Chapter 16
Rate of reactions / Kinetics: Rates and Mechanisms
of Chemical Reactions
16.1 Meaning of Reaction Rate
16.2 Reaction Rate and Concentration
Lecture 1
16.3 Reactant Concentration and Time
16.4 Models for Reaction Rate
Lecture 2
16.5 Reaction Rate and Temperature
Lecture 3
16.6 Catalyst
16.7 Reaction Mechanisms
Lecture 4
Highlights: Rates and Mechanisms of Chemical Reactions
Expression of the Reaction Rate: average, instantaneous and initial reactions rates
Factors That Influence Reaction Rates: concentration, physical state, temperature, use
of catalyst
The Rate Law and Its Components: initial rate, reaction order terminology, reaction
orders
Integrated Rate Laws: Concentration Changes over Time
First, second, and zero-order reactions, reaction order determination
The Effect of Temperature on Reaction Rate: Arrhenius equation, activation energy
Explanation of the Effects of Concentration and Temperature: collision theory temperature effect, molecular structure effect
Catalyst: Speeding Up a Chemical Reaction: reaction energy, homogeneous, and
heterogeneous catalyst
Reaction Mechanisms: Steps in the Overall Reaction: elementary reactions &
molecularity, rate-determining step, mechanism correlation with the rate law
Meaning of the reaction rate
Reaction: aA + bB
cC + dD
This is a balanced equation.
If the reactants mix, the products will form.
Definition of reaction rate or chemical kinetics:
1. Is the study of a reaction and the reaction rate is a positive quantity.
2. Is the change in concentration for reactants as a function of time.
3. Is the change in concentration for products as a function of time.
4. Therefore, the reaction rate is the central focus of the chemical kinetics.
0.18
0.16
N2O 5 concentration
0.14
0.12
0.1
0.08
0.06
0.04
0.02
0
0
10
20
30
time
40
50
mins
Chemical reaction of N 2O5 decomposition
60
Meaning of the reaction rate
aA + bB
cC + dD balanced equation
1 [ A]
1 [ B]
1 [C ]
1 [ D]



Rate = 
a t
b t
c t
d t
1. The reactant concentration decreases while the
product concentration increases.
2. The change in reactant concentration is always
negative,  0.
3. The change in product concentration is always
positive,  0.
4. [ ] square bracket expresses the concentration in
moles per liter (mol/L = M).
Expressing the reaction rate
cC + dD
1. Average reaction rate
The change in concentration
of a reactant/product over a
change in time [A] / t.
2. Instantaneous reaction rate
at a certain time, the slope of
a tangent to [A] vs time
a balanced equation
0.18
Point 1
0.16
0.14
N2O 5 concentration
aA + bB
0.12
Point 3
0.1
0.08
Point 2
0.06
0.04
0.02
3. Initial reaction rate
t = 0, the reactants just mix,
no products accumulate.
0
0
10
20
30
time
40
50
mins
Chemical reaction of N2O5 decomposition
60
Measurement of reaction rate
Average reaction rate
O3 (g) + C2H4 (g)  C2H4O (g) + O2 (g)
Within 10 sec
Time
sec
[O3]
M
0
0.165
10
0.124
20
0.093
30
0.071
Between 50 and 60 secs
40
0.053
Rate3 = -(0.029-0.039)/60-50) = 1.010-3 (M/sec)
50
0.039
60
0.029
During the reaction course, the average rate decreases
since the reactants were used up and there were fewer
of them present.
Rate1 = -(0.124-0.165)/(10-0) = 4.110-3 (M/sec)
Between 40 and 50 secs
Rate2 = -(0.039-0.053)/(50-40) = 1.4 10-3 (M/sec)
Instantaneous reaction rate
O3 (g) + C2H4 (g)  C2H4O (g) + O2 (g)
O3 concentration
0.18
0.16
[O3]1
0.14
0.128 M
If we choose the time interval
infinitely small, the average rate
becomes the instantaneous rate.
0.0600 M
[O3]2
The slope of line A is the
instantaneous rate for 20 sec.
33 sec
Slope = - ([O3]2-[O3]1)/(t2-t1) =
0.12
0.1
0.08
0.06
0.04
6 sec
0.02
- (0.0600-0.128)/(33-6) =
0
0
10
20
30
40
time mins
Chemical reaction of O3 decomposition
50
60
2.7210-3 (M/s)
Initial reaction rate
O3 (g) + C2H4 (g)  C2H4O (g) + O2 (g)
0.18
0.165 M
0.16
time = 0 sec, instantaneous
reaction rate is called the initial
reaction rate.
O3 concentration
0.14
0.12
0.100 M
0.1
0.08
0.06
Slope = -([O3]2-[O3]1)/(t2-t1)
0.04
0.02
=-(0.100-0.165)/(12-0)
12 sec
0
0
10
20
30
40
time mins
Chemical reaction of O3 decomposition
50
60
=5.4110-3 (M/s)
Four factors which influence on the reaction rate
-- Each reaction has its own characteristic rate.
-- The reaction rate is determined by the chemical
nature of the reactants.
-- Four factors will influence the reaction rate.
1. Concentration of reactants
2. Physical state of reactants
3. Temperature at which the reaction occurs
4. The Use of catalyst
A. Concentration influence
NOx
O3
1. Reactants crash into each other.
2. Reactants collide the vessel
walls.
3. Only the reaction happens
when the O3 and NOx
molecules collide.
4. The more molecules present,
the more frequently the collide,
the more often reaction occurs.
• The reaction rate is proportional to the reactants concentration.
• The energetic collision leads to the reaction.
• Rate  collision frequency  concentration of the reactants
B. Physical state influence
1. Reactant molecules must mix to collide.
2. Collision frequency between molecules depends on the state of reactants.
3. In aqueous solution, the thermal motion brings the reactants contact.
4. Heterogeneous phase– the contact mainly occurs at the interface.
5. Agitation and grinding will favor the reaction.
6. Enough surface area is needed for solid chemicals to react
It is safe to say:
the finer the solid/liquid reactant is,
the greater the surface is,
the more contact reactants make,
the faster the reaction occurs.
C. Temperature influence
1. Energetic collision (effective collision) results in the reaction.
2. Enough energy is required for the effective collision.
3. Temperature is the major effect on the energy.
Temperature affects the kinetic energy of a molecule.
Thus, temperature influences the collision.
It is safe to say:
At a higher temperature, more collision occurs.
Rising temperature will increase the number of
collisions and the energy of collisions.
So rising temperature enhance the reaction rate.
Rate  collision energy  temperature
Reaction rate and concentration
Rate expression and rate constant
Order of reaction
For a balanced reaction: aA + bB  cC + dD
The rate law (rate expression) indicates:
Rate = k [A]m [B]n
1. In the rate law, we assume the product does not appear.
2. The proportionality k is the rate constant.
3. k is specific for a given reaction at a certain temperature.
4. k is NOT influenced by proceed but temperature.
5. m, n is the individual reaction order with respect to the A and B reactants.
6. m, n is NOT necessarily related to the a, and b.
7. The exponent m and n determines how the concentration influences on the
rate.
Three components for the rate law:
Rate = k [A]m [B]n
Reaction rate, reaction orders, and rate constant.
The order of a reaction involving a single reactant:
Zero order :
m=0
First order :
m=1
Rate = k [A]m
Second order : m = 2
The order of a reaction involving a multi-reactants: Rate = k [A]m [B]n
Zero order : m + n = 0
First order : m + n = 1
Overall reaction order: m + n
Second order : m + n = 2
We use the experimental approach to determine the component of the rate law:
1. Using concentration measurement to find initial rate.
2. Using initial rate to find the reaction order.
3.
Using initial rate and reaction orders to calculate the rate constant.
Determine reaction order from the rate law
Example:
NO (g) + O3 (g)  NO2 (g) + O2 (g)
Rate = k [NO]m[O3]n
• The individual reaction order in NO : m
• The individual reaction order in O3 : n
• The overall reaction order :
m+n
From the experiment, the rate law :
Rate = k [NO][O3].
Q: Determine the individual order in NO and O3, as well the over all order.
The individual order in NO : 1
The individual order in O3 : 1
The overall reaction order : 1+1 = 2
Determining Reaction Order from Rate Laws
PROBLEM:
For each of the following reactions, determine the reaction order with
respect to each reactant and the overall order from the given rate law.
(a) 2NO(g) + O2(g)
(b) CH3CHO(g)
2NO2(g); rate = k[NO]2[O2]
CH4(g) + CO(g); rate = k[CH3CHO]3/2
(c) H2O2(aq) + 3I-(aq) + 2H+(aq)
I3-(aq) + 2H2O(l); rate = k[H2O2][I-]
PLAN: Look at the rate law and not the coefficients of the chemical reaction.
SOLUTION:
(a) The reaction is 2nd order in NO, 1st order in O2, and 3rd order overall.
(b) The reaction is 3/2 order in CH3CHO and 3/2 order overall.
(c) The reaction is 1st order in H2O2, 1st order in I- and zero order in H+,
while being 2nd order overall.
Determining Reaction Orders from the experimental data
• Write the rate law for each experiment: rate = k [O2]n[NO]m
• Take the ratio of rate law for expt 1 and 2 to determine m.
• Take the ratio of rate law for expt 1 and 3 to determine n.
• Sum of n and m is the overall reaction order m + n.
• Use the rate law for one of the expts to determine k.
Experiment
Initial Reactant Concentrations
(mol/L)
O2
NO
Initial Rate
(mol/L*s)
1
1.10  10-2
1.30  10-2
3.21  10-3
2
2.20  10-2
1.30  10-2
6.40  10-3
3
1.10  10-2
2.60  10-2
12.8  10-3
Determining Reaction Orders using initial rates
O2(g) + 2NO(g)
rate = k [O2]m[NO]n
2NO2(g)
Initial Reactant Conc. (M)
O2
NO
Experiment
Initial Rate (mol/L*s)
1
1.10  10-2
1.30  10-2
3.21  10-3
2
2.20  10-2
1.30  10-2
6.40  10-3
3
1.10  10-2
2.60  10-2
12.8  10-3
Compare 2 experiments in which the concentration of one reactant varies and the concentration
of the other reactant(s) remains constant.
k [O2]2m[NO]2n
rate2
=
rate1
6.40x10-3mol/L*s
3.21x10-3mol/L*s
=
k [O2]1m[NO]1n
[O2]2m
[O2]1m
2.20x10-2mol/L
=
1.10x10-2mol/L
=
[O2]2
m
[O2]1
m
Do a similar calculation
for the reactant, NO.
n = 2.
Overall reaction order:
; 2 = 2m
m=1
m+n =1+2 = 3
Use any expt result to determin k : k = initial rate/([O2][NO]2 ) = 1.73 103 L2/(mol2  s)
Unit for k depends on the reaction order.
Reactant Concentration and Time
Integrated rate law:
-- Expresses the concentration change for reactants with time.
Rate = - [A]/ t
Rate = k [A]m
From the rate law:
The expression for rate equals.
[A]
-
t
= k [A]m
By using calculus, the equation [1] can be integrated.
[1]
Integrated Rate Laws:
[A]t is the
concentration at any
time
express the concentration change with time
[A]
rate = -
t
= k [A]
ln
[A]0 is the initial
concentration
first order rate equation
[A]0
=-kt
[A]t
ln [A]t = ln [A]0 -k t
[A]
rate = -
t
= k [A]2
1
[A]t
[A]
rate = -
t
= k [A]0
second order rate equation
-
1
[A]0
=kt
1
1
=kt +
[A]t
[A]0
zero order rate equation- rate will not
change with the concentration
[A]t - [A]0 = - k t
Important information
Units of the Rate Constant k for Several Overall
Reaction Orders
Overall Reaction Order
Units of k (t in seconds)
0
mol/L*s (or mol L-1 s-1)
1
1/s (or s-1)
2
L/mol*s (or L mol -1 s-1)
3
L2 / mol2 *s (or L2 mol-2 s-1)
General formula:
Unit of k =
(L/mol)order-1
Unit of time
Integrated rate laws and reaction order
ln[A]t = -kt + ln[A]0
1/[A]t = kt + 1/[A]0
[A]t = -kt + [A]0
Important information
An Overview of Zero-Order, First-Order, and
Simple Second-Order Reactions
Zero Order
First Order
Second Order
Rate law
rate = k
rate = k [A]
rate = k [A]2
Units for k
mol/L*s
1/s
L/mol*s
Integrated rate law in
straight-line form
[A]t =
k t + [A]0
ln[A]t =
-k t + ln[A]0
1/[A]t =
k t + 1/[A]0
Plot for straight line
[A]t vs. t
ln[A]t vs. t
1/[A]t = t
Slope, y-intercept
k, [A]0
-k, ln[A]0
k, 1/[A]0
Half-life
[A]0/2k
ln 2/k
1/k [A]0
Determining Reaction Concentration at a Given Time
PROBLEM: At 10000C, cyclobutane (C4H8) decomposes in a first-order reaction,
with the very high rate constant of 87 s-1, to two molecules of ethylene
(C2H4).
(a) If the initial C4H8 concentration is 2.00M, what is the concentration
after 0.010 s?
(b) What fraction of C4H8 has decomposed in this time?
PLAN:
Find the [C4H8] at time, t, using the integrated rate law for a 1st order
reaction. Once that value is found, divide the amount decomposed by
the initial concentration.
SOLUTION:
ln
(a)
[C4H8]0
[C4H8]t
= k t ; ln
2.00
= (87 s-1)(0.010s)
[C4H8]
[C4H8] = 0.83mol/L
(b)
[C4H8]0 - [C4H8]t
[C4H8]0
2.00M - 0.87M
=
2.00M
= 0.58
Reaction Half-life
Reaction Half-life (t1/2) :
1. is defined as the time required for the reactant to reach half its
initial concentration.
2. Is expressed in time units for a give reaction.
3. Is characteristic of that reaction at a certain temperature.
The equation to calculate half-life for reactions:
Zero order:
First order:
Second order
[ A]o
2k
Ln2
k
1
k[ A]o
From the equation, it can be seen the half-life for first order reaction is not
dependent on the initial concentration.
Determining Reaction Concentration at a Given Time
PROBLEM: At 10000C, cyclobutane (C4H8) decomposes in a first-order reaction,
with the very high rate constant of 87 s-1, to two molecules of ethylene
(C2H4).
(a) If the initial C4H8 concentration is 2.00M, what is the concentration
after 0.010 s?
(b) What fraction of C4H8 has decomposed in this time?
PLAN:
Find the [C4H8] at time, t, using the integrated rate law for a 1st order
reaction. Once that value is found, divide the amount decomposed by
the initial concentration.
SOLUTION:
ln
(a)
[C4H8]0
[C4H8]t
= k t ; ln
2.00
= (87 s-1)(0.010s)
[C4H8]
[C4H8] = 0.83mol/L
(b)
[C4H8]0 - [C4H8]t
[C4H8]0
2.00M - 0.87M
=
2.00M
= 0.58
Reaction Half-life
Reaction Half-life (t1/2) :
1. is defined as the time required for the reactant to reach half its
initial concentration.
2. Is expressed in time units for a give reaction.
3. Is characteristic of that reaction at a certain temperature.
The equation to calculate half-life for reactions:
Zero order:
First order:
Second order
[ A]o
2k
Ln2
k
1
k[ A]o
From the equation, it can be seen the half-life for first order reaction is not
dependent on the initial concentration.
A plot of [N2O5] vs. time for three half-lives.
for a first-order process
t1/2 =
ln 2
k
=
0.693
k
Determining the Half-Life of a First-Order Reaction
PROBLEM:
Cyclopropane is the smallest cyclic hydrocarbon. Because its 60 0 bond
angles allow poor orbital overlap, its bonds are weak. As a result, it is
thermally unstable and rearranges to propene at 1000 0C via the
following first-order reaction:
CH2

H3C CH CH2 (g)
H2C CH2 (g)
The rate constant is 9.2 s-1, (a) What is the half-life of the reaction?
(b) How long does it take for the concentration of cyclopropane to reach one-quarter
of the initial value?
PLAN:
Use the half-life equation, t1/2 =
0.693
k
, to find the half-life.
One-quarter of the initial value means two half-lives have passed.
SOLUTION:
(a)
t1/2 = 0.693/9.2s-1 = 0.075 s
(b)
2 t1/2 = 2(0.075s) = 0.150 s second half life
Determining the rate constant for a First-Order Reaction
from half-life.
PROBLEM:
Iodine-123 is used to study thyroid gland function. This radioactive isotope breaks down in
a first order process with a half-life of 13.1 hr. What is the rate constant?
PLAN: use the equation:
t
½
=
ln 2
k
SOLUTION:
k=
ln 2
t½
=
0.693
13.1
= 5.2910-2 hr-1
= ? S-1 (please convert unit from hr-1 to s-1)
The effect of temperature on reaction rate
• Temperature does affect on the reaction rate.
• Higher temperature gives higher reaction rate.
• Temperature influences the rate by affecting the rate constant.
• A plot of k vs temperature gives exponential equation.
• The relationship between temperature and rate constant is expressed by
Arrhenius equation.
k = A e-Ea/RT
Lnk = LnA -
Ea
R
( )
k - rate constant,
A - frequency factor,
R - ideal gas constant,
Ea - activation energy
Higher temperature, larger k, increased rate
1
T
Activation energy:
1. Ea is the minimum amount of energy for a molecule to have for
reaction.
2. A chemical reaction between two substances occurs only when an
atom, ion, or molecule collides with others.
3. Only a fraction of the total collisions result in a reaction, because
usually only a small percentage of the substances interacting have
the minimum amount of kinetic energy a molecule must possess
for it to react.
k
= A e-Ea/RT
Ea 1
ln k = ln A ( )
R T
Dependence of the rate constant on temperature
k = A e-Ea/RT
Graphical determination of the activation energy
Ea 1
ln k = + ln A
R T
Ea

k  Ae
RT
The Effect of Temperature on Reaction Rate
Ea

k  Ae
RT
The Arrhenius Equation
ln k = ln A - Ea/RT
Re-arrange
ln
k2
k1
= -
Ea
R
1
T2
where k is the kinetic rate constant at T
Ea is the activation energy
R is the energy gas constant
T is the Kelvin temperature
A is the collision frequency factor
1
T1
Determining the Energy of Activation
PROBLEM:
The decomposition of hydrogen iodide,
2HI(g)
H2(g) + I2(g)
has rate constants of 9.51x10-9 L/mol*s at 500 K and 1.10x10-5 L/mol*s
at 600. K. Find Ea.
PLAN:
Use the modification of the Arrhenius equation to find Ea.
SOLUTION:
ln
k2
k1
= -
Ea
1
R
T2
Ea = - (8.314J/mol*K)
-
1
T1
Ea = - R ln
1.10x10-5L/mol*s
k2
1
k1
T2
1
ln
9..51x10-9L/mol*s
Ea = 1.76x105 J/mol = 176 kJ/mol
600K
-
-
1
500K
1
T1
-1
Determining the rate constant from the known
Energy of Activation
PROBLEM:
The reaction 2NOCl (g)  2NO (g) + Cl2 (g) has an activation
energy Ea of 1.00  102 kJ/mol and rate constant 0.286 L/mols at 500
K. What is the rate constant for 600K?
PLAN:
use the Arrhenius equation:
ln
k2
k1
= -
Ea
R
1
T2
SOLUTION:
Lnk2/k1 = 0.2406
k2/k1 = e0.2406
k2 = 0.364 L/mol  s (second order)
1
T1
11.4
Models for Reaction Rate
The dependence of possible collisions on the product of reactant concentrations.
•
•
A
B
•
4 collisions
A
B
•
•
•
A
Collision theory: reactants must collide to react.
The number of collision depends on the reactant
concentration.
At higher T, more collision have enough energy
to exceed the Ea.
The relative size for the Ea depends on whether
the overall reaction is exothermic or endothermic.
An effective orientation for reaction is important.
The structure complexity decreases the reaction
rate.
B
Add another
molecule of A
A
B
A
Add another
molecule of B
6 collisions
A
B
A
B
A
B
Energy-level diagram for a reaction
Ea (forward)
Ea (reverse)
REACTANTS
PRODUCTS
The forward reaction is exothermic because the
reactants have more energy than the products.
Collision Energy
Collision Energy
ACTIVATED STATE
The importance of molecular orientation to an effective collision.
NO + NO3
A = pZ where
2 NO2
Z is the collision frequency
p is the orientation probability factor
A is the frequency factor
Transition state theory
•
If the potential energy is less than activation energy, the molecules recoil. The repulsion
decreases, speed increases, and molecules move apart without reacting.
•
Kinetic energy of molecules push them together with enough force to overcome the
repulsion and then molecules react. Molecules are oriented effectively and moving at high
speed.
•
Nuclei in one atom attracts the electrons from another, so the atomic orbital will overlap,
electron density shifts.
•
Some bonds lengthen and weaken, some bonds start to form. The reactant molecules
gradually change their bonds and shapes, while turning into the products.
•
There exists a smooth transformation, neither a reactant nor a product, but a transitional
species with partial bonds.
•
Transition state theory depicts the kinetic energy of particles changing to the potential
energy during a collision.
•
Transition state reaches when a sufficiently energetic collision and effective molecular
orientation is given.
•
The reaction energy diagram indicates the changing energy of chemical system as it
progresses from reactants through transition state to products.
Reaction energy diagrams and possible transition states.
Drawing Reaction Energy Diagrams and Transition States
PROBLEM:
A key reaction in the upper atmosphere is
O3 (g) + O (g)
2O2 (g)
The Ea(fwd) is 19 kJ, and the  Hrxn for the reaction is -392 kJ. Draw a reaction
energy diagram for this reaction, postulate a transition state, and calculate Ea(rev).
PLAN:
Consider the relationships among the reactants, products and transition
state. The reactants are at a higher energy level than the products and the
transition state is slightly higher than the reactants.
SOLUTION:
Ea(rev)= (392 + 19) kJ
Potential Energy
Ea= 19kJ
O3+O
= 411 kJ
transition state
Hrxn = -392kJ
2O2
Reaction progress
O
breaking
bond
O
O
forming
bond
O
Reaction energy diagram of a catalyzed and an uncatalyzed process.
Generic graph indicates
the effect of a catalyst in
an hypothetical
exothermic chemical
reaction.
Catalysts provide an (alternative) mechanism involving a different transition state and
lower activation energy.
More molecular collisions have the energy needed to reach the transition state.
Catalysts can perform reactions much faster, more specific, or at lower temperatures.
Catalysts reduce the amount of energy needed to start a chemical reaction.
Catalysts cannot make energetically unfavorable reactions possible
The net free energy change of a reaction is the same whether a catalyst is used or not.
CATALYSTS
• Each catalyst has its own specific way of functioning.
• In general a catalyst lowers the energy of activation.
• Lowering the Ea increases the rate constant, k, and
thereby increases the rate of the reaction
• A catalyst increases the rate of the forward AND the
reverse reactions.
• A catalyzed reaction yields the products more quickly,
but does not yield more product than the uncatalyzed
reaction.
• A catalyst lowers Ea by providing a different mechanism,
for the reaction through a new, lower energy pathway.
REACTION MECHANISMS
A sequence of single reaction steps that sum to the overall reaction.
The overall rate of a reaction is related to the rate of the slowest, or rate-determining step (RDS).
Rate Laws for General Elementary Steps
Elementary Step is defined as an individual step, which together makes up the proposed
reaction mechanism.
Elementary step is NOT made of single step and must be physically reasonalbe.
Elementary step is characterized by its molecularity involving the reactant particles in the
steps.
Elementary Step
Molecularity
Rate Law
product
Unimolecular
Rate = k [A]
2A
product
Bimolecular
Rate = k [A]2
A+B
product
Bimolecular
Rate = k [A][B]
2A + B
product
Termolecular
Rate = k [A]2[B]
A
Determining Molecularity and Rate Laws for Elementary Steps
PROBLEM:
(1)
(2)
The following two reactions are proposed as elementary steps in the
mechanism of an overall reaction:
NO2Cl(g)
NO2(g) + Cl (g)
NO2Cl(g) + Cl (g)
NO2(g) + Cl2(g)
(a) Write the overall balanced equation.
(b) Determine the molecularity of each step.
(c) Write the rate law for each step.
PLAN: (a) The overall equation is the sum of the steps.
(b) The molecularity is the sum of the reactant particles in the step.
SOLUTION
:
(a) (1)
NO2Cl(g)
NO2(g) + Cl (g)
(2) NO2Cl(g) + Cl (g)
NO2(g) + Cl2(g)
2NO2Cl(g)
2NO2(g) + Cl2(g)
(b) Step(1) is unimolecular.
Step(2) is bimolecular.
(c) rate1 = k1 [NO2Cl]
rate2 = k2 [NO2Cl][Cl]
Models for Reaction Rate (Optional)
Two primary models:
• Collision model – Activation energy
• Transition-state model – Activation energy diagrams
• Collision theory: reactants must collide to react.
• The number of collision depends on the reactant concentration.
• At higher T, more collision have enough energy to exceed the Ea.
• The relative size for the Ea depends on whether the overall
reaction is exothermic or endothermic.
• An effective orientation for reaction is important.
• The structure complexity decreases the reaction rate.