Kinetics Kotz Chap 15

How fast are chemical reactions (the rate of a reaction)?


Hydrolysis of proteins (connected through amide linkages)

In your body, this reaction is mostly controlled by enzymes.

The rate in neutral water is about rate constant 10-6 / s

(J. Am. Chem. Soc. 1988: 7529), i.e.: ½ life 7 years

By contrast, reactions occur 103 in vivo at 107 / sec with enzymes (Lehninger)

i.e.: ½ life 10-5 sec

Some interesting statistics

(Deepak Chopra, Ageless Body, Timeless Mind), 1993

Skin completely replaced each month

Stomach lining each 5 days

Liver 6 weeks

Skeleton 3 months

By the end of this year, 98% of the atoms in your body will have been replaced!

N.B. Thre are only very few "errors" per millions of reactions when amino acids incorporated into protein

Your body does a lot of chemistry: it controls 6 x 1012 reactions/cell/s

insert graph(From notes 1996)

Thermodynamics tell you about the relative stability / energy of reactants & products Hreaction H(initial) - H(final)

That is, will a reaction actually occur (if a mechanistic pathway exists)

On the other hand, Kinetics tells you about the reaction rate / reaction pathway

The rates depend not on the stability of the product, but on the height of the barrier to the reaction H (barrier to reaction)

We are answer the following questions:

I) How to measure the rate of a reaction?

II) How do we interpret the measured reaction rate in molecular terms?

III) Can we make use of rates to understand reaction pathway? (YES)

Typical classes of reactivity

Ionic compounds - react very, very rapidly (e.g. Na+ + Cl-NaCl)

- react at the diffusion controlled limit (as fast as molecules can migrate to one another)

Radicals e.g., CH3 + CH H3CCH3

- similarly react at the diffusion controlled limit (as fast as molecules can migrate to one another)

Covalent compounds react very slowly, by comparison

Range of reaction

Millions of years -- femtosecond (10-12 sec) scale



I) Fe3+ + SCN ® Fe(SCN)2+

II) Bu2+ + CO32- ® BaCO3

III) Cu2+ + 4NH3 ® Cu(NH3)42+


IV) NaI + EtBr  CH3CH2I + NaBr(s)

V) 2 H2 g + O2 g ® 2 H2O (note, in the absence of a catalyst, this reaction is very slow)]

Experimental measurement of Reaction Rates.

Do we expect the rate of reaction to depend on the concentration of reagents? Normally yes - the higher the concentration, the more likely molecules will come into contact so that bond redistribution can take place.

We shall compare, in this course, only bimolecular (two molecules come together to make products) and unimolecular reaction (a molecule spontaneously does something)

To follow the reaction, can use either

I) the rate at which reagents disappear [NaI] or [CH3CH2Br] (reagents)


II) or the rate at which products are formed [NaBr] or [CH3CH2I]

NaI + CH3CH2Br  NaBr + CH3CH2I ()

If you want to measure a rate - how to do it? Need some independent news to determine instantaneous concentration (in this reaction, NaBr is insoluble in the solvent, we can follow the reaction by watching precipitate formation)

Measure: [CH3CH2I] + [CH3CH2Br]

or rate of appearance of new products [NaBr] or [CH3CH2I]

An inaccurate, "back of the envelope" of the instantaneous slope can be had by taking the difference between two adjacent peaks.

C1,2 (concentration at time t) t1,2

By convention

-reaction rate of the forward reaction (as drawn) is positive

A more accurate way to look at the rate is to take the slope of the tangent to the curve - this is the instantaneous rate

Typically reactions start fast and then slow down - why?

What is happening

A + B ® C

as [A] ¯ statistical chance of hitting B ¯

The Experimental Rate Law:

The rate law is a mathematical expression that relates the reaction RATE and the CONCENTRATION of reactants (we should expect this works, as indicated above - more reagents, more reaction)

Use NaI + EtBr  NaBr(s) + EtI

Rate of reaction may be defined in terms of any of the reagents/products. Since only complete molecules can react (can't have only part molecules, unlike part marks), the reagents and products are related by interger values, for the reaction shown above

The rate = -d[I-] / dt = -d[EtBr] / dt (rate of loss of reagents)

= +d[NaBr] / dt = +d[EtI] / dt (= rate of production of products)

One way to determine the integer relationships between the reagents/products is to run several different experiments keeping one species CONSTANT and we vary the concentration of the other

e.g. [EtBr] constant, & vary [NaI] or vice versa

Keep temperature constant (why?)(see below)
Expt Conc. Initial Rate
[EtBr] [NaI]  +d[NaBr(s)] / dt
1 1.0 M 1.0 M  3.0 X 10-2 mol/LS
2 1.0 M 0.5 M  1.5 x 10-2
3 2.0 M 1.0 M  6.0 x 10-2
The difference betwen reaction 1 and 2 is that the [NaI] was halved. The difference between 1 and 3 is that [EtBr] was doubled.

It can be seen that the rate is directly proporational to both [NaI] and [EtBr].

We can thus write a rate law that shows how the rate is related to concentration.

Reaction Rate = k [NaI]1 [EtBr]1 = k [NaI] [EtBr]

k (note small k, not K the equilibrium constant) is the "specific rate constant" a number that is specific to a given reaction at a given temperature. The bigger the number, the faster the reaction. At the same time, the equation shows that the higher the concentration the faster the concentration.

The exponent to the right of the concentration is called the "order" of reaction for that reagent.

i.e. Reaction is first order is NaI, Reaction in first order is EtBr

The overall Reaction Order is the sum of the exponents.

Take sum of exponents in the concentrations in rate law and add

-d[NaI] / dt = k [NaI]1 [EtBr]1

Overall order = 2 (1+1) a second reaction order may define the order for each species.

Since we have give the law in terms of disappearing reactants, the rate is preceeded by -

ie. -d [EtBr] / dt

Products appearing are +ve rates

ie. +d [NaBr] / dt

e.g. NaBr appearing

Different proportions

In the inconvenient case that a 1:1 relationship is not observed, have to change the proportionality.

For example:

* We also choose d[ ] / dt term for one mole

ie. If A + 3B ® 4C

-d[A] / dt = - 1/3 × d[B] / dt = 1/4 d[C] / dt

The loss of B goes three time faster than the loss of A. The loss of A is four times faster than the production of C.

How to treat reactions of different order

A molecule may spontaneously decompose (blow-up) without the need for other molecules.

Rate = -d[D] / dt = k [D] (i.e., a first order reaction)

This reaction actually is unimolecular (first order) (Since the rate is in mol/Ls, the units of k will be /s)

Does stoichimoetry always reflect rate equation??? NO!!!!!!!

Take a different reaction. The stoichimetry shows a 1:1 relationship between reactants, but a 1:2 relationship of products and reactants. What is going on? Look at the initial rates.

We would predict that rate = +d[N2] / dt = -1/2 d[NO] / dt = k [NO]2[H2]2

2NO(g) + 2H2(g) ® N2(g) + 2H2O
[NO] [H2 Rate of Decrease of [NO]
1 0.1 0.01  0.062 mol/L×sec 
2 0.1 0.04  0.246
3 0.3 0.01  0.558
The rates between 1 and 2 show that the reaction rate is directly proportional to [H2] not to [H2]2

The rates between 1 vs 3 show that a tripling of [NO] leads to a 9 fold increase [3]2 . Thus, the Rate is proportional to [NO]2

The rate law is then clear

Rate = k[NO]2[H2]

ie. The rate law is not necessarily the same as the stoichiometry of the reaction - why? Rate depends on mechanism (the observed transformation may involve a complex sequence of steps rather than a "one-shot" transformation)

Another e.g.

CH3CHO(g) ® CH4(g) + CO(g)

Our prediction from the stoichiometry is

Rate = k[CH3CHO]1 not dependent upon products
entry [CH3CHO] rate (mol/L×s)
1 0.1M 0.085 
2 0.2M 0.34 
3 0.3M 0.76 
4 0.4M 1.4 
From entry 1 and 2, a doubling of the concentration leads to a quadrupling of the rate, thus the rate law

Rate = -d[CH3CHO] / dt = k [CH3CHO]2

Since the units of rate are still mol/L s, the units of k must be L/mol s (L/mol s x mol2/L2 = mol/L s)

Here's an ugly one:

CHCl3(g) + Cl2(g) ® CCl4(g) + HCl(g)


Rate law found = -d[CHCl3] / dt = k [CHCl3]1[Cl2]1/2

overall order = 3/2

Another cutie:

2 NO(g) + Cl2(g) ® 2NOCl(g)

nitrosyl chloride

Rate Law: can't write expression down - must determine by experiment

It was found to be

Reaction Rate = -1/2 d[NO] / dt = -d[Cl2] / dt = k [NO]2 [Cl2]

Order = 2 + 1 = 3

2nd order in [NO]

1st order in [Cl2]

From this series of reactions we see that the: stoichiometric equation may not relate to overall reaction order. If the rate law does match the stoichiometry often the reaction is occurring in a single "primary reaction step" This is a step in which the reagents shown do go directly to product by direct interaction (see below).

With gases, can use partial pressures

÷ PNO           ÷PCl2

2 NO (g) + Cl2(g) ® 2 NOCl(g)

conc in terms of partial pressure

Reaction Rate = k [PNO]2[Pcl]1 but note change in units for k (not mol L-1 s-1 but atm-2 s-1)

-d[Pcl]/dt atm s-1= k [PNO]2[PCl]1

k -d[Pcl]/dt atm s-1 / [PNO]2[PCl]1 atm2 atm = atm-2 × s-1

First order

There are alternative method for testing a rate law -use entire reaction course, not just initial rate.

For example, for a reaction you think is first order

A ® B + C

Assume first order

-d[A] / dt = k[A]1

(k is a constant), A0 = [A] at t=0, At = [A] at time t=t


ie. ln[At] - ln[Ao] = -k (t - 0)


ln [At] / [Ao] = -kt ( y = mx + b)

ie. Plot ln [At] / [Ao] vs t

k = slope - y straight line if it is 1st order

eg. Hydrolysis of alkyl halide

(H3C)3CCl + H2O ® (H3C)3COH + HCl

Alcohol measure rate of HCl appearance (pH meter)

Find +d[HCl] / dt = k [C4HgCl]

In this case, the plot WAS a straight line, therefore the reaction was first order. What is the mechanism of this reaction? See below.

For first order reactions one can also use the Half-Life as a measure of First Order Reactions

Half life = time for the initial concentration A0 to be reduced by ½ (i.e., A0/2)

For a first order reaction t1/2

ie: ln([At]/[A0]) = -k t

At t = t1/2 At = A0/2

So ln[A0/2]/[A0] = -k t1/2 @ t = -t1/2

ln ½ = -kt1/2 or t1/2 = ln2 / k k = 0.693/t1/2

t ½ therefore gives directly value of k

Note that t1/2 not is not related to concentration

Who cares about this, you may ask: See problems in Tutorials in the book

Here's one example

Radio Carbon Dating

Carbon 14 decays in a first order process. While objects are living, they absorb 14C from the atmosphere. This ceases for some reason once something dies and the 14C concentration starts to decrease

In this case, we know that the number of counts is proportional to the [14C]

ln Ao/At = kt

t ½ = 5730 years. Thus, by comparison of the [14C] in living objects, and in the object to be dated, and knowing the t1/2 we can know k and calculate the concentration.

Temperature Effects

Reaction rates are highly sensitive to changes in T

Specific rate constant changes are responsible ie. k T

More or less, for T increase by 10 C there is a change of reaction rate by 2x (faster)

We can understand this in terms of the amount of energy


The theory of reactivity of molecules suggests that, for reaction to occur:

i) molecules must have sufficient energy to overcome an intrinsic reaction barrier (the activation energy - this will be different for every reaction)

ii) molecules must collide with one another for reaction to occur (e.g., bimolecular reactions)

iii) not every collision will lead to successful reaction (some collisions are non-productive)

As the temperature of the environment increases, the number of molecules with more energy than the intrinsic barrier increases - hence, the reaction goes faster.

If number of molecules with E > Ea is na (area under the curve)

(Ea = energy required for something (reaction) to occur)


na/ntotal = e-Ea/RT exponential function

The Reaction rate = k [A]x[B]y....with k = A e-Ea/RT (called Arrhenius equation)

A = constant with T ("Preexponential Term"). It reflects the frequency of collisions with correct geometry for reaction when the concentration = 1M

Ea = Activation Energy (depends ONLY on reaction) (the factor e-Ea/RT is always <1 (never have all collisions with E > Ea lead to successful reaction)

Then Ea is Barrier to chemical reaction


To determine Ea, determine k at 2 or more temperatures, then

Take k = Ae-Ea/RT

ln k = Ea/R x 1/T + ln A (constant)

y = M x + b

Plot ln k vs 1/T Þ straight line (especially if only two points)

Here's an example

H2(g) + I2 (g) ® 2HI (g)

observed rates:

k1 = 2.7 x 10-4 L/mol s @ 600 K

k2 = 3.5 x 10-3 L/mol s @ 650 K

1) What is Ea

k = Ae-Ea/RT

Do ratio 2 rates

k1 / k2 = Ae-Ea/R600 / Ae-Ea/R650

ln (k1 / k2 ) = -(Ea/R600)+Ea/R650

ln (k2 / k1 ) = Ea/R(1/600  -  1/650)

log (3.5 x 10-3 / 2.7 x 10-4) = [Ea / (2.303 x 8.31 J/mol K)] x (1/600 - 1/650)

(2.303 converts ln®log)

Ea = 1.11 x 2.303 x 8.31 J/mol / 1.28 x 10-4

Another example

The reaction of -IO3 and HSO3- gives a purple colour - assume we don't know the reaction products aside from the colour

The rate law was shown to be Rate = k [-IO3][HSO3-]

Ea is usually LARGE (kJ)

R is small in J

(8.314 J/kmol)

at 15C, colour change took 4.4 s

at 43C, colour change took 3.3 s (the concentrations were the same for both reactions)

To calculate Ea do a 2 point Arrhenius plot (or just take ratio of rate constants)

We know.

k1 = 1 / 4.4 (rate) x 1 / [IO3-] [H3O3-] for T1 = 15 C = 288 K

k2 = 1 / 3.3 (rate) x 1 / [IO3-] [HSO3-] for T2 = 43 C = 316 K

ln (1 / 4.4) (1 / [IO3-] [HSO3-]) = Ea/R 1/288 + ln / A

ln (1 / 3.3) (1 / [IO3-] [HSO3-]) = Ea/R 1/316 + ln / A

Both Ea and A are temperature independant

If take ratio of k1/k2

ln 3.3/ 4.4 = Ea/R [1 / 288 - 1 / 316]

Ea = 7.7 (7) x 103 J/mol

= 7.7 (7) kJ / mol

NO(g) + ½ Cl2 (g) ® ClNO nitrosyl chloride

We can draw a REACTION COORDINATE. This is a picture of the energy of the molecule(s) as the reactants are converted into products. The energy from the starting materials, via a barrier represented by the activation energy A, leads to (in this case) lower energy products. The DELTA Hreact is represented by B. The top of the curve is called the transition state, or activated complex. It has NO finite lifetime (< 10-13s)

[C......N = O] means transition state

On way to transition state, Cl moves closer starts to form bond.

Cl more away

NO + Cl reactants Cl - N= O product

Reaction Mechanism

Take place in series of simple steps involving only 1 or 2 molecules

If one molecule - UNIMOLECULAR-- if 2 moleucles, BIMOLECULAR

Three are at a time is very rare -- especially in gas phase (termolecular)

Each reaction step is called an elementary step. The series of elementary steps that leads from products to reactants is called the reaction mechanism. FOR AN ELEMENTARY STEP, THE RATE IS DETERMINED BY THE STOICHIOMETRY.

Once know the elementary steps, can deduct rate law.

For unimolecular

A ® B + C Raw Law -d[A] / dt = k(A)


A + B ® AB -d[A] / dt = k [A][B]

eg. 2NO2(g) + F2(g) ® 2 FNO2

Observed Rate Law -1/2 d[NO2]/dt = k [NO2][F2]

Recall no termolecular steps \ must be sequence of steps

Propose the following mechanism

1. NO2 + F2 FNO2 + F·      Rxn

2. F· + NO2 FNO2

if assume k1 << k2 i.e. Step 1 very slow with respect to Step 2. The observed rate for the overall reaction is the rate of the slowest step, also known as Rate Determining Step.

ie. Subsequent steps don't matter in this case, can use step, to write down rate law

\ Rxn Rate = -d[NO2]/dt = l[NO2][F2]

The two proposed steps are consistent with rate law and called a reaction mechanism. Can't prove by kinetics that this is the reaction mechanism, but can disprove proposed mechanisms by kinetics.

Another Ex

2 NO(g) + O2(g) ® 2NO2(g)

Observed Rate Law = -d[O2] / dt = k[O2][NO]2

But no termolecular reactions

A Possible Mechanism


2. N2O2 + O2 2NO2 -------------- slow

Assume k2 << k1/k-1

That is, propose step 1 is a fast equilibrium

k1 [NO]2 = k-1[N2O2]

ie. [N2O2]/ [NO]2 = k1 / k-1 = Kequi

if the equibrium is faster, then the second step is RDS

Then can write rate law

Rate = k[N2O2][O2], we can substitute the unknown [N2O2] by the equilibrium

Keq = [N2O2]/ [NO]2

[N2O2] = Keq [NO]2

The reaction rate = k2 [N2O2] [O2]

=k2 Keq [NO]2 [O2] = kobs [NO]2 [O2]

k2 and Keq are both const @ given T

Our proposed mechanism is thus consistent with this rate law.

Why do we care? What is value of knowing mechanism?

Fundamental understanding of mechanisms allows one to develop new reactions (to make plastics), interfere with reactions (design drugs) etc.

Chain Reactions

Nuclear fission, radical chain reactions to make plastics. The mechanism of these reactions has three components. Initiation, propagation and termination.

Initiate (for example by light) - generate reaactive species

H2(g)+ Cl2(g) ® 2HCl(g)

DELTA Hfo for HCl is ~ 92 kJ/mol. The reaction does not proceed unless we initiate it. Light and heat.


Initiation 1. Cl2® 2 Cl·

Propagation 2. Cl· + H2® HCl + H·

3. H· + Cl2® HCl + Cl


NOTE SUM: H2 + Cl2 ® 2HCl

is equal to the stoichiometry of the reaction

Termination: 4. Cl· + Cl·® Cl2

5. H· + Cl·® HCl

6. H· + H·® H2

all above are very exothermic usually need a third body. E.g. A wall

Other examples:


1) 2H2 + O2 ® 2H2O


Initiation: E.g. A flame ® HO· Hydroxyl radical

Propagation: E.g. HO· + H - H ® HO-H + H·

H· + O=O ® HO-O· peroxy rad

HO-O· + H-H ® HOOH + H·


Chains increase in number very rapidly - EXPLOSIVE

Another example,

2) Methane & chlorine

CH4(g) + Cl2 HCl + CH3Cl choromethane

Initiation Cl2 2Cl·

Prop. 1) Cl·  + CH4® CH3·  + HCl

2) CH3·  + Cl2 ® CH3Cl + Cl·

Termination you do it...........

3) Polymerization (like dominoes)




A catalyst is something that LOWERS to the Ea of the reaction and it is regenerated (or not consumed) at END of reaction.

B is the reaction coordinate without a catalyst, A is the reaction coordinate with a catalyst.


1) Homogenous catalyst where it is in the same phase as the reaction

eg. Both liquids

2) Heterogeneous catalyst where catalysts and compounds are in different phases

e.g. catalytic converter in auto exhaust catalyst is solid Pt/Pd/Rh on a support reagent reaction are gases

e.g. 2 H2O2 + I_ ® 2 H2O + O2 + I-

Rate Law

Rate = d[O2] / dt = k[H2O2][I-] (NOTE - ONLY 1 H2O2 in the rate law I- is the catalyst)

Proposed Mechanism


2. IOH + OH-® IO_ + H2O

3. IO_ + HO -- OH  I- + O2 + H2O

From Step 1, reaction rate = k[H2O2][I-]

ii) Heterogeneous catalysis

a) oxidation of acetone on a Cu surface

b) SO2 oxidation with O2 + V2O5(s)

2SO2(g) + O2(g) ® 2SO3(g) ® H2SO4

"Contact" process for H2SO4 manufacture

What did we learn?

A) measurement reaction rate

A + 2B ® 3C

-d[A]/dt= -1/2 d[B]/dt = 1/3 d[C]/dt

Rate law from initial reaction rates and concentration changes, specific rate constant k

For first order reaction; can integrate

ln Ao/At = kt t1/2 = ln 2/k

B) Temperature effects

Activeation energy Ea

Arrhenius equation k = A e-Ea/RT

C) Reaction mechanism

Transition states


Elementary Reaction Step

Slow Rate Determining Steps relation to rate law

Proposed Mech

With fast equilibrium steps - steady state

Chain Reactions

Initiation propagation & termination

Catalysis lowering Ea