1. Chemical Kinetics: Rate Laws ORDER OF REACTION
rate (= d[A] /dt) = k[A]x[B]y
Overall order of reaction = x + y
Example: rate = k[A]2[B]
The reaction is
second order in A
first order in B
overall reaction order = 1+ 2= 3
(sum of the exponents)

2. DETERMINING THE RATE LAW
Method of initial rates:
Initially, we know [A] and [B] (and [C] = [D] = 0)
initial rate = k1[A]ox[B]oy “o”  “initial”
(t = 0)
Vary [A]0 and [B]0, measure initial rates

3. Sample problem
Given the following data for the reaction
A + B  Z
What is the overall reaction order?
[A]0 [B]0 initial rate
1) 1.0M 1.0M 0.8x102 M/s
x102
x102
x102
Rate = k [A]a [B]b
a = overall order = a + b b= 1 2 3 4

4. 2N2O5 4NO2 +O2 [N2O5]i Initial rate 0.01 M 0.018 0.02 M 0.036
What is the rate law?
Rate = k [N2O5]x
x = ???

5. INTEGRATED RATE LAW or ln[A] = ln[A]0  k t (2 forms of same equation)
So far we have used differential rate laws
e.g., for
aA  products
rate = -[A] = k[A]x (a differential eqn)
 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)

6. GRAPHICAL ANALYSIS 1st 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

7. CH3NC:  CH3  C  N: [A] = [A]0 e-kt ln[A] = ln[A]0  kt
(1st Order Reaction)
Figure: 14-07a,b
[A] = [A]0 e-kt ln[A] = ln[A]0  kt

8. INTEGRATED RATE LAW 2nd order reaction
2nd order reaction (x = 2):
differential
rate law
Integrate to get:
y = m x b

9. Figure: 14-08a,b
Reaction: 2NO2(g)  2NO + O2
Rate = k[NO2]2
Rate Law:

10. Half lives Half life: t1/2 e.g. A  products For 1st order reactions:
time it takes for the concentration of a reactant to drop to half of its initial value.
e.g. A  products
t1/2 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)

11. Figure: 14-09

12. Half Life Problem He nucleus 239Pu  235U +  “safe”   5x10-5g left
Plutonium – very toxic
(lethal dose ≈ 5x10-5g)
If we bury 1 lb. spent nuclear fuel containing
1g 239Pu, how long until it’s safe to dig it up?
“safe”   5x10-5g left
t1/2 = 24,400 yr

13. Temperature dependence of reaction rates
As T increases, reaction rates increase
Why?
Look at energy profile for a typical reaction 

14. Fraction of molecules with enough energy  e-Ea/RT
Figure: 14-16
Fraction of molecules with enough energy  e-Ea/RT

15. Arrhenius Equation k = rate constant A = frequency factor
Related to collision frequency and orientation
Ea = Activation energy
R = gas constant (usually J/mol-K)
T = temperature in K

16. ARRHENIUS PLOT Arrhenius plot ln k = ln A – Ea/RT
plot of ln k vs 1/T is a straight line
slope = -Ea/R
intercept = ln A
Arrhenius plot
The bigger Ea is, the more the rate varies with T

17. RATE VS TEMPERATURE k2/k1 = e0.65 = 1.9
Most reactions have Ea = 20 – 200 kJ/mol
A “typical” Ea might be 50 kJ/mol
How does rate vary over a 10° temperature range? (e.g. from 300 to 310 K)
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)

18. Rxn: CH3NC  CH3CN Using the plot, find Ea for the reaction.
Figure: 14-17
Using the plot, find Ea for the reaction.

19. SAMPLE PROBLEM
For the same reaction,
CH3NC  CH3CN
if k = 2.52x10-5 s-1 at 189.7C, what is the rate constant at 430K?

20. REACTION MECHANISMS Reaction Mechanism:
process by which a reaction occurs
For elementary reaction steps
Reaction 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.

21. 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:
If step 1 is slow or
If step 2 is slow
Compare with experiment to determine which is correct.

22. 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 slowest step
Rate determining step = slow step

23. NO2 +CO  NO + CO2
Elementary steps in a mechanism must add up to give the balanced overall reaction.
NO3 is produced in step 1 and consumed in step 2
Intermediate: a stable molecule
Note: it is NOT the same as the activated complex.
Intermediates do not (should not) appear in the rate law

24. To find mechanisms Find the experimental rate law
Postulate elementary steps
Find the rate law predicted by the mechanism and compare to experiment.
No rate can be written in terms of intermediates

25. Example Cl2 + CHCl3  HCl + CCl4 Postulate the following mechanism -
Observed rate = kobs [Cl2]1/2 [CHCl3]
Postulate the following mechanism -
is it consistent with the experimental rate law??
Cl Cl fast
Cl + CHCl3  HCl + CCl slow
Cl + CCl3 CCl fast

26. REACTION MECHANISMS involve only unimolecular or bimolecular processes
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)

27. Explaining the rate law for 2NO(g) + Br2(g)  2ONBr(g)k1
NO + Br ONBr (fast)
ONBr2 + NO ONBr (slow)
2 NO(g) + Br2(g) ONBr(g) (overall)
k-1 k2
step (2) is slow (rate determining)
 rate = k2[ONBr2][NO] 
intermediate
need to express [ONBr2] in terms of [reactants]

28. 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:
Conclusion: more than one mechanism may fit the rate law

29. Catalysis
Catalyst: substance that speeds up a reaction without undergoing permanent change.
How: changes the mechanism
lowers the activation energy
Example: 2H2O2  2H2O + O2
Catalyst: Br2, MnO2, or catalase (enzyme)
Homogeneous catalysis: catalyst is in the same phase as the reactant. (Br2, catalase)
Heterogeneous catalysis: catalyst is in a different phase from the reactants (MnO2)
usually a solid catalyst and gas or solution reactants
Reaction happens on the surface of the catalyst

30. Energy profiles for catalyzed and uncatalyzed H2O2 decomposition
Figure: 14-20
Energy profiles for catalyzed and
uncatalyzed H2O2 decomposition

31. CATALYSIS Thermodynamic state functions
(E,  H,  G,  S…) are unaffected by catalysis
(Changes are path-independent)

32. HETEROGENOUS CATALYSIS
EXAMPLE: H2 + 1/2 O2  H2O
reaction requires breaking
strong H-H and O=O bonds
 
435 kJ kJ
negligible rate without catalyst
Usually, the stronger the bonds in reactants, the more we need a catalyst.
e.g. 3H2 + N2  2NH G° = 33 kJ/mol at 298 K
NN triple bond (D = 946 kJ)

33. Nitrogen fixation N2 + 3H2  2 NH3 NN triple bond (D = 946 kJ)
must break NN triple bond (difficult)
Important in biological systems (proteins, nucleic acids) & industrially (fertilizer, polymers, explosives, …)
Beans, bacteria, etc: nitrogenase
enzyme reduces N2 to NH3 at room temp, 1 atm pressure
Haber process: uses Fe/Al2O3 catalyst at
°C, 300 atm.
N2 + 3H2  2 NH3

34. CATALYTIC CONVERTER CATALYTIC CONVERTER O2 1) CO  CO2(g) + H2O
Hydrocarbons
2) NO, NO2  N2(g)
Catalysts: CuO, Cr2O3, Pt, Rh

35. HOW DOES LOWERING Ea AFFECT RATES?
EXAMPLE.
H2O2  H2O + 1/2 O2
hydrogen 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?)

36. CATALYZED VS UNCATALYZED
kcat
-(Ea,cat - Ea,uncat)/RT
= e
kuncat
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!

37. ENZYMES Enzymes are biological catalysts.
Enzymes are produced by organisms to accelerate and to control reaction rates.
Enzymes are typically large protein molecules or combinations of proteins with other molecules. The region where the substrate/s (reactant/s) bind is called the active site.
Enzymes differ from man-made catalysts:
More efficient.
More specific.
Rate can be controlled by changing enzyme activity.

38. ENZYME CATALYSIS
Nature’s catalysts – big organic molecules specifically designed for certain reactions.
Rate acceleration by > 1010 (how?)
Enzyme active sites are ideally suited for transition state binding (lowers Ea)
Juxtaposition of reactants 
high effective concentration (increases A)
Each factor enhances rate by ≥ 105

39. CONTROL OF ENZYMES
Some enzymes wait in the “off” state, such as blood-clotting and digestive proteins
They 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 of the protein sitting in close proximity.
If this shape is altered, the active site no longer functions and the enzyme is “turned off.”
Molecular shape depends on pH, temperature, and reactions of the enzyme.

40. ENZYME ACTIVE SITES
Denatured enzyme – parts of the active site are no longer in close proximity.
Representation of an active site in an enzyme.
Competitive Inhibition: Another way to inhibit an enzyme is to bind a molecule to its active site, blocking any catalytic activity.
Many drugs and poisons work by this mechanism.

41. DRUGS
Penicillin (antibiotic) blocks an enzyme that bacteria use to build cell walls.
People do not have this enzyme
Bacterial cells only are poisoned.
HIV-protease inhibitors bind to the active site of an enzyme that releases the viral coat proteins, preventing the production of the HIV virus.
HIV protease
Active site
Ritonavir (inhibitor)

42. ENZYMES
Metal ions are often bound at the active site and serve as the reaction center of the enzyme.
The enzyme carbonic anhydrase uses a Zn2+ ion at its active site to accelerate the reaction:
CO2 + H2O  H2CO3
In red blood cells, CO2 is converted to H2CO3 which deprotonates 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!

43. VITAMINS Vitamins are non-protein parts of enzymes, called co-enzymes.
When combined with the protein part they make enzymes.
Enzymes derived from vitamins play critical roles in redox chemistry in the body, which is the source of heat and energy.