reaction of ethylene oxide (EO) with ammonia to form mono-,
di-and triethanolamine
1 eqn AA CkdtdC
1=
2 eqn - RAR CkCkdtdC
21=
3 eqn RS CkdtdC
2=
A R Sk1 k2
Initial conditions: CA0, CR0 = CS0 = 0
Integrating Eqn 1 gives
tkCC
A
A1
0
ln =
CA = CA0 exp(-k1t) Eqn 4
This equation is a first order differential equation of the
form:
QPydxdy
=+ with y = CR and Q = f(x)
Substituting into Eqn 2 gives
)exp( 1012 tkCkCkdtdC
ARR =+ Eqn 5
Boundary conditions: t=0, CR0 = 0
Use an integrating factor
)exp()exp( 2tkPdx :here
Since CA0 = CA + CR + CS by stoichiometry
Then CS = CA0 CA - CR
The solution is:
[ ]tktkAR eeCkkkC 210
12
1
= Eqn 6
+
+= tktkAS ekkke
kkkCC 21
12
1
21
20 1 Eqn 7
If k2 is much larger than k1 then Eqn 7 reduces to:
( ) 120 ,1 1 kkeCC tkAS >>=
The rate is determined by k1 i.e. the 1st step
If k1 is much larger than k2 then Eqn 7 reduces to:
( ) 210 ,1 2 kkeCC tkAS >>=
The rate is determined by k2 i.e. the 2nd step
In general, for any number of reactions in series it is the
slowest step that has the greatest influence on the overall
reaction rate:
Rate Determining Step
t
Ci
CR
CS CA
To find CRmax differentiate and set dCR/dt = 0
12
12max
)/ln(kkkkt
= Eqn 8
The maximum concentration of R is found by inserting Eqn into
the expression for CR (Eqn 6)
)/(
2
1
0
max,122 kkk
A
R
kk
CC
=
Eqn 9
t
Ci
CR
CS CA
A R Sk1 k2
tmax
Eqn 8
CRmax
Eqn 4Eqn 7
Eqn 6
Eqn 9
Lecture 8Slide Number 2Slide Number 3Slide Number 4Slide Number
5Slide Number 6Slide Number 7Slide Number 8Slide Number 9Slide
Number 10Slide Number 11Slide Number 12
