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 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 reaction A + B Z

What is the overall reaction order?

[A]0 [B]0 initial rate1) 1.0M 1.0M 0.8x102 M/s• 2.0 1.0 1.6x102

• 2.0 2.0 6.4x102

• 1.0 2.0 3.2x102

Rate = k [A]a [B]b

a = overall order = a + bb=

• 0• 1• 2• 3• 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

−Δ

Δ−

l

INTEGRATED RATE LAWSo far we have used differential rate lawse.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)

€

rate =−Δ A[ ]

Δt= 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

CH3NC: CH3 C N:

[A] = [A]0 e-kt ln[A] = ln[A]0 kt

(1st Order Reaction)

INTEGRATED RATE LAW 2nd order reaction

2nd order reaction (x = 2):

differentialrate law

Integrate to get:

y = m x + b

€

rate =−Δ A[ ]

Δt= 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 of its 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 + 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

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

Fraction of molecules with enough energy e-Ea/RT

Arrhenius Equation

k = rate constant

A = frequency factorRelated to collision frequency and orientation

Ea = Activation energy

R = gas constant (usually 8.314 J/mol-K)

T = temperature in K

RTEa

eAk−

=

RTEAk 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

Arrhenius plot

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/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)

⎥⎦

⎤⎢⎣

⎡−=

122

1 11

TTR

E

k

k aln

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 189.7C, what is the rate constant at 430K?

⎥⎦

⎤⎢⎣

⎡−=

122

1 11

TTR

E

k

k aln

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 =−Δ NO2[ ]

Δt= k NO2[ ]

2

[ ] [ ][ ]CONOktCO

rate 3=Δ

Δ−=

Compare with experiment to determine which is correct.

or

If 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 slowest step

Rate determining step = slow step

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 2Intermediate: a stable moleculeNote: it is NOT the same as the activated

complex.

• Intermediates do not (should not) appear in the rate law

To find mechanisms

1. Find the experimental rate law

2. Postulate elementary steps

3. Find the rate law predicted by the mechanism and compare to experiment.

No rate can be written in terms of intermediates

Example

Cl2 + 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 for 2NO(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

[ N O ] [ B r ]

2k

2

2

r a t e = k [ O N B r ] [ N O ]

2 2

= k

k

k

1

- 1

[ N O ] [ B r ]

2

( ) [ N O ]2

k

( o b s . r a t e l a w )

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 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

Energy profiles for catalyzed anduncatalyzed H2O2 decomposition

Thermodynamic state functions (E, H, G, S…) are unaffected by catalysis

(Changes are path-independent)

CATALYSIS

HETEROGENOUS CATALYSISHETEROGENOUS CATALYSISEXAMPLE: H2 + 1/2 O2 H2O

reaction requires breaking strong 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 G° = 33 kJ/mol at 298 K

NN triple bond (D = 946 kJ)

Pt surface

H-H H-H O=O

gas molecules

solid

H H

=

O

=

O

adsorbed atoms

diffusion

on surface

H H

=

O

H H

O

Nitrogen fixation

must break NN triple bond (difficult)

Important in biological systems (proteins, nucleic acids) & industrially (fertilizer, polymers, explosives, …)

Beans, bacteria, etc: nitrogenaseenzyme reduces N2 to NH3 at room temp, 1 atm pressure

Haber process: uses Fe/Al2O3 catalyst at 400-500°C, 300 atm.

N2 + 3H2 2 NH3

NN triple bond (D = 946 kJ)

CATALYTIC CONVERTERCATALYTIC CONVERTER

O2

1) CO CO2(g) + H2O Hydrocarbons

2) NO, NO2 N2(g) Catalysts: CuO, Cr2O3, Pt, Rh

HOW DOES LOWERING Ea AFFECT RATES?

k

cat

k

uncat

=

A

cat

e

-E /RT

a,cat

A

uncat

e

-E /RT

a,uncat

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?)

CATALYZED VS UNCATALYZED

-(28 - 72 kJ/mol)

ln

k

cat

k

uncat

=

(.0083 kJ/mol K)(310 K)

= 17.1

>

k

cat

k

uncat

= 3 x 10

7

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

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.

ENZYME CATALYSIS

enzyme

binding sites

reactant molecules

enzyme-substrate complex

products

k = A e-E /RTa

1. Enzyme active sites are ideally suited for transition state binding (lowers Ea)

2. Juxtaposition of reactants high effective concentration (increases A)

Nature’s catalysts – big organic molecules specifically 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 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.

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.

DRUGS

Penicillin (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 releases the viral coat proteins, preventing the production of the HIV virus.

Active site

HIV protease

Ritonavir (inhibitor)

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!

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.