Transcript Chemical Kinetics
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 of Reaction
A
B Rate = Δ[B]/Δt = -Δ[A]/Δt Rate = Δ[product]/Δt = -Δ[reactant]/Δt
Defining Rate
• Rate is defined arbitrarily by one pdt or rxt • To be self consistent,
Stoichiometry
important!
• Example: 2 N 2 O 5 (g) 4 NO 2 (g) + O 2 (g)
2
N 2 O 5
Rate
= Δ[O 2 ]/ Δt (g)
4
NO 2 (g) + O 2 (g)
Rate
= Δ[NO 2 ]/
4
Δt = - Δ[N 2 O 5 ]/
2
Δt
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 NO 2 (g) + F 2 (g) 2 NO 2 F (g) • Rate = Δ[NO 2 F]/ 2Δt = -Δ[F 2 ]/ Δt = -Δ[NO 2 ]/ 2Δt • Rate α [NO 2 ] and [F 2 ] • Rate = k [NO 2 ] x • k = rate constant [F 2 ] y
x
and
y
are the
orders of reaction
, these are determined experimentally –
not
from stoichiometry!
• 2 NO 2 (g) + F 2 (g) 2 NO 2 F (g) • Rate = k [NO 2 ] x [F 2 ] y • From experiment, x = 1 , y = 1 • Rate = k [NO 2 ] [F 2 ]
=
• 1 st , 1 st Rate Law order in F 2 , 2 nd order in NO 2 order overall • One way to determine the rate law is from initial rates.
H 2 O 2 (aq) + 3 I (aq) + 2 H + (aq) I 3 (aq) + 2 H 2 O (l) Rate = k [H 2 O 2 ] x [I ] y [H + ] z [I ] [H + ] Expt # [H 2 O 2 ] 1 0.010
2 0.020
3 0.010
4 0.010
0.010
0.010
0.020
0.010
0.0005
0.0005
0.0005
0.001
Initial rate M/s 1.15 x 10 -6 2.30 x 10 -6 2.30 x 10 -6 1.15 x 10 -6
H 2 O 2 (aq) + 3 I (aq) + 2 H + (aq) I 3 (aq) + 2 H 2 O (l) Rate = k [H 2 O 2 ] x [I ] y [H + ] z Expt # [H 2 O 2 ] Rel 1 0.010
1 [I ] 0.010
Rel 1 [H + ] 0.0005
Rel 1 Initial rate M/s 1.15 x 10 -6 Rel 1 2 0.020
2 0.010
1 0.0005
1 2.30 x 10 -6 2 3 4 0.010
0.010
1 0.020
2 0.0005
1 2.30 x 10 -6 2 1 0.010
1 0.001
2 1.15 x 10 -6 1 Rate = k [H 2 O 2 ] [I ]
Example Problem
• 2 NO 2 (g) 2 NO (g) + O 2 (g) Expt # 1 2 [NO 2 ] 0.010
0.020
Rate, M/s 7.1 x 10 -5 2.8 x 10 -4
Relationship Between Concentration and Time
• We want to use a single determine the rate law.
experiment to • 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 – – Rate = -d[A]/dt k = -d[A]/dt – Rearrange and integrate from time = 0 to time = t [A] t – [A] o = k = -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
line ( not 0 order ) give straight • 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 N 2 O 5 to NO 2 is first order with k = 4.80 x 10 -4 s -1 and O at 45 2 o C.
– 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 N 2 O 5 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 N 2 O 5 4 NO 2 + O 2
First Order Reactions
Determine Order of Reaction by plotting data!
2 N 2 O 5 (g) 4 NO 2 (g) + O 2 (g) T (sec) 0 4000 8000 12000 16000 [N 2 O 5 ] (M) 2.15 x 10 -3 1.88 x 10 -3 1.64 x 10 -3 1.43 x 10 -3 1.25 x 10 -3
0.27
0.24
0.18
0.0022
0.002
0.0018
0.0016
0.0014
0.0012
0.001
0 5000 10000 time (sec)
NOT Zero order!
15000
T (sec) 0 4000 8000 12000 16000 [N 2 O 5 ] (M) ln [N2O 5 ] 0.00215
-6.14
0.00188
0.00164
0.00143
0.00125
-6.28
-6.41
-6.55
-6.68
0.14
0.13
0.14
0.13
-6.1
-6.2
-6.3
-6.4
-6.5
-6.6
-6.7
-6.8
time (sec) 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 1 st ) • A products • Rate = k[A] 2 • k[A] 2 = -Δ[A]/ Δt Rate = -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
2 HI (g) H 2 (g) + I 2 (g) @ 580K Time, sec 0 1000 2000 3000 4000 [HI], M 1.000
0.112
0.061
0.041
0.031
1.2
1 0.8
0.6
0.4
0.2
0 0 1000 2000 time (sec) 3000 4000 5000
Not Zero Order
Time, sec 0 1000 2000 3000 4000 [HI], M 1.000
0.112
0.061
0.041
0.031
ln [HI] 0 -2.19
-2.80
-3.19
-3.47
0 -0.5
-1 -1.5
-2 -2.5
-3 -3.5
-4 time (sec)
Not First Order
Time, sec 0 1000 2000 3000 4000 [HI], M 1.000
0.112
0.061
0.041
0.031
ln [HI] 0 -2.19
-2.80
-3.19
-3.47
1/[HI] 1 8.93
16.4
24.4
32.3
15 10 5 0 35 30 25 20 0 1000 2000 time (sec) 3000 4000 5000
Second Order Rate = k[HI] 2
Reactions Involving Gases
• A (g) products • PV = n A RT [A] = n A • ln[A]/[A] 0 = -kt • ln(P/RT) /(P 0 /RT) = -kt • ln P/P 0 = -kt • Can use the pressures / V = P/RT 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 t ½ , 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] • t ½ = (1/k) ln[A] 0 /([A] 0 /2) • t ½ = (1/k) ln2 = 0.693/k
Comparison of Half-Lives
• Use same procedure to derive each half-life
Order
t 1/2
Zero
[A] o / 2k
First
0.693 / k
Second
1/ k[A] o • 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
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.
NO 2 + CO NO + CO 2 2 elementary steps NO 2 + NO 2 + NO 3 + CO NO 3 + NO + NO 2 + CO 2 NO 2 + CO NO + CO 2 NO 3 is an intermediate
Cl 2 2 Cl Cl + CHCl 3 HCl + CCl 3 Cl + CCl 3 CCl 4 Cl 2 + CHCl 3 HCl + CCl 4 (overall) Rate laws for elementary steps
can
be written from stoichiometry. (unlike overall) 1. Rate = k 1 [Cl 2 ] unimolecular 2. Rate = k 2 [Cl][CHCl 3. Rate = k 3 [Cl][CCl 3 ] 3 ] bimolecular 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.
O 3 + 2 NO 2 O 2 + N 2 O 5 1. O 3 + NO 2 2. NO 3 + NO 2 NO 3 + O 2 N 2 O 5 slow fast Rate = k[O 3 ][NO 2 ] This explains why stoichiometry and rate law are independent
2 NO
2
+ O
3
N
2
O
5
+ O
2 Rate = k[NO 2 ][O 3 ]
2 NO + Cl
2
2 NOCl
Rate = k[NO] 2 [Cl 2 ]
Temperature Changes Rate
• 2NO (g) + Cl 2 • Rate = k[NO] 2 (g) [Cl 2 ] • k @ 25 o C = 4.9 x 10 -6 2NOCl (g) M -1 s -1 • k @ 35 o C = 1.5 x 10 -5 M -1 s -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 = E a = 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 = 𝑒 −𝐸𝑎 𝑅𝑇
Steric Factor
• Rate constant
k
depends on three things: – 1. Need to collide (z = collision rate) – 2. Need E > E a (fraction of collisions is 𝑒 −𝐸𝑎 𝑅𝑇 – 3. Need to be oriented in the right way to react!
• Steric factor (orientation factor) = p 0>p>1 • k = zp 𝑒 −𝐸𝑎 𝑅𝑇 • k = A 𝑒 −𝐸𝑎 𝑅𝑇 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 ln
k
E R a
1
T
ln
A
• y = m x + b • If you experimentally determine ___ and ____, then you can graphically determine ____ and ____. • Can also be used to determine k at any ____.
ln k vs. 1/T
m = -E a /R b = ln A
Useful Form of Arrhenius Equation
• In principle, best to do many experiments, graph line, and determine E a • In practice, can get decent value from only two experiments • Rearrange Arrhenius equation to get
ln k 1 k 2 = E a R 1 T 2 1 T 1
Example Problem
H 2 + I 2 2 HI k = 2.7 x 10 -4 M -1 s -1 @ 600 K k = 3.5 x 10 -3 M -1 s -1 @ 650 K a. Find E a 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 E a • Homogeneous – catalyst the same phase as the reactants • Heterogeneous solid) – catalyst in different phase (usually a
Ammonia Formation with a Catalyst
N 2 + 3 H 2
2 NH 3
Catalytic Hydrogenation
• Food processing • Trans fatty acids