chapter 14 kinetics...box 1: rate = k(5)(5)2 = 125k box 2 contains 7 red spheres and 3 purple...

Chemical

Kinetics

© 2015 Pearson Education, Inc.

Chapter 14

Chemical Kinetics

James F. Kirby

Quinnipiac University

Hamden, CT

Lecture Presentation

Chemical

Kinetics


Chemical Kinetics• In chemical kinetics we study the rate (or

speed) at which a chemical process occurs.

• Besides information about the speed at

which reactions occur, kinetics also sheds

light on the reaction mechanism, a

molecular-level view of the path from

reactants to products.

Chemical

Kinetics


14.1 Factors that Affect

Reaction Rates

1) Physical state of the reactants

2) Reactant concentrations

3) Reaction temperature

4) Presence of a catalyst

Chemical

Kinetics


Physical State of the Reactants

• The more readily the reactants collide,

the more rapidly they react.

• Homogeneous reactions are often

faster.

• Heterogeneous reactions that involve

solids are faster if the surface area is

increased; i.e., a fine powder reacts

faster than a pellet or tablet.

Chemical

Kinetics


Reactant Concentrations

• Increasing reactant

concentration generally

increases reaction rate.

• Since there are more

molecules, more

collisions occur.

Chemical

Kinetics


Temperature

• Reaction rate generally increases with

increased temperature.

• Kinetic energy of molecules is related to

temperature.

• At higher temperatures, molecules

move more quickly, increasing numbers

of collisions and the energy the

molecules possess during the collisions.

Chemical

Kinetics


Presence of a Catalyst

• Catalysts affect rate without being in the

overall balanced equation.

• Catalysts affect the kinds of collisions,

changing the mechanism (individual

reactions that are part of the pathway

from reactants to products).

• Catalysts are critical in many biological

reactions.

Chemical

Kinetics


14.2 Reaction Rate• Rate is a change in concentration over

a time period: Δ[ ]/Δt.

• Δ means “change in.”

• [ ] means molar concentration.

• t represents time.

• Types of rate measured:

average rate

instantaneous rate

initial rate

Chemical

Kinetics


From the data in Figure 14.3, calculate the

average rate at which A disappears over the

time interval from 20 s to 40 s.

Sample Exercise 14.1 Calculating an Average Rate of Reaction (pg. 2 of notes)

Chemical

Kinetics


Solution

Analyze We are given the concentration of A at 20 s (0.54 M) and at 40 s (0.30 M) and asked to calculate

the average rate of reaction over this time interval.

Plan The average rate is given by the change in concentration, Δ[A], divided by the change in time, Δt.

Because A is a reactant, a minus sign is used in the calculation to make the rate a positive quantity.

Solve

From the data in Figure 14.3, calculate the average rate at which A disappears over the time interval from

20 s to 40 s.

Chemical

Kinetics


Following Reaction Rates

Rate of a reaction

is measured using

the concentration

of a reactant or a

product over time.

In this example,

[C4H9Cl] is

followed.

Chemical

Kinetics


Following Reaction Rates

The average rate

is calculated by the

–(change in

[C4H9Cl]) ÷

(change in time).

The table shows

the average rate

for a variety of time

intervals.

Chemical

Kinetics


Plotting Rate Data• A plot of the data gives more

information about rate.

• The slope of the curve at one

point in time gives the

instantaneous rate.

• The instantaneous rate at

time zero is called the initial

rate; this is often the rate of

interest to chemists.

• This figure shows

instantaneous and initial rate

of the earlier example.

Note: Reactions

typically slow down

over time!

Chemical

Kinetics


Using Figure

14.4, calculate the

instantaneous

rate of

disappearance of

C4H9Cl at

t = 0 s (the initial

rate).

Sample Exercise 14.2 Calculating an Instantaneous Rate of Reaction (pg.2)

Chemical

Kinetics


Solution

Analyze We are asked to determine an

instantaneous rate from a graph of reactant

concentration versus time.

Using Figure 14.4, calculate the instantaneous rate

of disappearance of C4H9Cl at

t = 0 s (the initial rate).

Plan To obtain the instantaneous rate at t = 0 s, we must

determine the slope of the curve at t = 0. The tangent is

drawn on the graph as the hypotenuse of the tan triangle.

The slope of this straight line equals the change in the

vertical axis divided by the corresponding change in the

horizontal axis (which, in the case of this example, is the

change in molarity over change in time).

Chemical

Kinetics


Solve The tangent line falls

from [C4H9Cl] = 0.100 M to

0.060 M in the time change

from 0 s to 210 s. Thus, the

initial rate is

Chemical

Kinetics


Relative Rates

• As was said, rates are followed using a reactant or

a product. Does this give the same rate for each

reactant and product?

• Rate is dependent on stoichiometry.

• If we followed use of C4H9Cl and compared it to

production of C4H9OH, the values would be the

same. Note that the change would have opposite

signs—one goes down in value, the other goes up.

Chemical

Kinetics


Relative Rates and Stoichiometry

• What if the equation is not 1:1?

• What will the relative rates be for:

2 O3 (g) → 3 O2 (g)

Chemical

Kinetics


(a) How is the rate at which ozone

disappears related to the rate at which

oxygen appears in the reaction

2 O3(g) → 3O2(g)?

(b) If the rate at which O2 appears,

Δ[O2]/Δt, is 6.0 × 10–5 M/s at a particular

instant, at what rate is O3 disappearing at

this same time, –Δ[O3]/Δt?

Sample Exercise 14.3 Relating Rates at Which Products Appear and Reactants Disappear (pg. 3)

Chemical

Kinetics


Solution

Analyze We are given a balanced chemical

equation and asked to relate the rate of appearance

of the product to the rate of disappearance of the

reactant.

Plan We can use the coefficients in the chemical

equation as shown in Equation 14.4 to express the

relative rates

of reactions.

(a) How is the rate at which ozone disappears related to the rate at

which oxygen appears in the reaction

2 O3(g) → 3O2(g)?

Chemical

Kinetics


Solution

Solve

(a) Using the coefficients in the balanced equation and the

relationship given by Equation 14.4, we have:

(a) How is the rate at which ozone disappears related to the rate at which oxygen appears in the

reaction

2 O3(g) → 3O2(g)?

Chemical

Kinetics


Check We can apply a stoichiometric factor to convert the

O2 formation rate to the O3 disappearance rate:

(b)If the rate at which O2 appears, Δ[O2]/Δt, is 6.0 × 10–5

M/s at a particular instant, at what rate is O3 disappearing

at this same time, –Δ[O3]/Δt? Solving the equation from

part (a) for the rate at which O3 disappears, –Δ[O3]/Δt, we

have:

Chemical

Kinetics


14.3 Determining

Concentration Effect on Rate• How do we determine what effect the

concentration of each reactant has on

the rate of the reaction?

• We keep every concentration constant

except for one reactant and see what

happens to the rate. Then, we change a

different reactant. We do this until we

have seen how each reactant has

affected the rate.

Chemical

Kinetics


An Example of How

Concentration Affects Rate• Experiments 1–3 show how [NH4

+] affects rate.

• Experiments 4–6 show how [NO2−] affects rate.

• Result: The rate law, which shows the

relationship between rate and concentration for all

reactants:

Rate = k [NH4+] [NO2

−]

Chemical

Kinetics


An Example of How Concentration Affects Rate

rate law Rate = k [NH4+] [NO2

−]

Chemical

Kinetics


More about Rate Law

• The exponents tell the order of the reaction with

respect to each reactant.

• In our example from the last slide:

Rate = k [NH4+] [NO2

−]

• The order with respect to each reactant is 1. (It is

first order in NH4+ and NO2

−.)

• The reaction is second order (1 + 1 = 2; we just

add up all of the reactants’ orders to get the

reaction’s order).

• What is k? It is the rate constant. It is a

temperature-dependent quantity.

Chemical

Kinetics


Consider a reaction A + B → C for which rate =

k[A][B]2. Each of the following boxes represents

a reaction mixture in which A is shown as red

spheres and B as purple ones. Rank these

mixtures in order of increasing rate of reaction.

Sample Exercise 14.4 Relating a Rate Law to the Effect of Concentration on Rate (pg.5)

Chemical

Kinetics


Solution

Analyze We are given three boxes containing

different numbers of spheres representing mixtures

containing different reactant concentrations. We are

asked to use the given rate law and the

compositions of the boxes to rank the mixtures in

order of increasing reaction rates.

Plan Because all three boxes have the same

volume, we can put the number of spheres of each

kind into the rate law and calculate the rate for each

box.

Consider a reaction A + B → C for which rate = k[A][B]2. Each of the following boxes represents

a reaction mixture in which A is shown as red spheres and B as purple ones. Rank these mixtures

in order of increasing rate of reaction.

Chemical

Kinetics


Solution

Solve Box 1 contains 5 red spheres and 5 purple spheres,

giving the following rate:

Box 1: Rate = k(5)(5)2 = 125k

Box 2 contains 7 red spheres and 3 purple spheres:

Box 2: Rate = k(7)(3)2 = 63k

Box 3 contains 3 red spheres and 7 purple spheres:

Box 3: Rate = k(3)(7)2 = 147k

The slowest rate is 63k (Box 2), and the highest is 147k (Box

3). Thus, the rates vary in the order 2 < 1 < 3.

Consider a reaction A + B → C for which rate = k[A][B]2. Each of the following boxes represents

a reaction mixture in which A is shown as red spheres and B as purple ones. Rank these mixtures

in order of increasing rate of reaction.

Chemical

Kinetics


Check Each box contains 10 spheres. The rate law

indicates that in this case [B] has a greater

influence on rate than [A] because B has a larger

reaction order. Hence, the mixture with the highest

concentration of B (most purple spheres) should

react fastest. This analysis confirms the order 2 < 1

< 3.

Chemical

Kinetics


2N2O5(g) → 4NO2(g) + O2(g) [14.9]

Rate = k [N2O5]

H2(g) + I2(g) → 2HI2(g) [14.10]

Rate = k [H2] [I2]

CHCl3(g) + Cl2(g) → CCl4(g) + HCl(g) [14.11]

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

The following are some additional examples of

experimentally determined rate laws:

Chemical

Kinetics


Solution

Analyze We are given two rate laws and asked to express

(a) the overall reaction order for each and (b) the units for

the rate constant for the first reaction.

Plan The overall reaction order is the sum of the exponents

in the rate law. The units for the rate constant, k, are found

by using the normal units for rate (M/s) and concentration

(M) in the rate law and applying algebra to solve for k.

(a)What are the overall reaction orders for the reactions described in

Equations 14.9 and 14.11?

(b)What are the units of the rate constant for the rate law in Equation

14.9?

Sample Exercise 14.5 Determining Reaction Orders and Units for Rate Constants (pg. 5)

Chemical

Kinetics


Solution

Solve

(a)The rate of the reaction in Equation 14.9 is first

order in N2O5 and first order overall. The reaction

in Equation 14.11 is first order in CHCl3 and one-

half order in Cl2. The overall reaction order is

three halves.

(a) What are the overall reaction orders for the reactions described in Equations 14.9 and 14.11?


Chemical

Kinetics


Solution

Solve

(b) For the rate law for Equation 14.9, we have

Units of rate = (units of rate constant)(units of concentration)

so…

Notice that the units of the rate constant change as

the overall order of the reaction changes.

(b)What are the units of the rate constant for the rate law in Equation 14.9?


Chemical

Kinetics


Solution

Analyze We are given a table of data that relates concentrations of reactants with initial rates of reaction

and asked to determine (a) the rate law, (b) the rate constant, and (c) the rate of reaction for a set of

concentrations not listed in the table.

Plan (a) We assume that the rate law has the following form: Rate = k[A]m[B]n. We will use the given data to

deduce the reaction orders m and n by determining how changes in the concentration change the rate. (b)

Once we know m and n, we can use the rate law and one of the sets of data to determine the rate constant

k. (c) Upon determining both the rate constant and the reaction orders, we can use the rate law with the

given concentrations to calculate rate.

The initial rate of a reaction A + B → C was measured for several different starting concentrations of A

and B, and the results are as follows:

Using these data, determine (a) the rate law for the reaction, (b) the rate constant, (c) the rate of the

reaction when [A] = 0.050M and [B] = 0.100 M.

Sample Exercise 14.6 Determining a Rate Law from Initial

Rate Data

Chemical

Kinetics


Solve

(a) If we compare experiments 1 and 2, we see that [A] is held constant and [B] is doubled. Thus, this pair

of experiments shows how [B] affects the rate, allowing us to deduce the order of the rate law with

respect to B. Because the rate remains the same when [B] is doubled, the concentration of B has no

effect on the reaction rate. The rate law is therefore zero order in B (that is, n = 0).

In experiments 1 and 3, [B] is held constant, so these

data show how [A] affects rate. Holding [B] constant

while doubling [A] increases the rate fourfold. This

result indicates that rate is proportional to [A]2 (that is,

the reaction is second order in A). Hence, the rate law is

(b) Using the rate law and the data from experiment 1,

we have

(c) Using the rate law from part (a) and the

rate constant from part (b), we have

Because [B] is not part of the rate law, it is irrelevant to the rate if there is at least some B present to react

with A.

Check A good way to check our rate law is to use the concentrations in experiment 2 or 3 and see if we

can correctly calculate the rate. Using data from experiment 3, we have

Rate = k[A]2 = (4.0 × 10–3 M –1 s–1)(0.200 M)2 = 1.6 × 10–4 M/s

Thus, the rate law correctly reproduces the data, giving both the correct number and the correct units for

the rate.

Continued

Sample Exercise 14.6 Determining a Rate Law from Initial

Rate Data

Chemical

Kinetics


First Order Reactions

• Some rates depend only on one

reactant to the first power.

• These are first order reactions.

• The rate law becomes:

Rate = k [A]

Chemical

Kinetics


Relating k to [A] in a

First Order Reaction• rate = k [A]

• rate = −Δ [A] / Δt

• So: k [A] = −Δ [A] / Δt

• Rearrange to: Δ [A] / [A] = − k Δt

• Integrate: ln ([A] / [A]o) = − k t

• Rearrange: ln [A] = − k t + ln [A]o

• Note: this follows the equation of a line:

y = m x + b

• So, a plot of ln [A] vs. t is linear.

Chemical

Kinetics


An Example: Conversion of Methyl

Isonitrile to Acetonitrile• The equation for the reaction:

CH3NC → CH3CN

• It is first order.

Rate = k [CH3NC]

Chemical

Kinetics


Finding the Rate Constant, k

• Besides using the rate law, we can find

the rate constant from the plot of ln [A]

vs. t.

• Remember the integrated rate law:

ln [A] = − k t + ln [A]o

• The plot will give a line. Its slope will

equal −k.

Chemical

Kinetics


The decomposition of a certain insecticide in water at

12 °C follows first-order kinetics with a rate constant

of 1.45 yr–1. A quantity of this insecticide is washed

into a lake on June 1, leading to a concentration of

5.0 × 10–7 g/cm3. Assume that the temperature of

the lake is constant (so that there are no effects of

temperature variation on the rate). (a) What is the

concentration of the insecticide on June 1 of the

following year? (b) How long will it take for the

insecticide concentration to decrease to 3.0 × 10–7

g/cm3?

Sample Exercise 14.7 Using the Integrated First-Order Rate Law (pg.7 notes)

Chemical

Kinetics


Analyze We are given the rate constant for a reaction that

obeys first order kinetics, as well as information about

concentrations and times, and asked to calculate how much

reactant (insecticide) remains after 1 yr. We must also

determine the time interval needed to reach a particular

insecticide concentration. Because the exercise gives time

in (a) and asks for time in (b), we will find it most useful to

use the integrated rate law, Equation 14.13.

Plan

(a)We are given k = 1.45 yr–1, t = 1.00 yr, and [insecticide]0 =

5.0 × 10–7 g/cm3, and so Equation 14.13 can be solved for

[insecticide]t .

The decomposition of a certain insecticide in water at 12 °C follows first-order kinetics with a rate

constant of 1.45 yr–1. A quantity of this insecticide is washed into a lake on June 1, leading to a

concentration of 5.0 × 10–7 g/cm3. Assume that the temperature of the lake is constant (so that there are

no effects of temperature variation on the rate). (a) What is the concentration of the insecticide on June 1

of the following year? (b) How long will it take for the insecticide concentration to decrease to 3.0 × 10–7

g/cm3?

Sample Exercise 14.7 Using the Integrated First-Order Rate Law

Chemical

Kinetics


Solve

(a)Substituting the known quantities into Equation 14.13, we have

We use the ln function on a calculator to evaluate the

second term on the right [that is, ln(5.0 × 10–7)], giving

To obtain [insecticide]t = 1 yr, we use the inverse

natural logarithm, or ex, function on the calculator:

Note that the concentration units for [A]t and [A]0 must be the

same.

ln[insecticide]t = 1 yr = –(1.45 yr–1)(1.00 yr) + ln(5.0 × 10–7)

ln[insecticide]t = 1 yr = –1.45 + (–14.51) = –15.96

[insecticide]t = 1 yr = e–15.96 = 1.2 × 10–7g/cm3

Chemical

Kinetics


Solution

Analyze We are given the rate constant for a reaction that

obeys first order kinetics, as well as information about

concentrations and times, and asked to calculate how much

reactant (insecticide) remains after 1 yr. We must also

determine the time interval needed to reach a particular

insecticide concentration. Because the exercise gives time

in (a) and asks for time in (b), we will find it most useful to

use the integrated rate law, Equation 14.13.

Plan

(b)We have k = 1.45 yr–1, [insecticide]0 = 5.0 × 10–7 g/cm3,

and [insecticide]t = 3.0 × 10–7 g/cm3, and so we can solve

Equation 14.13 for time, t.

The decomposition of a certain insecticide in water at 12 °C follows first-order kinetics with a rate

constant of

1.45 yr–1. A quantity of this insecticide is washed into a lake on June 1, leading to a concentration of 5.0 ×10–7 g/cm3. Assume that the temperature of the lake is constant (so that there are no effects of temperature

variation on the rate). (a) What is the concentration of the insecticide on June 1 of the following year? (b)

How long will it take for the insecticide concentration to decrease to 3.0 × 10–7 g/cm3?

Chemical

Kinetics


Solve

(b) Again substituting into Equation 14.13, with

[insecticide]t = 3.0 × 10–7 g/cm3, gives

Solving for t gives

Check In part (a) the concentration remaining after 1.00 yr (that is, 1.2 × 10–7

g/cm3) is less than the original concentration (5.0 × 10–7 g/cm3), as it should be.

In (b) the given concentration (3.0 × 10–7 g/cm)2 is greater than that remaining

after 1.00 yr, indicating that the time must be less than a year. Thus, t = 0.35 yr is

a reasonable answer.

ln(3.0 × 10–7) = –(1.45 yr–1)(t) + ln(5.0 × 10–7)

t = –[ln(3.0 × 10–7) – ln(5.0 × 10–7)]/1.45 yr–1

= –(–15.02 + 14.51)/1.45 yr–1 = 0.35 yr


14.2 Reaction Rates (review) Average rates could represent the average rate of the

disappearance of a reactant (let’s say [A])

Change in [A] - ∆[A]

change in time ∆ t

Average rates could represent the average rate of the

appearance of a product (let’s say [B])

Change in [B] ∆[B]

change in time ∆ t=

=


14.2 Change in Rate with Time Instantaneous Rate (or Initial Rate): Rate at a

particular instant during the reaction

Also the slope at any particular instant

Slope also ∆Y

∆ X

There is a better way to calculate the

slope, but it involves calculus, which

we will not do in this class.

=

Note: Reactions typically slow down over time!


14.2 Reaction Rates and Stoichiometry

The rate of disappearance of a reactant is equaled to the

rate of appearance of a product because the

stoichiometric ratio is 1:1

What if it’s not 1:1

2 HI(g) H2(g) + I2(g)

Relative Reaction Rate Equation

-1 ∆[HI] ∆[H2] ∆[I2]

2 ∆ t ∆ t ∆ t= =


14.2 Reaction Rates and Stoichiometry

2 HI(g) H2(g) + I2(g)

Relative Reaction Rate Equation

-1 ∆[HI] ∆[H2] ∆[I2]

2 ∆ t ∆ t ∆ t

Relative reaction rate equations let us calculate how

molecules’ relative concentrations change over time.

These equations do not let us calculate actual

concentrations: only proportional concentrations of

reactants relative to products.

= =


14.3 Concentration and Rates Laws

Reaction Orders (General Rate Law Equation)

Rate k[A]m[B]n

The terms m and n are called reaction orders

and have to be determined experimentally.

Note that m and n are not always the same as

the coefficients in front of A and B in the

balanced chemical reaction.

=


14.3 Concentration and Rates Laws

Reaction Orders (General Rate Law Equation)

Rate k[NH4+][NO2

-]

The exponents above are both 1. We say, then, that

the reaction is first order in NH4+ and first order in NO2

-.

These numbers are called the individual reaction

orders for each reactant. The overall reaction order is

obtained by adding these together (1+1=2). This

reaction, then, is second order overall.

=


14.4 The Change of Concentration with Time

The general rate law does not let us

calculate how molecules’ concentrations

change over time. It only tells us how the

rate will be affected by altering [A] and [B].

So what if we want to be able to calculate

the actual concentrations of reactants and

products for a reaction over time? Can we

do that?



Yes, we can.

The equations we use are called

integrated rate laws, and they

combine elements from our relative

reaction rate equations and our

general rate laws.



These equations vary, depending on

whether the reaction is first order,

second order or third order and so

forth. We will only discuss first and

second order in this class.



A simple reactions looks like A products, it can sometimes

be (not always) first order. For such a reaction

- ∆[A] K[A]∆ t

=

We can take this

equation and use

calculus…… calculus

…to transform it

into this equation ln [A] = − kt + ln [A]o

This is the

integrated rate

law for 1st order

reactions



You should notice that the integrate rate law for first

order reactions…..

…..Has the formula y = mx + b: Y = mx + b

ln [A] = − kt + ln [A]o

So the graph of a first order reaction will show a

straight line if you plot ln [A] vs. time. It’s slope is − k

and it’s y-intercept is ln [A]o


Integrate Rate Law for Second Order Reactions

In contrast, a simple reaction that looks like A

products, A + B products can sometimes be (but

isn’t always) second order. For such a reaction

- ∆[A] K[A]2

∆ t

Or if both [A] and [B] contribute to the reaction rate

then, Rate = k[A]m[B]n . For this class we will consider

rate = k[A]2

=



For this type of reaction

- ∆[A] K[A]2

∆ t=

We can transform this

using calculus……calculus

…to obtain this

[A] = kt + [A]o

This is the

integrated rate

law for 2nd order

reactions

1 _ 1 _



This equation also has the formula y = mx + b:

Y = mx + b

[A] = − kt + [A]o

So the graph of a second order reaction will show a

straight line if you plot 1/[A] vs. time. It’s slope is k and

it’s y-intercept is 1/[A]o If you need to distinguish

between 1st & 2nd graph both.

1 _ 1 _

Chemical

Kinetics


Half-life• Definition: The amount of time

it takes for one-half of a

reactant to be used up in a

chemical reaction.

• First Order Reaction:

ln [A] = − k t + ln [A]o

ln ([A]o/2) = − k t½ + ln [A]o

− ln ([A]o/2) + ln [A]o = k t½

ln ([A]o / [A]o/2) = k t½

ln 2 = k t½ or t½ = 0.693/k

Chemical

Kinetics


Second Order Reactions

• Some rates depend only on a reactant

to the second power.

• These are second order reactions.

• The rate law becomes:

Rate = k [A]2

Chemical

Kinetics


Solving the Second Order

Reaction for A → Products

• rate = k [A]2

• rate = − Δ [A] / Δ t

• So, k [A]2 = − Δ [A] / Δ t

• Rearranging: Δ [A] / [A]2 = − k Δ t

• Using calculus: 1/[A] = 1/[A]o + k t

• Notice: The linear relationships for first

order and second order reactions differ!

Chemical

Kinetics


An Example of a Second

Order Reaction:

Decomposition of NO2A plot following NO2

decomposition shows

that it must be second

order because it is

linear for 1/[NO2], not

linear for ln [NO2].

Equation:

NO2 → NO + ½ O2

Chemical

Kinetics


Half-Life and Second Order

Reactions

• Using the integrated rate law, we can

see how half-life is derived:

1/[A] = 1/[A]o + k t

1/([A]o/2) = 1/[A]o + k t½

2/[A]o −1/[A]o = k t½

t½ = 1 / (k [A]o)

• So, half-life is a concentration

dependent quantity for second order

reactions!

Chemical

Kinetics


The reaction of C4H9Cl

with water is a first-order

reaction. (a) Use Figure

14.4 to estimate the half-

life for this reaction. (b)

Use the half-life from (a) to

calculate the rate

constant.

Sample Exercise 14.9 Determining the Half-Life of a First-Order

Reaction (pg. 10)

Chemical

Kinetics


Solution

Analyze We are asked to estimate the

half-life of a reaction from a graph of

concentration versus time and then

to use the half-life to calculate the

rate constant for the reaction.

The reaction of C4H9Cl with water is a first-order reaction. (a) Use

Figure 14.4 to estimate the half-life for this reaction.

Sample Exercise 14.9 Determining the Half-Life of a First-Order Reaction

Plan

(a)To estimate a half-life, we can select a

concentration and then determine the time required

for the concentration to decrease to half of that

value.

Chemical

Kinetics


Solve

(a) From the graph, we

see that the initial value of

[C4H9Cl] is 0.100 M. The

half-life for this first-order

reaction is the time

required for [C4H9Cl] to

decrease to 0.050 M, which

we can read off the graph.

This point occurs at

approximately 340 s.


Chemical

Kinetics


(b) Use the half-life from (a) to calculate the rate constant.

Solve

(a) 340 s.

(b) Solving Equation 14.15 for k, we have

Check At the end of the second half-life, which should

occur at 680 s, the concentration should have decreased by

yet another factor of 2, to 0.025 M. Inspection of the graph

shows that this is indeed the case.


Chemical

Kinetics


Zero Order Reactions

• Occasionally, rate is

independent of the

concentration of the

reactant:

• Rate = k

• These are zero order

reactions.

• These reactions are

linear in

concentration.

Chemical

Kinetics


Factors That Affect Reaction

Rate

1. Temperature

2. Frequency of collisions

3. Orientation of molecules

4. Energy needed for the reaction to take

place (activation energy)

Chemical

Kinetics


Factors That Affect Reaction Rate

1. Temperature? Most reactions speed up with

increased temperature.

Why? Let’s look at the rate law equation again:

Rate = k [A]m[B]nDoes changing the temperature change [A]? Does

changing the temperature change [B]? So why does

increasing the temperature increase a reactions' rate?

No

No

Because it increases the rate constant k

So how does it do that? The answer is explained

by using something called the collision model.

Chemical

Kinetics



Molecules have to collide to react. The greater

the number of collisions per second, the greater

the chance that molecules will get together in a

way that leads them to react.

Increasing a reaction’s temperature increases

individual molecules’ speeds. This increases the

total number of collisions they experience and,

hence, the overall likelihood that molecules will

get together in a way that leads them to react.

Chemical

Kinetics



But colliding by itself, isn't enough to

automatically cause a reaction to occur. For

example, in a mixture of H2 and I2, the molecules

undergo about 1010 collisions per second, but the

reaction still proceeds very slowly. In fact, only

one in every 1013 collisions causes a reaction.

Why?

The answer comes down to two additional factors:

orientation and activation energy.

Chemical

Kinetics


Temperature and Rate

• Generally, as

temperature increases,

rate increases.

• The rate constant is

temperature dependent:

it increases as

temperature increases.

• Rate constant doubles

(approximately) with

every 10 ºC rise.

Chemical

Kinetics


Frequency of Collisions• The collision model is based on the kinetic

molecular theory.

• Molecules must collide to react.

• If there are more collisions, more reactions

can occur.

• So, if there are more molecules, the

reaction rate is faster.

• Also, if the temperature is higher, molecules

move faster, causing more collisions and a

higher rate of reaction.

Chemical

Kinetics


The Collision Model

• In a chemical reaction, bonds are

broken and new bonds are formed.

• Molecules can only react if they collide

with each other.

Chemical

Kinetics


Orientation of Molecules

• Molecules can often collide without

forming products.

• Aligning molecules properly can lead to

chemical reactions.

• Bonds must be broken and made and

atoms need to be in proper positions.

Chemical

Kinetics


Energy Needed for a Reaction to

Take Place (Activation Energy)

• The minimum energy needed for a reaction to take

place is called activation energy.

• An energy barrier must be overcome for a reaction

to take place, much like the ball must be hit to

overcome the barrier in the figure below.

Chemical

Kinetics


Transition State

(Activated Complex)

• Reactants gain energy as the reaction

proceeds until the particles reach the

maximum energy state.

• The organization of the atoms at this

highest energy state is called the

transition state (or activated complex).

• The energy needed to form this state is

called the activation energy.

Chemical

Kinetics


Reaction Progress

• Plots are made to show

the energy possessed

by the particles as the

reaction proceeds.

• At the highest energy

state, the transition state

is formed.

• Reactions can be

endothermic or

exothermic after this.

transition

state

transition state – Very

unstable (transition

products are also

called intermediates)

Chemical

Kinetics


Reaction Progress• Suppose you could

measure the rates for

both the forward and

reverse reactions of the

process in the figure. In

which direction would

the rate be larger?

• Why? The energy barrier is lower in the forward

direction than in the reverse. Thus more

molecules will have energy sufficient to

cross the barrier in the forward direction.

The forward rate will be greater.

Chemical

Kinetics


Reaction Progress

• Suppose we have two

reactions A B and

B C. You could isolate

B, and it is stable. Is B

the transition state for

the reaction A C?

No. If B can be isolated, it can‘t correspond

to the top of the energy barrier. There would

be transition states for each of the individual

reactions above.

Chemical

Kinetics


Distribution of the Energy

of Molecules

• Gases have an average temperature, but

each individual molecule has its own energy.

• At higher energies, more molecules possess

the energy needed for the reaction to occur.

Chemical

Kinetics


The Relationship Between

Activation Energy & Temperature

• Like everything else in chemistry, there exists an

equation we can use to calculate activation

energy (Ea). It is called the Arrhenius Equation:

k = Ae−Ea/RT

• Where k is the rate constant, R is the ideal gas

constant (8.314 J/mol-K), and yes, you have to

use this value of R, instead of the others in your

book, for this particular equation), T is the reaction

temperature, Ea is its activation energy, and A is a

constant called a frequency factor.

Chemical

Kinetics


The Relationship Between Activation Energy & Temperature

• Reaction constant k changes with temperature.

By manipulating the Arrehenius equation in some

ways we can calculate what the new k value will

be if we changed temperature, by using the

following equation:

• Allows us to calculate the temperature if we have

rate constants at two different temperatures.

Chemical

Kinetics


Try this problem:

A certain first-order reaction has a rate constant of

2.75x10-2 s-1 at 20 C. What is the k at 60 C if

Ea = 75.5 kj/mol?

Chemical

Kinetics


The Relationship Between

Activation Energy & Temperature

• Activation energy can be determined graphically

by reorganizing the equation: ln k = −Ea/RT + ln A

Chemical

Kinetics


Law vs. Theory

• Kinetics gives what happens. We call

the description the rate law.

• Why do we observe that rate law? We

explain with a theory called a

mechanism.

• A mechanism is a series of stepwise

reactions that show how reactants

become products.

Chemical

Kinetics


Reaction Mechanisms

• Reactions may occur all at once or

through several discrete steps.

• Each of these processes is known as an

elementary reaction or elementary

process.

Chemical

Kinetics


Molecularity

The molecularity of an elementary reaction

tells how many molecules are involved in that

step of the mechanism.

Chemical

Kinetics


Termolecular?

• Termolecular steps require three

molecules to simultaneously collide with

the proper orientation and the proper

energy.

• These are rare, if they indeed do occur.

• These must be slower than

unimolecular or bimolecular steps.

• Nearly all mechanisms use only

unimolecular or bimolecular reactions.

Chemical

Kinetics


What Limits the Rate?

• The overall reaction cannot occur faster than

the slowest reaction in the mechanism.

• We call that the rate-determining step.

Chemical

Kinetics


What is Required of a

Plausible Mechanism?

• The rate law must be able to be devised

from the rate-determining step.

• The stoichiometry must be obtained when

all steps are added up.

• Each step must balance, like any equation.

• All intermediates are made and used up.

• Any catalyst is used and regenerated.

Chemical

Kinetics


A Mechanism With a Slow

Initial Step

• Overall equation: NO2 + CO → NO + CO2

• Rate law: Rate = k [NO2]2

• If the first step is the rate-determining step,

the coefficients on the reactants side are the

same as the order in the rate law!

• So, the first step of the mechanism begins:

NO2 + NO2 →

Chemical

Kinetics



Initial Step (continued)

• The easiest way to complete the first

step is to make a product:

NO2 + NO2 → NO + NO3

• We do not see NO3 in the stoichiometry,

so it is an intermediate, which needs to

be used in a faster next step.

NO3 + CO → NO2 + CO2

Chemical

Kinetics



Initial Step (completed)

• Since the first step is the slowest step, it

gives the rate law.

• If you add up all of the individual steps

(2 of them), you get the stoichiometry.

• Each step balances.

• This is a plausible mechanism.

Chemical

Kinetics


A Mechanism With a

Fast Initial Step

• Equation for the reaction:

2 NO + Br2 ⇌ 2 NOBr

• The rate law for this reaction is found to be

Rate = k [NO]2 [Br2]

• Because termolecular processes are rare,

this rate law suggests a multistep

mechanism.

Chemical

Kinetics


A Mechanism With a Fast

Initial Step (continued)

• The rate law indicates that a quickly

established equilibrium is followed by a

slow step.

• Step 1: NO + Br2 ⇌ NOBr2

• Step 2: NOBr2 + NO → 2 NOBr

Chemical

Kinetics


What is the Rate Law?

• The rate of the overall reaction depends

upon the rate of the slow step.

• The rate law for that step would be

Rate = k2[NOBr2] [NO]

• But how can we find [NOBr2]?

Chemical

Kinetics


[NOBr2] (An Intermediate)?

• NOBr2 can react two ways:

– With NO to form NOBr.

– By decomposition to reform NO and Br2.

• The reactants and products of the first

step are in equilibrium with each other.

• For an equilibrium (as we will see in the

next chapter):

Ratef = Rater

Chemical

Kinetics


The Rate Law (Finally!)

• Substituting for the forward and reverse rates:

k1 [NO] [Br2] = k−1 [NOBr2]

• Solve for [NOBr2], then substitute into the rate

law:

Rate = k2 (k1/k−1) [NO] [Br2] [NO]

• This gives the observed rate law!

Rate = k [NO]2 [Br2]

Chemical

Kinetics


Catalysts

• Catalysts increase the rate of a reaction by

decreasing the activation energy of the

reaction.

• Catalysts change the mechanism by which

the process occurs.

Chemical

Kinetics


Types of Catalysts

1) Homogeneous catalysts

2) Heterogeneous catalysts

3) Enzymes

Chemical

Kinetics


Homogeneous Catalysts

• The reactants and catalyst are in the

same phase.

• Many times, reactants and catalyst are

dissolved in the same solvent, as seen

below.

Chemical

Kinetics


Heterogeneous Catalysts• The catalyst is in a

different phase than

the reactants.

• Often, gases are

passed over a solid

catalyst.

• The adsorption of

the reactants is

often the rate-

determining step.

Chemical

Kinetics


Enzymes

• Enzymes are

biological catalysts.

• They have a region

where the reactants

attach. That region

is called the active

site. The reactants

are referred to as

substrates.

Chemical

Kinetics


Lock-and-Key Model

• In the enzyme–substrate model, the

substrate fits into the active site of an

enzyme, much like a key fits into a lock.

• They are specific.

chapter 14 kinetics...box 1: rate = k(5)(5)2 = 125k box 2 contains 7 red spheres and 3 purple...

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