15 Chemical Kinetics 1

15 Chemical Kinetics

Chemical kinetics: the study of reaction rate, a quantity conditions affecting it,

the molecular events during a chemical reaction (mechanism), and presence of other components (catalysis).

Factors affecting reaction rate:Concentrations of reactantsCatalystTemperatureSurface area of solid reactants or catalyst

What quantities do we study regarding chemical reactions?

15 Chemical Kinetics 2

Reaction Rate Defined

Reaction rate: changes in a concentration of a product or a reactant per unit time.

[ ] concentrationReaction rate = ——

t

[ ]

t

[ ]

t

change

Define reaction rate and explainAverage reaction rateInstantaneous reaction rate(2 tangents shown)Initial reaction rate

15 Chemical Kinetics 3

Expressing reaction ratesFor a chemical reaction, there are many ways to express the reaction rate. The relationships among expressions depend on the equation.

Note the expression and reasons for their relations for the reaction

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

[O2] 1 [NO] 1 [NO2] Reaction rate = – ——— = – — ———— = — ———

t 2 t 2 t

Make sure you can write expressions for any reaction and figure out the relationships. For example, give the reaction rate expressions for

2 N2O5 = 4 NO2 + O2

How can the rate expression be unique and universal?

15 Chemical Kinetics 4

Calculating reaction rateThe concentrations of N2O5 are 1.24e-2 and 0.93e-2 M at 600 and 1200 s after the reactants are mixed at the appropriate temperature. Evaluate the reaction rates for

2 N2O5 = 4 NO2 + O2

Solution: (0.93 – 1.24)e-2 – 0.31e-2 MDecomposition rate of N2O5 = – ———————— = – ——————

1200 – 600 600 s

= 5.2e-6 M s-1.

Note however, rate of formation of NO2 = 1.02e-5 M s-1.rate of formation of O2 = 2.6e-6 M s-1.

The reaction rates are expressed in 3 forms

Be able to do this type problems

15 Chemical Kinetics 5

Determine Reaction RatesTo measure reaction rate, we measure the concentration of either a reactant or product at several time intervals.

The concentrations are measured using spectroscopic method or pressure (for a gas). For example, the total pressure increases for the reaction:

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

Because 5 moles of gas products are produced from 2 moles of gas reactants. For the reaction

CaCO3 (s) CaO(s) + CO2 (g)

The increase in gas pressure is entirely due to CO2 formed.

barometer

15 Chemical Kinetics 6

Differential Rate LawsDependence of reaction rate on the concentrations of reactants is called the rate law, which is unique for each reaction.

For a general reaction, a A + b B + c C products

the rate law has the general form

order wrt A, B, and C, determined experimentally reaction rate = k [A]X [B]Y [C]Z

the rate constant

For example, the rate law is

rate = k [Br-] [BrO3-] [H+] for

5 Br- + BrO3- + 6 H+ 3Br2 + 3 H2O

The reaction is 1st order wrt all three reactants, total order 3.

Use differentials to express rates

15 Chemical Kinetics 7

Variation of Reaction rates and Order

First order, rate = k [A]

k = rate, 0th order

[A]

rate

2nd order, rate = k [A]2

The variation of reaction rates as functions of concentration for various order is interesting.

Mathematical analysis is an important scientific tool, worth noticing.

[A] = ___?

15 Chemical Kinetics 8

Differential Rate Law determinationEstimate the orders and rate constant k from the results observed for the reaction? What is the rate when [H2O2] = [I-] = [H+] = 1.0 M?

H2O2 + 3 I- + 2 H+ I3- + 2 H2O

Exprmt [H2O2] [I-] [H+] Initial rate M s-1

1 0.010 0.010 0.0050 1.15e-6 2 0.020 0.010 0.0050 2.30e-6 3 0.010 0.020 0.0050 2.30e-6

4 0.010 0.010 0.0100 1.15e-6

Solution next

Learn the strategy to determine the rate law from this example.

Figure out the answer without writing down anything.

15 Chemical Kinetics 9

Differential Rate Law determination - continue

Estimate the orders from the results observed for the reaction

H2O2 + 3 I- + 2 H+ I3- + 2 H2O

Exprmt [H2O2] [I-] [H+] Initial rate M s-1

1 0.010 0.010 0.0050 1.15e-6 2 0.020 0.010 0.0050 2.30e-6 1 for H2O2 3 0.010 0.020 0.0050 2.30e-6 1 for I-

4 0.010 0.010 0.0100 1.15e-6 0 for H+

1.15e-6 = k [H2O2]x [I-]y [H+]z

1.15e-6 k (0.010)x(0.010)y(0.0050)z exprmt 1 1 1 ----------- = ------------------------------------- ---- = --- 2.30e-6 k (0.020)x(0.010)y(0.0050)z exprmt 2 2 2

x = 1

( )x

15 Chemical Kinetics 10

Differential Rate Law determination - continue

Other orders are determined in a similar way as shown before.Now, lets find k and the rate

Thurs, rate = 1.15e-6 = k (0.010)(0.010) from exprmt 1

k = 1.15e-6 M s-1 / (0.010)(0.010) M3 = 0.0115 M-1 s-1

And the rate law is therefore,

– d [H2O2] k rate = ————— = 0.0115 [H2O2] [I-] a differential rate law

d t total order 2

The rate when [H2O2] = [I-] = [H+] = 1.0 M:

The rate is the same as the rate constant k, when concentrations of reactants are all unity (exactly 1), doesn’t matter what the orders are.

15 Chemical Kinetics 11

Differential Rate Law determination – continue

The reaction rate –d[H2O2]/dt = 0.0115 [H2O2] [I–], forH2O2 + 3 I- + 2 H+ I3

- + 2 H2OWhat is – d[I–]/dt when [H2O2] = [I–] = 0.5?

Solution: Please note the stoichiometry of equation and how the rate changes.– d[I–]/dt = – 3 d[H2O2]/dt = 3* 0.0115 [H2O2] [I–]

= 0.0345 * 0.5 * 0.5 = 0.0086 M s-1

In order to get a unique rate constant k, we evaluate k for the reaction

a A + b B product this way

rate = -1/a d[A]/dt = -1/b d[B]/dt = k [A]x [B]y

Note the reaction rate expression and the stoichiometry of equation.

15 Chemical Kinetics 12

Differential Rate Law determination - continueFrom the following reaction rates observed in 4 experiments, derive the rate law for the reaction A + B + C products where reaction rates are measured as soon as the reactants are mixed.

Expt 1 2 3 4 [A]o 0.100 0.200 0.200 0.100[B]o 0.100 0.100 0.300 0.100 [C]o 0.100 0.100 0.100 0.400 rate 0.100 0.800 7.200 0.400

Solution nextThis example illustrates the strategy to determine, and a reliable method to solve rate-law experimentally.

15 Chemical Kinetics 13

Differential Rate Law determination - continueFrom the following reaction rates, derive the rate law for the reaction A + B + C products where reaction rates are measured as soon as the reactants are mixed.

Expt 1 2 3 4 order[A]o 0.100 0.200 0.200 0.100 3 from expt 1 & 2

[B]o 0.100 0.100 0.300 0.100 2 expt 1, 2 & 3 [C]o 0.100 0.100 0.100 0.400 1 expt 1 & 4 rate 0.100 0.800 7.200 0.400

0.800 k 0.2x 0.1y 0.1z

----- = ----------------------0.100 k 0.1x 0.1y 0.1z

Therefore 8 = 2x

log 8 = x log 2x = log 8 / log 2 = 3

Assume rate = k [A]x[B]y[C]z

15 Chemical Kinetics 14

Integrated Rate Laws concentrations as functions of time

One reactant A decomposes in 1st or 2nd order rate law.

Differential rate law Integrated rate law

– d[A] / dt = k [A] = [A]o – k t

d[A] – —— = k [A] [A] = [A]o e – k t or ln [A] = ln [A]o – k t

d t

d[A] 1 1 [A] conc at t – —— = k [A]2 —— – —— = k t

d t [A] [A]o [A]o conc at t = 0

Describe, derive and apply the integrated rate laws

Learn the strategy to determine rate-law

15 Chemical Kinetics 15

Concentration and time of 1st order reaction

Describe the features of plot of [A] vs. t and ln[A] vs. t for 1st order reactions. Apply the technique to evaluate k or [A] at various times.

[A]

t

ln[A]

t

[A] = [A]o e – k t ln [A] = ln [A]o – k t

t½

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Half life & k of First Order Decomposition

The time required for half of A to decompose is called half life t1/2.

Since [A] = [A]o e – k t or ln [A] = ln [A]o – k t

When t = t1/2, [A] = ½ [A]o

Thus ln ½ [A]o = ln [A]o – k t1/2

– ln 2 = – k t1/2

k t1/2 = ln 2 = 0.693 relationship between k and t1/2

Radioactive decay usually follow 1st order kinetics, and half life of an isotope is used to indicate its stability.

Evaluate t½ from k or k from t½

15 Chemical Kinetics 17

1st order reaction calculationN2O5 decomposes according to 1st order kinetics, and 10% of it decomposed in 30 s. Estimate k, t½ and percent decomposed in 500 s.

Solution next

If the rate-law is known, what are the key parameters?

15 Chemical Kinetics 18

1st order reaction calculationN2O5 decomposes according to 1st order kinetics, and 10% of it decomposed in 30 s. Estimate k, t½ and percent decomposed in 500 s.

Solution: Assume [A]o = [N2O5]o = 1.0, then [A] = 0.9 at t = 30 s or 0.9 = 1.0 e – k t apply [A]o = [A] e– k t ln 0.9 = ln 1.0 – k 30 s– 0.1054 = 0 – k * 30 k = 0.00351 s – 1

t½ = 0.693 / k = 197 s apply k t ½ = ln 2

[A] = 1.0 e – 0.00351*500 = 0.173Percent decomposed: 1.0 – 0.173 = 0.827 or 82.7 %

After 2 t½ (2*197=394 s), [A] = (½)2 =¼, 75% decomposed.After 3 t½ (3*197=591 s), [A] = (½)3 =1/8, 87.5% decomposed.

Apply integrated rate law to solve problems

15 Chemical Kinetics 19

Typical Problem wrt 1st Order Reaction

The decomposition of A is first order, and [A] is monitored. The following data are recorded:

t / min 0 2 4 8[A]/[M] 0.100 0.0905 0.0819 0.0670

Calculate k (What is the rate constant? k = 0.0499)Calculate the half life (What is the half life? Half life = 13.89)Calculate [A] when t = 5 min. (What is the concentration when t = 5 min?)Calculate t when [A] = 0.0100. (Estimate the time required for 90% of A to decompose.)

Work out all the answers

15 Chemical Kinetics 20

A 2nd Order ExampleDimerization of butadiene is second order:

2 C4H6(g) = C8H12(g).

The rate constant k at some temperature is 0.100 /min. The initial concentration of butadiene [B] is 2.0 M.

Calculate the time required for [B] = 1.0 and 0.5 M

Calculate concentration of butadiene when t = 1, 5, 10, and 30.

Solution next

If the rate-law is known, what are the key parameters?

Apply the right model and work out all the parameters.

15 Chemical Kinetics 21

A 2nd Order ExampleDimerization of butadiene is second order:

2 C4H6(g) = C8H12(g).

The rate constant k at some temperature is 0.100 /min. The initial concentration of butadiene [B] is 2.0 M.

Calculate the time t required for [B] = 1.0 and 0.5 M

Calculate concentration of butadiene when t = 1, 5, 10, and 30.

1 1—— – —— = k t[B] [B]o

1 1—— – ——[B] [B]o

t = ————————k

[B]o

[B] = ——————[B]o k t + 1t = 1 5 10 15 30 35

[B] = 1.67 1.0 0.67 0.50 0.29 0.25

Work out the formulas and then evaluate values

15 Chemical Kinetics 22

Half life of 2nd Order Chemical Kinetics

t = 1 5 10 15 30 35

[B] = 1.67 1.0 0.67 0.50 0.29 0.25

0

0.5

1

1.5

2

2.5

3

3.5

4

4.5

1/[B]

How does half life vary in 2nd order reactions?

15 Chemical Kinetics 23

Plot of [B] vs. t & 1/[B] vs. t for 2nd Order Reactions

t = 1 5 10 15 30 35

[B] = 1.67 1.0 0.67 0.50 0.29 0.25

[B] 1— [B]

t t

1 1—— – —— = k t[B] [B]o

[B]o

[B] = ——————[B]o k t + 1

What kind of plot is linear for 1st and 2nd reactions?

15 Chemical Kinetics 24

Chemical Reaction and Molecular CollisionMolecular collisions lead to chemical reactions. Thus, the reaction constant, k is determined by several factors.

k = Z f p

Z: collision frequencyp, the fraction with proper orientationf, fraction of collision having sufficient energy for reactionf is related to the potential energy barrier called activation energy, Ea.

f e – Ea / RT or exp (– Ea / R T)

Thus, k = A e – Ea / RT

constant

Ea

Potential energy

reaction

constant

How does temperature affect reaction rates?

Explain energy aspect in a chemical reaction

15 Chemical Kinetics 25

Energy in chemical reactions

Potential energy

Progress of reaction

R + A

P + DH exothermic

Ea Ea for reverse reaction

Endothermic rxn

RA-PDactivated complex

Explain the various terms and energy changes in a chemical reaction

15 Chemical Kinetics 26

The Arrhenius Equation

1903 Nobel Prize citation” …in recognition of the extraordinary services he has rendered to the advancement of chemistry by his electrolytic theory of dissociation”

The temperature dependence of the rate constant k is best described by the Arrhenius equation:

k = A e – Ea / R T

or ln k = ln A – Ea / R T

If k1 and k2 are the rate constants at T1 and T2 respectively, then

k1 Ea 1 1 ln —— = – — — – — k2 R T1 T2

How does temperature affect reaction rates?

Derive and apply these relationship to solve problems, and recall the Clausius-Clapeyron equation.

15 Chemical Kinetics 27

Application of Arrhenius EquationFrom k = A e – Ea / R T, calculate A, Ea, k at a specific temperature and T.

The reaction:

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

The rate constant k = 1.0e-10 s-1 at 300 K and the activation energy Ea = 111 kJ mol-1. What are A, k at 273 K and T when k = 1e-11?

Method: derive various versions of the same formula

k = A e – Ea / R T

A = k e Ea / R T

A / k = e Ea / R T

ln (A / k) = Ea / R T

Make sure you know how to transform the formula into these forms.

15 Chemical Kinetics 28

Application of Arrhenius Equation (continue)

The reaction:

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

The rate constant k = 1.0e-10 s-1 at 300 K and the activation energy Ea = 111 kJ mol-1. What are A, k at 273 K and T when k = 1e-11?

Use the formula derived earlier:

A = k eEa / R T = 1e-10 s-1 exp (111000 J mol-1 / (8.314 J mol-1 K –1*300 K)) = 2.13e9 s-1

k = 2.13e9 s-1 exp (– 111000 J mol-1) / (8.314 J mol-1 K –1*273 K) = 1.23e-12 s-1

T = Ea / [R* ln (A/k)] = 111000 J mol-1 / (8.314*46.8) J mol-1 K-1 = 285 K

15 Chemical Kinetics 29

The Effect of Temperature on Reaction Rates

Reaction rate = k [A}x[B]y[C]z (Concentration effect at constant T)

k = A exp ( – Ea / RT) (Temperature effect)

Use graphic method to discuss the variation of k vs. Tvariation of k vs 1 / Tvariation of ln(k) vs Tvariation of ln(k) vs 1 / T

See a potential multiple choice question in an exam?

15 Chemical Kinetics 30

Elementary Reactions and MechanismElementary reactions are steps of molecular events showing how reactions proceed. This type of description is a mechanism.

The mechanism for the reaction between CO and NO2 is proposed to be

Step 1 NO2 + NO2 NO3 + NO (an elementary reaction)

Step 2 NO3 + CO NO2 + CO2 (an elementary reaction)

Add these two equations led to the overall reaction

NO2 + CO = NO + CO2 (overall reaction)

A mechanism is a proposal to explain the rate law, and it has to satisfy the rate law. A satisfactory explanation is not a proof.

Explain terms in red

15 Chemical Kinetics 31

Molecularity of Elementary ReactionsThe total order of rate law in an elementary reaction is molecularity.

The rate law of elementary reaction is derived from the equation. The order is the number of reacting molecules because they must collide to react.

A molecule decomposes by itself is a unimolecular reaction (step);two molecules collide and react is a bimolecular reaction (step); &three molecules collide and react is a termolecular reaction (step).

O3 O2 + O rate = k [O3]

NO2 + NO2 NO3 + NO rate = k [NO2]2

Br + Br + Ar Br2 + Ar* rate = k [Br]2[Ar]

Caution: Derive rate laws this way only for elementary reactions.

15 Chemical Kinetics 32

Molecularity of elementary reactions - Example

Some elementary reactions for the reaction between CH4 and Cl2 are

Cl2 2 Cl

2 Cl Cl2

2Cl + CH4 Cl2 + CH4*

Cl + CH4 HCl + CH3

CH3 + Cl CH3Cl

CH3 + CH3 CH3-CH3

CH3Cl + Cl HCl + CH2Cl

CH2Cl + Cl CH2Cl2

* * * (more)

Write down the rate laws and describe them as uni- bi- or ter-molecular steps yourself, please.

15 Chemical Kinetics 33

Elementary Reactions are Molecular Events

N2O5 NO2 + NO3

NO + O2 + NO2 NO2 + NO3

Explain differences between elementary and over reaction equations

15 Chemical Kinetics 34

Rate Laws and MechanismsA mechanism is a collection of elementary steps devise to explain the the reaction in view of the observed rate law. You need the skill to derive a rate law from a mechanism, but proposing a mechanism is task after you have learned more chemistry

For the reaction, 2 NO2 (g) + F2 (g) 2 NO2F (g), the rate law is,

rate = k [NO2] [F2] .

Can the elementary reaction be the same as the overall reaction?

If they were the same the rate law would have been

rate = k [NO2]2 [F2],

Therefore, they the overall reaction is not an elementary reaction. Its mechanism is proposed next.

15 Chemical Kinetics 35

Rate-determining Step in a MechanismThe rate determining step is the slowest elementary step in a mechanism, and the rate law for this step is the rate law for the overall reaction.

The (determined) rate law is, rate = k [NO2] [F2],for the reaction, 2 NO2 (g) + F2 (g) 2 NO2F (g), and a two-step mechanism is proposed:

i NO2 (g) + F2 (g) NO2F (g) + F (g)ii NO2 (g) + F (g) NO2F (g)

Which is the rate determining step?

Answer:The rate for step i is rate = k [NO2] [F2], which is the rate law, this suggests that step i is the rate-determining or the s-l-o-w step.

Explain rate determining step in a mechanism and use it to derive the rate law.

15 Chemical Kinetics 36

Deriving a Rate Law From a Mechanism - 0

The decomposition of H2O2 in the presence of I– follow this mechanism,

i H2O2 + I– k1 H2O + IO– slow ii H2O2 + IO– k2 H2O + O2 + I– fast

What is the rate law? Energy

Solve the problem

Eai

Eaii

reaction

15 Chemical Kinetics 37

Deriving a rate law from a mechanism - 1

The decomposition of H2O2 in the presence of I– follow this mechanism,

i H2O2 + I– k1 H2O + IO– slow ii H2O2 + IO– k2 H2O + O2 + I– fast

What is the rate law?

Solution

The slow step determines the rate, and the rate law is:

rate = k1 [H2O2] [I –]

Since both [H2O2] and [I –] are measurable in the system, this is the rate law.

15 Chemical Kinetics 38

Deriving a rate law from a mechanism - 2Derive the rate law for the reaction, H2 + Br2 = 2 HBr, from the proposed mechanism:

i Br2 2 Br fast equilibrium (k1, k-1)ii H2 + Br k2 HBr + H slow

iii H + Br k3 HBr fast

Solution:The fast equilibrium condition simply says that

k1 [Br2] = k-1 [Br]2

and [Br] = (k1/k-1 [Br2])½

The slow step determines the rate law,rate = k2 [H2] [Br] Br is an intermediate = k2 [H2] (k1/k-1 [Br2])½

= k [H2] [Br2] ½; k = k2 (k1/k-1)½ M-½ s -1

total order 1.5

explain

15 Chemical Kinetics 39

Deriving a rate law from a mechanism - 3The decomposition of N2O5 follows the mechanism:

1 N2O5 NO2 + NO3 fast equilibrium 2 NO2 + NO3 —k2 NO + O2 + NO2 slow3 NO3 + NO —k3 NO2 + NO2 fast

Derive the rate law.

Solution:

The slow step determines the rate,rate = k2 [NO2] [NO3] NO2 & NO3 are intermediate

From 1, we have [NO2] [NO3] —————— = K K, equilibrium constant [N2O5] K differ from k

Thus, rate = K k2 [N2O5]

15 Chemical Kinetics 40

Deriving rate laws from mechanisms – steady-state approximation

The steady-state approximation is a general method for deriving rate laws when the relative speed cannot be identified.It is based on the assumption that the concentration of the intermediate is constant.

Rate of producing the intermediate, Rprod, is the same as its rate of consumption, Rcons.

Rprod = Rcons

[Intermediate]

time

Rprod < RconsRprod > Rcons

Be able to apply the steady-state approximation to derive rate laws

15 Chemical Kinetics 41

Steady-state approximation - 2

Let’s assume the mechanism for the reaction. H2 + I2 2 HI

as follows.Step (1) I2 —k1 2 IStep (1) 2 I —k-1 I2

Step (2) H2 + 2 I —k2 2 HIDerive the rate law.Derivation:

rate = k2 [H2] [I] 2 (‘cause this step gives products)but I is an intermediate, this is not a rate law yet.Since k1 [I2] (= rate of producing I)

= k-1 [I]2 + k2 [H2] [I]2 (= rate of consuming I)

Thus, k1 [I2] [I]2 = ——————

k-1 + k2 [H2]

rate = k1 k2 [H2] [I2] / {k-1 + k2 [H2] }

Steady state

15 Chemical Kinetics 42

Steady-state approximation - 3

From the previous result:

k1 k2 [H2] [I2]rate = ———————

{k-1 + k2 [H2] }

Discussion:

(i) If k-1 << k2 [H2] then {k-1 + k2 [H2]} = k2 [H2] ,then rate = k1 k2 [H2] [I2] / {k2 [H2] } = k1 [I2] (pseudo 1st order wrt I2)using large concentration of H2 or step 2 is fast (will meet this condition).

(ii) If step (2) is slow, then k2 << k1, and if [H2] is not large, we have {k-1 + k2 [H2]} = k-1 and rate = k1 k2 [H2] [I2] / k1 = k2 [H2] [I2]

15 Chemical Kinetics 43

Steady-state approximation - 4In an alkaline solution, peroxydisulfate oxidizes sulfite to yield sulfate, S2O8

2- + SO32- + 2 OH- 3 SO4

2- + H2O.The following mechanism has been proposed:

i S2O82- + SO3

2- —k1 S2O72- + SO4

2- ii S2O7

2- + H2O —k2 2 SO42- + 2 H+

iii H+ + OH- —k3 H2O (fast equilibrium to be discussed)

Steady-state approximation follows these steps:

What is or are the intermediates I?Use which step to give the rate law that may involve [I]? Express the rates of producing and consuming intermediate(s)Express [I] of intermediate(s) in terms of [Reactants] Derive the rate law in terms of [Reactants]

Discuss See page 607 PHH Text

15 Chemical Kinetics 44

Catalysis

A catalyst is a substance that changes the rate of a reaction by lowing the activation energy, Ea. It participates a reaction in forming an intermediate, but is regenerated.

Enzymes are marvelously selective catalysts.

A catalyzed reaction,NO (catalyst)

2 SO2 (g) + O2 — 2 SO3 (g)via the mechanism i 2 NO + O2 2 NO2 (3rd order) ii NO2 + SO2 SO3 + NO

Uncatalyzed rxn

Catalyzed rxn

rxn

Energy

15 Chemical Kinetics 45

Catalyzed decomposition of ozoneR.J. Plunkett in DuPont discovered carbon fluorine chlorine compounds.

The CFC decomposes in the atmosphere:CFCl3 CFCl2 + ClCF2Cl3 CF2Cl + Cl.

The Cl catalyzes the reaction via the mechanism: i O3 + h v O + O2, ii ClO + O Cl + O2

iii O + O3 O2 + O2.The net result or reaction is

2 O3 3 O2

Scientists sound the alarm, and the CFC is banned now.

15 Chemical Kinetics 46

Homogenous vs. heterogeneous catalysts

A catalyst in the same phase (gases and solutions) as the reactants is a homogeneous catalyst. It effective, but recovery is difficult.

When the catalyst is in a different phase than reactants (and products), the process involve heterogeneous catalysis. Chemisorption, absorption, and adsorption cause reactions to take place via different pathways.

Platinum is often used to catalyze hydrogenation

Catalytic converters reduce CO and NO emission.

15 Chemical Kinetics 47

Heterogeneous catalystsCeryx's vision is to design, produce, and commercialize advanced systems that balance Cost, Performance, Emissions Reduction, and Fuel Penalty to make the economics of pollution control viable.

We explore new ways to look at the air quality challenges faced by industry and search for potential solutions by combining proven technologies with state-of-the-art science.

Catalyzed reactions:

CO + O2 CO2

2 NO N2 + O2

15 Chemical Kinetics 48

15 Chemical Kinetics 49

Enzymes – selective catalysts

Enzymes are a long protein molecules that fold into balls. They often have a metal coordinated to the O and N sites.

Molecules catalyzed by enzymes are called substrates. They are held by various sites (together called the active site) of the enzyme molecules and just before and during the reaction. After having reacted, the products P1 & P2 are released.

Enzyme + Substrate ES (activated complex)

ES P1 + P2 + E

Enzymes are biological catalysts for biological systems.

15 Chemical Kinetics 50

                                                                                      

      

X-ray 3-D structure of fumarate reductase. It reduces fumerate, an important role in the metabolism of anaerobic bacteria, from Max Planck Inst.

15 Chemical Kinetics 51

Chemical Kinetics - SummaryExplain how the various factors affect reaction rates.

Define reaction rates, average rates, initial rates and rate constants.Evaluate rate law from experiments

Properly apply 1st and 2nd differential rate laws and integrated rate laws.

Interpret elementary reactions and mechanisms. Derive rate laws from a given mechanism. Apply the steady-state method to derive the rate law of a given mechanism, and discuss the results.

Explain the action of catalysts in terms of chemistry and in terms of energy of activation.