4.
1 Kinetics of Homogenous Simple Reaction in Batch Reactor
Variation of Concentration (Conversion) with Time
eg A B
t 0s t 10s t 20s t 30s t 40s t 50s t 60s
Case 1: Irreversible, Isothermal Reactions at Constant Volume
• Isothermal - the temperature of the reacting mixture is maintained constant
through out the course of the reaction.
• Constant volume - the volume of the reacting mixture within the system remains
constant through out the course of reaction.
Cont…
• To make the volume of the reaction mixture to remain constant, it
is necessary for the mass & density to remain constant
• V = V0 for constant volume
• V= V0(1+εXA) for variable volume
Cont…
Unimolecular reaction: an elementary reaction that involves
one reactant molecule
Bimolecular reaction: an elementary reaction that involves
two reactant molecules
Ter-molecular reaction: an elementary reaction that involves
three reactant molecules.
Unimolecular:
AP with -d[A]/dt = k[A]=kCA
Bimolecular:
2A P with -d[A]/dt = k[A]2=kCA2
or
A + B P with -d[A]/dt = k[A][B]=kCACB
(a) First-order reaction
• Consider Uni-molecular irreversible, isothermal and constant volume reaction
𝑨 𝑩, −𝑟𝐴 = 𝑘𝐶𝐴 𝑓𝑟𝑜𝑚 𝑟𝑎𝑡𝑒 𝑙𝑎𝑤
−𝑑𝐶𝐴
−𝑟𝐴 = 𝑓𝑟𝑜𝑚 𝑚𝑎𝑡𝑒𝑟𝑖𝑎𝑙 𝑏𝑎𝑙𝑎𝑛𝑐𝑒
𝑑𝑡
𝑑𝐶𝐴
− = 𝑘𝐶𝐴 𝐶𝑜𝑚𝑏𝑖𝑛𝑖𝑛𝑔
𝑑𝑡
• Ordinary differential equation
𝑑𝐶𝐴
= −𝑘𝑑𝑡
𝐶𝐴
CA = CA0 at t = 0
CA = CA at t = t
• Integrating from CA = CA0 at t = 0 to CA at t = t
𝐶𝐴 𝑡
𝑑𝐶𝐴
= −𝑘 𝑑𝑡
𝐶𝐴
𝐶𝐴0 𝑡=0
The solution of CA(t) for the first-order irreversible reaction is plotted in
Figure above .
For a first-order irreversible reaction CA decreases from CA0 at t = 0 to CA0/e
(0.368CA0) at t = l/k and to CA0/e2 (0.135CA0) at t = 2/k.
• Half-life (t1/2) = time needed to reduce a concentration to half
its value (CA0 to CA0/2).
• Therefore; t1/2 is constant in first order reactions.
• It is independent in initial concentration of reactants
Example 4-1
The reaction A B has k = 0.01 sec-1. For CA0 = 2.0 moles/liter,
what time is required for 90% conversion in a constant-volume batch
reactor? For 99%? For 99.9%? Calculate t1/2?
b) Second-order reaction
Case1. Consider Bi-molecular(the same molecules), irreversible, isothermal, constant
volume system
2𝐴 𝑃𝑟𝑜𝑑𝑢𝑐𝑡𝑠, −𝑟𝐴 = 𝑘𝐶𝐴 2
1 𝑑𝑛𝐴 𝑑𝐶𝐴
−𝑟𝐴 = − =− From rate law
𝑉 𝑑𝑡 𝑑𝑡
From material balance
𝑑𝐶𝐴
− = 𝑘𝐶𝐴 2 Constant volume
𝑑𝑡
Combination
𝑑𝐶𝐴
= −𝑘𝑑𝑡 ODE
𝐶𝐴 2
𝐶𝐴 𝑑𝐶𝐴 𝑡
𝐶𝐴0 𝐶𝐴 2
= −𝑘 0
𝑑𝑡 𝐶𝐴 = 𝐶𝐴0 𝑎𝑡 𝑡 = 0
Integration 𝐶𝐴 = 𝐶𝐴 𝑎𝑡 𝑡 = 𝑡
𝐶𝐴0
𝐶𝐴 =
1 + 𝐶𝐴0 𝑘𝑡
Initial & final points
1 1 1
𝑡= ( − )
𝑘 𝐶𝐴 𝐶𝐴0
• Half-life (t1/2) = time needed to reduce a concentration to half
its value (CA0 to CA0/2).
𝐶𝐴0
𝐶𝐴 =
1 + 𝐶𝐴0 𝑘𝑡
𝐶𝐴0
𝐶𝐴 = 2
𝐶𝐴0 𝐶𝐴0
=
2 1 + 𝐶𝐴0 𝑘𝑡1
2
1
𝑡1 =
2 𝑘𝐶𝐴0
• In second order reaction the half-life is not constant but gets
longer with decreasing initial concentration of reactants.
Case2. Consider Bi-molecular(different molecules), irreversible,
isothermal and constant volume system
A+B Products
𝑑𝐶𝐴 𝑑𝐶𝐵
− =− = 𝑘𝐶𝐴 𝐶𝐵
𝑑𝑡 𝑑𝑡
𝑑𝐶𝐴
− = 𝑘𝐶𝐴 𝐶𝐵
𝑑𝑡
𝑑𝑋𝐴
𝐶𝐴0 = 𝑘(𝐶𝐴0 − 𝐶𝐴0 𝑋𝐴 )(𝐶𝐵0 − 𝐶𝐴0 𝑋𝐴 )
𝑑𝑡
𝑑𝑋𝐴 2 𝐶𝐵0
𝐶𝐴0 = 𝑘𝐶𝐴0 1 − 𝑋𝐴 𝑀 − 𝑋𝐴 , 𝑀=
𝑑𝑡 𝐶𝐴0
𝑑𝑋𝐴
= 𝐶𝐴0 𝑘𝑑𝑡
(1 − 𝑋𝐴 )(𝑀 − 𝑋𝐴 )
𝑋𝐴 𝑑𝑋𝐴 𝑡
0 (1−𝑋𝐴 )(𝑀−𝑋𝐴 )
= 𝐶𝐴0 𝑘 0
𝑑𝑡
𝑀 − 𝑋𝐴
𝑙𝑛 = 𝐶𝐴0 𝑀 − 1 𝑘𝑡, 𝑀≠1
𝑀(1 − 𝑋𝐴 )
Example 4-2
The reaction A B has k = 0.01 litermol-1sec-1. For CA0 = 2.0
moles/liter, what time is required for 90% conversion in a constant-volume
batch reactor? For 99%? For 99.9%? Calculate t1/2?
Homework!!!
The reaction A + B C has k = 0.01 litermol-1sec-1. For CA0 = 2.0
moles/liter, CB0 = 4 moles/liter what time is required for 90% conversion of
A in a constant-volume batch reactor? For 99%? For 99.9%? Calculate t1/2?
iii. Zero-Order Reactions
Consider an irreversible, isothermal, constant volume, zero order reaction
A Products
𝑑𝐶𝐴
− =𝑘
𝑑𝑡
𝐶𝐴 𝑡
Integration 𝐶𝐴0
−𝑑𝐶𝐴 =𝑘 0
𝑑𝑡
𝐶𝐴0 − 𝐶𝐴 = 𝑘𝑡
𝐶𝐴 = 𝐶𝐴0 − 𝑘𝑡 In terms of concentration
From definition of conversion, 𝐶𝐴 = 𝐶𝐴0 − 𝐶𝐴0 𝑋𝐴
𝐶𝐴0 − 𝐶𝐴0 𝑋𝐴 = 𝐶𝐴0 − 𝑘𝑡
𝑘
𝑋𝐴 = 𝑡
𝐶𝐴0
Rate is independent of concentration. In terms of conversion
Linear decrease of concentration with time.
• Half-life (t1/2) = time needed to reduce a concentration to half its value
1
(𝐶𝐴0 to 𝐶 )
2 𝐴0
𝐶𝐴0
𝐶𝐴0 − 𝐶𝐴 = 𝑘𝑡 , 𝐶𝐴 =
2
1
𝐶𝐴0 − 𝐶𝐴0 = 𝑘𝑡1
2 2
𝐶𝐴0
𝑡1 =
2 2𝑘
• Half life is only dependent of initial concentration
Example 2.3
The reaction A B has k = 0.01 molliter-1sec-1. For CA0 = 2.0
moles/liter, what time is required for 90% conversion in a constant-volume
batch reactor? For 99%? For 99.9%? Calculate t1/2?
Case2: Irreversible, isothermal, variable volume reaction
For variable volume reaction
𝑉 = 𝑉0 (1 + 𝜀𝑋𝐴 )
Consider the isothermal gas-phase reaction
𝐴 4𝐵
Assume pure reactant A
∆𝜐 𝝊𝑃 − 𝝊𝑅 4 − 1
= = =3
𝝊𝐴 𝝊𝐴 1
𝑛𝐴0 𝑛𝐴0
𝑦𝐴0 = = =1
𝑛 𝑇0 𝑛𝐴0
∆𝝊
𝜀= 𝑦𝐴0 = 3 ∗ 1 = 3
𝝊𝐴
Cont.…
Assume 50% inert present at the start
∆𝜐 𝜐𝑃 −𝜐𝑅 4−1
= = =3
𝜐𝐴 𝜐𝐴 1
𝑛𝐴0 𝑛𝐴0 1
𝑦𝐴0 = = =
𝑛𝑇0 2𝑛𝐴0 2
∆𝝊 1
𝜀= 𝑦 = 3 ∗ = 1.5
𝝊𝐴 𝐴0 2
𝑛𝐴 = 𝑛𝐴0 (1 − 𝑋𝐴 )
𝑛𝐴
𝐶𝐴 =
𝑉
𝑛𝐴0 (1−𝑋𝐴 )
𝐶𝐴 =
𝑉0 (1+ε𝑋𝐴 )
Cont.
a) First order reaction
• Consider Uni-molecular, irreversible, isothermal, variable volume reaction
A P
1 𝑑𝑛𝐴 𝐶𝐴0 (1−𝑋𝐴 )
− = 𝑘𝐶𝐴 , 𝐶𝐴 = , V = 𝑉0 (1 + 𝜀𝑋𝐴 )
V 𝑑𝑡 1+𝜀𝑋𝐴
Substitute nA = nA0(1-XA) and differentiate it
𝑑𝑋𝐴
= 𝑘(1 − 𝑋𝐴 )
𝑑𝑡
𝑑𝑋𝐴
= 𝑘𝑑𝑡
(1 − 𝑋𝐴 )
Use integration
𝑋𝐴 = 1 − 𝑒 −𝑘𝑡
• Similar with constant volume system for first order
Cont.
b) Second order reaction
• Consider Bi-molecular, irreversible , isothermal, variable volume, second order reaction
2A B
1 𝑑𝑛𝐴 2
𝐶𝐴0 1 − 𝑋𝐴
− = 𝑘𝐶𝐴 , 𝐶𝐴 = , 𝑉 = 𝑉0 (1 + 𝜀𝑋𝐴 )
𝑉 𝑑𝑡 1 + 𝜀𝑋𝐴
𝑛𝐴0 𝑑𝑋𝐴 𝑘𝐶𝐴0 ^2 1 − 𝑋𝐴 ^2
=
𝑉0 1 + ε𝑋𝐴 𝑑𝑡 (1 + ε𝑋𝐴 )^2
(1 + ε𝑋𝐴 )
𝑑𝑋𝐴 = 𝑘𝐶𝐴0 𝑑𝑡
(1 − 𝑋𝐴 )^2
• After integration
1+ε
ε ln 1 − 𝑋𝐴 + − 1 − ε = 𝑘𝐶𝐴0 𝑡
1 − 𝑋𝐴
Cont.
c) Zero order reaction
• Consider an irreversible, isothermal, variable volume zero order reaction
A 3B
1 𝑑𝑛𝐴
− =𝑘
𝑉 𝑑𝑡
𝐶𝐴0 (1 − 𝑋𝐴 )
𝐶𝐴 =
1 + ε𝑋𝐴
𝑛𝐴0 𝑑𝑋𝐴
=𝑘
𝑉0 1 + ε𝑋𝐴 𝑑𝑡
𝑑𝑋𝐴 𝑘
= 𝑑𝑡
(1 + ε𝑋𝐴 ) 𝐶𝐴0
• After integration
1 ε𝑘 𝑡
𝑋𝐴 = (𝑒 𝐶𝐴0 − 1)
ε
Case3: Reversible, isothermal, constant volume reactions
Reversible Reaction(First-Order in both directions)
• Consider first-order, isothermal, reversible reaction in both directions and
constant volume reaction
• k1 and k−1 are forward- and reverse-rate constants respectively
• In terms of net rate of disappearance of A
𝑛𝑒𝑡 𝑟𝑎𝑡𝑒 𝑑𝑖𝑠𝑎𝑝𝑒𝑟𝑎𝑛𝑐𝑒 𝑜𝑓 𝐴 = 𝑟𝑎𝑡𝑒 𝑜𝑓 𝑑𝑖𝑠𝑎𝑝𝑒𝑟𝑎𝑛𝑐𝑒 𝑜𝑓 𝐴 𝑖𝑛 𝑓𝑜𝑟𝑤𝑎𝑟𝑑 − 𝑟𝑎𝑡𝑒 𝑜𝑓 𝑓𝑜𝑟𝑚𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝐴 𝑖𝑛 𝑟𝑒𝑣𝑒𝑟𝑠𝑒
−𝑟𝐴 = −𝑟𝐴,1 − 𝑟𝐴,−1
−𝑟𝐴,1 = 𝑘1 𝐶𝐴 , 𝑟𝐴,−1 = 𝑘−1 𝐶𝑃
−𝑟𝐴 = 𝑘1 𝐶𝐴 − 𝑘−1 𝐶𝑃
𝑑𝐶𝐴
− 𝑑𝑡
= 𝑘1 𝐶𝐴 − 𝑘−1 𝐶𝑃 From stoichiometric law
𝑑𝐶𝐴
= −𝑘1 𝐶𝐴 + 𝑘−1 𝐶𝑃 𝐶𝑃 = 𝐶𝑃0 + 𝐶𝐴0 − 𝐶𝐴
𝑑𝑡
𝐶𝑃 = 𝐶𝐴0 − 𝐶𝐴 if initially pure A (CP0=0)
• At equilibrium, −𝑟𝐴, 1 = 𝑟𝐴,−1 , 𝑠𝑜 𝑡𝑎𝑡 −𝑟𝐴 = 0
0 = k1 CA,eq − k −1 CP,eq
k1 CA,eq = k −1 CP,eq
k1 CP,eq
=
k −1 CA,eq
CA,eq
k −1 = k1
CP,eq
k1 CP,eq
K eq = =
k−1 CA,eq
• After integration
𝑘−1 −(𝑘1 +𝑘−1 )𝑡 𝑘−1
𝐶𝐴 = 𝐶𝐴0 − 𝐶𝐴0 + 𝐶𝑃0 𝑒 + (𝐶𝐴0 + 𝐶𝑃0 )
𝑘1 +𝑘−1 𝑘1 +𝑘−1
If we have pure A initially, CP0 = 0
𝑘−1 −(𝑘 +𝑘 )𝑡
𝑘−1
𝐶𝐴 = 𝐶𝐴0 1 − 𝑒 1 −1 + 𝐶𝐴0
𝑘1 + 𝑘−1 𝑘1 + 𝑘−1
4.2. Kinetics of Simple Reaction in Continuous Reactors
Variation of Conversion with Volume of plug flow reactor
Case1: Liquid Phase Irreversible Reactions
• Consider the first-order reaction,
A Products
• Rate equation −𝑟𝐴 = 𝑘𝐶𝐴
𝑑𝐹𝐴
• Material balance equation −𝑟𝐴 = −
𝑑𝑉
𝑑𝐹𝐴
• Combine both equations − = 𝑘𝐶𝐴
𝑑𝑉
𝐹𝑉 = 𝐹𝑉0 (liquid phase)
CA = 𝐶𝐴0 − 𝐶𝐴0 𝑋𝐴 (liquid phase)
𝐹𝐴 = 𝐹𝐴0 − 𝐹𝐴0 𝑋𝐴
Cont…
𝑑𝑋𝐴
Substituting 𝐹𝐴0 = 𝑘𝐶𝐴0 (1 − 𝑋𝐴 )
𝑑𝑉
𝐹𝐴0
𝐹𝐴0 = 𝐶𝐴0 𝐹𝑉0 𝑜𝑟 𝐶𝐴0 =
𝐹𝑉0
𝑑𝑋𝐴
𝐹𝑉0 = 𝑘(1 − 𝑋𝐴 )
𝑑𝑉
𝑉 𝑋𝐴 𝑑𝑋𝐴
0
𝑑𝑉 = 𝑘𝐹𝑉0 0
1−𝑋𝐴
𝑉 = 𝑘𝐹𝑉0 ∗ −ln(1 − 𝑋𝐴 )
𝑉
= 𝜃𝐹 = 𝑘 ∗ −ln(1 − 𝑋𝐴 )
𝐹𝑉0
• Space time (θF ) is the time necessary to process one reactor volume of fluid
based on entrance condition(or the time it takes for the fluid equal to volume
of reactor to enter the reactor completely).
• θF is obtained by dividing reactor volume by the volumetric flow rate entering
the reactor
Case2: Gas Phase Irreversible Reactions
Consider a first-order irreversible gas phase reaction
A 2P
• Combining rate law and material balance
𝑑𝐹𝐴
− = 𝑘𝐶𝐴
𝑑𝑉
• Express in terms of Conversion
𝑑𝑋𝐴 1−𝑋𝐴
𝐹𝐴0 = 𝑘𝐶𝐴0
𝑑𝑉 1+𝜀𝑋𝐴
1+𝜀𝑋𝐴
𝑑𝑉 = 𝑘𝐹𝑉0 ( )𝑑𝑋𝐴
1−𝑋𝐴
𝑋𝐴
1 + 𝜀𝑋𝐴
𝑉 = 𝑘𝐹𝑉0 𝑑𝑋𝐴
1 − 𝑋𝐴
0
𝑋𝐴
𝑉 1 + 𝜀𝑋𝐴
= 𝜃𝐹 = 𝑘 𝑑𝑋𝐴
𝐹𝑉0 1 − 𝑋𝐴
0
−𝑑𝐹𝐴 𝐹 𝑑𝑋
Material balance equation, −𝑟𝐴 = = 𝐴0 𝐴
𝑑𝑉 𝑑𝑉
𝐹 𝐹 (1−𝑋𝐴 )
Rate equation, −𝑟𝐴 = 𝑘𝐶𝐴 = 𝑘 𝐴 = 𝑘 𝐴0
𝐹𝑉 𝐹𝑉
𝐹𝐴0 𝑑𝑋𝐴 𝐹𝐴0 (1−𝑋𝐴 )
Combine them, =𝑘
𝑑𝑉 𝐹𝑉
𝐹 +𝐹 𝑋
From ideal gas law, 𝐹𝑉 = 𝑇0 𝐴0 𝐴 𝑅𝑇
𝑃
𝐹𝐴0 𝑑𝑋𝐴 𝑘𝐹 (1−𝑋 ) 𝑘𝐹𝐴0 1−𝑋𝐴 𝑃 𝑘𝑦𝐴0 1−𝑋𝐴 𝑃
= 𝐹𝑇0+𝐴0𝐹𝐴0𝑋𝐴𝐴 = 𝐹𝐴0 =
𝑑𝑉 𝑅𝑇 𝐹 1+ 𝑋𝐴 𝑅𝑇 1+𝑦𝐴0 𝑋𝐴 𝑅𝑇
𝑃 𝑇0 𝐹 𝑇0
𝑘𝑦𝐴0 𝑃 1+𝑦𝐴0 𝑋𝐴
𝑑𝑉 = 𝑑𝑋𝐴
𝐹𝐴0 𝑅𝑇 1−𝑋𝐴
𝑘𝑦𝐴0 𝑃 𝑋𝐴 1+𝑦𝐴0 𝑋𝐴
𝑉 = 0
𝑑𝑋𝐴 = − ln 1 − 𝑋𝐴 − 𝑦𝐴0 𝑙𝑛 1 − 𝑋𝐴 − 𝑦𝐴0 𝑋𝐴
𝐹𝐴0 𝑅𝑇 1−𝑋𝐴
𝐹𝐴0 𝑅𝑇 1
𝑉= 1 + 𝑦𝐴0 ln − 𝑦𝐴0 𝑋𝐴
𝑘𝑦𝐴0 𝑃 1−𝑋𝐴
𝐹𝑇0 𝑅𝑇 1 𝐹𝐴0
𝑉= 1 + 𝑦𝐴0 ln − 𝑦𝐴0 𝑋𝐴 𝑤𝑒𝑟𝑒 = 𝐹𝑇0
𝑘𝑃 1−𝑋𝐴 𝑦𝐴0
Cont…
• Residence time(𝑡)of an element of fluid leaving a reactor is the length of time
spent by that element within the reactor
𝑉
𝑡=
𝐹𝑉
• For variable volume system, 𝐹𝑉 = 𝐹𝑉0 (1 + 𝜀𝑋𝐴 )
𝑉
𝑡=
𝐹𝑉0 (1+𝜀𝑋𝐴 )
𝜃𝐹
𝑡=
1+𝜀𝑋𝐴
• For a constant volume system (𝜀 = 0), residence time is identical to space
time
𝐹𝑉 = 𝐹𝑉0
𝑉
𝑡= = 𝜃𝐹
𝐹𝑉0
Cont…
Variation of conversion with Volume of Perfect Mixed Flow Reactor
Case1: Liquid Phase Irreversible Reactions
• Consider the first-order reaction
A 2P
• Rate equation −𝑟𝐴 = 𝑘𝐶𝐴
𝐹𝐴0 −𝐹𝐴
• Material balance equation −𝑟𝐴 =
𝑉
𝐹𝐴0 −𝐹𝐴
• Combine them = 𝑘𝐶𝐴
𝑉
𝐹𝐴0 −𝐹𝐴0 (1−𝑋𝐴 )
= 𝑘𝐶𝐴0 (1 − 𝑋𝐴 )
𝑉
𝐹𝑉0 𝑋𝐴
𝑉=
𝑘 1 − 𝑋𝐴
𝑉 1 𝑋𝐴
= 𝜃𝐹 =
𝐹𝑉0 𝑘 1 − 𝑋𝐴
Cont..
Consider first order a gas phase reaction
A 2R
𝐹𝐴0 −𝐹𝐴
Combine rate and material balance equation, = 𝑘𝐶𝐴
𝑉
𝐹𝐴0 −𝐹𝐴0 (1−𝑋𝐴 ) 𝐶𝐴0 (1−𝑋𝐴 )
Express in terms of conversion, = 𝑘
𝑉 1+𝜀𝑋𝐴
𝑉 1 1 + 𝜀𝑋𝐴
= 𝜃𝐹 = 𝑋𝐴
𝐹𝑉0 𝑘 1 − 𝑋𝐴
Cont…
𝐹 𝐹𝐴0(1−𝑋𝐴)
Rate equation, −𝑟𝐴 = 𝑘𝑃 𝑃𝐴 = 𝑘𝑃 𝑦𝐴 𝑃 = 𝑘𝑃 𝐴 𝑃 = 𝑘𝑃 𝑃
𝐹𝑇 𝐹𝑇0 +𝐹𝐴0 𝑋𝐴0
𝐹𝐴0 (1−𝑋𝐴 ) 𝑦𝐴0 (1−𝑋𝐴 )
= 𝑘𝑃 𝑃 = 𝑘𝑃 𝑃
𝐹𝑇0 (1+𝑦𝐴0 𝑋𝐴 ) 1+𝑦𝐴0 𝑋𝐴
𝐹𝐴0 −𝐹𝐴 𝐹𝐴0 𝑋𝐴
Material balance equation, −𝑟𝐴 = =
𝑉 𝑉
𝑦𝐴0(1−𝑋𝐴 ) 𝐹𝐴0 𝑋𝐴
Combine them, 𝑘𝑃 𝑃=
1+𝑦𝐴 𝑋𝐴 𝑉
𝐹𝐴0 (1+𝑦𝐴0 𝑋𝐴 )𝑋𝐴
V=
𝑘𝑃 𝑦𝐴0 1−𝑋𝐴 𝑃
𝐹𝐴0 = 𝑦𝐴0 𝐹𝑇0
𝐹𝑇0 (1+𝑦𝐴0 𝑋𝐴0 )𝑋𝐴
𝑉=
𝐾𝑃 1−𝑋𝐴 𝑃
Summary
Kinetics of reaction
Variation XA with reaction time for specified order kinetics of Batch reactor
Variation of XA with size of the reactor for specified order kinetics of continuous
reactors(CSTR and PFR)
Design equation
The general design or performance equation for a reactor is developed by combining the
rate equation and the material balance equation
The performance equations for unspecified rate equation are:
1. Batch reactor
𝑋𝐴
𝑑𝑋𝐴
𝑡 = 𝑛𝐴0
𝑉(−𝑟𝐴 )
0
2. CSTR reactor
𝑉 𝑋𝐴
=
𝐹𝐴0 −𝑟𝐴
3. PFR reactor
𝑉 𝑋𝐴 𝑑𝑋𝐴
= 0 −𝑟𝐴
𝐹𝐴0
Example1: The rate constant for the reaction CH3COOC2H5 (aq) + OH− (aq) → CH3CO2
−(aq) + CH CH OH(aq) is 0.11 dm3 mol−1 s−1. What is the concentration of ester after (a) 10
3 2
s, (b) 10 min when ethyl acetate is added to sodium hydroxide so that the initial
concentrations are [NaOH] = 0.050 mol dm−3 and [CH3COOC2H5] = 0.100 mol dm−3?
Example2: For a given isothermal, variable volume reaction, A 4B with initial
concentration CA0 = 1 mole dm-3, CB0 = 0 and, the rate constant k = 0.25 (hr-1, mol dm-3 hr-1,
liter mole-1 hr-1). Find XA (t) in a batch reactor for these initial conditions at reaction times of
1, 2, 3 & 4 hr and, then find the concentrations CA (t) and CB (t).
Example 3: An aqueous feed of A and B (100 mol/liter, 200 mol/liter respectively) is to be
converted . The kinetics of the reaction is represented by
𝐦𝐨𝐥
𝐀+𝐁 𝐑 , −𝐫𝐀 = 𝟐𝟎𝟎𝐂𝐀 𝐂𝐁 [ ]
𝐥𝐢𝐭𝐞𝐫.𝐦𝐢𝐧
a. Find the time needed to 99% conversion of A to product in a batch reactor?
b. Find the size of reactor required to achieve 99% conversion of A to product in mixed flow
reactor if the feed rate is 25 liter/min.?
c. Find the size of reactor required to achieve 99% conversion of A to product in plug flow
reactor if the feed rate is 25 liter/min.?
