ADVANCED THERMODYNAMICS ENGINEERING phần 8 doc

Direction of Heat Transfer Prior to discussing the direction of a chemical reaction, we will consider the direction of heat transfer.. Likewise, the direction of a chemical reaction unde

1 Steady State System

Under steady state conditions dNk/dt = 0, dN/dt = 0,dmk /dt = 0, dm/dt = 0 and from

Eq (28),m˙i = m˙ e Likewise, from Eq.(33), N˙

The reaction equation is written as follows:

2 CO + 3 O2→ NCO,e CO + NCO2,e CO2 + NO2,e O2 (A)Since mass does not accumulate within the reactor, the atom balance for C and O at-oms yields

With the three equations Eqs (B) through (D), we can solve for the three unknowns

NCO,e, NCO2,e and NO2,e, i.e.,

NCO,e = 1.6, NCO2,e = 0.4, and NO2,e = 2.8

The mass conservation implies

dNCO2 /dt = 0 + N˙

CO2 gen - 0.8,where N˙

CO2, gen > 0, since CO2 is a product that is generated In the initial periodswhen the combustor is being fired, the CO2 concentration gradually increases due tothe term dNCO2 /dt

Similarly for 90% efficiency,

NCO,e = 0.2 NCO2,e = 1.8, and NO2,e = 2.1

dmCO /dt = 2×28+ m˙CO,gen - 0.2×28×28 = 50.4 +m˙CO,gen.The combustor is operating steadily so that dmcO/dt = 0,

˙

mCO,gen = - 50.4 kg/sSimilarly ,

dNCO /dt = 2 - 0.2 + N˙

CO, gen = 1.8+ N˙

CO, gen.Since dNCO/dt = 0,

˙

N CO, gen = - 1.8 k mole s-1, and

A combustor is fired with 2 kmole of CO and 3 k mole of O2 When the combustor isjust started, very little CO burns As it warms up, more and more CO are burnt As-sume that mass does not accumulate within the reactor At the point when the com-bustor achieves 40% efficiency, write the mass conservation, mole balance and en-ergy conservation equations If the combustor reaches a steady state with combustion

com-Chapter 12

A INTRODUCTION

Nature is inherently heterogeneous and, consequently, natural processes occur in such

a direction so as to create homogeneity and equilibrium (which is a restatement of the SecondLaw) In the previous sections, we assumed that hydrocarbon fuels react with oxygen to pro-duce CO2, H2O, and other products We now ask the question whether these products, e.g.,

CO2, and H2O, can react among themselves to produce the fuel and molecular oxygen If not,then why not? What governs the direction of reaction? Now we will characterize the parame-ters that govern the predominant direction of a chemical reaction We will also discuss thecomposition of reaction products under equilibrium conditions

B REACTION DIRECTION AND CHEMICAL EQUILIBRIUM

1 Direction of Heat Transfer

Prior to discussing the direction of a chemical reaction, we will consider the direction

of heat transfer Heat transfer occurs due to a thermal potential, from a higher to a lower perature Thermal equilibrium is reached when the temperatures of the two systems (one that istransferring and the other that is receiving heat) become equal Due to the irreversible heattransfer from a warmer to a cooler system, δσ > 0

tem-2 Direction of Reaction

The direction of heat transfer is governed by a thermal potential For any infinitesimalirreversible process δσ > 0 (For heat transfer to take place from a lower to a higher tempera-ture δσ < 0, which is impossible.) Likewise, the direction of a chemical reaction under speci-fied conditions is also irreversible and occurs in such a direction such that δσ > 0.The direction

of a chemical reaction within a fixed mass is also such that δσ > 0 due to chemical bility For instance, consider the combustion of gaseous CO at low temperatures and highpressures, i.e.,

gCO+(1/2))g

O2) for the reactants and FP (=)g

CO2) for the products govern the direction ofchemical reaction at specified values of T and P If FR>FP, Reaction 1a dominates and vice versa if FR<FP, just as the thermal potential T governs the direction of heat transfer

Consider a premixed gaseous mixture that contains 5 kmole of CO, 3 kmole of O2 and

4 kmole of CO2, placed in a piston–cylinder–weight assembly (PCW) at a specified constanttemperature and pressure It is possible that the oxidation of CO within the cylinder releasesheat, in which case heat must be transferred from the system to an ambient thermal reservoir

An observer will notice that after some time the oxidation reaction (Reaction (1a)) ceases whenchemical equilibrium is reached (at the specified temperature and pressure) Assume that theobserver keeps an experimental log that is reproduced in Table 1

In the context of Reaction 1a, if 0.002 kmole of CO (dNC O) are consumed, then1/2×(0.002) = 0.001 kmole of O2 (dNO

2) are also consumed and 0.002 kmole of CO2 (dNCO2)are produced Assigning a negative sign to the species that are consumed and using the associ-ated stoichiometric coefficients,

Table 1: Experimental log regarding the oxidation of CO at specified conditions

Time, sec CO, kmole CO2, kmole O2, kmole

2 denotes the increase in the number of moles of CO2 as a result of the reaction and

νCO2 the stoichiometric coefficient of CO2 in the reaction equation During times 0 < t < tA,

3 Mathematical Criteria for a Closed System

i Specified Values of U, V, and m

For a closed, fixed mass system (operating at specified values of U, V and m) going an irreversible process (cf Chapter 3)

Thus for adiabatic reactions within a rigid vessel, the entropy S reaches a maximum

ii Specified Values of S, V, and m

Recall from Chapter 3 that

dU = TdS – P dV – T δσ (5a)Note that Eq (5a) is valid for a process where irreversible process (δσ > 0) or reversible proc-ess (δσ =0) occurs For a system operating at specified values of S, V, and m,

(dUS,V = – T δσ) ≤ 0, with “<0” for irreversible, “=0” for reversible process (5b)iii Specified Values of S, P, and m

vi Specified Values of T, P, and m

dG = – S dT + V dP

In order to validate equation (8b) for an irreversible process of a fixed mass, we must mine the value of either δσ or dG during an irreversible process (cf Example 8) As wasshown in Chapter 3 when we considered an irreversible mixing process and in Chapter 7 when

deter-we considered an irreversible evaporation process at specified (T,P), deter-we must determine thevalue of G for the reacting system as the reaction proceeds

4 Evaluation of Properties During an Irreversible Chemical Reaction

The change in entropy between two equilibrium states for an open system is

repre-sented by the relation (cf Chapter 3 and Chapter 8)

Similarly,

dU = T dS – P dV + Σµk dNk (9b)Equation (9b) can be briefly explained as follows For a fixed mass closed system, dNk =0.Thus, dU = TdS- PdV, and the change in internal energy ≈ heat added - work performed How-ever, if mass crosses the system boundary and the system is no longer closed (e.g., pumping ofair into a tire), chemical work is performed for the species crossing the boundary (=Σµk dNk).Likewise,

It is apparent from Eqs (9a) to (9e) that

-T (∂S/∂Nk)U,V =-T (∂S/∂Nk)H,P = (∂U/∂Nk) S,V= (∂A/∂Nk)T,V = (∂G/∂Nk)T,P = ˆgk =µk, and (9f)

-TdSU,V,,m = - TdSH,,P,m = dUS,V,m= dHS,P,m = dAT,V,,m= dGT,P,,m = Σµk dNk (9g)

a Nonreacting Closed System

In a closed nonreacting system in which no mass crosses the system boundary dNk =

0 Therefore for a change in state along a reversible path,

b Reacting Closed System

Assume that 5 kmole CO, 3 kmole of O2 and 4 kmole of CO2 (with a total mass equal

to 5×28+3×32+4×44= 412 kg) are introduced into two identical piston–cylinder–weight semblies A and B We will assume that system A contains anti-catalysts or inhibitors whichsuppress any reaction while system B can engage in chemical reactions which result in thefinal presence of 4.998 kmole of CO, 2.999 kmole of O2 and 4.002 kmole of CO2 The specieschanges in system B are dNCO= -0.002, dNO2= -0.001 and dNCO2 = 0.002 kmole, respectively.The Gibbs energy change dG of system B is now determined by hypothetically injecting 0.002kmole of CO2 into system A and withdrawing 0.002 kmole of CO and 0.001 kmole of O2 from

as-it (so that total mass is still 412 kg) so as to simulate the final condas-itions in system B TheGibbs energy GA = G + dGT,P System A is open even though its mass has been fixed Thechange dGT,P, during this process is provided by Eq (9e) Thus,

dGT,P;A = (–0.002)µCO + (–0.001) µO2 + (+0.002) µCO2 (12)Since the final states are identical in both systems A and B, the Gibbs energy change dGT,P;Bduring this process must then equal dGT,P;A Therefore,

dGT,P;B = – Tδσ = dGT,P;A, i.e., (Σµk dNk)A < 0 (13)The sum of the changes in the Gibbs energy associated with the three species CO, CO2, O2 are

dGT,P;B = – Tδσ = (µCO dNCO +µO2 dNO2 + µCO2 dNCO2 )< 0, i.e., (14a)

dGT,P,B =(–0.002)µCO + (–0.001) µO2 + (+0.002) µCO2= -Τδσ < 0 (14b)Note that system B is a chemically reacting closed system of fixed mass

Recall that for irreversible processes involving adiabatic rigid closed systems dSU,V,m

> 0 and from Eq (9g), Σµk dNk < 0 This inequality involves constant (T,V) processes with Abeing minimized or constant (T,P) processes with G being minimized If (S,V) are maintainedconstant for a reacting system (e.g., by removing heat as a reaction occurs), then dUS,V,m =

ΣµkdNk < 0 In this case S is maximized at fixed values of U, V and m, while U is minimized

at specified values of S, V, and m The inequality represented by Eq (13) is a powerful tool fordetermining the reaction direction for any process

c Reacting Open System

If in one second, a mixture of 5 kmole of CO, 3 kmole of O2, and 4 kmole of CO2

flows into a chemical reactor and undergoes chemical reactions that oxidize CO to CO2, thesame criteria that are listed in Eqs (4) through (8) can be used as long we follow a fixed mass

As reaction proceeds inside the fixed mass, the value of G should decrease at specified (T, P)

so that dGT,P≤ 0

5 Criteria in Terms of Chemical Force Potential

The reaction CO+ 1/2O2→ CO2, must satisfy the criterion provided by Eq (14b) to ceed In the context of the above discussion, dividing Eq (15) by 0.002 or the degree of reac-tion,

pro-dGT,P /0.002 = (–1) µCO + (–1/2) µO2 + µCO2 < 0 (14c)Recall that µ κ = ˆgk, i.e.,

where k and νk represent the reacting species and its stoichiometric coefficient for a reaction

In case of the reaction CO+ 1/2O2→ CO2, νCO = -1, νO2 = -1/2, νCO2 = 1 Equation (14c) can

be alternately expressed in the form

µCO + νO2µO2 > µCO2

Defining the chemical force for the reactants and products as

FR = µCO + (1/2) µO2, and FP = µCO2 for reaction CO+ 1/2O2 →CO2 (15)The criterion dGT,P< 0 leads to the relation FR > FP This criterion is equally valid for an adia-batic closed rigid system (U, V, m specified), and adiabatic and isobaric system (H, P, m speci-fied), isentropic rigid closed system (S, V, m specified), isentropic and isobaric systems (S, P,

m specified), and, finally, isothermal and isovolume systems (T, V, m specified)

From Eq (2),

dξ = dN/ν = dN /(-1) = dN /(+1) = 0.002

Replacing 0.002 in Eq (14c) by dξ, the stoichiometric coefficients by νk, and generalizing forany reaction

In the CO oxidation example, the values of F for the reactants and products are

FR = µCO + (1/2) µO2, and FP = µCO2 (21)Since (FR – FP) > 0 for oxidation to proceed,

which is similar to the inequality Thot > Tcold that allows heat transfer to occur from a hotter to acolder body In a manner similar to the temperature (thermal potential), FR and FP are analo-gous intensive properties called chemical force potentials The chemical potential µk is thesame as partial molal Gibb’s function g)

k, (= )

hk - T )s

k), which is a species property Each cies has a unique way of distributing its energy and, thus, fixing the entropy A species distrib-uting energy to a larger number of states has a low chemical potential and is relatively morestable During chemical reactions, the reacting species proceed in a direction to form morestable products (i.e., towards lower chemical potentials) The physical meaning of the reactionpotential is as follows: For a specified temperature, if the population of the reacting species(e.g., CO and O2) is higher (i.e., higher value of FR) than the product molecules (i.e., CO2 atlower FP), then there is a high probability of collisions amongst CO and O2 resulting in a reac-tion that produces CO2 On the other hand, if the population of the product molecules (e.g.,

spe-CO2) is higher (larger FP value) as compared to the reactant molecules CO and O2 (i.,e., lower

FR), there is a higher probability of collisions amongst CO2 molecules which will break into

CO and O2 If the temperature is lowered, the molecular velocities are reduced and the tional energy may be insufficient to overcome bond energy among the atoms in the moleculesthat is required to the potential F(T, P Xi)

transla-a Example 1

Five kmole of CO, three of O2, and four of CO2 are instantaneously mixed at 3000 Kand 101 kPa at the entrance to a reactor Determine the reaction direction and the val-

Solution

We assume that if the following reaction occurs in the reactor:

gCO(T,P) = hCO(T,P) – T sCO(T,P) = (hf,CO0 + (ht,3000K – ht,298K)CO)– 3000×( sCOo

(3000) – 8.314(ln×P/1)) = (–110530+93541) –3000×273.508–8.314×ln 1)

Similarly, at 3000K and 1 bar,

gO2 = –755099 kJ kmole–1, and gCO2= –1242910 kJ kmole–1 (G)The species mole fractions

XCO = 5÷(5+3+4) = 0.417, XO2 = 3÷(5+4+3) = 0.25, and XCO2 = 0.333 (H)Further,

µCO = ˆgCO (3000K, 1 bar, XCO = 0.417) = gCO(3000K, 1 bar) + 8.314×3000×ln(0.417) = –837513 + 8.314 × 3000 × ln 0.467

= –856504 kJ kmole–1 of CO in the mixture (I)Similarly,

µO2 = (3000K, 1 bar, XO

2=0.25) = –789675 kJ per kmole of O2 (J)

µCO2 = (3000K, 1 bar, XCO

2=0.333) = –1270312 kJ per kmole of CO2 (K)Therefore, based on the oxidation of 1 kmole of CO,

ues of FR, FP, and G What is the equilibrium composition of the gas leaving the actor? How is the process altered if seven kmole of inert N2 is injected into the reac-

re-FR > FP, (N)which implies that assumed direction is correct and hence CO will oxidize to CO2.The oxidation of CO occurs gradually As more and more moles of CO2 are produced,its molecular population increases, increasing the potential FP Simultaneously, the

CO and O2 populations decrease, thereby decreasing the reaction potential FR until thereaction ceases when chemical equilibrium is attained Thus chemical equilibrium isachieved when FR = FP, i.e., dGT,P=0 This is illustrated in Figure 1 The correspond-ing species concentrations are

The Gibbs energy at any section

mini-Nitrogen does not participate in the reaction Therefore, dNN

2= 0 and, so, the sions for FR and FP are unaffected However, the mole fractions of the reactantschange so that the values of FR and FP are different, as is the equilibrium composition.The G expression for this case is

< FP, the reverse reaction CO2→ CO + 1/2 O2 becomes possible.

6 Generalized Relation for the Chemical Potential

Recall from Chapter 8 that

µk = ˆgk = gk(T,P) + RT ln ˆαk, where (23)

ˆak = ˆfk/fk is the ratio of the fugacity of species k in a mixture to the fugacity of the same cies in its pure state Equation (15) can be generalized for any reaction in the form

spe-Σˆgk dNk = Σgk(T,P) + RT ln ˆαk) dNk≤ 0, (24)where the activity coefficient ˆak equals the species mole fraction for ideal mixtures and theequality applies to the equilibrium state

b Example 2

Consider the reactions

Figure 2: Illustration of the minimization of the Gibbs energy at equilibrium with respect to thenumber of moles of carbon dioxide produced

O2 in a reactor at 1 bar and 298 K Assume that cp,C/

T–86700/T2 in SI units and T is in K Assume ideal mixture

Solution

If |(FR –FP)|I > |(FR– FP)|II, then the first reaction dominates and vice versa Note thatthe reaction potentials are functions of the species populations and hence vary as a re-action proceeds Using Eq (23),

For solids and liquids, g T Pk( , )≈ gko

(T) Assume that 0.001 moles of C(s) react with0.0005 moles of O2 to produce 0.001 moles of CO Hence,

pO

2 = XO

2P = (50– 0.0005) ÷(0.001+(50–0.0005)) = 0.9999 P = 0.9999 bar.Therefore,

Figure 3: The reaction potentials for reactions I and II with respect to the number

of moles of carbon that are consumed

Which of the two reactions is more likely when 1 kmole of C reacts with 50 kmole of

R = 1.771+0.000877

gO2(298K, 1 bar) = 0 – 298 × 205.03 = –61099 kJ kmole–1 (E)Similarly,

FR = –1710+(–61099) = –62810 kJ, and FP = = –484048 kJ kmole–1, i.e., (G)

(dG/|dNC|)II = (dG/dξ)II = - (FR – FP)II = –394390 kJ

The variations in the reaction potentials for reactions I and II with respect to the ber of moles of carbon that are consumed at a reactant temperature of 298 K are pre-sented in Figure 3, and the corresponding variation in GI and GII in Figure 4 At 298 K

num-CO2 production dominates The analogous variations in GI and GII at 3500 K are sented in Figure 5 At the higher temperature CO formation is favored

pre-Remarks

Since,

gk(T, P, Xk) = hk– Tsk = hk – T (sko–Rln P Xk/1) = hk – T {sko–R ln (P/1)} + RT ln Xk = gk(T,P)+ RT ln Xk,

in general, the values of gk(T, P, Xk) are a function of the species mole fractions If

we assume that |gk(T,P)| » |RT ln Xk,|, then gk(T, P, Xk)≈ gk(T,P)

This offers an approximate method of determining whether reaction I or II is favored.For instance, if the reactions are assumed to go to completion, ∆GI = gCO – (gC +1/2gO2) Likewise, we can evaluate ∆GII to determine whether |∆GII| > |∆GI| ;if so,the CO2 production reaction is favored Values of ∆G(T,P) at 1 bar, i.e., ∆Go(T) aretabulated (Tables 27A and 27B at T= 298 K)

In addition to reactions I and II, consider the following reactions:

which a positive number or FR =gCO2 + gC < FP =2gCO This implies that the tion cannot proceed in the indicated direction In reaction III, the reaction potential ofthe products (FP) is initially low and the value of FR is higher However, the equilib-rium state is reached at a very low CO concentration when dGT,P = 0, i.e., FR = FP.Thereafter, FP > FR or dGT,P > 0, and the reaction does not proceed In other words,

reac-∆Go>0 implies that

C + CO2→ large amounts of leftover C and CO2 + small amounts of CO2 (H)

On the other hand for reaction II, ∆Go

< 0 implies that

C + O2→ small amounts of leftover C and O2 + large amounts of CO2 (I)Generally, the value of ∆Go for a reaction indicates the extent of completion of thatreaction A relatively large negative value of ∆Go

implies that FR » FP, and this quires the largest decrease in the reactant population (or extent of completion of reac-tion) before chemical equilibrium is reached Normally, a positive value for ∆G im-plies that the reaction will produce an insignificant amount of products (reaction III)

re-We will now show that the value of ∆Go for reaction IV can be obtained in terms ofthe corresponding values for reactions I and II For reactions I, II, and IV, respectively

II- ∆Go

I =gCO2(298 K) – gCO(298 K) = –394390 + 137137 = –257253 kJ

We can arbitrarily set gko= 0 for the elemental species C and O2 at T=298 K so that

C CHEMICAL EQUILIBRIUM RELATIONS

For the reaction CO2→ CO + 1/2 O2 to occur, FR (= ˆgCO

2) > FP (=ˆgCO+ 1/2ˆgO

2) Ingeneral,

1 Nonideal Mixtures and Solutions

The physical meaning of this relation is as follows Consider an ideal gas mixture of 5 kmole

of CO, 3 kmole of O2 and 4 kmole of CO2 in a PCW assembly in which a reaction proceeds atfixed (T,P) Here, ˆak = Xk and K(T,P) remains constant while Xk changes Equation (28)must be satisfied as the reaction proceeds and the equality holds good at chemical equilibrium

Unlike the superheated steam tables in which properties are tabulated as functions of(T,P), K(T,P) is tabulated typically only at P0, the standard pressure, since simple relations areavailable (particularly for ideal gases) to relate K(T,P) to K(T,P0) Such a relation is providedbelow Recall from Chapter 8 that for a species k at a state characterized by specified (T,P)

gk (T,P) = gk (T,Po) + (RT ln (f k(T,P)/f k(T,Po)), (29)where f. is the fugacity of species k If that species is an ideal gas, fk(T,P) = P and fk(T,P0) = P0.The second term on the RHS represents the deviation from this behavior at Po Selecting Po = 1bar,

gk(T,P) = gko(T) + (RT ln (f k(T,P)/f k(T, 1bar)), so that (30)using this relation in Eq (28),

-5.0E+05 -4.0E+05 -3.0E+05 -2.0E+05 -1.0E+05 0.0E+00 1.0E+05

CO+1/2 O2=CO

H2+1/2 O2=H2

Figure 6: The variation in the value of ∆Go with respect to the temperature for severalreactions

K(T,P) = exp(–Σνkgko(T)/(RT)) exp(– Σνk ln (f k(T,P)/f k(T, 1bar))) (31)Now let us define

tabu-K(T,P) = Ko(T) Π (f k(T,P)/f k(T, 1bar))(−νk) (34)Subsequently, using Eq (34) in Eq (28b), as the reaction proceeds, the following inequalitymust be satisfied:

Ko(T) ≥ Π (f k(T,P) ˆαk/f k(T, 1bar))νk (35’)

At chemical equilibrium

Ko(T) = Π (f k(T,P) ˆαk/f k(T, 1bar))νk (35a)

In order to evaluate f k(T, 1bar) and the corresponding Ko(T) it is convenient to define the state

of each species k at a standard state corresponding to a pressure of 1 bar

a Standard State of an Ideal Gas at 1 Bar

If one considers the state of species k to be an ideal gas at 1 bar, then f k(T, 1bar) = 1.Hence, using Eq (30),

gk(T,P) = gko (T) + RT ln (fk(T,P)/1) (35b)where gko (T) is evaluated at the temperature T assuming the species k to be an ideal gas.Since the substance is non-ideal at the state (T,P), the second term in Eq (35b) accounts forthe correction due to non-ideal behavior

The values of gko(T) can be determined using the expression

where the term hk0(T) includes chemical and thermal enthalpies The gk0 (T) can also be timated using Gibbs function of formation (Chapter 11) Unless otherwise stated, Eq (35c)will be for determining the values of gk0(T)

es-b Standard State of a Nonideal Gas at 1 Bar

Consider the following chemical reaction

f k(l) (T, P) = f k(l) (T,Psat) POYk≈ f k(l) (T,Psat),

and the Poynting correction factor is (cf Chapter 8)

POY k = exp {∫PPsat vk( )ldP / RT ( )} ≈ 1

Since POY ≈1 for most liquids and solids, Eq (36a) assumes the form

0

g ( )l T , P ( ) ≈ g ( )l T ( ) (36b)

c Example 3

Determine the value of gH O2 ( )l (40ºC, 10 bar) and

standard state Also determine the value of

mixture in which water constitutes 85% on mole basis

= Psat = 0.07384 bar and

Using Eqs (A) and (E),

2 ( )(313 K, 1 bar) + 8.314×313×ln (0.07384/1) (F)Now,

gH O go

2 ( )(313 K, 1 bar) = (hH O go

2 ( ) – TsH O go

2 ( ))313 K, 1 bar = –241321 – 313 × 190.33 = –300894 kJ kmole–1 (G)Using Eqs (A), (E), and (F),

gH Oo

2 ( ) l (313 K,10 bar) = –300894 + 8.314 × 313 × ln (0.07384/1)

= –307675 kJ kmole–1 (H)

If the liquid state is selected at 1 bar instead of an ideal gas state, Eq (A) becomes

gH O g2 ( )(40ºC, 10 bar) for water por Select the standard state to be at 1 bar at 40ºC Assume ideal gas behavior at the

gH O

2 ( ) l (T,P)=gH O

2 ( ) l(T,1 bar) + 8.314 ln(fH O

2 ( ) l (T,P)/fH O

2 ( ) l (T,1 bar)), where (I)

At given T, RT d ln (f) = v dP (Chapter 8) and hence

2 ( ) l – Ts

H O o

( )

gidH O

2 l (313 K, 10 bar) = gH O

2 ( ) l (313 K, 10 bar) + 8.314 × 313 × ln 0.85 = –308159 kJ kmole–1

Remarks

We have selected the standard state with regard to both the liquid and gaseous states.For liquids or solids, g(T,P) ≈ go(T) fk(T,P) ≈ fk(T,1) and hence K(T,P) ≈ Ko

(T)

2 Reactions Involving Ideal Mixtures of Liquids and Solids

For ideal mixtures of liquids and solids αˆk = Xk , and Eq (28b) assumes the formK(T,P) ≥ Π Xkν k, k : liquid and solid phases, (37)where Xk denotes the mole fraction of solid or liquid species k (e.g., CaSO4(s) in a mixture of

Fe2O3(s) , CaSO4(s), and CaO(s)) For a solid,

3 Ideal Mixture of Real Gases

For an ideal mixture of real gases

k k

where ˆfk denotes the fugacity of component k inside the mixture and fk the pure species gacity (cf Chapter 8) Selecting the standard state for all species to be that for an ideal gas at 1bar, using Eq (35a)

fu-Ko(T) ≥ Π(Xkfk(T,P)/1), ideal mix of real gas mixtures, (40)where fk(T,P) are in units of pressure (bar) Recall that fk(T,P) = φk P where φk is the fugacitycoefficient

4 Ideal Gases

For ideal gases, fk(T, P) = P and ˆαk (T,P,Xk) = Xk Therefore, fk (T, 1 bar) = 1, and

Eq (35a) assumes the form

-∆G0

(T)/¯RT ≥ ln (pCO2/1)1 - ln (pCO/1)1 - ln (pO2/1)1/2), where

∆G0(T) = ¯g0CO2 (T) - ¯g0CO (T)- 1/2 ¯g0O2 (T) For convenience the ¯g0 values are tabulated

in Tables A-8 to A-19 Equation (41b) stipulates that at any specified temperature, Ko(T) isconstant and while the reaction occurs, the partial pressure terms on the left hand side of Eq.(41b) keep increasing to approach the value of Ko(T)

Since

Some texts use the nomenclature Kp instead of Ko to indicate that partial pressures are involved

on the left hand side of Eq (43)

gOo2 = 51660 - 1800 x 264.701 = -424800 kJ kmole–1, and from Table A-19

Ko(T) = exp (–127305÷ (8.314×1800)) = 0.00020

From Tables A-28 B log 10 (K0(T)) = -3.696, i.e., K0(T) = 0.0002

For the reaction CO2 → CO + 1/2 O2 to occur, Eq (46) must be satisfied For thespecified composition,

Therefore, CO will oxidize to CO, i.e., the reverse path is favored

Consider a mixture with the following composition at 1800 K and 2 MPa, i.e., N

O2= 0.6 kmole, NCO = 3.6 kmole, and Ntion will the following reaction proceed: CO + 1/2 O2→ CO2, or CO2→ CO + 1/2 O2

We can solve for x and select the root, such that x>0, and y>0, i.e.,

x = 1.8875 kmole, and y = 4 – 2x = 4 – 2 × 1.8875 = 0.225 kmole

Remarks

This problem can also be solved by minimizing G (= xˆg O2 (T,P, XO2) + y ˆg N2 (T,P,

XN2)) at the specified values of T and P, subject to restriction given by Eq (A) Whenthere are a large number of species, say O, O2, O3, etc., we resort to the minimization

of the Gibbs energy and the LaGrange multiplier method can be used See later parts

Determine the NO and NO2 concentrations at chemical equilibrium, assuming the

One kmole of air is in a closed piston–cylinder–weight assembly placed at 298 K and

1 bar Trace amounts of NO and NO2 are generated according to the overall reaction

ontains 2 kmole of O2 A weight is placed on the top ofpiston such that the pressure is 1 bar The gas is then instantaneously heated to 3000

K and maintained at this temperature We find that O atoms are formed and the

con-The values of Gibbs function of formation

gjo(298 K) = 0 kJ kmole–1 Assume that the gases are ideal

KoNO = 4.54×10–31,Since NO exists in trace quantities the partial pressures of N2 and O2 in Eq (D) arevirtually unaffected by reactions (B) and (C) Hence,

KoNO2 = 1.03×10–18,(pNO

g (298 K) for NO and NO2 are, tively 86550 and 51310 kJ kmole–1, and for the elements “j” in their natural forms

respec-There are four unknowns NCO, NCO2 and NO2) and we have three atom balanceequations.

Oxygen atoms: NO= 5 ×1 + 3×2 + 4 ×2 = NCO×1 + NCO2× 2 + NO2× 2 (D)

The fourth equation is given by equilibrium condition at 3000 K: CO2⇔ CO + 1/2

O2 reaction, log10K = –0.48 Using this value in Eq (A)

2 = 1.52 kmole, and N = 17.52 kmole

When the feed stream is altered to react 6 kmole of CO, 3 kmole of CO2, 3.5 kmole of

O2, and 7 kmole of N2, the respective inputs of C, O and N atoms remain unaltered at

9, 19 and 14 respectively Therefore, the equilibrium composition is unchanged Thisindicates that it does not matter in which form the atoms of the reacting species enterthe system The same composition, for instance, could be achieved by reacting a feedstream containing 9 kmole of C(s) (solid carbon, such as charcoal), 9.5 kmole of O2and 7 kmole of N2 (which is treated as an inert in this problem)

From Eq (A) we note that for a specified temperature, the value of Ko(T) is unique.Therefore, if the pressure changes, the temperature does not Eq (E) dictates that thecomposition is altered and more CO2 is produced as the pressure is increased

h Example 8

pressure What is the resulting equilibrium composition and Gibbs energy?

Solution

We leave it to the reader to show that at equilibrium

Consider a PCW assembly that is immersed in an isothermal bath at 3000 K It tially consists of 9 kmole of C atoms and 19 kmole of O atoms (total mass =

ini-9×12.01+ 19×16 = 412 kg) is allowed to reach chemical equilibrium at 3000 K and 1bar What is the equilibrium composition? What is the value of the Gibbs energy? If

we keep placing sand particles one at a time on the piston to a final pressure of 4 bar,i.e., we have allowed sufficient time for chemical equilibrium to be reached at that

At a temperature of 3000 K and a pressure of 4 bar, the equilibrium compositionchanges to

NCO

2= 6.4 kmole, NCO = 2.6 kmole, and NO

Remarks

There is no entropy generated since there is no irreversibility The difference in theminimum Gibbs free energies (i.e., at the equilibrium states) between the two states(3000 K,10 bar) to (3000 K,4 bar) is

If we compress the products very slowly from 1 to 4 bar isothermally at 3000 K, thereaction tends to produce more CO2, i.e., NCO2 increases from 5.25 to 6.4 kmole In-stead, rapid compression to 4 bar produces an insignificant change from the composi-tion at 1 bar, i.e., the products will be almost frozen at NCO2 = 5.25, NCO = 3.75, NO2 =2.37 even though the state is now at 3000 K and 4 bar The products during this initialtime are in a nonequilibrium state The value of G at this state is GFrozen =5.25׈gCO2(3000, 4 bar, XCO2=5.25/11.37) + ^gCO(3000, 4 bar, XC O=3.75/11.37) +

^gO2(3000, 4 bar, XO2=3.75/11.37) = 5.25 × (-1,227,400) + 3.75 × (-830,600) + 2.37

× (-759,800) = -11,360,000 kJ, which is higher than G = -11,374,000 kJ at the rium composition corresponding to 3000 K, 4 bar If we allow more time at the state(3000 K, 4 bar) and examine the composition after a long while, then chemical equi-librium will have been reached, and G will have approached its minimum value of–11,360,000 kJ

equilib-A similar phenomenon occurs when these reacting gases flow at the slowest possiblevelocity through a diffuser where the pressure at the diffuser exit is 4 bar If we fol-low the 412 kg mass when it flows through the diffuser, it will reach its equilibriumcomposition given by Eq (D) However, if the same mass flows at high velocity, thecomposition at the exit of the diffuser can be almost the same as at the inlet.

5 Gas, Liquid and Solid Mixtures

Consider the following chemical reactions

( , ) ˆ( , )

( , ) ˆ(

SO3( ) g = –371060 kJ kmole–1 at 298 K and 1 bar

SO3( ) g) at 298 K for pure H2SO4 (l) and for an ideal mixture of

Further, since the liquid is pure,

ˆ

α H SO2 4= 1, andˆ

2 ( ))(pSO g

3 ( )) = 1.44×10–16,which indicates that the partial pressures are very low, i.e., there is negligible disso-ciation

Ideal Liquid Mixture:

Since the liquid phase is in an ideal mixture, the activity of H2SO4(l) = XH SO2 4, and

Since CO, H2 and CO2 are treated as ideal gases, fk = P, ˆaH2O = 1, fH2O(l) (T,P) ≈ fH2O(l)

(T,1) and for others αˆk = Xk so that Eq (35a) transforms to

∆Go

= –38910 – 457203 – (–169403 – 306675)Consider the water gas shift reaction H2O(l) + CO → H2(g) + CO2(g) Determine the

= –20035 kJ kmole of CO, and

pSO

3 remains unchanged, but XSO

3decreases, i.e., more of the sulfate will be formed

In many instances, in power plants SO2 released due to coal combustion is allowed toreact with lime in order to produce sulfates according to the following reaction

CaO(s) + SO2 + 1/2O2→ CaSO4 (s)

The equilibrium relation for this reaction is

)) (P/1)–1/2.Increasing the pressure at constant T, causes XSO

2to decrease so that a lesser amount

of SO2 will be emitted, i.e., more SO2 is captured from the combustion gases

l Example 12

CO? Also explain what happens if the outlet contains C(s), CO, and O

One kmole each of C(s) and O2 enter a reactor at 298 K The species CO, CO2, and O2

leave the reactor at 3000 K and 1 bar at equilibrium Find the value of the equilibriumcomposition at the exit What is the heat transfer across the boundary? What will hap-pen if the inlet stream is altered to contain 1/2 kmole of oxygen and one kmole of

Consider the reaction of SO3(g) with CaO(s), a process that is used to capture the SO3released during the combustion of coal, i.e., CaO(s) + SO3(g) → CaSO4(s) Determinethe equilibrium relation assuming that the sulfates and CaO are mixed at the molecu-lar level (i.e., they are mutually soluble) and are unmixed What is the partial pressure

The total moles leaving the reactor are

2(298 K) = 121426 kJ per kmole of C(s) consumed (J)

If the inlet stream is altered to contain 1/2 kmole of oxygen and one kmole of CO, theatom balance remains unchanged Therefore, the outlet composition will remain un-altered However, the heat transfer will change, since the inlet stream containing COand O2 has a lower enthalpy as compared to the mixture of C(s) and O2 Hence, thevalue of Q will be lower

If CO,O2 and C(s) are present at the outlet, the overall chemical reaction is

There are two atom balance equations, and we will also consider the following tion to be in equilibrium, i.e.,

K(T) = pCO/1/((pO

2/1) (fC(s)(T,P)/fC(s)(T,1)))Since (fC(s)(T,P) ≈ fC(s)(T,1)),

Ko(T) = pCO/1/((pO

2/1)1/2) = (XCO/XO

Hence, from O atom and C atom balances

2= b+2c, b + d = 1; thus c = (2-b)/2, d = (1-b), and N = b+c = (b + 2)/2, so that

(b = 2/{(P/2)1/2/Ko(T) + 1})< 1, and d = 1 – b

For the C(s) + 1/2 O2→ CO,

Ko(3000 K) = 106.4.Since Ko(3000 K) is relatively large, Eq (M) suggest that XO

2≈ 0, i.e., almost all ofthe oxygen is consumed and converted into product

X can be similarly expressed The required carbon atom input is

Let the number of kmole of O2 entering a reactor equal the value a, while the number

of moles of O2, CO, CO2, and C(s) leaving that reactor equal N

NC(s) Assume the following reactions (respectively, A and B) to be in equilibrium:C(s)+ 1/2 O2→ CO, and CO2→ CO + 1/2 O2 Determine NCO(T) and minimumamount of carbon(s) that should enter the reactor so as to maintain equilibrium at the

6 van’t Hoff Equation

The van’t Hoff Equation for Ko(T) is due to Jacobus Henricus van't Hoff(1852–1911) It presents a relation between the equilibrium constant Ko(T) and the enthalpy ofreaction ∆HR

Eq (66) is a linear relationship Figure 7 presents plots of ln K(T) vs 1/T If Tref = To, we cansimplify the constant A so that

tem-perature Conversely, if ∆HR < 0 (e.g., exothermic combustion reactions), B < 0 At the sition temperature Ttrans the value of K equals unity Applying Eqs (66), (69), and (70),

tran-Ko(T) = 1= exp(∆SR

o/R) exp (–∆HR

o/RTtrans)

Derive the Clausius–Clapeyron Relation from the van’t Hoff equation by consideringDetermine the equilibrium constant for the reaction NH4HSO4(l) → NH3(g) + H2O(g)+ SO3(g) At 298 K, ∆Ho = 336500 kJ kmole–1 and ∆So = 455.8 kJ kmole–1 K Deter-

If the equilibrium constant is known at a reference temperature Tref,

∆VR≈ constant, and

ln (K(T,P)) = –(1/RT) P ∆VR

o + constant

If the value of K(T,Pref) is known, then,

equilib-where ∆HR > 0 for the reaction Hence, d(ln K)/dT > 0, and the value of K increaseswith an increase in the pressure Consequently, since

Ko(T) = (pCO/1) (pO

2/1).0.5/(pCO

The value of pCO

2 decreases (as does that of XCO

2) The effect of increasing the perature is to dissociate more CO2

com-where A = 9.868 and B = 33742.4 in the appropriate units Assume that CO is in trace

Consider the combustion of CH4 in air at 1 bar Determine the number of moles ofproducts produced (and, in particular, the CO concentration) per kmole of fuel con-sumed if the oxygen mole fraction in the products is 3% on a dry basis Assume that

7 Equilibrium for Multiple Reactions

This concept will be illustrated through the following example

0.0003 0.0004 0.0005 0.0006 0.0007 0.0008 0.0009 0.001

1/T, K

CH 4 Carbon

Figure 8: The CO emission due to the combustion of carbon and CH4 at a 20%excess air level

Consider the stoichiometric combustion of one kmole of CH4 with air The productsare at 2250 K and 1 bar stream and contain CO2, CO, H2O, H2, O2, N2, and OH De-

Assume for example NCO, NO2, NCO2 Solve for other species from Eqs (B) to (E).Then check whether Eqs (F) to (H) are satisfied If not iterate We have developed aspreadsheet based program which presents solution for all species at any given T and

P, and any given air composition The results (in terms of one kmole of methane sumed) are

8 Adiabatic Flame Temperature with Chemical Equilibrium

The energy balance equation applicable during combustion in a reactor is

dEcv/dt = ˙Q cv – W˙

cv + Σ,k N˙ik )

h i,k – ΣkN˙e,k)

a Steady State Steady Flow Process

At steady state dEcv/dt = 0 For an adiabatic reactor, ˙Q cv = 0 Since W˙

There are six species of unknown composition The atom conservation equations for

C, H, N, and O atoms are:

Consider the stoichiometric combustion of 1 kmole of CH4 with air at 1 bar The cies enter an adiabatic reactor at 298 K (state 1) and the products leaving it are deter-mined to be the species CO2, CO, H2O, H2, O2, and N2 at equilibrium Determine the

an adiabatic reactor for selected fuels.

9 Gibbs Minimization Method

a General Criteria for Equilibrium

At a specified temperature and pressure, for any composite system, (dGT,P = ΣµkdNk)

≤ 0, where the equality holds at equilibrium The Gibbs energy for a chemical reaction creases as the reaction progresses (cf dGT,P <0) until it reaches a minimum value Since, G =G(T,P,N1, N2 ), the criteria for the G minima are

∂F/∂NO2 = ∂G/∂NO2 + 2λ = 0, i.e., µO2 + 2λ = 0 (and µO + λ = 0) (E)Assuming the ideal mixture model to apply,

2= –755102 kJ kmole–1, and gOo= –323359 kJ kmole–1.

We can solve for NO

2 and selecting the root, such that NO

…, NK) subject to the atom balance equations is as follows

Formulate the atom balance equations for each element “j”

Σkdjk Nk = Aj, j = 1, …, J, k = 1, …, K, (81)where djk denotes the number of atoms of an element “j” in species k (e.g., for the element j=O

in species k =CO2, d = 2) and Aj is the number of atoms of type j entering the reactor Thisrelation can be expressed using the La Grange multiplier method, i.e.,

λj(Σk djk Nk – Aj), j = 1, …, J, k = 1, …, K (82)The Gibbs energy at equilibrium must be minimized subject to this condition

We create a function

F = G + Σjλj(Σk djk Nk – Aj), j = 1, …, J, k = 1, …, K, (83)such that

∂F/∂Nk = (∂G/∂Nk)T,P + (Σjλjdjk) T,P = 0, j = 1, …, J, k = 1, …, K, i.e., (84)Next, we minimize F with respect to Nk, k = 1, …, K, and for each species,

µk + Σjλjdjk = 0, j = 1, …, J, k = 1, …, K (85)