kinetics ch 15 kinetics thermodynamics and kinetics are not directly related investigate the rest of...
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Kinetics
Ch 15
Kinetics
• Thermodynamics and kinetics are not directly related
• Investigate the rest of the reaction coordinate
• Rate is important!
Chemical Kinetics
• Kinetics – the study of the rates of chemical reactions
• Rate of reaction – change in concentration per unit time rate = Δ conc / Δ time
• Rate is generally not constant. It changes over the course of a reaction
• A B
A B
10 18 24 28 31 33
What is happening to the rate of the reaction as time progresses? Why?
Rate = Δ[B]/Δt = -Δ[A]/Δt
Rate = Δ[product]/Δt = -Δ[reactant]/Δt
A B
Rate of Reaction
2 N2O5 (g) 4 NO2 (g) + O2 (g)
Rate = Δ[O2]/ Δt
Stoichiometry important!
Rate = Δ[NO2]/ 4Δt = - Δ[N2O5]/ 2Δt
2 N2O5 (g) 4 NO2 (g) + O2 (g)
Defining Rate• Rate is defined arbitrarily by one pdt or rxt• To be self consistent,• Example:
Another Example
Data
Calculated Rates
• Collect concentration data for reactants and products, then graph
• Effect of stoichiometry
• Average rate• Instantaneous
rate
Rate Law
• Study rates to understand mechanism of reaction
• True rate depends on forward and reverse reactions (remember equilibrium?)
• But we can write rate law based on reactants– Many reactions functionally irreversible– Use initial rates (reverse rate is negligible)
Rate Laws
• Two forms of rate law• Differential Rate Law (Rate Law)
– How rate depends on concentration of reactants– Experiment: Initial Rates of multiple trials
• Integrated Rate Law– How concentrations of species depend on time– Experiment: One trial sampled at multiple times
Relationship Between Rate and Concentration
• 2 NO2 (g) + F2 (g) 2 NO2F (g)
• Rate = Δ[NO2F]/ 2Δt = -Δ[F2]/ Δt
= -Δ[NO2]/ 2Δt
• Rate α [NO2] and [F2]
• Rate = k [NO2]x [F2]y
• k = rate constant• x and y are the orders of reaction, these are determined
experimentally – not from stoichiometry!
• 2 NO2 (g) + F2 (g) 2 NO2F (g)• Rate = k [NO2]x [F2]y
• From experiment, x = 1 , y = 1• Rate = k [NO2] [F2] = Rate Law• 1st order in NO2 , 1st order in F2, 2nd
order overall• One way to determine the rate law is from
initial rates.
H2O2 (aq) + 3 I- (aq) + 2 H+ (aq) I3- (aq) + 2 H2O (l)
Expt #
[H2O2] [I-] [H+] Initial rateM/s
1 0.010 0.010 0.0005 1.15 x 10-6
2 0.020 0.010 0.0005 2.30 x 10-6
3 0.010 0.020 0.0005 2.30 x 10-6
4 0.010 0.010 0.001 1.15 x 10-6
Rate = k [H2O2]x [I-]y [H+]z
H2O2 (aq) + 3 I- (aq) + 2 H+ (aq) I3- (aq) + 2 H2O (l)
Expt #
[H2O2] Rel [I-] Rel [H+] Rel Initial rateM/s
Rel
1 0.010 1 0.010 1 0.0005 1 1.15 x 10-6 1
2 0.020 2 0.010 1 0.0005 1 2.30 x 10-6 2
3 0.010 1 0.020 2 0.0005 1 2.30 x 10-6 2
4 0.010 1 0.010 1 0.001 2 1.15 x 10-6 1
Rate = k [H2O2]x [I-]y [H+]z
Rate = k [H2O2] [I-]
Example Problem
• 2 NO2 (g) 2 NO (g) + O2 (g)
Expt # [NO2] Rate, M/s
1 0.010 7.1 x 10-5
2 0.020 2.8 x 10-4
Relationship Between Concentration and Time
• We want to use a single experiment to determine the rate law.
• We will do this by plotting concentration versus time.
• We will deal with simplest cases initially—only one reactant, generally “A”
Zero Order Reaction
• How can a reaction rate be concentration independent?
Zero Order Reactions
• Zero order reaction A B– Rate = k[A]0 = k – Rate = -d[A]/dt– k = -d[A]/dt– Rearrange and integrate from time = 0 to time = t– [A]t – [A]o = -kt– [A]t = -kt + [A]o
Graphing Zero Order
– [A]t = -kt + [A]o – y = mx + b– Plot of conc. vs. time
gives straight line with slope of -k
– Units of k are M/s
First Order Reactions• Plot of conc vs time does not give straight
line (not 0 order)• Rate changes over time: Doubling
concentration of A doubles the rate• A products• Rate = k[A] Rate = -d[A]/ dt• k[A] = -d[A]/ dt • UTMOC ln[A]/[A]0 = - kt• ln[A] = -kt + ln[A]0
Example Problem
• The decomposition of N2O5 to NO2 and O2 is first order with k = 4.80 x 10-4 s-1 at 45 oC.– If the initial concentration is 1.65 x 10-2 M,
what is the concentration after 825 sec?– How long would it take for the concentration of
N2O5 to decrease to 1.00 x 10-2 M?
Graphing First Order
• Plot of ln[A] vs t gives straight line with a slope of -k and a y-intercept of ln[A]0
• Units of k = s-1
2 N2O5 4 NO2 + O2
First Order Reactions
Determine Order of Reaction by plotting data!
T (sec) [N2O5] (M)
0 2.15 x 10-3
4000 1.88 x 10-3
8000 1.64 x 10-3
12000 1.43 x 10-3
16000 1.25 x 10-3
2 N2O5 (g) 4 NO2 (g) + O2 (g)
0.27
0.24
0.18
0.001
0.0012
0.0014
0.0016
0.0018
0.002
0.0022
0 5000 10000 15000
time (sec)
[N2O
5] (
M)
NOT Zero order!
T (sec) [N2O5] (M) ln [N2O5]
0 0.00215 -6.14
4000 0.00188 -6.28
8000 0.00164 -6.41
12000 0.00143 -6.55
16000 0.00125 -6.68
0.14
0.13
0.14
0.13
-6.8
-6.7
-6.6
-6.5
-6.4
-6.3
-6.2
-6.1
time (sec)
ln [
N2O
5]
This is First Order
Second Order Reactions• Plot of [conc] vs t and plot of ln[conc] vs t
do not give straight lines. (not 0 or 1st )• A products• Rate = k[A]2 Rate = -Δ[A]/ Δt• k[A]2 = -Δ[A]/ Δt • UTMOC 1/[A] = kt + 1/[A]0
Graphing Second Order
• Plot of 1/[A] vs t gives straight line, with a slope of k and a y-intercept of 1/[A]0
Time, sec [HI], M
0 1.000
1000 0.112
2000 0.061
3000 0.041
4000 0.031
2 HI (g) H2 (g) + I2 (g) @ 580K
0
0.2
0.4
0.6
0.8
1
1.2
0 1000 2000 3000 4000 5000
time (sec)
[HI]
(M
)
Not Zero Order
Time, sec [HI], M ln [HI]
0 1.000 0
1000 0.112 -2.19
2000 0.061 -2.80
3000 0.041 -3.19
4000 0.031 -3.47
-4
-3.5
-3
-2.5
-2
-1.5
-1
-0.5
0
time (sec)
ln [
HI]
(M
)
Not First Order
Time, sec [HI], M ln [HI] 1/[HI]
0 1.000 0 1
1000 0.112 -2.19 8.93
2000 0.061 -2.80 16.4
3000 0.041 -3.19 24.4
4000 0.031 -3.47 32.3
0
5
10
15
20
25
30
35
0 1000 2000 3000 4000 5000
time (sec)
1/[H
I]
Second Order Rate = k[HI]2
Reactions Involving Gases
• A (g) products• PV = nART [A] = nA / V = P/RT
• ln[A]/[A]0 = -kt
• ln(P/RT) /(P0/RT) = -kt
• ln P/P0 = -kt
• Can use the pressures of gases for the concentrations.
Half Life• k = describes speed of the reaction
– Large k = fast reaction• Another way to describe speed is to use t½,
the half life.• This is the time needed to decrease to ½
[A]0.• For a first order reaction,• t = (1/k)ln[A]0/[A]• t½ = (1/k) ln[A]0/([A]0/2)• t½ = (1/k) ln2 = 0.693/k
Comparison of Half-Lives
• Use same procedure to derive each half-life
• For zero order, each half life is half as long as previous one• For first order, each half-life is the same• For second order, each half life is twice as long as the
previous one
Order Zero First Second
t1/2 [A]o / 2k 0.693 / k 1/ k[A]o
Application Question
• Kinetic data were plotted for A 2B + C
• What can these data tell you about this reaction?
More Complicated System
• So far we have assumed one reactant• How do we study A + 2B C + 3D• Run experiment with B in huge excess
– Rate = k [A]x[B]y but [B] remains constant– Called “psuedo” kinetics—can be used to
determine order of A
• Repeat with excess A to get pseudo-order of B• Combine experiments to get real rate law
Reaction Mechanisms
• A reaction may be more complex that 1 simple collision – may form intermediates.
• It is unlikely that 3 or more molecules will collide simultaneously.
• Elementary steps – describe a molecular event.
NO2 + CO NO + CO2
2 elementary steps
NO2 + NO2 + NO3 + CO NO3 + NO + NO2 + CO2
NO2 + CO NO + CO2
NO3 is an intermediate
Cl2 2 Cl
Cl + CHCl3 HCl + CCl3
Cl + CCl3 CCl4
Cl2 + CHCl3 HCl + CCl4 (overall)
Rate laws for elementary steps can be written from stoichiometry. (unlike overall)
1. Rate = k1[Cl2] unimolecular
2. Rate = k2[Cl][CHCl3] bimolecular
3. Rate = k3[Cl][CCl3] bimolecular
Rate Determining Step
• Rate Determining Step (rate limiting step) – is the slowest step leading to the formation of the products (slow step).
• The rates of any steps after the slow step are not important.
O3 + 2 NO2 O2 + N2O5
1. O3 + NO2 NO3 + O2 slow
2. NO3 + NO2 N2O5 fast
Rate = k[O3][NO2]
This explains why stoichiometry and rate law are independent
2 NO2 + O3 N2O5 + O2
Rate = k[NO2][O3]
2 NO + Cl2 2 NOCl
Rate = k[NO]2[Cl2]
Temperature Changes Rate
• 2NO (g) + Cl2 (g) 2NOCl (g)
• Rate = k[NO]2 [Cl2]
• k @ 25 oC = 4.9 x 10-6 M-1s-1
• k @ 35 oC = 1.5 x 10-5 M-1s-1
• This is more than 3x increase!• Why is there a temperature dependence on k?• Can k and temperature be related
theoretically and quantitatively?
Dependence of Rate Constant on Temperature
• Exponential of rate constant on absolute temperature
• Every curve different• This one represents
double rate every 10 K
Collision Theory• Collision Theory – molecules must collide
in order to react.• Rate α # collisions / sec α [reactants]• From KMT
• Increase temp, increase speed• Accounts for higher rate• Kinetic energy made into
potential energy to break
bonds
Activation Energy• Arrhenius – expanded collision theory (1888)• Molecules must collide with enough energy
to rearrange bonds.• If not, they just bounce off.• Activation Energy = Ea = minimum amount
of energy required to initiate a chemical reaction.
• Activated Complex (transition state) – temporary species in reaction sequence, least stable, highest energy, often undetectable.
Activation Energy and Transition States
Molecules with Enough KE to Overcome Activation Energy
• Boltzmann distribution
• Doubling temperature more than doubles fraction of molecules with enough energy
• Fraction =
Steric Factor• Rate constant k depends on three things:
– 1. Need to collide (z = collision rate)– 2. Need E > Ea (fraction of collisions is
– 3. Need to be oriented in the right way to react!
• Steric factor (orientation factor) = p 0>p>1• k = zp • k = A A is pre-exponential factor, or
frequency factor, for the Arrhenius equation
Orientation of Molecules
Arrhenius Equation• k = Ae-Ea/RT • ln k = ln Ae-Ea/RT
• y = m x + b• If you experimentally determine
___ and ____, then you can graphically determine ____ and ____.
• Can also be used to determine k at any ____.
ATR
Ek a ln
1ln
m = -Ea/R
b = ln A
ln k vs. 1/T
lnk1
k2
Ea
R
1 1
T1T2-=
Useful Form of Arrhenius Equation
• In principle, best to do many experiments, graph line, and determine Ea
• In practice, can get decent value from only two experiments
• Rearrange Arrhenius equation to get
Example Problem
H2 + I2 2 HI
k = 2.7 x 10-4 M-1s-1 @ 600 K
k = 3.5 x 10-3 M-1s-1 @ 650 K
a. Find Ea
b. Calculate k @ 700 K
Catalysis• To increase the rate of a reaction
– 1. Increase temperature– 2. Add a catalyst
• Catalyst – a substance which increases the rate of a reaction but is not consumed.
• Catalysts are involved in the course of a reaction.• Usually by lowering Ea
• Homogeneous – catalyst the same phase as the reactants
• Heterogeneous – catalyst in different phase (usually a solid)
Ammonia Formation with
a Catalyst
N2 + 3 H2 2 NH3
Catalytic Hydrogenation
• Food processing• Trans fatty acids
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