ocw chapter 13
TRANSCRIPT
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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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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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! 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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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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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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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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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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