2/21/2012
Ch. 14 - Kinetics
Unit 6 Kinetics & Equilibria
Outline: Kinetics
• the study of the speed of chemical
reactions
• Reaction Rate – the speed of the
reaction
Factors That Affect Reaction Rates
• Concentration of Reactants
Reaction Rates
how rates are measured
Rate Laws
how rates depend on the amount of
reactants
Integrated Rate Laws
how to calculate the amount left or
the time needed to reach a given
amount
Half-life
the time needed for half of the
reactants to be used up
Arrhenius Equation
how rate constant changes with T
Mechanisms
the process by which a reaction
occurs
 As the concentration of reactants increases, so
does the likelihood that reactant molecules will
collide
• Temperature
 At higher temperatures, reactant molecules have
more kinetic energy, move faster, and collide more
often and with greater energy
• Catalysts
 changes the speed of rxn by changing mechanism
• Surface Area
 increased surface area allows more reactant
molecules to be available for reaction
aA + bB  cC + dD
For the reaction: AB
[A] = concentration of A (M)
Rate = - [A] / t = + [B] / t
Rate
1 A
a t
1 B
b t
1 C
c t
1 D
d t
Rate is negative because [A] is
decreasing
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for H2 + I2  2HI
Rate Law
• shows how rate depends on the concentration
of reactants
• Rate = k[A]m[B]n…
• k = rate constant
• m, n, … = rxn order (determined
experimentally)
• overall rxn order = m + n + …
Rate = k[H2][I2]
1st order with respect to H2 & I2
2nd order overall
1st order reactions are most common
Units of k depend on rxn order
• 0th order
• 1st order
• 2nd order
k = M/s
k = 1/s
k = 1/Ms
Reaction Rates
• concentration vs. time
• slope of a line tangent
to the curve at any point
is the instantaneous
rate at that time.
• steeper slope = faster
rate
Using Initial Rates to
Determine Rate Laws
Reaction Rates
• reaction slows down
with time because the
concentration of the
reactants decreases
• A+BC
Trial
[A] (M)
[B] (M)
Initial Rate (M/s)
1
0.100
0.100
4.0 x 10-5
2
0.100
0.200
4.0 x 10-5
3
0.200
0.100
16 x 10-5
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Change of Concentration with
Time
• Rate = k[A]m[B]n
• Notice when [B] doubles, rate does not
change
n = 0
• When [A] doubles, rate is squared
• [A]t = concentration at time “t”
• [A]0 = concentration at t = 0 (initial)
• Use graphs to determine reaction order
m = 2
• Rate = k[A]2[B]0
• Use data to solve for k
Zeroth Order
First Order
• [A]t = [A]0 = -kt
• Plot [A] vs. time to get a straight line
• Slope = -k
• Plot ln[A] vs. time to get a straight line
• Slope = -k
Second Order
Half-Life
• the time required for
one-half of a
reactant to react
• Plot 1/[A] vs. time to get a straight line
• Slope = k
• Because [A] at t1/2 is
one-half of the
original [A],
[A]t = 0.5 [A]0.
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Half-Life
0th order
[A]
2k
Temperature and Rate
t1
• Generally, as temperature
increases, so does the
reaction rate.
• This is because k is
temperature dependent.
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1st order
2nd order
The Collision Model
• In a chemical reaction, bonds are
broken and new bonds are formed.
• Molecules can only react if they collide
with each other.
• Anything that increases the number of
collisions will increase the rate
The Collision Model
Furthermore, molecules must collide with the
correct orientation and with enough energy to
cause bond breakage and formation.
 [reactants], T, P, V
Activation Energy
• Ea – the minimum amount of energy
needed for a reaction to proceed
Reaction Coordinate Diagrams
• Activated complex
(transition state) –
arrangement of
atoms with the most
energy
• Ea – activation
energy
• - H  exothermic
• + H  endothermic
• Note: Smaller Ea
results in faster rxn
( E has no effect)
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Maxwell–Boltzmann Distributions
• Temperature is
defined as a
measure of the
average kinetic
energy of the
molecules in a
sample.
• At any temperature there is a wide
distribution of kinetic energies.
Maxwell–Boltzmann Distributions
• If the dotted line represents the activation
energy, as the temperature increases, so does
the fraction of molecules that can overcome
the activation energy barrier.
Maxwell–Boltzmann Distributions
• As the temperature
increases, the curve
flattens and
broadens.
• Thus at higher
temperatures, a
larger population of
molecules has
higher energy.
Maxwell–Boltzmann Distributions
This fraction of molecules can be found through the expression:
where R is the gas constant and T is the temperature in Kelvin .
• As a result, the
reaction rate
increases.
Arrhenius Equation
• A is the frequency factor (likelihood that
collisions would occur with the proper
orientation for reaction)
• When 2 trials differ only in temperature…
k
ln 1
k2
Ea 1
R T2
1
T1
Arrhenius Equation
Taking the natural
logarithm of both
sides, the equation
becomes
1
RT
y = mx + b
When k is determined experimentally at
several temperatures, Ea can be calculated
from the slope of a plot of ln k vs. 1/T.
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Reaction Mechanisms
• sequence of steps a rxn takes in
proceeding from reactants to products
• Elementary step – individual step
• Molecularity – the number of reactant
molecules involved in an elementary
step
Reaction Mechanisms
unimolecular – 1 molecule
bimolecular – 2 molecules
termolecular – 3 molecules
• intermediate – substance formed in one
step and consumed in another
Step 1:
Step 2:
Overall:
O3  O2 + O
O3 + O  2 O2
2 O3  3 O2
What is the intermediate?
O
What is the molecularity?
Step 1: unimolecular
Step 2: bimolecular
Slow Initial Step
NO2 (g) + CO (g)
NO (g) + CO2 (g)
• The rate law for this reaction is found
experimentally to be
Rate = k[NO2]2
• CO is necessary for this reaction to occur, but the
rate of the reaction does not depend on its
concentration.
• This suggests the reaction occurs in two steps.
Rate determining step
• In a multistep process, one of the steps will
be slower than all others
• The overall reaction cannot occur faster than
the slowest step
Slow Initial Step
Step 1 (slow)
Step 2 (fast)
Overall Rxn
NO2 + NO2
NO3 + CO
NO3 + NO
NO2 + CO2
NO2 + CO  NO + CO2
• The NO3 intermediate is consumed in the second
step.
• CO is not involved in the slow, rate-determining step, it
does not appear in the rate law.
• Rate = k [NO2]2
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Fast Initial Step
Step 1 (fast) NO + Br2  NOBr2
Step 2 (slow) NOBr2 + NO  2 NOBr
Overall
2 NO + Br2  2 NOBr
Rate 2 determines rate of reaction
Rate 2 = k2[NOBr2][NO]
but this includes an intermediate
Catalysts – change the speed of
rxn
homogeneous catalyst – present in the
same phase as the reactants
heterogeneous catalyst – different phase
than reactant
inhibitor – slows the reaction
catalyst – term usually associated with
speeding up the reaction
active site – location where reactant
molecules are adsorbed
Catalysts
One way a
catalyst can
speed up a
reaction is by
holding the
reactants together
and helping bonds
to break.
Rate 1 = k1[NO][Br2] = k-1[NOBr2]
Solve for [NOBr2] = (k1/k-1)[NO][Br2]
and substitute…
Rate 2 = (k2k1/k-1)[NO]2[Br2]
Catalysts
• increase the rate of a reaction by decreasing
the activation energy
• change the mechanism by which the process
occurs
Enzymes
• Enzymes are
catalysts in
biological systems.
• The substrate fits
into the active site of
the enzyme much
like a key fits into a
lock.
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Ch. 15 - Equilibrium
• Equilibrium occurs when the rate of a
forward reaction is the same as the rate
of the reverse reaction
• DOES NOT mean that the amount of
reactants and products are the same, but
their concentrations WILL stay constant
• Equilibrium is a DYNAMIC process,
reactions do not stop
• Shown in a reaction as a double arrow 
Law of Mass Action
• mass of reactants consumed = mass of
products formed
• Forward Reaction

AB
Rate = kf[A]
• Reverse Reaction

BA
Rate = kr[B]
• At equilibrium, kf[A] = kr[B]
• kf/kr = [B]/[A]
• = the equilibrium constant
Equilibrium Constant
Expressions
For Gaseous Equilibrium
• For the following reaction:
Kp
aA + bB  cC + dD
Kc
(Pc ) c (Pd ) d
(Pa )a (Pb ) b
[C]c [D]d
[A ]a [B]b
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Relationship between Kc & Kp
• Equilibrium constant for reverse
reaction is the reciprocal of the forward
reaction
• For A  B
• Kcf = [B]/[A]
• Kcr = [A]/[B]
•
•
•
•
PV = nRT
P = nRT / V
P = (n/V)RT
note: [A] = na/V
• Substitute this in for each substance
• Kp = Kc(RT) n
• n = ngas products – ngas reactants
Value of K
• If K>>1
products are favored
• If K<<1
reactants are favored
• Homogeneous equilibria – all
substances are in the same phase (i.e.
gaseous, aqueous)
• Heterogeneous equilbria – substances
present exist in more than one phase
NOTE: concentration of SOLIDS and
LIQUIDS remain constant, therefore they
are EXCLUDED from the equilibrium
expression
Write Kc and Kp for the
following reactions
• CO2 (g) + H2 (g)  CO (g) + H2O (l)
• 3 Fe (s) + 4 H2O (g)  Fe3O4 (s) + 4 H2
(g)
Calculating Equilibrium
Constants
• Tabulate known initial and equilibrium
concentrations of all substances
• Calculate [x] for all substances where
both the initial and equilibrium
concentrations are known
• Use stoichiometry to calculate [x] for
other substances
• Determine all equilibrium concentrations
and use them to find Kc
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Work Sample Exercises 15.8,
15.9
Predicting Direction of Rxn
• Q = reaction quotient
found in the same manner as K, but uses initial
concentrations
• If Q = K
reaction is already at equilibrium
• If Q > K
there are too many products, rxn needs to
move left
• If Q < K
there are too many reactants, rxn needs to
move right
Calculating Equilibrium
Concentrations
• Sample & Practice Exercises 15.11
Change in Concentration of
Reactants or Products
• Adding reactants / removing products
more products will be produced
• Removing reactants / adding products
more reactants will be produced
Le Chatlier’s Principle
• a system at equilibrium that is disturbed
will shift its equilibrium position so as to
counteract the effect of the disturbance
Change in P or V
• If volume is reduced, or pressure
increases
system will shift to the side with the least
number of moles of gas
• N2 (g) + 3 H2 (g)  2 NH3 (g)
• increase in P will cause rxn to shift right
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Changing T
• Endothermic reaction
 R + heat  P
• Exothermic reaction
 R  P + heat
• Adding heat will move the reaction in
the direction that absorbs heat (just like
adding other reactants or products)
What happens to K when…
• heat is added to an exothermic reaction
reaction shifts left
K decreases
• The pressure increases during the
synthesis of water in the gaseous phase
reaction shifts right
K increases
Effect of Catalysts
• NONE on K!
• Catalysts do not change the conditions
of equilibrium
• Equilibrium will be achieved faster
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