2 Combustion and Thermochemistry

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2 Combustion and

Thermochemistry



Review of Property Relations

Extensive Properties

V (m3), U (J), H (J)(=U+PV )

Intensive Properties

v (m3/kg), u (J/kg),h =(J/kg)(=u+Pv )

P,T

V=mv;U=mu;H=mh



Equation of State

For Ideal-gas Behavior:

PV=NR u T

PV=mRT

Pv=RT

P= RT where, R=R u /MW

R u = 8315J/kmol-K; MW is the gas

molecular weight



Calor i f ic Equations of State

u=u(T,v)

h=h(T,P)du=

dh=

dvv

udT

T

uT v )()(

dP P

hdT T

hT P )()(



Constant-volume specific heats

Constant-pressure specific heats

P p

vv

T

hc

T uc

)(

)(



For Ideal Gas

T

T

pref

T

T

vref

T

T

ref

ref

dT chT h

dT cuT u

P

h

v

u

)(

)(

0)(

0)(



Translation

(a)Monatomic species

Translation Rotation

Vibration

(b)Diatomic Species



Fig.2.2



Ideal-Gas Mixtures

Constituent mole fraction and mass fraction

Mole fraction of species i,xi

tot

i

i

ii

N

N

N N N

N x

21



Mass fraction of species i, Y i

tot

i

i

ii

m

m

mmm

mY

21



By definition

1

1

i i

i

i

y

x



Relation of x i and Y i

imixii

mixiii

MW MW Y x

MW MW xY

/

/



Mixture molecular weight MW mix

i

iimix

i

iimix

MW Y MW

MW x MW

)/(

1



Partial pressure of ith species,P i

For ideal gas:

P x P

P P

ii

i

i



Mass-(or Molar-) specific

mixture properties

),(),(

),(),(

ii

i

imix

ii

i

imix

i

iimix

i

iimix

P T s x P T s

P T sY P T s

h xh

hY h

The mixture entropy is calculated:



The consti tuent entropies

Standard-state( P ref P 0=1 atm):

ref

iuref iii

ref

iref iii

P P R P T s P T s

P P R P T s P T s

ln),(),(

ln),(),(



Latent heat of Vapor ization

It is also called the enthalpy of vaporization

Clausius-Clapeyron Equation

),(),(),( P T h P T h P T h liquid vapor fg

2

sat

sat fg

sat

sat

T

dT

R

h

P

dP



First Law of Thermodynamics

First Law-Fixed Mass

212121 E W Q

Heat added to

system in going

fgom state1 to

state 2

Work done by system

on surroundings in

going from state 1 to

state 2

Change in total

system energy in

going from state

1 to state 2



E1-2(E2-E1)

) 21 ( 2 gz vum E

Mass-specific

system internal

energy

Mass-specific

kinetic energy

Mass-specific

system potential

energy

12212121 eeewq



dt dE W Q /

Instantaneousrate of heat

transferred into

system

Instantaneous rateof work done by

system, or power

Instantaneoustime rate of

change of

system energy

dt dewq /

e E/m



First Law-Control Volume-SSSF

)( 000 iiicvcv v P v P mememW Q

Rate of

heat

transferred

across the

control

surface

from the

surroundings, to the

control

volume

Rate of

all work

done by

the

control

volume,in

cluding

shaftwork, but

excluding

flow

work

Rate of

energyflowing

out of

the

control

volume

Rate of

energy

flowing

into the

control

volume

Net rate of

workjassociated

with

pressure

forces

where fluidcrosses the

control

surface,

flow work



The principal assumptions

The control volume is fixed relative to the

coordinate system

The properties of the fluid at each pointwithin the control volume,or the control

surface, do not vary with time.

Fluid properties are uniform over the inletand outlet flow areas

There are only inlet and one exit stream



2

1 2 gz vue

Totalenergy

per unit

mass

Internalenergy

per unit

mass

Kinetic

energy per

unit mass

Potentialenergy per

unit mass

/ P u Pvuh



F inal form of energy conservation

for a control volume

iiicvcv

z z g vvhhmW Q0

22

00 2

1

iiicvcv z z g vvhhwq

0

22

00 2

1



Reactant and Product Mixtures

Stoichiometry

3.76=79/21( by volume)

The stoichiometric air-fuel ratio

22222 76.3)2/()76.3( aN O H y xCO N Oa H C y x

fuel

air

stoic fuel

air stoic

MW

MW a

m

m F A

1

76.4)/(

S b ti ti f



Some combustion properties of

methane, hydrogen, and solid carbon

for reactants at 298K Δh R Δh R (O/F)stoic Tad,eq

(kJ/kgfuel)(kJ/kgmix) (kg/kg) (K)

CH4+air -55,528 -3,066 17.11 2226

H2+O2 142,919 -15,880 8.0 3079

C(s)+air -32,794 -2,645 11.4 2301



The equivalence ratio

stoic

stoic

A F

A F

F A

F A

)/(

)/(

)/(

)/(

>1, fuel rich mixtures

<1, fuel lean mixtures

=1, stoichiometric mixture



Percent stoichiometr ic air

100% % tric stoichiome

Percent excess air

%100)-(1

%

excess



Example 2.1

A small, low-emission, stationary gas-

turbine engine operates at full

load(3950kW) at an equivalence ratio of 0.286 with an air flowrate of 15.9kg/s.

The equivalent composition of the

fuel(natural gas) is C1.16H4.32. Determinethe fuel mass flowrate and the operating

air-fuel ratio for the engine.



Solution

Given: =0.286, MWair =28.85,

mair =15.9 kg/s,

MWfuel=1.16(12.01)+4.32(1.008)=18.286

Find: mfuel and (A/F)

We will proceed by first finding (A/F) and

then mfuel .The solution requires only the

application of definitions expressed in

Equs above



where a=x+y/4=1.16+4.32/4=2.24. Thus,

and, from Equ. above

,76.4)/( fuel

air stoic

MW

MW a F A

,82.16

286.18

85.28)24.2(76.4)/( stoic F A

8.58286.0

82.16)/(

stoic F A

(A/F)



Since ( A/F ) is the ratio of the air flowrate to

the fuel flowrate,

Comment

Note that even at full power, a large quantity

of excess air is supplied to the engine.

skg skg F A

mm air /270.08.58

/9.15)/(



Example 2.2

A natural gas-fired industrial boiler

operates with an oxygen

concentration of 3 mole percent in

the flue gases. Determine the

operating air-fuel ratio and the

equivalence ratio. Treat the natural

gas as methane.



Solution

Given: x O 2 =0.03, MW fuel =16.04

MW air =28.85.

Find : (A/F) and .

We can use the given O2 mole fraction to

find the air-fuel ratio by writing an

overall combustion equation assuming

“complete combustion,” i.e., no

dissociation (all fuel C is found in CO2

and all fuel H is found in H2O):



CH4+a(O2+3.76N2)CO2+2H2O+bO2+3.76aN2

where a and b are related from conservation of

O atoms.

2a=2+2+2b

From the definition of a mole fraction

.76.41

2

76.321

2

2 a

a

ab

b

N

N x

mix

O

O



Substituting the known value of xO2(=0.03)

and then solving for a yields

or a=2.368

The mass air-fuel ratio, in general, is

expressed as

a

a

76.41

203.0



3.2004.16

)85.28)(368.2(76.4)/(

1

76.4)/(

,)/(

F A

MW

MW a F A

so

MW

MW

N

N F A

fuel

air

fuel

air

fuel

air



To find , we need to determine( A/F )stoic.

For a=2,hence,

Apply the definition of ,

84.03.20

1.17

)/(

)/(

1.1704.16

85.28)2(76.4)/(

F A

F A

F A

stoic

stoic

Ab l t ( St d di d)



Absolute(or Standardized)

Enthalpy and Enthalpy of

Formation)( )( )( ,

0

, T hT hT h i sref i f j

Absoluteenthalpy at

temperature T

Enthalpy of

formation at

standard

reference

state(Tref ,P0)

Sensibleenthalpy

change in

going from

Tref to T

Where,

)()()( 0

,, ref i f ii s T hT hT h



A standard state

A standard-state temperature:

Tref =25C(298.15K)

A standard-state pressure:

Pref

=P0=1 atm(101,325Pa)



The enthalpy of formation are zero for

the elements in their naturally

occurring state at the reference statetemperature and pressure.

For example, at 25ºC and 1 atm, oxygen

exists as diatomic molecules; hence,

0)( 298

0

, 2O f h



To form oxygen atoms at sandard state

requires the breaking of a rather strong

chemical bond. The bond dissociationenergy for O2 at 298K is 498,390 kJ/kmol O2

OO f kmol kJ h /195,249)( 0

,



Enthalpy of formation have a clear

physical interpretation as the net change in

enthalpy associated with breaking thechemical bonds of the standard state

elements and forming new bonds to create

the compound of interest.



Example 2.3

A gas stream at 1 atm contains a misture of

CO, CO2, and N2 in which the CO mole

fraction is 0.10 and the CO2 mole fractionis 0.20. The gas-stream temperature is

1200K. Determine the absolute enthalpy of

the mixture on both a mole basis(kJ/kmol)and a mass basis(kJ/kg). Also determine

the



Solution

Given X co=0.10, T =1200K, X co2=0.20, P =1 atm

Find:hmix, hmix , Y co, Y co2 , and Y N2

Finding hmix requires the straightforwardapplication of the ideal-gas mixture law, thus,

and

7.0122

COCO N X X X

222

222

0

298,

0

,

0

298,

0

,

0

298,

0

,

)(

)(

)(

N f N f N

CO f CO f CO

CO f CO f COiimix

hT hh x

hT hh x

hT hh xh xh



Substitute values from Appendix A

hmix=0.1[-110,540+28,440]

+0.2[-393,546+44,488]

+0.7[0+28,118]

=-55,339.1kJ/kmolmix

To find hmix, we need to determine the

molecular weight of the mixture:



MW mix= xi MW i=0.1(28.01)+0.20(44.01)+0.7

(28.013)=31.212

Then,mix

mix

mixmix kg kJ

MW

hh /12.1869

212.31

1.339,55

6282.0

212.31

013.2870.0

2820.0212.31

01.4420.0

0897.0

212.31

01.2810.0

2

2

N

CO

CO

Y

Y

Y

E h l f C b i d



Enthalpy of Combustion and

Heating Values

For the steady-flow reactor, complete

combustion:

All C CO2

All H H2O

Fig.2.7

reac prod icv hhhhq 0



The definition of the enthalpy of reaction,

or the enthalpy of Combustion, hR (per

mass of mixture), is

or, in terms of extensive properties,

reac prod cv R hhqh

reac prod R H H H





The enthalpy of combustion is shown in

fig2.8

Consistent with the heat transfer being

negative, the absolute enthalpy of the

products lies below that of the reactants.

For example: stoichiometric mixture of

CH4 and air, Hreac=-74,831kJ. At the same

conditions H prod=-877,236kJ,Thus

HR =-877,236-(-74,831)=-802,405kJ



To a per-mass-of-fuel basis

for the above example:

fuel R

fuel

R MW H kg

kJ h /)(

016,50)043.16/405,802()(4

CH

R kg

kJ

h

T it f i t



To a per-unit -mass-of-mixture

basis

where,

mix

fuel

fuel

R

mix

Rm

m

kg

kJ h

kg

kJ h )()(

1)/(1

F Ammm

mm

fuel air

fuel

mix

fuel



The stoichiometric air-fuel ratio for CH4 is

17.11; thus,

Note:The value of the enthalpy of

combustion depends on the temperature

chosen for its evaluation.

8.2761111.17

016,50)(

mixkg

kJ h

Th h t f b ti h ti



The heat of combustion-heating

value hc

hc = hR

The upper or higher heating value,HHV ,is

the heat of combustion calculatedassuming that the most amount of energy,

hence the designation “upper”.

The lowering heating value, LHV,corresponds to the case where none of the

water none of the water is assumed to

condense.



Example 2.4

A. Determine the upper and lower heating

values at 298 K of gaseous n-decane,

C10H22, per kilomole of fuel and per kilogram of fuel. The molecular weight of

n-decane is 142.284.

B. If the enthalpy of vaporization of n-decane is 359 kJ/kgfuel at 298K,what are

the upper and lower heating values of

liquid n-decane?



Solution A. For 1 mole of C10H22, the combustion

equation can be written as

For either the upper or lower heating value,

where the numerical value of Hproddepends on whether the H2O in the

products is liquid(determine HHV)or

gaseous(LHV).

222222210 )76.3(5.15)(1110)76.3(5.15)( N g or l O H CO N O g H C

prod reac RC H H H H



The sensible enthalpy for all species

involved are zero since we desire H C at

the reference state(298K). Furthermore, the

enthalpy of formation of the O2 and N2 are

also zero at 298K. Recognizing that

and

prod

ii prod

reac

iireac h N H h N H



We obtain

We can calculate the enthalpy of

formation for the liquid water:

]1110[)1( 0

)(,

0

,

0

,)(, 2222102 l O H f CO f H C f l O H C hhh HHV H

kmol kJ hhh fg g O H f l O H f /857,285010,44847,2410

)(,

0

)(, 22



Adiabatic Flame Temperature

There are two definitions of adiabatic

flame temperature.

One for constant-pressure combustionOne for constant-volume combustion

Or on a per-mass-of-mixture basis

),(),( P T H P T H ad prod ireac

),(),( P T h P T h ad prod ireac



T ad is defined from this first-law

statement, which is called the

constant-pressure adiabatic flametemperature.

T ad is simple and this quantity

requires knowledge of thecomposition of the combustion

products.



The flame temperature

are typically several

thousand kelvins.

Calculating the complex

composition by invokingchemical equilibrium is the

subject of the next section.



Example 2.5

Estimate the constant-pressure adiabatic

flame temperature for the combustion of a

stoichiometric CH4-air mixture. The

pressure is 1 atm and the initial reactant

temperature is 298 K.

Use the following assumptions:

1.“Complete combustion”(no dissociation)

i.e., the product mixture consists of only

CO2,H2O,and N2.



2. The product mixture enthalpy is estimated

using constant specific heats evaluated at1200K(~0.5(Ti+Tad),where Tad is guessed

to be about 2100K)



Solution

Mixture composition:

Properties(Appendices A and B)

52.7 ,2 ,1

52.721)76.3(2

222

222224

N O H CO N N N

N O H CO N OCH



Species Enthalpy of

Formation @298K

h0f,i (kJ/kmol)

Specific Heat

@1200K

cP,i(kJ/kmol-K)

CH4 -74,831 -----

CO2 -393,546 56.21

H2O -241,845 43.87

N2 0 33.71

O2 0 ----



From F irst Law

298)]-33.71((7.52)[0

)]298(87.43845,241)[2(

)298(21.56546,393)[1(

)]298([

831,74)0(52.7)0(2)831,74)(1(

,

0

,

ad

ad

ad

ad i pi f i prod

react

i

prod

i prod

react

iireact

T

T

T

T ch N H

kJ H

h N H h N H



Equating Hreact to Hprod and

solving for T ad yields

T ad =2318K

Constant-volume adiabatic



Constant-volume adiabatic

f lame temperature

Ideal Otto-cycle analysis:

where U is the absolute(or standardized)internal energy of the mixture.

),(),( f ad prod init init reac P T U P T U

0)( f init prod reac P P V H H



For the ideal-gas law

0)(

,

ad prod init reacu prod reac

prod

ad u prod ad ui f

reac

init ureacinit uiinit

T N T N R H H

Thus

T R N T R N V P

T R N T R N V P



On a per-mass-of-mixture basis

0)(

obtainthusWe

/

/

prod

ad

reac

init u prod reac

prod prod mix

reacreacmix

MW T

MW T Rhh

MW N m

or

MW N m



Example 2.6

Estimate the constant-volume

adiabatic flame temperature for

a stoichiometric CH4-air

mixture using the same

assumtions as in Example 2.5.Initial conditions are T i =298K,

P =1 atm(101,325Pa).



Solution

The same composition and properties used in

Example 2.5 apply here. We note, however,

that the c p,i values should be evaluated at atemperature somewhat greater than 1200K,

since the constant-volume T ad will be

higher than the constant-pressure T ad

.

Nevertheless, we will use the same values

as before.



First Law:

0)(

0)(

ad prod init reacu

prod

ii

reac

ii

ad prod init reacu prod reac

T N T N Rh N h N

or

T N T N R H H

Substitute numerical values we



Substitute numerical values, we

have

kJ T

T T

T H

kJ H

ad

ad

ad

ad prod

reac

)298(5.397236,877

)]29833.71((7.52)[0 )]298(87.43845,241)[2(

)]298(21.56546,393)[1(

831,74)0(52.7)0(2)831,74)(1(



And

where N reac =N prod =10.52kmol.

Reassembling and solving for T ad yields

T ad =2889K

)298)(52.10(315.8)( ad ad prod init reacu T T N T N R



Chemical Equilibrium

In high temperature combustion

processes, the products of

combustion are not a simple

mixture of ideal products, as may

be suggested by the simple

atombalance used to determinestoichiometry. Rather, the major

species dissociate, producing a

minor species.



For example

The ideal combustion products for

burning a hydrocarbon with air are

CO2,H2O,O2, and N2. Dissociation of these species and reactions among

the dissociation products yields the

followingspecies:H2,OH,CO,H,O,N,NO, and

possibly others.



The problem we address here is the

calculation of the mole fractions of all of

the product species at a given temperature

and pressure,subject to the constraint of conserving the number of moles of each of

the elements present in the initial mixture.

This element constraint merely says thatthe number of C, H, O, and N atoms is

constant, regardless of how they are

combined in the various species.

The equilibrium-constant



The equilibrium constant

approach

There are several ways to approach the

calculation of equilibrium composition. To

be consistent with the treatment of equilibrium in most undergraduate

thermodynamics courses, we focus on the

equilibrium-constant approach and limit

our discussion to the application of ideal

gases.



Second-Law Considerations

The concept of chemical equilibrium has its

roots in the second law of thermodynamics.

Consider a fixed-volume, adiabatic reaction

vessel in which a fixed mass of reactants

form products. As the reactions proceed,

both the temperature and pressure rise until

a final equilibrium condition is reached.This final state(temperature,pressure, and

composition) is not governed solely by first

law considerations----second law

Consider the combustion



Consider the combustion

reaction

222

1COOCO

If the final temperature is high enough, the CO2

will dissociate. Assuming the products to consist

only of CO2, CO, and O2, we can write:

productshot

tsreaccold

OCOCOOCO

22

tan

22

)1(2

1



is the fraction of the CO2 dissociated.

Tad is the function of the dissociation fraction

: =1, no heat released and unchanged.=0, the maximum amount of heat release

occurs and the temperature and pressure

would be the highest possible allowed bythe first law.



What constraints are imposed by the

second law on this thought experiment

where we vary ? The entropy of the product mixture can be calculated by

summing the product species entropies, i.e.,

22 2)1(),(),(

3

1

OCOCO

i

i f ii f mix s s s P T s N P T S



What constraints are imposed by the

second law on this thought experiment

where we vary ? The entropy of the product mixture can be calculated by

summing the product species entropies, i.e.,

22 2)1(),(),(

3

1

OCOCO

i

i f ii f mix s s s P T s N P T S



Where N i is the number of moles of species

i in the mixture. The individual species

entropies are obtained from

0,

0 ln)(

P

P R

T

dT cT s s i

u

T

T

i pref ii

f

ref



Where ideal-gas behavior is assumed, and P i is the partial pressure of the ith species.

Plotting the mixture entropy as a function of

, we see a maximum at about 1- =0.5

for CO+0.5O2=CO2



For our choice of conditions(constant

U,V, and m , which implies no heat or

work interactions), the second lawrequires that the entropy change

internal to the system:

dS 0



Thus, we see that the composition of the

system will spontaneously shift toward the

point of maximum entropy when

approaching from either side, since dS is

positive. Once the maximum entropy is

reached, no further change in composition

is allowed, since this would required thesystem entropy to decrease in violation of

the second law. Formally, the condition for

equilibrium can be written:



In summary, if we fix the internal energy,

volume, and mass of an isolated system,

the application of second law, first law andequation of state define the equilibrium

temperature, pressure, and chemical

composition.

0)( ,, mV U dS

Gibb F ti



Gibbs Function

Although the foregoing was useful in

illustrating how the second law

comes into play in establishingchemical equilibrium, the use of an

isolated(fixed-energy) system of

fixed mass and volume is notparticularly useful for many of the

typical problems involving chemical

equilibrium.



For example, there is frequently a

need to calculate the composition of

a mixture at a give temperature,pressure, and stoichiometry. For

this problem, the Gibbs free energy,

G, replaces the entropy as theimportant thermodynamic property.



The Gibbs free energy is defined in terms

of other thermodynamic properties as:

The second law can then be expressed as

TS H G

0)( ,, m P T dG

Which state that the Gibbs f nction al a s



Which state that the Gibbs function always

decreases for a spontaneous, isothermal,

isobaric change of a fixed-mass system inthe absence of all work effects except

boundary(P-dV) work. This principle

allows us to calculate the equilibriumcomposition of a mixture at a given

temperature and pressure. The Gibbs

function attains a minimum in equilibrium,

in contrast to the maximum in entropy we

saw for the fixed-energy and fixed-volume

case. Thus, at equilibrium,

0)( dG



For a mixture of ideal gases, the Gibbs

function for the ith species is given by

where is the Gibbs function of the pure

species at the standard-state pressure( P i

= P 0)

and Pi is the partial pressure. The standard-

state pressur, P 0 by convention taken to be

1 atm, appears in the denominator of the

lo arithm term.

0)( ,, m P T dG

)/ln(

00

,, P P T R g g iuT iT i

0

,T i g

Gibb f ti f f ti



Gibbs function of formation

)()()( 0

elements

00, T g vT g T g j

j

jii f

Where the v j

’ are the stoichiometric coefficients

of the elements required to form one mole of the

compound of interest. For example, the

coefficients are vO2’ =1/2 and vC

’=1 for forming

a mole of CO from O2 and C, respectively. Aswith enthalpies, the Gibbs functions of

formation of the naturally occurring elements

are assigned values of zero at the reference state.

The Gibbs function for a



f f

mixture of ideal gases

where Ni is the number of moles of the ithspecies.

For fixed temperature and pressure, the

equilibrium condition becomes:

or

)]/ln([ 00

,, P P T R g N g N G iuT iiT imix

0mixdG



The second term can be shown to be zero by

reconizing that d(ln P i)=d P i/ P i and that

d P i=0, since all changes in the partial

pressures must sum to zero because the

total pressure the total pressure is constant.

Thus,

0]/ln([)]/ln([ 00

,

00

, P P T R g d N P P T R g dN iuT iiiuT ii

)]/ln([0 00

, P P T R g dN dG iuT iimix



For the general system, where

the change in the number of moles of

each species is directly proportionalto its stoichiometric coefficient, i.e.,

fF eE bBaA



kf dN

kedN

kbdN

kadN

F

E

B

A

S b tit ti d li th



Substituting and canceling the

proportionality constant k , we obtain

Equation can be rearranged and the log terms

grouped together to yield:

0)]/ln([)]/ln([

)]/ln([)]/([

00

,

00

,

00

,

00

,

P P T R g f P P T R g e

P P T R g b P P T R g a

F uT F E uT E

BuT B AuT A

.)/()/(

.)/()/(ln

...)...(

00

00

0

,

0

,

0

,

0

,

etc P P P P

etc P P P P T R

g b g a g f g e

b

B

a

A

f F

e E

u

T BT AT F T E



The term in parentheses on the left-hand-side

is called the standard-state Gibbs function

change

GT

0

,i.e.,

T B f A f F f E f

T BT AT F T E T

g b g a g f g e

g b g a g f g eG

...)...(g

y,alternatelor

...)...(

0,

0,

0,

0,

0T

0

,

0

,

0

,

0

,

0

The equilibrium constant K



The equilibrium constant K P

.)/()/(

.)/()/(00

00

etc P P P P

etc P P P P K

b

B

a

A

f F

e E

p

With these definitions, our statement of chemical

equilibrium at constant temperature and

pressure, is given by

)/exp(

ln

0

0

TRGK

or

K T RG

uT p

puT



From definition of K p and its relation to , we

G0T can obtain a qualitative indication of

whether a particular reaction favors

products(goes strongly to completion) or reactants(very little reaction occurs) at

equilibrium. If G0T is positive, reactants

will be favored since ln K p is negative,which requires that K p itself is less than

unity. Similarly, if G0T is negative, the

reaction tends to favor products



Physical insight to this behavior can be

obtained by appearing to the definition of

G in terms of the enthalpy and entropychanges associated with the reaction. We

can write:

000 S T H GT



Which can be substituted into equ. 2.66b

For K p to be greater than unity,which favors

products, the enthalpy change for the

reaction, H 0 ,should be negative, i.e., the

reaction is exothermic and the systemenergy is lowered. Also, positive change in

entropy, which indicate greater molecular

chaos lead to values of K >1

uu RS T R H

P ee K //0

Example 2 7



Example 2.7

Complex Systems



Complex Systems

2 Combustion and Thermochemistry

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