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Chapter 14-Chemical Kinetics. Tour of the Ozone Hole http://www.atm.ch.cam.ac.uk/tour/. 16 September INTERNATIONAL DAY FOR THE PRESERVATION OF THE OZONE LAYER The Ozone Hole of 2008 is larger than in 2007

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slide1

Chapter 14-Chemical Kinetics

Tour of the Ozone Hole

http://www.atm.ch.cam.ac.uk/tour/

slide2

16 September INTERNATIONAL DAY FOR THE PRESERVATION OF THE OZONE LAYER

The Ozone Hole of 2008 is larger than in 2007

Geneva, 16 September 2008 (WMO) - “After decades of chemical attack, it may take another 50 years or so for the ozone layer to recover fully. As the Montreal Protocol has taught us, when we degrade our environment too far, nursing it back to health tends to be a long journey, not a quick fix”, said Ban Ki-moon, Secretary-General of the United Nations, on the occasion of the International Day for the Preservation of Ozone Layer today. According to the World Meteorological Organization (WMO), the 2008 Antarctic ozone hole will be larger than the one of 2007. The observed changes in the stratosphere could delay the expected recovery of the ozone layer. It is therefore vital that all Member States with stratospheric measurement programmes continue to support and enhance these measurements.

slide3

O3 concentration is reported in Dobson Units (DU)

1DU =2.69E15 O3 molecules/cm2

In September, 2006 the largest ozone hole ever was observed.

slide4

Temperature inversions and wind direction changes dramatically influence air quality in cities like Salt Lake…

slide5

Photo-chemical Smog

NO(g) + O3(g) → NO2(g) + O2(g)

slide6

In 1998, a total of 24.5 megatons of nitrogen oxide (NOx) compounds were released to the atmosphere.

NOx compound are produced primarily in the combustion of hydrocarbons

N2(g) + O2(g) → 2 NO(g) DH° = + 180.6 kJ

2 NO(g) + O2(g) → 2 NO2(g) DH° = - 114.2 kJ

NO2(g) + hn→ NO(g) + O(g)photochemical

O2(g) + O(g) → O3(g) Ozone production

slide8

Chemical Kinetics looks at both the chemical mechanism as well as the rate at which reactions occur.

slide9

Chemical Reactions usually involve molecular collisions that result in bonds being broken or made.

Both of these reactions are bimolecular…i.e. two molecules collide in order for the reaction to occur.

slide11

The Lewis structure of NO2 indicates that the nitrogen atom has an unpaired electron. Two NO2 molecules combine by using their unpaired electrons to form an

N-N bond.

slide12

Chemical Reaction Mechanism

In the reverse of the previous reaction, a unimolecular elementary reaction may involve bond breakage. If an N2O4 molecule possesses enough energy (from heat or light), molecular vibrations can break the N-N bond to produce two NO2 molecules.

When 3 molecules collide to form chemical product the reaction mechanism would be called ter-molecular.

slide13

The reaction between O3 and NO is believed to occur by a mechanism that consists of the single bimolecular step illustrated here in a molecular

view.

slide14

A chemical reaction rate is the change in concentration of a reactant or product during a particular time interval for the reaction

N2(g) + O2(g) → 2 NO(g)

reaction rate tutorial

»PC version

Learn to calculate the average and instantaneous rate from the rate expression and concentration vs. time data. Includes practice exercises.

Reaction Rate Tutorial
slide18

(a) To calculate the average rate of reaction, determine how many moles are consumed during the time interval and divide by the time:

  • n = (0.25 mol/L) – (0.50 mol/L) = – 0.25 mol/L;
  • Rate = M/s = 0.25 M/30. s = 8.33E-3
  • Round to two significant figures: Rate = 8.3E-3 M/s
  • NOTE UNITS of REACTION RATES
  • (b) The rate for any particular reagent is the coefficient for that reagent times the rate of reaction: Rate(NH3) = (2/3)(8.3-3 M/s) = 5.6 x 10-3 M/s
  • (c) The concentration of any particular reagent is its initial concentration minus the change during the time interval:
  • Change = (Coeff)(Rate)(time) =
  • –(1/3)(8.3E-3M/s)(30 s) = –8.3 x 10-2 M
  • Concentration = (1.25 M) – (0.083 M) = 1.17 M
slide19

Problem

An engineer is studying the rate of the Haber synthesis:

N2(g) + 3 H2(g) → 2 NH3(g)

Starting with a closed reactor containing 1.25 mol/L of N2 and 0.5 mol/L of H2, the engineer find that the H2 concentration has fallen to 0.25 M after 30 seconds.

(a) What is the average rate of reaction over this time.

(b) What is the ave. rate of production of NH3.

(c) What is the N2 concentration after 30s?

slide20

(a) What is the average rate of reaction over this time.

(b) What is the ave. rate of production of NH3.

slide22

2 NO2→ 2 NO + O2

A plot of [NO2], [O2], and [NO] as a function of time (seconds) for the decomposition reaction of NO2. The concentration data are shown in the table.

slide24

Instantaneous Rates

The rate at t=0 is the instantaneous initial rate. This is the most common way of reporting rates in the laboratory

slide26

Instantaneous Rates

Tropospheric ozone is rapidly consumed in many reactions, including the following. O3 + NO → NO2 + O2

Use the following data to calculate the instantaneous rate of the preceding reaction at t = 0.000 s and t = 0.102 s.

slide29

The rate of a chemical reaction increases with increasing concentration because the reactants (NO + O2) are more likely to collide.

It follows: reaction rate ~ [reactants]

Or that reaction rate = k[reactants]n

Note that the units of “k” (the rate constant) are different depending on n, the “order” of the reaction

slide31

14.48. Compounds A and B react to give a single product, C. Write the rate law for each of the following cases and determine the units of the rate constant by using the units M for concentration and s for time:

a. The reaction is first order in A and second order in B.

b. The reaction is first order in A and second order overall.

c. The reaction is independent of the concentration of A and second order overall.

d. The reaction is second order in both A and B.

slide33

14.53.Rate Laws for Destruction of Tropospheric Ozone

The reaction of NO2 with ozone produces NO3 in a second-order reaction overall:

NO2(g) + O3(g) → NO3(g) + O2(g)

a. Write the rate law for the reaction if the reaction is first order in each reactant.

b. The rate constant for the reaction is 1.93  104M–1s–1 at 298 K. What is the rate of the reaction when

[NO2] = 1.8  10–8M and [O3] = 1.4  10–7M?

c. What is the rate of the appearance of NO3 under these conditions?

d. What happens to the rate of the reaction if the concentration of O3(g) is doubled?

slide34

b. The rate constant for the reaction is 1.93  104M–1s–1 at 298 K. What is the rate of the reaction when

[NO2] = 1.8  10–8M and [O3] = 1.4  10–7M?

c. What is the rate of the appearance of NO3 under these conditions?

d. What happens to the rate of the reaction if the concentration of O3(g) is doubled?

slide37

The single experiment approach

For reactions such as: O3(g) → O2(g) + O(g)

The rate is found to be first order in ozone

rate = k[O3]

The mathematical solution can be found by integrating this expression.

What are the units for “k”?

slide40

Problem

The decomposition of N2O5 (g) is 1st Order.

(a) Write the rate equation.

(b) If 2.56 mg of N2O5 is present initially, and 2.50 mg remain after 4.26 min at 55 C. Calculate the rate constant, k, for this process.

slide42

N2O(g) → N2(g) + 1/2 O2(g)

Rate = k[N2O]

t½ = 1 s

slide44

Half-life Problem

The half-life for a first order reaction at 550°C is 85 seconds. How long would it take for 23% of the reactant to decompose?

slide45

At high temperatures, the reaction:

2 NO2(g) → 2NO(g) + O2(g)

is second-order in NO2, i.e.

Rate = k[NO2]2

The integrated rate law for a 2nd order reaction is:

[NO2]-1t = kt + [NO2]-10

What are the units for “k”?

slide47

14.64. Two structural isomers of ClO2 are shown:

The isomer with the Cl–O–O skeletal arrangement is unstable and rapidly decomposes according to the reaction

2ClOO(g) → Cl2(g) + 2O2(g).

The following data were collected for the decomposition of ClOO at 298 K:

Determine the rate law for the reaction and the value of the rate constant at 298 K. What is the half-life for the reaction?

slide48

Is the reaction 1st or 2nd order in Cl-O-O?

Plot of Concentration (M) versus Time

[COO]

Time (us)

slide51

Problem Chapter 14

The condensation reaction of butadiene has a rate constant of 0.93 L/mol۰min. If the initial concentration of C4H6 is 0.240 M, find:

(a) the time at which the concentration will be 0.100 M

(b) the concentration after 3.5 hr.

slide53

(a) The units of the rate constant indicate that this is a second-order reaction, so Rate = k[C4H6]2 and

  • 1/[A] – 1/[A]o = kt ;
  • The problem states that k = 0.93 M-1 min-1.
  • (a) kt = (10.0 – 4.17) M-1 = 5.83 M-1;
  • t = 6.3 min;
  • (b) 1/[A] = 1/0.24M + (0.93 M-1 min-1)(25 min) = (4.17 + 23.25) M-1
  • = 27.42 M-1
  • [A] = 3.6E-2 M.
reaction order tutorial

»PC version

Use interactive graphs to explore how reaction rate varies with concentration of reactant. Example problems outline the steps for determining the rate law and rate constant from concentration and initial rate data. Includes practice problems.

Reaction Order Tutorial
slide55

58. Hydroperoxyl radicals (HO2) react rapidly with ozone in this elementary reaction

HO2 + O3 → OH + 2 O2

Determine the pseudo-first-order rate constant and the second-order rate constant for this reaction from the data

What is the molecularity of the reaction?

slide58

Problem Solution

From the plot, k’ = 1.03E-2 ms-1

And the pseudo first-order rate equation is:

Rate = k’[HO2]

The overall rate must be”

rate = k[O3][HO2]

So that

k’ = k[O3]

And

k = k’/[O3] = 1.03E-2 ms-1/1.0E-3M

= 1.0E+1 ms-1M-1

slide59

Reaction Mechanism

Overall reaction: 2 NO2(g) → 2 NO(g) + O2(g)

slide60

Reaction mechanisms…elementary reactions

Step 1) NO2 + NO2 → NO + NO3

Step 2) NO3 →NO + O2

Overall) 2 NO2 → 2 NO + O2

NO3 is a reaction intermediate as it does not

appear in the overall reaction

slide61

Reaction mechanisms…rate determining steps

Step 1) NO2 + NO2 → NO + NO3(slow)

Step 2) NO3 →NO + O2 (fast)

Overall) 2 NO2 → 2 NO + O2

Rate (1) = k1[NO2]2

Rate (2) = k2[NO3]

The slowest step in the mechanism is usually

Rate Determining.

For this example, the mechanism predicts that

the experimentally observed rate will be second

order in NO2.

slide62

LINKING MECHANISMS AND RATE LAWS

1.  The mechanism is one or more elementary reactions that describes how the chemical reaction occurs.

These elementary reactions may be unimolecular, bimolecular, or (very rarely) termolecular.

2.  The sum of the individual steps in the mechanism must give the overall balanced chemical equation.

3. The reaction mechanism must be consistent with the experimental rate law.

If the rate law predicted by the proposed mechanism differs from the experimental rate law, the proposed mechanism is wrong.

If the rate law predicted by the proposed mechanism matches the experimental rate law, the proposed mechanism is a possible description of how the reaction proceeds, but must be verified by experiments.

reaction mechanisms tutorial

»PC version

Learn to calculate the rate expression of a multi-step reaction from its elementary steps by identifying the rate-determining step. Includes practice exercises.

Reaction Mechanisms Tutorial
slide64

Problem 70. The rate laws for the thermal and photochemical decomposition of NO2 are different. Which of the following mechanisms can be attributed to thermal versus photochemical decomposition. Given:

Rate (thermal) = k[NO2]2

Rate (photo) = k[NO2]

a. NO2 + NO2 → N2O4 slow

N2O4 → N2O3 + O fast

N2O3 + O → N2O2 + O2 fast

N2O2 → 2 NO fast

slide65

Problem 103. The rate laws for the thermal and photochemical decomposition of NO2 are different. Which of the following mechanisms can be attributed to thermal versus photochemical decomposition. Given:

Rate (thermal) = k[NO2]2

Rate (photo) = k[NO2]

a. Step 1: NO2(g) → NO(g) + O(g) slow

Step 2: O(g) + NO2(g) + O2(g) fast

b.

c.

slide66

Problem 69b

step1 NO2 + NO2 → N2O4 fast

step2 N2O4 → NO + NO3slow

step3 NO3 → NO + O2 fast

For this case, the second step is Rate Determining, so that the overall rate would be:

rate = k2[N2O4]

However, since N2O4 is a reaction intermediate and not generally observed, we must substitute NO2 for it. We can do this by assuming that step1 is both fast and reversible. Then the rate forward for step1 is:

rate forward = k1[NO2]2

And the rate of the reverse reaction is:

rate reverse = k-1[N2O4]

Note that the forward and reverse reactions have different rate constants.

slide67

step1 NO2 + NO2 → N2O4 fast

step2 N2O4 → NO + NO3slow

step3 NO3 → NO + O2 fast

If it is then assumed that both the forward and back reactions quickly reach equilibrium, i.e. we set the rate forward equal to the rate reverse,

rate forward= k1[NO2]2 = k-1[N2O4] = rate reverse

Then we can algebraically solve for the concentration of the intermediate…N2O4

[N2O4] = k1/k-1[NO2]2

And substitute this value into the rate-determining step2

rate = k2[N2O4]

= k2(k1/k-1)[NO2]2

= kobserved[NO2]2

Now the mechanism and rate law are consistent with thermal decomposition.

slide68

Reaction Rates and Temperature

The rate constant, k, depend on temperature as:

k = Ae-(Ea/RT)

Where the parameter “A” is a frequency factor, and Ea is the activation energy for the reaction.

The frequency factor is the number of collisions per sec times the probability that a collision has an effective orientation

slide69

The Arrhenius Equation

k = Ae-(Ea/RT)

Was developed from the observed relationship between rate (k) and temperature.

The theory postulates an Activation Energy (Ea) which is the energy barrier that must be overcome before two molecules can react.

Svante August Arrhenius

slide70

Activation energy

Activated complex

R*

R

P

slide71

Reaction Rates and Temperature. Effect of molecular orientation: O3 + NO → NO2 + O2

slide72

For reactions that proceed via more than one step, there may be more than one Ea linked to forming the activated complex

slide73

Reaction Mechanisms

2 NO2(g) → 2 NO(g) + O2(g)

The elementary reaction step with the larger activation energy will be rate determining.

slide74

NO + O3→ NO2 + O2DH < 0

For the reverse reaction:

NO2 + O2 → NO + O3DH > 0

Which process has the larger rate constant, k?

Exothermic

Endothermic

slide75

To determine Ea, the rate of reaction must be measured over several temperatures.

Ln(k) = -Ea/(RT) + lnA

slide77

Ammonium cyanate (NH4NCO) decomposes to urea(NH2CONH2). The following data were obtained at 50°C.

At 25°C the concentration falls from 0.500 M to 0.300 M in 6.0 hrs.

A) Determine the rate Law

B) Determine the rate constant at 50°C.

C) Determine the activation energy.

slide80

Problem

Values of the rate constant of the reaction:

N2 + O2→ 2NO

Are as follows:

A) Calculate Ea

B) Calculate A (freq. factor)

c) Calculate k at T=300K

arrhenius equation tutorial

»PC version

This tutorial explores how energy, rate constant, and the effects of temperature and orientation are related. Includes practice exercises.

Arrhenius Equation Tutorial
collision theory tutorial

»PC version

Explore the effects of temperature, orientation of reactant molecules, and catalysts on reaction rates. Includes practice exercises.

Collision Theory Tutorial
slide85

Rate of reaction versus reaction “spontaneity”.

The rate of an endergonic reaction (DG<0) may be greater than that for an exergonic reaction (DG<0)

slide86

Catalysts increase the rate of reaction by lowering the activation energy. Note: a catalyst will not change DE (or DG) for the reaction.

2O3(g) ↔ 3O2(g)

slide88

Catalytic converters reduce emissions of NO (and CO) by lowering the activation energy for decomposition to N2 and O2. That is adsorption of NO on the Pt/Pl surface weakens the N-O bond.

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This concludes the Norton Media Libraryslide set for chapter 14ChemistryThe Science in Context byThomas Gilbert,Rein V. Kirss, &Geoffrey DaviesW. W. Norton & CompanyIndependent and Employee-Owned