Chemical Reaction Engineering
Lecture 2
General algorithm of Chemical Reaction Engineering
• Mole balance• Mole balance
• Rate laws• Rate laws
• Stoichiometry• Stoichiometry
• Energy balance• Energy balance
•Combine and Solve•Combine and Solve
Classification of reactions
• Phases involved:– Homogeneous reaction – reaction that occur in one phase – Heterogeneous reaction – reaction that involves more
than one phase and usually occurs at the interface between the phases (e.g. heterogeneous catalysis)
• Equilibrium position– Reversible reaction – reaction that can proceed in either
direction depending on the concentration of reagents and products
– Irreversible reaction – reaction that at given conditions can be assumed to proceed in one direction only (i.e. reaction equilibrium involves much smaller concentration of the reagents)
Elementary reactions• Kinetics of chemical reactions determined by the
elementary reaction steps.• Molecularity of an elementary reaction is the number
of molecules coming together to react in one reaction step (e.g. uni-molecular, bimolecular, termolecular)
• Probability of meeting 3 molecules is very small, so uni-molecular and bimolecular reaction are the only two to consider
Elementary reactions
Uni-molecular: first order in the reactant
[ ] [ ]d AA P k Adt
⎯⎯→ = −
Bimolecular: first order in the reactant
[ ] [ ][ ]d AA B P k A Bdt
+ ⎯⎯→ = −
Proportional to collision rate
2H Br HBr Br+ ⎯⎯→ +
! rate of disappearance of individual components can be calculated as:
1 i
i
dn dvdt dt
ξν
= =
Relative Rates of Reaction
• if we are interested in species A we can use A as the basis of calculation and define the reaction rate with respect to A
aA bB cC dD+ ⎯⎯→ +
b c dA B C Da a a
+ ⎯⎯→ +
• Reaction rate is not a constant, it depends- on concentration of the reagents- on the temperature- on the total pressure in the reactions involving gas phase- ionic strength and solvent in liquid state- presence of a catalyst
,...)],()][([ BAAA CCfnTkr =−
dr
cr
br
ar DCBA ==
−=
−
Rates of Reaction
• units of the reaction rate constant
,...)],()][([ BAAA CCfnTkr =−
timeionConcentratk
n−
=1)(][
algebraic function of concentrations
Arrhenius law/( ) E RTk T Ae−=
...⋅⋅ βαBA CC
• reaction order: ...++= βαn
• reaction rate is found experimentally, data on frequency factor (A), activation energy (E) and the order of the reaction can be found in relevant handbooks.
Non-Elementary Reaction rates
• The reaction rate dependence on the concentration and temperature will become more complicated when a reaction comprising several elementary steps is considered (incl. catalytic and reversible reactions)
Reversible reaction• Let’s consider a reaction of diphenyl formation
1
16 6 12 10 22 k
kC H C H H
−
⎯⎯→ +←⎯⎯
• rate of change for benzene
2
26 61 1
[ ] 2 2B B D Hd C H r k C k C C
dt −= = − +
B D
Equilibrium constant
2
26 61 1
[ ] 2 2B B D Hd C H r k C k C C
dt −= = − +
• noticing that the equilibrium constant: 1 1CK k k−=
in terms of concentration⎟⎟
⎠
⎞⎜⎜⎝
⎛−=−
c
HDBB K
CCCkr 22
12
• reaction rates in terms of other reagents:
⎟⎟⎠
⎞⎜⎜⎝
⎛−=
−=
c
HDB
BD
KCC
Ckrr 22121
Few more facts of K• From thermodynamics 0ln rRT K G=−Δ
• for gases, K can be defined in terms of pressures or concentrations
/ /
/
c a d aC D
p b aA B
p pKp p
=/ /
/ ( ) ; 1c a d aC D
C pb aA B
C C c d bK K RTC C a a a
δ δ−= = = + − −
• temperature dependence:
2
( )ln rH Td KdT RT
Δ=
0
1 21
( ) 1 1( ) ( )exp rp p
H TK T K TRT T T
⎡ ⎤⎛ ⎞Δ= −⎢ ⎥⎜ ⎟
⎝ ⎠⎣ ⎦
0
2
( ) ( )ln r R p RH T c T Td KdT RT
Δ +Δ −=
Reaction rate
226 61
[ ] 2 D HB B
C
C Cd C H r k Cdt K
⎡ ⎤= = − +⎢ ⎥
⎣ ⎦
Dependence on the concentration can be calculated knowing reaction mechanism, as before
Temperature dependence: Arrhenius equation
/( ) E RTk T Ae−=
• Below, we will consider only the concentration and temperature dependence of the reaction rate
Reaction rate• From the collision theory, only molecules
with the energy higher than the activation energy can react:
• The usual “rule of thumb that the reaction rate doubles with every 10ºC is not always true:
1 2
1 1/
1 2( ) ( )E R
T Tk T k T e⎛ ⎞
− ⋅ −⎜ ⎟⎝ ⎠=
Summary of the Reactor design equationsIn the previous lecture we found design equations for various reactors:
To solve them we need to find disappearence rate as a function of conversion:
)(XgrA =−
Relative Rates of Reaction
• if we are interested in species A we can define A as the basis of calculation
aA bB cC dD+ ⎯⎯→ +
b c dA B C Da a a
+ ⎯⎯→ +
• conversion: Moles of A reactedMoles of A fedAX =
• Let’s see how we can relate it to the reaction rate for various types of reactors.
Batch reactorb c dA B C Da a a
+ ⎯⎯→ +
• conversion:Moles of A reacted
Moles of A fedAX =
• number of moles A left after conversion:
( )0 0 0 1A A A AN N N X N X= − = −
• number of moles B left after conversion:
0 0B B AbN N N Xa
= −
( )0 1AAA
N XNCV V
−= =
( ) ( )( )00 0 A BB ABB
N b a XN b a N XNCV V V
Θ −−= = =
Batch reactors• For every component in the reactor we can write after
conversion X is achieved:
δ
δ - the total molar increase per mole of A reacted XNNN ATT ⋅⋅+= 00 δ
Batch reactor• Now, if we know the number of moles of every component
we can calculate concentration as a function of conversion. ( )
( )( )
( )( )
( )( )V
XadNVNC
VXacN
VNC
VXabN
VXNabN
VNC
VXN
VNC
DADD
CACC
BAABBB
AAA
+Θ==
+Θ==
−Θ=
−==
−==
0
0
000
0 1
• However, generally V can be a function of X as well...• In a constant volume reactor (e.g. batch reactor, liquid
reactor): ( )( )( ) etc. ,;1
0
0
XabCCXCC
BAB
AA
−Θ=−=
Flow reactors
• Equations for flow reactors are the same with number of moles N changed for flow rate F [mol/s].
Flow reactors
• Stoichiometric table for a flow system
Flow reactors
• For a flow system a concentration at any point can be obtained from molar flow rate F and volumetric flow rate
• For reaction in liquids, the volume change is negligible (if no phase change occurred):
( )0 1AA A
FC C Xv
= = − 0B A BbC C Xa
⎛ ⎞= Θ −⎜ ⎟⎝ ⎠
timelitertimemoles
vFC A
A ==
Reactions involving volume change• In a gas phase reaction a molar flow rate might
change as the reaction progresses
2 2 33 2N H NH⎯⎯→+ ←⎯⎯4 moles 2 moles
• In a gas phase reaction a molar flow rate might change as the reaction progresses
Batch reactor with variable volume• As such it would be a rare case (e.g. internal combustion
engine), but a good model case:
• If we divide the gas equation at any moment in time by the one at moment zero:
RTZNPV T=compressibility factor
⎟⎟⎠
⎞⎜⎜⎝
⎛⎟⎟⎠
⎞⎜⎜⎝
⎛⎟⎟⎠
⎞⎜⎜⎝
⎛⎟⎠⎞
⎜⎝⎛=
000
00
T
T
NN
ZZ
TT
PPVV
XXNN
NN
T
A
T
T εδ +=⋅⎟⎟⎠
⎞⎜⎜⎝
⎛+=⎟⎟
⎠
⎞⎜⎜⎝
⎛11
0
0
0
( )XTT
PPVV ε+⎟⎟
⎠
⎞⎜⎜⎝
⎛⎟⎠⎞
⎜⎝⎛= 1
0
00 function V=g(X)
Flow reactor with variable flow rate
TT
F PCv ZRT
= =
• Using the gas equation we can derive the total concentration as:
0 00
0 0
TT
F PCv Z RT
= =at the entrance:total concentration:
00
0 0 0
T
T
PF Z Tv vF P Z T
=total volume rate:
0 000 0
0 0 0 0
j jTj j T
T T
F F Z TPPF Z TC F v CF P Z Tv F P Z T
⎛ ⎞= = =⎜ ⎟
⎝ ⎠
=1
Flow reactor with variable flow rate• In a gas phase reaction a molar flow rate might
change as the reaction progresses
( )( )
0
00
0
1
A j jj
F XC P Tv X
P T
ν
ε
Θ +=
+
0T T AF F F Xδ= + ⋅ ⋅
00
0 0 0
T
T
PF Z Tv vF P Z T
= ( ) ( )0 00 0 0
0 0
1 1AP PT Tv v y X v XP T P T
δ ε= + = +
Example 3.6 (p.118)
• For reaction above calculate– equilibrium conversion in a constant batch reactor;– equilibrium conversion in a flow reactor;– assuming the reaction is elementary, express the rate of
the reaction– plot Levenspil plot and determine CSTR volume for 80%
conversion• assume the feed is pure N2O4 at 340K and 202.6kPa.
Concentration equilibrium constant: KC=0.1mol/l; kA=0.5 min-1.
2 4 22N O NO⎯⎯→←⎯⎯
Example1. Batch reactor
( )2 2 2 2
0 0
0
4 41 1
Be A e A eC
Ae A e e
C C X C XKC C X X
= = =− −
30 00
0
0.071mol/dm .AA
y PCRT
= =
0.44eX =
equilibrium conversion:
Example2. Flow reactor
( ) ( )( )2 2 2 2
0 0
0
4 41 1 1
Be A e A eC
Ae A e e e
C C X C XKC C X X Xε
= = =− − +
30 00
0
0.071mol/dm .AA
y PCRT
= =
( )( )
( )0 0
0
0
1 11 1
1
A AAA
AB
F X C XFCv v X XC XC
X
ε ε
ε
− −= = =
+ +
=+
0.51eX =
Example3. Rate laws 2
BA A A
C
Cr k CK
⎡ ⎤− = −⎢ ⎥
⎣ ⎦
– Constant volume ( )2 2
00
41 AA A A
C
C Xr k C XK
⎡ ⎤− = − −⎢ ⎥
⎣ ⎦
– Flow( )
( )
2 20 0
2
1 41 1A A
A AC
C X C Xr kX K Xε ε
⎡ ⎤−− = −⎢ ⎥
+ +⎢ ⎥⎣ ⎦
– Levenspiel plot
Example4. CSTR volume for X=0.4, feed of 3 mol/min
– Constant volume
( )2 2
00
41 AA A A
C
C Xr k C XK
⎡ ⎤− = − −⎢ ⎥
⎣ ⎦
– PFR – in the next lecture ☺
40.4
30
| 7 10
1714dm|
A X
A
A X
rF XVr
−=− = ⋅
= =−
Problems
• Class: P3-15• Home: P3-7, P3-13