Reaction Rate equation

▪ A rate equation characterizes the rate of reaction

▪ Its form may either be suggested by theoretical considerations
or simply be the result of an empirical curve-fitting procedure.
▪ The value of the constants of the equation can only be found
by experiment; predictive methods are inadequate at present.
▪ The determination of the rate equation is usually a two-step
Interpretation of Batch Reactor Data procedure:
✓ first the concentration dependency is found at fixed
temperature
✓ then the temperature dependence of the rate constants is
found, yielding the complete rate equation
1 𝑑𝑁 /
𝑟 = = 𝒌𝐶 = 𝑘 𝑒 𝐶
𝑉 𝑑𝑡
3

AIMS OF CHAPTER 2 MONITORING THE BATCH REACTOR

Students are able to:

➢ Two types of equipment: batch and flow reactors.
➢Determine the reaction rate ➢ The batch reactor is simply a container to hold the contents
➢Examine the temperature effect on while they react. Monitoring the batch reactor:
reaction rate 1. By following the concentration of a given component.
2. By following the change in some physical property of the fluid,
➢Interpret the Batch Reactor Data such as the electrical conductivity or refractive index.
3. By following the change in total pressure of a constant-volume
system.
4. By following the change in volume of a constant-pressure
system.

2 4
 ANALYZING KINETIC DATA ANALYZING KINETIC DATA

❖ PART A: CONSTANT-VOLUME BATCH REACTOR

1. Integral methods constant-density reaction system
2. Differential methods
3. Half-life time
4. Intial reaction rate ❖ PART B: VARYING-VOLUME BATCH REACTOR

5 7

không thuận nghịch

Irreversible- bậc
1st order

ANALYZING KINETIC DATA

1, loại 1 phânA→P
reaction tử A→C

Irreversible
không – 2ndbậc
thuận nghịch
2order
loại 2reaction
phân tử

Integral methods Zero-Order

baäc Reactions
0
The integral method is easy to use and is recommended when
testing specific mechanisms, or relatively simple rate expressions, nthbậOrder
cn

or when the data are so scattered that we cannot reliably find the
Overall Order of
derivatives needed in the differential method. The integral method Irreversible Reactions from
the Half-Life
can only test this or that particular mechanism or rate form. Integral methods

CONSTANT-VOLUME
BATCH REACTOR
Irreversible Reactions in
Parallel
Differential methods Differential methods

More complicated situations but requires more accurate or Interpretation of kinetic

data Irreversible Reactions in
larger amounts of data; the differential method can be used to Series
Integral methods A→R→S
develop or build up a rate equation to fit the data. VARYING-VOLUME
BATCH REACTOR
First-Order Reversible
(ε0) Reactions :
In general, it is suggested that integral analysis be attempted first. Differential methods A ↔ R

6 8
PART A: CONSTANT-VOLUME BATCH REACTOR

p N N − a.x N a N − N
C = = = A0 = A0 −
RT V V V Δn V
❖ V = const or
1 dN i 1 d(C i V) 1 C i dV + V dC i dC i p = C RT = pA0 − (P − P ) (2.3)
ri = = = = (2.1)
V dt V dt V dt dt r
for R p = C RT = pR0 + (P − P ) (2.4)
• For ideal gases, where C = p / R T Δn
1 dpi where P0: initial total pressure of the system
ri = (2.2)
RT dt P: total pressure at time t
pA0: initial partial pressure of A
pR0: initial partial pressure of R
If the precise stoichiometry is not known, or if more than one stoichiometric equation is
needed to represent the reaction, then the "total pressure" procedure cannot be used

11

Analysis of Total Pressure Data Obtained in a

A.1. Integral Method of Analysis of Data
Constant-Volume System

General Procedure
aA+ bB +.. = rR + sS +.. 1) Assume the rate equation
dC A dC A
At t=0 NA0 NB0 +.. = NR0 + NS0 +.. Ntr − rA = − = kf(C) − = kdt
dt f(C A )
At t NA= NA0 -ax NB= NB0 -bx NR =NR0 + rx NS =NS0 + sx, Ntr CA t
dC A
2) Integrate −  = F (C A ) = k  dt = kt
Initially the total number of moles C A0
f(C A ) 0
N0= NA0+ NB0+ …+NR0+ NS0+ NA0+.. Ntr
At time t it is
3) Calculate F(CA) by experimental data
N = N0 + x (r+ s+ … – a –b – …) = N0+ x n
4) Plot F(CA) vs. t
where n = r+ s+ … – a –b – …
𝑁−𝑁
x= 5) If the fit is unsatisfactory, another rate equation is
∆𝑛
10 guessed and tested
 không thuận nghịch
Irreversible- bậc
1st order
1, loại 1 phânA→P
tử A→C
reaction
(1) Irreversible Unimolecular-Type First-Order Reactions A P
Irreversible
không – 2ndbậc
thuận nghịch
2order
loại 2reaction
phân tử

Zero-Order
baäc Reactions
0
A PRODUCTS dC A
− = k CA
nthbậOrder
cn dt
Overall Order of
Irreversible Reactions from
the Half-Life CA
Integral methods t
dC A
CONSTANT-VOLUME
BATCH REACTOR −  = k  dt
Differential methods
Irreversible Reactions in
Parallel
C A0
C A 0
Interpretation of kinetic
data Irreversible Reactions in
Series
A→R→S
Integral methods
CA
VARYING-VOLUME
BATCH REACTOR
First-Order Reversible
− ln = kt
(ε0) Reactions :
A ↔ R
C A0
Differential methods

13 15

PART A: CONSTANT-VOLUME BATCH REACTOR Conversion XA

constant-density reaction system N = NA𝑜 (1 − X )

Irreversible – 2nd order Empirical Rate
Irreversible- 1st order Equations of nth Order N N (1 − X ) 𝑪𝑨𝒐 − 𝑪𝑨
reaction
reaction
dC C = = A𝑜 = CA𝑜 (1 − X ); ⇒ 𝑿𝑨 =
A+B→P −r = − V V 𝑪𝑨𝒐
A→P dt dX
2A→P A + 2B→P = kC − dC = CA𝑜 dX ⇒ = k(1 − X )
dt
Overall Order of Irreversible Reactions
Irreversible Reactions in Parallel dC
Zero-Order Reactions from the Half-Life t1/2 − = kC
A → R, k1 dt
aA + bB → P A → S, k2

First-Order dX
Irreversible
Reversible Reactions ⇒ = k dt ⇒ − ( − 𝐀) =
Reactions in Series 1 − X
A R
A→R→S
14 16
 (1) Irreversible Unimolecular-Type First-Order Reactions (2) Irreversible Bimolecular-Type Second-Order
A P Reactions A + B→ Products
Caution. - dCA/dt = kCA 0,6.. CB0,4 are first order but are not enable to this kind
of analysis; hence, not all first- order reactions can be treated as shown above. dC dC
− = − = k C .C (2.12)
dt dt
the amounts of A and B that have reacted at any time t are equal and given by CAoXA

CAo X = CBo X

dX
CA𝑜 = k ( CA𝑜 − CA𝑜 . X )(CB𝑜 − CA𝑜 . X )
dt
( CA = CA𝑜 − CA𝑜 . X
CB = CB𝑜 − CB𝑜 . X = CB𝑜 − CA𝑜 . X )
𝐂 𝒐
letting =
𝐂 𝒐

17 20

(1) Irreversible Unimolecular-Type First-Order Reactions ❖ (2) Irreversible Bimolecular-Type Second-Order

A P Reactions A + B→ Products
dX
CA𝑜 = k ( CA𝑜 − CA𝑜 . X )(CB𝑜 − CA𝑜 . X
EXAMPLE A.1.1 Liquid A decomposes by first- dt
dX 𝐂
order kinetics, and in a batch reactor 50% of A is ⇒ CAo = k CA𝑜 ( 1 − X )( M − X ) =
𝐂
𝒐

dt 𝒐

converted in a 5-minute run. How much longer dX

would it take to reach 60% conversion? = kCA𝑜 dt
(1 − X )(M − X )

− 𝐁 − 𝐀 𝐂𝐁 . 𝐂𝐁
= = =
− 𝐀 ( − 𝐀) 𝐂 𝒐 𝐀 𝐀

= ( − ) = (𝐂 𝒐 −𝐂 𝒐) ≠

• If CB0 >> CA0 then CB remains approximately constant at all

times a pseudo first-order reaction
18
 (2) Irreversible Bimolecular-Type Second-Order Caution 2: A + 2B→ Products
Reactions A + B→ Products first order with respect to both A and B, hence second order overall

𝐂 𝒐
=
𝐂 𝒐
−𝑑𝐶
= 𝑘𝐶 𝐶 = 𝑘𝐶 1 − 𝑋 𝑀 − 2𝑋 ; 𝑴 ≠ 𝟐
𝑑𝑡

C 𝐶 𝑀 − 2𝑋
ln = ln = 𝐶 (𝑀 − 2)𝑘𝑡; 𝑴≠𝟐
CBo 𝐶 𝑀(1 − 𝑋 )

When a stoichiometric reactant ratio is used:

1 1 1 X
− = . = 2kt 𝑴=𝟐
C CAo CA𝑜 1 − X
Hình 2.3
22 24

Caution 1: Equal initial concentrations of A and B (M=1) (2) Irreversible Bimolecular-Type Second-Order
or: Reactions A + B→ Products
a) 2A → products EXAMPLE A.1.2 (3.22)
dC
− = kC = kC A𝑜 (1 − X ) (2.14𝑎) For the reaction A →R, second-order
dt
kinetics and CA0 = 1mol/liter, we get 50%
− = . = kt (2.14𝑏) conversion after 1 hour in a batch reactor.
Ao Ao 1 X
What will be the conversion and
concentration of A after 1 hour if CA0 = 10
2.14b mol/liter?
2.14b

23 25
 (3) Empirical Rate Equations of nth Order (4) Zero-Order Reactions
dC The rate of reaction is independent of the concentration of materials
− = kC
dt dC
− = k
dt
1 n = ( n − 1)kt, n ≠ 1
C 1 n − CA (2.21)
𝑜 − = CAo X = , t<CA0/k (2.24)
𝒐 𝐀
CA=0, t ≥ CA0/k
1 n
CA0 1 − X n − 1 = (n − 1)kt (2.22)

The order n cannot be found explicitly from Eq. 2.21, so a trial-and-error

solution must be made. 2.24
2.24
→ Select a value for n → calculate k
→ The value of n which minimizes the variation in k is the desired value of n.
2.24

27 31

(3) Empirical Rate Equations of nth Order (5) Overall Order of Irreversible Reactions from the Half-Life
t1/2

EXAMPLE A.1.3 (3.23) aA + bB → Products

dC
− = k C .C
For the decomposition A → R, CA0 = 1 dt
If the reactants are present in their stoichiometric ratios, they will remain at that
mol/liter, in a batch reactor conversion is ratio throughout the reaction. Thus, for reactants A and B at any timeCB/ CA = b/a
75% after 1hour, and is just complete after 2 dC b b ..
hours. Find a rate equation to represent −
dt
= k C .( C ) ...= k
a a
....𝐶

these kinetics. −
dC
= k C
dt

𝟐
⇒ = (𝟐. 𝟐𝟔) (n ≠ 1 )
𝐤 ( )

Plot (log t ½) vs (log CAo)→ slope: 1-n

29
(5) Overall Order of Irreversible Reactions from the Half-Life At least two components must be followed. we can find the
t1/2 concentration of the third component: CA + CR + CS = const
dC
− = k C + 𝑘 C = (𝑘 + 𝑘 ) C
𝟐 − dt
⇒ = (𝟐. 𝟐𝟔)
𝐤 ( − )
CA
− ln = ( k1 + k 2 ) t
Plot (log t ½) vs (log CAo)→ slope: 1-n C A0
Plot -ln(CA/CAo) vs t → k1+k2 Slope k1+k2
2.26
the fractional conversion in a given time
rises with increased concentration for rR dC R k1
orders greater than one, drops with = =
increased concentration for orders less rS dCS k 2
than one, and is independent of initial
concentration for reactions of first order.

CR − CR 0 k1
= Slope k1/k2
C S − C S0 k2
Plot CR vs Cs → k1/k2
35

(6) Irreversible reactions in parallel A → R, k1

A → S, k2
A → R, k1
A → S, k2
dC
−r = − = k C +k C = (k + k ) C
dt

dC
+r = + = k C
dt

dC
+r = + = k C
dt

Typical concentration-time curves for competing reactions

CRo=CSo=0, k1>k2
34 36
 (7) Irreversible reactions in series The time at which the maximum concentration of R occurs

A → R → S , k1 & k2 e k − e k
C = k CAo
k − k
No R or S present, initial concentration of A: CAo
C
𝑑𝐶
=0  k2 
ln 
− ln = k t 𝑑𝑡
dC A C
= − k1 C A
dt C =C e k 1  k1 
t max = =
dC
+ k C = k CA0 e k
k log tb k 2 − k1
dC R dt
→ The maximum concentration
= + k1 C A − k 2 C R dy/dx+Py=Q
dt of R:
k 2 / ( k 2 − k1 )
e k − e k
C R, max  k1 
dC S
= + k 2 CR
C = k CAo
k − k
=   Typical concentration-time curves

dt C A0  k2 
for consecutive first-order reactions.
37 2.42 39

CA0 = CA + CR + CS (7) Irreversible reactions in series

EXAMPLE A.1.7 (3.14)

k k
C = CAo 1 + e + e For the elementary reactions in series, find the
k − k k − k
maximum concentration of R and when it is reached.

k ⟩⟩ k C = CAo 1 − e
→ the rate is determined by k1, or the first step of the two-
step reaction.
k ⟩⟩ k C = CAo 1 − e A → R → S , k1 , k2

→ the rate is determined by k2, or the second step of the e k − e k

C = k CAo
two-step reaction. k − k

38 40
 (8) First-Order Reversible Reactions: A ↔ R
k1
k (8) First-Order Reversible Reaction
KC = K = equilibrium constant
2

dC R − dC A dX A EXAMPLE A.1.8 (3.9)

= = C A0 = k1C A − k 2 C R
dt dt dt The first-order reversible liquid reaction:
= k1 (C A0 − C A0 X A ) − k 2 (C R0 + C A0 X A )
At equilibrium dCA/dt = 0 takes place in a batch reactor. After 8 minutes, conversion of
C R0 A is 33.3% while equilibrium conversion is 66.7%.Find the
KC − rate equation for this reaction.
C A0 C Ao − C Ae C Re
 X Ae = = =
KC + 1 C Ao C Ao

C Re C Ro + C Ao X A e k
KC = = = 1
C Ae C Ao − C Ao X A e k2

41 43

không thuận nghịch bậc

(8) First-Order Reversible Reaction
Irreversible- 1st order
1, loại 1 phânA→P
reaction tử A→C

Irreversible
không – 2ndbậc
thuận nghịch
2order
loại 2reaction
phân tử

dX A
= (k 1 + k 2 )(X Ae − X A ) Zero-Order
baäc Reactions
0

dt
nthbậOrder
cn

 X   C − CA e   1  Overall Order of

 = − ln A  = k 1 1 +  t = (k 1 + k 2 )t
Irreversible Reactions from
− ln1 − A  C −C   K Integral methods the Half-Life

 X Ae   A0 Ae   C  CONSTANT-VOLUME
BATCH REACTOR

Plot – ln (1 – XA/XAe) vs t → straight line→ slope:

Irreversible Reactions in
Parallel
Differential methods

k1(1 + 1/KC). Interpretation of kinetic

data Irreversible Reactions in
Series

the irreversible reaction is simply the special case of Integral methods A→R→S
VARYING-VOLUME

the reversible reaction in which CAe=0 or XAe=1 or

BATCH REACTOR
First-Order Reversible
(ε0) Reactions :
Kc=∞ Differential methods A ↔ R

42 45
 A.2. Differential Method of Analysis of Data CONSTANT-VOLUME
BATCH REACTOR

Procedure:
p N N − a.x N a N − N
1. Plot the CA vs. t data, and then by eye carefully draw a C = = = A0 = A0 −
smooth curve to represent the data. This curve most likely will RT V V V Δn V
not pass through all the experimental points. or
2. Determine the slope of this curve at suitably selected p = C RT = pA0 − (P − P ) (2.3)
concentration values. These slopes dCA/dt = rA are the rates of
reaction at these compositions. r
for R p = C RT = pR0 + (P − P ) (2.4)
3. Search for a rate expression to represent this rAvs. CAdata Δn
where P0: initial total pressure of the system
P: total pressure at time t
pA0: initial partial pressure of A
pR0: initial partial pressure of R
If the precise stoichiometry is not known, or if more than one stoichiometric equation is
needed to represent the reaction, then the "total pressure" procedure cannot be used

46 48

không thuận nghịch

Irreversible- bậc
1st order
1, loại 1 phânA→P
reaction tử A→C CONSTANT-VOLUME
BATCH REACTOR
Irreversible
không – 2ndbậc
thuận nghịch
2order
loại 2reaction
phân tử

Zero-Order
baäc Reactions
0 Example A.1.9
nthbậOrder
cn
Consider the following gas phase reactions,
express the relationship between the total
Overall Order of
Irreversible Reactions from pressure of the system and partial pressure of the
the Half-Life
Integral methods
reacting materials:
CONSTANT-VOLUME
BATCH REACTOR
Irreversible Reactions in a. N2O4→2 NO2
Parallel
Differential methods
b. N2O →N2 + ½ O2
Interpretation of kinetic
data Irreversible Reactions in
Series
c. N2O5 →N2O4 + ½ O2 with 60 mole % N2O5 and
VARYING-VOLUME
Integral methods A→R→S
40 mole % inerts initially.
BATCH REACTOR
First-Order Reversible
(ε0) Reactions :
Differential methods A ↔ R

47 49
 không thuận nghịch bậc
1, loại 1 phân tử A→C

không thuận nghịch bậc

2 loại 2 phân tử

Example A.1.10 baäc 0

The thermal decomposition of dimethyl ether in gas phase

bậc n
was studied by measuring the increase in pressure in a
constant volume reactor. At 504 °C and initial pressure of khoâng thuaän nghòch baäc
312 mmHg, the following results are obtained: toång quaùt theo thôøi gian
baùn sinh t1/2
Integral methods
t (s) 390 777 1195 3155
CONSTANT-VOLUME
P, mmHg 408 488 562 779 731 BATCH REACTOR Irreversible- 1st order
songreaction
song khoâA→P
ng thuaän
nghòch
Find the rate equation to represent this decomposition Differential methods
Irreversible – 2nd
order reaction
assuming only ether was present initially ans Interpretation of kinetic
data noái tieáp khoâng thuaän
decomposition reaction is: nghòch
Zero-Order
A→R→S
Integral methods Reactions
(CH3)2O →CH4 + H2 +CO VARYING-VOLUME
BATCH REACTOR
thuaänth Order
(ε0) n nghòch baäc 1:
A ↔ R
Differential methods
…
51 54

Zero-Order Reactions
CONSTANT-VOLUME
BATCH REACTOR PART B: VARYING-VOLUME BATCH REACTOR

A reaction is of zero order when the rate of reaction is = 𝟏 + 𝜺𝐀 𝐀 → Volume of reaction system
𝟎
independent of the concentration of the reactant. 𝐕𝐗𝐀 − 𝐗𝐀
dC 𝜺𝐀 = eA is the fractional change in volume of the
−
dt
= k 𝐕𝐗 𝐀 system between no conversion and complete
conversion of reactant A.
𝒐 − 𝐀 = CAo X = , t<CA0/k (2.24)
CA=0, t ≥ CA0/k
Example A.1.11
EX: Considering isothermal gas phase reaction A 4R
The composition of gas A →2.7 R is zero order reaction, occurs in a constant volume by starting with pure A reactant:
reactor (following with the results in the table). The initial mixture contains 80 mole % 4 − 1
A and 20 mole % inerts. 𝜀 = =3
a. If the initial total pressure is 10 at, with pure A, find the toal pressure after 1 h. 1
b. If the initial partial pressure of A is 1 at, of inert is 9 at, find the total pressure after
1h. with 50% inerts present at the start:
t (h) 0 1
5 − 2
Total pressure (at) 1 1.5 𝜀 = = 1.5
2
53 55
PART B: VARYING-VOLUME BATCH REACTOR PART B: VARYING-VOLUME BATCH REACTOR

Consider a reaction A→2R

with 50 mole % A and 50 mole % inerts initially has a rate:
1 dN i 1 d(C i V) 1 VdCi + Ci dV
ri = = = -rA= k CA2
V dt V dt V dt For constant volume system: -rA= k CA2= k CAo2(1-XA)2
dCi C i dV 1.5 − 1
 ri = + For varying volume system: 𝜀 =
1
= 0.5
dt V dt
V = V0 (1 + e A X A ) -rA= k CA2=
CAo2(1−XA)2
( . )

VX A = 1 − VX A = 0 fractional change in
eA = volume of the system
VX A = 0
56 58

PART B: VARYING-VOLUME BATCH REACTOR PART B: VARYING-VOLUME BATCH REACTOR

C =
N N
= A0
1 − X
𝐶𝐴 = CAo
1 − X 1- Zero order: the rate of reaction is independent of
1 + 𝜀 X
V V 1 + 𝜀 X
the concentration of the reactant.
1 dN CA0 dX
−r = − = =𝑘
V dt 1 + 𝜀 X dt

𝐶 ∫ = ln( 1 + 𝜀 𝑋 ) = ln = 𝑘𝑡 (2.63)
𝑽 − 𝑽𝒐 ∆𝑽
1 dN 1 NAo d 1 − X = 𝟎 𝟏 + 𝜺𝑨 𝑨 ⇒ 𝑨 = =
𝑽 𝒐 𝜺𝑨 𝑽𝒐 𝜺𝑨
−r = − = −
V dt V 1 + 𝜀 X dt
(2.63)

CAo dX
=
1 + 𝜀 X dt (2.63)

57 59
 PART B: VARYING-VOLUME BATCH REACTOR PART B: VARYING-VOLUME BATCH REACTOR

2- Irreversible- 1st order reaction : 3- Irreversible – 2nd order reaction:

−1 dN A 2A → saã
P n phêím
= kC A
V dt A + B → Psaã
n phêím C A = CB
1 − X o o
𝐶𝐴 = CAo CA kC A (1 − X A )
1 + 𝜀 X o dX A = o Phûúng tr ònh vêå
n t öë
c phaã
n ûá
n g laâ
:
1 + e A X A dt 1 + eA XA
CA 2
o dX A  1 − XA 
Daå
n g t ñch phên laâ
: = kC 2A  
1 + e A X A dt o  1 + eA X A 
–ln (1 − X A ) = kt
𝑽−𝑽 𝒐 ∆𝑽
hay=biï𝟎 íu𝟏 di
+ ï𝜺𝑨în 𝑨t heo
⇒ 𝑨t hï
= í t ñch =
t öín g cöå
n g cuã
a hï å: XA 1 + eA XA (1 + e A ) X A
𝑽 𝒐 𝜺𝑨

𝑽𝒐 𝜺𝑨
V 
o (1 − X A )2
dX A =
1 − XA
+ e A ln (1 − X A ) = kC A t
o
− ln  1 −  = kt
 e V 
 A o
60 63

PART B: VARYING-VOLUME BATCH REACTOR PART B: VARYING-VOLUME BATCH REACTOR

Example B.1 (3.30) 4- nth order reaction :

Find the first-order rate constant for the disappearance of
n
A in the gas reaction A → 1.6R if the volume of the reaction 1 dN A  1 − XA 
mixture, starting with pure A increases by 50% in 4 min. − = kC nA = kC nA  
The total pressure within the system stays constant at 1.2 V dt o  1 + eA X A 
atm, and the temperature is 25°C.

XA (1 + e A X A )n −1
o (1 − X A )n
dX A = C nA−1 k t
o

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 PART C: TEMPERATURE AND REACTION RATE PART C: TEMPERATURE AND REACTION RATE

E 1
lnk = ln A −   Example C.2
Arrheùnius law RT
-E/R
ln k
The activation energy of a chemical reaction
k = k0 e- E / RT
is 17982 cal/mol in the absence of a
whereas:
1/T
catalyst, and 11980 cal/mol with a catalyst.
k0: frequency factor By how many times will the rate of the
E : activation energy, J/mol reaction will grow in the presence of a
R: 8.27 J/mol.K catalyst, if a reaction proceeds at 25°C?
T: K

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PART C: TEMPERATURE AND REACTION RATE

Example C.1
A common rule of temperature is that the rate
of the reaction doubles for each 10°C rise in
temperature. What activation energy would
this suggest at a temperature of 25 °C?

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