Chemical Kinetics Final

8/12/2019 Chemical Kinetics Final

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Chemical Kinetics


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Contents

1 The Rate of a Chemical Reaction

2 Measuring Reaction Rates

3 Effect of Concentration on Reaction Rates:

The Rate Law

4 Zero-Order Reactions

5 First-Order Reactions

6 Second-Order Reactions

7 Reaction Kinetics: A Summary


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Contents

8 Theoretical Models for Chemical Kinetics

9 The Effect of Temperature on Reaction Rates

10 Reaction Mechanisms

11 Catalysis

Focus On Combustion and Explosions


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1 The Rate of a Chemical Reaction

Rate of change of concentration with time.

2 Fe3+(aq) + Sn2+ 2 Fe2+(aq) + Sn4+(aq)

t = 38.5 s [Fe2+] = 0.0010 M

t = 38.5 s [Fe2+] = (0.00100) M

Rate of formation of Fe2+= = = 2.610-5M s-1[Fe2+]

t

0.0010 M

38.5 s


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Rates of Chemical Reaction

[Sn4+]

t

2 Fe3+(aq) + Sn2+ 2 Fe2+(aq) + Sn4+(aq)

[Fe2+]

t=

1

2

[Fe3+]

t= -

1

2


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General Rate of Reaction

a A + bB cC + dD

Rate of reaction = rate of disappearance of reactants

=[C]

t1

c=

[D]t

1

d

[A]t

1a

= - [B]t

1b

= -

= rate of appearance of products


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2 Measuring Reaction Rates

H2O2(aq) H2O(l) + O2(g)

2 MnO4-(aq) + 5 H2O2(aq) + 6 H

+

2 Mn2++8 H2O(l) + 5 O2(g)


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H2O2(aq) H2O(l) + O2(g)

Example 15-2

-(-1.7 M / 2600 s) =

6 10-4M s-1

-(-2.32 M / 1360 s) = 1.7 10-3M s-1

Determining and Using an Initial Rate of Reaction.

Rate =-[H2O2]

t


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Example 2

-[H2O

2] = -([H

2O

2]

f- [H

2O

2]

i) = 1.7 10-3M s-1t

Rate = 1.710-3M s-1t

=- [H2O2]

[H2O2]100 s2.32 M = -1.7 10-3M s-1100 s

= 2.17 M

= 2.32 M - 0.17 M[H2O2]100 s

What is the concentration at 100s?

[H2O2]i= 2.32 M


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3 Effect of Concentration on Reaction

Rates: The Rate Law

a A + bB. gG + hH .

Rate of reaction = k [A]m[B]n.

Rate constant = k

Overall order of reaction = m+n+.


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Example 3 Method of Initial Rates

Establishing the Order of a reaction by the Method of Initial

Rates.

Use the data provided establish the order of the reaction with

respect to HgCl2and C2O22-and also the overall order of the

reaction.


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Example 3

Notice that concentration changes between reactions are by a

factor of 2.

Write and take ratios of rate laws taking this into account.


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Example 3

R2= k[HgCl2]2m[C2O4

2-]2n

R3= k[HgCl

2]

3

m[C2O

4

2-]3

n

R2

R3

k(2[HgCl2]3)m[C2O42-]3n

k[HgCl2]3m[C2O4

2-]3n

=

2m= 2.0 therefore m = 1.0

R2

R3

k2m[HgCl2]3m[C2O4

2-]3n

k[HgCl2]3m[C2O42-]3n= = 2.0=

2mR3

R3

= k(2[HgCl2]3)m[C2O4

2-]3n


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Example 3

R2= k[HgCl2]21[C2O42-]2n= k(0.105)(0.30)n

R1= k[HgCl2]11[C2O4

2-]1n= k(0.105)(0.15)n

R2

R1

k(0.105)(0.30)n

k(0.105)(0.15)n=

7.110-5

1.810-5 = 3.94

R2

R1

(0.30)n

(0.15)n= = 2n

=

2n= 3.98 therefore n = 2.0


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+ = Third Order

R2= k[HgCl2] [C2O4

2-]

First order

Example 3

1

Second order

2


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Intrepretasi Data Eksperimen untuk

Menentukan Pers. Kecepatan Reaksi

Ekperimen dilakukan dilakukan di laboratoriumsecara batch

Misal untuk reaksi: A B

Data yang diambil/diperoleh berupa: [A] vs waktu atau

[B] vs waktu


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Definisi Konversi :

Konversi (X) dinyatakan sebagai:

Jumlah mol reaktan yang bereaksi x 100%

Jumlah mol reaktan mula-mula

Hubungan antara konsentrasi reaktan pada suatu saat dengan

konversi:

[A] = [A]o (1-X)

Jika reaksi melibatkan 2 reaktan atau lebih, konversi bisa dinyatkan

terhadap reaktan 1 atau2. Jika tidak ada keterangan, konversi

mengacu pada limiting reactant

Intrepretasi Data Eksperimen untuk

Menentukan Pers. Kecepatan Reaksi (2)


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4 Zero-Order Reactions

A products

Rrxn= k [A]0

Rrxn= k

[k] = mol L-1s-1


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Integrated Rate Law

-[A]t+ [A]

0= kt

[A]t= [A]0- kt

t-[A]

dt= k-d[A]

Move to the

infinitesimal= k

And integratefrom 0to time t


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5 First-Order Reactions

H2O2(aq) H2O(l) + O2(g)

= -k [H2O2]d[H2O2]

dt

= - k dt[H2O2]

d[H2O2][A]0

[A]t

0

t

= -ktln [A]t[A]0

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

[k] = s-1


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First-Order Reactions


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6 Second-Order Reactions

Rate law where sum of exponents m + n +

= 2.A products

= kt +1

[A]0[A]t

1

dt= -k[A]2

d[A][k] = M-1 s-1 = L mol-1 s-1


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Second-Order Reaction


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Pseudo First-Order Reactions

Simplify the kinetics of complex reactions Rate laws become easier to work with.

CH3CO2C2H5+ H2O CH3CO2H + C2H5OH

If the concentration of water does not changeappreciably during the reaction.

Rate law appears to be first order.

Typically hold one or more reactants constant byusing high concentrations and low concentrations

of the reactants under study.


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Testing for a Rate Law

Plot [A] vs t

Utk Reaksi order 0

Plot ln[A] vs t

Utk Reaksi Order 1

.

Plot 1/[A] vs t

Utk Reaksi order 2.


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7 Reaction Kinetics: A Summary

Calculate the rate of a reaction from a known ratelaw using:

Determine the instantaneous rate of the reactionby:

Rate of reaction = k [A]m[B]n.

Finding the slope of the tangent line of [A] vs t or,

Evaluate[A]/t, with a short t interval.


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Soal UAS TM-TI 2010

Data berikut ini diperoleh dari percobaan dekomposisiHI menurut reaksi:

HI (g) 0,5 H2 (g) + 0,5 I2 (g)

Tentukanlah:a) Order reaksi dan konstanta kecepatan reaksinya!

b) Konsentrasi HI pada saat reaksi berlangsung selama

60 menit!


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Summary of Kinetics

Determine the order of reaction by:

Using the method of initial rates.

Find the graph that yields a straight line.

Test for the half-life to find first order reactions.

Substitute data into integrated rate laws to find

the rate law that gives a consistent value of k.


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Summary of Kinetics

Find the rate constant k by:

Find reactant concentrations or times for certainconditions using the integrated rate law after

determining k.

Determining the slope of a straight line graph.

Evaluating k with the integrated rate law.

Measuring the half life of first-order reactions.


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8 Theoretical Models for

Chemical Kinetics

Kinetic-Molecular theory can be used to calculatethe collision frequency.

In gases 1030 collisions per second.

If each collision produced a reaction, the rate would beabout 106M s-1.

Actual rates are on the order of 104

M s-1

. Still a very rapid rate.

Only a fraction of collisions yield a reaction.

Collision Theory


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Activation Energy

For a reaction to occur there must be aredistribution of energy sufficient to break certain

bonds in the reacting molecule(s).

Activation Energy is: The minimum energy above the average kinetic energy

that molecules must bring to their collisions for a

chemical reaction to occur.


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Activation Energy


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Collision Theory

If activation barrier is high, only a few moleculeshave sufficient kinetic energy and the reaction is

slower.

As temperature increases, reaction rate increases.

Orientation of molecules may be important.


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Collision Theory


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Collision Theory


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Transition State Theory

The activated complexis ahypothetical species lyingbetween reactants and

products at a point on the

reaction profilecalled thetransition state.


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Prentice-Hall 2002 General Chemistry: Chapter 15 Slide 38 of 55

Transition State Theory


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9 Effect of Temperature on

Reaction Rates

Svante Arrhenius demonstrated that many rateconstants vary with temperature according to the

equation:

k = Ae-Ea/RT

ln k = + ln AR

-Ea

T

1

ln = -R

-Ea

T2

1

k2

k1

T1

1+


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Arrhenius Plot

N2O5(CCl4) N2O4(CCl4) + O2(g)

= -1.2104KR

-Ea

-Ea= 1.0102kJ mol-1


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Arrhenius Equation

k = Ae-Ea/RT ln k = + ln AR

-Ea

T

1

ln k2ln k

1= + ln A - - ln A

R

-Ea

T2

1

R

-Ea

T1

1

ln = -R

-Ea

T2

1

k2

k1

T1

1


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10 Reaction Mechanisms

A step-by-step description of a chemical reaction. Each step is called an elementary process.

Any molecular event that significantly alters a

molecules energy of geometry or produces a newmolecule.

Reaction mechanism must be consistent with: Stoichiometry for the overall reaction.

The experimentally determined rate law.


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Elementary Processes

Unimolecularorbimolecular. Exponentsfor concentration terms are the same as

the stoichiometric factorsfor the elementary

process.

Elementary processes are reversible.

Intermediatesare produced in one elementaryprocess and consumed in another.

One elementary step is usually slower than all theothers and is known as the rate determining step.


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Slow Step Followed by a Fast Step

H2(g) + 2 ICl(g) I2(g) + 2 HCl(g) dt = k[H2][ICl]

d[P]

Postulate a mechanism:

H2(g) + 2 ICl(g) I2(g) + 2 HCl(g)

slowH2(g) + ICl(g) HI(g) + HCl(g)

fastHI(g) + ICl(g) I2(g) + HCl(g)

dt= k[H2][ICl]

d[HI]

dt= k[HI][ICl]

d[I2]

dt= k[H2][ICl]

d[P]


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Slow Step Followed by a Fast Step


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Fast Reversible Step Followed by a Slow Step

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

= -kobs[NO2]2[O2]

d[P]

Postulate a mechanism:

dt= k2[N2O2][O2]

d[NO2]

fast 2NO(g) N2O2(g)k1

k-1

slow N2O2(g) + O2(g) 2NO2(g)k2

dt= k2 [NO]

2[O2]d[NO2]

k-1

k12NO(g) + O2(g) 2 NO2(g)

K =k-1

k1=

[NO]

[N2O2]

= K [NO]2k-1

k1= [NO]2[N2O2]


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Catalysis

Alternative reaction pathway of lower energy. Homogeneous catalysis.

All species in the reaction are in solution.

Heterogeneous catalysis. The catalyst is in the solid state.

Reactants from gas or solution phase are adsorbed.

Active sites on the catalytic surface are important.


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Catalysisa. Homogeneous


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b. Heterogeneous

l i f


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Catalysis on a Surface

C l i


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Enzyme Catalysis

E + S ESk

1

k-1ES E + P

k2

S i Ki i


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Saturation Kinetics

E + S ES

k1

k-1 E + Pk2

dt= k1[E][S]

k-1[ES]k2[ES]= 0d[P]

dt = k2[ES]

d[P]

k1[E][S] = (k-1+k2 )[ES]

[E] = [E]0[ES]

k1[S]([E]0[ES]) = (k-1+k2 )[ES]

(k-1+k2 ) + k1[S]

k1[E]0[S][ES] =

Mi h li M


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Michaelis-Menten

dt=

d[P]

(k-1+k2 ) + k1[S]

k1k2[E]0[S]

dt=d[P]

(k-1+k2 ) + [S]

k2[E]0[S]

k1

dt=

d[P]

KM+ [S]

k2[E]0[S]

dt=d[P] k2[E]0

dt

=d[P]

KM

k2[E]0[S]

Q i


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Questions

Develop problem solving skills and base your strategy noton solutions to specific problems but on understanding.

Choose a variety of problems from the text as examples.

Practice good techniques and get coaching from people who

have been here before.

o

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