1
-:
-: .1 .2
1
CHAP . 1
MOLE BALANCES The rate of of the reaction , -rA
The rate of the reaction tell us how fast a number of moles of one chemical species are being consumed to form f orm another chemical species . The term chemical species refers to any chemical component or element with a given identity. The identity of a chemical species is determine by 1.kind atoms " .
2. Number
3.configuration of that species "
Any change ( 1.2.3 ) identity change Ar : - chemical species species ( change in A identity per unit unit ti me per unit volume ) * The symbol r j -r j : - ( mole j consumed / reacted / disappeared ) / time * volume
Various type of of industrial industrial reactors :-
1. continuous reactor 2. batch reactor 3. semi batch reactor
1 – continuous continuous flow reactor a. CSTR :- continuous stirred tank reactor b. PFR :- plug flow reactor c. packed fed reactor
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HAYTHAM ALZOUBI
2
CHAP . 1
MOLE BALANCES
2.batch reactor t0=0
t 1 = final time
NA0 :- initial mole of A CA0 =
CA =
Batch reactor balance In – out out +generation = accumulation acc umulation In =out = 0 Generation = accumulation acc umulation ∗ = ∗ ∗
t= ∫
assume v = v0 = constant
Time necessary to reduce the number of moles of A from NA0 to NA. −
X=
=
NA :- MOLES OF A UNREACTED
-rA= K CAα CBβ
α+β=n α th order with respect to A β order with respect to B n order overall K : constant rate of chemical reaction = ((volume / mole ) n-1)/time
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3
CHAP . 1
MOLE BALANCES
EX1 What the time needed so that the final moles of A become 10% of its initial value in 10 L batch reactor Where –rA = 0.1 min-1 CA ∗
t= ∫
t= ∫ .
∗∗
→ CA = NA / V
* → NA=0.1 NA0
t=-10
t = 23 min −
X=
=
−. =
X = 0.9
PROBLEM 1/15 /d The reaction A →
calculate the time necessary to consume 99.9% of species A in a 1000 dm3 constant volume batch reactor with CAo = 0.5 mol/dm3
(a) -rA = k with k = 0.05
. ( −)
− =∫ = ∗ . .
t= ∫ t= -20
. * ( 0.001 CA – CA CA )
t= -20
. *-0.999 *0.5 = 9.99 h
(b) -rA = kcA
→ k = 0.0001 s-1
c) -rA = kcA2
→ = .
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HAYTHAM ALZOUBI
4
CHAP . 1
MOLE BALANCES
CSTR( molar flow rate)
In – out out +generation=accumulation +g eneration=accumulation Assumption a. good mixing ( ci = cf ) ) b. steady state * for liquid phase v=v0 *for gas phase v≠v0 (
(
FA0-FA+rA*v=0
CSTR volume necessary to reduce the molar flow rate from FA0 to FA. − V= −
→ FA = CA * V
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HAYTHAM ALZOUBI
5
CHAP . 1
MOLE BALANCES
Problem 1/15 The reaction A→
is to be carried out isothermally in a continuous-flow reactor. Calculate the CSTR reactor volume necessary to consume 99% of A (CA= 0.01CA,) when the entering molar flow rate is 5 mol/h, assuming the reaction rate -rA is: (a) -rA = k with k = 0.05 mol/L.h (b) -rA = kCA with k = 0.0001 s-1 (c) -rA = kCA with k = 3 dm 3/mol.h *The entering volumetric flow rate is 10 dm3/h For a constant volumetric flow rate rate v=v0 a:-
V=
− −
CA0 =
→ FA = CA * V → FA0 = CA0 * V
= 0.5 mol / dm ∗
3
− − V= − −
→ V=
. .
∗.
∗ (. )
V= 99 L or 99 dm3
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HAYTHAM ALZOUBI
6
CHAP . 1
MOLE BALANCES Plug Flow Reactor
( Tubler , PFR , plug ) CA decreasing as L increasing X increasing as L increasing X = 1 C A0 = 0 when reversible and complete reaction X PFR > X CSTR X or CA change axially or longitudinally( l ongitudinally( ) ( no change in r direction ) In tubluar reactor when CA decrease V increase and X increase
Assumption
Steady state X or C A change in L direction X or C A doesn't change in r direction
V = ∫
what is the volume of tubler reactor so that F A= 0.1 FA0
∗ * -0.9 * 10 = 90
V = -10
L
given FA0= 10 mol / min
v0= 0.1L/S
b) -rA=k CA
a) –rA=k when k is 0.1
2
V = ∫
N=0
−∗ −
1. K= 0.1 mol / L.S
V = ∫
→ V = ∫ − → V =
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− −
HAYTHAM ALZOUBI
k = 0.1/s
= ∫
∗
=
−∗
7
CHAP . 1
MOLE BALANCES
C ) -rA=k CA 2 V = (v2/k ) * (
k = 0.1 L / mole . s
)
→ FA = 0.1 * 10 = 1
MOL / S
= rA
→
( ∗ )
=
= -k CA
=∫ ∗
∫
V = ( v0/k) ln
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HAYTHAM ALZOUBI
8
CHAP . 1
MOLE BALANCES
Example 1. A certain reaction has a rate given by -r, = 0.005C 2, mol/cm3 . min If the concentration is to be expressed in mol/liter and time in hours, what would be the value and units of the rate constant?
2. consider a liquid phase
A→
The first order (-rA = kcA) reaction is carried out in a tubular reactor in which the Volumetric flow rate , v is constant constant a) Determine the reactor volume necessary to reduce the exiting concentration to 10% of the entering concentration when the volumetric flow rate is 10 dm3/min 0.23 mi n -1 and the specific reaction rate k is 0.23 b) Calculate the volume of a CSTR for the conditions in a. Which volume is larger ANS ( 391.3 L ) c) Calculate the time to reduce the number of moles moles of A to 1% of its initial value in a constant-volume batch reactor for the reaction and data in a. ANS ( 20 MIN )
3) A 10 miute experimental run shows that 75% of liquid reactant is converted to product by 0.5 order rate . what would be the fraction fraction converted in a half- hour run ?
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HAYTHAM ALZOUBI
9
CHAP . 1
MOLE BALANCES
Reactor Mole Balances Summary
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10
CHAP . 1
TYSIR SARHAN
MOLE BALANCES
HAYTHAM ALZOUBI
11
CHAP . 1
TYSIR SARHAN
MOLE BALANCES
HAYTHAM ALZOUBI
12
CH2:- CONVERSION AND FREACTOR SIZING Chapter 2 Conversion and reactor sizing
batch reactor -rA = f ( X ) NA = NA0 - NA0 X NA :- MOLE OF A IN REACTOR AT t NA0 :- MOLE OF A INITIALLY FED AT t=0 NA0 X :- MOLE OF A THAT HAVE BEEN CONSUMED BY CHEMICAL REACTION ( RACTED ) NA0 = NA0 ( 1-X ) dNA = - NA0 dX − ∗
t= ∫
Ex :- In a liquid phase A → of calculate time taken to achieve 90 % conversion in a 200 dm 3 (constant volume )batch reactor with CA0=1M after mixing at temperature 77 ℃ Given :- K = 0.1 min-1 ( * first order )
t= ∫
= -k -1 ln ( 1- X ) = -10 min ln 0.1 = 23 min ( − )
t in this case depend on k,x K = 0.1 MOL / MIN . L ( zero order ) = =
t = ∫
∗.∗
. .
=
how would your answer change if – rA= k cA2
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13
CH2:- CONVERSION AND FREACTOR SIZING Ideal gas law CA0 = ( yA0 * P0 ) / ( R*T 0) yA0=
=
=
Example 2 – 1 1 A gas of pure A at 830 kpa enters a reactor reactor with a volumetric flow rate of 2 dm3 / s at at 500 k , calculate 1. entering concentration of A . CA0 = ( yA0 * P0 ) / ( R*T 0) ∗
CA0 = . . .
∗
= . mol/dm3
2. entering molar flow rate FA0 = CA0 * V0 FA0 = ( 0.2 mol / dm3 ) * 2 dm3 / s = 0.4 mol / s
Note
( equimolar flow of A & inert )
→ yA0 = 0.5
3. calculate time ( in hour ) taken taken to consume 50% of A in 2 dm 3 (constant volume ) batch reactor with C A0=0.20 mol / L Given :- -rA= k CA02 ( 1-X )2 when k = 0.001 L / MOL. S
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14
CH2:- CONVERSION AND FREACTOR SIZING
CSTR REACTOR V=
∗ –
Example ( the second order liquid phase reaction A
→ )
Calculate the volume ( in m 3 ) to achieve 80 % conversion in a CSTR reactor , A enters the reactor at a molar flow rate 0.4 mol / s and volumetric flow rate 1 L/ S Given k = 0.003 L/MOL . S V = ( v 02*X)/ ( k*F A0)(1-X) 2 V = ( 1 (L 2 / S2) * 0.8 ) / ( 0.003 (L/MOL.S) * 0.4 (MOL/S) * 0.2 2) V= 16.6 m3
How would your answer change if
1. The reaction is zero order 2. The reaction is first order
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15
CH2:- CONVERSION AND FREACTOR SIZING PFR REACTOR −∗
= / / → = ∫
The first order reaction A → B i s c a r r i e d o u t i n a t u b u l a r r e a c t or or i n w h i c h v o l u m e t r i c f l o w rate is constant v de te rm in e the reactor reactor volume volume necessary necessary to achieve achieve 90 % conversion conversion , w he n t he volumetric flow rate is 10 liters/min. and specific specific reaction rate k is 0.23/ min V =
∗ −
ANS V = 100 L Example The irreversible irreversible liquid phase second order reaction 2A →
Is carried out in CSTR reactor , the entering concentration concentration of A , C A0 is 2 molar and the exit concentration Of a is 0.1 molar . the entering and exiting volumetric flow rate v 0 is constant at 3 dm 3/s Given k = 0.03 1. what is the UNIT of k ? 2. what is the molar flow rate of A ? 3. the conversion ( X ) = ……. 4. what is the volume of the reactor reactor ? ANS ( V=47.5 L )
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16
CH2:- CONVERSION AND FREACTOR SIZING Example A→ + A 200 dm3 dm3 constant volume batch reactor is prussurized to 20 atm with amixture of 75% A and 25% inert The gas phase reaction is carried out isothermally at 227
℃ . .
( how many a) Assuming that the ideal gas law is valid ( many mol of A are in the reactor initially ? what is the initiall concentration of A ?
NT0 = /
→
NT0 = ( 20 atm * 200 dm3 ) / ( 0.082
. * 500 K ) = 97.6 mol .
NA0 = NT0 * yA0 = 97.6 mol * 0.75 = 73.2 mol CA0 = NA0 / V0 = 73.2 mol / 200 dm 3 = 0.366 mol / dm 3 b) is the reaction is first order when k = 0.1 min -1 calculate the time to consume 99% of A − = ∫ = 46 .1 min ∗ (−)
t= ∫
c) if the reaction is second order ( k = 0.7 dm 3 / mol . min ) calculate the time to consume 80 % of A − ∗
t= ∫
→ t = ∫
= 15.6 min ∗∗( ∗∗( −)( −)(− −))
d) calculate the pressure in the reactor at the time in the part c if the temperature 227 ℃ ( ) NT=NA+NB+NC+NI NT=0.25 NT0 +0.2 NT0 +0.8 NA0 +0.8 NA0 = ( 0.25*97.6 mol ) + ( 1.8 * 73.2 mol ) = 156.1 mol P = ( N T R T / V ) = ( 156.1 mol *
. . * 500 K ) / 200 L ) .
P = 25.6 atm
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17
CH2:- CONVERSION AND FREACTOR SIZING
R eacto ctor siz si zi ng 4
X
PFR CSTR
X
X
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18
CH2:- CONVERSION AND FREACTOR SIZING
Example 2-2 The reation described by the data in table 2-2 A→ Is to be carried out in a CSTR reactor . species A enters the reactor at amolar flow rate of 0.4 mol / s
a) calculate the volume necessary yo achieve achieve 80 % conversion in a CSTR
−
V=
∗ –
→ V = 8 m3 * 0.8 = 6.4 m 3
b) shade the area area that would give the CSTR volume necessary to achieve 80 % conversion X,
−
) CSTR
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HAYTHAM ALZOUBI
(
19
CH2:- CONVERSION AND FREACTOR SIZING
f( NA )
f( CA )
f(X)
PFR
5 POINT
X1=0.2
X2=0.4
X3=0.6
X1=0.1
X2=0.2
X3=0.3 5 POINT
5
4 POINT
4
1/-rA
PFR
∗ ∆ = { f ( X0 ) + 4f ( X1 ) + 2f ( X2 ) +4f ( X3 ) + f ( X4 ) } −
V= ∫
X0
∆ =
1/-rA
−
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f( x0 )
20
CH2:- CONVERSION AND FREACTOR SIZING Example 2-3 The reaction described by the data in tables below is to be carried out in a PFR . the entering molar flow rate of A is 0.4 mol/s
a) use the most accurate of the integration integration formula to determine determine the PFR reactor volume necessary to achieve 80 % conversion . ∗ ∆ = { f ( X0 ) + 4f ( X1 ) + 2f ( X2 ) +4f ( X3 ) + f ( X4 ) } −
V= ∫
X0=0 X1=0.2 X2=0.4 X3=0.6 X4=0.8 V=
∆ =
− = ( 0.8 – 0 ) / 4 = 0.2
. { 0.89 + 4(1.33) +2(2.05) + 4(3.54) + 8 } = 2.165 m3
b) shade the area that would give the PFR PFR the volume necessary to achieve achieve 80 % conversion .
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21
CH2:- CONVERSION AND FREACTOR SIZING
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22
CH2:- CONVERSION AND FREACTOR SIZING Reactor in series In series the conversion is increase but in parallel arrangement conversion is the same . Given – rA rA as a function of conversion one can also design any sequence of reactors : Volume of CSTR in series V n = ( FA0/-rAn ) * ( X n – Xn-1 ) → -rA exist at Xn Volume of PFR in series −∗
= ∫
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23
CH2:- CONVERSION AND FREACTOR SIZING
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24
CH2:- CONVERSION AND FREACTOR SIZING
Example 2-5 For the two CSTERs in sreies . 40% conversion is achieved in the first reactor . what is the volume of each of the two two reactors necessary to achieve 80 % overall conversion of the entering species A
For reactor 1 when X=0.4 V1= ( FA0 / -rA1 )X1 * X1 = 2.05 m3 * 0.4 = 0.82 m 3 For reactor 2 when X 2 = 0.8 then V 2 = ( FA0/-rA2 ) * ( X 2 – X1) V2 = ( 8 m3 ) * ( 0.8 -0.4 ) = 3.2 m 3 Vt = V1+V2 = 0.82+3.2=4.02 m 3
The volume necessary to achieve 80 % conversion in one CSTR V = 8*0.8 = 6.4 m3 The sum of the two CSTR reactor volumes in series is less than the volume of one CSTR to achieve the same conversion
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25
CH2:- CONVERSION AND FREACTOR SIZING Example 2-7 n – C4H10 C4H10 →
The isomerization of butane was carried out adiabatically in the liquid phase and the data in table below were obtained
Calculate the volume of each of the reactor for an entering molar flow rate of n-bentane of 50 Kmol / h
1:- ( CSTR reactor at X=0.2 ) V1 = ( FA0 / -rA ) * X1 = 0.94 m3 * 0.2 = 0.188 m3 2:- ( PFR reactor when X 1 = 0.2 and X2 = 0.6 ) . ∗ –
V2 = ∫.
3 POINT . ∗ = ( ∆ /) ( FA0 ) { f ( X0) + 4 f ( X1 ) + f ( X 2 ) } –
V2 = ∫.
∆ =
( .−. )
V2 = ( 0.2 / 3 )( 50 kmol / h ){ 0.01887 +4*0.017 + 0.02632 )( kmol.m 3/h) = 0.38m3 V3 ( CSTR ) = 2m3 * ( 0.65-0.6 ) = 0.1 m 3
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26
CH2:- CONVERSION AND FREACTOR SIZING Space time is obtained by dividing the reactor volume by the volumetric flow rete entering the reactor
= ( V/v0 ) space time is the time necessary to process one volume of reactor fluid at the entrance conditions. This is the time it takes for the amount of fluid that takes up the entire volume of the reactor to either completely enter or completely exit the reactor.
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27
CH2:- CONVERSION AND FREACTOR SIZING Example 1 – calculate calculate the reactor volume V1 and V2 for the plug flow sequence shown below when the intermediate conversion is 40 % and the final conversion is 80 %
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28
CH2:- CONVERSION AND FREACTOR SIZING
2 – A first order liquid phase and irreversible reaction A→ + Is carried out in a batch reactor over the period of an 60 minute a conversion of 90 % is achieved . the stirred tank reactor that is currently used in batch mode is under consideration consideration for conversion to a CSTR in order to increase the tons of B and C that can be produced per year what is the space time required to achieve 90 % conversion in a CSTR . -rA= k CA For batch reactor t = ( 1/k ) ln ( 1 / ( 1-X ) ) 60 min = ( 1/ k ) ln ( 1/ ( 1-0.90 ) )
→ k = 2.3 hr -1
For CSTR reactor V= ( F A0 * X ) / -r A → CA = CA0 ( 1-X )
= X / ( ( k * ( 1-X ) ) = 3.9 hr
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HAYTHAM ALZOUBI
→CA0 = FA0/V0
29
CH2:- CONVERSION AND FREACTOR SIZING 3 – A A CSTR is fed with a solution of reactant A of concentration 10 k mole/m
3
At a volumetric flow flow rate is 0.02 m 3/s . The product mixture mixture from the 1 st CSTR CSTR is fed into a 2 nd CSTR of twice the volume of the first CSTR . the two CSTRs in series are required to achieve an overall conversion of 80 % . The reaction ( A
→ ) is first order in A with a reaction rate constant
Is 0.2 s-1 . what is the volume of the 1 st CSTR ??
V1 = ( v0 * X1 ) / ( k ( 1-X 1 ) ) ) X1 , V 1 )
CA1 CA2
CA0
CA1
X2 = ( CA0 – C CA2 ) / CA0 = 0.80 when CA0 = 10 k mol / m 3 CA2 = 2 k mol/m3 V2 =
( − ) ∗
FA0 = 10* 0.02 = 0.2 K mol /s V2 = 2V1 ( − ) ∗ =2* ∗ ( − )
X1 = 0.459 V1 =
∗ ( − )
=
.∗. = 0.0848 m3 .∗( .∗( −. )
P ( 2-4/7/8/9)
TYSIR SARHAN
PROBLEMS
HAYTHAM ALZOUBI
30
CH3 :- RATE LAWS AND STOICHIOMETRY
A homogeneous reaction is one that involves one phase but a heterogeneous involves more than one phase . A irreversible reaction is one that process only on one direction and continuous in that direction until the reaction exhausted ( ). A reversible direction can proceed in each direction Depending on the concentration of reaction and product relative the corresponding equilibrium on the concentration . In irreversible reaction no chemical completely . The molecularity of reaction is the number of atoms / ions and molecules aA +b B→ cC+dD
A is limiting reactant ) limiting reactant (
limiting reactant ) a ,b ( stoichiometry coefficient C+o2 →co2 2 mol of c and 1 mol of o 2 fed to the reactor Determine the limiting reactant O2 is the limiting reactant A + ( b/a ) B → ( c/a ) C + ( d/a ) D if A is the limiting reactant ( -rA / a ) = ( -r B / b ) = ( r C / c ) = ( r D / d ) (
TYSIR SARHAN
)
HAYTHAM ALZOUBI
31
CH3 :- RATE LAWS AND STOICHIOMETRY
Example 2NO+O2↔2NO2 If NO2 is formed at a rate 4 mole / m 3.s Find the rate of disappearance of NO ( -rNO / 2 ) = ( rNO2 / 2 ) rNO2 = -4 mol / m3.s
Power rate law aA +b B→ cC+dD -rA = f ( CA , T , P , type of catalyst ) -rA = k cAα cBβ α + β= n ( overall order )
as T increase k increase α , β :- from experimental
k = f( T ) = A −/ E :- activation energy
A :- preexponential factor/ frequency factor
If ( a = α ) and ( b = β ) called elementary reaction
Example 0.5 A + B → 2D + C What is the rate of D information α=a = 0.5
β=b = 1
( -rA /0.5 ) = ( rD /2 )
rD= -4 rA
rD = 4 kA CA0.5 CB
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32
CH3 :- RATE LAWS AND STOICHIOMETRY Reversible reaction aA + bB
↔ cC+ dD
at equilibrium the rate of reaction is identically zero for all species rA = -rA = 0 the concentration at equilibrium are related by the thermodynamic relationship for the equilibrium K C KC =( CCec * CDed ) / ( CAeα * CBeb ) e :- equilibrium The unit of K C are ( mol / dm 3 ) d+c-a-b Example
2B
↔ D + H2
what is the net rate of disappearance of B For reversible reaction a. forward reaction 2B → D + H2 -rB ( forward ) = kB * CB2 b. backward reaction D + H2 → 2B rB ( backward ) = k -B * CD * CH2 the net rate of formation rB = rB forward + rB backward rB = - kB * CB2 + k -B * CD * CH2 -rB = - kB * CB2 + k -B * CD * CH2 At equilibrium ( – rA = rA = 0 ) and elementary elementary reaction KC = ( kB / k-B )e this is called thermodynamically consistent
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33
CH3 :- RATE LAWS AND STOICHIOMETRY Example A ↔ B + C Given – rAf = = k CA
-rAb = k-1 CB2 CC
Is the reaction thermodynamically ??? At equilibrium r A = 0 rA = -k CA + k-1 CB2 Cc 0 = -k CA + k-1 CB2 CC
( k / k-1 ) = ( CB2 * Cc ) / C A
KC = ( CB CC ) / CA
KC ≠ ( k / k -1 -1 )
This reaction isn't thermodynamically cconsistent onsistent The reaction rate constant k = f( T ) = A −/
Extrapolation Ln k = ln A - ( E / RT )
unit of A is the same unit of k Y
ln k
X
E = -slope * R
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HAYTHAM ALZOUBI
T-1
34
CH3 :- RATE LAWS AND STOICHIOMETRY
Example 3-1-1
A K ( s-1 ) 0.00043 0.00103 0.0018 0.00355 0.00717
1 2 3 4 5 6
B Ln k -7.75 -6.88 -6.32 -5.64 -4.94
C T-1 K-1 0.0032 0.00314 0.0031 0.00305 0.00300
SLOPE 0 -1 -2
ln k = -14017T-1 + 37.12
-3 -4 -5 -6 -7 -8 -9
-14017
SLOPE
E= -slope * R = -14017 K * 8.314 j/mol . K = 116.5 kj/mol intercept 37.12 Ln A = intercept = 37.12 A = 1.32 * 10 16 s-1 k =1.32 * 10 16 s-1 exp { -14017 K/T}
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35
CH3 :- RATE LAWS AND STOICHIOMETRY A→
-rA = 0.1 CA
A → -rA = 0.1 CA – rA
Determine the frequency factor from the following data if activation energy equal 1000 j / mol k ( min -1 ) T(c) k1= A −/
0.01 50
?? 100
k2= A −/
( k1 / k2 ) = exp { ( -E/R ) * ( T 1-1 – T T1-1 } 0.1 min-1 / k2 = exp { ( -1000 / 8.314 ) * (323 -1 – 373 373-1) } =0.013644 min -1
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36
CH3 :- RATE LAWS AND STOICHIOMETRY Stoichiometry table for a batch reactor at v constant aA + bB →cC + dD if A is a limiting reactant reactant
Species A B C D I ( INERT ) Total
Initial mol NA0 NB0 NC0 ND0 NI0 NT0
Change mol -NA0 X -(b/a) NA0 X (c/a) NA0 X (d/a) NA0 X ---------
Remaining mol NA=NA0-NA0X NB= NB0 -(b/a) NA0 X NC= NC0+(c/a) NA0 X ND=ND0+ (d/a) NA0 X
----NT=NT0 = ((d/a)+ (c/a) -(b/a)-1)* N A0 X
General equation CJ = CA0 (
+ : product FJ = FA0 (
± ℧ X )
Stoichiometry coefficient (b/a) ( d/a)(c/a)(a/a) ℧ :- Stoichiometry
- : reactant
± ℧ X )
V=V0 Example :- In a liquid phase elementary reaction A+B
→
calculate time taken to achieve 60 % conversion in a 200 dm 3 (constant volume ) batch reactor with N A0=20 mol , N B0=20 mol and NC0 = 50 mol Given :- k = 0.1 L/mol.min
n=2
-rA=k CA CB CA= CA0 ( 1-X )
CB=CA0 ( 1-X )
CA0 = 0.1 mol / L
-rA = k CA02 ( 1-X ) 2 .
t= ∫
= 100 min * (( 1/ ( 1-X )) -1) (−)( −)(− − )
= 150 min
what is the final mol of C ? N C = 20 mol ((50/20) +3*0.6 ) = 86 mol is the reaction thermodynamically consistent ( show your work ) what is the unit of K C TYSIR SARHAN
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37
CH3 :- RATE LAWS AND STOICHIOMETRY
V is constant 1 ) in liquid phase ( compressible ) 2 ) in gas phase if a) given ( constant volume ) b) = 0
= ( c/a ) + ( d/a ) – ( ( b/a ) – 1 1
In a gas phase P V = NRT if ( T P V ) is constant N will change due to the chemical reactant If V variable
≠
V=V0 ( 1+ X ) ( T/T0) (P/P0)
V :- volume of the reaction mixture at any time V0 :- initial volume of the reaction mixture
:- yA0 *
yA0 = NA0 / NT
T=T0 isothermal CJ = ( CA0 / 1+ ) ( (
TYSIR SARHAN
p=p 0 isobaric
± ℧ X ) at isothermal and isobaric
HAYTHAM ALZOUBI
38
CH3 :- RATE LAWS AND STOICHIOMETRY Problem 3/13 NO2CL + 2NH3 → NO2NH2+NH4CL The liquid phase reaction is first order in both ONCB ( NO 2CL ) and ammonia with k = 0.0017 m 3 /kmol.min /kmol.min at 188 ℃ with E = 11273 cal / mol . the initial entering concentration of ONCB and ammonia are 1.8 kmol /m 3 and 6.6 kmol / m 3 , respectively a . write the rate law for the rate of disappearance of ONCB in terms of concentration ONCB is a limiting reactant A + 2B → C+D Assume A and B both first order -rA = k CA CB CA = CA0 (
1* X )
CA =1.8 * ( 1-X )
CB= CA0 ( CB= 1.8 (
. .
2X)
2X)
-rA= k * 3.24 kmol 2 /m6 ( 1-X )( 3.67 – 2X 2X ) f . what is the rate of the reaction when X = 0.9 at T = 25
℃
k = k0 exp ( ( E/R ) ( T 0-1 – T T-1 )) k = 0.0017 m3 /kmol . min exp ( (11273 (11273 cal / mol ) / ( 1.987 cal /mol /mol .K ) ( 461-1 -298-1) k-1
k = 2.027 * 10 -6 m3 /kmol.min /kmol.min -rA= 2.027 * 10 -6 m3 /kmol.min /kmol.min * 1.82 kmol2 /m6 * ( 1-0.9 ) * ( 3.67-2*0.9 ) -rA = 1.228 *10 -6 kmol/m3.min g . what would be the concentration CSTR reactor volume at 25 90% conversion for fed rate of 2 dm 3 / min V=
℃ to achieve
∗ ∗∗ = = ( 0.9 * 1.8 kmol/m 3*0.002 m3 / min ) / ( 1.228*10 1.228*10-6 kmol /m3.min) – –
V= 2638.436 m3
TYSIR SARHAN
HAYTHAM ALZOUBI
39
CH3 :- RATE LAWS AND STOICHIOMETRY Stoichiometric table for a flow system Species A
Fed rate FA0
Change -FA0X
Effluent/exist
B
FB0
-(b/a)FA0X
FB= FA0 (
(b/a) X )
C
FC0
(c/a)FA0X
FC= FA0 (
+ (c/a) X )
D
FD0
+(d/a)FA0X
FC= FA0 (
+ (d/a) X )
I
FI0
------
FI= FA0 ( )
FA= FA0 (
X)
FT = FT0 + ( (d/a)+ (c/a)-(b/a)-1) *F A0*X
= (d/a)+ (c/a)-(b/a)-1 FT = FT0 + FA0X
= ( change in total number of moles moles ) / mole mole of A reacted The gas phase reaction 0.5 N2 +1.5 H2 →NH3 Is to be carried out isothermally . the molar fed is 50% H 2 and 50% N 2 , at a pressure 16.4 atm and 227 ℃ a) what are CA0 , , ? calculate the concentration of ammonia and hydrogen when the conversion of H 2 is 60 % H2 is the limiting reactant b) if the reaction is elementary with k N2 = 40 dm3 /mol . s write the rate of reaction as a function of conversion for a flow system and a constant volume batch system
TYSIR SARHAN
HAYTHAM ALZOUBI
40
CH3 :- RATE LAWS AND STOICHIOMETRY
KC
conversion at equilibrium
Example 3-6 The reversible gas phase decomposition of nitrogen tetroxide ,N 2O4 to dioxide NO2 N2O4 ↔2NO2 is to be carried out at constant temperature . the feed consists of pure N2O4 at 340 K and 202.6 kpa ( 2 atm ) . the concentration equilibrium constant at 340 K is 0.1 mol / dm 3 a) calculate the equilibrium conversion of N2O4 in a constant volume batch reactor A ↔2B KC = ( CBe2 /CAe ) CA = CA0 ( 1-Xe )
CB= 2CA0 Xe
CA0 = ( yA0 * p ) / ( RT ) yA0 = 1 CA = 2 atm / ( 0.0821 atm . dm 3 / mol . k ) * 340K = 0.07174 mol/dm 3 KC = (4CA02 Xe2) / ( C A0 (1-Xe)) = (4CA0 Xe2)/ (1-Xe) = 0.1 mol / dm3 0.1 mol / dm 3 = ( 4*0.07174 mol/dm 3 * Xe2 ) / ( 1- Xe) 0.28696 X2+0.1 X -0.1=0 Xe = 0.44 b) calculate the equilibrium conversion of N 2O4 in a flow reactor v is not constant calculate
,
= 2-1 = 1 v = v0 ( 1+
= yA0 * = 1*1 = 1 Xe )
TYSIR SARHAN
HAYTHAM ALZOUBI
41
CH3 :- RATE LAWS AND STOICHIOMETRY CA = (CA0(1-X))/(1+X) KC = 4X2 /(1-X)2
CB = (2CA0X)/(1+X) ( 0.28696+0.1 )X2 -1 =0
Xe=0.51 c) assuming the reaction is elementary , express the rate of the reaction solely as a function of conversion for a flow system and for a batch system for a batch system ( assume v is constant ) rA = rAf + + rAb = -k CA + k-1 CB2 -rA = k ( CA – ( ( CB2 /KC )) = k ( C A0 ( 1-X ) – ( ( 4CA02X2)/KC ) For a flow system Assume the volume is not constant -rA = k (CA0 ( 1-X/1+X) – ( ( 4CA02X2 /( 1+X)2 .KC)) d) determine the CSTR CSTR volume necessary to achieve 80% of equilibrium conversion for a fed rate 3 mol/min and k A = 0.5min-1 X=0.8 Xe = 0.8 * 0.51 = 0.4 From b c at flow system – rA = 0.00070 mol / dm 3 . min V= ( FA0 * X ) / -rA = (3 mol/min) * ( 0.4 ) / (0.00070 mol / dm 3 . min)=1.71 m 3
: + expansion -
Shrinking
-
0 constant
TYSIR SARHAN
HAYTHAM ALZOUBI
42
CH3 :- RATE LAWS AND STOICHIOMETRY
TYSIR SARHAN
HAYTHAM ALZOUBI
43
Ch4 :- isothermal reactor design Example 4-2 page 163 It is desired to produce 200 million pounds per year of EG ( ethylene glycol ) . The reactor is to be operated isothermally . A 1 lb mol/ft3 solution of ethylene oxide ( EO ) in water water is fed to the reactor ( show in figure below ) together with equal volumetric solution of water water containing 0.9 wt % of catalyst H 2SO4 . the specific reaction rate constant is 0.311 min -1 , determine 1. if 80 % conversion is is to be achieved , determine the necessary CSTR volume 2. if two 800-gal reactor were arranged in parallel , what is the corresponding conversion . 3. if two 800-gal reactor were arranged is series , what is the corresponding conversion . C2H4O + H2O
→C2H6O2
V = ( F A0.X )/(-rA) β
-rA = k CA CA =CA0 ( 1-X ) FA0
v0
CA0
FC = FA0.X FC = 6.137 lbm-mol/min 200*106 lbm year
1 year
1 days
1hour
lb-mol
365 days
24 hour
60 min
62 lbm
FA0 = 7.67 lbm-mol/min
TYSIR SARHAN
HAYTHAM ALZOUBI
44
Ch4 :- isothermal reactor design
) DILUTION (
CA1 = 1lb-mol /Ft3 V1= ( FA0)/CA1 = ( 7.67/1) = 7.67 Ft 3 MW CH2O = 62.4/18 = 3.4667 lb -mol/Ft3
V0 = VA +VB = 2 * 7.67 Ft 3/min = 15 .34 Ft3/min Dilution CA1 * VA1 = CA0*V0 CA0 = 0.5 lb-mol/Ft 3 CA = 0.5 ( 1-0.8 ) -rA = 0.311 min-1 * 0.1 lb-mol/Ft3 = 0.031 lb-mol /Ft3.min V CSTR = 197.34 Ft 3
TYSIR SARHAN
HAYTHAM ALZOUBI
46
Ch4 :- isothermal reactor design X Da
Da X
-rA =( k CA0 ) / ( 1 + k)n The rate of disappearance of A in the n reactor CSTR in parallel V = ( F A0 .X ) / -rA The conversion achieved in any one of the reactor in parallel is identical X1 = X2 = Xn
For a second order liquid phase reaction V = FA0.X / -rA = FA0.X/k CA02 CA = CA0(1-X) FA0=CA0.v0 V = ( CA0.v0.X )/kCA02( 1-X ) 2
= X / kC A0( 1-X )2 X = k CA0 ( X2-2X+1) Da = k CA0 Da X2 -2Da X – X X + Da =0
+−√(+) X= TYSIR SARHAN
HAYTHAM ALZOUBI
47
Ch4 :- isothermal reactor design
2 / In parallel X = ( k ) / ( 1+ ) I Ft3=7.48 gal V = v0 /2
= 800 gal
=
1 Ft3*2 7.48 gal
1 15.34 Ft 3 / min
13.94 min
X= 13.94 min * 0.311 min -1 1+ ( 13.94 min * 0.311 min -1 ) X=0.812
TYSIR SARHAN
HAYTHAM ALZOUBI
48
Ch4 :- isothermal reactor design 3/ in series
X1 = ( k ) / ( 1+ )
= V1/v01 800 gal
1 Ft3 7.48 gal
1 15.34 Ft3/min
= 6.97 min Da = k = 6.97 min * 0.311 min -1 = 2.168 X1 = ( 2.168) / 3.168 = 0.684 ch2 V = FA0 ( ( X2-X1)/-rA ) -rA2 = kCA2 CA2 = CA0 ( 1-X2) V/v0 = ( X 2-X1) / ( k ( 1-X 2 )) X2 = ( Da + X 1 ) / (1+Da ) = 0.90
TYSIR SARHAN
HAYTHAM ALZOUBI
49
Ch4 :- isothermal reactor design Tubular reactor −
V = FA0 ∫
second order
Liquid phase reaction
-rA = k CA2
C A= CA0 ( 1-X )
V = FA0/kCA02 ( X/1-X ) FA0 = CA0 .v0 Da = k CA0 X = Da / ( 1+Da ) Tubular reactor ( v is variable ) −
V = FA0 ∫
second order
-rA = k CA2
CA = FA / v = FA / ( v0 ( 1+ X) ) CA = CA0 ( 1-X ) CA = CA0 ( 1-X ) / ( 1+ X )
TYSIR SARHAN
HAYTHAM ALZOUBI
50
Ch4 :- isothermal reactor design
Example 4/3 Determine the plug flow reactor volume necessary to t o produce 300 million pound of ethylene a year from cracking a feed stream of pure ethane . the reaction is irreversible and follows an elementary rate law . we want achieve 80 % conversion of ethane . operating the reactor isothermally isothermally at 1100 K at pressure of 6 atm Given k = 0.072 s -1 at 1000 K C2H6 → C2H4 + H2 A → B+ C FB = 300 * 10 6 lb / year 300 * 106 lb Year
1 Year 365 days
1 days 24 h
1h 3600 s
FB = 0.340 lb-mol / s FB = FA0 .X FA0 =0.34/0.8 = 0.424 lb-mol / s
TYSIR SARHAN
HAYTHAM ALZOUBI
1 lb-mol 28 lb
51
Ch4 :- isothermal reactor design −
V = FA0 ∫
-rA = k CA
CA = CA0 ( ( 1-X ) / ( 1+
)
= 1+1-1=1 yA0 = 1
= * yA0 = 1
1 Rankin = 1.8 Kelvin Temperature = 1980 R R = 0.73 Ft3.atm/lb-mol. R CA0 = yA0. P / RT 0 Pressure = 6 atm CA0 = 4.15 * 10 -3
E = 82000 cal R = 1.987 cal / mol.K T1 = 1000 K T2 = 1100 K k ( T2 ) = k ( T 1 ) exp (
( T1-1
T2-1 ) )
k = 3.07 s -1 V = 80.5 Ft 3
TYSIR SARHAN
HAYTHAM ALZOUBI
52
Ch4 :- isothermal reactor design
TYSIR SARHAN
HAYTHAM ALZOUBI
53
CH 5 : - COLLECTION AND ANALYSIS OF RATE DATA Kinetics of chemical reaction Chemical kinetics, also known as reaction kinetics, is the study of rates of chemical processes. Chemical kinetics includes investigations of how different experimental e xperimental conditions can influence the speed of a chemical reaction and yield information about the reaction's mechanism and transition states, as well as the construction of mathematical models that can describe the characteristics of a chemical reaction The rate of reaction is the change in the amount of reactants or products over a time interval .The rate of reaction in chemistry, is usually expressed in moles/second (mol/s) or molarities/second(mol/L∙s molarities/seco nd(mol/L∙s)) The speed of a chemical reaction is affected by factors such as the temperature, concentration ,volume, surface area, and orientation. When the temperature is greater, there is a greater fraction of particles that have more energy than the activation energy, enabling them to collide and react. These particles also have more kinetic energy. At the end of this chapter we can determine deter mine the reaction order and specific reaction rate from experimental date
Analysis of data to find rate law
1) di ffer entia ntial me method thod of analysi analysiss 2) i ntegra gr al method of analy lyssi s
TYSIR SARHAN
HAYTHAM ALZOUBI
54
CH 5 : - COLLECTION AND ANALYSIS OF RATE DATA Differential method A→ product -rA = ( -dC A /dt ) = k CAα
slope = ( ∆ CA /
∆ t
)
extrapolation Ln ( -rA ) = Ln k + α Ln C A Slope = α
intercept = Ln k
Example The liquid phase reaction A -rA ( mol / L.s ) CA ( M)
→ B
0.002 0.1
0.008 0.2
What is the general form of power rate lat for this reaction at constant temperature ? -rA1 = k CA1 α
-rA2 = k CA2 α
( 0.002 / 0.008 ) = ( 0.1/0.2) α
α = 2
-rA = k CA 2 What is the value of power rate constant 0.002( mol / L.s ) = k * ( 0.1 ) 2 ( mol/L)2 k= 0.2 L / mol .s
TYSIR SARHAN
HAYTHAM ALZOUBI
55
CH 5 : - COLLECTION AND ANALYSIS OF RATE DATA INTEGRAL METHOD To determine the reaction order ( liquid phase phase batch reactor )
Need to guess reaction order
Integrate the differential form of equation used to model the reactor used
If the right reaction order is assume the plot of concentration – concentration – time time data should be linear .
TYSIR SARHAN
HAYTHAM ALZOUBI
56
CH 5 : - COLLECTION AND ANALYSIS OF RATE DATA C :- intercept = CA0 m :- slope = -k
TYSIR SARHAN
HAYTHAM ALZOUBI
57
CH 5 : - COLLECTION AND ANALYSIS OF RATE DATA
Example A + B → C + D Time 0 50 100 150 200 (min) CA 0.0500 0.0380 0.00306 0.00256 0.00222 3 (mol/dm ) it was reported that the reaction is second order w.r.t.A. GIVEN CB0= 0.5 mol/dm 3 1.using the data confirm the reaction is second order 2.what is the value and unit of k ? 3.what is the value of C A0
TYSIR SARHAN
HAYTHAM ALZOUBI
250
300
0.00195
0.00174
58
CH 5 : - COLLECTION AND ANALYSIS OF RATE DATA 1. Time 0 (min) CA 0.0500 (mol/dm3) CA-1 20
50
100
150
200
250
300
0.0380
0.00306
0.00256
0.00222
0.00195
0.00174
26.32
326.8
390.63
450.45
512.82
574.71
2. k=slope = 0.1248 dm 3/mol.min 3.intercept = CA0-1 = 20.118
TYSIR SARHAN
CA0 = 0.0497 mol/dm 3
HAYTHAM ALZOUBI
59
CH 5 : - COLLECTION AND ANALYSIS OF RATE DATA The plot in terms of conversion At zero order X = ( k/C A0) *t Slope = ( k/C A0) at first order ln ( 1-(1-X) ) = k t slope = k at second order ( X / (X-1 )) = C A0 k t Slope = CA0 k
Half time Is defined as the time it takes for the concentration of reactant to t o fall to half of its initial value
-rA = kCAα
A → PRODUCT
Slope = 1-α 1-α
α≠1
half time TYSIR SARHAN
HAYTHAM ALZOUBI
60
CH 5 : - COLLECTION AND ANALYSIS OF RATE DATA Example The half period of a first order reaction is 50 min , in what time will it go to 90% completion . CA = 0.5 C A0
X = ( CA0 – C CA ) /CA0
X= 0.5
Time of of batch reactor at first order t = ( -1/k ) * ln ( 1-X ) X= 0.5
t = 50 min
t1/t2 = ln(1-X1)/ln(1-X2)
t 90%=166 min Example A first order reaction is 75 % completed in 72 min how will time will will it take for 1) 50 % completion
2) 87.5 % completion
Problem 5/10 The gas phase decomposition A
→ B
+ 2C is carried carried out out in a constant volume
batch reactor . Runs 1 through 5 were carried out at 100 carried out at 110 ℃
℃ .
while run 6 was
A ) from the data in the the table below , determine the reaction order and specific reaction rate constant run 1 2 3 4 5 6
CA0 0.025 0.0133 0.01 0.05 0.075 0.025
TYSIR SARHAN
Half life 4.1 7.7 9.8 1.96 1.3 2
HAYTHAM ALZOUBI
61
CH 5 : - COLLECTION AND ANALYSIS OF RATE DATA Run
Ln CA0
LN ( t1/2 )
1
-3.6888795
1.410987
2
-4.3199912
2.0412203
3
-4.6051702
2.282382
4
-2.9957323
0.67294447
5
-2.5902672
0.26236426
6
-3.6888795
0.693147
2.5
2
1.5
1
y = -1.0129x -1.0129x - 2.3529 2.3529
0.5
0 -5
-4
-3
-2
-1
ln CA0
Slope = 1-α 1- α = -1.012 Intercept = ln ( ( 2
α-1
α = 1+1.012 ≅ 2 -1) /( k ( α-1) ) = -2.352
k= 10.516 L/gmol.min L/gmol.mi n
at T = 100 ℃
B ) what is the activation energy for this reaction at T = 110
℃
CA0 = 0.025 from the table table
0.693147 = ln ( 1/ k(2-1)) + ( 1-2 ) ln 0.025 k= 20 L/gmol.min
TYSIR SARHAN
HAYTHAM ALZOUBI
0
62
CH 5 : - COLLECTION AND ANALYSIS OF RATE DATA k1= A e
( -E/RT1)
T1 = 373 K k1/k2 = e
k2= A e ( -E/RT2) T2=383 K
( -E/RT1)
+
R = 8.314 J/mol.K e (E/RT2)
E=76.53 kJ/mol A second second order order liquid reaction is 20% complete in 10 min . calculate calculate 1.the specific rate constant . 2.the time taken for the reaction go to half life. life . Given CA0 = 1 mol / L
TYSIR SARHAN
HAYTHAM ALZOUBI
63
CH 5 : - COLLECTION AND ANALYSIS OF RATE DATA
TYSIR SARHAN
HAYTHAM ALZOUBI
