A diamond is forever.

Is It So?

Consider:

Gibb’s Free Energy

C

diamond ?



ΔG = ∑ ΔG

products

C

graphite

- ∑ ΔG

reactants

ΔG = ΔG

graphite

- ΔG

diamond

ΔG = (0) - (3 kJ/mol) ΔG = -3 kJ/mol

Look quick, before it turns into graphite.

While it’s true her diamond is spontaneously turning into graphite before her eyes, it’s happening very slowly. Don’t hold your breath waiting to see any change. It takes billions of years.

While thermodynamics answers the question as to whether or not a reaction or event is spontaneous, it DOES NOT tell how fast a reaction goes. This is what kinetics does....describes the rate of the reaction.

Chemical Kinetics are about the only tool a chemist has to probe the actual mechanism of a chemical reaction. It is the study of reaction rates.

Reaction Rates

Reaction Rate: The slope of a Concentration vs Time graph.

NO 2  NO + O 2

Reaction Rates

Factors affecting reaction rates: • The nature of the reaction mechanism • The concentration of the reactants • • The temperature at which the reaction occurs • The presence of a catalyst The surface area of solid or liquid reactants or catalysts

Red  Blue

Rate Laws describe reaction rates mathematically



d

   

dt A

 Rate = k[reactant 1] m [reactant 2] n …

Rate Laws

To simplify the rate laws, we assume conditions where only the forward reaction is important. This produces rate laws that only contain reactant concentrations.

There are two forms of the rate law…

Differential rate laws: Shows how the rate depends on concentration. Sometimes called just the “rate law”.

Integrated rates laws: Shows how the concentration depends on time.

The choice of which rate law to use depends on the type of data that can be collected conveniently and accurately. Once you know one type, the other can be calculated.

Integrated Differential

The choice of which rate law to use depends on the type of data that can be collected conveniently and accurately. Once you know one type, the other can be calculated.

Integrated Differential

This of course requires the use of my calculus.

Other Stuff to Know: Reaction Order

The differential rate law for most reactions has the general form….

Rate = k[reactant 1] m [reactant 2] n … The exponents m and n are called reaction orders. Their sum (m + n) is called the overall reaction order.

Other Stuff to Know: Reaction Order

Usually the values of the reaction order must be determined experimentally, they cannot be predicted by looking at the chemical reaction. In most rate laws, reaction orders are 0, 1, or 2. However, occasionally they are a fraction or even negative.

The rate of a reaction depends on concentration; however the rate constant does not.

Finding The Differential Rate Law

The most common method for experimentally determining the differential rate law is the method of initial rates.

In this method several experiments are run at different initial concentrations and the instantaneous rates are determined for each at the same value of time (as near t = 0 as possible)

Example:

Using Initial Rates to Determine the Form of the Rate Law

A + B



C

2 3 Exp # [A] [B] Initial Rate (M/s) 1 .100M .100M 4x10 -5 .100M .200M 4x10 -5 .200M .100M 16x10 -5 From this data, find the form of the rate law..

Rate = k[A]

m

[B]

n

Exp # [A] [B] Initial Rate (M/s) 1 2 3 .100M .100M 4x10 .100M .200M 4x10 -5 -5 .200M .100M 16x10 -5

Rate = k[A]

m

[B]

n

Rate

2

Rate

1  4  10  5 4  10  5 

k

[.

100 ]

m k

[.

100 ]

m

 [.

200 ] [.

100 ]

n n

1  [.

200 ] [.

100 ]

n n

1 

[B]

0

= 1

2

n

n = 0

Exp # [A] [B] Initial Rate (M/s) 1 2 3 .100M .100M 4x10 .100M .200M 4x10 -5 -5 .200M .100M 16x10 -5

Rate

3

Rate

1  16  10  5 4  10  5 

k

[.

200 ]

m k

[.

100 ]

m

 [.

100 ]

n

[.

100 ]

n

4  [.

200 ]

m

[.

100 ]

m

4  2

m

Rate = k[A]

2

[B]

0

= k [A]

2

m = 2

Exp # [A] [B] Initial Rate (M/s) 1 2 3 .100M .100M 4x10 .100M .200M 4x10 -5 -5 .200M .100M 16x10 -5 Now, solve for k…

k



rate

[

A

] 2 [

B

] 0  4  10  5 [.

100 ] 2  4  10  3

Rate = 4x10

-3

[A]

2

First Order Reactions

Rate = 

d

   

dt A

 Integrating, etc. leads to… [A] = [A] o e -kt Ln [A] t – ln [A] 0 = -kt ln  

A

0

t

   

kt

Derive the Integrated First Order Rate Law:

Given a first order differential rate law…

Rate

  [

A

] 

t



k

[

A

] That is, the rate of change of the concentration of reagent A is proportional to how much A there is.

Let x = [A] (the concentration of A), Then…



x



t

 

kx

The minus sign is because the value of [A], or x, is decreasing.

Transforming to calculus notation…

dx

 

kx dt

Separate the variables…

dx

 

k dt x

Integrate both sides…  1

x dx

  

k dt

 1

x dx

  

k dt

Integrating… Solving for x… ln

x

 

kt



C e

ln

x



e



kt



C x



e



kt e C x



e C e



kt

x



e



kt e C x



e C e



kt

Now use initial conditions to determine C.

At t = 0, x = [A] o [A] o = initial concentration of A [

A

]

o



e C e



k

( 0 ) ln

C

[

A

]

o

 ln  ln [

A

]

o e C



e C e C

 [

A

]

o

Recall that..

x



e C e



kt x

 [

A

]  [

A

]

o e



kt

x

 [

A

]  [

A

]

o e



kt

Take ln’s of both sides… ln [

A

]  ln  [

A

]

o e



kt

  ln [

A

]

o

 ln

e



kt

So… ln [

A

]  ln [

A

]

o

ln [

A

]  ln[

A

]

o

 

kt

First Order Reactions

Rate

 

d

   

dt A

 Ln [A] t – ln [A] 0 = -kt ln  

A

0

t

   

kt

These equations can be used to determine … • The concentration of a reactant remaining at any time after the reaction has started.

• The time required for a given fraction of a sample to react.

• The time required for a reactant concentration to reach a certain level.

Half Life (First Order Process)

Half Life of a Reaction

The half life of a reaction, t of its initial value. ½ , is the time required for the concentration of a reactant to drop to one half ln    

A t

  0    

kt

ln    1 2

A

0   0     

kt

1 2 ln (½) = -kt ½

t

1 2   ln   2

k

T ½ = 0.693/k Half life for a first order rate law is independent of the initial concentration of reactant. Thus, the half life is the same at any time during the reaction.

Half Life for a First Order Reaction (Garden Variety Half Life – Radioactivity)

First order reaction example

The first order rate constant for the decomposition of a certain insecticide in water at 12 12 o C.

o C is 1.45 1, leading to a concentration of 5.0x10

-7 g/mL  yr . A quantity of this insecticide is washed into a lake on June g/ml water. Assume that the effective temperature of the lake is a. What is the concentration of the insecticide on June 1 of the following year?

b. How long will it take for the concentration to drop to 3.0x10

-7 g/ml?

a. What is the concentration of the insecticide on June 1 of the following year?

Ln [A] t – ln [A] 0 = -kt Ln [A] t = -kt + ln [A 0 ] t=0 ln [insecticide] t=1 year g/ml) = - (1.45g/mL  yr)(1 yr) + ln (5.0x10

-7 ln [insecticide] t=1 year = - (1.45g/mL) + (-14.51g/mL) ln [insecticide] t=1 year = -15.96 g/mL [insecticide] t=1 year = e -15.96

[insecticide] t=1 year = 1.2x10

-7 g/mL

b. How long will it take for the concentration of the insecticide to drop to 3.0x10

-7 g/ml?

ln [A] t = -kt + ln [A 0 ] t=0 ln (3.0x10

-7 g/ml) = - (1.45 yr -1 )(t) + ln (5.0x10

-7 g/ml) t = -[ln (3.0x10

-7g /ml) – ln (5.0x10

-7 g/ml)]/1.45yr

-1 t = - [-15.02 + 14.51]/1.45yr

-1 t = 0.35 yr

Second Order Reactions

A second order reaction is one whose rate depends on the reactant concentration raised to the second power, or on the concentrations of two different reactants, each raised to the first power. For a reaction that is second order in just one reactant, A, the rate law is given by…

Rate = k[A]

2

Second Order Reactions

Rate = k[A] 2 

   

2 

t

Integrating and stuff leads to the Integrated Rate Law… 1  

t



kt

 1   0

Second Order Half Life’s

1  

t



kt

 1   0 Setting [A] = ½[A] o and solving for time gives…

t

1 2 

k

1   0 Unlike the half life of first order reactions, the half life of a second order reaction is dependent on the initial concentration of the reactant.

Half Life for a Second Order Reaction

Zero Order Reaction

A zero order reaction is one whose rate depends on the reactant concentration raised to the zero power, or in other words, it does not depend on the concentration of the reactant as long as some reactant is present. For the reaction that is zero order in just one reactant, A, the rate law is given by… Rate = k[A] 0 Integrating… [A] t = -kt + [A] 0 Solving for t ½ T ½ = [A] 0 /2k

Half Life for a Zero Order Reaction: (Rate doesn’t depend on concentration)

Rate Laws: A Summary

To simplify the rate laws, we assume conditions where only the forward reaction is important. This produces rate laws that only contain reactant concentrations.

There are two types of rate laws… Differential rate laws: Shows how the rate depends on concentration.

Integrated rate laws: Shows how concentration depends on time.

Rate Laws: A Summary

The most common method for experimentally determining the differential rate law is the method of initial rates. In this method several experiments are run at different initial concentrations and the instantaneous rates are determined for each at the same value of time (as near t=0 as possible).

The point is to evaluate the rate before the concentrations change significantly from the initial values. From a comparison of the initial rates and the initial concentrations the dependence of the rate on the concentrations of various reactants can be obtained – that is the order of each reactant can be determined.

Rate Laws: Summary

To experimentally determine the integrated rate law for a given reaction, concentrations are measured at various values of time as the reaction proceeds. Then the job is to see which integrated rate law correctly fits the data. Typically this is done visually, by ascertaining which type of plot gives a straight line. Once the correct straight line plot is found, the correct integrated rate law can be chosen and the value of the rate constant, k, can be obtained from the slope of the plot. Also, the differential rate law for the reaction can be calculated.

All the Good Stuff

Reaction Order Zeroth First Differential Rate Law 

d dt



k

Integrated Rate Law Plot for Straight Line    

o



kt vs t

Slope of Plot -k -k 

d

   

dt A

    

o e



kt

ln

vs t

Second 

d

   

dt A

 2  1   

kt

 

o

1  

vs t

k

We can use this information to experimentally find the rate law for a reaction.

Reaction Order Zeorth First Differential Rate Law 

d dt



k

Integrated Rate Law Plot for Straight Line    

o



kt vs t

Slope of Plot -k -k 

d

   

dt A

    

o e



kt

ln

vs t

Second 

d

   

dt A

 2  1   

kt

 

o

1  

vs t

k

The gas phase decomposition of NO 2 383 o C giving the following data.

is studied at NO 2  NO + O 2 Time (sec) 50 100 150 200 250 Conc. NO 2 Molar 0.1000

0.0170

0.0090

0.0062

0.0047

Time 50 100 150 200 250 Graph the ln (concentration) vs Time. It the plot is a straight line, the reaction is first order.

Original Data

Molar Conc.

0.1000

ln Conc.

-2.3026

0.0170

0.0090

-4.0745

-4.7105

0.0062

0.0047

-5.0832

-5.3602

0.0000

0 -1.0000

-2.0000

-3.0000

-4.0000

-5.0000

-6.0000

ln Concentration vs Time

5 10 15 20 25

Time (sec)

Graph

Original Data Graph the 1/(concentration) vs Time. It Reaction constant k = 10.16

second order. Time 50 100 150 200 250 Molar Conc.

0.1000

Conc.-1 10.0000

0.0170

58.8235

0.0090

111.1111

0.0062

161.2903

0.0047

212.7660

concentration -1 vs Time

250.0000

200.0000

150.0000

100.0000

50.0000

0.0000

0 y = 10.16x + 9.1984

5 10

Time

15 20 Straight line  Order Reaction Second 25

Remember that the rate equation between two substances A and B looks like this… Rate = k[A] m [B] n The equation shows the effect of changing the concentrations of the reactants on the rate of the reaction. What about all the other things like temperature and catalysts, for example which also change rates of reaction? Where do these fit into this equation?

These are all included in the so-called rate constant – which is only actually constant if all you are changing is the concentration or the reactants. If you change the temperature or the catalyst the rate constant changes. These changes of the rate constant are shown mathematically in the Arrhenius Equation.

Svante August Arrhenius

Svante August Arrhenius was a Swedish physical chemist best known for his theory that electrolytes, certain substances that dissolve in water to yield a solution that conducts electricity, are separated, or dissociated, into electrically charged particles, or ions, even when there is no current flowing through the solution.

In 1903 he was awarded the Nobel Prize for Chemistry.

1859-1927

The Arrhenius Equation

k



Ae



E a RT

K = rate constant A = frequency factor E a = activation energy R = the gas constant (8.314 J/mol·K) T = temperature in Kelvin

The Arrhenius Equation

k



Ae



E activation RT Or the logarithmic form…

ln

k

 ln

A



E activation RT

The expression…

e



E activation RT

For reasons that are beyond the scope of any course at this level, this expression counts the fraction of the molecules present in a gas which have energies equal to or in excess of the activation energy at a particular temperature.

k



Ae



E activation RT

The frequency factor, A is sometimes called the pre-exponential factor of the steric factor.

It is a term which includes factors like the frequency of collisions and their orientation. It varies slightly with temperature, although not much. It is often taken as constant across small temperature ranges.

Using the Arrhenius Equation: The effect of a change of temperature.

k



Ae



E activation RT

You can use the Arrhenius equation to show the effect of a change of temperature on the rate constant – and therefore on the rate of the reaction. If the rate constant doubles, for example, so will the rate of the reaction.

Example: What happens to the rate of a reaction if the temperature increases from 20 o C to 30 o C?

The frequency factor, a, is approximately constant for such a small temperature change. We need to look at how e excess of the activation energy.

-(Ea/RT) changes --- the fraction of molecules with energies equal to or in Assume an activation energy of 50 kJ/mol. In the equation, we have to write that as 50,000 J/mol. The value of the gas constant R is 8.31 J/K·mol Now raise the temperature just a little bit to 30 o C (303k)

e



E activation RT

 1 .

2  10  9 

e

 50 , 000 8 .

31  293

e

 

E activation RT

2 .

4   10  9

e

 50 , 000 8 .

31  303

e

At 20 

E activation RT

 1 .

2  10  9 o C 

e

 50 , 000 8 .

31  293 At 30 o C

e



E activation RT



e

 50 , 000 8 .

31  303  2 .

4  10  9 The fraction of the molecules able to react has almost doubled by increasing the temperature by 10 o C. That causes the rate of the reaction to almost double.

This is the basis for the old rule of thumb that a reaction rate doubles for every 10 o C rise in temperature.

At the higher temperature, more molecules have energy greater than the activation energy Ea.

Catalysts:

A substance that changes the rate of a reaction without being consumed in the reaction.

• • Provides an easier way to react.

Lowers activation energy Cattleist?

• Enzymes are biological catalysts.

• Inhibitor: A substance that decreases the rate of reaction (a negative catalyst)

The effect of a catalyst is to lower the activation energy.

Lower activation energy means more molecules will react.

Uncatalyzed Catalyzed Energy Energy

The effect of a catalyst

A catalyst provides an alternate reaction mechanism, or route for the reaction. This alternate route necessarily has a lower activation energy. The overall effect of a catalyst is to lower the activation energy.

Continuing our example of a reaction with an activation energy of 50,000 J/mol… What is the effect on the rate of lowering the activation to 25,000 J/mol?

With Catalyst

e



E activation RT

  3 .

5  10  5

e

 25 , 000 8 .

31  293 Without Catalyst

e



E activation RT

 1 .

2  10  9 

e

 50 , 000 8 .

31  293 No wonder catalysts speed up reactions!

The Collision Model

Molecules must collide to react. The greater the number of collisions occurring per second, the greater the reaction rate.

Only a small fraction of collisions actually lead to a reaction.

In 1888 the Swedish chemist Svante Arrhenius suggested that molecules must possess a certain minimum amount of energy in order to react. In order to react, colliding molecules must have a total kinetic energy equal to or greater than some minimum value. This minimum energy is called the activation energy (E a ). The value of E a varies from reaction to reaction.

Factors Influencing Reaction Rates

The rate of a given reaction can be affected by: • The physical states of the reactants • The concentration of the reactants (for gases, pressure is equivalent to concentration) • The reaction temperature • The presence of a catalyst

The greater the concentration, the greater the frequency of collisions.

The Greater the Temperature the Greater the Frequency of Collisions.

An increase in temperature generally increases the rate of a reaction.

According to kinetic theory, at higher temperature the atoms and molecules move faster.

1. They collide more frequently 2. They collide with greater energy and each collision has a greater chance of being successful.

Note: The activation energy does NOT get any lower!

Not all collisions produce a reaction

Energy Diagrams

Intermediaries – Activated Complexes Reactants Products

Molecularity

Chemical reactions take place as a result of collisions between molecules. Each type of collision in a given reaction is called a single event or an elementary step.

The number of molecules that participate as reactants in an elementary step defines the molecularity of the step.

Unimolecular: A single molecule is involved in the elementary step Bimolecular: Two molecules are involved Termolecular: Three molecules are involved. Termolecular steps are far less probable than unimolecular or bimolecular processes and are rarely encountered.

The chances that four or more molecules will collide simultaneously with any regularity is extremely remote, consequently, such collisions are never proposed as a part of a reaction mechanism.

Example:

The reaction between NO 2 overall reaction: NO 2 + CO  and CO has the NO + CO 2 A study of the kinetics of this reaction revealed the rate law for the reaction is.

Rate = k[NO 2 ] 2 This Rate Law requires that the slow step of the reaction involves a collision between two NO simple reaction above?

2 molecules. How can this be a step in the seemingly

NO 2 + CO  NO + CO 2 Further study of this reaction established that two NO 2 molecules can react as follows… NO 3 NO 2 + NO 2  NO 3 + NO is a highly reactive material which is capable of transferring an oxygen atom.

NO 3 + CO  NO 2 + CO 2 2NO 2 + NO 3 + CO  NO 2 + NO 3 + NO + CO 2 NO 2 + CO  NO + CO 2

NO 2 + CO  NO + CO 2 The first equations sets the rate law, so it must be the slow one.

NO 2 NO 3 + NO 2 + CO   NO 3 NO 2 + NO + CO 2 NO 2 + CO  NO + CO 2 slow fast Remember, the rate law only provides information about the slowest reaction in the mechanism.

Experimental evidence for reaction mechanisms: Consider the reaction of methyl acetate with water….

Methyl Acetate + water  H O H H-C-C-O-C-H + H 2 O  H H Acetic Acid + Methanol H H H-C-C-OH + H-C-OH H H The methyl acetate molecule can break in one of two places to complete the reaction…

Methyl Acetate + water  H O H H-C-C-O-C-H + H 2 O  H H Acetic Acid + Methanol H H H-C-C-OH + H-C-OH H H H O H-C-C H H H O-H O-C-H H -OR-

Methyl Acetate + water  H O H H-C-C-O-C-H + H 2 O  H H Acetic Acid + Methanol H H H-C-C-OH + H-C-OH H H H O H-C-C H H O-H H O-C-H H -OR H O H-C-C-O H H O-H How can we determine experimentally which is correct?

H -C-H H

Use a radioactive isotope of oxygen, O 18 To make up the water H O H 18 H O H-C-C O 18 -H H

Methyl Acetate + water  H O H H-C-C-O-C-H + H 2 O  H H Acetic Acid + Methanol H H H-C-C-OH + H-C-OH H H H O H-C-C H H H O-H O-C-H H -OR H O H-C-C-O H H O-H H -C-H H

Elementary Steps

Experimental studies of reaction mechanisms begin with rate measurements. Next this data is analyzed to determine the rate constant and order of the reaction. The rate law is then written, and finally a plausible mechanism for the reaction in terms of elementary steps is formulated.

The elementary steps must satisfy two requirements…

1. The sum of the elementary steps must give the overall balanced equation for the reaction.

2. The rate-determining step, which is the slowest step in the sequence of steps leading to product formation, should predict the same rate law as is determined experimentally.

Rate Laws of Multistep Mechanisms

• The slowest elementary step is the rate determining step.

• The rate determining step governs the rate law for the overall reaction.

• For elementary steps, the exponent in its rate are the same as the coefficients of the reactants in the chemical equation. Thus, the rate law for an elementary process can be predicted from the chemical equation for the process.

Rate Laws of Elementary Steps

A  Products A+A  Products A+B  Products A+A+A  Products A+A+B  A+B+C  Products Products Rate = k[A] Rate = k[A] 2 Rate = k[A][B] Rate = k[A] 3 Rate = k[A] 2 [B] Rate = k[A][B][C]

Measuring the rate of a reaction.

Formulating the Rate Law Postulating a reasonable reaction mechanism Sequence of steps in the study of a reaction mechanism

Example:

The decomposition of Hydrogen Peroxide is facilitated by iodide ions.

The overall reaction is… 2H 2 O 2  2H 2 O + O 2 By experiment, the rate law is found to be… rate = k[H 2 O 2 ][I ] Thus the reaction is first order with respect to both H 2 O 2 and I .

Decomposition of Hydrogen Peroxide

2H 2 O 2  2H 2 O + O 2 rate = k[H 2 O 2 ][I ] You can see that H equation.

2 O 2 decomposition does not occur in a single elementary step corresponding to the overall balanced If it did, the reaction would be second order in H 2 O 2 (as a result of the collision of two H 2 O 2 I molecules). What’s more, the ion, which is not even part of the overall equation, appears in the rate law expression.

We can account for the observed rate law by assuming that the reaction takes place in two separate elementary steps, each of which is bimolecular: step 1: H 2 O 2 step 2: H 2 O 2 + I + IO   H 2 O + IO H 2 O + O 2 + I -

step 1: H 2 O 2 step 2: H 2 O 2 + I + IO   H 2 O + IO H 2 O + O 2 Slow + I The first step must be the rate determining step, because it matches the rate law. Thus, the rate of the reaction can be determined from the first step alone:

rate = k[H

2

O

2

][I

-

]

Note that the IO ion is an intermediate because it does not appear in the overall balanced equation. Although the I ion also does not appear in the overall equation, I former is present at the start of the reaction and at its completion. The function of I differs from IO in that the is to speed up the reaction --- that is, it is a catalyst.

Another Example

: Consider the Reaction… 2NO 2 Cl  2NO 2 + Cl 2 This is found to be a first order reaction, it’s experimentally determined rate law is… rate = k[NO 2 Cl] Could the overall reaction occur in a single step by the collision of two NO 2 Cl molecules? This would lead to… rate = k[NO 2 Cl] 2 Chemists believe the actual mechanism is… NO 2 Cl  NO 2 Cl + Cl  NO 2 NO 2 + Cl + Cl 2 Slow Fast

Still Another Example:

Consider the reaction… 2NO + 2H 2  N 2 + 2H 2 O Experimentally, the rate law has been found to be… rate = k[NO] 2 [H 2 ] A chemically reasonable mechanism that yields the correct form for the rate law is… 2NO + H 2 N 2 O + H 2   N 2 O + H 2 O (slow) N 2 + H 2 0 (fast) However, there is a serious flaw in the proposed mechanism -- it has a elementary process that involves the simultaneous collision between three molecules, two of which must be NO and one H 2 .

Consider the reaction… 2NO + 2H 2  N 2 + 2H 2 O An alternative proposal is… 2NO ↔ N 2 O 2 (fast) N 2 O 2 + H 2 N 2 O + H 2   N N 2 2 O + H + H 2 2 O O (slow) (fast) Since the second step is rate determining, the rate for the reaction should be… rate = k[N 2 O 2 ][H 2 ] However, the experimental rate law does not contain the species N 2 O 2 , an intermediate.

2NO ↔ N 2 O 2 (fast) N 2 O 2 + H 2 N 2 O + H 2   N N 2 2 O + H + H 2 2 O O (slow) (fast) The first elementary step is an equilibrium reaction, so consider both forward and reverse reactions.

Rate (forward reaction) = k f [NO] 2 Rate (reverse reaction) = k r [N 2 O 2 ] If, however, we view the first equation as a dynamic equilibrium reaction, then the rate of the forward and reverse reactions are equal. Then… K f [NO] 2 = k r [N 2 O 2 ] [N 2 O 2 ] = k f /k r [NO] 2

K f [NO] 2 = kr[N 2 O 2 ] [N 2 O 2 ] = k f /k r [NO] 2 Substituting into the proposed rate law (rate = k[N 2 O 2 ][H 2 ]) gives….

rate = k (k f /k r )[NO] 2 [H 2 ] Or rate = k’ [NO] 2 [H 2 ] The rate law derived from the proposed mechanism matches the experimental rate law, thus the three step mechanism does appear to be reasonable on the basis of kinetics.