2 = 0.3334, X 2 = 0.6664 (sigificat recombiatio) h = -4811 kj/kg, u = -5184 kj/kg, s = 7382 J/kgK Work doe (cotrol volume, steady flow) = +1027 kj/kg (1.6 higher) Work doe (cotrol mass) = 787 kj/kg (3.3 higher) Moral: let your molecules recombie! AME 513 - Fall 2012 - Lecture 3 - Chemical Thermodyamics 2 26 13
Summary - Lecture 3" I order to uderstad what happes to a chemically reactig mixture if we wait a very log time, we eed to apply 1st Law of Thermodyamics (coservatio of eergy) - but this does t tell us what the allowable directio of the reactio is; A B or B A are equally valid accordig to the 1st Law 2d Law of Thermodyamics (icreasig etropy) - ivokes restrictios o the directio of allowable processes (if A B is allowed the B A is t) Equilibrium occurs whe the worst possible ed state is reached; oce this poit is reached, o further chemical reactio is possible uless somethig is chaged, e.g. T, P, V, etc. The statemets of the 2d law are ds 0 (costat u v, e.g. a rigid, isulated box) dg = d(h - Ts) 0 (costat T ad P, e.g. isothermal pisto/ cylider) AME 513 - Fall 2012 - Lecture 3 - Chemical Thermodyamics 2 27 Summary - Lecture 3" The applicatio of the 2d law leads to complicated lookig expressios called equilibrium costraits that ivolve the cocetratios of each type of molecule preset i the mixture These equilibrium costraits are coupled with coservatio of eergy, coservatio of each type of atom, ad the ideal gas law (or other equatios of state, ot discussed i this class) to obtai a complete set of equatios that determie the ed state of the system, e.g. T, P ad compositio of the combustio products AME 513 - Fall 2012 - Lecture 3 - Chemical Thermodyamics 2 28 14