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 = ???

[ ].

52

hr

mol

t

ON

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l

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?

!"

#$%

&'=

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!

VITAMINS

Vitamins are non-protein parts of enzymes, calledco-enzymes.

When combined with the protein part they makeenzymes.

Enzymes derived from vitamins play critical rolesin redox chemistry in the body, which is thesource of heat and energy.