Chemical Kinetics: Rates and Mechanisms of Chemical Reactions

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Learning Objectives:. Reaction Rate. Expressing the Reaction Rate. **The Rate Law and Its Components. **Integrated Rate Laws: Concentration Changes over Time. Catalysis: Speeding Up a Reaction. Theories of Chemical Kinetics. ID: 708139

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Chemical Kinetics: Rates and Mechanisms of Chemical Reactions

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Presentations text content in Chemical Kinetics: Rates and Mechanisms of Chemical Reactions

Chemical Kinetics:
Rates and Mechanisms of Chemical Reactions
Learning Objectives:
Reaction Rate
Expressing the Reaction Rate
**The Rate Law and Its Components
**Integrated Rate Laws: Concentration Changes over Time
Catalysis: Speeding Up a Reaction
Theories of Chemical Kinetics
*Reaction Mechanisms: The Steps from Reactant to Product
Chemical reactions: Fast or Slow
Explosion
Ripening of fruits/vegetables
Digestion
Fossilization from ancient vegetation to fossil fuel
Microscopic view between fast and slow reactions
Factors That Influence
Reaction Rate
Particles must collide in order to react: Greater collision rate, faster rate.Higher the concentration of reactants, reaction rateGreater number of collisions. Household bleach (CloroxTM
).
Physical state
of the reactants influences reaction rate. -
Substances must mix in order for particles to collide. Example:
Surface area (powder form vs. chunky form)Higher the temperature, reaction rate
At higher temperatures particles have more energy and therefore collide more often and more effectively. Milk spoils faster at higher temperature.
The effect of
Surface Area
on reaction rate
A hot steel nail glows feebly when placed in O
2
.
The same mass of steel wool (
more surface area
) bursts into flame, aka reacts much faster
Sufficient collision energy is required for a reaction to
occur
What is the
Reaction Rate
Reaction rate: the changes in concentrations of reactants or products per unit time.
For the general reaction A
→ B, we measure the concentration of A at
t
1
and at
t2:
“-” : because the concentration of A is decreasing. This gives the rate a positive value.
D[A]Dt
Rate =
change in concentration of A
change in time
conc A
2
- conc A
1
t
2
-
t
1
= -
= -
“[xxx]” = concentration in moles per liter.
Average rate
Instantaneous rate
= slope of tangent line at tInitial rate = instantaneous rate at t = 0
C
A
B
0
Question
: point
A or B?______ has the higher [O3]______ has higher reaction rate After reaction begins, the rate (increase or decrease)
A
B
1
Plots of [reactant] and [product] vs. time.
C
2H4 + O3 → C2
H
4
O + O
2
[O
2] increases just as fast as [C2H4] decreases.
Rate = -
D[C2H4]Dt
= -
D
[O
3
]
D
t
D
[C
2
H
4
O]
D
t
D
[O
2
]
D
t
=
=
Quick Question
:
a
or
b
?
Which has higher
D
[O
2
]
?
Which has higher reaction rate?
A
a
b
Key: b; a
2
Plots of [reactant] and [product] vs. time.
H
2 + I2 → 2HI
[HI] increases twice as fast as [H
2
] decreases.
Rate = -
D
[H
2
]Dt
= -
D
[I
2
]
D
t
=
D
[HI]
D
t
1
2
D
[I
2
]
D
t
Rate =
D
[H
2
]
D
t
D
[IH]
D
t
= -2
= -2
The expression for the rate of a reaction and its numerical value depend on which substance serves as the reference.
3
In general, for the reaction
a
A + bB → cC + dD
where
a
,
b
,
c, and d are the coefficients for the balanced equation, the rate in terms of each species is expressed as:
The Reaction Rates depends on the stoichiometry of Reaction
Analogy: For the same person, heart rate and breathing rate is not the same at the same moment.
_________________________________________
4
Example:
The balanced equation for the reaction of bromate ion with bromide in acidic solution is given by
BrO3- + 5Br- + 6H+ → 3Br2 + 3H2OAt a particular instant in time, the value of -△[Br-]/△t is 1.0 × 10-2 mol/L ∙ s. What is the value of △[Br2
]/
△
t
in the same unit
?
6.0 × 10-3 mol/L ∙ sPractice: What is the value of -△[H+]/△
t in the same unit?
5
The Rate Law:
How Rate Depends on Concentrations
For any general reaction occurring at a fixed temperature
a
A +
b
B
→
cC + dD
Rate = k[A]m[B]n
k = Rate Constant
m and n : Reaction Orders, determined by experiment.k depends on temperature and catalyst, not affected by concentrations.
Order:
zero order;
first
order;
second
order
6
Individual and Overall Reaction Orders
For the reaction 2NO(
g) + 2H2(g) → N2(
g
) + 2H
2
O(
g):
The rate law is rate = k[NO]2[H2]
The above reaction is 2nd order for NO, 1st order with respect to H2; 3rd order overall.
If the concentration of a reactant does NOT affect the rate, the order of this reactant is ______ order.
7
Plots of
Reactant concentration, [A], vs. T
ime
8
Plots of
Rate vs. Reactant
Concentration [A]
9
Determining Reaction Orders: A. Initial Rate Method
For the general reaction
aA + bB → d
C
+
d
D
, rate law as Rate = k[A]m[B]n
To determine the values of m and n, we run a series of experiments and find the order ONE by ONE
To find n, only ___ is changed, measure new initial rate3. Set first trial ([A]1 and [B]1
measure initial rate1 at time = 0).To find m, only [A] is changed while [B] unchanged, measure new initial rate2
.
0
Initial Rate Method to Determine the order
Experiment
Initial Rate
(mol/L·s)
Initial [A]
(mol/L)
Initial [B]
(mol/L)
1
1.75x10-32.50x10-1
3.00x10-12
3.50x10-35.00x10-1
3.00x10
-1
3
3.95x10
-3
2.50x10
-1
4.50x10
-1
The mathematical way to find the order:
order = ______________
1
Finding the order with respect to A
compare Trial# 1 and ___, where [B]
is kept
constant but [A]
changes:
m
=
Rate =
k
[A]
m[B]nTrial
Initial Rate(mol/L·s)
Initial [A]
(mol/L)
Initial [B]
(mol/L)
1
1.75x10
-3
2.50x10
-1
3.00x10
-1
2
3.50x10
-3
5.00x10
-1
3.00x10
-1
3
3.95x10
-3
2.50x10
-1
4.50x10
-1
2
Practice: Find
the order with respect to
B
compare Trial# 1 and ___, where
[A] is kept
constant but
[B] changes:
n
=
Rate = k[A]m[B]nTrial
Initial Rate(mol/L·s)
Initial [A]
(mol/L)
Initial [B]
(mol/L)
1
1.75x10
-3
2.50x10
-1
3.00x10
-1
2
3.50x10
-3
5.00x10
-1
3.00x10
-1
3
3.95x10
-3
2.50x10
-1
4.50x10
-1
3
Practice: Finding the rate constant
k
Choose
any trial to determine the rate
constant
k
k
=
__________
Trial 1: k
= 0.0778 mol-2 L2 s-1
Rate = k[A]1[B]2
Trial
Initial Rate
(
mol
/L·s)
Initial [A]
(
mol
/L)
Initial [B]
(
mol
/L)
1
1.75x10
-3
2.50x10
-1
3.00x10
-1
2
3.53x10
-3
5.00x10
-1
3.00x10
-1
3
3.92x10
-3
2.50x10
-1
4.50x10
-1
Your k
= __________________
4
Easier Way to Find Reaction Orders?
Double the concentration(s)
For the simple reaction A → products:
If the rate
doubles
when [A]
doubles
, rate
 [A]1, reaction is _____ order with respect to A.
If the rate quadruples when [A] doubles, rate  [A]2, reaction is __________ order with respect to [A].
If the rate does not change when [A] doubles, rate  [A]0, reaction is _______ order with respect to A.
5
Units of the Rate Constant
k for Common Reaction Orders
Overall Reaction OrderUnits of k
(
t
in seconds)
0
mol/L·s
(or mol L-1s-1)
11/s (or s-1)
2L/mol·s (or L mol-1s
-1)3L
2
/mol
2
·s
(or L
2
mol
-2
s
-1
)
General formula:
L
mol
unit of
t
order-1
Units of
k
=
The value of
k
is easily determined from experimental rate data. The
units
of
k
depend on the overall reaction order.
6
Information sequence to determine the kinetic parameters of a reaction.
Series of plots of concentration vs. time
Initial rates
Reaction orders
Rate constant (
k
) and actual rate law
Determine slope of tangent at t
0
for each plot.
Compare initial rates when
[A]
changes and
[B]
is held constant (and vice versa).
Substitute initial rates, orders, and concentrations into
rate =
k
[A]
m
[B]
n
,
and solve for k.
7
Integrated Rate Laws
An integrated rate law includes
time as a variable.
First-order rate equation:
rate = -

[A]

t
=
k
[A]
Second-order rate equation:
rate = -

[A]

t
=
k
[A]
2
Zero-order rate equation:
rate = -

[A]

t
=
k
[A]
0
=
k
Remember both rate and concentration depends on time
Integrated Rate Law:
_____________________
_____________________
_____________________
8
Graphical
presentation of 1st
order reactionFirst-order reaction
ln
[A]
0
[A]
t
=
kt
integrated rate law
ln[A]
t
= _______ + _____
straight-line form
A plot of
ln[A]
vs.
time
gives a straight line for a first-order reaction.
Rate constant
k
= - slope
9
Graphical
presentation of 2nd
order reactionSecond-order reaction
integrated rate law
1
[A]
t
1
[A]
0
-
=
kt
straight-line form
1
[A]
t
=
kt
+
P
lot
of
1/[A]
vs.
time: linear.
Rate
constant
k
= slope
y = mx + b
0
Graphical
presentation of zero order reaction
Zero-order reaction
P
lot
of
[A] vs. time
gives a straight line for a first-order reaction.
Rate constant k = - slope
integrated rate law
[A]t - [A]0 = - kt
straight-line form
[A]
t
= _____ + ____
1
Example:
Determining the Reactant Concentration after a Given Time
At 1000
o
C,
cyclobutane
(C
4
H8) 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 2.00 M, what is the concentration after 0.010 s?
0.83 mol/L
Given: k = 87 s-1, t = 0.010 s, [A]0 = 2.00 M, ln([A]t/[A]0) = - ktFind: [A]t
2
Example:
Apply Integrated rate law in Problem solving
At a certain temperature,
cyclobutane
(C
4
H
8
) decomposes in a first-order reaction, with a rate constant of 1.0 s-1.
(b) What fraction of C4H8 has decomposed after 3.0 sec?
0.95
Given: k = 1.0 s-1, t = 3.0 s, ln([A]t/[A]0) = - kt
Find: 1 - [A]
t
/[A]
0
3
Example:
Apply Integrated rate law in Problem solving
At a certain temperature,
cyclobutane
(C
4
H
8
) decomposes in a first-order reaction. After 10.0 seconds, 20.0% of cyclobutane decomposed. Determine the rate constant at this temperature.
0.0223 s-1
Given: t = 10.0 s, 1 - [A]t/[A]0) = 0.200Find: k
4
Graphical determination of
Reaction order: Decomposition of N
2O5
Three ways to graph:
[A] vs. time
ln[A] vs. time
1/[A] vs. time
Since the plot of ln[N2O5] vs. time gives a straight line, the reaction is first order.
5
Reaction Half-life
The
half-life (t1/2) for a reaction is the time taken for the concentration of a reactant to drop to half its initial value.
For a
first-order
reaction,
t
1/2
does not depend on the starting concentration.
t1/2 =
ln 2
k
=
0.693
k
The half-life for a first-order reaction is a
constant
, independent on the concentration of reactant
6
[
N
2O5] vs. time for three reaction half-lives.
t
1/2
=
for a first-order process
ln 2
k
0.693
k
=
7
Half-life Equations
For a
first-order reaction, t1/2 does not depend on the initial concentration.
For a
second-order
reaction,
t
1/2
is inversely proportional to the initial concentration:
1k[A]0
t1/2 =
For a
zero-order
reaction,
t
1/2
is
directly
proportional to the initial concentration:
[A]
0
2k
0
t
1/2
=
8
Practice:
Determining the Half-Life of a First-Order Reaction
At 1000
°
C, the rearrangement reaction of cyclopropane has rate constant of 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-tenth of the initial value?
a. Half life t1/2 = 0.693/k = 0.075 s
b. 0.25 s
9
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 kmol/L·s1/s
L/mol·sHalf-life
Integrated rate law
in straight-line form
[A]
t
= -
kt
+ [A]
0
ln[A]
t
= -
kt
+ ln[A]
0
Plot for straight line
[A]
t
vs.
t
ln[A]
t
vs.
t
vs.
t
Slope,
y
intercept
-
k
, [A]
0
-k
, ln[A]
0
k
,
[A]
0
2
k
ln 2
k
1
k[
A]
0
1
[
A]
t
=
kt +
1
[
A]
0
1
[
A]
t
1
[
A]
0
0
Collision Theory and Concentration
The basic principle of
collision theory is that particles must collide in order to react.
An increase in the concentration of a reactant leads to
higher frequency
of collisions, hence increasing reaction rate.
The number of collisions depends on the
product
of the numbers of reactant particles, not their sum.
Concentrations are multiplied in the rate law, not added.
1
Temperature Affects Rate Constant
Temperature has a dramatic effect on reaction rate.
For many reactions, an increase of 10°C will double or triple the rate.
Experimental data shows that
k increases exponentially as T increases
: the
Arrhenius equation
k
=
Ae
-Ea/RT
k
= rate constant
A
= frequency factor
E
a
= activation energy
Higher
T
increased rate
larger
k
k
=
Ae
-E
a
/RT
k
= rate constant
A
= frequency factor
E
a
= activation energy
2
Increase of the rate constant with temperature for the hydrolysis of an ester.
Expt
[Ester]
[H
2
O]
T
(K)
Rate
(mol/L·s)
k(L/mol·s)
10.100
0.200
288
1.04x10
-3
0.0521
2
0.100
0.200
298
2.20x10
-3
0101
3
0.100
0.200
308
3.68x10
-3
0.184
4
0.100
0.200
318
6.64x10
-3
0.332
Reaction rate and
k
increase exponentially as
T
increases.
3
Activation Energy
In order to be
effective, collisions between particles must exceed a certain energy threshold.
The
lower
the activation energy, the
faster
the reaction.
When particles collide effectively, they reach an
activated state. The energy difference between the reactants and the activated state is the activation energy (Ea) for the reaction.
Smaller Ea
increased rate
larger
f
larger
k
4
Energy-level diagram for a
reaction
Collisions must occur with sufficient energy to reach an activated state.
This particular reaction is reversible and is exothermic in the forward direction.
5
Temperature and Collision Energy
An increase in temperature causes an increase in the kinetic energy of the particles. This leads to
more frequent collisions
Also, at a higher temperature, higher fraction of molecules has the kinetic energy above the activation energy
E
a
6
The Effect of
Ea and T
on the Fraction (f) of Collisions with Sufficient Energy to Allow ReactionEa (kJ/mol)
f
(at
T
= 298 K)
50
1.70x10-9
757.03x10-14100
2.90x10-18
Tf (at E
a
= 50 kJ/
mol
)
25°C (298 K)
1.70x10
-9
35°C (308 K)
3.29x10
-9
45°C (318 K)
6.12x10
-9
7
Calculating Activation Energy
k
= Ae
-E
a
/RT
Arrhenius equation: Exponential form vs. Logarithm form
ln
k
= ln
A -
Ea 1R T
Use rates at two different temperatures:
At T
1
, ln
k
1
= ln
A
-
E
a
1
R
T
1
At T
2
, ln
k
2
= ln
A
-
E
a
1
R
T
2
E
a
= ___________________
8
Graphical determination of the Activation energy:
ln
k (as y) vs. 1/T (as x)
ln
k
= ln
A
-
Ea 1R T
Ea
= (-8.314 J/mol) x slope
Unit: Joule
A
= exp(
y
intercept)
Unit: same unit as rate constant
9
Find Activation Energy and A factor from graph
Expt
[Ester]
[H
2
O]
T
(K)
Rate
(mol/L·s)k
(L/mol·s)
10.100
0.200
288
1.04x10
-3
0.0521
2
0.100
0.200
298
2.20x10
-3
0101
3
0.100
0.200
308
3.68x10
-3
0.184
4
0.100
0.200
318
6.64x10
-3
0.332
Ea
= (-8.314 J/mol) x slope
A
=
exp
(
y
intercept
)
46.9 kJ/
mol
1.6 x 10
7
L/
mol
s
0
Practice:
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.
Ea = 1.76x105 J/mol = 1.76x102 kJ/mol
1
Molecular Structure and Reaction Rate
For a collision between particles to be effective, it must have both sufficient
energy and the appropriate relative orientation between the reacting particles.
k
=
Ae
-E
a
/RT
A
in the Arrhenius equation is the
frequency factor for the reaction.
A
=
p
Z
p
= orientation probability factor
Z = collision frequency
p
depends on the specific reaction and is related to the structural complexity of the reactants.
2
Molecular orientation for an Effective Collision
NO(
g
) + NO
3
(
g
) → 2NO2(g)
Only one relative orientation of these two molecules, Effective Orientation, that leads to an effective collision.
Think: Which bonds are broken? Which bonds are formed?
3
Transition State Theory
E
ffective collision between particles leads to the formation of a transition state or activated complex.
The
transition state
is an
unstable
species that contains
partial bonds. It is a transitional stage between reactants and products. Transition states cannot be isolated.
The transition state exists at the point of maximum potential energy. The energy required to form the transition state is the activation energy.
4
Transition state: Reaction
between BrCH3
and OH-
The transition state contains partial bonds (
dashed
) between C and Br and between C and O. It has a trigonal bypyramidal shape.
BrCH
3
+ OH- → Br- + CH3OH
5
Quantum mechanical Study of the reaction between BrCH
3
and OH-.
6
Reaction Energy Diagram: Endothermic vs. Exothermic Reaction
7
Example
Drawing Reaction Energy Diagrams and Transition States
For reaction: O
3
(
g
) + O(
g
) → 2O2(g)The Ea(fwd) is 19 kJ, and the DHrxn for the reaction as written is -392 kJ. Draw a reaction energy diagram, label the reactant, product, activation energy, reaction enthalpy change, and transition state. Predict a structure for the transition state, and calculate Ea(rev).
8
Reaction
Mechanism: Elementary Step
Two Step Mechanism: Decomposition of Ozone
Unlike a balanced equation, many chemical reactions do not occur with
all
the reactant molecules together turning into final products.
The
mechanism
of a reaction is the sequence of single reaction steps (elementary steps) that make up the overall equation. Driving direction from OC to LAX.
9
Reaction intermediates
Reaction
intermediate
is the species (molecules, atoms, ions, etc.) produced in the initial elementary
steps
,
typically
unstable and reactive, later consumed in the subsequent elementary steps.
0
Molecularity of Elementary Steps
Each elementary step is characterized by its
molecularity, the number of particles involved in the reaction: Unimolecular; Bimolecular; Termolecular
(three molecules)
1
Elementary Step
Molecularity
Rate Law
A product
2A product
A + B product
2A + B product
Unimolecular
Bimolecular
Bimolecular
Termolecular
Rate = [A]
Rate =
k
[A]
2
Rate =
k
[A][B]
Rate =
k
[A]
2
[B]
The rate law for an
elementary step
can
be deduced from the reaction stoichiometry –
reaction order equals molecularity
for an elementary step only
.
2
Example:
Reaction mechanism, Rate law for Elementary Steps, Reaction intermediate
The following elementary steps are proposed for a reaction mechanism:
NO
2
Cl(
g
)
→ NO2(g) + Cl(g) NO2Cl(g) + Cl(
g) → NO2(g) + Cl2(g)
Write the overall balanced equation.Write the rate law for each step.
Identify the reaction intermediate.
3
The Rate-Determining Step of a Reaction
The
slowest elementary step in a reaction is the rate-determining or rate-limiting step. Analogy: Traffic between Orange County and Riverside
The rate law for the rate-determining step becomes the rate law for the overall reaction.
The reaction NO
2
(
g
) + CO(
g) → NO(g) + CO2(g) has been proposed to occur by a two-step mechanism:
NO2(g) + NO2(g) → NO3(
g) + NO(g) [slow; rate-determining](2) NO3(g) + CO(g) → NO2(g) + CO2(g) [fast]
Observed rate law: rate =
k
[NO
2
]
2
4
Correlating Mechanism with
Rate Law
A valid mechanism must meet three criteria:
The elementary steps must add up to the overall balanced equation.
The elementary steps must be reasonable.
The mechanism must correlate with the
observed
rate law.
If first elementary step is the rate-determining, its rate law should match the experimental rate law
A mechanism is a
hypothesis
–we cannot prove it is correct, but if it is consistent with the data, and can be used to predict results accurately, it is a useful model for the reaction.
5
Example: Propose a Mechanism with a Slow Initial Step
The overall reaction 2NO
2(g) + F2(g) →2NO2F(
g
) has an
experimental rate law Rate =
k
[NO2][F2].
In the first elementary step is rate-determining step, the mechanism would be:
For step 1 (the slow step), rate1 = k1 x ______ which matches the experimental rate law
Step 1:
Step 2: The second NO2 molecule needs to be reactant so that the overall reaction has two NO2 molecules
6
Reaction energy
diagrams for
Two-step reaction(A) NO2 and F2 (B) NO and O2.
Each step
in multi-step mechanism
has its own transition
state
7
Catalysis: Speeding up a Reaction
A
catalyst is a substance that increases the reaction rate without itself being consumed in the reaction.
In general, a catalyst provides an alternative reaction pathway that has a
lower total activation energy
than the
uncatalyzed
reaction.
A catalyst will speed up
BOTH the forward and the reverse reactions.
A catalyst does not affect either DH or the overall yield for a reaction.
8
E
nergy diagram:
Catalyzed vs. Uncatalyzed
9
Decomposition
of H
2O2 catalyzed by bromide ionA small amount of NaBr is added to a solution of H
2
O
2
.
Oxygen gas forms quickly as Br
-(aq) catalyzes the H2O2 decomposition; the intermediate Br2 turns the solution orange.
0
Metal-catalyzed
hydrogenation of ethene
A heterogeneous catalyst is in a different phase than the reaction mixture.
1
Enzyme: Protein as Biological Catalyst
Enzyme active site (“Lock”) has
fixed shape, which matches the shape of its substrate(“Key”) and be specific to its substrate.
2
Induced-fit model for Enzyme
The enzyme
active site changes shape in order to bind its substrate(s) more effectively. The active site is
selective towards certain substrates.
3
Chlorofluorocarbons (CFC) destroy ozone layer over the Antarctic: Satellite images of the increasing size of the Antarctic ozone hole (
purple
).
4
Reaction energy diagram for breakdown of O
2
by Cl atoms. (Not drawn to scale.)
5
Concentration of O
3
at Various Times in its Reaction with C2H4 at 303 K
Time (s)
0.0
20.0
30.0
40.0
50.0
60.0
10.0
Concentration of O
3
(mol/L)
3.20x10
-5
2.42x10
-5
1.95x10
-5
1.63x10
-5
1.40x10
-5
1.23x10
-5
1.10x10
-5

[C
2
H
4
]

t
rate = -

[O
3
]

t
= -
Practice:
Estimate the rate between 10.0s and 20.0s.
Key:
4.7x10
-6
mol/L s
6
In general, for the reaction
a
A + bB → cC + dD
1
a
D
[A]
D
t
1
b
D
[B]
D
t
rate = -
= -
1
c
D
[C]
D
t
=
1
d
D
[D]
D
t
=
where
a
,
b
,
c
, and
d
are the coefficients for the balanced equation, the rate is expressed as:
The Reaction Rates depends on the stoichiometry of Reaction
Analogy: For the same person, heart rate and breathing rate is not the same at the same moment.
7
Rate 2
Rate 1
=
k
[A] [B]
k
[A] [B]
m
2
n
2
m1
n
1
Finding
m
, the order with respect to A:
We compare experiments 1 and 2, where [B] is kept constant but [A] doubles:
=
[A]
[A]
m
2
m
1
=
[A]
2
[A]
1
m
3.50x10
-3
mol/L
·s
1.75x10
-3
mol/L
·s
=
5.00x10
-2
mol/L
2.50x10
-2
mol/L
m
Dividing, we get 2.00 = (2.00)
m
so
m
= 1
Rate =
k
[A]
m
[B]
n
8
Rate 3
Rate 1
=
k
[A] [B]
k
[A] [B]
m
3
n
3
m1
n
1
Finding
n
, the order with respect to B:
We compare experiments 3 and 1, where [A] is kept constant but [B] doubles:
=
[B]
[B]
n
3
n
1
=
[B]
3
[B]
1
n
3.50x10
-3
mol/L
·s
1.75x10
-3
mol/L
·s
=
6.00x10
-2
mol/L
3.00x10
-2
mol/L
m
Dividing, we get 2.00 = (2.00)
n
so
n
= 1
9
Practice:
Determining Reaction Orders from Rate Data
PROBLEM:
For the following reaction:
NO
2
(
g
) + CO(
g) → NO(g) + CO2(g) rate = k[NO2]m[CO]nUse the following data to determine the individual and overall reaction orders:
ExperimentInitial Rate (mol
/L·s)Initial [NO2](mol/L)
Initial [CO]
(mol/L)
1
0.0050
0.10
0.10
2
0.080
0.40
0.10
3
0.0050
0.10
0.20
rate =
k
[NO
2
]
2
[CO]
0
or rate =
k
[NO
2
]
2
0
Group Practice: What is the rate law for the following reaction?
Initial Rate
Initial Reactant Concentrations (
mol
/L)
Experiment
(
mol
/L·s)[O2
][NO]1
3.21x10-31.10x10
-21.30x10-2
2
6.40x10
-3
2.20x10
-2
1.30x10
-2
3
28.8x10
-3
1.10x10
-2
3.90x10
-2
O
2
(
g
) + 2NO(
g
)
→ 2NO
2
(
g
)
Rate law: rate =
k
[O
2
]
m
[NO]
n
rate =
k
[O
2
][NO]
2
1
Integrated Rate Laws
An integrated rate law includes
time as a variable.
First-order rate equation:
rate = -

[A]

t
=
k
[A]
ln
[A]
0
[A]
t
=
kt
Second-order rate equation:
rate = -

[A]

t
=
k
[A]
2
1
[A]
t
1
[A]
0
-
=
kt
Zero-order rate equation:
rate = -

[A]

t
=
k
[A]
0
[A]
t
- [A]
0
= -
kt
2
Practice:
The reaction A → B + C
is second order in A. When [A]0 = 0.100 M, the reaction is 20.0% complete in 40.0 minutes. Calculate the value of the rate constant (in L/min ∙ mol).
[A]
t
= 0.080 M
k = 0.61 L/min mol
3
Practice:
The half-life of a certain first-order reaction is 25 minutes. What fraction of the original reactant concentration will be consumed after 2.0 hours?
k
= 0.02
7
7 min
-1
[A]
t/[A]0 = exp(-kt) = 0.036, 1 - [A]t
/[A]0 = 0.964
4
Calculating Activation Energy
k
= Ae
-E
a
/RT
E
a
can be calculated from the Arrhenius equation:
so ln
k = ln A -
Ea 1R T
straight-line form
k
2
k
1
ln
= -
E
a
R
−
1
T
1
1
T
2
If data is available at two different temperatures:
5
Group Practice:
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. Predict the rate constant at 700C.
Ea = 1.76x102 kJ/mol
First, find the A factor in Arrhenius equation; Then use Arrhenius equation to find k
6
SOLUTION:
rate
1 = k1[NO2
Cl]
rate
2
=
k2[NO2Cl][Cl]
(b)
(a) Writing the overall balanced equation:
(1) NO2Cl(g) → NO2(g) + Cl(g)
(2) NO2Cl(g) + Cl(g) → NO2(g) + Cl2(g)
2NO
2
Cl(
g
)
→ 2NO
2
(
g
) + Cl
2
(
g
)
The species does NOT appear in the overall reaction is the reaction intermediate.
(c)
PROBLEM:
The following elementary steps are proposed for a reaction mechanism:
(1) NO
2
Cl(
g
)
→ NO
2
(
g
) + Cl(
g
)
(2) NO
2
Cl(
g
) + Cl(
g
) → NO
2
(
g
) + Cl
2
(
g
)
(a)
Write the overall balanced equation.
(b)
Write the rate law for each step.
Identify the reaction intermediate
.
7
Mechanisms with a Slow Initial Step
The overall reaction 2NO
2(g) + F2(g) →2NO2
F(
g
) has an experimental rate law Rate =
k
[NO2][F2].
The accepted mechanism isNO2(g) + F2(g) → NO2F(g) + F(g
) [slow; rate determining]NO2(g) + F(g) → NO2F(g) [fast]
The elementary steps sum to the overall balanced equation:
2NO2(g) + F2(g)
→2NO
2
F(
g
)
Both steps are bimolecular and are therefore reasonable.
rate
1
=
k
1
[NO
2
[F
2
]
rate
2
=
k
2
[NO
2
][F]
Step 1 is the slow step, and rate
1
correlates with the
observed
rate law.
The mechanism is therefore reasonable.
8