ocw chapter 13

8/13/2019 Ocw Chapter 13

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Chemical Engineering Thermodynamics

!"#$%!&' )#&!*%+, #-.%'%/)%.$

Mohammad Fadil Abdul Wahab


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1. Chemical reaction is the heart of chemical processes.

2. Take place in a reactor.

3. A value-added process.

Transform raw materials into products of greater value.

Economic potential or Gross Profit must be positive.i.e. Main products have a higher price than

the raw materials (reactants).

Gross Profit is based solely on price of reactants and products , excluding the equipment and operating costs .


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2. Chemical Reaction Equilibrium

Determination of maximum possible conversionin a chemical reaction.This chapter will cover this part.

1. Reaction Kinetics

The study of rates of reactioni.e. How fast is the reaction?You will learn this in your Chemical Reaction Engineering

class.


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)5;2C3@ I:@5C26 G6 !3@G586:3@

1. Both reaction kinetics and equilibrium conversionare function of T, P and composition

3. So both kinetics and equilibrium conversion mustbe considered for optimum reactor design.

2. Example: Exothermic reaction,

An increase in reaction T

will increase in rates of reaction

but decreases the conversion.


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Single direction (from reactants to products).i.e. forward reaction only.

These reactions highly favor formation of the products.

L + R P + S

Only an extremely small quantity of limitingreactants (if any) remains in the system at equilibrium.

Usually 100% single-pass conversion (of limiting reactant)is considered.


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Appreciable quantities of all reactants and productsspecies can coexist at equilibrium.

Hence the extent of reaction (also conversion)is limited by the chemical equilibrium.

A + B C + D

Forward and reverse reactions.

Eventually equilibrium is reached where rate of forwardreaction is equal to rate of reverse reaction


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P13:29:34518:2 #H0;C3@

! 1 A 1 + ! 2 A 2 " ! 3 A 3 + ! 4 A 4

! 1 ,! 2 are stoic. coefficients of reactant ( - value)

! 3 ,! 4 are stoic. coefficients of product ( + value)

A1 , A 2 , A 3 , A 4 are molecules or atoms

Example ,

CH4

+ 2O2 ! CO 2 + 2 H 2O

" CH4 = # 1, " O2 = # 2

" CO2

= 1, " H2O

= 2


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#B;4


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Reaction coordinate (!) characterizes the extentor degree to which a reaction has taken place.

Also known as,

the extent of reaction (as used in Felder and Rousseau),progress variable,degree of advancement,degree of reaction.


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R38 P:@JD5 )5;2C3@For species i, integrate eqn 13.3 from initial state ( ! =0)

to a state where! =

! , so

dni

nio

ni

! = " i d # 0

#

!

ni $ n

io = "

i(# $ 0)

ni

= nio

+ " i#

ni! = n = n io! + " # i!

= no

+ #"

Summation over all species,

Note: At initial state prior to reaction, ! = 0


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So mole fraction of species i,

y i or x i = ni

n

= nio + ! i"

no + !"

.......... mole fraction of species i is a function of "


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#B;4


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R38 $0DC


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So mole fraction of species i,

y i or x i = n in =

n io + ! ij " j j

#

n ioi

# + ! ij " j j

#i

#

Example of multiple rxns, Main reaction,

C 2 H

6! C

2 H

4 + H

2

Side reactions,

C 2 H

6 + H

2! 2CH

4

C 2 H

4 + C

2 H

6! C

3 H

6 + CH

4


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Yield

=moles of desired product formed

moles that would have been formed if there were no side reactions and

the limiting reactant had reacted completely

Selectivity

=moles of desired product formed

moles of undesired product formed

Also for multiple reactions,


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!8:158:; 37 !954 )B@ #H0:D:F8:04

From FPR for homogenous system of variable composition,

d (nG ) = (nV )dP ! (nS )dT + i dn ii

" (11.2)

For system with single chemical reaction,substitute eqn 13.3,

d (nG ) = (nV )dP ! (nS )dT + " i i d # i

$

Apply the criterion of exactness,

d (nG )d !

"

#$

%

&'

T , P

= ( i i

i

)


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Using 1st law, 2nd Law and FPR, we could show that atequilibrium (see chapter 14 and next slide),

(dnG )T , P = 0 (14.68)

So, at chemical rxn equilibrium,

Hence,

d (nG )d !

"

#$

%

&'

T , P

= 0

!

i ii

" = 0 (13.8) This is the criteria ofChemical Rxn Equilibrium


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All irreversible process occurring at constant T & P proceed insuch a direction as to cause a decrease in the Gibbs energyof the system.


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Rearrange:

! iG io

i" + RT ln f i

f io

# $ %

& ' (

! i

i" = ! iG io

i" + RT ln f i

f io

# $ %

& ' (

! i

i) = 0

i

! signifies the product over all species i.

e. g . "i = 5 a i =

a1a 2a 3a 4a 5 .

Let f i

f io

! " #

$ % &

' i

i( = K (13.10)

K is known as equilibrium constant.


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So,!

iG

i

o

i

" + RT ln K = 0

ln K =

# ! iG

i

o

i

"

RT =

#$ G o

RT

where ,!

iG

i

o

i

" = # G o (13.12)

= The Std Gibbs Energy Change of Rxn at equi T

The data for G i o is available in the form of " G of,i,298

See Table C.4 pg 686


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! G o = " iG

i

o

i

#

= " H 2

G H 2

o + " CO

GCO

o+ "

CH 4G

CH 4

o+ "

H 2 OG

H 2 O

o

Calculate " G o at 298K,

CH 4 + H 2O ( g ) ! CO +3 H 2EXAMPLE

= ! H 2 " G f , H 2,298o +! CO " G f ,CO ,298

o + ! CH 4

" G f ,CH 4,298o + !

H 2O" G f , H 2O ,298

o

=3(0) + 1(-137169) + (-1)(-50460) + (-1)(-228572)

=141863 Joules/mol CH 4 reacted #


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ln K =!" G o RT

(13.11b)

K = exp !" G o

RT

# $ %

& ' (

(13.11a)

for T=T 0

K 0 = exp!" G 0

o

RT 0

# $ %

& ' (

(13.21)

Note: Data for standard state isusually available at T 0=298.15Kor 25 oC and P o=1 bar


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#M521 37 * 3@ I

To calculate K at T other than the standard state Tof 298K. Lets use van Hoffs eqn,

d ln K =! H o

RT 2dT (13.14)

If ! H o (std heat of rxn) could be assumed CONSTANT,integration gives,

ln K

K ' =

" ! H o

R

1T "

1T '

# $ %

& ' ( (13.15)


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K = K ' exp ! " H o

R

1T

! 1T '

# $ %

& ' (

)*+

,-.

Let's use " H o = " H 0o and T ' = T o

We could rearrange,

K = K 0

exp ! " H

0

o

R

1

T ! 1

T 0

#

$ %

&

' (

)

*++

,

-..

= K 0

exp" H 0

o

RT 01 !

T 0

T

#

$ %

&

' (

)

*++

,

-..

= K 0 K 1

where,

K 1 = exp! H 0

o

RT 0

1 "T

0

T

# $ %

& ' (

# $ %

& ' (

(13.22)


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Eqn (13.15) can also be written as,

ln K = ! " H o

R

1T

+" H o

R

1T '

+ ln K '# $ %

& ' (

(13.15a)

Exothermic reaction, slope positive,

(K decrease with increasing T)

Endothermic reaction, slope negative,(K increase with increasing T)

! " H o

R= slope

ln K

K ' = ! "

H o

R

1

T !

1

T '# $ %

& ' (

Plot of ln K vs1

T

is a straightline as shown

in Figure 13.2.


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If ! H o could not be assumed constant,

K = K 0 K

1 K

2 Note: Data for standard state isusually available at T 0=298.15Kor 25 oC and P o=1 bar

where, ! =TT0

For heat capacity constant, " A = # i A ii

$ etc.

where,

K 0 = exp!" G 0

o

RT 0

#

$ %

&

' ( K 1 = exp

" H 0o

RT 01 !

T 0

T

#

$ %

&

' (

#

$ %

&

' (

K 2

= exp

! A [ln " # (" #1

" )] +

12

! BT 0

(" # 1) 2

" +

16 ! CT 0

2(" # 1) 2 (" + 2)

" +

12

! DT

02

(" #1) 2

" 2

$

%&&

'&&

(

)&&

*&&

(13.24)


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)5D;C3@ 37! 13 5H0:D:F8:04234


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also ! i" = !

K = P

P o# $ %

& ' (

!

) i yi( )! i

i*

so

! i y

i( )" i

i# =

P

P o$ % &

' ( )

*"

K (13.26)


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If the mixture is an ideal mixture,

yi! i( )" i

i# = P

P o$ % &

' ( )

*"

K (13.27)

If the mixture is an ideal mixture at low pressure,

it becomes an ideal-gas mixture, so

yi( )!

i

i" =

P

P o# $ %

& ' (

) !

K (13.28)


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Remember, we can express y i in terms of ! ,

For single rxn,

y i =n io + !" inio# + ! " i#

For multiple rxns,

y i =

n io + ! ij " j j

#

n ioi

# + ! ij " j j

#i

#


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#M521 37 * ;@> N 3@ #H0:D:F8:04!3@G586:3@

Consider an ideal gas reaction,

yi( )!

i

i

" = P P

o

#

$ %

&

' (

) !

K (13.28)

ln K = !

" H o R

1

T +" H o

R

1

T ' + ln K '

# $ %

& ' (

And eqn 13.15a gives therelation of K wrt. T ,


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Effect of TEMPERATURE

For endothermic rxn , an

increase in T will result inan increase in K,therefore an increase in,

yi( )!

i

i" =

yc!

c yd !

d

ya|! a | yb

|! b |

yi( )

! i

i" = P

P o# $ %

& ' (

) !

K

The composition or fractionof products will be higher.

An increase in ! e.

Shift of rxn to the right.

Higher equilibrium conversion.


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Effect of TEMPERATURE

For exothermic rxn , an increase

in T will result in a decrease inK, therefore a decrease in,

yi( )!

i

i" = P

P o# $ %

& ' (

) !

K

yi( )!

i

i" =

yc!

c yd !

d

ya|! a | yb

|! b |

The composition or fractionof products will be reduced.

A decrease in ! e.

Shift of rxn to the left.

Lower equilibrium conversion.


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Effect of PRESSURE

If ! is negative* value, an

increase in P (at constant T)causes an increase in,

yi( )!

i

i" = P

P o# $ %

& ' (

) !

K

yi( )!

i

i" =

ycV c yd

V d

ya|V a | yb

|V b |

The composition or fractionof products will be higher.

An increase in ! e.

Shift of rxn to the right.Higher equilibrium conversion.

*reduction in mole number


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Effect of PRESSURE

If ! is positive* value, an

increase in P (at constant T)will result in a decrease in,

yi( )!

i

i" = P

P o# $ %

& ' (

) !

K

yi( )!

i

i" =

ycV c yd

V d

ya|V a | yb

|V b |

The composition or fractionof products will be reduced.

A decrease in ! e.

Shift of rxn to the left.

Lower equilibrium conversion.*increase in mole number


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)5D;C3@ 37! 13 5H0:D:F8:04234


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To find in term of measured variable,

G i = ! i (T ) + RT ln f i (11.31)

Apply at T and std state pressure of P=1 bar,

f i f i

o

Gio = !

i (T ) + RT ln f io

The difference,

Gi " G io = RT ln

f i f i

o


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From FPR, dG = VdP ! SdT at constant T

Integrate at constant T for pure liquid i from P o to P,

Gi ! G

i

o= V

i dP

P o

P

"

Combine ,

RT ln f i f i

o = V i dP

P o

P

!

For V i =V iliq " constant

RT ln f i f i

o = V i ( P # P

o )


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So,ln

f i f i

o =

V i ( P ! P o ) RT

f i f i

o =expV i ( P ! P o )

RT

" # $

% & '

K =! i xi f i f i

o

" # $

% & '

( i

i) = ! i xiexp V i ( P * P

o ) RT

" # $

% & '

" # $

% & '

( i

i)

= ( ! i xi )( i exp

( iV i ( P * P o ) RT

" # $

% & ' i)

= exp( P * P o )

RT (( iV i

i+ )" # $

% & '

(! i xi )( i

i)

So,

Substitute and rearrange,

(! i x i )" i

i# = K exp ( P

0 $ P ) RT

(" iV ii% )& ' (

) * +

(13.31)


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The exponent term is usually 1, except for highpressure system, so,

(! i x i )" i

i

# = K (13.32)

For ideal liquid solution,

( x i )!

i

i

" = K (13.33) Known as the law of mass action

As shown earlier, x i can be written in term of !


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ocw chapter 13

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