Post on 29-Mar-2018
Chemical Kinetics: Rate Laws
ORDER OF REACTIONrate (= −d[A] /dt) = k[A]x[B]y
Overall order of reaction = x + y
Example: rate = k[A]2[B]
The reaction issecond order in Afirst order in Boverall reaction order = 1+ 2= 3
(sum of the exponents)
DETERMINING THE RATE LAW
Method of initial rates:
Initially, we know [A] and [B] (and [C] = [D] = 0)
initial rate = k1[A]ox[B]o
y “o” ⇒ “initial” (t = 0)
Vary [A]0 and [B]0, measure initial rates
Sample problem
Given the following data for the reactionA + B → Z
What is the overall reaction order?
[A]0 [B]0 initial rate1) 1.0M 1.0M 0.8x10−2 M/s2) 2.0 1.0 1.6x10−2
3) 2.0 2.0 6.4x10−2
4) 1.0 2.0 3.2x10−2
Rate = k [A]a [B]b
a = overall order = a + bb=
1. 02. 13. 24. 35. 4
2N2O5→ 4NO2 +O2
[N2O5]i Initial rate
0.01 M 0.0180.02 M 0.0360.04 M 0.072
What is the rate law?Rate = k [N2O5]x
x = ???
[ ].
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hr
mol
t
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INTEGRATED RATE LAWSo far we have used differential rate lawse.g., for
aA → products
rate = -Δ[A] = k[A]x (a differential eqn) Δ t
Integration gives [A] as a function of t (more useful)
1st order reaction (x = 1):
Integrate to get: [A] = [A]0 e-kt
or ln[A] = ln[A]0 − k t
(2 forms of same equation)
!
rate ="# A[ ]#t
= k A[ ]1
GRAPHICAL ANALYSIS1st order:
ln [A] = ln [A]0 − k t
y = b + m x
Plot of ln [A] vs. t gives a straight line slope = −k
intercept = ln [A]0
CH3N≡C: → CH3 C ≡ N:
[A] = [A]0 e-kt ln[A] = ln[A]0 − kt
(1st Order Reaction)
INTEGRATED RATE LAW2nd order reaction
2nd order reaction (x = 2):
differentialrate law
Integrate to get:
y = m x + b
!
rate ="# A[ ]#t
= k A[ ]2
!
1
A[ ]= k t +
1
A[ ]0
Reaction: 2NO2(g) → 2NO + O2
Rate = k[NO2]2
Rate Law:
!
1
NO2[ ]
= k t +1
NO2[ ]0
Half livesHalf life: t1/2
time it takes for the concentration of a reactant to drop to half ofits initial value.
e.g. A → productst1/2 is where [A] = 1/2[A]0
For 1st order reactions: t1/2 doesn’t depend on concentration
E.g. nuclear decays: 14C → 14N + e−
t1/2 = 5730 years(carbon dating)
!
lnA[ ]A[ ]
.0
= " k t
!
t1/ 2
=ln 2
k=0.693
k
Half Life Problem He nucleus
239Pu → 235U + α ↑ Plutonium – very toxic (lethal dose ≈ 5x10-5g)
If we bury 1 lb. spent nuclear fuel containing1g 239Pu, how long until it’s safe to dig it up? “safe” ⇒ ≤ 5x10-5g left
t1/2 = 24,400 yr
Temperature dependence of reaction rates
As T increases, reaction rates increase
Why?
Look at energy profile for a typical reaction
rxn progress
products
reactants !E rxn
E ainternal
energy (E)
E a= activation barrier
Δ
Fraction of molecules with enough energy ∝ e-Ea/RT
Arrhenius Equation
k = rate constantA = frequency factor
Related to collision frequency and orientation
Ea = Activation energyR = gas constant (usually 8.314 J/mol-K)T = temperature in K
RT
Ea
eAk!
=
RT
EAk a!= lnln
or
ln k = ln A – Ea/RT
plot of ln k vs 1/T is a straight line
slope = -Ea/Rintercept = ln A
Arrheniusplot
The bigger Ea is, the more the rate varies with T
ARRHENIUS PLOT
1/T(high T) (low T)
-E aln k
ln A
R
(K )-1
RATE VS TEMPERATURE
Most reactions have Ea = 20 – 200 kJ/molA “typical” Ea might be 50 kJ/mol
How does rate vary over a 10° temperature range? (e.g.from 300 to 310 K)
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122
111
TTR
E
k
ka
ln
k2/k1 = e0.65 = 1.9
(about twice as fast at 310 K compared to 300 K)
rule of thumb: reaction rates double for every 10°rise in temperature
(assumes Ea ≈ 50 kJ/mol)
Rxn: CH3NC → CH3CN
Using the plot, find Ea for the reaction.
SAMPLE PROBLEM
For the same reaction,CH3NC → CH3CN
if k = 2.52x10-5 s-1 at 189.7°C, what is the rateconstant at 430K?
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#$%
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122
111
TTR
E
k
ka
ln
Reaction Mechanism:process by which a reaction occurs
For elementary reaction stepsReaction proceeds as written
NO + O3 → NO2 +O2
Rate = k [NO][O3]
Most reactions do not occur as a single elementary step.They occur as the result of several elementary steps.
REACTION MECHANISMS
Example: NO2 +CO → NO + CO2
Probable mechanism:NO2 + NO2 → NO3 + NO 1
NO3 + CO → NO2 + CO2 2
NO2 + CO → NO + CO2
Rate Law for multi-step mechanism:
!
rate ="# NO
2[ ]#t
= k NO2[ ]2
[ ] [ ][ ]CONOkt
COrate
3=
!
!"=
Compare with experiment to determine which is correct.
orIf step 1 is slow
If step 2 is slow
Example: NO2 +CO → NO + CO2
According to experiment:Rate ∝ [NO2]2
Consistent with step 1 as the RATE DETERMINING(or slowest) step
Rate cannot proceed any faster then the sloweststep
Rate determining step = slow step
NO2 +CO → NO + CO2
• Elementary steps in a mechanism must add up togive the balanced overall reaction.
• NO3 is produced in step 1 and consumed in step 2Intermediate: a stable moleculeNote: it is NOT the same as the activated
complex.
• Intermediates do not (should not) appear in therate law
To find mechanisms1. Find the experimental rate law2. Postulate elementary steps3. Find the rate law predicted by the
mechanism and compare to experiment.
No rate can be written in terms ofintermediates
ExampleCl2 + CHCl3 → HCl + CCl4
Observed rate = kobs [Cl2]1/2 [CHCl3]
Postulate the following mechanism -is it consistent with the experimental rate
law??
Cl2 2Cl fast
Cl + CHCl3 → HCl + CCl3 slow
Cl + CCl3→ CCl4 fast
REACTION MECHANISMS Another example:
2NO(g) + Br2(g) → 2ONBr(g) observed rate = k[NO]2[Br2]
Does this mean mechanism is:
No! 3 body collisions are very rare (unlikely) compared to 2-body (bimolecular)
In general: All elementary reaction steps involve only unimolecular or bimolecular processes
(exception – solute reactions with solvent molecules)
NO
NO
Br2
ON
Br
Br
ON
ONBr
+
ONBr
?
Explaining the rate law for2NO(g) + Br2(g) → 2ONBr(g)
step (2) is slow (rate determining)
⇒ rate = k2[ONBr2][NO] ↑ intermediate
need to express [ONBr2] in terms of [reactants]
NO + Br2 ONBr2 (fast)
ONBr2 + NO 2 ONBr (slow)
2 NO(g) + Br2(g) 2 ONBr(g) (overall)
k1
k-1
k2
=k
k
1
-1
[NO] [Br ]2k 2
2
rate = k [ONBr ][NO]2 2
= kk
k
1
-1[NO][Br ]
2( )[NO]2
k(obs. rate law)
Conclusion: more than one mechanism may fit the rate law
step (1): fast equilibrium ⇒ forward rate = back rate
k1[NO][Br2] = k-1[ONBr2]
[ONBr2] = k1/k-1 [NO][Br2]
Now plug this into the rate law:
CatalysisCatalyst: substance that speeds up a reaction without
undergoing permanent change.
How: changes the mechanismlowers the activation energy
Example: 2H2O2 → 2H2O + O2
Catalyst: Br2, MnO2, or catalase (enzyme)
Homogeneous catalysis: catalyst is in the same phaseas the reactant. (Br2, catalase)
Heterogeneous catalysis: catalyst is in a different phasefrom the reactants (MnO2)
• usually a solid catalyst and gas or solution reactants• Reaction happens on the surface of the catalyst
Energy profiles for catalyzed anduncatalyzed H2O2 decomposition
Thermodynamic state functions(ΔE, Δ H, Δ G, Δ S…) are unaffected by catalysis
(Changes are path-independent)
CATALYSIS
HETEROGENOUS CATALYSISHETEROGENOUS CATALYSISEXAMPLE: H2 + 1/2 O2 → H2O
reaction requires breakingstrong H-H and O=O bonds
↑ ↑ 435 kJ 498 kJ
negligible rate without catalyst
Usually, the stronger the bonds in reactants, the more we need a catalyst.e.g. 3H2 + N2 → 2NH3 ΔG° = −33 kJ/mol
at 298 K
N≡N triple bond (D = 946 kJ)
Pt surface
H-H H-H O=O
gas molecules
solid
H H =O =O
adsorbed atoms
diffusionon surface
H H
=O
H H
O
Nitrogen fixation
must break N−N triple bond (difficult)
Important in biological systems (proteins, nucleicacids) & industrially (fertilizer, polymers, explosives, …)
Beans, bacteria, etc: nitrogenaseenzyme reduces N2 to NH3 at room temp, 1 atmpressure
Haber process: uses Fe/Al2O3 catalyst at 400-500°C, 300 atm.
N2 + 3H2 → 2 NH3
N≡N triple bond (D = 946 kJ)
CATALYTIC CONVERTERCATALYTIC CONVERTER
O21) CO → CO2(g) + H2O Hydrocarbons
2) NO, NO2 → N2(g) Catalysts: CuO, Cr2O3, Pt, Rh
HOW DOES LOWERING Ea AFFECTRATES?
kcat
kuncat
= Acat
e-E /RTa,cat
Auncat
e-E /RTa,uncat
EXAMPLE.H2O2 → H2O + 1/2 O2hydrogen peroxide (toxic)
Uncatalyzed reaction has Ea = 72 kJ
Catalase (enzyme in liver) lowers Ea to 28 kJ
What is the ratio of kcat/kuncat at 37°C?(body temperature)
Assume Acat = Auncat(Q: is this a good assumption?)
CATALYZED VS UNCATALYZED
-(28 - 72 kJ/mol)ln
kcat
kuncat
=(.0083 kJ/mol K)(310 K)
= 17.1
>kcat
kuncat
= 3 x 107
speeds up by a factor of 30 million!
Peptidase enzymes – break up proteins into amino acids(in your stomach)
similar effect on Ea
Without these it would take ~ 300 years to digest a steak!
kcat
kuncat= e -(Ea,cat - Ea,uncat)/RT
ENZYMESEnzymes are biological catalysts.
Enzymes are produced by organisms to accelerate and tocontrol reaction rates.
Enzymes are typically large protein molecules orcombinations of proteins with other molecules. Theregion where the substrate/s (reactant/s) bind is called theactive site.
Enzymes differ from man-made catalysts:More efficient.More specific.Rate can be controlled by changing enzyme activity.
ENZYME CATALYSIS
enzyme
binding
sites
reactant molecules
enzyme-substrate
complex
products
k = A e-E /RTa
1. Enzyme active sitesare ideally suited fortransition statebinding (lowers Ea)
2. Juxtaposition ofreactants ⇒high effectiveconcentration(increases A)
Nature’s catalysts – big organic moleculesspecifically designed for certain reactions.Rate acceleration by > 1010 (how?)
Each factor enhances rate by ≥ 105
CONTROL OF ENZYMESSome enzymes wait in the “off” state, such as blood-
clotting and digestive proteinsThey are activated (reacted to make the active form)when needed.
The active site depends on the enzyme conformation(shape):
• Metal ions are held in place by different sections ofthe protein sitting in close proximity.
• If this shape is altered, the active site no longerfunctions and the enzyme is “turned off.”
• Molecular shape depends on pH, temperature, andreactions of the enzyme.
ENZYME ACTIVE SITES
Denatured enzyme –parts of the active siteare no longer in closeproximity.
Representation of anactive site in anenzyme.
Competitive Inhibition: Another way to inhibit an enzyme is tobind a molecule to its active site, blocking any catalytic activity.
Many drugs and poisons work by this mechanism.
DRUGSPenicillin (antibiotic) blocks an enzyme that bacteria use to build cell
walls.People do not have this enzymeBacterial cells only are poisoned.
HIV-protease inhibitors bind to the active site of an enzyme that releasesthe viral coat proteins, preventing the production of the HIV virus.
Active site
HIV protease
Ritonavir (inhibitor)
ENZYMESMetal ions are often bound at the active site and serve as thereaction center of the enzyme.
The enzyme carbonic anhydrase uses a Zn2+ ion at its activesite to accelerate the reaction:
CO2 + H2O → H2CO3
In red blood cells, CO2 is converted to H2CO3 whichdeprotonates to form HCO3
-.HCO3
- leaves the cell and serves as a buffer for blood plasma.
In the lungs, HCO3- is re-protonated to form H2CO3.
Carbonic anhydrase converts H2CO3 back to CO2(g) and H2O.
Exhale!