Chapter 13: Chemical Kinetics
Principles of Chemistry
A Molecular Approach,1st Ed.
Nivaldo Tro
Dr. Azra Ghumman
Memorial University of Newfoundland
Chemical Kinetics
13.2
13.3
The Rate of a Chemical Reaction
The rate Law: The effect of concentration on
reaction rate
Temperature and Rate: Collision and TransitionState Theories
Recation Mechanism (Excluding mechanisms
with a fast initial step) Arrhenius Equation
Catalysis
13.5
13.6
13.7
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The Rate of a Chemical Reaction
Reaction rate -The speed of a chemical reaction
A measure of how fast the reaction consumes the reactants
or makes products
Analogous to the speed of your car
» Units = (mi/hr)
Rate of a reaction can only be determined experimentally
Chemical kinetics- The study of the factors that affect reaction
rates and the mechanism by which a reaction proceeds
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Defining Reaction Rate
Rate of Reaction- is the increase in molar concentration of
product of a reaction per unit time or the decrease in molar
concentration of reactant per unit time
Units of rate = units of conc./unit of time(mol L-1.s-1)
The change in concentration of a reactant is negative
The negative sign makes the overall rate positive
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Reaction Rate
The reaction rate is a positive quantity that
expresses how the concentration of a reactant or
product changes with time
For a reaction
A
B
[B] increases to [B]max theoretical yield
[A] decreases to [A]min
The average rate of reaction = - [A]/ t = [B]/ t
Note that [A] is -ve since [A] is decreasing
The rate is equal to the rate of loss of A and the
rate of gain of B
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Factors Effecting the rate of a
Reaction
Concentration of reactants
Catalyst- A substance that increase the rate of reaction
without being consumed in the reaction
by decreasing the activation energy of the reaction
Homogeneous and heterogenous
Temperature at which the reaction occurs
Increase in temperature
increase in speed of molecules
increase in no.of effective collisions
increase in the
rate of reaction.
A good “rule of thumb” is that a reaction rate approximately
doubles with a 10oC rise in temperature
Surface area of a solid reactant or catalyst
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Average Rate
The average rate is the change in measured concentrations
within the given time interval t .
e.g.
2 N 2O 5 ( g )
4NO2 (g ) O 2 (g )
rate of formation of oxygen
= [O2] f-[O2] I / t f-t i
Rate of formation of oxygen
[O 2 ]
t
The larger the time interval, the more the average rate deviates
from the instantaneous rate
Instantaneous rate – rate at any instant or at t=0
Initial rate: Instantaneous rate at t =0 is called initial rate
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Average Rate
The average rate is equal to the slope of concentration vs
time graph for a given time interval
For any value of time interval the slope
will be equal to
[x]/ t = [x]f -[x]I /(t f -t i)
• Thus, the rate of a reaction
decreases as the reaction proceeds.
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H2
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I2
8
2 HI
Assuming a 1 L container, at 10 s, we used 0.181 moles of H2.
Therefore the amount of HI made is 2(0.181 moles) = 0.362 moles
Ave.rate decreases as the reaction progresses.
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Instantaneous Rate
The instantaneous rate is the change in
concentration at any one particular time.
slope at one point of a curve
determined by taking the slope of a line tangent to
the curve at that particular point
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Instantaneous Rate
H2(g) + I2(g)
2 HI(g)
Using [H2], the instantaneous
rate at 50 s is:
Using [HI], the
instantaneous rate
at 50 s is
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Sample Problem
Calculate the average rate of decomposition of N2O5, by the
reaction 2 N2O5 (g)
4NO2 (g) + O2 (g)
during the time interval from t= 600s to t = 1200s , use the
following data
Time (s)
600
1200
[N2O5] mol.L-1
1.24 x 10-2
0.93 x 10-2
Average rate of decomposition of N2O5: rate = -
(0.93 1.24) x10 2 mol.L 1
(1200 600)s
.
5.2x10 6 mol.L 1 s
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[ N2O5]/ t
0.31x10 2 mol.L 1
600s
1
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Reaction Rate and Stoichiometry
In most reactions, the coefficients of the balanced equation are
not all the same
H2(g) + I2(g)
2 HI(g)
The change in the number of moles of one substance is a
multiple of the change in the number of moles of another
For the above reaction, for every 1 mole of H2 used, 1 mole
of I2 will also be used and 2 moles of HI will be made.
Therefore, the rate of change will be different
To equate the rates the change in the concentration of each
substance is divided by its coefficient in balanced equation.
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Expressing Reaction rates
We can express the rate in terms of any species in the reaction
as the amounts of products and reactants are related by
stoichiometry
For a hypothetical reaction
aA + bB
cC + dD
The rate of reaction is defined as:
Rate
1
a
A
t
1
b
B
t
1 C
c t
1
d
D
t
Knowing the rate change in the concentration of any one
reactant or product at a point in time allows you to determine
the rate of change in concentration of any other reactant or
product
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Example 13.1
For the reaction given, the [I ] changes from 1.000 M
to 0.868 M in the first 10 s.
a. Write the rate equation for the reaction
b. Calculate the average rate in the first 10 s and
the rate of change in the concentration of [H+ ] during this time
interval ( [H+ ]/ t)
H2O2(aq) + 3 I (aq) + 2 H+ (aq)
I 3 (aq) + 2 H2O(l)
Solution: Write the rate equation in terms of reactants and
products. First calculate the rate and then rate of change in
[H+ ]
Rate
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H 2O2
t
1
3
I
t
1
2
H
t
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I3
t
1
2
H 2O
t
15
b. Use the rate equation to calculate the average rate
of the reaction
c.
Solve the Rate equation for the unknown value.
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Experimental Determination of
Reaction Rates
To obtain the rate of a reaction you must determine the
concentration of a reactant or product during the
course of the reaction.
A method for slow reactions is to withdraw samples
from the reaction vessel at various times and analyze
them.
More convenient techniques- that continuously
monitor the progress of a reaction by observing the
change in some physical property like colour or
absorption of light by some species or pressure of a
gas.
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13.3 The Rate Law: The effect of
Concentration on reaction rates
The Rate Law-An equation that relates the rate of a reaction to
the concentrations of reactants ( and catalyst) raised to various
powers.
The rate of a reaction is directly proportional to the
concentration of each reactant raised to a power.
For a general rxn:
aA + bB
products
n and m are called the orders for each reactant.
k is called the rate constant.
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Rate Constant k
Rate constant k
It has a fixed value at any given temperature, but
it varies with temperature
Units of k depends on the form of rate law
For the reaction 2NO2 (g) +F2(g) 2NO2F(g)
Rate law is
rate = k[NO2 ][F2 ]
and k = rate /[NO2 ][F2 ]
Units for rate constant k are:
k = mol L-1 .s-1 /(mol L- 1 ) 2 = L.mol- 1 .s-1
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Reaction order
Reaction order- The exponent of the concentration of a
species in the rate law . are determined experimentally
For the equation
Rate = k[A]n
n= order determines how the rate depends on the
concentration of the reactant
Zero order (n=0)-The rate is independent of concentration of
the reactant
Doubling [A] will have no effect on the reaction rate
Note that [A]0 =1
First order(n=1)- The rate is directly proportional to the
reactant concentration
Doubling [A] will double the rate of the reaction.
Second order(n=2)- the rate is directly proportional to the
square of the reactant concentration.
Doubling [A] will Chapter
quadruple
the rate of the reaction.
13: Chemical Kinetics
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Reactant Concentration vs. Time
A
Products
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Reaction Order
Negative orders are also possible.
When the rate decreases to half with doubling
concentration of the reactant, the reaction order is –1
T a b le 1 3 .1
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C o p y rig h t © H o u g h t o n M ifflin C o m p a n y . A ll rig h ts re s e rv e d .
P r e s e n t a t io n o f L e c t u r e O u t l in e s , 1 3 a – 8
22
Order of reaction
Overall order of a reaction equals the sum of the
orders of the reactants species in the rate law
For example the experimentally determined rate law
for the following reaction is
2 NO (g) 2H2 (g)
Rate
N2 (g) 2H2O(g)
k [NO] 2 [H 2 ]
the reaction is second order in NO, first order in H2,
overall order = 2+1=3
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Reaction Order
Consider the general reaction
aA + bB
dD + eE
Rate law for this reaction can be written as
Rate = k[A]m[B]n[C]p
a. The exponents m, n, and p are always determined
experimentally.
b. They are usually but not always integers.
c. They have no relationship with the coefficients in the balanced
equation ( a,b,d, and e).
By knowing the rate law and rate constant for a reaction you
can calculate the rate of reaction for any value of the
reactant concentration.
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Calculating rate constant k
The rate equation for the reaction of NO with ozone is
Rate = k[NO][O3]
If the rate is 6.60×10- 5 M/sec when [NO] is 1.00×10- 6 M and [O3]
is 3.00×10- 6 M, calculate the rate constant.
Solution:
Given: [NO] =1.00×10- 6 M,
[O3] = 3.00 × 10- 6 M,
Rate = 6.60 x10- 6 M/s
Find: k, rearrange the rate equation for the value of k
k = rate /[NO][O3] (mol L-1.s-1/mol2L-2)
=(mol L-1) –1s-1
Substitute the given values to calculate the value of k
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Finding the Order of a reaction—The
Initial Rate Method
The Initial Rate Method-for determining the order
of a reaction with respect to each reactant, involves
performing a series of experiment, each time varying
the initial concentration of only one reactant and
measuring its initial rate
The resulting change in rate indicates the order
with respect to that reactant
By changing the initial concentration of one reactant
at a time, the effect of each reactant’s concentration
on the rate can be determined
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Graphing Kinetic Data (Optional)
In addition to the method of initial rates, rate laws
can be deduced by graphical methods.
If we rewrite the first-order concentration-time
equation in a slightly different form, it can be
identified as the equation of a straight line.
This means if you plot ln[A] versus time, you
will get a straight line for a first-order
reaction.
Slope will give you the rate constant ‘k’
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Sample Problem
Iodide ion is oxidized in acidic solution to tri-iodide
ion, I3- , by hydrogen peroxide.
H2O2 (aq) 3I (aq) 2H (aq)
I 3 (aq) 2H2O(l)
A series of four experiments was run at different
concentrations, and the initial rates of I3- formation
were determined.
a. From the following initial rate data, obtain the
reaction orders with respect to H2O2, I-, and H+ .
b. Calculate the numerical value of the rate constant.
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Sample Problem
Initial Concentrations (mol/L)
Exp. 1
Exp. 2
Exp. 3
Exp. 4
H2O2
I-
H+
Initial Rate [mol/(L.s)]
0.010
0.020
0.010
0.010
0.010
0.010
0.020
0.010
0.00050
0.00050
0.00050
0.00100
1.15 x 10-6
2.30 x 10-6
2.30 x 10-6
1.15 x 10-6
•Comparing experiment 1 and 2, you see that when the
H2O2 concentration doubles (with other concentrations
constant), the rate doubles.
•This implies a first-order dependence with respect to
H2O2
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Sample Problem
Initial Concentrations (mol/L)
Exp. 1
Exp. 2
Exp. 3
Exp. 4
H2O2
I-
H+
Initial Rate [mol/(L.s)]
0.010
0.020
0.010
0.010
0.010
0.010
0.020
0.010
0.00050
0.00050
0.00050
0.00100
1.15 x 10-6
2.30 x 10-6
2.30 x 10-6
1.15 x 10-6
•Comparing experiment 1 and 3, you see that when the
I- concentration doubles (with other concentrations
constant), the rate doubles.
•This implies a first-order dependence with respect
to I-.
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Sample Problem
Initial Concentrations (mol/L)
Exp. 1
Exp. 2
Exp. 3
Exp. 4
H2O2
I-
H+
Initial Rate [mol/(L.s)]
0.010
0.020
0.010
0.010
0.010
0.010
0.020
0.010
0.00050
0.00050
0.00050
0.00100
1.15 x 10-6
2.30 x 10-6
2.30 x 10-6
1.15 x 10-6
•Comparing experiment 1 and 4, you see that when the
H+ concentration doubles (with other concentrations
constant), the rate is unchanged.
•This implies a zero-order dependence with respect
to H+.
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The experimental rate law is
Rate
k[ H 2O 2 ][I ]
The reaction orders with respect to
H2O2 = 1
I- = 1
and H+ = 0
Overall order of the reaction= 1+1 +0 = 2
You can now calculate the rate constant by
substituting values from any of the experiments.
Using Experiment 1 you obtain:
1 .15 10
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6
mol
L s
k 0 .010
mol
L
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0 .010
mol
L
33
k
1.15 10 6 s 1
0.010 0.010mol/L
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1.2
10
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2
L/(mol
s)
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