© 2008 Brooks/Cole 1 © 2008 Brooks/Cole 2

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© 2008 Brooks/Cole 1

Chapter 13: Chemical Kinetics: Rates of Reactions


Chemical Kinetics

“The study of speeds of reactions and the nanoscale pathways or rearrangements by which atoms and molecules are transformed to products”


Reaction Rate


Reaction Rate Combustion of Fe(s) powder:


Reaction Rate

Change in [reactant] or [product] per unit time.

rate = = change in concentration of Cv+

elapsed time Δ [Cv+ ] Δt

Cresol violet (Cv+; a dye) decomposes in NaOH(aq):

Cv+(aq) + OH-(aq) → CvOH(aq)


Time, t [Cv+] Average rate (s) (mol / L) (mol L-1 s-1)

0.0 5.000 x 10-5

10.0 3.680 x 10-5 20.0 2.710 x 10-5

30.0 1.990 x 10-5 40.0 1.460 x 10-5

50.0 1.078 x 10-5

60.0 0.793 x 10-5 80.0 0.429 x 10-5

100.0 0.232 x 10-5

13.2 x 10-7

9.70 x 10-7

7.20 x 10-7

5.30 x 10-7

3.82 x 10-7

2.85 x 10-7

1.82 x 10-7

0.99 x 10-7

Average rate of the Cv+ reaction can be calculated:

Reaction Rate

2


Reaction Rates and Stoichiometry

Stoichiometry: Loss of 1 Cv+ → Gain of 1 CvOH

Rate of Cv+ loss = Rate of CvOH gain

Another example: 2 N2O5(g) 4 NO2(g) + O2(g)

Negative rate Positive rate

Rate of loss of N2O5 divided by -2, equals rate of gain of O2

Cv+(aq) + OH-(aq) → CvOH(aq)


For any general reaction: a A + b B c C + d D

The overall rate of reaction is:

Reactants decrease with time. Negative sign.

Products increase with time. Positive sign


Rate = = = = − 1 a Δ[A] Δt − 1

b Δ[B] Δt + 1

c Δ[C] Δt

+ 1 d Δ[D] Δt



H2 (g) + I2 (g) → 2 HI (g) the rate of loss of I2 is 0.0040 mol L-1 s-1. What is the rate of formation of HI ?

Rate = = = − Δ[H2] Δt − Δ[I2]

Δt + 1 2 Δ[HI] Δt

Rate = − = − (-0.0040) = Δ[H2] Δt

+ 1 2 Δ[HI] Δt

For:

Δ[HI] Δt

So = +0.0080 mol L-1 s-1


Average Rate and Instantaneous Rate

5.0E-5

4.0E-5

3.0E-5

2.0E-5

1.0E-5

0

[Cv+

] (m

ol/L

)

0 20 40 60 80 100 t (s)

Graphical view of Cv+ reaction:


Average Rate and Instantaneous Rate


Rate may change when [reactant] changes.

t [Cv+] Rate of Cv+ Rate/[Cv+] (s) (M) loss (M / s) (s-1)

0 5.00 x 10-5 1.54 x 10-6 0.0308

80 4.29 x 10-6 1.32 x 10-7 0.0308

•  Cv+ example shows this. •  For Cv+ the rate is proportional to concentration.

Effect of Concentration on Reaction Rate

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Rate Law and Order of Reaction A general reaction will usually have a rate law:

rate = k [A]m [B]n . . .

The orders are usually integers (-2, -1, 0, 1, 2…), but may also be fractions (½, ⅓…)


Determining Rate Laws from Initial Rates


Data for the reaction of methyl acetate with base:

CH3COOCH3 + OH- CH3COO- + CH3OH

Rate law: rate = k [CH3COOCH3]m [OH-]n



Dividing the first two data sets: 4.5 x 10-4 M/s = k (0.040 M)m(0.080 M)n

2.2 x 10-4 M/s = k (0.040 M)m(0.040 M)n

1 raised to any power = 1



Use experiments 2 & 3 to find m:

9.0 x 10-4 M/s = k (0.080 M)m(0.080 M)n

4.5 x 10-4 M/s = k (0.040 M)m(0.080 M)n




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If a rate law is known, k can be determined:

2.2 x 10-4 M/s (0.040 M)(0.040 M) k =

k = 0.1375 M-1 s-1 = 0.1375 L mol-1 s-1

rate [CH3COOCH3 ][OH-]

k =

Could repeat for each run, take an average… But a graphical method is better.

Using run 1:



The Integrated Rate Law

rate = – = k [A] Δ[A] Δt

(as a differential equation) = – = k [A] d [A] dt

Calculus is used to integrate a rate law.

Consider a 1st-order reaction: A products

Integrates to: ln [A]t = −k t + ln [A]0

y = m x + b (straight line)


Order Rate law Integrated rate law Slope

The Integrated Rate Law The reaction:

0 rate = k [A]t = -kt + [A]0 -k 1 rate = k[A] ln[A]t = -kt + ln[A]0 -k

A products doesn’t have to be 1st order. Some common integrated rate laws:

1 [A]t

2 rate = k[A]2 = kt + +k 1 [A]0

y x

The most accurate k is obtained from the slope of a plot.

ln[A

] time t

First-order reaction slope = -k

1/[A

]

time t

Second-order reaction

slope = k

[A]

time t

Zeroth-order reaction

slope = -k


The Integrated Rate Law

•  The reaction is first order (the only linear plot) •  k = -1 x (slope) of this plot.

Rate data for the decomposition of cyclopentene C5H8(g) → C5H6(g) + H2(g)

were measured at 850°C. Determine the order of the reaction from the following plots of those data:


Half-lives are only useful for 1st -order reactions. Why?

Half-Life


For a 1st-order reaction: ln[A]t = -kt + ln[A]0

Half-Life

When t = t1/2 [A]t = ½[A]0

Then: ln(½[A]0) = -kt1/2 + ln[A]0

ln(½[A]0/[A]0) = -kt1/2 {note: ln x – ln y = ln(x/y)} ln(½) = -ln(2) = -kt1/2 {note: ln(1/y) = –ln y }

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Half Life

0.010

0.008

0.006

0.004

0.002

0

[cis

plat

in]

(mol

/L)

0 400 800 1200 1600 2000 t (min)

•  ½ the cisplatin lost after 475 min. •  (0.0100 M → 0.0050 M)

•  [cisplatin] halves every 475 min

t1/2 of a 1st-order reaction can be used to find k. For cisplatin (a chemotherapy agent):

k = = 0.693 475 min

ln 2 t1/2

= 1.46 x 10-3 min-1


Use an integrated rate equation.

a)  [reactant],1600.s after initiation. b)  t for [reactant] to drop to 1/16th of its initial value. c)  t for [reactant] to drop to 0.0500 mol/L.

Calculating [ ] or t from a Rate Law


In a 1st-order reaction, [reactant]0 = 0.500 mol/L and t1/2 = 400.s (a) Calculate [reactant] ,1600.s after initiation.

1st order:

k = ln 2/ t½ = 0.6931/(400. s) = 1.733x10-3 s-1

and ln [A]t = -kt + ln [A]0

so ln[A]t = -(0.001733 s-1)(1600 s) +ln(0.500)

ln[A]t = -2.773 + -0.693 = -3.466

[A]t = e-3.466 = 0.0312 mol/L



In a 1st-order reaction, [reactant]0 = 0.500 mol/L and t1/2 = 400.s (b) Calculate t for [reactant] to drop to 1/16th of its initial value.

Note: part (a) could be solved in a similar way. 1600 s = 4 t1/2 so 0.500 → 0.250 → 0.125 → 0.0625 → 0.0313 M.

[reactant]0 [reactant]0 t1/2 1 2


1 2


1 4

[reactant]0 [reactant]0 t1/2 1

16 1 8

4 t1/2 = 4 (400 s) = 1600 s



From part (a): k = 1.733 x 10-3 s-1

ln [A]t = -kt + ln [A]0

then ln (0.0500) = -(0.001733 s-1) t + ln(0.500)

-2.996 = -(0.001733 s-1) t – 0.693

t = 1.33 x 103 s

In a 1st-order reaction, [reactant]0 = 0.500 mol/L and t1/2 = 400.s (c) Calculate t for [reactant] to drop to 0.0500 mol/L ?

t = -2.303 -0.001733 s-1



Nanoscale View: Elementary Reactions

6


Elementary reactions

unimolecular

bimolecular

unimolecular

unimolecular


Unimolecular Reactions 2-butene isomerization is unimolecular:

C=C CH3

H H

H3C (g) C=C

CH3

H

H

H3C (g)


Final state Initial state

Reaction Progress (angle of twist)

ΔE = -7 x 10-21 J

E a =

435

x 1

0-21

J

500

400

300

200

100

0 Pot

entia

l ene

rgy

(10-

21 J

)

-30° 0 30° 60° 90° 120° 150° 180° 210°

cis-trans conversion twists the C=C bond. •  This requires a lot of energy (Ea= 4.35x10-19J/molecule = 262 kJ/mol) •  Even more (4.42x10-19J/molecule) to convert back.

transition state or activated complex

Ea is the activation energy, the minimum E to

go over the barrier.

Exothermic overall

Unimolecular Reactions


Transition State


Bimolecular Reactions

I- must collide with enough E and in the right location to cause the inversion.

e.g. Iodide ions reacting with methyl bromide:

I-(aq) + CH3Br(aq) ICH3(aq) + Br-(aq)

transition state



I- must collide in the right location to cause the inversion.

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•  Also has an activation barrier (Ea). •  Forward and back Ea are different. •  Here the forward reaction is endothermic.

Products (final state)

Reactants (initial state)

Reaction Progress (changing bond lengths and angles)

E a =

126

x 1

0-21

J

ΔE = 63 x 10-21 J

150

120

90

60

30

0 Pot

entia

l ene

rgy

(10-

21 J

)

transition state



Temperature and Reaction Rate

Increasing T will speed up most reactions.

Higher T = higher average Ek for the reactants. = larger fraction of the molecules can

overcome the activation barrier.

25°C

kinetic energy

num

ber o

f mol

ecul

es

Ea

75°C Many more molecules have enough E to react at 75°C, so the reaction goes much faster.


Temperature and Reaction Rate Reaction rates are strongly T-dependent. Data for the I- + CH3Br reaction:

250 300 350 400 T (K)

0.00

0.1

0

0

.20

0.30

k

(L m

ol-1

s-1

)

T (K) k (L mol-1 s-1) 273 4.18 x 10-5

290 2.00 x 10-4

310 2.31 x 10-3

330 1.39 x 10-2 350 6.80 x 10-2 370 2.81 x 10-1


-Ea / RT k = A e

Quantity Name Interpretation and/or comments A Frequency factor How often a collision occurs with

the correct orientation. Ea Activation energy Barrier height. e-Ea/RT Fraction of the molecules with

enough E to cross the barrier. T Temperature Must be in kelvins. R Gas law constant 8.314 J K-1 mol-1.

Temperature and Reaction Rate


Determining Activation Energy Take the natural logarithms of both sides:

-Ea / RT ln k = ln A e

1 T ln k = + ln A Ea

R −

A plot of ln k vs. 1/T is linear (slope = −Ea/R).

-Ea / RT ln k = ln A + ln e

ln k = ln A + ln e Ea RT −

ln e = 1

ln ab = ln a + ln b


Determining Activation Energy The iodide-methyl bromide reaction data:

intercept = 23.85

slope = -9.29 x 10 3 K

0 0.001 0.002 0.003 0.004

28

18

8

-2

-12

ln k

1/T (K-1)

Ea = -(slope) x R

= -(-9.29 x103 K) 8.314 J K mol

= 77.2 x 103 J/mol = 77.2 kJ/mol

A = eintercept = e23.85 A = 2.28 x 1010 L mol-1 s-1 (A has the same units as k)

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Rate Laws for Elementary Reactions


Reaction Mechanisms

When [H3O+] is between 10-3 M and 10-5 M, rate = k [I-][H2O2]


Reaction Mechanisms

2 I-(aq) + H2O2(aq) + 2 H3O+(aq) I2 (aq) + 4 H2O(l)

slow

fast

fast

overall 2 I- + H2O2 + 2 H3O+ I2 + 4 H2O

Shows the bonding in H2O2


Reaction Mechanisms


Reaction Mechanisms

A good analogy is supermarket shopping: •  You run in for 1 item (~1 min = fast step), but… •  The checkout line is long (~10 min = slow step). •  Time spent is dominated by the checkout-line wait. •  In a reaction, a slow step may be thousands or

even millions of times slower than a fast step.


Reaction Mechanisms

The overall rate is expected to be rate = k [H2O2][ I- ] as observed!

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Consider: 2 NO (g) + Br2 (g) 2 NOBr (g)

Mechanisms with a Fast Initial Step

The generally accepted reaction mechanism is:

2 NO + Br2 2 NOBr



NO + Br2 NOBr2 fast, reversible



NO + Br2 NOBr2 reversible, equilibrium k1

k-1

rate forward = rate back

k1[NO][Br2] = k-1[NOBr2]

[NOBr2] = k1[NO][Br2] k-1



rate = k2 [NOBr2] [NO]

The earlier rate law:

k1[NO][Br2] k-1

rate = k2 [NO]

k1k2 k-1

rate = [Br2][NO]2

Now only contains starting materials - can be checked against experiment.

becomes:


Summary Elementary reactions: the rate law can be written down from the stoichiometry.

unimolecular rate = k[A] bimolecular rate = k[A]2 or rate = k[A][B]


Catalysts and Reaction Rate

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2-butene isomerization is catalyzed by a trace of I2.

No catalyst: rate = k [cis-2-butene]

A trace of I2(g) speeds up the reaction, and: rate = k [ I2 ]½ [cis-2-butene]

C=C CH3

H H

H3C (g) C=C

CH3

H

H

H3C (g)

k ≠ uncatalyzed k


I2 splits into 2 atoms. Each has an unpaired e-. (shown by the dot)

I• attaches and breaks one C-C bond

C=C CH3

H H

H3C C–C

H

CH3

H

H3C

I •

I2 is not in the overall equation, and is not used up. The mechanism changes!



Loss of I• and formation of C=C

C–C CH3

H

H

H3C

I • C=C CH3

H

H

H3C + I•

C–C H

CH3

H

H3C

I • C–C

CH3

H

H

H3C

I •

Rotation around C-C

I2 is regenerated





Reaction Progress

ΔE = -4 kJ/mol Ea = 115 kJ/mol

I• leaves; double bond reforms

I2 dissociates to I• + I•

Transition state for the uncatalyzed reaction

I• + I• regenerates I2

I• adds to cis-2-butene, (double → single bond)

Rotation around C-C

Ea = 262 kJ/mol





Enzymes: Biological Catalysts

11


Enzyme Activity and Specificity


Enzyme Activity and Specificity


Enzyme Activity and Specificity Enzymes are effective catalysts because they: •  Bring and hold substrates together while a reaction occurs.

•  Hold substrates in the shape that is most effective for reaction.

•  Can donate or accept H+ from the substrate (act as acid or base)

•  Stretch and bend substrate bonds in the induced fit so the reaction starts partway up the activation-energy hill.


Enzyme kinetics



Reaction Progress

ΔE

Ea

E'a

Transformation of the substrate to

products Activation energy E'a is much smaller than Ea and so the enzyme makes the reaction

much faster

Potential energy, E

Formation of the enzyme-substrate

complex

Transition state for the uncatalyzed reaction


Enzyme Activity and Specificity Enzyme catalyzed reactions:


Catalysis in Industry

CH3OH(l) + CO(g) CH3COOH(l) RhI3

  auto exhausts are cleaned by catalytic converters: 2 CO(g) + O2(g) 2 CO2

2 C8H18(g) + 25 O2 16 CO2(g) + 18 H2O(g) 2 NO(g) N2(g) + O2(g)

Pt-NiO

Pt-NiO

catalyst

12


…they form N2 and O2 and leave the surface.

Controlling Automobile Emissions

…N and O migrate on the surface until they get close to like

atoms…

…dissociates into N and O atoms (each

bonded to Pt)…

…forms a bond with the Pt surface…

NO approaches the Pt surface…


Converting Methane to Liquid Fuel Methane is hard to transport. It can be converted to methanol:

CH4(g) + ½ O2(g) → CO(g) + 2 H2(g)

CO(g) + 2 H2(g) → CH3OH(l)

A Pt-coated ceramic catalyst allows the 1st reaction to occur at low T.

© 2008 Brooks/Cole 1 © 2008 Brooks/Cole 2

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Transcript of © 2008 Brooks/Cole 1 © 2008 Brooks/Cole 2