Transcript Ch #16: Kinetics

```Ch #16: Kinetics
• Kinetics: Rates and Mechanisms
of Chemical Reactions
Chemical kinetics:
• Study of reaction rates, changes in
concentrations of reactants or products as a
function of time.
Factors that influence reaction
rate:
1.Concentration - molecules must collide to
react;
Rate α collision frequency α conc
2.Physical state - molecules must mix to collide;
Smaller the particle size, greater the surface area
and more the collisions.
3.Temperature - molecules must collide with
enough energy to react;
Rate α collision frequency α conc.
Figure 16.3
The effect of surface area on reaction rate.
Figure 16.4
Collision energy and reaction rate.
Factors that influence reaction rate:
4.The use of a catalyst.
Expressing the Reaction Rate
• reaction rate - changes in the concentrations of
reactants or products per unit time
• reactant concentrations decrease while product
concentrations increase
• For a reaction A → B
• Rate of reaction = -∆[A]/∆t
• [A]= conc of A in mol/L
• ∆=change
Expressing the Reaction Rate
• Units= moles per liter per second.
• Mol L-1s-1 or mol/ L.s
• Change I product conc is positive so the rate is
Rate of reaction = ∆[B]/∆t
• Rate decreases during course of reaction.
• Instantaneous rate: Rate at a particular instant:
considering closer values.
Expressing the Reaction Rate
• Use initial rate as soon as the reactants are
present( no products at this time)
• For a reaction aA + bB →cC+ dD,
• Rate =-1/a∆[A]/∆t=1/b∆[B]/∆t==1/c∆[C]/∆t==1/d∆[A]/∆t
Table 16.1 Concentration of O3 at Various Time in its
Reaction with C2H4 at 303K
C 2 H 4 (g) + O
Time (s)
-
 (conc A)
t
3 (g)
C
2 H 4 O(
g) + O
2 (g)
Concentration of O3 (mol/L)
0.0
3.20x10-5
10.0
2.42x10-5
20.0
1.95x10-5
30.0
1.63x10-5
40.0
1.40x10-5
50.0
1.23x10-5
60.0
1.10x10-5
Figure 16.5
The concentrations of O3 vs. time during its reaction with C2H4
C 2 H 4 (g) + O
3 (g)
rate =
 [C2H4]
t
-
-
 [O3]
t
=
C
2 H 4 O(
g) + O
2 (g)
Figure 16.6
Tools of the
Laboratory
Plots of [C2H4] and [O2] vs. time.
Problem 1
• P.1.2H2 (g) +O2 (g)→ 2H2O (g) Write the rate
equation in terms of reactants and products. If
[O2] is decreasing at 0.23 mol/L .s at what rate
is [H2O] increasing?
The Rate Law and its components:
• Rate = k[A]m [B]n
• K= rate constant(does not change as reaction
proceeds)
• M and n are reaction orders.
• Coefficients are not related to reaction orders.
• Rate constant and orders can only be found by
experimental data.
Determining the Initial Rate:
• By performing expt and collecting data.
Reaction order:
• Rate=k[A]- 1st order
• Rate = k[A]2- 2nd order
• Rate = k[A]0- Zero order OR Rate=k(1)=k
(not dependent on conc of A)
Problem 2
• P.2.For each of the following reactions,
determine the reaction order with respect to
each reactant and the overall order from the
given rate law.
• 2NO(g) + O2(g) → 2NO2(g); rate =
k[NO]2[O2]
• CH3CHO(g) →
CH4(g) + CO(g); rate =
k[CH3CHO]3/2
Problem 2
• H2O2(aq) + 3I-(aq) + 2H+(aq) →
(aq) + 2H2O(l); rate = k[H2O2][I-]
I3-
Determining Reaction Orders:
• P.3.NO2(g) + CO (g) → NO (g) + CO2(g)
rate=k[NO2]m [CO]n
Sample Problem 16.3 Determining Reaction Order from Initial Rate Data
PROBLEM:
Many gaseous reactions occur in a car engine and exhaust
system. One of these is
N O 2 (g) + C O (g)
N O (g) + C O 2 (g)
rate = k[NO2]m[CO]n
Use the following data to determine the individual and overall reaction orders.
Experiment
1
Initial Rate(mol/L*s) Initial [NO2] (mol/L) Initial [CO] (mol/L)
0.10
0.40
0.10
2
0.0050
0.080
3
0.0050
0.10
0.20
PLAN:
0.10
Solve for each reactant using the general rate law using the
method described previously.
SOLUTION:
rate = k [NO2]m[CO]n
First, choose two experiments in which [CO] remains
constant and the [NO2] varies.
Sample Problem 16.3 Determining Reaction Order from Initial Rate Data
continued
rate 2
=
rate 1
0.080
0.0050
=
rate 3
rate 1
=
0.0050
0.0050
k [NO2]m2[CO]n2
k [NO2
]m
0.40
m
=
1
m
The reaction is
2nd order in NO2.
[NO2] 1
16 = 4m and m = 2
;
0.10
k [NO2]m3[CO]n3
k [NO2]m1 [CO]n1
0.20
=
1
[CO]n
[NO2] 2
=
[CO] 3
n
[CO] 1
n
;
1 = 2n and n = 0
0.10
rate = k [NO2]2[CO]0 = k [NO2]2
The reaction is
zero order in CO.
Determining the Rate constant:
• Calculated from collected data.
• Units of Rate constant depend on order.
Table 16.4 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
Figure 16.7
Integrated rate laws and reaction order
1/[A]t = kt + 1/[A]0
ln[A]t = -kt + ln[A]0
[A]t = -kt + [A]0
Problem 5
• P.5At 1000 o C, cyclobutane (C4H8)
decomposes in a first-order reaction, with the
very high rate constant of 87s-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?
Rules for orders through graphs
•
•
•
•
If a straight line is produced with the foll:
Ln[reactant] vs time-1st
1/[reactant] vs time-2nd
[reactant]vs time -0 order.
Half Life
• Half Life: Time required for the reactant
concentration to reach half of its initial value.
Figure 16.9
A plot of [N2O5] vs. time for three half-lives.
for a first-order process
t1/2 =
ln 2
k
=
0.693
k
Problem 6
• P.6 Cyclopropane is the smallest cyclic hydrocarbon.
Because its 600 bond angles allow poor orbital
overlap, its bonds are weak. As a result, it is
thermally unstable and rearranges to propene at
10000C via a first-order reaction:
• The rate constant is 9.2s-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?
Effect of Temperature:
• Temperature affects rate by affecting rate
constant.
• Arrhenius equ: k=Ae-Ea/RT
• K=rate const, e=base of natural logs,
T=absolute temp, R= Universal gas const,
Ea=activation energy (minimum energy that
molecules must have to react)
Effect of Temperature:
•
•
•
•
•
•
As T increases, k increases, rate increases.
Ln k =ln A- Ea/R(1/T)
A plot of ln k vs 1/T gives a straight line
Slope=-Ea/R
Y-int= ln A
Ln K2/k1=-Ea/R(1/T2-1/T1)
The Effect of Temperature on Reaction Rate
The Arrhenius Equation
k  Ae

Ea
where k is the kinetic rate constant at T
RT
Ea is the activation energy
R is the energy gas constant
ln k = ln A - Ea/RT
ln
k2
k1
Ea
= RT
T is the Kelvin temperature
A is the collision frequency factor
1
T2
1
T1
Figure 16.11 Graphical determination of the activation energy
ln k = -Ea/R (1/T) + ln A
Problem 7
P.7.The decomposition of hydrogen iodide
2HI(g)
→ H2(g) + I2(g)
• has rate constants of 9.51x10-9L/mol*s at 500.
K and 1.10x10-5 L/mol*s at 600. K. Find Ea.
Figure 16.12
Information sequence to determine the kinetic parameters of a reaction.
Series of plots
of concentration vs. time
Initial
rates
Determine slope
of tangent at t0 for
each plot
Plots of
concentration
vs. time
Reaction
Rate constant
orders
(k) and actual
Compare initial
rate law
rates when [A]
Substitute initial rates,
changes and [B] is
orders, and concentrations
Find k at
held constant and
into general rate law:
varied T
m
n
vice versa
rate = k [A] [B]
Integrated
rate law
(half-life,
t1/2)
Use direct, ln or
inverse plot to
find order
Rate constant
and reaction
order
Rearrange to
linear form and
graph
Activation
energy, Ea
Find k at
varied T
Effects of concentration and
temperature:
• Collision Theory: Reactant particles must
collide with each other.
• Concs are multiplied in rate law.
• Ea: Energy required to activate the molecules
into a state from which reactant bonds can
change into product bonds.
• Only those collisions with enough energy to
exceed Ea can lead to reaction.
Figure 16.13
The dependence of possible collisions on the product
of reactant concentrations.
A
B
4 collisions
A
B
A
molecule of A
B
6 collisions
A
B
A
molecule of B
A
B
A
B
A
B
Effects of concentration and
temperature:
• Rise in temp enlarges the fraction of collisions
with enough energy to exceed Ea.
• F= e-Ea/RT e-base of natural log, T=temp, R=gas
const.
• In exothermic reaction: Eaf > Ear
• In endothermic reaction: Eaf < Ear
• : k=Ae-Ea/RT A is the frequency factor,
A=pZ, Z=collision frequency and
p=orientation probability factor.
Figure 16.14
The effect of temperature on the distribution of collision energies
Figure 16.15
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
Figure 16.17
The importance of molecular orientation to an effective collision.
NO + NO3
2 NO2
A is the frequency factor
A = pZ where
Z is the collision frequency
p is the orientation probability factor
Transition state theory:
• Transition state/activated complex: neither
reactant nor product present, a transitional
species with partial bonds is present.
• Ea is the energy required to stretch and deform
bonds in order to reach transition state.
• Reaction energy diagram : TB. Pg.699
Figure 16.18
Nature of the transition state in the reaction between CH3Br and OH-.
CH3Br + OH-
CH3OH + Br -
transition state or activated complex
Figure 16.19 Reaction energy diagram for the reaction of CH3Br and OH-.
Figure 16.20
Reaction energy diagrams and possible transition states.
Molecularity of a reaction:
• Elementary steps make up the reaction
mechanism.
• An elementary step is not made up of simpler
steps.
• Molecularity means the number of reactant
particles involved in the step.
Molecularity of a reaction:
•
•
•
•
•
In an elementary step equation coeff= order.
Reaction order= molecularity
A→ 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]
REACTION MECHANISMS
Table 16.6 Rate Laws for General Elementary 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
Problem 8
• P.8. 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) + Cl(g)
(a) Write the overall balanced equation.
• NO2Cl(g) → NO2(g) + Cl(g)
• NO2Cl(g) + Cl(g)→ NO2(g) + Cl(g)
Problem 8
(a) Write the overall balanced equation.
(b) Determine the molecularity of each step.
(c) Write the rate law for each step
Rate-Determining step:
•
•
•
•
Also called as rate limiting step.
It represents rate law for overall reaction.
Slow reaction is a rate determining reaction.
The elementary steps must add up to the
overall equation.
• The elementary steps must be physically
reasonable.
• The mechanism must correlated with the rate
law.
Catalysis:
• 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.
Catalysis:
• 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
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