3. Determination of Rates of
Reaction.
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3.1 Simple Reactor Types & Rate Equation
Ideal reactors have three ideal flow or contacting patterns. These three flow or reacting
are easy to treat (it is simple to find their performance equations) and one of them
often is the best pattern possible (it will give the most of whatever it is we want).
Suppose a single-phase reaction aA + bB → cC + dD. The most useful measure of
reaction rate for reactant A is then
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Experience shows that the rate of reaction is influenced by the composition and the
energy of the material. By energy we mean the temperature (random kinetic energy of
the molecules), the light intensity within the system (this may affect the bond energy
between atoms), the magnetic field intensity, etc. Ordinarily we only need to consider
the temperature, thus, the rate equation may be expressed as follows;
wmulewa@tum.ac.ke
3/19
3.2 CONSTANT Volume Batch Reactor
When we mention the constant-volume batch reactor we are really referring to the
volume of reaction mixture, and not the volume of reactor. Thus, this term actually
means a constant-density reaction system.
In a constant-volume system the measure of reaction rate of component i becomes
1 𝑑𝑑𝑁𝑁𝑖𝑖 𝑑𝑑 𝑁𝑁𝑖𝑖 ⁄𝑉𝑉
𝑑𝑑𝐶𝐶𝑖𝑖
𝑟𝑟𝑖𝑖 =
=
=
𝑉𝑉 𝑑𝑑𝑑𝑑
𝑑𝑑𝑑𝑑
𝑑𝑑𝑑𝑑
Or for ideal gases, where C = p/RT
1 𝑑𝑑𝑝𝑝𝑖𝑖
𝑟𝑟𝑖𝑖 =
𝑅𝑅𝑅𝑅 𝑑𝑑𝑑𝑑
Thus, the rate of reaction of any component is given by the rate of change of its
concentration or partial pressure. For gas reactions with changing numbers of moles, a
simple way of finding the reaction rate is to follow the change in total pressure 𝜋𝜋 of the
system.
This equation gives the concentration or partial pressure of reactant
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A as a function of the total pressure 𝜋𝜋 at time t, initial partial
pressure of A, 𝑝𝑝A0 , and initial total pressure of the system, 𝜋𝜋0 .
a is S. coefficient, Δn = sum of S. coefficients for products subtract S.
coefficients for reactants.
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3.3 Conversion
The fractional conversion, or the fraction of any reactant, say A, converted to
something else, or the fraction of A reacted away is simply called the conversion of A,
with symbol XA.
Suppose that NA0 is the initial amount of A in the reactor at time t = 0, and that NA is
the amount present at time t. Then the conversion of A in the constant volume system
is given by
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5/19
3.4 Integral Method of Analysis of Data
The integral method of analysis always puts a particular rate equation to the test by
integrating and comparing the predicted C versus t curve with the experimental C
versus t data. If the fit is unsatisfactory, another rate equation is guessed and tested.
3.4.1 Irreversible Unimolecular-Type First-Order Reactions
Consider the reaction A → Products;
𝑑𝑑𝐶𝐶𝐴𝐴
−𝑟𝑟𝐴𝐴 = −
= 𝑘𝑘𝐶𝐶𝐴𝐴
𝑑𝑑𝑑𝑑
For this reaction. Separating and integrating we obtain
𝐶𝐶𝐴𝐴
𝑡𝑡
𝑑𝑑𝐶𝐶𝐴𝐴
−�
= 𝑘𝑘 � 𝑑𝑑𝑑𝑑
𝐶𝐶𝐴𝐴𝐴 𝐶𝐶𝐴𝐴
0
Or
In terms of conversion,
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− ln
𝐶𝐶𝐴𝐴
�𝐶𝐶 = 𝑘𝑘𝑘𝑘
𝐴𝐴𝐴
𝑑𝑑𝑋𝑋𝐴𝐴�
𝑑𝑑𝑑𝑑 = 𝑘𝑘 1 − 𝑋𝑋𝐴𝐴
6/19
which on rearranging and integrating gives
𝑋𝑋𝐴𝐴
𝑡𝑡
𝑑𝑑𝑋𝑋𝐴𝐴
�
= 𝑘𝑘 � 𝑑𝑑𝑑𝑑
0 1 − 𝑋𝑋𝐴𝐴
0
Or
− ln 1 − 𝑋𝑋𝐴𝐴 = 𝑘𝑘𝑘𝑘
A plot of In (1 - XA) or In (CA/CA0) vs. t, gives a straight line through the origin for
this form of rate of equation i.e. 1st Order.
wmulewa@tum.ac.ke
7/19
3.4.2 Irreversible Second-Order Reactions
For the second-order equimolar reaction (with equal initial concentrations of A and B;
A + B → Products), or for the reaction (2A → Products), the defining second-order
differential equation becomes
Which on integration yields,
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8/19
3.4.3 Empirical Rate Equations of nth Order
When the mechanism of reaction is not known, we often attempt to fit the data with
an nth-order rate equation of the form
𝑑𝑑𝐶𝐶𝐴𝐴
−𝑟𝑟𝐴𝐴 = −
= 𝑘𝑘𝐶𝐶𝐴𝐴𝑛𝑛
𝑑𝑑𝑑𝑑
which on separation and integration yields
1−𝑛𝑛
𝐶𝐶𝐴𝐴1−𝑛𝑛 − 𝐶𝐶𝐴𝐴𝐴
= 𝑛𝑛 − 1 𝑘𝑘𝑘𝑘,
𝑛𝑛 ≠ 1
The order n cannot be found explicitly from this equation, so a trial-and-error solution
must be made. This is not too difficult, however. Just select a value for n and calculate
k. The value of n which minimizes the variation in k is the desired value of n.
3.4.4 Zero Order Reactions
A reaction is of zero order when the rate of reaction is independent of the concentration
of materials; thus
𝑑𝑑𝐶𝐶𝐴𝐴
−𝑟𝑟𝐴𝐴 = −
= 𝑘𝑘
𝑑𝑑𝑑𝑑
As a rule, reactions are of zero order only in certain concentration ranges-the higher
concentrations.
wmulewa@tum.ac.ke
9/19
If the concentration is lowered far enough, we usually find that the reaction becomes
concentration-dependent, in which case the order rises from zero.
Integrating the zero order rate equation and noting that CA can never become negative,
we obtain directly
𝐶𝐶
𝐶𝐶𝐴𝐴𝐴 − 𝐶𝐶𝐴𝐴 = 𝐶𝐶𝐴𝐴𝐴 𝑋𝑋𝐴𝐴 = 𝑘𝑘𝑘𝑘 for 𝑡𝑡 < 𝐴𝐴𝐴�𝑘𝑘
𝐶𝐶
𝐶𝐶𝐴𝐴 = 0 𝑓𝑓𝑓𝑓𝑓𝑓 𝑡𝑡 ≥ 𝐴𝐴𝐴�𝑘𝑘
Which means that the conversion is proportional to time.
wmulewa@tum.ac.ke
10/19
3.5 Overall Order of Irreversible Reactions from the Half-Life t1/2
For nth order reactions, the rate law is given by;
1−𝑛𝑛
𝐶𝐶𝐴𝐴1−𝑛𝑛 − 𝐶𝐶𝐴𝐴𝐴
= 𝑛𝑛 − 1 𝑘𝑘𝑘𝑘,
𝑛𝑛 ≠ 1
Defining the half-life of the reaction, t1/2, as the time needed for the concentration of
reactants to drop to one-half the original value, we obtain
0.5 1−𝑛𝑛 − 1 1−𝑛𝑛
𝑡𝑡1⁄2 =
𝐶𝐶𝐴𝐴𝐴
𝑘𝑘 𝑛𝑛 − 1
From this expression a plot of log t1/2 vs. log CA0 gives a straight line of slope 1 – n.
The half-life method requires making a series of runs, each
at a different initial concentration, and shows that the
fractional conversion in a given time rises with increased
concentration for orders greater than one, drops with
increased concentration for orders less than one, and is
independent of initial concentration for reactions of first
order.
wmulewa@tum.ac.ke
11/19
3.6 Fractional Life Method tF
The half-life method can be extended to any fractional life method in which the
concentration of reactant drops to any fractional value F = CA/CA0 in time tF. The
derivation is a direct extension of the half-life method giving
F 1−𝑛𝑛 − 1 1−𝑛𝑛
𝑡𝑡F =
𝐶𝐶𝐴𝐴𝐴
𝑘𝑘 𝑛𝑛 − 1
Thus, a plot of log tF versus log CA0 will give the reaction order.
a constant
F1−𝑛𝑛 − 1
log 𝑡𝑡F = log
+ 1 − 𝑛𝑛 log 𝐶𝐶𝐴𝐴𝐴
𝑘𝑘 𝑛𝑛 − 1
y
wmulewa@tum.ac.ke
c
mx
12/19
Example 1
Liquid A decomposes by first-order kinetics, and in a batch reactor 50% of A is
converted in a 5-minute experiment. How much longer would it take to reach 75%
conversion?
Example 2
After 8 minutes in a batch reactor, reactant (CA0 = 1 mol/liter) is 80% converted; after
18 minutes, conversion is 90%. Find a rate equation to represent this reaction.
Example 3
In a homogeneous isothermal liquid polymerization, 20% of the monomer disappears in
34 minutes for initial monomer concentration of 0.04 and also for 0.8 mol/liter. What
rate equation represents the disappearance of the monomer?
Example 4
Reactant A decomposes in a batch reactor; A → Products.
The composition of A in the reactor is measured at various times with results shown in
the following rows. Find a rate equation to represent the data.
Time t, s
Concentration CA, mol/liter
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0
10
20
8
40
6
60
5
120
3
180
2
300
1
13/19
3.7 Method of Initial Rates
The method of initial rates is a commonly used technique for deriving rate laws. As the
name implies, the method involves measuring the initial rate of a reaction. The
measurement is repeated for several sets of initial concentration conditions to see how
the reaction rate varies. This might be accomplished by determining the time needed to
exhaust a particular amount of a reactant.
The analysis of the data involves taking the ratios of rates measured where one of the
concentrations does not change.
Example
For the reaction A + B → Products, the following data were collected:
Experiment
[A] (mol/L)
[B] (mol/L)
Rate (mol/L∙s)
1
0.01
0.01
0.0347
2
0.02
0.01
0.0694
3
0.02
0.02
0.2776
For this data, determine the order of reaction with respect to species A and B, the
overall order of the reaction and the rate constant.
wmulewa@tum.ac.ke
14/19
3.8 Isolation Method
The isolation method is a technique for simplifying the rate law in order to determine
its dependence on the concentration of a single reactant. The concentrations of all the
reactants except one are made to be in large excess. Thus for example, if for the
reaction: aA + bB + cC → dD + …. with a rate law of the form;
−𝑟𝑟 = 𝑘𝑘 A
𝑎𝑎
B
𝑏𝑏
C
𝑐𝑐
we let B and C to be in large excess with concentrations [B0] and [C0], then it may be
assumed that their concentration is constant throughout the reaction. Thus we may
write:
−𝑟𝑟 = 𝑘𝑘 ′ A 𝑎𝑎
Where
𝑘𝑘 ′ = 𝑘𝑘 B 𝑏𝑏 C 𝑐𝑐
This means that if we take logs of both sides we get:
ln −𝑟𝑟 = ln 𝑘𝑘 ′ + 𝑎𝑎 ln A
wmulewa@tum.ac.ke
15/19
3.9 VARIABLE Volume (Constant Pressure) Batch Reactor
These reactors are much more complex than the simple constant-volume batch reactor.
The progress of the reaction is followed by noting the changes in volume.
This kind of reactor can be used for isothermal constant pressure operations, of
reactions having a single stoichiometry. For such systems the volume is linearly related
to the conversion, or
𝑉𝑉 − 𝑉𝑉0
𝑑𝑑𝑑𝑑
𝑉𝑉 = 𝑉𝑉0 1 + 𝜀𝜀𝐴𝐴 𝑋𝑋𝐴𝐴 or 𝑋𝑋𝐴𝐴 =
or 𝑑𝑑𝑉𝑉𝐴𝐴 =
𝑉𝑉0 𝜀𝜀𝐴𝐴
𝑉𝑉0 𝜀𝜀𝐴𝐴
where 𝜀𝜀𝐴𝐴 is the fractional change in volume of the system between no conversion and
complete conversion of reactant A. Thus
𝑉𝑉𝑋𝑋𝐴𝐴=1 − 𝑉𝑉𝑋𝑋𝐴𝐴=0
𝜀𝜀𝐴𝐴 =
𝑉𝑉𝑋𝑋𝐴𝐴=0
For example, consider the isothermal gas-phase reaction A → 4R. By starting with pure
reactant A, 𝜀𝜀𝐴𝐴 = (4 – 1)/1 = 3. But with 50% inert gas present at the start, two
volumes of reactant mixture yield, on complete conversion, five volumes of product
mixture. In this case 𝜀𝜀𝐴𝐴 = (5 – 2)/2 = 1.5.
Sometimes, 𝜀𝜀𝐴𝐴 is also referred to as the “expansion factor”.
wmulewa@tum.ac.ke
16/19
Noting that
Then
𝑁𝑁𝐴𝐴 = 𝑁𝑁𝐴𝐴𝐴 1 − 𝑋𝑋𝐴𝐴
𝑁𝑁𝐴𝐴
𝑁𝑁𝐴𝐴 1 − 𝑋𝑋𝐴𝐴
1 − 𝑋𝑋𝐴𝐴
𝐶𝐶𝐴𝐴 =
=
= 𝐶𝐶𝐴𝐴𝐴
𝑉𝑉
𝑉𝑉0 1 + 𝜀𝜀𝐴𝐴 𝑋𝑋𝐴𝐴
1 + 𝜀𝜀𝐴𝐴 𝑋𝑋𝐴𝐴
Thus
𝐶𝐶𝐴𝐴
1 − 𝑋𝑋𝐴𝐴
1 − 𝐶𝐶𝐴𝐴 ⁄𝐶𝐶𝐴𝐴𝐴
=
𝑜𝑜𝑜𝑜 𝑋𝑋𝐴𝐴 =
𝐶𝐶𝐴𝐴𝐴 1 + 𝜀𝜀𝐴𝐴 𝑋𝑋𝐴𝐴
1 + 𝜀𝜀𝐴𝐴 𝐶𝐶𝐴𝐴 ⁄𝐶𝐶𝐴𝐴𝐴
3.9.1 Zero Order (Variable Volume Batch Reactor)
For a homogeneous zero-order reaction the rate of
change of any reactant A is independent of the
concentration of materials, or
𝐶𝐶𝐴𝐴𝐴 𝑑𝑑 ln 𝑉𝑉
−𝑟𝑟𝐴𝐴 =
= 𝑘𝑘
𝜀𝜀𝐴𝐴
𝑑𝑑𝑑𝑑
Integrating gives
𝐶𝐶𝐴𝐴𝐴 𝑉𝑉
ln = 𝑘𝑘𝑘𝑘
𝜀𝜀𝐴𝐴 𝑉𝑉0
wmulewa@tum.ac.ke
17/19
3.9.2 First Order (Variable Volume Batch Reactor)
For a unimolecular-type first-order reaction the rate of change of reactant A is
𝐶𝐶𝐴𝐴𝐴 𝑑𝑑 ln 𝑉𝑉
1 − 𝑋𝑋𝐴𝐴
−𝑟𝑟𝐴𝐴 =
= 𝑘𝑘𝐶𝐶𝐴𝐴 = 𝑘𝑘𝐶𝐶𝐴𝐴𝐴
𝜀𝜀𝐴𝐴
𝑑𝑑𝑑𝑑
1 + 𝜀𝜀𝐴𝐴 𝑋𝑋𝐴𝐴
Replacing XA, by V and integrating gives
∆𝑉𝑉
− ln 1 −
= 𝑘𝑘𝑘𝑘,
∆𝑉𝑉 = 𝑉𝑉 − 𝑉𝑉0
𝜀𝜀𝐴𝐴 𝑉𝑉0
3.9.3 Second Order
For a bimolecular-type second-order reaction;
2A → Products or A + B → Products,
with CA0 = CB0 (equimolar), the rate is given by
2
𝐶𝐶𝐴𝐴𝐴 𝑑𝑑 ln 𝑉𝑉
1
−
𝑋𝑋
𝐴𝐴
2
−𝑟𝑟𝐴𝐴 =
= 𝑘𝑘𝐶𝐶𝐴𝐴2 = 𝑘𝑘𝐶𝐶𝐴𝐴𝐴
𝜀𝜀𝐴𝐴
𝑑𝑑𝑑𝑑
1 + 𝜀𝜀𝐴𝐴 𝑋𝑋𝐴𝐴
Integrating gives, after much algebraic manipulation
1 + 𝜀𝜀𝐴𝐴 ∆𝑉𝑉
∆𝑉𝑉
+ 𝜀𝜀𝐴𝐴 ln 1 −
= 𝑘𝑘𝐶𝐶𝐴𝐴𝐴 𝑡𝑡
𝑉𝑉0 𝜀𝜀𝐴𝐴 − ∆𝑉𝑉
𝑉𝑉0 𝜀𝜀𝐴𝐴
wmulewa@tum.ac.ke
18/19
Revision Question 1
Find the overall order of the irreversible reaction
2H2 + 2NO → N2 + 2H2 O
from the following constant-volume data using equimolar amounts of hydrogen and
nitric oxide:
Total Pressure, mmHg
200
240
280
320
360
Half-Life, sec
265
186
115
105
67
Revision Question 2
Aqueous A reacts to form R (A → R) and in the first minute in a batch reactor its
concentration drops from CA0 = 2.03 mol/L to CAf = 1.97 mol/L. Find the rate
equation for the reaction if the kinetics are second order with respect to A.
Revision Question 3
Find the first-order rate constant for the disappearance of A in the gas reaction
2A → R
if, on holding the pressure constant, the volume of the reaction mixture, starting with
80% A, decreases by 20% in 3 min.
wmulewa@tum.ac.ke
19/19