Chapter 13Rates of

Reactions

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Reaction Rates

1. Definition of Reaction Rate

2. Experimental Determination of Rate

3. Dependence of Rate on Concentration

4. Change of Concentration with Time

5. Temperature and Rate; Collision and Transition-State Theories

6. Arrhenius Equation

Contents and Concepts

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Reaction Mechanisms

7. Elementary Reactions

8. The Rate Law and the Mechanism

9. Catalysis

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Learning Objectives

Reaction Rates

1. Definition of a Reaction Rate

a. Define reaction rate.

b. Explain instantaneous rate and average rate of a reaction.

c. Explain how the different ways of expressing reaction rates are related.

d. Calculate average reaction rate.

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2. Experimental Determination of Rate

a. Describe how reaction rates may be experimentally determined.

3. Dependence of Rate on Concentration

a. Define and provide examples of a rate law, rate constant, and reaction order.

b. Determine the order of reaction from the rate law.

c. Determine the rate law from initial rates.

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4. Change of Concentration with Timea. Learn the integrated rate laws for first-

order, second-order, and zero-order reactions.

b. Use an integrated rate law.c. Define half-life of a reaction.d. Learn the half-life equations for first-order,

second-order, and zero-order reactions.e. Relate the half-life of a reaction to the rate

constant.f. Plot kinetic data to determine the order of a

reaction.

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5. Temperature and Rate; Collision and Transition-State Theoriesa. State the postulates of collision theory.

b. Explain activation energy (Ea).c. Describe how temperature, activation

energy, and molecular orientation influence reaction rates.

d. State the transition-state theory.e. Define activated complex.f. Describe and interpret potential-energy

curves for endothermic and exothermic reactions.

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6. Arrhenius Equationa. Use the Arrhenius equation.

Reaction Mechanisms7. Elementary Reactions

a. Define elementary reaction, reaction mechanism, and reaction intermediate. Determine the rate law from initial rates.

b. Write the overall chemical equation from a mechanism.

c. Define molecularity.d. Give examples of unimolecular, bimolecular, and

termolecular reactions.e. Determine the molecularity of an elementary

reaction.f. Write the rate equation for an elementary reaction.

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8. The Rate Law and the Mechanisma. Explain the rate-determining step of a mechanism.b. Determine the rate law from a mechanism with an

initial slow step.c. Determine the rate law from a mechanism with an

initial fast, equilibrium step.

9. Catalysisa. Describe how a catalyst influences the rate of a

reaction.b. Indicate how a catalyst changes the potential-

energy curve of a reaction.c. Define homogeneous catalysis and

heterogeneous catalysis.d. Explain enzyme catalysis.

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Chemical kinetics is the study of reaction rates, including how reaction rates change with varying conditions and which molecular events occur during the overall reaction.

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The following questions are explored in this chapter:

• How is the rate of a reaction measured?

• Which conditions will affect the rate of a reaction?

• How do you express the relationship of rate to the variables affecting the rate?

• What happens on a molecular level during a chemical reaction?

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Which conditions will affect the rate of a reaction?

Four variables affect the rate of reaction:

1. The concentrations of the reactants

2. The concentration of the catalyst

3. The temperature at which the reaction occurs

4. The surface area of the solid reactant or catalyst

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Concentrations of Reactants

Often the rate of reaction increases when the concentration of a reactant is increased. In some reactions, however, the rate is unaffected by the concentration of a particular reactant, as long as it is present at some concentration.

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Concentration of Catalyst

Hydrogen peroxide, H2O2, decomposes rapidly in the presence of HBr, giving oxygen and water. Some of the HBr is oxidized to give the orange Br2, as can be seen above.

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Temperature at Which Reaction Occurs

Usually reactions speed up when the temperature increases. It takes less time to boil an egg at sea level than on a mountaintop, where water boils at a lower temperature. Reactions during cooking go faster at higher temperature.

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Each test tube contains potassium permanganate, KMnO4, and oxalic acid, H2C2O4, at the same concentrations. Permanganate ion oxidizes oxalic acid toCO2 and H2O. Top: One test tube was placedin a beaker of warm water (40°C); the other was kept at room temperature (20°C).Bottom: After 10 minutes, the test tube at 40°C showed a noticeable reaction, whereas the other did not.

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Surface Area of a Solid Reactant or Catalyst

If a reaction involves a solid with a gas or liquid, then the surface area of the solid affects the reaction rate. Because the reaction occurs at the surface of the solid, the rate increases with increasing surface area. For example, a wood fire burns faster if the logs are chopped into smaller pieces. Similarly, the surface area of a solid catalyst is important to the rate of reaction.

Right: The photo shows a powder igniting.

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Reaction rate is the increase in molar concentration of product of a reaction per unit time or the decrease in molar concentration of reactant per unit time. The unit is usually mol/(L • s) or M/s.

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The rate of this reaction can be found by measuring the concentration of O2 at various times. Alternatively, the concentration of NO2 could be measured. Both of these concentrations increase with time. The rate could also be determined by measuring the concentration of N2O5, which would decrease over time.

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2N2O5(g) 4NO2(g) + O2(g)

We can express these changes in concentration as follows. Square brackets mean molarity.

Rate of formation of O2 =

Rate of formation of NO2 =

Rate of decomposition of N2O5 =

tΔ

]Δ[NO2

tΔ

]Δ[O2

tΔ

]OΔ[N 52

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We can relate these expressions by taking into account the reaction stoichiometry.

tΔ

]Δ[NO

4

1 2 tΔ

]Δ[O2

tΔ

]OΔ[N

2

1 52

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These equations give the average rate over the time interval t. As t decreases and approaches zero, the equations give the instantaneous rate.

The next slides illustrate this relationship graphically for the increase in concentration of O2.

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The average rate is the slope of the hypotenuse of the triangle formed.

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As t gets smaller and approaches zero, the hypotenuse becomes a tangent line at that point.

The slope of the tangent line equals the rate at that point.

?

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Peroxydisulfate ion oxidizes iodide ion to triiodide ion, I3

-. The reaction is

S2O82-(aq) + 3I-(aq)

2SO42-(aq) + I3

-(aq)

How is the rate of reaction that is expressed as the rate of formation of I3

- related to the rate of reaction of I-?

tΔ

]Δ[I

3

1 -

tΔ

]Δ[I -3

?

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Calculate the average rate of formation of O2 in the following reaction during the time interval from 1200 s to 1800 s. At 1200 s, [O2] = 0.0036 M; at 1800 s, [O2] = 0.0048 M.

Δt

]Δ[O2 s 1200-1800

0.0036-0.0048 M

s 600

0.0012 M

/s 10 2.0 -6 M

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a.The slope at point A is greater than the slope at point B, so the instantaneous rate at point A is greater than the instantaneous rate at point B.

b.No. If it were, the graph would be linear.

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Rates are determined experimentally in a variety of ways.

For example, samples can be taken and analyzed from the reaction at several different intervals.

Continuously following the reaction is more convenient. This can be done by measuring pressure, as shown on the next slide, or by measuring light absorbance when a color change is part of the reaction.

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The reaction rate usually depends on the concentration of one or more reactant. This relationship must be determined by experiment.

This information is captured in the rate law, an equation that relates the rate of a reaction to the concentration of a reactant (and catalyst) raised to various powers. The proportionality constant, k, is the rate constant.

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For the generic reaction

aA + bB +cC dD + eE

the rate law is

Rate = k[A]m[B]n[C]p

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The reaction order with respect to a specific reactant is the exponent of that species in the experimentally determined rate law.

The overall order of a reaction is the sum of the orders of the reactant species from the experimentally determined rate law.

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For the reaction

the rate law is

Rate = k[NO2][F2]

In this case, the exponent for both NO2 and F2 is 1 (which does not need to be written).

We say the reaction is first order in each. The reaction is second order overall.

?

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Hydrogen peroxide oxidizes iodide ion in acidic solution:

H2O2(aq) + 3I-(aq) + 2H+(aq)

I3-(aq) + 2H2O(l)

The rate law for the reaction is

Rate = k[H2O2][I-]

a. What is the order of reaction with respect to each reactant species?

b. What is the overall order?

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H2O2(aq) + 3I-(aq) + 2H+(aq) I3-(aq) + 2H2O(l)

Given: Rate = k[H2O2][I-]

a. The reaction is first order in H2O2.

The reaction is first order in I-.

The reaction is zero order in H+.

b. The reaction is second order overall.

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a. The concentration of Q does not affect the rate. The rate will depend only on [R].

b. Rate = k[R]2

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The rate law for a reaction must be determined experimentally.

We will study the initial rates method of determining the rate law. This method measures the initial rate of reaction using various starting concentrations, all measured at the same temperature.

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Returning to the decomposition of N2O5, we have the following data:

2N2O5(g) 2N2(g) + 5O2(g)

Note that when the concentration doubles from experiment 1 to 2, the rate doubles.

Initial N2O5 Concentration

Initial Rate of Disappearance of N2O5

Experiment 1 1.0 × 10-2 M 4.8 × 10-6 M/s

Experiment 2 2.0 × 10-2 M 9.6 × 10-6 M/s

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Examining the effect of doubling the initial concentration gives us the order in that reactant.

In this case, when the concentration doubles, the rate doubles, so m = 1.

When rate is multiplied by m is

½ -1

1 0

2 1

4 2

?

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2NO(g) + O2(g) 2NO2(g)

Obtain the rate law and the rate constant using the following data for the initial rates of disappearance of NO:

Exp. [NO]0, M [O2]0, M

Initial Rate of Reaction of NO, M/s

1 0.0125 0.0253 0.0281

2 0.0250 0.0253 0.112

3 0.0125 0.0506 0.0561

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To find the order in NO, we will first identify two experiments in which the concentration of O2 remains constant. Then, we will divide the rate laws for those two experiments. Finally, we will solve the equation to find the order in NO.

To find the order in O2, we will repeat this procedure, this time choosing experiments in which the concentration of NO remains constant.

Once the order in NO and the order in O2 are known, data from any experiment can be substituted into the rate law. The rate law can then be solved for the rate constant k.

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[O2] remains constant in experiments 1 and 2.

The generic rate law is rate = k[NO]m[O2]n. We will put the result from experiment 2 in the numerator because its rate is larger than that of experiment 1.

nm

nm

MM

MMM

M

0.0253 0.0125

0.0253 0.0250

s0.0281

s0.112

m2 4

NO. in order second is reaction The

2m

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[NO] remains constant in experiments 1 and 3.

The generic rate law is rate = k[NO]m[O2]n. We will put the result from experiment 3 in the numerator because its rate is larger than that of experiment 1.

nm

nm

MM

MMM

M

0.0253 0.0125

0.0506 0.0125

s0.0281

s 0.0.0561

n2 2

.O in order first is reaction The

1

2

n

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It is also possible to find the order by examining the data qualitatively.

Again choosing experiments 1 and 2, we note that the concentration of NO is doubled and the rate is quadrupled. That means the reaction is second order in NO.

Choosing experiments 1 and 3, we note that the concentration of O2 is doubled and the rate is doubled. That means the reaction is first order in O2.

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The rate law is rate = k[NO]2[O2].

22 ONO

rate

obtain we, for Solving

k

k

s

10 7.11

0.0253 0.0125s

0.0281

1: experiment from data Using

2

3

2

MMM

M

k

Using data from experiment 2: k = 7.08 × 103/M2 s

Using data from experiment 3: k = 7.10 × 103/M2 s

These values are the same to 2 significant figures.

?

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Iron(II) is oxidized to iron(III) by chlorine in an acidic solution:

2Fe2+(aq) + Cl2(aq)

2Fe3+ (aq) + 2Cl-(aq)

What is the reaction order with respect to Fe2+, Cl2, and H+? What is the rate law and the relative rate constant?

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Exp. [Fe2+], M [Cl2], M [H+], M Rate, M/s

1 0.0020 0.0020 1.0 1.0 × 10-5

2 0.0040 0.0020 1.0 2.0 × 10-5

3 0.0020 0.0040 1.0 2.0 × 10-5

4 0.0040 0.0040 1.0 4.0 × 10-5

5 0.0020 0.0020 0.5 2.0 × 10-5

6 0.0020 0.0020 0.1 1.0 × 10-4

The following data were collected. The rates given are relative, not actual.

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The generic rate law is

Rate = k[Fe2+]m[Cl2]n[H+]p

To find the order in Fe2+, we choose two experiments in which both [Cl2] and [H+] remain constant: 1 and 2, or 3 and 4.

Exp [Fe2+] [Cl2] [H+] Initial Rate1 0.0020 0.0020 1.0 1.0 × 10-5

2 0.0040 0.0020 1.0 2.0 × 10-5

3 0.0020 0.0040 1.0 2.0 × 10-5

4 0.0040 0.0040 1.0 4.0 × 10-5

In each case, the concentration of Fe2+ doubles and the rate doubles.

The reaction is first order in Fe2+; m = 1.

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To find the order in Cl2, we choose two experiments in which both [Fe2+] and [H+] remain constant: 1 and 3, or 2 and 4.

Exp [Fe2+] [Cl2] [H+] Initial Rate1 0.0020 0.0020 1.0 1.0 × 10-5

3 0.0020 0.0040 1.0 2.0 × 10-5

2 0.0040 0.0020 1.0 2.0 × 10-5

4 0.0040 0.0040 1.0 4.0 × 10-5

In each case, the concentration of Cl2 doubles and the rate doubles.

The reaction is first order in Cl2; n = 1.

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To find the order in H+, we choose two experiments in which both [Fe2+] and [Cl2] remain constant: 1 and 5, 1 and 6, or 5 and 6.

1 0.0020 0.0020 1.0 1.0 × 10-5

5 0.0020 0.0020 0.5 2.0 × 10-5

From experiment 5 to experiment 1, the concentration doubles while the rate is halved.

1 0.0020 0.0020 1.0 1.0 × 10-5

6 0.0020 0.0020 0.1 1.0 × 10-4

From experiment 6 to experiment 1, the concentration is increased 10 times, the rate is cut to one-tenth of the original rate.

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5 0.0020 0.0020 0.5 2.0 × 10-5

6 0.0020 0.0020 0.1 1.0 × 10-4

From experiment 6 to experiment 5, the concentration is increased 5 times; the rate is cut to one-fifth of the original rate.

Each of these describes the order in H+ as p = -1.

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The rate law is rate = k[Fe2+][Cl2][H+]-1 .

H

ClFe Rate

is this express to way Another

22k

s

2.50

0.002 0.002

1.0 s

10 1.0

ClFe

H rate

1: experiment from data Using

5

22

MMM

MM

k

The same value for k is obtained using each experiment.

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Note that the units on the rate constant are specific to the overall order of the reaction.

• For a zero-order reaction, the unit is M/s or mol/(L s).

• For a first-order reaction, the unit is 1/s or s-1.

• For a second-order reaction, the unit is 1/(M s) or L/(mol s).

If you know the rate constant, you can deduce the overall rate of reaction.

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a. These factors might include things such as genetics, diet, exercise, weight, and stress.

b. You would need to run controlled studies quantifying each of the factors and measures of aging.

c. When smoking doubles, aging increases fourfold. That is a second-order reaction, so m = 2.

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The rate law tells us the relationship between the rate and the concentrations of reactants and catalysts.

To find concentrations at various times, we need to use the integrated rate law. The form used depends on the order of reaction in that reactant.

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?

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Cyclopropane is used as an anesthetic. The isomerization of cyclopropane to propene is a first-order reaction with a rate constant of 9.2/s. If an initial sample of cyclopropane has a concentration of 6.00 M, what will the cyclopropane concentration be after 1.00 s?

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The rate law is

Rate = k[cyclopropane]

k = 9.2/s

[A]0 = 6.00 M

t = 1.00 s

[A]t = ?

The reaction is first order, so the integrated rate law is

kt- A

A ln

0

t

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49.2

0

10 1.01A

A et

Mt 6.0010 1.01A 4

Mt410 6.1A

9.2- s 1.00

s

9.2-

A

Aln

0

t

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The half-life of a reaction, t½, is the time it takes for the reactant concentration to decrease to one-half of its initial value.

By substituting ½[A]0 for [A]t, we solve the integrated rate law for the special case of t = t½.

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?

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Ammonium nitrite is unstable because ammonium ion reacts with nitrite ion to produce nitrogen:

NH4+(aq) + NO2

-(aq) N2(g) + 2H2O(l)

In a solution that is 10.0 M in NH4+, the

reaction is first order in nitrite ion (at low concentrations), and the rate constant at 25°C is 3.0 × 103/s.

What is the half-life of the reaction?

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The rate is first order in nitrite, NO2-:

k = 3.0 × 10-3/s

For first-order reactions:

kt½ = 0.693

s10 3.0

0.6930.6933-1/2

k

t

t½ = 2.3 × 102 s = 3.9 min

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Let’s look again at the integrated rate laws:

In each case, the rate law is in the form of y = mx + b, allowing us to use the slope and intercept to find the values.

y = mx + b

y = mx + b

y = mx + b

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For a zero-order reaction, a plot of [A]t versus t is linear. The y-intercept is [A]0.

For a first-order reaction, a plot of ln[A] t versus t is linear. The graph crosses the origin (b = 0).

For a second-order reaction, a plot of 1/[A] t versus t is linear. The y-intercept is 1/[A]0.

0[A] - [A] :Order Zero ktt

0[A] ln - [A] ln :OrderFirst ktt

0[A]

1

[A]

1 :Order Second kt

t

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Graphs of concentration and time can also be used to determine the order of reaction in that reactant.

• For zero-order reactions, [A] versus t is linear.• For first-order reactions, ln[A] versus t is

linear.• For second order reactions, 1/[A] versus t is

linear.

This is illustrated on the next slide for the decomposition of NO2.

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Left: The plot of ln[NO2] versus t is not linear, so the reaction is not first order.

Right: The plot of 1/[NO2] versus t is linear, so the reaction is second order in NO2.

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The initial concentration decreases in each time interval. The only equation that results in a larger value for t½ is the second-order equation.

The reaction is second order.

tkt

kt

kt

[A]

1 :Order Second

0.693 :Order First

2

[A] :Order Zero

1/2

1/2

01/2

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Rate Constant and Temperature

The rate constant depends strongly on temperature. How can we explain this relationship?

Collision theory assumes that reactant molecules must collide with an energy greater than some minimum value and with the proper orientation.

The minimum energy is called the activation energy, Ea.

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We will now explore the effect of a temperature increase on each of the three requirements for a reaction to occur.

The rate constant can be given by the equation

k = Zfp

where

Z = collision frequency

f = fraction of collisions with the minimum energy

p = orientation factor

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Certainly, Z will increase with temperature, as the average velocity of the molecules increases with temperature.

However, this factor alone cannot explain the dramatic effect of temperature changes.

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The fraction of molecular collisions having the minimum energy required is given by f:

Now we can explain the dramatic impact of temperature. The fraction of collisions having the minimum energy increases exponentially with temperature.

RT

ERT

E

e

e fa

a 1

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The reaction rate also depends on p, the proper orientation for the collision. This factor is independent of temperature.

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Importance of Molecular Orientation inthe Reaction of NO and Cl2

(A) NO approaches with its N atom toward Cl2, and an NOCl bond forms. Also, the angle of approach is close to that in the product NOCl.

(B) NO approaches with its O atom toward Cl2. No NOCl bond can form, so NO and Cl2 collide and then fly apart.

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Transition-state theory explains the reaction resulting from the collision of two molecules in terms of an activated complex (transition state), an unstable grouping of atoms that can break up to form products.

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The potential energy diagram for a reaction visually illustrates the changes in energy that occur.

The next slide shows the diagram for the reaction

NO + Cl2 NOCl2‡ NOCl + Cl

where NOCl2‡ indicates the activated complex.

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The reaction of NO with Cl2 is an endothermic reaction.

The next slide shows the curve for a generic exothermic reaction.

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?

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Sketch a potential energy diagram for the decomposition of nitrous oxide.

N2O(g) N2(g) + O(g)

The activation energy for the forward reaction is 251 kJ; Ho = +167 kJ.

Label your diagram appropriately.

What is the reverse activation energy?

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Reactants

Products

Ea= 251 kJ

H = 167 kJ

Ea(reverse) = 84 kJ

Energy

per

mol

Progress of reaction

N2O

N2 + O

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Rate constants for most chemical reactions closely follow an equation in the form

This is called the Arrhenius equation. A is the frequency factor, a constant.

RT

E

e A ka

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A more useful form of the Arrhenius equation is

21

a

1

11 ln

TT

R

E

k

k2

?

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In a series of experiments on the decomposition of dinitrogen pentoxide, N2O5, rate constants were determined at two different temperatures:

• At 35°C, the rate constant was 1.4 × 10-4/s.

• At 45°C, the rate constant was 5.0 × 10-4/s.

What is the activation energy?

What is the value of the rate constant at 55°C?

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This is actually two problems.

First, we will use the Arrhenius equation to find Ea.

Then, we will use the Arrhenius equation with Ea to find the rate constant at a new temperature.

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T1 = 35°C = 308 K T2 = 45°C = 318 K

k1 = 1.4 × 10-4/s k2 = 5.0 × 10-4/s

21

a

1

11 ln

TT

R

E

k

k2

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We will solve the left side and rearrange the right side.

EK 10

K) K)(308 (318Kmol

J 8.3151.273

a

E J/mol 10 1.0 54a

K) K)(308 (318

KmolJ

8.315

K 10 1.273

a E

K 318

1

K 308

1

KmolJ

8.315/s10 1.4

/s10 5.0 ln a

4-

-4

E

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T1 = 35°C = 308 K T2 = 55°C = 328 K

k1 = 1.4 × 10-4/s k2 = ?

Ea = 1.04 × 105 J/mol

21

a

1

11 ln

TT

R

E

k

k2

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K 328K 308

K 20

KmolJ

8.315

molJ

10 1.04

/s10 1.4 ln

5

4-

2 k

k /s10 1.7 -32

K 328

1

K 308

1

KmolJ

8.315

molJ

10 1.04

/s10 1.4 ln

5

4-

2 k

2.476 /s10 1.4

ln4-

2

k

11.90 /s10 1.4 4-

2 k

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Reaction Mechanism

A balanced chemical equation is a description of the overall result of a chemical reaction. However, what actually happens on a molecular level may be more involved than what is represented by this single equation. For example, the reaction may take place in several steps. That set of steps is called the reaction mechanism.

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Each step in the reaction mechanism is called an elementary reaction and is a single molecular event.

The set of elementary reactions, which when added give the balanced chemical equation, is called the reaction mechanism.

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Because an elementary reaction is an actual molecular event, the rate of an elementary reaction is proportional to the concentration of each reactant molecule. This means we can write the rate law directly from an elementary reaction.

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The slowest step in the reaction mechanism is called the rate-determining step (RDS). The rate law for the RDS is the rate law for the overall reaction.

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Rate Law and Reaction Mechanism

A reaction mechanism cannot be directly observed.

We can, however, determine the rate law by experiment and decide if the reaction mechanism is consistent with that rate law.

The rate of a reaction is determined completely by the slowest step, the rate-determining step.

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For example, the reaction of NO2 with F2 is believed to occur in the following elementary steps:

NO2 + F2 NO2F + F (slow step)

F + NO2 NO2F (fast step)

The overall reaction is

2NO2 + F2 2NO2F

What is the rate law for this mechanism?

Rate = k[NO2][F2]

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A reaction intermediate is a species produced during a reaction that does not appear in the net equation because it reacts in a subsequent step in the mechanism.

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For example, we look again at the mechanism for the reaction of NO2 with F2:

NO2 + F2 NO2F + F (Slow step)

F + NO2 NO2F (Fast step)

The overall reaction is

2NO2 + F2 2NO2F

F does not appear in the overall reaction equation. It is produced in the first step and used in the second step. Thus F is a reaction intermediate.

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Catalysis

Catalysis is an increase in the rate of reaction that results from the addition of a catalyst.

Enzymes are biological catalysts.

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A catalyst is not consumed in a reaction. Rather, it is present in the beginning, is used in one step, and is produced again in a subsequent step. As a result, the catalyst does not appear in the overall reaction equation.

A catalyst increases the reaction rate by providing an alternative reaction path with a lower activation energy. When Ea is reduced, k increases exponentially.

This relationship is illustrated on the potential energy diagram for the decomposition of ozone.

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Consider the following reaction:

NO2(g) + CO(g) NO(g) + CO2(g)

At 500 K, the reaction mechanism is believed to involve two elementary reactions:

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

NO3(g) + CO(g) NO2(g) + CO2(g)The overall reaction equation is

NO2(g) + CO(g) NO(g) + CO2(g)

NO3 is a reaction intermediate: It is produced in the first step and used in the second step. There is no catalyst.

Rate = k[NO2][NO2] = k[NO2]2

(Slow step)

?

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Write the rate law for the first step in the mechanism for the decomposition of ozone.

Cl(g) + O3(g) ClO(g) + O2(g)

ClO(g) + O(g) Cl(g) + O2(g)

Rate = k[Cl][O3]

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Chlorofluorocarbons, such as CCl2F2, decompose in the stratosphere as a result of their irradiation with the short-wavelength ultraviolet light present at those altitudes. The decomposition yields chlorine atoms that catalyze the decomposition of ozone in the presence of oxygen atoms (present in the stratosphere) to give oxygen molecules.

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The mechanism of the decomposition is

Cl(g) + O3(g) ClO(g) + O2(g) (Slow step)

ClO(g) + O(g) Cl(g) + O2(g)

What is the overall reaction equation?

O3(g) + O(g) 2O2(g)

Is there a reaction intermediate? If so, what is it?Yes. ClO.

Is there a catalyst? If so, what is it?Yes. Cl.

What is the rate law for the overall reaction?

Rate = k[Cl][O3]

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The decomposition of hydrogen peroxide is catalyzed by iodide ion. The reaction mechanism is thought to be

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

At 25°C, the first step is slow relative to the second step. What is the rate law predicted by the reaction mechanism? Is there a reaction intermediate?

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Rate = k[H2O2][I-]

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

Overall reaction equation:2H2O2(aq) 2H2O(l) + O2(g)

I- was present in the beginning, used in the first step, and produced in the second step. It is a catalyst.

IO- is the intermediate.

IO- was produced in the first step and used in the second step. It is an intermediate.