Chapter 13 Chemical Reaction Equilib

7/31/2019 Chapter 13 Chemical Reaction Equilib

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Chapter 13: Chemical Reaction Equilibrium

Extent of a Chemical Reaction (j):Consider the generic chemical reaction:

aA + bB + yY + zZ or AA + BB + YY + ZZ

In this particular reaction i is called the stochiometric coefficient of

species i; positive (+) for product species,negative (-) for reactant species

For each independent chemical reaction, j, the extent (also called degreeof advancement, degree of reaction, or progress variable) is defined by:

dj = dni,j/ i

For example:Reaction 1 (j = 1): CH4 (g) +2H2O (g) CO2 (g) + 4H2 (g)Reaction 2 (j = 2): CH4 (g) + H2O (g) CO (g) + 3H2 (g)

CH4,1= -1;

CH4,2= -1;

H2O,1= -2;

H2O,2= -1

CO2,1 = +1; CO2,2 = 0; CO,2 = +1; H2,2 = +3; N2,2 = 0


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i

oi,i

i

n

nn

i

0

nn/dnd

i

oi,i

nn iioi

n

n

n

n

][n

n

n

ny

o

iio

species#i

1i

io

iio

species#i

1i

iio

iio

species#i

1i

i

ii

reaction#j

1j

jji,ioi nn

reactions#j

1j

jjo

reactions#j

1j

jji,io

species#i

1i

ji,

reactions#j

1j

jo

reaction#j

1j

jji,io

jji,

reactions#j

1j

io

species#i

1i

jji,

reaction#j

1j

io

species#i

1i

i

i

i

n

n

n

n

][n

n

n

ny

For a single reaction the subscript j is dropped: d = dni/

For multiple reactions: dj = dni,j/ j


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21

21

N2H2COCO2H2OCH4

CH4CH4

225

1

nnnnnn

ny

21

21

N2H2COCO2H2OCH4

H2OH2O

225

22

nnnnnn

ny

21

21

N2H2COCO2H2OCH4

H2H2

225

340

nnnnnn

ny

21

1

N2H2COCO2H2OCH4

CO2CO2

2251

nnnnnnny

21

2

N2H2COCO2H2OCH4

COCO

225

0

nnnnnn

ny

21N2H2COCO2H2OCH4

N2N2

225

1

nnnnnn

n

y

For example:Reaction 1 (j = 1): CH4 (g) +2H2O (g) CO2 (g) + 4H2 (g)Reaction 2 (j = 2): CH4 (g) + H2O (g) CO (g) + 3H2 (g)If initially we have the following moles of the given species:

nCH4,o = 1; nH2O,o = 2; nCO2,o = 1; nN2,o = 1; nH2,o = 0; nCO,o=0Then, in terms of1 and 2:nCH4 = 1-1- 2 ; nH2O = 2 - 21 - 2; nCO2= 1 + 1; nCO= 0 + 2


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Chemical Reaction Equilibrium

e

Gt

(dGt)T,P = 0 or Gt is minimum

Constant T and P

t

T,PT,P

G nG0


5/41

i

i i i i

i i

i i e

iT,P

o o o o

i i i i i i i i i

i i i

o o

i i

d nG = nV dP - nS dT + dn = nV dP - nS dT + d

nG = 0 at = see the previous slide

G +RTln f / f = 0 G + RTln f / f = 0

-G= ln f / f

RT

i

1 n

i

o

i

f(T,P,y ,..y

o

f(T) )only

=-G

K ExpRT

f

f

Standard or Reference State Properties

Are defined for each species at an arbitrary pressure, P0, but at the system

temperature, T.

Standard state need not be the same for all species taking part in a reaction

For a particular species the standard state chosen for must be the same as

that for

For a gaseous species, Gio is usually taken as that of an ideal gas at P0 = 1

bar and the system T. Hence fio = 1 bar

For a liquid species, Gio is usually taken as that of a pure liquid at at P0 = 1

bar and the system T. Hence fio

= fi(T,1 bar)

o

i

Go

if


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The Chemical Reaction Equilibrium Problem

Given:Known initial mole numbers (nio) of chemicalspecies are reacted at T and P, through knownchemical reaction(s) occurring in a homogeneousphase of gas or liquid, until equilibrium is attained

at T and P.

Required:

a) At equilibrium, what is the composition (e.g.,

mole fraction) of each chemical species in thereaction mixture?

b) How do these compositions change if thereaction T and/or P are changed


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ProcedureIdentify the independent, balanced, chemical reactions involved (J) and write

down these reactions (see example 13-11 for the case when only species prevailing

at equilibrium are kno

O1 2 J

o

PT rxn i

o

T

wn)

Compute the equilibrium constant for each reaction, K , K , ..K , at the given

T from Gibbs free change for each reaction at T G and C values

eq 13-18 when H =f(T) or

o

o oT 298

T rxn ii

1 2 N

eq 13-15 when H = H f(T)

-GThat is: K Exp i = 1,.., JRT

Express mole fractions of each species in terms of the reaction extents, , ,..,

R

o

,

i1 2 J j o

i

elate each equilibrium constant to species mole fractions expressed in terms of

freactions extents, , ,.., , i.e., Kf

The results are J algebraic equation

i j

i

j

s in J reactions extents (each independent

reaction has its own extent, ). Solve for these extents and find the corresponding

mole fractions of each speciesConvert mole fractions

into other concentration units, like partial pressure, molarities,...


8/41

Thermodynamic Equilibrium Constant, K

The equilibrium constant of a reaction, j, like:aA + bB yY + zZ, is given by

Where aA, aB, aY, and aZ are the activities of speciesA, B, Y, and Z, respectively, in the equilibriumreaction mixture at its T, P, and composition

.

Activity is a function of T, P and composition of the

equilibrium reaction mixture.

,

i,j

i i

y zo y z a b Y Z

j i Y Z A B a b

A B

a aK f /f a a a a a

a a

i j

ii o

i

fa =

f


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Thermodynamic Equilibrium Constant, K (Continued)aA + bB yY + zZ

Real Cases: ai = yiPi(T,P,y1,y2,,yN)/Po

total pressure

mole fraction of species iin the reaction mixture

fugacity coefficient of species

i in the reaction mixture

reference pressure (1 bar)

Z Y Z Y

A A

z y a bz y z y

T,j a b a b

B B o

y y PKy y P

The equilibrium constant of reaction j at T


10/41

Thermodynamic Equilibrium Constant, K (Continued)

Ideal Cases: ai = yiP/Po

total pressure

mole fraction of species i

in the reaction mixture

Z Y

A

z y a bz y

T,j a b

B o

y y PK

y y P

reference pressure (1 bar)


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Thermodynamic Equilibrium Constant (Continued), K

Liquid reactions in real solutions:

ai = fi/fio = xifii(T,P,x1,x2,,xN)/fi

o

Liquid reactions in ideal solutions: set all s to 1

mole fraction of species i

in the liquid mixtureactivity coefficient of species

i in the liquid mixture

Z Y Z YA A

oz y z y

T,j i ia b a biB B

P-Px x K exp Vx x RT


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Z Y

A

z y a bz y

a b

B o

y y P

y y P

K

RT

Gexp

o

T

Z Y Z Y

A A

z y a bz y z y

a b a b

B B o

y y Py y P

b

B

a

yz

b

B

a

yz

xx

xx

A

YZ

A

YZ

b

B

a

yz

xx

xx

A

YZ

real gas mixture

gas mixture assumed ideal solution

ideal gas mixture

Real liquid mixtures

f(T) only. Why?

f(T, P , composition (extents))

Z Y Z Y

A A

z y a bz y z y

a b a b

B B o

y y P

y y P

ideal liquid mixtures

aA + bB yY + zZ


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Note that:

For an exothermic reaction, i.e., is negative, then as T goesup, K goes down

For an endothermic reaction, i.e., is positive, then as T goesup, K goes up too.

ln(K) goes linearly with (1/T) with a slop of =

Typical variation of ln(K) with (1/T) for some reactions are shown inthe coming slide (can be used to get K also)

To

1

T

1

R

HlnKlnK

o

jTo,

jTo,jT,

oH

R

Ho j298,

oH


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what?tellsLinearity

TatslopeRHR

H

d(1/T)

dlnK oT

o

T

298

1

T

1

R

HlnKlnK

Appendix):(TableGR298.15

1lnK

)ToftindependenisH(i.e.,HHIf

o

j298,

j298,jT,

o

if,298,ji,j298,

o

jT,

o

j298,

o

jT,

T

To

T

To

dTRT

CdT

R

C

T

1

RT

H

RT

HG

RT

GlnK-

)TondependsH(i.e.,f(T)HIf

0

jP,

0

jP,

0

T

o

0

T

0

T0

jT,

jT,

o

jT,

o

jT,

jo,jo,jo,


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Consider the gas phase reaction: aA + bB yY + zZ

Assume further that the reaction mixture is an ideal gas.

Therefore, at equilibrium:

At a given T, K is constant (why?). So, if (y+z-a-b) is > 0 and P

increases, the amount must decrease.

But if (y+z-a-b) is < 0 and P increases, the amountmust increase.

Z Y

A

z y a bz y

a bB o

y y P

y y P

KRT

G

exp

o

T

Z Y A

z y a b

B(y y )/(y y )

Z Y A

z y a b

B(y y )/(y y )

Effect of Pressure on Equilibrium CompositionGas Reactions


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Example 2:

Consider the chemical reaction:

2H2O(g) + C2H6(g) = 2CO(g) + 5H2(g)

Find Ho298. Is the reaction exothermic or endothermic at298 K?

Calculate the equilibrium constant (K) at 25 oC

Assuming Ho to be temperature independent, find K at 750 K

If a reactor is initially charged with 4 moles H2O, 1 mole C2H6,1 mole N2 what are the mole fractions and the mole numbers ofof all species if the reaction reaches equilibrium at 750 Kand 1 bars (assume ideal gas mixtures).


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Example 2 (Continued)

Solution:

From Appendix C, obtain the following data:

Species fHo298(Jmol

-1) fGo298(Jmol

-1)

C2H6(g) -83,820 -31,855

H2 O(g) -241,818 -228,572

CO(g) -110,525 -137,160

H2(g) 0 0

(1)

Ho298 = 5(0) + 2(-110,525) -1(-83,820) -2(-241,818)

= 346,406 Jmol-1

Since Ho298is positive, the reaction is endothermic

(2)

Go298 = 5(0) + 2(-137,160) -1(-31,855) -2(-228,572)

= 214,679 Jmol-1


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Since Go298 is positive, then the reaction as written in not spontaneous,

i.e., for its occurrence work must be done on the reaction system.

K298 = exp(-Go298/RT) = exp[-214,679/(8.314*298)] = 2.33810

-38

extremely low K at 298 K means that almost no products will form

(3)

To find K750, and because Ho is constant;

lnK750 = ln(2.33810-38)(346,406/8.314)*(750-1-298-1) = -2.3861

K750

= exp(-2.3861) = 0.092

(4) Now relate K750 to mole fractions through reaction extent

To

1

T

1

R

HlnKlnKoTo

ToT

Z Y

A

z y a bz y

T a b

B o

y y PK

y y P


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with Po = 1bar, P = 1 bars, y+z-a-b = 5+2-1-2 = 4, and

nH2 = 0 + 5 xH2 = 5/(6+4 )

nCO = 0 + 2 xCO = 2/(6+4 )nC2H6 = 1 - xC2H6 = (1) /(6+4 )

nH2O = 4 - 2 xC2H6 = (4 - 2) /(6+4 )

nN2 = 1 xN2 = 1 /(6+4 )

0.092 = 12500 7

/[(1- )(4-2 )2

(6+4 )4

] [1/1]4

7.36 10-6 - 7/[(1- )(4-2 )2(6+4 )4] [1/1]4=0

Solution (using Goal-Seek in Excel) gives: = 1.0

Substituting for = 1 to get:

nH2 = 0 + 5 = 5 xH2 = 5/(6+4 ) = 0.5

nCO = 0 + 2= 2 xCO = 2/(6+4 ) = 0.2nC2H6 = 1=0 xC2H6 = (1) /(6+4 ) = 0

nH2O = 4 - 2 = 2 xC2H6 = (4 - 2) /(6+4 ) = 0.2

nN2 = 1 xN2 = 1 /(6+4 ) = 0.1

xi = = 1.00


20/41

Example 3

Based on thermodynamic considerations only, which is better to

carry out the previous reaction in example 2 at lower or higher

pressures? What about temperature?

Since y + z a b = 4, which is > 1, then increasing pressure

will shift the reaction to the left (lower products) and vise versa.

As to Temperature, since the reaction is endothermic,

decreasing T will decrease K, hence, the reaction will shift to

the left by lowering T.


21/41

Example 13.6

Consider producing ethanol (C2H5OH) by the direct hydrationof ethylene (C2H4) phase at 250

oC (523 K) and 35 bar (vapor

phase reaction) for an initial steam-to-ethylene ratio of 5

(1) Find the equilibrium constant for the above reaction at thespecified temperature of 250 oC.

(2) Estimate the maximum conversion of ethylene to ethanolassuming the reaction mixture as a real gas mixture

(3) Repeat part (2) assuming the reaction mixture to be an ideal gas

Solution:

(1) Following the same procedure used in example 2 before, youcan show that for the reaction: C

2

H(g) + H2

O(g) = C2

H5

OH(g)K523 = 10.0210

-3 (verify this answer)


22/41

Example (Continued)

Po = 1bar, P = 35 bars, = 1-1-1 = -1, and

nC2H4 = 1 - yC2H4 = (1) /(6 - )nH2O = 5 - yH2O = (5 - ) /(6 - )

nC2H5OH = yC2H5OH = 1 /(6 - )

Lets assume the gas as an ideal solution. In this case

Therefore, we need to compute fugacity coefficients for all

species. Let us the generalized correlation based on virial EoS,i.e.

C2H5OH

H2O

111

C2H5OH

1 1 1 1

H2O C2H4 C2H4 o

y PK

y y P

4.2

r

11.6

r

o

1o

r

ri

0.172/T0.139Band0.422/T0.083B

:withBBT

Pexp


23/41

Example (Continued)

Tc/K Pc/bar i Tri Pri Bo B1 i

C2H4 282.3 50.40 0.087 1.853 0.694 -0.074 +0.126 0.977

H2O 647.1 220.55 0.345 0.808 0.159 -0.511 -0.281 0.887

C2H5OH 513.9 61.48 0.645 1.018 0.569 -0.327 0.021 0.827


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Example (Continued)

Substituting for K, i, P, Po, the xi,s in terms of in the K-

expression below

and solving for yields: = 0.233Now go back and solve for all mole fractions

C2H5OH

H2O

111

C2H5OH

1 1 1 1

H2O C2H4 C2H4 o

y PK

y y P


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35/41 Problem 13.29


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Chapter 13 Chemical Reaction Equilib