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1

Lecture 8, Kinetics

rates of chemical reactions

the factors that affect rate of reactions

the mechanisms (the series of steps) by which

reactions occur

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Chemical kinetics studies:

We are all familiar with processes in which some quantity changes with time

Car travels at 40 km/hour (miles/hour)

Faucet delivers water at 30 l/min (gallons/minute)

Factory produces 32,000 tires/day

Each of these ratios describe changes whichhappened with time (a rate)

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Rate

The speed with which the reactants disappear and the products form is called the rate of the reaction.

A study of the rate of reaction can give detailed information about how reactants change into products.

The series of individual steps that add up to the overall observed reaction are called the reaction mechanism.

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Kinetics, rate of the reaction, reaction mechanism

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When blue dye is reacting with bleach, the latterconverts dye into a colourless products.The colour decreases and eventually disappears. The rate of the reaction could be determined by repeatedly measuring both the colour intensity and the elapsed time. The concentration of the dye could be calculated from the intensity of the blue colour.

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A UV-VIS spectroscopy can be used for determining reaction rates. Light of the wavelength that is absorbed by the investigated substance is passed through a reaction chamber. The changes in the reactant or product concentration as reaction progresses cause the decrease or increase of the light intensity. In this manner it is possible to determine the reactant or product concentration change.

6

Identify which of the following are rates:

A) 15 cm

B) 30 m / s2

C) 25oC

D) 5oC/min

E) 1.2 mol /min

F) 45 min

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QuizExperience tells us that different chemical

reactions occur at very different rates, e.g.

1. Some reactions proceed very rapidly, even

explosively i.e. combustion reactions –

burning methane (component of natural

gas) or isooctane (C8H18) in gasoline.

2. Other reactions carry on very slowly i.e. an

erosion of the rocks or an iron rusting

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We have learnt on the spontaneity of reaction from

thermodynamics. But it is hard to state whether all

spontaneous reactions are rapid ones. The reactions of

strong acids with strong bases are thermodynamically

favoured (spontaneous) and occur at very rapid rates.

1) 2HCl + Mg(OH)2 → MgCl2 + 2H2O

ΔG0rxn = -591.8 kJ/mol

Similarly reaction some compound with oxygen (e.g.

burning) is thermodynamically favoured and rapid e.g.

2) CH4 + 2O2 → CO2 + 2H2O

ΔG0rxn = -800.8 kJ/mol

9 10

But

3) C(diamond) + O2(g)→CO2(g) ΔG0rxn = -397 kJ/mol

This reaction does not occur at an observable rate.

4) C(graphite) + O2(g) →CO2(g) ΔG0rxn = -394 kJ/mol.

This reaction occurs rapidly.

The difference in the reaction speed of 3 and 4

reactions is explained by kinetics, not thermodynamics.

The rate of reaction describes how fast reactants are used up and products are formed.

Knowledge of the rate of a reaction can be an invaluable tool in helping us to understand how chemical compounds behave when they interact.

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Rate of reaction, v,

A rate, is always expressed as a ratio.

One way to describe a reaction rate is to select one component of the reaction and describe the change in its concentration per unit of time:

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Rate, formula

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Molarity (mol/L) is normally the concentration unit and the second (s) is the most often used unit of time.

Typically, the reaction rate has the units

t

X

tt

tXtXX

) of (conc.

) (

) at time of conc. at time of (conc. respect to with rate

12

12

1-1- s L molor s

mol/L

13 14

Consider the following reaction at a constant temperature in closed system:MgCO3(s) + 2HCl(aq) → CO2(g) + H2O(l) + MgCl2(aq)

Which of the following properties could be used to determine reaction rate?

A. Mass of the system

B. Pressure of the gas

C. Concentration of H2O

D. Concentration of MgCO3

E. Disappearing of solid

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Consider the following reaction:

2CaCrO4(s) +2H+(aq) 2Ca2+

(aq) + H2O(l) + Cr2O72-

(aq)

(orange)

The progress of the reaction could be followed by observing the rate of

A. mass loss

B. decrease in pH

C. precipitate formation

D. formation of orange colour in thesolution

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An 8.00 g piece of magnesium was placed into6.0 M HCl. After 25 s 3.50 g of unreacted magnesium remained. The average rate at which magnesium was consumed is:

A. 0.14 g/s

B. 0.18 g/s

C. 0.32 g/s

D. 4.50 g/s

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a A + b B → c C+ d D

rate= -Δ[A]/ Δt;

rate = -Δ[B]/ Δt, or

rate = Δ[C]/ Δt;

rate= Δ[D]/ Δt

The reaction rate must be positive because it describes the

forward reaction, which consumes A and B.

The concentration of reactants A and B decrease in time

interval Δt.

Δ[A]/ Δt and Δ[B]/ Δt are negative quantities.

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If no other reaction takes place, the changes in

concentration are related to one another. For every a

mol/L that described decrease of [A], [B] must

decrease by b mol/L, [C] must increase by c mol/L

and so on…

The number of moles of reactants or products that

occur per litter in a given time describe the rate of

reaction.

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a A+ b B → c C+ d D

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2A + B →3C +DIf the rate of disappearance of A is equal to -0.084 mol/L s at the start of the reaction what are the rates of change for B, C and D at this time?

Rate of change of B = Rate of change of C = Rate of change of D =

a) B= 0.042 M/s; C= 0.056 M/s; D= - 0.042 mol/L sb) B = -0.042M/s; C = 0.126 M/s; D = 0.042 mol/L sc) B= -0.042 M/s; C= - 0.126 M/s; D= 0.042 mol/L s

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Consider the following reaction:

N2H4(l) + 2H2O2(l) N2(g) + 4H20(l)

In 1.0 seconds, 0.015 mol of H2O2 is consumed.

The rate of production of N2 is

A. 1.5 x 10-3 mol/s

B. 7.5 x 10-3 mol/s

C. 6.0 x 10-3 mol/s

D. 1.5 x 10-2 mol/s

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Compared to the rate with respect to propane:◦ Rate with respect to oxygen is five times faster

◦ Rate with respect to carbon dioxide is three times faster

◦ Rate with respect to water is four times faster

Since the rates are all related any may be monitored to determine the reaction rate

)(4)(3)(5)( 22283 gOHgCOgOgHC

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Determine relative reaction rates of the four substrates involved in the following chemical reaction. Give the appropriate numbers instead w, x, y and z letters:

2C2H2(g) + 5O2(g) → 4CO2 + 2H2O(l)

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Instantaneous reaction

rate

A reaction rate is generally not constant throughout the reaction.

Since the most of reactions depend on the concentration of reactants, the rate changes as they are used up.

The rate at any particular moment of givenreaction is called the instantaneous rate.

It can be calculated from a concentration versus time plot.

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A plot of the concentration of HI versus time for the reaction:2HI(g) H2(g) + I2(g). The slope is negative because we are measuring the disappearance of HI. When used to express the rate it is used as a positive number. 29

Plot of [H2] vs time for the reaction of 1.000 M H2 with 2.000 M ICl. The instantaneous rate of reaction at anytime, t, equals the negative slope of the tangent to thiscurve at time t. The initial ratio of the reaction is equal to the negative of the initial slope (t=0). The determination of the instantaneous rate at t=2 second is illustrated.

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QuizBased on the graph below determine the instantaneous rate of change of [X] at 8 seconds

82 24 s

0.04

time

[X] =….. M/s

31

0.10

conce

ntr

ati

on

0.06

32

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A rate law is a mathematical equation that describes the

progress of the reaction.

In general, rate laws must be determined experimentally.

Unless a reaction is an elementary reaction, it is not

possible to predict the rate law from the overall chemical

equation.

There are two forms of a rate law for chemical kinetics:

the differential rate law and the integrated rate law.

34

For a generic reaction:

reactant 1 + reactant 2 → product

with no intermediate steps in this reaction

mechanism, the rate is given by

Rate = k [reactant 1]n [reactant 2]m

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Rate constant, k

rate = k [reactant 1]n [reactant 2]m

where k is the constant of proportionality

named rate constant. Its value is generally

constant provided that reaction is performed at

constant temperature T.

Values of k are always positive, although it

may be an exception of this rule.

Consider the following reaction:

From experiment, the rate law (determined from initial rates) is

At 0oC, k equals 5.0 x 105 L5 mol-5 s-1

Thus, at 0oC

OHISeHISeOH 2332 3246

231

32 ][][][rate HISeOHk

2332

1555 ][]][[)s mol L 100.5(rate HISeOH-

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The exponent in a rate law is called the order of reaction with respect to the corresponding reactant.

The exponents in the rate law are generally unrelated to the chemical equation’s coefficients.◦ Never simply assume that exponents and

coefficients are the same.

◦ The exponents must be determined from the results of experiments.

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For the rate law:

We can say◦ The reaction is first order with respect to H2SeO3

◦ The reaction is third order with respect to I-

◦ The reaction is second order with respect to H+

◦ The reaction is sixth order overall

Exponents (orders of reactions) in a rate law can be fractional, negative, and even zero.

231

32 ][][][rate HISeOHk

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Analyse the following rate equations, and determine the orders of reaction with respectto reactant and overall reaction order:

1. rate = k [Cu2+] [NH3]

2. rate = k [OH-]

3. rate = k [NO]2[O2]

4. rate = k [A]3[B]2

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Reaction order quiz

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Looking for patterns in experimental data provide way to determine the exponents in a rate law.

One of the easiest ways to reveal patterns in data is to form ratios of results using different sets of conditions.

This technique is generally applicable.

Consider the hypothetical reaction:

nm BAk

productsBA

][][rate

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Suppose the experimental concentration-rate data for five experiments is:

5.40 0.30 0.30 5

2.40 0.20 0.30 4

0.60 0.10 0.30 3

0.40 0.10 0.20 2

0.20 0.10 0.10 1

)s L (mol )L (mol )L (molExpt

Rate Initial ][ ][

Conc. Inital

1-1-1-1-

BA

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For experiments 1, 2, and 3 [B] is held constant, so any change in rate must be due to changes in [A]

The rate law says that at constant [B] the rate is proportional to [A]m

m

mm

m

A

A

A

A

2L mol 10.0

L mol 20.0

][

][

2s L mol 20.0

s L mol 40.0

rate

rate

][

][

rate

rate

1-

1-

1

2

1-1-

1-1-

1

2

1

2

1

2

Thus m=1

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For experiments 3, 4, and 5 [A] is held constant, so any change must be due to changes in [B]

The rate law says that at constant [A] the rate is proportional to [B]n

Using the results from experiment 3 and 4:

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The reaction is second order in respect to B and

rate=k [A]1[B]2

n

nn

n

B

B

B

B

2L mol 10.0

L mol 20.0

][

][

4s L mol .600

s L mol 40.2

rate

rate

][

][

rate

rate

1-

1-

3

4

1-1-

1-1-

3

4

3

4

3

4

Thus n=2

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The rate constant (k) can be determined using data from any experiment

Using experiment 1:

Using data from a different experiment might give a slightly different value of k

1-2-22

21-1-

-1-1

2

s mol L 10 2.0

)L mol )(0.10L mol (0.10

s L mol 20.0

]][[

rate

BA

k

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For 2NO + O2 --> 2NO2 , initial rate data are:

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Experiment [NO]M/L

[O2 ] M/L

Rate[mM/ s]

1 0.010 0.010 2.5

2 0.010 0.020 5.0

3 0.030 0.020 45.0

Determine the reaction rates in terms of [NO], [O2] and k

Quiz

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The reaction has the rate law: rate = k[C][D]2. What will happen to the reaction rate when the following change in conditions is performed?

doubling [C]

tripling [D]

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Quiz

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The relationship between concentration and time can be derived from the rate law and calculus.

Integration of the rate laws gives the integrated rate laws, which present concentration as a function of time.

Integrated laws can be very complicated, so only a few simple forms will be considered.

51 52

In this reaction the speed of the reaction doesnot depend on the reagent concentrations.

𝑣 = −∆ 𝑐

∆ 𝑡= 𝑘𝑐0 = 𝑘

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v

t

v

c

CC0

t

kc kv 0

k- a tg

•photochemical reactions

•heterophaseous reaction in which the slowest process

is connected with the phase change

•burning ethanol by living organism

• decomposition of ammonia

• synthesis of HCl from hydrogen and chlorine

performed on the sun light

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◦ Formula for the rate law is: rate = k [A]

◦ The integrate rate law can be expressed as:

[A]0 is [A] at t (time) = 0

[A]t is [A] at t = t

e = base of natural logarithms = 2.71828…

kt

t

eAAktA

A 0t0 ][][or

][

][ln

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First order reaction

Absorption, distribution, elimination rates

Microbial death kinetics

Photo dissociation of ozone with UV light

Decomposition of hydrogen peroxide at room temperature

Hydrolysis of sucrose (sugar) to glucose and fructose

e.g. SO2Cl2→ SO2 + Cl2 at 320 oC k = 2 x 10-5 kJ/ s

C2H6→ 2 CH3• at 700 oC k = 5.36 x 10-4 kJ/ s

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First order reaction examples

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For certain first-order reaction the initial

concentration of reactant A is equal 2M/L and

the rate constant of this reaction k = 0.15M/ L s.

What is the concentration of A after time 6s ?

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Example of calculation

Or ln𝐴 0

𝐴 𝑡= 𝑘𝑡; 𝑙𝑛 𝐴 0 − 𝑙𝑛 𝐴 6 = 𝑘𝑡 = 0.813 𝑀/𝐿

If a reaction is first order with a rate constant of 5.48 x 10-2 sec-1, how long is required for 1/4 of the initial concentration of reactant to be used up?

Assume that initial concentration of reactantis equal to 1 M.

58

Quiz

A plot of ln[A]t versus t gives a straight line with a slope of -k

The decomposition of N2O5. (a) A graph of concentration versus

time for the decomposition at 45oC. (b) A straight line is obtained

from a logarithm versus time plot. The slope is negative the rate

constant.59

The simplest second-order rate law has the form

The integrated form of this equation is

2][ rate Bk

tBB

BB

ktBB

t

t

at time ofion concentrat the][

ofion concentrat initial the][

][

1

][

1

0

0

60

Second order reaction

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Decomposition of HI without catalyst

Decomposition of NO2 to NO and O2

ClO- + Br-→ BrO- + Cl-

at 25 oC k = 4.2 x 10-7 kJ/l mol-1s-1

H+ + OH-→ H2O

at 25 oC k = 1.35 x 1011 kJ/l mol-1s-1

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Second order reaction examples Graphical methods can also be applied to

second-order reactions

A plot of 1/[B]t versus t gives a straight line with a slope of k

Second-order kinetics. A plot of 1/[HI] versus time

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The amount of time required for half of a reactant to disappear is called the half-life, t1/2

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Half life t1/2

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𝑡12

=𝐴0

2𝑘

For zero-order reactions, the half-life depends on the initial concentration of reactant and the rate constant.

65 66

It is not affected by the initial concentration

ktkt

A

A

AAtt

ktA

A

t

t

2lnor

][

][ln

ngsubstituti ,][2

1][ ,at

][

][ln :law rateorder -First

2/12/1

02

1

0

02/1

0

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◦ The half-life of a second-order reactions doesdepend on the initial concentration

0

2/12/1

0

2/1

002

1

02/1

0t

][

2lnor

][

1

][

1

][

1

ngsubstituti ,][2

1][ ,at

][

1

][

1 :law rateorder -Second

Bktkt

B

ktBB

BBtt

ktBB

t

69

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(2) Graphical method

to t1 t2 t3 t4 ...

co c1 c2 c3 c4 ...

n = 0 n = 1 n = 2

c

1

tc

1

k a tg

ln c

ln co

t

k- a tg

c

co

t

k- a tg

t kcc t kc lnc ln t kc

1

c

1

Reagent concentration

Substance A decomposes by a first-order reaction. Starting initially with [A] = 2.00 M, after 150 min [A] = 0.50 M. What is t1/2 for this reaction?

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Quiz

First-order radioactive decay of iodine-131. The

initial concentration is represented by [I]0.

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The table below presents plote of concentration of biologically active metabolite T-IDA vs time. Using of these data determine graphically the half-time life (t1/2) of this metabolite.

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Time[min]

0 10 20 30 40 50 60 70

Concentration[mol /L]

100 50 25 12,5 6.25 3.13 1.56 0.781

Quiz

75

0

20

40

60

80

100

120

0 10 20 30 40 50 60 70 80

co

nc

en

tra

tio

n[m

ol

/L]

Time [min]

T1/2 = 10 min

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There are five principle factors that

influence reaction rates:

1) Chemical nature of the reactants

2) Ability of the reactants to come in contact

with each other

3) Concentration of the reactants

4) Temperature

5) Availability of rate-accelerating agents called

catalysts

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Chemical nature of the reactants

◦ Bonds break and form during reactions

The most fundamental difference in reaction rates lies in the

reactants themselves.

Some reactions are fast by nature whereas others are slow

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C(diamond) + O2(g)→CO2(g) ΔG0rxn = -397 kJ/mol

This reaction one does not occur.

C(graphite) + O2(g) →CO2(g) ΔG0rxn = -394 kJ/mol.

This reaction occurs rapidly.

http://zsp1krosno.bloog.pl/id,3926917,title,Wlasciwosci-substancji-w-

zaleznosci-od-rodzaju-wiazan-

chemicznych,index.html?smoybbtticaid=61a302

Diamond structure Graphite structure

Ability of the reactants to meet

◦ Most reactions require that particles (atoms,

molecules, or ions) collide before the reaction can

occur.

◦ This depends on the phase of the reactants.

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◦ In a homogeneous reaction the reactants are in the

same phase:

For example both reactants in the gas (vapour) phase.

◦ In a heterogeneous reaction the reactants are in

different phases:

For example one reactant is present in the liquid

whereas the second is in the solid phase.

◦ In heterogeneous reactions the reactants meet only at

the intersection between the phases.

◦ Thus the area of contact between the phases

determines the rate of the reaction.

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Effect of crushing a solid.

When a single solid is

subdivided into much smaller

pieces, the total surface area on

all of the pieces becomes very

large.

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Concentration of the reactants

◦ Both homogeneous and heterogeneous reaction

rates are affected by reactant concentration

83 84

85

Volume

O2 [cm3]

Time [s]

0.5 mol/l of

H2O2

0.4 mol/l

0.3 mol/l

0.2 mol/l

0.1 mol/ l

H2O2(aq) → H2O(l) + ½ O2 (g)

◦ The rates for almost all chemical reactions

enhance as the temperature is increased

Cold-blooded creatures, such as insects and

reptiles, become sluggish at lower temperatures

as their metabolism slows down

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Temperature

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87

In experiments it was determined that in

homogenous reaction temperature

enhancement of 10o results 2-4 times

increasing of reaction speed.

Increase of temperature cause increase of

rate constant (k) in rate law.

Number describing how many times k increases is known as temperaturecoefficient (θ).

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According to the van't Hoff rule

89

𝑄 =𝑘2𝑘1

𝑇−𝑇010

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2. If temperature increases from 20oC to 50oC the velocity of reaction increases in 8 times. What the temperature coefficient is equal to?

A) 8; b) 4; c) 3; d) 2

1. The rate of a chemical reaction doubles for every 10°C rise of temperature. If the temperature is raised by 50°C, the rate of the reaction increases by about:

a) 10 Times; b) 24 Times; c) 32 Times; d) 64 times

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91

To sum up

Consider this reaction:

Zn(s) + 2HCl(aq) → ZnCl2(aq) + H2(g)

http://www.ausetute.com.au

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Condition Affect on Rate Explanation

ConcentrationIncreasing the

concentration of HCl will increase the reaCtion rate.

More HCl particles means there will be more collisions between

HCl and Zn.

TemperatureIncreasing temperature increases the reaction

rate.

HCl particles will gain more kinetic energy increasing the number of collisions with Zn

atoms. More Zn and HCl particles will have sufficient energy to react resulting in more successful

collisions.

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Particle Size

Reducing the size of Zn particles will

increase the rate of reaction.

Reducing the size of the Zn particles increases the surface area

available for reaction with HClmolecules resulting in more

collisions.

Stirring Rate

Increasing the stirring rate of this

mixture will increase the reaction rate.

Stirring will keep small Zn particles in suspension, increasing the

surface area available for collisions, resulting in an increased

reaction rate.

Condition Affect on Rate Explanation

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Student performed experiment on dissolution of Mg in HCl in four test tubes according to the following conditions.

94

Test tube Mg size HCl concentration

1 cube 1.0 M

2 cube 0.5 M

3 powder 1.0 M

4 powder 0.5 M

Determine the order of test tubes according to the

reaction time decreasing

Quiz

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A catalyst is a substance that changes the rate of a chemical reaction without itself being used up.◦ Positive catalysts speed up reactions

◦ Negative catalysts or inhibitors slow reactions

(Positive) catalysts speed reactions by allowing the rate-limiting step to proceed with a lower activation energy.

Thus a larger fraction of the collisions iseffective.

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(a) The catalyst provides an alternate, low-energy path from the

reactants to the products.

(b) A larger fraction of molecules have sufficient energy to react

when the catalyzed path is available.

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Catalysts can be divided into two groups◦ Homogeneous catalysts exist in the same phase

as the reactants.

◦ Heterogeneous catalysts exist in a separate phase.

NO2 is a homogeneous catalyst for the production of sulfuric acid in the lead chamber process.

The mechanism is:

98

The second step is slow, but when it is catalyzed by NO2 it speeds up:

4223

322

1

2

22

SOHOHSO

SOOSO

SOOS

222

1

322

NOONO

SONOSONO

99

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Heterogeneous catalysts are typically solids

Consider the synthesis of ammonia from hydrogen and nitrogen by the Haber process

322 2NHN3H

100

The Haber process. Catalytic formation of ammonia molecules

from hydrogen and nitrogen on the surface of a catalyst.

101

102

Mechanism theories

The reaction’s mechanism is the series of

simple reactions called elementary processes.

The rate law of an elementary process can be

written from its chemical equation.

103

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One of the simplest models explaining

reaction rates is collision theory.

According to collision theory, the rate of

reaction is proportional to the effective

number of collisions per second among the

reacting molecules.

An effective collision is one that actually gives

product molecules.

The number of all types of collisions increase

with concentration, including effective

collisions.

104

There are a number of reasons why only a

small fraction of all the collisions leads to

the formation of product:

◦ Only a small fraction of the collisions are

energetic enough to lead to products.

◦ Molecular orientation is important because a

collision on the “wrong side” of a reacting

species cannot produce any product.

This becomes more important as the complexity of

the reactants increases.

105

The key step in the decomposition of NO2Cl to NO2 and Cl2 is

the collision of a Cl atom with a NO2Cl molecules. (a) A poorly

orientated collision. (b) An effectively orientated collision.

106 107

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◦ The minimum of kinetic energy of the colliding particles must have is called the activation energy, Ea .

◦ In a successful collision, the activation energy changes to potential energy as the bonds rearrange to for products.

◦ Activation energies can be large, so only a small fraction of the well-orientated, colliding molecules have it.

◦ When temperature increases the average kinetic energy of the reacting particle alsoincreases.

108

Kinetic energy distribution for a reaction at two different

temperatures. At the higher temperature, a larger fraction of the

collisions have sufficient energy for reaction to occur. The shaded

area under the curves represent the reacting fraction of the

collisions.109

The potential-energy diagram for an exothermic

reaction. The extent of reaction is represented as

the reaction coordinate.110

A unsuccessful (a) and successful (b) collision for an exothermic

reaction.

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Activation energies and heats of reactions can be determined from potential-energy diagrams Potential-energy

diagram for an

endothermic

reaction. The heat

of reaction and

activation energy

are labeled.

112

Reactions generally have different activation energies in the forward and reverse direction

Activation energy barrier for the forward and reverse reactions.

113

The brief moment during a successful

collision that the reactant bonds are

partially broken and the product bonds are

partially formed is called the transition

state.

Transition state theory explains what

happens when reactant particles come

together.

114

Formation of the activated complex in the reaction between

NO2Cl and Cl.

NO2Cl+ClNO2+Cl2115

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116

The overall rate law determined for the mechanism must agree with the observed rate law.

The exponents in the rate law for an elementary process are equal to the coefficients of the reactants in chemical equation

2

2

32

]k[NO rate

NONO2NO

:process Elementary

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Multistep reactions are common.

The sum of the elementary processes must give the overall reaction.

The slowest set in a multistep reaction limits how fast the final products can be formed and is called the rate-determiningor rate-limiting step.

Simultaneous collisions between three or more particles are extremely rare.

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A reaction that depended on a three-body collision would be extremely slow.

Thus, reaction mechanism seldom includeselementary process that involves more than two-body or bimolecular collisions.

Consider the reaction

The mechanism is thought to be

tal)(experimen ][Hk[NO]rate

O2HN2H2NO

22

222

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The second step is the rate-limiting step, which gives

N2O2 is a reactive intermediate, and can be eliminated from the expression.

(fast) OH N H ON

(slow) OH ONHON

(fast) ON 2NO

2222

22222

22

]][HON[ rate 222k

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The first step reaches a fast equilibrium

At equilibrium, the rate of the forward and reverse reaction are equal

2

22

22

2

22

2

NO][]ON[

or ]ON[NO][

thus]ON[se)rate(rever

NO][rd)rate(forwa

r

f

rf

r

f

k

k

kk

k

k

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Substituting, the rate law becomes

Which is consistent with the experimental rate law.

]H[NO]['rate

or ]H[NO][rate

]H[]ON[rate

2

2

2

2

222

k

k

kk

k

r

f

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The activation energy is related to the rate constant by the Arrhenius equation

k = rate constantEa = activation energye = base of the natural logarithmR = gas constant = 8.314 J mol-1 K-1

T = Kelvin temperatureA = frequency factor or pre-exponential factor

RTEaAek/

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The activation energy can be related to the rate constant at two temperatures

121

2 11ln

TTR

E

k

k a

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The activation energy of the first order reaction is 50,2 kJ/mol at 25oC. At what temperature will the rate constant double?

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Quiz

126 127

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128129

True or false quiz1. A catalyst alters the rate of a chemical reaction by:

a) always providing a surface on which molecules react

b) changing the products formed in the reaction

c) inducing an alternate pathway for the reaction

with generally lower activation energy

d) changing the frequency of collisions between molecules

2. The Rate of a Chemical Reaction

a) usually is increased when the concentration of one of the

reactants is increased

b) is dependent on temperature

c) may be inhibited by certain catalytic agents

d) will be very rapid if the activation energy is large

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