chemical kinetics 2009-20101 chemical kinetics the area of chemistry concerned with the speeds, or...

Chemical Kinetics 2009-2010 1

Chemical Kinetics

The area of chemistry concerned with the speeds, or rates, at which

a chemical reaction occurs.


Reaction Rate

The reaction rate is the change in the concentration of a reactant or a product with time, (M/s or M . s-1), where M is molarity and s represents seconds.

Another way to represent rate is mol . L-1 s-1


Factors that Influence Reaction Rate

Under a given set of conditions, each reaction has its own characteristic rate, which is ultimately determined by the chemical nature of the reactants. (You will remember this from Chem I - potassium and water have a different rate of reaction than iron and oxygen.)

For a given reaction (using the same reactants), we can control four factors that affect its rate: the concentration of reactants, their physical state, the temperature at which the reaction occurs, and the use of a catalyst.


The Collision Theory of Chemical Kinetics

The kinetic molecular theory of gases postulates that gas molecules frequently collide with one another. Therefore, it seems logical to assume – and it is generally true - that chemical reactions occur as a result of collisions between reacting molecules.

In terms of the collision theory of chemical kinetics, then, we expect the rate of a reaction to be directly proportional to the frequency of the collisions (number of molecular collisions per second).

Rate collision frequency

Means “proportional to”


ConcentrationSince molecules must collide in order to

react, the more frequently they collide, the more often a reaction occurs. Thus, reaction rate is proportional to the concentration of reactant

Therefore, if we increase the concentration…we increase the collision

frequency, which…increases the rate

Rate collision frequency concentration


Physical State

Molecules must mix in order to collide. When reactants are in the same phase, as in aqueous solution, occasional stirring keeps them in contact. When they are in different phases, more vigorous mixing is needed. The more finely divided a solid or liquid reactant, the greater the surface are per unit volume, the more contact it makes with the other reactant, and the faster the reaction.


TemperatureMolecules must collide in order to react.

Since the speed of a molecule depends on its temperature, more collisions will occur if the temperature is increased.

Speed of a molecule

Number of collisions


TemperatureMolecules must also collide with enough

energy to react. Increasing the temperature increases the kinetic energy of the molecules, which in turn increases the energy of the collisions.

Therefore, at a higher temperature, more collisions occur with enough energy to react. Thus, raising the temperature increases the reaction rate by increasing the number and especially the energy of the collisions.


Temperature

Two familiar kitchen appliances employ this effect: a refrigerator slows down chemical processes that spoil food, whereas an oven speeds up other chemical processes to cook it.


Expressing Reaction RateBefore we can deal quantitatively with the effects of

concentration and temperature on reaction rate, we must be able to express the rate mathematically. A rate is a change in some variable per unit of time.

For example, the rate of motion of a car is the change of position of the car divided by time. A car that travels 57 miles in 60. minutes is traveling at…

57 miles/60. minutes = .95 miles/min

In the case of chemical reactions, the positions of the substances do not change over time, but their concentrations do.


We know that any reaction can be represented by the general equation

reactants products

This equation tells us that during the course of a reaction, reactants are consumed while products are formed. As a result, we can follow the progress of a reaction by monitoring either the decrease in concentration of the reactants or the increase in concentration of the products.


The following figure shows the progress of a simple reaction in which A molecules are converted to B molecules:

A B

A

B


The decrease in number of A molecules and the increase in the number of B molecules with time are shown below.

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In general, it is more convenient to express the reaction rate in terms of the change in concentration with time. Thus, for the reaction A B we can express the rate as:

Rate = -[A]

t

[A]final – [A]initial

Because the concentration of A decreases during the time interval, [A] is a negative quantity. Because the rate of a reaction is always a positive quantity, a minus sign is needed in the rate expression to make the rate positive.


or[B]

tRate =

The rate of product formation does not require a minus sign because [B], ([B]final – [B]initial) is a positive quantity (the concentration increases with time), so rate is a positive value already.


These rates are average rates because they are averaged over a certain time period (t).

Rate can be expressed as M/s, M . s-1, or mol . L-1 . s-1

Rate = -[A]

tor

[B]

tRate =


Reaction Rates and Stoichiometry

We have seen that for stoichiometrically simple reactions of the type A B, the rate can either be expressed in terms of the decrease in reactant concentration with time, -[A]/t, or the increase in product concentration with time, [B]/t.

For more complex reactions, we must be careful in writing the rate expressions.

Consider the reaction2A B

Two moles of A disappear for each mole of B that forms


Another way to think of this is to say that the rate of disappearance of A is twice as fast as the rate of appearance of B. We write the rate as either

Rate = -

1 [A]

2 tor Rate =

[B]

t

For the reaction

2A B


In general, for the reaction

aA + bB cC + dD

The rate is given by

Rate = - 1 [A]

a t

1 [B]

b t= - =

1 [C]

c t=

1 [D]

d t

1 [A]

a t

1 [C]

c t


Write the expression for the following reactions in terms of the disappearance of the reactants and the appearance of the products:

3O2(g) 2O3(g)

Rate = 1 [O3]

2 t=-

1 [O2]

3 t


If O2 is disappearing at .25 M/s, what is the rate of formation of O3?

1 [O3]

2 t=

1 -.25M

3 s- =

.50Ms = 3(x)s x = .17M

1 [O2]

3 t-

1 x

2 s


By definition, we know that to determine the rate of a reaction we have to monitor the concentration of the reactant (or product) as a function of time.

• For reactions in solution, the concentration of a species can often be measured by spectroscopic means.

• If ions are involved, the change in concentration can also be detected by an electrical conductance measurement.

• Reactions involving gases are most conveniently followed by pressure measurements.


Reaction of Molecular Bromine and Formic Acid

In aqueous solutions, molecular bromine reacts with formic acid (HCOOH) as follows:

Br2(aq) + HCOOH(aq) 2Br-(aq) + 2H+(aq) + CO2(g)

Reddish-brown

colorless colorless colorless colorless

The rate of Br2 disappearance can be determined by monitoring the color over time.As the reaction proceeds, the color of the solution … goes from brown to colorless


Reaction of Molecular Bromine and Formic Acid

As the reaction proceeded, the concentration of Br2 steadily decreased and the color of the solution faded.



Measuring the change (decrease) in bromine concentration at some initial time ([Br2]0) and then at some other time, ([Br2]t) allows us to determine the average rate of the reaction during that interval:

[Br2]

tAverage rate =

[Br2]t – [Br2]0

tfinal – tinitial

Average rate =


Use the data in the following table to calculate the average rate over the first 50 second time interval.

Time(s)

[Br2]

(M)

Rate(M/s)

0.0 0.0120 4.20 x 10-5

50.0 0.0101 3.52 x 10-5

100.0 0.00846 2.96 x 10-5

150.0 0.00710 2.49 x 10-5

200.0 0.00596 2.09 x 10-5

250.0 0.00500 1.75 x 10-5

300.0 0.00420 1.48 x 10-5

350.0 0.00353 1.23 x 10-5

400.0 0.00296 1.04 x 10-5


Time(s)

[Br2]

(M)

Rate(M/s)

0.0 0.0120 4.20 x 10-5

50.0 0.0101 3.52 x 10-5

100.0 0.00846 2.96 x 10-5

150.0 0.00710 2.49 x 10-5

200.0 0.00596 2.09 x 10-5

250.0 0.00500 1.75 x 10-5

300.0 0.00420 1.48 x 10-5

350.0 0.00353 1.23 x 10-5

400.0 0.00296 1.04 x 10-5

(0.0101 – 0.0120)M

50.0 – 0.0)sAverage rate =

= 3.80 x 10-5 M/sNotice, this average rate is slower than rate at time 0, but faster than the rate at time 50


Time(s)

[Br2]

(M)

Rate(M/s)

0.0 0.0120 4.20 x 10-5

50.0 0.0101 3.52 x 10-5

100.0 0.00846 2.96 x 10-5

150.0 0.00710 2.49 x 10-5

200.0 0.00596 2.09 x 10-5

250.0 0.00500 1.75 x 10-5

300.0 0.00420 1.48 x 10-5

350.0 0.00353 1.23 x 10-5

400.0 0.00296 1.04 x 10-5

Now use the data in the same table to calculate the average rate over the first 100 second time interval.


Time(s)

[Br2]

(M)

Rate(M/s)

0.0 0.0120 4.20 x 10-5

50.0 0.0101 3.52 x 10-5

100.0 0.00846 2.96 x 10-5

150.0 0.00710 2.49 x 10-5

200.0 0.00596 2.09 x 10-5

250.0 0.00500 1.75 x 10-5

300.0 0.00420 1.48 x 10-5

350.0 0.00353 1.23 x 10-5

400.0 0.00296 1.04 x 10-5

Avg rate = (0.00846 – 0.0120)M

100.0 – 0.0)s

= 3.54 x 10-5 M/sNotice, this average rate is slower than rate at time 0, but faster than the rate at time 100


These calculations demonstrate that the average rate of the reaction depends on the time interval we choose.

By calculating the average reaction rate over shorter and shorter intervals, we can obtain the rate for a specific instant in time, which gives us the instantaneous rate of the reaction at that time.


The figure below shows the plot of [Br2] versus time, based on the data table given previously. Graphically, the instantaneous rate at 100 seconds after the start of the reaction is the slope of the line tangent to the curve at that instant.

The instantaneous rate at any other time can be determined in a similar manner.

Unless otherwise stated, we will refer to the instantaneous rate as simply “the rate”.


rate [Br2]

rate = k[Br2]

At a specific temperature, a rate constant (k) is a constant of proportionality between the reaction rate and the concentrations of reactants.

k is specific for a given reaction at a given temperature; it does not change as the reaction proceeds.

k, the Rate Constant


Rearrange the equation rate = k[Br2]

To solve for k

k =

Rate

[Br2]

Since reaction rate has the units M/s, and [Br2] is in M, the unit of k for this first order reaction is 1/s or s-1.


Time(s)

[Br2]

(M)

Rate(M/s)

0.0 0.0120 4.20 x 10-5

50.0 0.0101 3.52 x 10-5

100.0 0.00846 2.96 x 10-5

150.0 0.00710 2.49 x 10-5

200.0 0.00596 2.09 x 10-5

250.0 0.00500 1.75 x 10-5

300.0 0.00420 1.48 x 10-5

350.0 0.00353 1.23 x 10-5

400.0 0.00296 1.04 x 10-5

k = rate/[Br2]

(s-1)

Calculate the rate constant for the following reaction


k = rate/[Br2]

(s-1)

Time(s)

[Br2]

(M)

Rate(M/s)

0.0 0.0120 4.20 x 10-5

50.0 0.0101 3.52 x 10-5

100.0 0.00846 2.96 x 10-5

150.0 0.00710 2.49 x 10-5

200.0 0.00596 2.09 x 10-5

250.0 0.00500 1.75 x 10-5

300.0 0.00420 1.48 x 10-5

350.0 0.00353 1.23 x 10-5

400.0 0.00296 1.04 x 10-5

Because k is a constant (for this reaction at this specific temperature), it doesn’t matter which row we consider, so let’s consider the data at time 0.0 seconds…

rate = k[Br2]

3.50 x 10-3

k = 4.20 x 10-5

0.0120= 3.50 x 10-3

rate

[Br2] k =


Time(s)

[Br2]

(M)

Rate(M/s)

0.0 0.0120 4.20 x 10-5

50.0 0.0101 3.52 x 10-5

100.0 0.00846 2.96 x 10-5

150.0 0.00710 2.49 x 10-5

200.0 0.00596 2.09 x 10-5

250.0 0.00500 1.75 x 10-5

300.0 0.00420 1.48 x 10-5

350.0 0.00353 1.23 x 10-5

400.0 0.00296 1.04 x 10-5

k = rate/[Br2]

(s-1)

3.50 x 10-3

To prove that k is a constant, calculate k at time 200.0 seconds

The slight variations in the values of k are due to experimental deviations in rate measurements.

3.51 x 10-3


Time(s)

[Br2]

(M)

Rate(M/s)

0.0 0.0120 4.20 x 10-5

50.0 0.0101 3.52 x 10-5

100.0 0.00846 2.96 x 10-5

150.0 0.00710 2.49 x 10-5

200.0 0.00596 2.09 x 10-5

250.0 0.00500 1.75 x 10-5

300.0 0.00420 1.48 x 10-5

350.0 0.00353 1.23 x 10-5

400.0 0.00296 1.04 x 10-5

k = rate/[Br2]

(s-1)3.50 x 10-3

3.49 x 10-3

3.50 x 10-3

3.51 x 10-3

3.51 x 10-3

3.50 x 10-3

3.52 x 10-3

3.48 x 10-3

3.51 x 10-3

Filling in the rest of the table…


As we will see in the next section, the label for k is determined by the overall

reaction order

Overall Reaction

Order

Units of k (when t is seconds)

0 mol/L . s (or mol L-1 s-1)

1 1/s (or s-1)

2 L/mol . s (or L mol-1 s-1)

3 L2/mol2 . s (or L2 mol-2 s-1)


It is important to understand that k is NOT affected by the concentration of Br2.

• The rate is faster at a higher concentration and slower at a lower concentration of Br2, but the ratio of rate/[Br2] remains the same provided the temperature doesn’t change.


The Rate LawThe rate law expresses the relationship of the rate of a reaction

to the rate constant (k) and the concentrations of the reactants raised to a power. For the general reaction

aA + bB cC + dD

The rate law takes the form

Rate = k[A]x[B]y

Where x and y are numbers that must be determined experimentally.

Note – in general, x and y are NOT equal to the stoichiometric coefficients a and b from the overall balanced chemical equation. When we know the values of x, y and k, we can use the rate equation shown above to calculate the rate of the reaction, given the concentrations of A and B.


The reaction orders define how the rate is affected by the concentration of each reactant.

This reaction is xth order in A, yth order in B.

Rate = k[A]x[B]y


The following rate law was determined for the formation of nitrogen (VI) oxide and molecular oxygen from nitrogen (IV) oxide and ozone

Rate = k[NO2][O3]

How would the rate of this reaction be affected if the concentration of NO2 increased from 1.0 M to 2.0 M?


Rate = k[NO2][O3]

This reaction is first order with respect to both NO2 and O3. This means that doubling the concentration of either reactant would double the rate of the reaction.

(2)1 = 2How many times greater the concentration is What the order is for

that reactant

How many times greater the rate of the reaction will be


The reaction between nitrogen monoxide and molecular oxygen is described by a different rate law.

Rate = k [NO]2[O2]

How would the rate of this reaction be affected if the concentration of NO increased from 1.0 M to 2.0 M?


Rate = k[NO]2[O2]

This reaction is second order with respect to NO and first order with respect to O2.

Doubling the concentration NO would increase the rate by a factor of four

(2)2 = 4How many times greater the concentration is What the order is for

that reactant

How many times greater the rate of the reaction will be


Rate = k [NO]2[O2]

How would the rate of this reaction be affected if the concentration of NO increased from 1.0 M to 5.0 M?

In this case, when the concentration of NO is multiplied by 5, the rate increases by a factor of 25!

(5)2 = 25


The exponents x and y specify the relationships between the concentrations of reactants A and B and the reaction rate. Added together, they give us the overall reaction order, defined as the sum of the powers to which all reactant concentrations appearing in the rate law are raised. For the equation

Rate = k[A]x[B]y

The overall reaction order is x + y.


For the following reaction

(CH3)3CBr(l) + H2O(l) (CH3)3COH(l) + HBr(aq)

The rate law has been found to be

rate = k[(CH3)3CBr]

This reaction is first order in 2-bromo-2-methylpropane. Note that the concentration of H2O does not even appear in the

rate law. Thus, the reaction is zero order with respect to H2O. This means that the rate does not depend on the concentration of H2O; we could also write the rate law for this reaction as

rate = k[(CH3)3CBr][H2O]0

What is the overall order of this reaction?

1st order overall (1+0=1)

49

Reaction orders are usually positive integers or zero, but they can also be fractional or negative. In the reaction

CHCl3(g) + Cl2(g) CCl4(g) + HCl(g)

A fractional order appears in the rate law:

rate = k[CHCl3][Cl2]1/2

This order means that the reaction depends on the square root of the Cl2 concentration. If the initial Cl2 concentration is increased by a factor of 4, for example, the rate increases by V4 (= 2), therefore the rate would double.


A negative exponent means that the reaction rate decreases when the concentration of that component increases. Negative orders are often seen for reactions whose rate laws include products. For example, in the atmospheric reaction

2O3(g) 3O2(g)

The rate law has been shown to be

Rate = k[O3]2[O2]-1 ; or [O3]2

[O2]rate = k

If the [O2] doubles, the reaction proceeds half as fast.


To see how to determine the rate law of a reaction, let us consider the reaction between fluorine and chlorine dioxide:

F2(g) + 2ClO2(g) 2FClO2(g)


One way to study the effect of reactant concentration on reaction rate is to determine how the initial rate depends on the starting concentrations. It is preferable to measure the initial rates because as the reaction proceeds, the concentrations of the reactants decrease and it may become difficult to measure the changes accurately. Also, as the reaction continues, the product concentrations increase,

products reactants

so the reverse reaction becomes increasingly likely. Both of these complications are virtually absent during the earliest stages of the reaction.


The following table shows three rate measurements for the formation of FClO2.

[F2]0

(M)

[ClO2]0

(M)

Initial Rate(M/s)

0.10 0.010 1.2 x 10-3

0.10 0.040 4.8 x 10-3

0.20 0.010 2.4 x 10-3

54

[F2]0

(M)

[ClO2]0

(M)

Initial Rate(M/s)

0.10 0.010 1.2 x 10-3

0.10 0.040 4.8 x 10-3

0.20 0.010 2.4 x 10-3

Looking at trials 1 and 3, we see that as we double [F2]0 while holding [ClO2]0 constant, the reaction rate doubles. Thus the rate is directly proportional to [F2], and the reaction is first order with respect to F2.

55

[F2]

(M)

[ClO2]

(M)

Initial Rate(M/s)

0.10 0.010 1.2 x 10-3

0.10 0.040 4.8 x 10-3

0.20 0.010 2.4 x 10-3

Similarly, the data in trials 1 and 2 show that as we quadruple [ClO2] while holding [F2] constant, the rate increases by four times, so the rate is also directly proportional to [ClO2], making the reaction 1st order with respect to [ClO2]


We can summarize our observations by writing the rate law as

Rate = k[F2][ClO2]

Because both [F2] and [ClO2] are raised to the first power, the reaction is second order overall.


What is the rate constant for this reaction at this temperature?


From the reactant concentrations and the initial rate from any of the trials, we can calculate the rate constant. Using the data from the previous table we can write

k = rate

[F2][ClO2]

1.2 x 10-3 M/s

(0.10M)(0.010M)= 1.2 L/mol.s

Rate = k[F2][ClO2]

Using data from the first trial

Notice the label for a 2nd order reaction.


The reaction of nitric oxide with hydrogen at 1280 oC is

2NO(g) + 2H2(g) N2(g) + 2H2O(g)

From the following data that was collected experimentally at this temperature, determine the rate law and calculate the rate constant.

Experiment [NO] [H2] Initial Rate (M/s)

1 5.0 x 10-3 2.0 x 10-3 1.3 x 10-5

2 10.0 x 10-3 2.0 x 10-3 5.0 x 10-5

3 10.0 x 10-3 4.0 x 10-3 10.0 x 10-5



1 5.0 x 10-3 2.0 x 10-3 1.3 x 10-5

2 10.0 x 10-3 2.0 x 10-3 5.0 x 10-5

3 10.0 x 10-3 4.0 x 10-3 10.0 x 10-5

We assume that the rate law will take the form

Rate = k[NO]x[H2]y

Experiments 1 and 2 show that when we double the concentration of NO at constant concentration of H2, the rate quadruples. Thus the reaction is second order in NO. Experiments 2 and 3 indicate that doubling [H2] at constant [NO] doubles the rate; the reaction is first order in H2. The rate law is therefore given by

Rate = k[NO]2[H2]

Which shows that it is a (1 + 2) or third-order reaction overall.



1 5.0 x 10-3 2.0 x 10-3 1.3 x 10-5

2 10.0 x 10-3 2.0 x 10-3 5.0 x 10-5

3 10.0 x 10-3 4.0 x 10-3 10.0 x 10-5

The rate constant can be calculated using the data from any one of the experiments. Since

k = rate

[NO]2[H2] k = 5.0 x 10-5 M/s

(10.0 x 10-3 M)2(2.0 x 10 -

3M)k = 2.5 x 102 L2/ mol2 . s Notice the label for a

3nd order reaction.


The following points summarize our discussion of the rate law:

• Rate laws are ALWAYS determined experimentally.

• Reaction order should be defined in terms of reactant (not product) concentrations.

• The order of a reactant is not related to the stoichiometric coefficient of the reactant in the overall balanced equation.


Relationship Between Reactant Concentration and Time

Rate law expressions enable us to calculate the rate of a reaction from the rate constant and reactant concentrations.

A rate law can also be used to determine the concentration of any reactant at any time during the course of a reaction.


First-order reactionsA first order reaction is a reaction whose rate

depends on the reactant concentration raised to the first power.

rate = k[A]

In a first-order reaction of the type

A product

Rate can be expressed as

rate = - [A]t

65

Through the methods of calculus, this expression is integrated over time to obtain the integrated rate law for a first-order reaction:

ln[A]0

[A]t

= kt

Since ln[A]0

[A]t

= ln[A]0 – ln[A]t

We can write the integrated rate law for a first order reaction as

ln[A]0 – ln[A]t = kt


Cyclobutane decomposes at 1000 oC to two moles of ethylene (C2H4) with a very high rate constant, 87 s-1.

If the initial concentration of cyclobutane is 2.00 M, what is the concentration after 0.010 s? (the label = 1st order!)

ln[C4H8]0 – ln[C4H8]t For a first order reaction kt =

Rearranging the equation so that ln [C4H8]t can be solved for

ln [C4H8]t = ln[C4H8]0 - kt

Substituting the data

ln [C4H8]t = ln(2.00 mol/L) - (87 s-1)(0.010 s)ln [C4H8]t = 0.69 – 0.87 = -0.18

Taking the antilog (inverse log) of both sides

[C4H8]t = e-0.18 = 0.84 mol/L C4H8 present at 0.010 s

87 s-1


What percent of the cyclobutane has decomposed in this time?

2.00 M - .84 M = 1.16 M

1.16 M/2.00 M x 100 = 58% has decomposed (42 % remains)

Chemical Kinetics 2009-2010

Reaction half-life (t1/2)

The t1/2 of a reaction is the time required to reach half the initial reactant concentration.

For a first order reaction, the formula for determining t1/2 is

t1/2 = ln 2 k

0.693 k

=

Note – t1/2 of a first order reaction is constant, it is independent of reactant concentration!


A plot of [N2O5] vs time for three half-lives.

0.060

0.050

0.040

0.030

0.020

0.010

0.000

[N2O

5]

0 24 48 72Time (min)


Determining t1/2 for a first-order reaction

Cyclopropane is the smallest cyclic hydrocarbon. Because its 60o bond angles allow only poor orbital overlap, its bonds are weak. As a result, it is thermally unstable and rearranges to propene at 100 oC via the following reaction:

H2C

CH2

CH2

CH3 – CH = CH2


The rate constant is 9.2 s-1. How long does it take for the initial concentration of cyclopropane to decrease by one-half?

t1/2 = ln 2 k

0.693 k

= = 0.075 s0.6939.2 s-1

=

It takes 0.075 s for half the cyclopropane to form propene.

The label tells you this is a _____ order reaction.1st


For a general second-order rate equation, the expression including time can become quite complex, so let’s consider only the simplest case, one in which the rate law contains only one reactant

2A product

The integrated rate law for a second-order reaction involving one reactant:

1[A]t

1[A]0

- = kt

Second-order reactions


At 25 oC, hydrogen iodide breaks down very slowly to hydrogen and iodine according to the following:

rate = k[HI]2 and k = 2.4 x 10-21 L/mol . s

If 0.0100 mol HI(g) is placed in a 1.0 L container, how long will it take for the concentration of HI to reach 0.00900 mol/L?


rate = k[HI]2 and k = 2.4 x 10-21 L/mol . s

1[A]t

1[A]0

- = kt

1 0.00900

1 0.0100

- = (2.4 x 10-21 )(t)

111 – 100 = (2.4 x 10-21)(t)

11/ 2.4 x 10-21 = t = 4.6 x 1021 seconds


In contrast to the half-life of a first-order reaction, the half-life of a second-order reaction DOES depend on reactant concentration:

t1/2 = 1 k[A]0


In the simple decomposition reaction

AB(g) A(g) + B(g)

rate = k[AB]2 and k = 0.20 L/mol . s

How long will it take for [AB] to reach half of its initial concentration of 1.50 M?


t1/2 = 1 k[A]0

t1/2 = 1 (0.20)(1.50)

t1/2 = 3.3 seconds


Determining the Reaction Order from the Integrated Rate

LawSuppose you don’t know the rate law

for a reaction and don’t have the initial rate data needed to determine the reaction orders. Another method for finding reaction orders is a graphical technique that uses concentration-time data directly.


To find the reaction order from the concentration-time data, some trial-and-error graphical plotting is required:

• If you obtain a straight line when you plot [reactant] vs. time, the reaction is zero order with respect to that reactant.


A plot of ln[A] vs t gives a straight line for a reaction that is first order in A

A plot of 1/[A] vs t gives a straight line for a reaction that is second order in A


Graphical determination of the reaction order for the decomposition of N2O5.

Time [N2O5]

0 0.0165

10 0.0124

20 0.0093

30 0.0071

40 0.0053

50 0.0039

60 0.0029

Directions for Creating Lists

1.Stat Edit then ENTER

2.Enter the data in L1 and L2



3. Move the cursor up to the top of L3, and enter the formula ln L2

4. Repeat making L4 1/L2

Time [N2O5] ln[N2O5] 1/[N2O5]

0 0.0165 -4.104 60.6

10 0.0124 -4.390 80.6

20 0.0093 -4.68 1.1 x 102

30 0.0071 -4.95 1.4 x 102

40 0.0053 -5.24 1.9 x 102

50 0.0039 -5.55 2.6 x 102

60 0.0029 -5.84 3.4 x 102


Directions for Graphing Lists

1.Push

2.Turn on Plot 1, then hit ENTER

3.Select the middle graph (line graph)

4.X list = __________

5.Y list = __________

6. ZoomStat (9)

7.Repeat process for L1 vs L3 and L1 vs L4

2nd

Y =

Will depend on the graph

L1

L2

ZOOM



Plotting of [N2O5] vs time and 1/[N2O5] vs time give curved lines, indicating that the reaction IS NOT zero order nor is it second order with respect to N2O5.



A plot of ln[N2O5] vs time gives a straight line, indicating the reaction IS first order in N2O5


Activation Energy According to Collision Theory of Chemical

Kinetics, molecules must collide to react. However, not all collisions lead to reactions. Energetically speaking, there is some minimum collision energy below which no reaction occurs. Any molecule in motion possesses kinetic energy; the faster it moves, the greater its kinetic energy. When molecules collide, part of their kinetic energy is converted to vibrational energy. If the initial kinetic energies are large enough, the colliding molecules will vibrate so strongly that some of the chemical bonds will break. This bond fracture is the first step toward product formation. If the initial kinetic energies are too small, the molecules will merely bounce off each other intact, and no change results from the collision.


We postulate that in order to react, the colliding molecules must have a total kinetic energy equal to or greater than the activation energy, (Ea), which is defined as the minimum amount of energy required to initiate a chemical reaction.


Molecules must also be oriented in a favorable position – one that allows the bonds to break and atoms to rearrange.


The picture to the right shows a few of the possible collision orientations for this simple gaseous reaction:

NO(g) + NO3(g) 2NO2(g)

Of the five collisions shown, only one has an orientation in which the N of NO collides with an O of NO3. Actually, the probability factor (p) for this reaction is 0.006; only 6 collisions in every thousand have an orientation that leads to a reaction.


Collisions between individual atoms have p values near 1: almost no matter how they hit, they react. In such cases, the rate constant depends only on the frequency and energy of the collisions.

At the other extreme are biochemical reactions, in which the reactants are often two small molecules that can react only when they collide with a specific tiny region of a giant molecule-a protein or nucleic acid. The orientation factor for such reactions is often less than 10-6: fewer than one in a million sufficiently energetic collisions leads to product formation.


When reactants collide at the proper angle with energy equal to the activation energy (Ea), they undergo an extremely brief interval of bond disruption and bond formation called a transition state.

During this transition state, the reactants form a short-lived complex that is neither reactant nor product, but has partial bonding characteristics of both. This transitional structure is called an activated complex.Endothermic/Exothermic (choose

one)


An activated complex is a highly unstable species with a high potential energy. (It was energized by the particle collision.) Once formed, it will break up almost immediately.

The activated complex exists along the reaction pathway at the point where the energy is greatest – at the peak indicated by the activation energy.

Activation energy is the energy required to achieve the transition state and form the activated complex


We can think of activation energy, (Ea), as a barrier that prevents less energetic molecules from reacting…

…because only the molecules who have enough kinetic energy to exceed the activation energy can take part in the reaction.


Maxwell Boltzmann DiagramBecause the number of reactant molecules in an ordinary reaction is very large, the speeds, and hence also the kinetic energies of the molecules, vary greatly.


Normally, only a small fraction of the colliding molecules -- the fastest-moving ones – have enough kinetic energy to exceed the activation energy.

At higher temperature, more molecules can surpass the activation energy, therefore, the rate of product formation is greater at the higher temperature.

As temperature increases, the probability of finding molecules at higher temperature increases.


With very few exceptions, reaction rates increase with increasing temperature.

As a general rule of thumb, you can expect a 10 oC increase in temperature to result in a doubling of the reaction rate.


Arrhenius Equation

The dependence of the rate constant of a reaction on temperature can be expressed by the following equation, known as the Arrhenius equation:

k = Ae-Ea/RT

Where Ea is the activation energy (in J/mol), R the gas constant (8.314 J/K . mol), T the absolute temperature, and e the base of the natural logarithm scale. The quantity A represents the collision frequency and is called the frequency factor.


The frequency factor is the product of the collision frequency (Z) and an orientation factor (p) which is specific for each reaction. The factor p is related to the structural complexity of the colliding particles. You can think of it as the ratio of effectively oriented collisions to all possible collisions.

In the activation energy problems we are solving, the actual value of A need not be known because A can be treated as a constant for a given reacting system over a fairly wide temperature range.

Frequency Factor (A)


• As the activation energy increases, k

k = Ae-Ea/RT

decreases,

•As the temperature increases, kincreases

and as k decreases, ratedecreases

• And as k increases, rateincreases


You can derive the following equation from the Arrhenius equation:

-Ea 1 R T

ln k =

+ ln A


-Ea 1 R T

ln k =

+ ln A

Y =

mx + b

Thus, a plot of ln k versus 1/T gives a straight line whose

•slope (m) is equal to –Ea/R and whose

•intercept (b) with the y-axis is ln A.

m

slope (m) is equal to –Ea/R


Given k and temperature, you can use your graphing calculator to determine the activation energy of a reaction.

See the following example…


T (K)

700

730

760

790

810

k

0.011

0.035

0.105

0.343

0.789

Enter these values into L1 and L2 of your graphing calculator.


1/T

1.43 x 10-3

1.37 x 10-3

1.32 x 10-3

1.27 x 10-3

1.23 x 10-3

ln k

-4.51

-3.35

-2.254

-1.070

-0.237

The slope of the line can be determined using linear regression.

To determine the activation energy, we need to graph 1/T on the x axis (L3) and ln k on the y-axis (L4).


Directions for Linear Regression

1.Stat Calc LinReg(ax + b) then ENTER

2.At the prompt enter L3,L4 then

ENTER

a = slope

This value can then be put in the equation

slope = -Ea/R


If you did the linear regression correctly, slope for this reaction = -21,880

therefore…

Ea = 1.82 x 102 kJ/mol

Ea = -(-2.19 x 104 K)(8.314 J/K .

mol) Ea = 1.82 x 105 J/mol


ln k2

k1

= Ea

R

1 1

T2 T1

You can also determine activation energy if you are given the two different rate constants at two different temperatures. Or, you can use this formula to find the rate constant at another temperature if the activation energy is known.


The rate constant of a first-order reaction is 3.46 x 10-2 s-1 at 298 K. What is the rate constant at 350 K if the activation energy for the reaction is 50.2 kJ/mol?

k1

=

k2

= T1 = T2 =

3.46 x 10-2s-1

298 350

ln k2

3.46 x 10-2 = 3.01 Taking the antilog of both sides k2

3.46 x 10-2= e 3.01 = 20.3 k2 =

0.702 s-1

k2

k1

Ea

Rln

1 1

T2 T1T2 T1

=

k2

k1

Ea

Rln

1 1

1 1350 298

ln k2

3.46 x 10-2=50,200 J/mol8.314 J/K . mol


Reaction MechanismsAs we mentioned earlier, an overall

balanced chemical equation does not tell us much about how a reaction actually takes place. In many cases, it merely represents the sum of several elementary steps, or elementary reactions, that represent the progress of the overall reaction at the molecular level. The term for the sequence of elementary steps that leads to product formation is reaction mechanism.

The reaction mechanism is comparable to the route of travel followed during a trip; the overall chemical equation specifies only the origin and the destination.


As an example of a reaction mechanism, let us consider the reaction between nitrogen monoxide and oxygen:

2NO(g) + O2(g) 2NO2(g)We know that the products are not formed directly from the collision of two NO molecules with an O2 molecule because N2O2 is detected during the course of the reaction.


Let us assume that the reaction actually takes place via two elementary steps as follows:

NO(g) + NO(g) N2O2(g)

N2O2(g) + O2(g) 2NO2(g)

Elementary Step 1

Elementary Step 2

Overall reaction 2NO + N2O2 + O2 N2O2 + 2NO2Each of the elementary steps listed above is called a

bimolecular reaction because each step involves two reactant molecules. A step that just involves one reactant molecule is a unimolecular reaction. Very few termolecular reactions, reactions that involve the participation of three reactant molecules in one elementary step, are known, because the simultaneous encounter of three molecules is a far less likely event than a bimolecular collision. (There are no known examples of reactions involving the simultaneous encounter of four molecules.)


NO(g) + NO(g) N2O2(g)

N2O2(g) + O2(g) 2NO2(g)

Elementary Step 1

Elementary Step 2

Overall reaction 2NO + N2O2 + O2 N2O2 + 2NO2

Species such as N2O2 are called intermediates because they appear in the mechanism of the reaction (that is, in the elementary steps) but not in the overall balanced equation. Keep in mind that an intermediate is always formed in an early elementary step and consumed in a later elementary step. Note – an intermediate ≠ activated complex!


You can propose a mechanism for a reaction if you consider…

• The elementary steps in a multistep reaction mechanism must always add to give the balanced chemical equation of the overall process. (Any intermediates that are formed in earlier steps must be consumed in later steps.)

• Unimolecular and bimolecular reactions are more common than termolecular reactions.

• The rate of the overall reaction is limited by the rate of the slowest elementary step, (For that reason, the slowest elementary step is typically called the rate-determining step.


Gaseous nitrogen monoxide reacts with fluorine gas to produce nitrogen hypofluorite (NOF). The intermediate product NOF2(g) has been isolated as an intermediate in this reaction. Propose a two step mechanism consistent with this intermediate product.

Step 1: NO(g) + F2(g) NOF2(g)

Step 2: NOF2(g) + NO(g) 2NOF(g)

•The elementary steps add to give the overall balanced chemical equation

•The first and the second step are bimolecular


Even when a proposed mechanism is consistent with the rate law, later experimentation may show it to be incorrect or only one of several alternatives.


Knowing the elementary steps of a reaction enables us to propose a rate law.

Suppose we have the following elementary step:

A products

Because there is only one reactant molecule present, this is a/n ___molecular reaction. It follows that the larger the number of A molecules present, the faster the rate of product formation.

Thus the rate of a unimolecular reaction is directly proportional to the concentration of A, or is first order in A:

Rate = k[A]

uni


For a bimolecular elementary reaction involving A and B molecules

A + B product

the rate of product formation depends on how frequently A and B collide, which in turn depends on the concentration of A and B. Thus we can express the rate as

Rate = k[A][B]


Similarly, for a bimolecular elementary reaction of the type

A + A products or

2A products

the rate becomesRate = k[A]2


Rate Laws for Elementary Steps

Elementary Step Molecularity Rate Law

A product

2A product

A + B product

2A + B product

unimolecularbimolecular

rate = k[A]

rate = k[A]2

bimolecular rate = k[A][B]termolecular rate = k[A]2[B]


Remember, when we study a reaction that has more than one elementary step, the rate law for the overall process is given by the rate-determining step, which is the slowest step in the sequence of steps leading to product formation.


Example: The gas-phase decomposition of dinitrogen monoxide (N2O) is believed to occur via two elementary steps:

Step 1: N2O N2 + O

Step 2: N2O + O N2 + O2

Experimentally the rate law is found to be Rate = k[N2O]

(a) Write the equation for the overall reaction.(b) Identify the intermediates. (c) What can you say about the relative rates

of steps 1 and 2?


(a) Adding the equations for steps 1 and 2 gives the overall reaction

2N2O 2N2 + O2

(b) Since the O atom is produced in the first elementary step and it does not appear in the overall balanced equation, it is an intermediate.

(c) Step 1 must be the rate-determining (slower) step because that step is also first order with respect to N2O. (The rate law for the rate-determining step should match the rate law that is determined experimentally.)


Example 2: Hydrogen Peroxide Decomposition

Does the decomposition of hydrogen peroxide occur in a single step?

The overall reaction is

2H2O2(aq) 2H2O(l) + O2(g)

By experiment, the rate law is found to be

Rate = k[H2O2][I-]

I-


From this alone you can see that H2O2 decomposition does not occur in a single elementary step corresponding to the overall balanced equation. If it did, the rate law would be

Rate = [H2O2]2

or in other words, the reaction would be second order in H2O2. Remember, the experimentally determined rate law for this reaction was shown to be

Rate = k[H2O2][I-]


Catalysis

For the decomposition of hydrogen peroxide we see that the reaction rate depends on the concentration of iodide ions even though I- does not appear in the overall equation. I- is a catalyst for this reaction, a substance that increases the rate of a chemical reaction without itself being consumed.

2H2O2(aq) 2H2O(l) + O2(g)I-

Rate = k[H2O2][I-]


A catalyst exists before the reaction occurs and can be recovered and reused after the reaction is complete. This is the opposite of intermediates, which are produced in one step of a mechanism and consumed in another.


In other words, a catalyst typically lowers the activation energy for the reaction by forming an activated complex that has less potential energy.

In many cases, a catalyst increases the rate by providing a set of elementary steps with more favorable kinetics than those that exist in its absence.


A + B

C + D

A + B

C + D

Forward reaction

Reverse reaction

Because the activation energy for the reverse reaction is also lowered, a catalyst enhances the rates of the forward and reverse reaction equally.


There are three general types of catalysis: heterogeneous catalysis, homogeneous catalysis, and enzyme catalysis.


In heterogeneous catalysis the reactants and the catalyst are in different phases. Usually the catalyst is a solid and the reactants are either gases or liquids. Heterogeneous catalysis is by far the most important type of catalysis in industrial chemistry, especially in the synthesis of many key chemicals.


Ammonia is an extremely valuable inorganic substance used in the fertilizer industry and many other applications.

N2(g) + 3H2(g) 2NH3(g) H = -92.6 kJ

This reaction is extremely slow at room temperature, and although raising the temperature accelerates the above reaction, it also promotes the decomposition of NH3 molecules into N2 and H2, thus lowering the yield of NH3.

In 1905, after testing literally hundreds of compounds at various temperatures and pressures, Fritz Haber discovered that iron plus a few percent of oxides of potassium and aluminum catalyze the reaction. This procedure is known as the Haber process.


Haber ProcessFirst the H2 and the N2 molecules bind to the surface of the catalyst. This interaction weakens the covalent bonds within the molecules and eventually causes the molecules to dissociate. The highly reactive H and N atoms combine to form NH3 molecules, which then leave the surface.


Nitric acid is one of the most important inorganic acids. It is used in the production of fertilizers, dyes, drugs, and in many other products. The major industrial method of producing nitric acid is the Ostwald process. The starting materials, ammonia and molecular oxygen, are heated in the presence of a platinum-rhodium catalyst to about 800 oC.


In homogeneous catalysis, the reactants are dispersed in a single phase, usually liquid. Acid and base catalyses are the most important types of homogeneous catalysis in liquid solution.

Homogeneous catalysis can also take place in the gas phase. A well-known example of catalyzed gas-phase reactions is the lead chamber process, which for many years was the primary method of manufacturing sulfuric acid.


Homogeneous catalysis has several advantages over heterogeneous catalysis. For one thing, the reactions can often be carried out under atmospheric conditions, thus reducing production costs and minimizing the decomposition of products at high temperatures. In addition, homogeneous catalysts can be designed to function selectively for a particular type of reaction, and homogeneous catalysts cost less than the precious metals (for example, platinum and gold) used in heterogeneous catalysis.


Of all the intricate processes that have evolved in living systems, none is more striking or more essential than enzyme catalysis.

Enzymes are biological catalysts. Enzymes can increase the rate of a biochemical reaction by a factor ranging from 106 to 1012 times!

An enzyme acts only on certain molecules, called substrates, while leaving the rest of the system unaffected. It has been estimated that an average living cell may contain some 3000 different enzymes, each of them catalyzing a specific reaction in which a substrate is converted into the appropriate products.


An enzyme is typically a large protein molecule that contains one or more active sites where interactions with substrates takes place. These sites are structurally compatible with specific substrate molecules, in much the same way as a key fits a particular lock. In fact, the notion of a rigid enzyme structure that binds only to molecules whose shape exactly matches that of the active site was the basis of an early theory of enzyme catalysis, the so-called lock-and-key theory.


This theory accounts for the specificity of enzymes, but it contradicts research evidence that a single enzyme binds to substrates of different sizes and shapes.

Chemists now know that an enzyme molecule (or at least its active site) has a fair amount of structural flexibility and can modify its shape to accommodate more than one type of substrate.


The End

chemical kinetics 2009-20101 chemical kinetics the area of chemistry concerned with the speeds, or...

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