Chemical kinetics

Chemical Kinetics What is the definition of “chemical kinetics?”

What is chemical kinetics?

• Chemical kinetics is the study of the speed of a chemical reaction • It includes a study of the time it takes for a chemical reaction to occur, and • It includes factors that affect the speed of a reaction.

In addition, data related to chemical kinetics provide information about the pathway (mechanism) of a reaction. The pathway or mechanism is the series of elementary steps that result in the formation of product(s). We can use a simple illustration to demonstrate what we mean by the pathway (mechanism) that is taken during the formation of product (or products). If we want to travel from one city (Miramar, FL) in the United Sates to another city (Los Angeles, CA) in the United States, then there are several transportation modes we may use. We could take a car, a bus, a plane, or a train; however, regardless of the transportation mode, we have a variety of different routes we can take, and one particular route may be better than all the others. A similar situation exists when a chemical reaction occurs. There may be a variety of pathways that could be taken for reactants to form products, and our task is to use experimental data to determine the specific pathway that reactants take to form products. This process uses chemical kinetic experiments to collect data that could be used to suggest the series of elementary steps that result in the formation of products, and the mechanism of the reaction is the series of elementary steps that result in the formation of the product or products. Generally, the elementary steps are either unimolecular (that means the step involves only one molecule) or bimolecular (that means the step involves two molecules).

In order to begin our understanding of chemical kinetics, let’s consider the following general reaction:

Where A and B represent reactants and C represents a product Now, let’s develop a hypothetical table that demonstrates the decrease in the concentrations of A and B versus time, and the increase in the concentration of C versus time.

1

time, seconds 0

Concentration A, mol/L 0.76

Concentration B, mol/L 0.38

Concentration C, mol/L 0

1

0.31

0.16

0.20

2

0.13

6.5 x 10-2

0.40

3

5.2 x 10-2

2.6 x 10-2

0.58

4

2.1 x 10-2

1.1 x 10-2

0.73

5

8.8 x 10-3

4.4 x 10-3

0.86

6

3.6 x 10-3

1.8 x 10-3

0.95

7

1.4 x 10-3

7.0 x 10-4

1.02

8

6.1 x 10-4

3.1 x 10-4

1.07

9

2.5 x 10-4

1.3 x 10-4

1.07

10

1.0 x 10-4

5.0 x 10-5

1.07

The concentrations of A and B (the reactants) decrease during the ten seconds of the experiment; whereas, the concentration of C increases during the ten seconds of the experiment.

Following is a graph of the experiment where the concentrations versus time for A and B (the reactants) decrease, and concentration of C (the product) increases:

2

Concentration (mol/L)

1.2 1

0.8 0.6 0.4 0.2 0 0

2

4

6 8 time (seconds)

10

12

The above graph demonstrates the decrease of the concentrations of the reactants, A and B, during the ten seconds of the experiment, and an increase of the concentration of the product, C, during the ten seconds of the experiment. Therefore, during a chemical reaction, we can generalize that the concentrations of the reactants decrease with increase in time, and the concentrations of products increase with increase in time.

For a chemical reaction, the speed of a reaction (the rate of the reaction) can be studied by examining the decrease in reactants or the increase in products over time. For example, we can experimentally determine the rate for the general reaction A + b B → c C + d D. This rate is referred to as the “Rate Law,” and the Rate Law is represented by:

Rate = k [A] [B] m

n

Where “m” and “n” are determined experimentally, and are not necessarily equal to “a” and “b” (the stoichiometry of the chemical reaction). We will learn how to experimentally determine the values of “m” and “n” later. But what must be emphasized is that “m” and “n” are not necessarily the same as “a” and “b,” and “m” and “n” must be experimentally determined.

In order to begin to understand this concept, let’s examine the following diagram in where “A” represents 1 mole of a substance, and “B” represents 2 moles of a substance when A is converted to B:

3

In a 1.00 liter container total of 3.011 x 1024 molecules 5 moles

In a 1.00 liter container total of 6.022 x 1024 molecules 10 moles

Therefore, the mole ratio is 1:2 In this diagram, we have a reactant, A, with 3.011 x 1024 molecules (5 mole) producing a product, 2 B, 6.022 x 1024 molecules (10 moles). Therefore, there are twice as many molecules of B, the product, produced from A, the reactant.

If the symbol Δ represents the change of the concentration of a substance and the change of time, then Δ A represents the change in concentration of A and Δ t would be the change in time; therefore, the change in concentration of A divided by the change in time would be

Δ[A] Δt and the decrease in concentration of A versus time could be represented as

Δ[A] Δt

Then Δ B represents the change in concentration of B and Δ t would be the change in time; therefore, the change in concentration of B divided by the change in time would be: Δ[B] Δt -

4

and the increase in concentration of B versus time could be represented as

+

Δ[B] Δt

Since twice as many “B” molecules appear as “A” molecules disappear, then the decrease in the concentration of “A” with change in time with respect to the increase in concentration of “B” with the same interval of time could be represented by:

−2

ΔA ΔB =+ Δt Δt or

-

Δ [A ] 1 Δ [B] = Δt 2 Δt

The rate of [A] disappearing is

Δ[A] Δt Therefore, let’s suppose that after twenty (20) seconds, ½ of [A] disappears, then the rate of change in [A] with respect to the change of time would be: [A] f - [A]i Δ[A] (0.50 mol/L - 1.00 mol/L) mol === 2.5 x 10 −2 or 2.5 x 10 −2 M/s Δt 20 s - 0 s 20 s L-s

-

and the Rate of B appearing would be: 1 Δ[B] 1 [B] f - [B]i 1 (1.00 mol/L - 0 mol/L) mol = = = 2.5 x 10 −2 or 2.5 x 10 −2 M/s 2 Δt 2 20 s - 0 s 2 20 s L-s

Notice that the two results are equal demonstrating that -

Δ [A ] 1 Δ [B] = Δt 2 Δt

or

5

If you desire twice as much “A,” then the expressions could be written as:

-2

[A] f - [A]i Δ[A] (0.50 mol/L - 1.00 mol/L) mol = -2 = -2 = 5.0 x 10 −2 Δt 20 s - 0 s 20 s L-s

[B] f - [B]i Δ[B] (1.00 mol/L - 0 mol/L) mol = = = 5.0 x 10 −2 or 5.0 x 10 −2 M/s Δt 20 s - 0 s 20 s L-s Therefore by doubling the concentration of “A,” you double the rate of the reaction.

What effect do you think tripling the concentration of “A” would have on the rate of the reaction?

−3

Δ[A] 3 Δ[B] = Δt 2 Δt

Δ[A] = −3 Δt 3 Δ[B] 3 + = 2 Δt 2 −3

[A] f - [A] i

(0.50 mol/L -1.00 mol/L) = 7.5 x 10 −2 M/s 20 s - 0 s 20 s - 0 s [B] f - [B] i 3 (1.00 mol/L -0 mol/L) = = 7.5 x 10 −2 M/s 20 s - 0 s 2 20 s - 0 s = -3

Therefore, tripling the initial concentration results in tripling the rate.

Now, we can write the relationship of the average rate change of a reaction from the balanced chemical formula. So, for the general reaction: aA+bB→cC + dD The Average Rate Change would be:

-

1 Δ [A] 1 Δ [B] 1 Δ [C] 1 Δ [D] x = - x = x = x a Δt b Δt c Δt d Δt

or If we multiple the concentration of “A” by “a,” then the Average Rate Change would be:

6

-

Δ[A] a Δ[B] a (Δ[C] a Δ[D] = - x = = Δt b Δt c Δt d Δt

For example, let’s consider the following reaction: N2O5 (g) → 2 NO2 (g) + ½ O2 (g) The Average Rate Change would be: -

Δ ⎡⎣ N 2O5 (g) ⎤⎦ Δt

=

Δ ⎡O2 (g) ⎤⎦ 1 Δ ⎡⎣ NO2 (g) ⎤⎦ x =2x ⎣ 2 Δt Δt

or If we double the concentration of N2O5 (g), then the Rate Change would be:

−2 x

Δ[N 2O 5 (g) ] Δt

=

Δ[NO 2 (g) ] Δt

=4x

Δ[O 2 (g) ] Δt

Determination of the Rate of a Reaction The rate of a chemical reaction must be obtained from a set of well-executed experiments. The rate begins with obtaining the concentrations of reactants after 1% or 2% of the limiting reagent has been consumed.

For example, let’s consider the reaction of 1-chlorobutane (I) with water to produce 1-butanol (II) and hydrochloric acid: CH3CH2CH2CH2Cl (aq) + H2O (l) → CH3CH2CH2CH2OH (aq) + HCl (aq) I

II

Let’s assume we were able to study the concentration of 1-chlorobutanol over time, and the result were tabulated as follows:

7

time, seconds 0

[1-chlorbutane], mol/L 0.10

50

9.05 x 10-2

100

8.2 x 10-2

150

7.41 x 10-2

200

6.71 x 10-2

300

5.49 x 10-2

400

4.48 x 10-2

500

3.68 x 10-2

800

2.00 x 10-2

The reaction of 1-chlorobutane is with water; therefore, the reaction is a solvolysis reaction, and 1-chlorobutane is the limiting reagent.

The amount of 1-chlorobutane consumed at an average rate between 0 and 50 seconds is 9.50%.

time, seconds

[CH3CH2CH2CH2Cl]

Average Rate, M/s

0.0

0.1000

1.90 x 10

50.0

0.0905

-4

The amount of 1-chlorobutane consumed at an average rate between 50 and 100 seconds is 18.0%.

time, seconds

[CH3CH2CH2CH2Cl]

Average Rate, M/s

50.0

0.0905

1.70 x 10

100.0

0.0820

-4

8

The amount of 1-chlorobutane consumed at an average rate between 100 and 150 seconds is 25.9%. time, seconds

[CH3CH2CH2CH2Cl]

Average Rate, M/s

100.0

0.0820

1.58 x 10

150.0

0.0741

-4

The amount of 1-chlorobutane consumed at an average rate between 150 and 100 seconds is 32.9%. time, seconds

[CH3CH2CH2CH2Cl]

Average Rate, M/s

150.0

0.0741

1.74 x 10

200.0

0.0671

-4

The amount of 1-chlorobutane consumed at an average rate between 200 and 300 seconds is 45.1%. time, seconds

[CH3CH2CH2CH2Cl]

Average Rate, M/s

200.0

0.0671

1.22 x 10

300.0

0.0549

-4

The amount of 1-chlorobutane consumed at an average rate between 300 and 400 seconds is 55.2%. time, seconds

[CH3CH2CH2CH2Cl]

Average Rate, M/s

300.0

0.0549

1.01 x 10

400.0

0.0448

-4

The amount of 1-chlorobutane consumed at an average rate between 400 and 500 seconds is 63.2%.

9

time, seconds

[CH3CH2CH2CH2Cl]

400.0

0.0448

500.0

0.0368

Average Rate, M/s 8.00 x 10

-5

The amount of 1-chlorobutane consumed at an average rate between 500 and 800 seconds is 80.0%. time, seconds

[CH3CH2CH2CH2Cl]

500.0

0.0368

800.0

0.0200

Average Rate, -5

5.60 x 10

The instantaneous rate can be determined from the first set of data, i.e., the data between 0 – 50 seconds.

The instantaneous rate at 0 seconds is

10

Instantaneous Rate at 0 s = Instantaneous Rate at 0 s =

0.10 M - 0.060 M 190 s - 0 s

0.10 M = 2.1 x 10 −4 M/s 190 s

The instantaneous rate at 500 seconds is

Instantaneous Rate at 500 s =

Instantaneous Rate at 500 s =

0.042 M - 0.020 M 800 s - 400 s 0.022 M = 5.5 x 10 −5 M/s 400 s

Order of Reactions • • • • •

Zero order means that the rate of the reaction is independent of the concentration of the reactants. For instance, the reaction rate may depend on light.

First order means that the rate of the reaction depends on a step in the mechanism that is unimolecular (involves only one molecule).

Pseudo first order reaction means that one of the reactants in the ratedetermining step is a solvent.

Second order means that the rate of the reaction depends on a step in the mechanism that is bimolecular (involves two molecules).

Rarely is the rate determining step a third order reaction, because that would mean that the rate of the reaction depends on the step in the mechanism that is termolecular (involves three molecules). Also, a termolecular reaction may involve interaction with a molecule in the activated state or the surface of the reaction vessel or some other third body phenomenon. You will not encounter single steps in this course or in organic chemistry that are termolecular.

11

Let’s return to the reaction between 1-chlorobutane and water. The data is:

time, seconds 0

[1-chlorbutane], mol/L 0.10

50

9.05 x 10-2

100

8.2 x 10-2

150

7.41 x 10-2

200

6.71 x 10-2

300

5.49 x 10-2

400

4.48 x 10-2

500

3.68 x 10-2

800

2.00 x 10-2

We are going to do a series of plots to determine if the reaction is zero order, first order, or second order. We do this by plotting data to determine which order would give a straight-line that fits the equation y = mx + b. The equations for zero order, first order, and second order will be derived later.

If the reaction were zero, the equation for a straight-line equation would be:

(a o - x) = -kt + a o In the equation for a straight line, y = mx + b, y = ao-x , x = t, m (slope) = -k, and b (the intercept) = ao And (ao-x) is the concentration of 1-chlorobutanol after x amount of 1chlorbutanol is converted to 1-butanol, and ao is the initial concentration of 1chlorobutanol. Therefore, if the reaction is zero order, i.e., independent* of the concentration of 1-chlorobutanol, then a plot of the concentration after x amount is converted to product versus time would give a straight-line. If a straight line is not obtained, then the reaction is not zero order.

12

time (seconds) 0

[CH3CH2CH2CH2Cl], mol/L 0.10

50

9.05 x 10-2

100

8.2 x 10-2

150

7.41 x 10-2

200

6.71 x 10-2

300

5.49 x 10-2

400

4.48 x 10-2

500

3.68 x 10-2

800

2.00 x 10-2

* Rate = k [CH3CH2CH2CH2Cl]0 or Rate = k

[n-butyl chloride]

60 50 40 30 20 10 0 0

100

200

300

400

500

600

700

800

900

time (seconds)

We didn’t get a straight line; therefore, the reaction is not zero order in 1chlorbutane (n-butyl chloride).

13

If the reaction were first order, the equation for a straight-line equation would be:

⎛ a ⎞ kt log10 ⎜ o ⎟ = 2.303 ⎝ a o -x ⎠ log10 a o - log10 (a o - x) =

log10 (a o - x) = -

kt 2.303

kt + log10 a o 2.303

In the equation of a straight line, y = mx + b, y = log10 (ao –x)

x = t

m (slope) = -k/2.303

b (intercept) = log ao

or

⎛ a ⎞ ln ⎜ o ⎟ = kt ⎝ a o -x ⎠

ln a o - ln (a o - x) = kt ln (a o - x) = - kt + ln a o y = ln (ao – x) x=t m (slope) = -k b (intercept) = ln ao

14

And (ao-x) is the concentration of 1-chlorobutanol after x amount of 1chlorbutanol is converted to 1-butanol, and ao is the initial concentration of 1chlorbutanol. Therefore, if the reaction is first order, i.e., dependent of the concentration of 1-chlorobutanol to the first power*, then a plot of the concentration after x amount is converted to product versus time would give a straight-line. If a straight line is not obtained, then the reaction is not first order. * Rate = k [CH3CH2CH2CH2Cl] time, seconds

log [CH3CH2CH2CH2Cl ]

ln [CH3CH2CH2CH2Cl]

0

-1

-2.3

50

-1.04

-2.4

100

-1.09

-2.51

150

-1.13

-2.60

200

-1.17

-2.69

300

-1.26

-2.90

400

-1.35

-3.11

500

-1.43

-3.29

800

-1.7

-3.92

A plot of the common log (log to the base 10) of 1-chlorobutane (n-butyl chloride) versus time gives a straight line; therefore, the reaction is first order in 1-chlorbutane (n-butyl chloride).

15

0 -0.2

0

100

200

300

400

500

600

700

800

900

log [n-butyl chloride]

-0.4 -0.6 -0.8 -1 y = -0.0009x - 0.9985

-1.2 -1.4 -1.6 -1.8

time (seconds)

Also, a plot of the natural log (log to the base e) of 1-chlorobutane (n-butyl chloride) versus time gives a straight line; therefore, supporting the fact that the reaction is first order in 1-chlorbutane (n-butyl chloride).

0 -0.5 0

200

400

600

800

1000

-1.5 -2

ln[C4H9Cl]

-1

-2.5 -3

y = -0.002x - 2.2987

-3.5 -4 -4.5

time (seconds)

We obtained a straight line; consequently, the reaction is first order in 1chlorobutane (n-butyl chloride).

If the reaction were second order, the equation for the straight-line equation would have been:

16

1 1 = kt + a o -x ao y = mx + b

y=

1 a o -x

x=t m=k

b (intercept) =

1 ao

And (ao-x) is the concentration of 1-chlorobutanol after x amount of 1chlorbutanol is converted to 1-butanol, and ao is the initial concentration of 1chlorbutanol. Therefore, if the reaction is second order, i.e., dependent on the concentration of 1-chlorobutanol to the second power*, then a plot of the concentration after x amount is converted to product versus time would give a straight-line. If a straight line is not obtained, then the reaction is not second order with regards to 1-chlorobutane (n-butyl chloride.

Rate = k [CH3CH2CH2CH2Cl]2 time 1/[CH3CH2CH2CH2 Cl] (seconds) 0 10

50

11.0

100

12.2

150

13.5

200

14.9

300

18.2

400

22.3

500

27.2

800

50

17

1/[n-butyl chloride]

60 50 40 30 20 10 0 0

200

400

600

800

1000

time (seconds)

We didn’t get a straight line; therefore, the reaction is not second order in 1-chlorbutane (n-butyl chloride).

Our experimental data indicate that the solvolysis of 1-chlorobutane (n-butyl chloride) CH3CH2CH2CH2Cl (aq) + H2O (l) → CH3CH2CH2CH2OH (aq) + HCl (aq) is first order with respect to 1-chlorobutane (n-butyl chloride). Since the reaction is first order in 1-chlorobutane, then the slope of the reaction is -k for the plot of ln [1-chlorobutane] versus t and -k/2.303 for log10 [1-chlorobutane] versus t.

The equation of the straight line for the first order reaction is: y = -0.002 x -2.2987 for the natural log plot slope = -k –k = -0.002 s-1 k = 2 x 10-3 s-1 and y = -0.009 x – 0.9985 for the common log plot

18

slope = - k/2.303 -k/2.303 = -0.0009 s-1 k = 2.303 x 0.009 s-1 = 2 x 10-3 s-1 Consequently, the rate constant for the solvolysis reaction is 2 x 10-3 s-1.

And The Rate Law for the reaction is Rate = 2 x 10-3 s-1 [CH3CH2CH2CH2Cl] The Rate of Decomposition of Dinitrogen Pentoxide 2 N2O5 (g) → 4 NO2 (g) + O2 (g) was experimentally determined to be Rate = k [N2O5] This means that the reaction is first order in dinitrogen pentoxide. The first order reaction for the decomposition of 2 moles of dinitrogen pentoxide gives insight into the series of elementary steps (the mechanism/pathway) that result in the formation of four moles of nitrogen dioxide and one mole of oxygen. However, it takes a little imagination to visualize the pathway, but with practice, you will be able to work through the process. The elementary steps are either unimolecular or bimolecular, but the rate determining step must be in agreement with the rate equation, i.e., the slow step would involve only the concentration of N2O5 undergoing a reaction; therefore, the rate determining step must be unimolecular in order to be in agreement with Rate = k [N2O5]. Notice that 2 moles of dinitrogen pentoxide decompose to form 4 moles of nitrogen dioxide and 1 mole of oxygen; therefore, our series of elementary steps must total the stoichiometry of the final equation: 2 N2O5 (g) → 4 NO2 (g) + O2 (g).

Let’s see if we can think of simple steps that would give these results where we are not violating rules for drawing Lewis structures that were studied in Chemistry 1210.

19

Step 1, a unimolecular step, is the slow step of the mechanism; therefore, it will be our rate-determining step.

..

O

+ N

..

O ..

:O : ..

-

..

O

..

O:

+ N :O : ..

-

.

+ N.

+

..

O ..

..

O:

:O : ..

-

:O : ..

-

Lewis Structure of N2O5 Intermediate 1

+ N

NO2

NO3,

NO2 and Intermediates 1 are formed from the homolytic cleavage of an N-O bond of N2O5.

Step 2, also a slow step of the mechanism, is a repeat of step 1. This allows us to account for the stoichiometry of the chemical reaction.

..

O

+ N

..

O ..

:O : ..

-

+ N

..

O

..

O:

:O : ..

-

.

+ N.

+

..

O ..

..

O:

:O : ..

-

:O : ..

-

N2O5

+ N

NO2

NO3

Intermediate 1 Step 3, a bimolecular step, is a fast step, and is the coupling of intermediate 1 formed in steps 1 and 2 to form intermediate 2. .. :O

+ N

.. .

O ..

.. .

O .. +

:O : ..

NO3, Intermediate 1

+ N

..

O:

:O : ..

NO3, Intermediate 1

.. :O

+ N :O : ..

-

..

O ..

..

O ..

+ N

..

O:

:O : ..

-

N2O6, Intermediate 2

20

Step 4, a unimoleclar step, is a fast step that involves the homolytic cleavage of two N-O bonds of intermediate 2 and the formation of an oxygen molecule:

N2O6, Intermediate 2

NO2

O2

NO2

The sum of the four elementary steps is 2 N2O5 (g) → 4 NO2 (g) + O2 (g).

Consequently, the mechanism occurs in four elementary steps (unimolecular or bimolecular). The intermediates are guesses, and we don’t know for certain if they are formed unless we devise some cleaver method to trap them. The proposed mechanism is in agreement with the rate law or the kinetics given for the decomposition of dinitrogen pentoxide, and we are consistent with the rules for drawing Lewis structures studied in Chemistry 1210. Therefore, we can say that we have a reasonable mechanism, but we don’t know if it is the definitive mechanism. If we use the definition that a mechanism is a series of elementary steps (unimolecular or bimolecular) that rationalizes the formation of product, then we can assume that we have suggested a pathway that leads to the formation of the desired products.

Applying the techniques we just discussed, let’s suggest a possible mechanism for the following reaction: NO2 (g) + CO (g) → NO (g) + CO2 (g) If the rate is described by the following rate equation:

Rate = k [NO2(g) ]2 The rate equation indicates that the reaction is second order in NO2. Notice that the order of the reaction is not necessarily equal to the stoichiometry of the chemical reaction.

So, we can conclude that the rate-determining step must be a bimolecular process in which two NO2 molecules form an intermediate. So, let’s start with that bit of information.

21

Step 1, a bimolecular step, is the rate-determining step, and is a bimolecular reaction between two NO2 molecules to form intermediate 1:

N2O4, Intermediate 1 Step 2, a bimolecular step, is a reaction between Intermediate1 and carbon monoxide in a fast step to form intermediate 2:

CN2O4, Intermediate 2 Step 3, a unimolecular step, is the decomposition of intermediate 2 into carbon monoxide and intermediate 3.

N2O3, Intermediate 3 Step 4, a unimolecular step, is the decomposition of intermediate 3 into NO2 and NO

22

Intermediate 3

NO2

NO

The sum of steps 1-4 is

NO2

CO

CO2

NO

Consequently, the pathway that leads to NO2 (g) + CO (g) → NO (g) + CO2 (g) has been adequately explained by the above series of four elementary steps.

Let’s suggest a mechanism for the reaction 2 NO2 (g) + F2 (g) → 2 NO2F (g) If the rate is described by the following rate equation:

Rate = k [NO 2 (g) ] [F2 (g) ] This is a second order reaction that is first order in NO2 and first order in F2. So, we can conclude that the rate-determining step must be a bimolecular process in which one molecule of NO2 reacts with one molecule of F2 to form an intermediate that ultimately results in the formation of product in a fast step. So, let’s start with that bit of information.

Step 1 (a bimolecular step), the rate-determining step (the slow step), is a bimolecular step that involves the collision of one molecule of NO2 with fluorine gas:

23

NO2

F2

NO2F

Step 2 (a bimolecular step) is a rapid step involving the formation of another NO2F molecule resulting from the reaction of another NO2 molecule with F.

The sum of steps 1 and 2 gives the desired equation:

NO2

F2

NO2F

Consequently, the pathway that leads to 2 NO2 (g) + F2 (g) → 2 NO2F (g) has been adequately explained by two elementary bimolecular steps.

24

Factors that Effect the Rate of a Chemical Reaction The factors that affect the rate of a reaction are: • • • •

The physical state of the reactants, i.e., the solid, liquid, or gaseous state The concentration of the reactants (if the rate of the reaction depends on the concentration of the reactants) Temperature Catalyst

Conditions Necessary for a reaction to occur The conditions necessary for a reaction to occur are: • • • •

Molecules must collide Molecules must have the appropriate orientation upon collision Molecules must have sufficient energy to overcome the energy barrier to the reaction Bonds must break and bonds must form

A Method that can be used to Determine the Values of “m” and “n” in the Equation Rate = k [A]m [B]n For the Reaction a A + b B → cC + dD Let’s illustrate this method by using the hypothetical reaction A + B → C at 25oC Rate = k [A]m [B]n Following are several experiments that determine the rate of the reaction with different concentrations using a method analogous to the discussion Part 1 under the heading “Determination of the Rate of a Reaction:”

25

Experiment

[A], moles/L

[B], moles/L

1

0.1000

0.1000

Initial Rate, M/s 5.500 x 10-6

2

0.2000

0.1000

2.200 x 10-5

3

0.4000

0.1000

8.800 x 10-5

4

0.1000

0.3000

1.650 x 10-5

5

0.1000

0.6000

3.300 x 10-5

From Experiments 1 and 2, we can determine the value of “m.�

(1) 5.5 x 10-6 M/s = k [0.1000 M]

[0.1000 M] m n k [0.2000 M] [0.1000 M] m

(2) 2.2 x 10-5 M/s =

n

Divide equation (1) into equation (2)

k [0.2000 M ] [0.1000 M ] 2.2 x 10-5 M/s = m n 5.5 x 10-6 M/s k [0.1000 M ] [0.1000 M ] m

n

4 = [ 2]

m

2=m Another approach is

(1) log (5.5 x 10-6 ) = log k + m log [0.1000] + n log [0.1000]

(2) log (2.2 x 10-5 ) = log k + m log [0.2000] + n log [0.1000]

Subtract equation (2) from equation (1)

log (5.5 x 10-6 ) -log (2.2 x 10-5 ) = m [log ( 0.1000 ) -log (0.2000 )] -5.3 - (-4.7) = m [-1 - (-0.7)] -0.6 = m [-0.3] -0.6 =m -0.3 2=m

26

From Experiments 4 and 5, we can determine “n.”

(1) 1.65 x 10-5 M/s = k [0.1000 M]

m

(2) 3.3 x 10-5 M/s = k [0.1000 M]

m

[0.3000 M]

n

[0.6000 M]

n

Divide equation (1) into equation (2)

k [0.1000 M ] [0.6000 M ] 3.3 x 10-5 M/s = m n 1.65 x 10-6 M/s k [0.1000 M ] [0.3000 M ] m

n

2 = [ 2]

n

1=n The other solution:

(1) log (1.65 x 10-5 ) = log k + m log [0.1000] + n log [0.3000]

(2) log (3.3 x 10-5 ) = log k + m log [0.1000] + n log [0.6000]

Subtract equation (2) from equation (1)

log (1.65 x 10-5 ) -log (3.3 x 10-5 ) = m [log ( 0.3000) -log ( 0.6000 )] -4.78 - (-4.5) = m [-0.5227 - (-0.2218)] -0.3 = m [-0.3] -0.3 =m -0.3 1=m Therefore, the Rate Equation for the hypothetical reaction A+B→C Is

Rate = k [A]

2

[B]

1

27

or

Rate = k [A]2 [B] The reaction is second order in A and first order in B And the overall order of the reaction is 2 + 1 = 3 (third order) Once we have the Rate Equation, we can calculate the rate constant k

Rate

=k

[ A ] [ B] 2

We can use any of the experiments to determine the rate constant. Let’s use Experiment 3

M s =k [0.1000 M ]

8.800 x 10-5

[0.4000 M ]

2

1 =k M2 s L2 -3 5.500 x 10 =k mol2 s 5.500 x 10-3

We get the same result if we use another experiment. So, let’s use the data from experiment 1.

Rate

[ A ] [ B] 2

=k

M s =k [0.1000 M ]

5.500 x 10-5

[0.1000 M ]

2

1 =k M2 s L2 5.500 x 10-3 =k mol2 s 5.500 x 10-3

28

Let’s see how this might work for a specific reaction. Suppose we do a set of kinetic experiments for the following reaction:

O CH3

C

+

O

_ OH

CH3

+

C

OCH3

O

methyl acetate

hydroxide ion

Rate = k [CH3CO 2CH 3 ]m ⎡⎣ - OH ⎤⎦

_

acetate ion

CH3OH methanol

n

Following is one set of kinetic data at various concentrations of methyl acetate and hydroxide ion for this reaction:

1

[CH3CO2CH3], mol/L 0.050

[-OH], mol/L 0.050

Initial Rate, M/s 0.00034

2

0.050

0.100

0.00069

3

0.100

0.100

0.00137

Experiment

We can calculate “n” using data from Experiments 1 and 2

(1) 3.4 x 10-4 M/s = k [0.050 M]

[0.050 M] m n k [0.50 M] [0.100 M] m

(2) 6.9 x 10-4 M/s =

n

Divide equation (1) by equation (2)

k [0.050 M ] [0.100 M ] 6.9 x 10-4 M/s = m n 3.4 x 10-5 M/s k [0.050 M ] [0.050 M ] m

n

2 = [ 2]

n

1=n We can calculate “m” using data from Experiments 2 and 3

29

(1) 6.9 x 10-4 M/s = k [0.050 M]

[0.050 M] m n k [0.100 M] [0.100 M] m

(2) 1.37 x 10-3 M/s =

n

Divide equation (1) by equation (2)

k [0.100 M ] [0.100 M ] 1.37 x 10-3 M/s = m n 6.9 x 10-5 M/s k [0.050 M ] [0.100 M ] m

n

2 = [ 2]

m

1=m Since “m” and “n” equal 1, the Rate Equation for the reaction between methyl acetate and hydroxide ion is

Rate = k [CH 3CO 2CH 3 ] [ - OH] We can now calculate the rate constant, k, given the rate equation.

Rate =k [CH 3CO 2 CH 3 ] [ - OH]

M s =k [0.100 M] [0.100 M] 1 0.137 =k Ms L 0.137 =k mol s 0.00137

Let’s determine the Rate Law for the reaction, i.e., the values for “m” and “n” in the equation:

Rate = k [NO]m [O 2 ]n 2 NO (g) + O2 (g) → 2 NO2 (g) given the following data:

30

Experiment 1

[NO (g)], moles/L 0.020

[O2 (g)], moles/L 0.010

Initial Rate, M/s 0.028

2

0.020

0.020

0.057

3

0.020

0.040

0.114

4

0.040

0.020

0.227

5

0.010

0.020

0.014

Using experiments 1 and 2, we can calculate “n.”

(1) 0.028 M/s = (0.020 M)m (0.010 M)n (2) 0.057 =(0.020)m (0.020)n Divide equation (2) by equation (1)

0.057 M/s (0.020 M)m (0.020 M)n = 0.028 M/s (0.020 M)m (0.010 M)n (0.020 M)n 2= (0.010 M)n

2 = 2n 1=n Using experiments 2 and 5, we can calculate “m.”

(1) 0.057 M/s = (0.020 M)m (0.020 M)n (2) 0.014 M/s = (0.010 M)m (0.020 M)n Divide equation (1) by equation (2)

31

0.057 M/s (0.020 M)m (0.020 M)n = 0.014 M/s (0.010 M)m (0.020 M)n

(0.020 M)m 4= (0.010 M)m

4 = 2m

2=m

The rate expression is

Rate = k [NO]2 [O2]

The reaction is second order in NO and first order in oxygen, and overall third order.

Would this Rate Law conflict with our rule that the series of elementary steps should be unimolecular or bimolecular? The answer is no, because we must think of a mechanism (pathway) that would be a series of elementary steps and match the Rate Law, Rate = k [NO]2 [O2]. Notice that the rate laws include the concentrations of reactants, because reactants are measureable quantities that can be graphed against time to give us average rates.

Well, let’s see if we can proposed a mechanism that would be consistent with the rate law.

Step 1, a bimolecular step, of our proposed mechanism is a fast equilibrium step to produce intermediate 1:

Intermediate 1

NO NO N2O2

Step 2, a bimolecular step and the rate-determining step, is a slow step resulting in the formation of intermediate 2.

32

Intermediate 2

N2O2 O2

N2O4

Step 3, a unimolecular step and the final step, is a fast step where intermediate 2, N2O4, decomposes to two NO2 molecules.

Intermediate 2

NO2 NO2

Nowhere in the reaction mechanism do you see a termolecular step! Each step is either unimolecular or bimolecular. The second step is the rate controlling step; therefore, the Rate = k [N2O2] [O2]; however, N2O2 is not a reactant, but there is a prior rapid equilibrium that relates N2O2 to NO (one of the reactants).

This prior equilibrium can be expressed as:

k1[NO]2 = k −1[N 2O 2 ]

k1[NO]2 = [N 2O 2 ] k -1 Substituting

k1k 2 [NO]2 k -1 for [N2O2] gives the rate constant in terms of the reactants where each step of the mechanism is either unimolecular or bimolecular (there is no termolecular step):

33

Rate =

k1k 2 [NO]2 [O 2 ] k -1

The Relationship between Concentration and Time for a First Order Reaction The relationship between concentration and time for a first order reaction involves calculus, and calculus may be beyond your mathematical skills; however, let’s develop the relationship anyhow. Let’s consider the reaction A → product as a first order reaction. If the initial concentration of the reactant, A, is ao and the concentration of the reactant after the passage of time is ao-x, where x is the amount of material that was converted to the product at time “t,” then the mathematical expression for the production of x at time “t” is given by the following differential equation:

dx = k (a o - x) dt Which means that the change in x as respect to time is equal to the rate constant times the amount of reactant remaining at some future time. This differential equation can be rearranged and solved by integration in the following manner: dx = k dt a o -x dx ∫ a o -x = ∫ k dt let s = ao-x ds = =dx then

−∫ -

ds = k ∫ dt s ln s = kt + C

34

-ln (a o -x) = kt + C at x = o and t = o, C = -ln a o -ln (a o -x) = kt - ln a o ln a o - ln (a o -x) = kt ln

ao = kt a o -x

or log

ao kt = a o -x 2.303

Therefore, a plot of

ao versus t a o -x

gives a straight line of the form y = mx + b where ln

y = ln

and

ao a o -x

m = k and b = 0

For example, given the following data for the thermal conversion of cyclopropane to propene (figure 1), show that a plot of

ao ln versus t a o -x gives a straight line

35

H C

H C H

H

H H C

H Δ

H

H

C

C

H C

H

cylcopropane

Figure 1

The data:

H

propene

ao , moles/L

x, moles/L

ao –x, moles/L

ln [ao/ ao –x]

t, seconds

0.050

0

0.050

0

0

0.050

0.0004

0.0496

9.0 x 10-3

600

0.050

0.0009

0.0491

0.0180

1200

0.050

0.0015

0.0485

0.0300

2000

0.050

0.0022

0.0478

0.045

3000

0.050

0.0036

0.0464

0.075

5000

0.050

0.0057

0.0443

0.120

8000

0.050

0.0070

0.0430

0.150

10000

0.050

0.0082

0.0418

0.180

12000

36

0.2

ln (ao /(ao – x))

0.18

y = 2E-05x

0.16 0.14 0.12 0.1 0.08 0.06 0.04 0.02 0 0

2000

4000

6000

8000

10000 12000 14000

time (seconds)

The slope, k (the rate constant), of the straight line is 2 x 10-5 s-1 The equation

ln

ao = kt a o -x

can be written as ln ao – ln (ao – x) = kt – ln (ao – x) = kt – ln ao ln (ao – x) = - kt + ln ao where y = ln (ao – x), x = t, m = -k, and b = ln ao

37

-k

ao, moles/L 0.050

x, moles/L 0

ao –x, moles/L 0.050

ln [ ao –x] -2.996

t, seconds 0

0.050

0.0004

0.0496

-3.0038

600

0.050

0.0009

0.0491

-3.014

1200

0.050

0.0015

0.0485

-3.026

2000

0.050

0.0022

0.0478

-3.0407

3000

0.050

0.0036

0.0464

-3.070

5000

0.050

0.0057

0.0443

-3.1167

8000

0.050

0.0070

0.0430

-3.1466

10000

0.050

0.0082

0.0418

-3.1748

12000

A plot of ln (ao – x) versus t gives the following straight line where –slope = -2.98 -3 0

2000

4000

6000

8000

10000

12000

14000

-3.02

Ln (ao – x)

-3.04 -3.06

y = -2E-05x - 2.9957

-3.08 -3.1

-3.12 -3.14

-3.16 -3.18 -3.2

time (seconds)

- k = - 2 x 10-5 s-1 k = 2 x 10-5 s-1

38

The Relationship between Concentration and Time for a Zero Order Reaction The relationship between concentration and time for a zero order reaction involves calculus as well. Let’s consider the reaction A → product as a zero order reaction. If the initial concentration of the reactant, A, is ao and the concentration of the reactant after the passage of time is ao-x, where x is the amount of material that was converted to the product at time “t,” then the mathematical expression for the production of x at time “t” is given by the following differential equation:

dx =k dt This means that the change in x as respect to time is equal to the rate constant only and independent. This differential equation can be rearranged and solved by integration in the following manner: dx = kdt

dx = k ∫ dt

∫

x = kt + C

at x = 0 and t = 0, C = 0

x = kt a o -x

∫

t

dx = k ∫ dt

o

ao – (ao – x) = kt

ao –x = -kt + ao

This equation is of the form y = mx + b where

y = ao – x x = t m = -k b = ao ao

39

Therefore, if the reaction is zero order, a plot of ao -x versus time would give a straight line.

Application of the Graphical Method for Determining the Order of a Reaction For the reaction, 2 N2O5 (g) → 4 NO2 (g) + O2 (g) Following is the experimental kinetic data for the formation of nitrogen dioxide gas and oxygen gas from dinitrogen pentoxide gas: The following data are obtained for the reaction [ N2O5 ], moles/L

t minutes

2.08

3.07

1.67

8.77

1.36

14.45

0.72

31.28

Let’s construct a matrix that will provide the data to plot a zero order reaction, a first order reaction, and a second order reaction. Our task will be to determine which plot versus time will give a straight line.

[ N2O5 ] moles/L (zero order) 2.08

ln [ N2O5 ]

t, minutes

(first order)

1/[ N2O5 ] L/mole (second order)

0.732

0.481

3.07

1.67

0.513

0.599

8.77

1.36

0.307

0.735

14.45

0.72

-0.329

1.390

31.28

40

Plot of [N2O5] versus time, a zero order reaction 2.5

[N2O5]

2

1.5 1 0.5 0 0

10

20

30

40

time (minutes)

The plot is not a straight line; therefore, the reaction is not zero order.

Plot ln [N2O5] versus time, a first order reaction 0.8 y = -0.0376x + 0.8462

ln[N2O5]

0.6 0.4 0.2 0 0

5

10

15

20

25

30

35

-0.2 -0.4

time (minutes)

Gives a straight line; therefore, the reaction is first order in dinitrogen pentoxide. where the slope = -0.0376 -0.0376 min-1 = -k 3.76 x 10-2 min-1 = k

41

1/[N2O5]

Plot 1/[NO2] versus time, a second order reaction

1.6 1.4 1.2 1 0.8 0.6 0.4 0.2 0 0

5

10

15

20

25

30

35

time (minutes)

Does not result in a straight line; therefore, the reaction is not second order.

Therefore, for the reaction 2 N2O5 (g) → 4 NO2 (g) + O2 (g) Rate = k [N2O5] Can you think of a mechanism for the reaction?

Let’s propose one. Well, we know that the reaction is first order in N2O5; therefore, the rate-controlling step (the slow step) must be a unimolecular step involving N2O5.

Step 1, the rate-determining step (rate-controlling step), is a unimolecular step involving the cleavage of an N-O bond.

N2O5

NO2

NO3, Intermediate 1

Step 2, a fast step, is a bimolecular reaction between intermediate 1 and another molecule of N2O5 to produce another molecule of NO2 and intermediate 2.

42

Step 3, the final step, is a fast step where Intermediate 2, in a unimolecular reaction, forms two molecules of NO2 and one molecule of oxygen.

The sum of steps 1-3 gives the equation and

2 N2O5 (g) → 4 NO2 (g) + O2 (g)

Rate = k [N2O5]

43

First order in N2O5 A Review of the Graphical Method for Determining the Order of a Reaction (1) For a first order reaction, y = ln (ao – x); x = t; slope = -k; and the intercept is ln ao or y = ln (ao /(ao – x)); x = t; slope = k; and the intercept =0 (2) For a second order reaction, y = 1/ (ao – x); x = t; slope = k; and the intercept is 1/ ao (3) For a zero order reaction, y = x; x = t; slope = k and the intercept = 0 or y = ao – x ; x = t; slope = -k and the intercept = ao Half Life The half-life is the time it takes for ½ of the reactant(s) to form the product(s).

The half-life of a first order reaction can be obtained from the relationship between concentration and time for a first order reaction.

ao = kt a o -x at the half-life, t1/2, ao-x = ½ ao ln

Therefore,

⎛ ⎞ ⎜a ⎟ ln ⎜ o ⎟ = kt 1 ao 2 ⎜⎝ ⎟⎠ 2 ln 2 = kt 1 2

0.693 = kt 1 2

0.693 = t1 k 2 (The half-life equation for a first order reaction)

44

The half-life of a second order reaction can be obtained from the relationship between concentration and time for a second order reaction.

1 1 = kt + a o -x ao at the half-life, t1/2, ao-x = ½ ao Therefore,

1 1 = kt 1 + ao ao 2 2 2 1 = kt 1 + ao ao 2 2 1 = kt 1 ao ao 2 1 = kt 1 ao 2 1 = t1 aok 2 (The equation for the half-life of a second order reaction) The half-life of a zero order reaction can be obtained from the relationship between concentration and time for a zero order reaction. ao –x = -kt + ao

where ao –x = ½ ao

Therefore,

ao = -kt 1 + a o 2 2

ao - a o = -kt 1 2 2

45

−

ao = -kt 1 2 2

ao = t1 2k 2 (The half-life of a zero order reaction) Application of Half Life Equation The rate constant, k, for transforming cyclopropane into propene is 0.054 hr-1 (a) Calculate the half-life of cyclopropane.

0.693 = t1 0.054 2 hr

13 hr = t 1 2

(b) Calculate the fraction of cyclopropane remaining after 18.0 hours.

ln

ao = kt a o -x

ln

a o -x = - kt ao

a o -x = e - kt ao a o -x = e - (0.054/h) x 18.0 h ao a o -x = e - 0.972 ao a o -x = 0.38 ao (c) Calculate the fraction of cyclopropane remaining after 51.5 hours.

46

ln

ao = kt a o -x

ln

a o -x = - kt ao

a o -x = e - kt ao a o -x = e - (0.054/h) x 51.5 h ao a o -x = e - 2.8 ao a o -x = 0.061 ao The Effect of Temperature (T) on the Reaction Rate Temperature affects the rate constant of a chemical reaction. An increase in temperature results in an increase in the numerical value of the rate constant.

The equation that shows this affect is the Arrhenius Equation that relates temperature to the rate constant, k.

k=Ae

-

E act RT

Where Eact is the energy of activation, i.e., the minimum energy required for the reaction to transcend the energy barrier to the reaction.

Following are examples of the energy of activation

47

https://www.chem.tamu.edu/class/majors/tutorialnotefiles/activation.htm

The equation can be put into the form of y = mx + b (the equation for a straight line):

ln k = -

E act + ln A RT

Where y = ln k

x=

1 T

m(slope) = b = ln A

E act RT

“A� is referred to as the frequency factor or pre-exponential factor and it represents the frequency of collisions between reactant molecules.

48

The decomposition of dinitrogen oxide to nitrogen and oxygen at various temperatures is an illustration of how the Arrhenius Equation (published by Svante Arrhenius, a Swedish Chemist in 1889) can be used to calculate the energy of activation of a reaction.

The experimental data for the rate constant, k, at four temperatures (the temperatures must be in kelvin) for the reaction

2 N2O (g) → 2 N2 (g) + O2 (g)

are listed in the following table:

T, K k, M-1/s ln k 104(1/T) 1125

11.5900

2.450

8.890

1053

1.6700

0.510

9.50

1001

0.3800

-0.968

9.99

838

0.0010

-6.810

11.9

Let’s clarify the table. The temperature (T) is given in kelvins, the units for the rate constant indicate that the reaction is second order, and the reciprocal of the temperature is multiplied by 104 to make the numbers easier to graph; therefore, you need to remember to multiply the slope by 104 in order to compensation for arbitrarily multiplying the actual values for “1/T” by 104.

A graph of ln k versus 104 (1/T) gives the following straight line:

49

5 0 0

5

10

15

20

-10

ln k

-5

-15 -20

y = -3.0712x + 29.721

-25 -30

104(1/T)

The slope is -3.07; therefore, you need to multiply it by 104 in order to compensate for arbitrarily multiplying the reciprocal of the kelvin temperatures by 104. Therefore, the actual value for the slope is -3.07 x 104 K. Now, we can calculate the energy of activation, because we know that the slope, m, is equal to –(Eact/R).

slope = -3.07 x 10 4 K = -

E act R

-3.07 x 10 4 K x - R = E act -3.07 x 10 4 K x - 8.314

J = E act K mol

2.55 x 105 J/mol = E act or Eact = 255 kJ/mol The energy of activation for our second order decomposition is 2.55 x 105 J/mol or 255 kJ/mol Based on this information, can you suggest a mechanism for the reaction?

Did you suggest a mechanism that may look like the following?

Since the order is 2, this suggests a rate-controlling (rate determining) step that is bimolecular.

50

Step 1, the rate-controlling (rate determining) step, is the bimolecular reaction between two dinitogen oxide molecules.

Intermediate Step 2 is a rapid decomposition of the intermediate to the nitrogen and oxygen.

The sum of the two steps is 2 N2O → 2 N2 + O2 Therefore, Rate = k [N2O]2

Effect of a Catalyst on the Rate of a Reaction (1) Lowers the energy barrier to the reaction by lowering the energy barrier to reactivity and lowering the energy of activation (2) There are two types of catalysts:

51

(a) A homogeneous catalyst, i.e., a catalyst in the same phase as the reacting molecules (b) Herterogeneous catalyst,i.e., a catalyst in a different phase from the reacting molecules An example of a homogeneous catalyst: 1. H 2O2 (aq) + Br(aq) →

2. H 2O2 (aq) +

1 Br2 (aq) 2

1 1 Br2 (aq) + H 2 O (l) + O2 (g) 2 2 1 → Br(aq) + H 2O (l) + O 2 (g) 2

The sum of equations 1 and 2

2 H 2O 2 (aq) → 2 H 2O (l) + O 2 (g) Br – serves as a homogeneous catalyst in this reaction. The bromide ion facilitates the lowering of the energy barrier so that the decomposition of peroxide can occur faster.

An example of a heterogeneous catalyst is the catalytic hydrogenation of an alkene in the presents of metals like platinum, rhodium, palladium, etc. The finely divided metal is placed in the reaction vessel with the alkene and hydrogen gas, under pressure.

52

Another Interesting Application of the Rate Law Svirbely and Roth reported the experimental data between the reaction of propionaldehyde (propanal) and hydrocyanic acid in the Journal of the American Chemical Society.

The Reaction:

The Experimental Data:

53

time, minutes 2.78

[HCN], moles/L 0.0990

[CH3CH2CHO], moles/L 0.0566

5.33

0.0906

0.0482

8.17

0.0830

0.0406

15.23

0.0706

0.0282

19.80m

0.0653

0.0229

∞

0.0424

0.0000

Check to determine if the reaction is first order in HCN

-2.35 -2.4

0

2

4

6

8

10

12

14

16

18

ln([HCN]-x)

-2.45 -2.5 -2.55 -2.6 -2.65 -2.7 -2.75

time (minutes)

A plot of ln ([HCN]-x) versus time does not give a straight line; therefore, the reaction is not first order.

Check to determine if the reaction is first order in propionaldehyde

54

0 0

2

4

6

8

10

12

14

16

18

ln ([propionaldehyde] -x)

-0.5 -1

time (minutes)

-1.5 -2 -2.5 -3 -3.5 -4

A plot of ln ([propionaldehyde]-x) versus time gives a straight line; therefore, the reaction is first order in propionaldehyde.

Let’s take another approach…

Let [HCN] = ao and [propionaldehyde] = bo Then,

dx = k (a o -x) (b o -x) dt dx = k dt (a o -x) (b o -x)

∫ (a

dx = k ∫ dt o -x) (b o -x)

Solution: (a -x) b 1 1 ln o = kt ln o (a o -b o ) b o -x (a o -b o ) a o

55

(a -x) 1 1 0.0566 ln o = kt ln 0.0424 M b o -x 0.0424 M 0.0990 23.6 M -1 ln

(a o -x) = kt - 23.6 M -1 ln 0.572 b o -x

23.6 M -1 ln

(a o -x) = kt - 23.6 M -1 (-0.559) b o -x

23.6 M -1 ln

(a o -x) = kt + 13.2 M -1 b o -x

This new equation is of the form y = mx + b Where

y = 23.6 M −1 ln

ao - x bo − x

x=t m=k b = 13.2 M Let’s reconstruct the data using the new second order equation:

23.6 M −1 ln time

ao – x

bo - x

ao - x bo − x

56

Consequently, the reaction is first order in HCN and first order in propionaldehyde, i.e., second order overall.

Rate = k [HCN] [CH3CH2CHO] The slope = 0.688 M-1min-1 Could you suggest a mechanism for the reaction between HCN and propionaldehyde (propanal)?

Let’s suggest a series of elementary steps that could rationalize the formation of products.

Step 1, a bimolecular reaction, is a rapid equilibrium protonation/deprotonation reaction.

Step 2, a bimolecular reaction, is a slow step involving nucleophilic attack by the cyanide anion on the potentially positive charge carbon atom of the carbonyl group to form the product.

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The sum of steps 1 and 2 gives the following balanced equation:

Therefore,

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k 2 k1 are ratio of rate constants that equal k k -1 Therefore,

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