Jaslin Kinetika Kimia
Kinetika Kimia
(Chemical Kinetics)
Jaslin Ikhsan, Ph.D.
Kimia FMIPA UNY
Course Summary.
• Dependence of rate on concentration.
• Experimental methods in reaction
kinetics.
• Kinetics of multistep reactions.
• Dependence of rate on temperature.
Recommended reading.
1. P.W. Atkins, Kimia Fisika (terjemahan), jilid 2, edisi keempat,
Jakarta: Erlangga, 1999.
2. P.W. Atkins, Physical Chemistry, 5th ed., Oxford:
Oxford University Press, 1994
4. Arthur M. Lesk, Introduction to Physical Chemistry
Edited by Foxit PDF Editor
Copyright (c) by Foxit Software Company, 2004
For Evaluation Only.
3. Gordon M. Barrow, Physical Chemistry
Chemical Kinetics.
Lecture 1.
Review of Basic concepts.
Reaction Rate: The Central Focus of
Chemical Kinetics
Chemical reaction kinetics.
•
•
•
Chemical reactions involve the
forming and breaking of
chemical bonds.
Reactant molecules (H2, I2)
approach one another and
collide and interact with
appropriate energy and
orientation. Bonds are
stretched, broken and formed
and finally product molecules
(HI) move away from one
another.
How can we describe the rate
at which such a chemical
transformation takes place?
reactants
products
H 2 ( g ) + I 2 ( g ) → 2HI ( g )
• Thermodynamics tells us all
about the energetic feasibility
of a reaction : we measure the
Gibbs energy ∆G for the chemical
Reaction.
• Thermodynamics does not tell us
how quickly the reaction will
proceed : it does not provide
kinetic information.
Basic ideas in reaction kinetics.
•
Chemical reaction kinetics deals with the rate of velocity of chemical
reactions.
•
We wish to quantify
– The velocity at which reactants are transformed to products
– The detailed molecular pathway by which a reaction proceeds (the reaction
mechanism).
•
These objectives are accomplished using experimental measurements.
•
Chemical reactions are said to be activated processes : energy (usually
thermal (heat) energy) must be introduced into the system so that
chemical transformation can occur. Hence chemical reactions occur
more rapidly when the temperature of the system is increased.
•
In simple terms an activation energy barrier must be overcome before
reactants can be transformed into products.
Reaction Rate.
•
What do we mean by the term
reaction rate?
–
•
How may reaction rates be
determined ?
–
–
•
•
•
The term rate implies that something
changes with respect to something
else.
The reaction rate is quantified in
terms of the change in concentration
of a reactant or product species with
respect to time.
This requires an experimental
measurement of the manner in which
the concentration changes with time
of reaction. We can monitor either
the concentration change directly, or
monitor changes in some physical
quantity which is directly proportional
to the concentration.
The reactant concentration
decreases with increasing time, and
the product concentration increases
with increasing time.
The rate of a chemical reaction
depends on the concentration of each
of the participating reactant species.
The manner in which the rate
changes in magnitude with changes in
the magnitude of each of the
participating reactants is termed the
reaction order.
Reactant
concentration
Product
concentration
d [R] d [P]
RΣ = −
=
dt
dt
Net reaction rate
Units : mol dm-3 s-1
[P]t
[R]t
time
Rate, rate equation and reaction order : formal
definitions.
• The reaction rate (reaction velocity) R is quantified in terms of
changes in concentration [J] of reactant or product species J
with respect to changes in time. The magnitude of the reaction rate
changes as the reaction proceeds.
∆[ J ] 1 d [ J ]
RJ =
lim
=
t
∆
→
0
∆t
υJ
υ J dt
1
2 H 2 ( g ) + O2 ( g ) → 2 H 2O( g )
R=−
d [O2 ] 1 d [H 2O ]
1 d [H 2 ]
=−
=
dt
2 dt
2 dt
• Note : Units of rate :- concentration/time , hence RJ has units mol dm-3s-1 .
υJ denotes the stoichiometric coefficient of species J. If J is a reactant υJ
is negative and it will be positive if J is a product species.
• Rate of reaction is often found to be proportional to the molar
concentration of the reactants raised to a simple power (which
need not be integral). This relationship is called the rate equation.
The manner in which the reaction rate changes in magnitude with
changes in the magnitude of the concentration of each participating
reactant species is called the reaction order.
Problem Example:
The rate formation of NO in the reaction:
2NOBr(g) ----> 2NO(g) + Br2(g)
was reported as 1.6 x 10-4 mol/(L s).
What is the rate of reaction and the rate of consumption
of NOBr ?
Log R
Log R
Slope = α
Slope = β
Log [A]
Log [B]
k
xA + yB
→
Products
empirical rate
equation (obtained
from experiment)
rate constant k
R=−
stoichiometric
coefficients
Rate equation can not in
general be inferred from
the stoichiometric equation
for the reaction.
1 d [ A]
1 d [B ]
α
β
=−
= k [ A] [B ]
x dt
y dt
α, β = reaction
orders for the
reactants (got
experimentally)
Different rate equations
imply different mechanisms.
H 2 + I 2 → 2 HI
d [HI ]
= k [H 2 ][I 2 ]
dt
H 2 + Br2 → 2 HBr
R=
d [HBr ] k [H 2 ][Br2 ]
R=
=
k ′[HBr ]
dt
1+
[Br2 ]
1/ 2
H 2 + Cl2 → 2 HCl
d [HCl ]
1/ 2
= k [H 2 ][Cl2 ]
R=
dt
H 2 + X 2 → 2 HX
X = I , Br , Cl
• The rate law provides an important guide
to reaction mechanism, since any proposed
mechanism must be consistent with the
observed rate law.
• A complex rate equation will imply a complex
multistep reaction mechanism.
• Once we know the rate law and the rate
constant for a reaction, we can predict the
rate of the reaction for any given composition
of the reaction mixture.
• We can also use a rate law to predict the
concentrations of reactants and products at
any time after the start of the reaction.
Zero order kinetics. The reaction proceeds at the same rate R
regardless of concentration.
Rate equation :
da
R=−
=k
dt
a = a0 when t = 0
0
units of rate constant k :
mol dm-3 s-1
[ A]
integrate
using initial
condition
a (t ) = −kt + a0
half life : t = τ 1/ 2 when a =
a
τ 1/ 2 =
slope = -k
a0
2k
a0
2
τ 1 / 2 ∝ a0
τ 1/ 2
a0
diagnostic
plot
R ∝ [ A]
R
slope =
t
a0
1
2k
k
A
→
products
First order kinetics.
Initial
concentration a0
rate = −
First order differential
rate equation.
da
−
= ka
dt
d [ A] t
dt
Initial condition
t = 0 a = a0
Solve differential
equation
Separation
of variables
For a first order reaction the relationship:
(rate) t ∝ [ A] t
(rate) t = k [ A] t
a (t ) = a0 e − kt = a0 exp[− kt ]
is valid generally for any time t.
k is the first order
rate constant, units: s-1
Reactant concentration
as function of time.
First order kinetics.
Half life τ1/2
t = τ 1/ 2
θ = θ1/ 2
τ 1/ 2 =
k → s −1
a (t ) = a0 e − kt = a0 exp[− kt ]
a (t )
= exp[− θ ]
u=
a0
θ = kt
a0
2
u = 1/ 2
a=
ln 2 0.693
=
k
k
Mean lifetime of reactant molecule
1
τ=
a0
∫
∞
0
1
a(t )dt =
a0
∫
∞
0
a0 e − kt dt =
1
k
First order kinetics: half life.
u=
a
a0
u =1
t = τ 1/ 2
In each successive period
of duration τ1/2 the concentration
of a reactant in a first order reaction
decays to half its value at the start
of that period. After n such periods,
the concentration is (1/2)n of its
initial value.
a = a0 / 2
u = 0 .5
u = 0.25
u = 0.125
τ 1/ 2
τ 1/ 2 =
a0
ln 2 0.693
=
k
k
half life independent of
initial reactant concentration
1st order kinetics
u (θ )
a (t ) = a0 e
u (θ ) = e −θ
θ = kt
2nd order kinetics
− kt
u (θ )
a (t ) =
a0
1 + ka0t
u (θ ) =
θ = ka0t
1
1+θ
Second order kinetics: equal
reactant concentrations.
k
2A
→
P
dm3mol-1s-1
slope = k
da
= ka 2
dt
t = 0 a = a0
−
separate variables
integrate
1
1
= kt +
a
a0
half life t = τ 1/ 2 a = a0
2
τ 1/ 2 =
1
a
1
ka0
τ 1/ 2
t
a0
a (t ) =
1+ ka0t
1
τ 1/ 2 ∝
a0
τ 1/ 2 ↓ as a0 ↑
a0
rate varies as
square of reactant
concentration
1st and 2nd order kinetics : Summary .
Reaction
k1
A →
Products
k2
2 A →
Products
Differential
rate equation
Concentration
variation with
time
da
−
= k1a
dt
a (t ) =
a0 exp[− k1t ]
a (t ) =
da
2
−
= k2 a
a0
dt
Diagnostic
Equation
ln a(t ) = −k1t + ln a0
1
1
= k 2t +
a (t )
a0
Half
Life
τ 1/ 2 =
ln 2
k1
τ 1/ 2 =
1
k 2 a0
1 + k 2 a0t
ln a(t)
Slope = - k1 1/a(t)
τ1/2
Slope = k2
t
t
1st order
2nd order
a0
Diagnostic
Plots .
k
nA
→
P
n th order kinetics: equal reactant
concentrations.
da
−
= ka n
dt
t = 0 a = a0
separate variables
integrate
a
n ≠1
n = 0, 2,3,…..
1
a n −1
1
n −1
= (n − 1)kt +
1
a0
n −1
rate constant k
obtained from slope
Half life
τ 1/ 2 =
n −1
2 −1
(n − 1)ka0 n−1
τ 1/ 2 ∝ a01− n
n > 1 τ 1/ 2 ↓ as a0 ↑
n < 1 τ 1/ 2 ↑ as a0 ↑
2 −1
ln τ 1/ 2 = ln
− (n − 1) ln a0
(
)
n
1
k
−
n −1
slope = (n − 1)k
t
ln τ 1/ 2
slope = −(n − 1)
reaction order n determined
from slope
ln a0
Summary of kinetic results.
t = 0 a = a0
k
nA
→
P
t = τ 1/ 2
a=
Rate equation
a0
2
Integrated
expression
Units of k
k
a(t ) = −kt + a0
Half life
τ1/2
mol dm-3s-1
a0
2k
1
ka
a
ln 0 = kt
a (t )
s-1
ln 2
k
2
ka
2
1
1
= kt +
a (t )
a0
dm3mol-1s-1
1
ka0
ka
3
1
1
2
=
kt
+
2
2
a(t )
a0
dm6mol-2s-1
3
2
2ka0
ka
n
1
Reaction
Order
0
3
n
da
R=−
dt
a n −1
= (n − 1)kt +
1
a0
n −1
1 2 n −1 − 1
n − 1 ka 0 n −1
Problem Examples:
1. A first order reaction is 40% complete at the end of 1 h.
What is the value of the rate constant?
In how long willl the reaction be 80 % complete ?
2. The half-life of the radioactive disintegration of radium is
1590 years. Calculate the decay constant. In how many
years will three-quarters of the radium have undergone
decay ?
3. Derive the rate equation for reaction whose orders are:
(a). one-half, (b). three and a-half, (c). 4, and (d). n !
Chemical Kinetics
Lecture 2.
Kinetics of more complex
reactions.
Jaslin Ikhsan, Ph.D.
Kimia FMIPA
Universitas Negeri Jogjakarta
Tujuan Perkuliahan:
1. Menurunkan Rate Law dari Kinetika
Reaksi yang lebih Kompleks,
2. Menjelaskan Consecutive Reactions:
a. Rate Determining Steps,
b. Steady State Approximation.
Reference:
1. P.W. Atkins, Physical Chemistry, 5th ed,
Oxford: 1994
2. Ira N. Levine, Physical Chemistry
Second order kinetics:
Unequal reactant concentrations.
A+ B
→ P
k
rate equation
R=−
da
db dp
=−
=
= kab
dt
dt dt
initial conditions
t = 0 a = a0
b = b0
a0 ≠ b0
integrate using
partial fractions
1
F (a, b ) =
b0 − a0
b b0
= kt
ln
a a0
dm3mol-1s-1
F (a, b )
slope = k
t
Consecutive Reactions .
•Mother / daughter radioactive
decay.
218
Mass balance requirement:
Po→ 214 Pb→ 214 Bi
k1 = 5 × 10 −3 s −1
k1
k2
A →
X →
P
k 2 = 6 × 10 − 4 s −1
p = a0 − a − x
The solutions to the coupled
equations are :
3 coupled LDE’s define system :
a (t ) = a0 exp[− k1t ]
da
= −k1a
dt
dx
= k1a − k 2 x
dt
dp
= k2 x
dt
x(t ) =
k1a0
{ exp[− k1t ] − exp[− k2t ] }
k 2 − k1
p(t ) = a0 − a0 exp[− k1t ] −
k1a0
{ exp[− k1t ] − exp[− k2t ] }
k 2 − k1
We get different kinetic behaviour depending
on the ratio of the rate constants k1 and k2
Consecutive reaction : Case I.
Intermediate formation fast, intermediate decomposition slow.
Case I .
κ=
k2
>
TS I
II : fast
∆GII‡
energy
I : slow rds
k2
κ=
k1
TS II
∆GI‡
∆GII‡
A
Step I rate determining
since it has the highest
activation energy barrier.
X
P
reaction co-ordinate
k1
k2
A →
X →
P
normalised concentration
Case
CaseIIII..
κ=
1.2
k2
>> 1
k1
k 2 >> k1
Intermediate X
is fairly reactive.
Concentration of
intermediate X
will be small at
all times.
1.0
Product P
0.8
u=a/a0
v =x/a0
0.6
w=p/a0
0.4
Reactant A
0.2
κ = k 2/k1 = 10
Intermediate X
0.0
0
2
4
6
8
10
τ = k1t
normalised concentration
1.2
u=a/a0
1.0
v=x/a0
Intermediate concentration
is approximately constant
after initial induction period.
w=p/a0
0.8
P
A
0.6
0.4
X
0.2
0.0
0.0
0.2
0.4
0.6
τ = k1t
0.8
1.0
k1 >> k 2
Rate Determining Step
A → X → P
k1
Fast
k2
Slow
• Reactant A decays rapidly, concentration of intermediate species X
is high for much of the reaction and product P concentration rises
gradually since X--> P transformation is slow .
k 2 >> k1
Rate Determining
Step
A → X → P
k1
k2
Slow
Fast
• Reactant A decays slowly, concentration of intermediate species X
will be low for the duration of the reaction and to a good approximation
the net rate of change of intermediate concentration with time is zero.
Hence the intermediate will be formed as quickly as it is removed.
This is the quasi steady state approximation (QSSA).
Parallel reaction mechanism.
k1
A →
X
k2
• We consider the kinetic analysis of a concurrent
A →
Y
or parallel reaction scheme which is often met in
k1, k2 = 1st order rate constants
real situations.
• A single reactant species can form two
We can also obtain expressions
distinct products.
for the product concentrations
st
We assume that each reaction exhibits 1 order x(t) and y(t).
kinetics.
• Initial condition : t= 0, a = a0 ; x = 0, y = 0
.• Rate equation:
da
= k1a + k 2 a = (k1 + k 2 )a = kΣ a
R=−
dt
a (t ) = a0 exp[− kΣt ] = a0 exp[− (k1 + k 2 )t ]
• Half life: τ 1/ 2 = ln 2 = ln 2
kΣ
k1 + k 2
dx
= k1a = k1a0 exp[− (k1 + k 2 )t ]
dt
x(t ) = k1a0
x(t ) =
∫ exp[− (k
t
1
0
+ k 2 )t ] dt
k1a0
{1 − exp[− (k1 + k2 )t ]}
k1 + k 2
dy
= k 2 a = k 2 a0 exp[− (k1 + k 2 )t ]
dt
y (t ) = k 2 a0
y (t ) =
∫ exp[− (k
t
0
1
+ k 2 )t ] dt
k 2 a0
{1 − exp[− (k1 + k2 )t ]}
k1 + k 2
x(t ) k1
• All of this is just an extension of simple Final product analysis
=
Lim
st
yields
rate
constant
ratio.
t
→
∞
y
(
t
)
k2
1 order kinetics.
Parallel Mechanism: k1 >> k2
normalised concentration
1.0
Theory
k
κ = 2 = 0.1
k1
k1
v(∞ )
= 10 =
k2
w(∞)
κ = 0.1
a(t)
0.8
v(∞ ) = 0.9079
x(t)
u(τ)
v(τ)
w(τ)
0.6
0.4
κ = k2 /k1
/
0.2
y(t)
w(∞ ) = 0.0908
0.0
0
1
2
3
τ = k1t
4
5
6
Computation
v(∞ ) 0.9079
≅
= 9.9989
w(∞ ) 0.0908
Parallel Mechanism: k2 >> k1
normalised concentration
1.0
Theory
u(τ)
v(τ)
w(τ)
0.6
0.4
a(t)
0.2
k2
= 10
k1
k1
v(∞ )
= 0.1 =
k2
w(∞)
v(∞) ≅ 0.0909
x(t)
0.0
0
κ=
w(∞) ≅ 0.9091
y(t)
κ = 10
0.8
1
2
3
τ = k1t
4
5
6
Computation
v(∞ ) 0.0909
≅
= 0.0999
w(∞ ) 0.9091
Reaching Equilibrium on the
Macroscopic and Molecular Level
N2O4
NO2
N2O4 (g)
colourless
2 NO2 (g)
brown
Chemical Equilibrium :
a kinetic definition.
•
•
Countless experiments with chemical
systems have shown that in a state of
equilibrium, the concentrations of
reactants and products no longer change
with time.
This apparent cessation of activity occurs
because under such conditions, all
reactions are microscopically reversible.
We look at the dinitrogen tetraoxide/
nitrogen oxide equilibrium which
occurs in the gas phase.
N2O4 (g)
colourless
Kinetic analysis.
r
R = k [N 2O4 ]
s
2
R = k ′[NO2 ]
Valid for any time t
2 NO2 (g)
brown
Concentrations vary
with time
[NO2 ] t ↑
[N 2O4 ] t
concentration
•
Equilibrium:
r s
t →∞
R=R
2
k [N 2O4 ]eq = k ′[NO2 ]eq
[NO2 ]eq2 k
= =K
[N 2O4 ] k ′
↓
Kinetic
regime
Concentrations time
invariant
[NO2 ] eq
[N 2O4 ] eq
Equilibrium
state
NO2
t →∞
N2O4
time
Equilibrium
constant
First order reversible reactions :
understanding the approach to chemical equilibrium.
Rate equation
k
A
B
k'
u + v =1
τ = 0 u =1 v = 0
da
= −ka + k ′b
dt
Rate equation in normalised form
Initial condition
t = 0 a = a0
du
1
+u =
dτ
1+θ
b=0
Mass balance condition
Solution produces the concentration expressions
∀t
a + b = a0
u (τ ) =
v(τ ) =
Introduce normalised variables.
u=
a
a0
v=
b
a0
τ = (k + k ′)t θ =
k
k′
1
{1 + θ exp[− τ ]}
1+θ
θ
1+θ
{1 − exp[− τ ]}
Q(τ ) =
Reaction quotient Q
v(τ )
=θ
u (τ )
1 − exp[− τ ]
[
]
+
−
1
θ
exp
τ
First order reversible reactions: approach to equilibrium.
Kinetic
regime
1.0
concentration
0.8
Equilibrium
Product B
K = Q(τ → ∞ )
0.6
u (τ)
v (τ)
0.4
=
Reactant A
0.2
0.0
0
2
4
τ = (k+k')t
6
8
v(∞ )
u (∞ )
Understanding the difference between reaction quotient Q and
Equilibrium constant K.
12
Approach to
Equilibrium
Q
(Chemical Kinetics)
Jaslin Ikhsan, Ph.D.
Kimia FMIPA UNY
Course Summary.
• Dependence of rate on concentration.
• Experimental methods in reaction
kinetics.
• Kinetics of multistep reactions.
• Dependence of rate on temperature.
Recommended reading.
1. P.W. Atkins, Kimia Fisika (terjemahan), jilid 2, edisi keempat,
Jakarta: Erlangga, 1999.
2. P.W. Atkins, Physical Chemistry, 5th ed., Oxford:
Oxford University Press, 1994
4. Arthur M. Lesk, Introduction to Physical Chemistry
Edited by Foxit PDF Editor
Copyright (c) by Foxit Software Company, 2004
For Evaluation Only.
3. Gordon M. Barrow, Physical Chemistry
Chemical Kinetics.
Lecture 1.
Review of Basic concepts.
Reaction Rate: The Central Focus of
Chemical Kinetics
Chemical reaction kinetics.
•
•
•
Chemical reactions involve the
forming and breaking of
chemical bonds.
Reactant molecules (H2, I2)
approach one another and
collide and interact with
appropriate energy and
orientation. Bonds are
stretched, broken and formed
and finally product molecules
(HI) move away from one
another.
How can we describe the rate
at which such a chemical
transformation takes place?
reactants
products
H 2 ( g ) + I 2 ( g ) → 2HI ( g )
• Thermodynamics tells us all
about the energetic feasibility
of a reaction : we measure the
Gibbs energy ∆G for the chemical
Reaction.
• Thermodynamics does not tell us
how quickly the reaction will
proceed : it does not provide
kinetic information.
Basic ideas in reaction kinetics.
•
Chemical reaction kinetics deals with the rate of velocity of chemical
reactions.
•
We wish to quantify
– The velocity at which reactants are transformed to products
– The detailed molecular pathway by which a reaction proceeds (the reaction
mechanism).
•
These objectives are accomplished using experimental measurements.
•
Chemical reactions are said to be activated processes : energy (usually
thermal (heat) energy) must be introduced into the system so that
chemical transformation can occur. Hence chemical reactions occur
more rapidly when the temperature of the system is increased.
•
In simple terms an activation energy barrier must be overcome before
reactants can be transformed into products.
Reaction Rate.
•
What do we mean by the term
reaction rate?
–
•
How may reaction rates be
determined ?
–
–
•
•
•
The term rate implies that something
changes with respect to something
else.
The reaction rate is quantified in
terms of the change in concentration
of a reactant or product species with
respect to time.
This requires an experimental
measurement of the manner in which
the concentration changes with time
of reaction. We can monitor either
the concentration change directly, or
monitor changes in some physical
quantity which is directly proportional
to the concentration.
The reactant concentration
decreases with increasing time, and
the product concentration increases
with increasing time.
The rate of a chemical reaction
depends on the concentration of each
of the participating reactant species.
The manner in which the rate
changes in magnitude with changes in
the magnitude of each of the
participating reactants is termed the
reaction order.
Reactant
concentration
Product
concentration
d [R] d [P]
RΣ = −
=
dt
dt
Net reaction rate
Units : mol dm-3 s-1
[P]t
[R]t
time
Rate, rate equation and reaction order : formal
definitions.
• The reaction rate (reaction velocity) R is quantified in terms of
changes in concentration [J] of reactant or product species J
with respect to changes in time. The magnitude of the reaction rate
changes as the reaction proceeds.
∆[ J ] 1 d [ J ]
RJ =
lim
=
t
∆
→
0
∆t
υJ
υ J dt
1
2 H 2 ( g ) + O2 ( g ) → 2 H 2O( g )
R=−
d [O2 ] 1 d [H 2O ]
1 d [H 2 ]
=−
=
dt
2 dt
2 dt
• Note : Units of rate :- concentration/time , hence RJ has units mol dm-3s-1 .
υJ denotes the stoichiometric coefficient of species J. If J is a reactant υJ
is negative and it will be positive if J is a product species.
• Rate of reaction is often found to be proportional to the molar
concentration of the reactants raised to a simple power (which
need not be integral). This relationship is called the rate equation.
The manner in which the reaction rate changes in magnitude with
changes in the magnitude of the concentration of each participating
reactant species is called the reaction order.
Problem Example:
The rate formation of NO in the reaction:
2NOBr(g) ----> 2NO(g) + Br2(g)
was reported as 1.6 x 10-4 mol/(L s).
What is the rate of reaction and the rate of consumption
of NOBr ?
Log R
Log R
Slope = α
Slope = β
Log [A]
Log [B]
k
xA + yB
→
Products
empirical rate
equation (obtained
from experiment)
rate constant k
R=−
stoichiometric
coefficients
Rate equation can not in
general be inferred from
the stoichiometric equation
for the reaction.
1 d [ A]
1 d [B ]
α
β
=−
= k [ A] [B ]
x dt
y dt
α, β = reaction
orders for the
reactants (got
experimentally)
Different rate equations
imply different mechanisms.
H 2 + I 2 → 2 HI
d [HI ]
= k [H 2 ][I 2 ]
dt
H 2 + Br2 → 2 HBr
R=
d [HBr ] k [H 2 ][Br2 ]
R=
=
k ′[HBr ]
dt
1+
[Br2 ]
1/ 2
H 2 + Cl2 → 2 HCl
d [HCl ]
1/ 2
= k [H 2 ][Cl2 ]
R=
dt
H 2 + X 2 → 2 HX
X = I , Br , Cl
• The rate law provides an important guide
to reaction mechanism, since any proposed
mechanism must be consistent with the
observed rate law.
• A complex rate equation will imply a complex
multistep reaction mechanism.
• Once we know the rate law and the rate
constant for a reaction, we can predict the
rate of the reaction for any given composition
of the reaction mixture.
• We can also use a rate law to predict the
concentrations of reactants and products at
any time after the start of the reaction.
Zero order kinetics. The reaction proceeds at the same rate R
regardless of concentration.
Rate equation :
da
R=−
=k
dt
a = a0 when t = 0
0
units of rate constant k :
mol dm-3 s-1
[ A]
integrate
using initial
condition
a (t ) = −kt + a0
half life : t = τ 1/ 2 when a =
a
τ 1/ 2 =
slope = -k
a0
2k
a0
2
τ 1 / 2 ∝ a0
τ 1/ 2
a0
diagnostic
plot
R ∝ [ A]
R
slope =
t
a0
1
2k
k
A
→
products
First order kinetics.
Initial
concentration a0
rate = −
First order differential
rate equation.
da
−
= ka
dt
d [ A] t
dt
Initial condition
t = 0 a = a0
Solve differential
equation
Separation
of variables
For a first order reaction the relationship:
(rate) t ∝ [ A] t
(rate) t = k [ A] t
a (t ) = a0 e − kt = a0 exp[− kt ]
is valid generally for any time t.
k is the first order
rate constant, units: s-1
Reactant concentration
as function of time.
First order kinetics.
Half life τ1/2
t = τ 1/ 2
θ = θ1/ 2
τ 1/ 2 =
k → s −1
a (t ) = a0 e − kt = a0 exp[− kt ]
a (t )
= exp[− θ ]
u=
a0
θ = kt
a0
2
u = 1/ 2
a=
ln 2 0.693
=
k
k
Mean lifetime of reactant molecule
1
τ=
a0
∫
∞
0
1
a(t )dt =
a0
∫
∞
0
a0 e − kt dt =
1
k
First order kinetics: half life.
u=
a
a0
u =1
t = τ 1/ 2
In each successive period
of duration τ1/2 the concentration
of a reactant in a first order reaction
decays to half its value at the start
of that period. After n such periods,
the concentration is (1/2)n of its
initial value.
a = a0 / 2
u = 0 .5
u = 0.25
u = 0.125
τ 1/ 2
τ 1/ 2 =
a0
ln 2 0.693
=
k
k
half life independent of
initial reactant concentration
1st order kinetics
u (θ )
a (t ) = a0 e
u (θ ) = e −θ
θ = kt
2nd order kinetics
− kt
u (θ )
a (t ) =
a0
1 + ka0t
u (θ ) =
θ = ka0t
1
1+θ
Second order kinetics: equal
reactant concentrations.
k
2A
→
P
dm3mol-1s-1
slope = k
da
= ka 2
dt
t = 0 a = a0
−
separate variables
integrate
1
1
= kt +
a
a0
half life t = τ 1/ 2 a = a0
2
τ 1/ 2 =
1
a
1
ka0
τ 1/ 2
t
a0
a (t ) =
1+ ka0t
1
τ 1/ 2 ∝
a0
τ 1/ 2 ↓ as a0 ↑
a0
rate varies as
square of reactant
concentration
1st and 2nd order kinetics : Summary .
Reaction
k1
A →
Products
k2
2 A →
Products
Differential
rate equation
Concentration
variation with
time
da
−
= k1a
dt
a (t ) =
a0 exp[− k1t ]
a (t ) =
da
2
−
= k2 a
a0
dt
Diagnostic
Equation
ln a(t ) = −k1t + ln a0
1
1
= k 2t +
a (t )
a0
Half
Life
τ 1/ 2 =
ln 2
k1
τ 1/ 2 =
1
k 2 a0
1 + k 2 a0t
ln a(t)
Slope = - k1 1/a(t)
τ1/2
Slope = k2
t
t
1st order
2nd order
a0
Diagnostic
Plots .
k
nA
→
P
n th order kinetics: equal reactant
concentrations.
da
−
= ka n
dt
t = 0 a = a0
separate variables
integrate
a
n ≠1
n = 0, 2,3,…..
1
a n −1
1
n −1
= (n − 1)kt +
1
a0
n −1
rate constant k
obtained from slope
Half life
τ 1/ 2 =
n −1
2 −1
(n − 1)ka0 n−1
τ 1/ 2 ∝ a01− n
n > 1 τ 1/ 2 ↓ as a0 ↑
n < 1 τ 1/ 2 ↑ as a0 ↑
2 −1
ln τ 1/ 2 = ln
− (n − 1) ln a0
(
)
n
1
k
−
n −1
slope = (n − 1)k
t
ln τ 1/ 2
slope = −(n − 1)
reaction order n determined
from slope
ln a0
Summary of kinetic results.
t = 0 a = a0
k
nA
→
P
t = τ 1/ 2
a=
Rate equation
a0
2
Integrated
expression
Units of k
k
a(t ) = −kt + a0
Half life
τ1/2
mol dm-3s-1
a0
2k
1
ka
a
ln 0 = kt
a (t )
s-1
ln 2
k
2
ka
2
1
1
= kt +
a (t )
a0
dm3mol-1s-1
1
ka0
ka
3
1
1
2
=
kt
+
2
2
a(t )
a0
dm6mol-2s-1
3
2
2ka0
ka
n
1
Reaction
Order
0
3
n
da
R=−
dt
a n −1
= (n − 1)kt +
1
a0
n −1
1 2 n −1 − 1
n − 1 ka 0 n −1
Problem Examples:
1. A first order reaction is 40% complete at the end of 1 h.
What is the value of the rate constant?
In how long willl the reaction be 80 % complete ?
2. The half-life of the radioactive disintegration of radium is
1590 years. Calculate the decay constant. In how many
years will three-quarters of the radium have undergone
decay ?
3. Derive the rate equation for reaction whose orders are:
(a). one-half, (b). three and a-half, (c). 4, and (d). n !
Chemical Kinetics
Lecture 2.
Kinetics of more complex
reactions.
Jaslin Ikhsan, Ph.D.
Kimia FMIPA
Universitas Negeri Jogjakarta
Tujuan Perkuliahan:
1. Menurunkan Rate Law dari Kinetika
Reaksi yang lebih Kompleks,
2. Menjelaskan Consecutive Reactions:
a. Rate Determining Steps,
b. Steady State Approximation.
Reference:
1. P.W. Atkins, Physical Chemistry, 5th ed,
Oxford: 1994
2. Ira N. Levine, Physical Chemistry
Second order kinetics:
Unequal reactant concentrations.
A+ B
→ P
k
rate equation
R=−
da
db dp
=−
=
= kab
dt
dt dt
initial conditions
t = 0 a = a0
b = b0
a0 ≠ b0
integrate using
partial fractions
1
F (a, b ) =
b0 − a0
b b0
= kt
ln
a a0
dm3mol-1s-1
F (a, b )
slope = k
t
Consecutive Reactions .
•Mother / daughter radioactive
decay.
218
Mass balance requirement:
Po→ 214 Pb→ 214 Bi
k1 = 5 × 10 −3 s −1
k1
k2
A →
X →
P
k 2 = 6 × 10 − 4 s −1
p = a0 − a − x
The solutions to the coupled
equations are :
3 coupled LDE’s define system :
a (t ) = a0 exp[− k1t ]
da
= −k1a
dt
dx
= k1a − k 2 x
dt
dp
= k2 x
dt
x(t ) =
k1a0
{ exp[− k1t ] − exp[− k2t ] }
k 2 − k1
p(t ) = a0 − a0 exp[− k1t ] −
k1a0
{ exp[− k1t ] − exp[− k2t ] }
k 2 − k1
We get different kinetic behaviour depending
on the ratio of the rate constants k1 and k2
Consecutive reaction : Case I.
Intermediate formation fast, intermediate decomposition slow.
Case I .
κ=
k2
>
TS I
II : fast
∆GII‡
energy
I : slow rds
k2
κ=
k1
TS II
∆GI‡
∆GII‡
A
Step I rate determining
since it has the highest
activation energy barrier.
X
P
reaction co-ordinate
k1
k2
A →
X →
P
normalised concentration
Case
CaseIIII..
κ=
1.2
k2
>> 1
k1
k 2 >> k1
Intermediate X
is fairly reactive.
Concentration of
intermediate X
will be small at
all times.
1.0
Product P
0.8
u=a/a0
v =x/a0
0.6
w=p/a0
0.4
Reactant A
0.2
κ = k 2/k1 = 10
Intermediate X
0.0
0
2
4
6
8
10
τ = k1t
normalised concentration
1.2
u=a/a0
1.0
v=x/a0
Intermediate concentration
is approximately constant
after initial induction period.
w=p/a0
0.8
P
A
0.6
0.4
X
0.2
0.0
0.0
0.2
0.4
0.6
τ = k1t
0.8
1.0
k1 >> k 2
Rate Determining Step
A → X → P
k1
Fast
k2
Slow
• Reactant A decays rapidly, concentration of intermediate species X
is high for much of the reaction and product P concentration rises
gradually since X--> P transformation is slow .
k 2 >> k1
Rate Determining
Step
A → X → P
k1
k2
Slow
Fast
• Reactant A decays slowly, concentration of intermediate species X
will be low for the duration of the reaction and to a good approximation
the net rate of change of intermediate concentration with time is zero.
Hence the intermediate will be formed as quickly as it is removed.
This is the quasi steady state approximation (QSSA).
Parallel reaction mechanism.
k1
A →
X
k2
• We consider the kinetic analysis of a concurrent
A →
Y
or parallel reaction scheme which is often met in
k1, k2 = 1st order rate constants
real situations.
• A single reactant species can form two
We can also obtain expressions
distinct products.
for the product concentrations
st
We assume that each reaction exhibits 1 order x(t) and y(t).
kinetics.
• Initial condition : t= 0, a = a0 ; x = 0, y = 0
.• Rate equation:
da
= k1a + k 2 a = (k1 + k 2 )a = kΣ a
R=−
dt
a (t ) = a0 exp[− kΣt ] = a0 exp[− (k1 + k 2 )t ]
• Half life: τ 1/ 2 = ln 2 = ln 2
kΣ
k1 + k 2
dx
= k1a = k1a0 exp[− (k1 + k 2 )t ]
dt
x(t ) = k1a0
x(t ) =
∫ exp[− (k
t
1
0
+ k 2 )t ] dt
k1a0
{1 − exp[− (k1 + k2 )t ]}
k1 + k 2
dy
= k 2 a = k 2 a0 exp[− (k1 + k 2 )t ]
dt
y (t ) = k 2 a0
y (t ) =
∫ exp[− (k
t
0
1
+ k 2 )t ] dt
k 2 a0
{1 − exp[− (k1 + k2 )t ]}
k1 + k 2
x(t ) k1
• All of this is just an extension of simple Final product analysis
=
Lim
st
yields
rate
constant
ratio.
t
→
∞
y
(
t
)
k2
1 order kinetics.
Parallel Mechanism: k1 >> k2
normalised concentration
1.0
Theory
k
κ = 2 = 0.1
k1
k1
v(∞ )
= 10 =
k2
w(∞)
κ = 0.1
a(t)
0.8
v(∞ ) = 0.9079
x(t)
u(τ)
v(τ)
w(τ)
0.6
0.4
κ = k2 /k1
/
0.2
y(t)
w(∞ ) = 0.0908
0.0
0
1
2
3
τ = k1t
4
5
6
Computation
v(∞ ) 0.9079
≅
= 9.9989
w(∞ ) 0.0908
Parallel Mechanism: k2 >> k1
normalised concentration
1.0
Theory
u(τ)
v(τ)
w(τ)
0.6
0.4
a(t)
0.2
k2
= 10
k1
k1
v(∞ )
= 0.1 =
k2
w(∞)
v(∞) ≅ 0.0909
x(t)
0.0
0
κ=
w(∞) ≅ 0.9091
y(t)
κ = 10
0.8
1
2
3
τ = k1t
4
5
6
Computation
v(∞ ) 0.0909
≅
= 0.0999
w(∞ ) 0.9091
Reaching Equilibrium on the
Macroscopic and Molecular Level
N2O4
NO2
N2O4 (g)
colourless
2 NO2 (g)
brown
Chemical Equilibrium :
a kinetic definition.
•
•
Countless experiments with chemical
systems have shown that in a state of
equilibrium, the concentrations of
reactants and products no longer change
with time.
This apparent cessation of activity occurs
because under such conditions, all
reactions are microscopically reversible.
We look at the dinitrogen tetraoxide/
nitrogen oxide equilibrium which
occurs in the gas phase.
N2O4 (g)
colourless
Kinetic analysis.
r
R = k [N 2O4 ]
s
2
R = k ′[NO2 ]
Valid for any time t
2 NO2 (g)
brown
Concentrations vary
with time
[NO2 ] t ↑
[N 2O4 ] t
concentration
•
Equilibrium:
r s
t →∞
R=R
2
k [N 2O4 ]eq = k ′[NO2 ]eq
[NO2 ]eq2 k
= =K
[N 2O4 ] k ′
↓
Kinetic
regime
Concentrations time
invariant
[NO2 ] eq
[N 2O4 ] eq
Equilibrium
state
NO2
t →∞
N2O4
time
Equilibrium
constant
First order reversible reactions :
understanding the approach to chemical equilibrium.
Rate equation
k
A
B
k'
u + v =1
τ = 0 u =1 v = 0
da
= −ka + k ′b
dt
Rate equation in normalised form
Initial condition
t = 0 a = a0
du
1
+u =
dτ
1+θ
b=0
Mass balance condition
Solution produces the concentration expressions
∀t
a + b = a0
u (τ ) =
v(τ ) =
Introduce normalised variables.
u=
a
a0
v=
b
a0
τ = (k + k ′)t θ =
k
k′
1
{1 + θ exp[− τ ]}
1+θ
θ
1+θ
{1 − exp[− τ ]}
Q(τ ) =
Reaction quotient Q
v(τ )
=θ
u (τ )
1 − exp[− τ ]
[
]
+
−
1
θ
exp
τ
First order reversible reactions: approach to equilibrium.
Kinetic
regime
1.0
concentration
0.8
Equilibrium
Product B
K = Q(τ → ∞ )
0.6
u (τ)
v (τ)
0.4
=
Reactant A
0.2
0.0
0
2
4
τ = (k+k')t
6
8
v(∞ )
u (∞ )
Understanding the difference between reaction quotient Q and
Equilibrium constant K.
12
Approach to
Equilibrium
Q