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Chemical Kinetics(4 lectures)

Dr. Paul T. Maragh

Tue. 5:00 p.m. / Wed. 9:00 a.m.

1 full question on C10K Paper 1

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What is Chemical Kinetics?

The study of the speed or RATE at which

a chemical reaction occurs.

What are some of the factors that affect

the RATE of a chemical reaction? The nature of the reactants and products

Temperature CatalystsThe concentrations of the reacting species.

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Homogeneous reactionsgas phase: H2(g) + I2(g) 2HI(g)

liquid phase: NaOH(aq) + HCl(aq) NaCl(aq) + H2O(l)

Heterogeneous reactionsFe(s) + 2H+(aq) Fe2+(aq) + H2(g)

Irreversible reaction

Reversible reaction (equilibrium) e.g. N2O4(g) 2NO2(g)

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The factors mentioned will affect the compositioncomposition of the reaction mixture at any given time.time.

Therefore

The change in composition of the reaction mixture with time

is the rate of reaction, denoted by R, r or .

R is the same whether monitoring reactants or products

Generally,

aA + bB cC + dD

dt

Dd

ddt

Cd

cdt

Bd

bdt

Ad

aR

][1][1][1][1

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Example: 2H2 + O2 2H2O

Then, dt

OHd

dt

Od

dt

HdR

][

2

1][][

2

1 222

R has fixed dimensionality:ratio of concentration upon time, i.e. [(amount of material)(volume)-1] [time]-1

common units: mol dm-3 s-1

Compare rate of loss of H2 and rate of loss of O2.

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Graphically

[C]

[A]

Time / s

Co

nc.

/ M

The tangents to the curve are the slopes = Rate

All reaction rates are positive

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Rate laws, rate constants, reaction order

• Consider the simple reaction A + B P

R = ƒ ([A][B])

And, R [A]m [B]n

With the use of a proportionality constant k, which is the rate constant (independent of conc. but dependent on temp.), R = -d[A]/dt = R = -d[A]/dt = k k [A][A]mm [B] [B]nn

Such an equation is called the Such an equation is called the rate lawrate law

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The exponents m and n are the order of the reaction with respect to reactant A and the order of the reaction with respect to reactant B respectively.The order of the reaction = m + n

If m = n =1, then the reaction is first-order in A and first-order in B, but second-order overall, therefore: R = k [A][B]

Hence, 11323

13

) (

]][[

smoldmdmmol

sdmmol

BA

Rk

Units for rate constant

for 2nd order reaction

If first-order overall????

13

13

][

sdmmol

sdmmol

A

Rk

Units for rate constant

for 1st order reaction

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Molecularity

• Molecularity is the number of molecules coming together to react in an elementary step.

• Elementary reactions are simple reactions (described by molecularity)

(a) A Products UNI-molecular reaction

e.g. H2C

H2C CH2

CH3CH

CH2

(b) A + A Products or A + B Products BI-molecular

e.g. CH3I + CH3CH2O- CH3OCH2CH3 + I-

(c) 2A + B P or A + B + C P Ter-molecular

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Molecularity = Order of reaction

• Reaction order is determined by experiment only

• Reaction order is an empirical quantity (values range -2 to 3).

• Can be fractional – found mainly in gas phase

• Can be negative,

m

nnm

A

BkBAkR

][

][][][

A is an inhibitor (decreases the rate)

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Mechanism, Rate – determining step and Intermediates

• Assembly of elementary steps (to give products(s)) is called the reaction mechanism.

e.g. H2 + Cl2 2HCl. HCl is NOT formed in this one step, but proceeds by a series of elementary steps:

Cl2 2Cl•

Cl• + H2 HCl + H •

Cl• + H • HCl

H2 + Cl2 2HCl Overall reaction

Mechanism – arrived at from

theory and experiment

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• Rate-determining step (RDS) is the slowest elementary reaction in the mechanism and controls the overall rate of the reaction.

e.g. A + 2B D + E

mechanism: A + B C + E fast

B + C D slow – rate determining step

A + 2B D + E

C is an intermediate – formed, and then used up in the reaction

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Intermediates –

A + B C ProductsK

K

k

Equilibrium is dynamic, this means Rf = Rr.

Assume k << kr, then slow step is:

A + B C

C Prod. (Slow: RDS)

k

R = k[C] ]][[][ ,]][[

][BAKCthen

BA

CK

Rate = kK[A][B]

Rate = Rate = kk’[A][B]’[A][B] k’ = kK

reverse

forwardeq k

kK

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Deriving the Integrated Rate Expressions

• First-order reactionsFirst-order reactions – –

A B, then the rate of disappearance of A is:k

][][

Akdt

AdR

Rearranging gives:

kdtA

Ad

][

][

At time t = 0, [A] = [A]0

And when t = t, [A] = [A]t

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Integrating:

tA

Adtk

A

Adt

0

][

][ 0 ][

][

ktA tA

A ]][ln[

][

][ 0

xdxx

that

call

ln1

Re

ktAA t )]ln[](ln[ 0

ln[A]ln[A]tt = ln[A] = ln[A]00 - - kktt

Integrated form of the 1st order rate expression

y = c + mx

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Other useful forms

ln[A]t

t / s

-slope = -k

Intercept = ln[A]0

ktA

A t

0][

][ln

-slope = -k

t / s

ln([A]t/[A]0)

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Recall ln[A]t = ln[A]o - kt

Antilog gives:

[A]t = [A]0 e-kt

[A]t

t / s

Intercept = [A]0

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• Second-order reactionsSecond-order reactions – –

Two possible cases:

Case Case II : A + A Products OR 2A Products

Case Case IIII : A + B Products

2][][

2

1Ak

dt

Adr

Rearranging gives:kdt

A

Ad2

][

][2

At time t = 0, [A] = [A]0

And when t = t, [A] = [A]t

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Integrating:

tA

Adtk

A

Adt

0

][

][ 22

][

][0

xx

xdxxdx

x

1

12

1 112

2

2

ktA

tA

A

2][

1][

][ 0

OR kt

A

tA

A

2][

1][

][ 0

ktAA t

2][

1

][

1

0

Integrated form of the2nd order rate expression

y = c + mx

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(1/[A]t) / dm3 mol-1

t / s

slope = 2k

Intercept = 1/[A]0

ktAA t

2][

1

][

1

0

y = c + mx

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What can we conclude about RATE LAWS

versus

INTEGRATED RATE EXPRESSSIONS??

a rate law can tell us the rate of a reaction,

once the composition of the reaction mixture is known

An integrated rate expression can give us the concentration

of a species as a function of time. It can also give us the

rate constant and order of the reaction by plotting the

appropriate graph

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The Study of Half-Lives

• The half-life, t½, of a reaction is the time taken for the concentration of a reactant to fall to half its initial value.

• It is a useful indication of the rate of a chemical reaction.

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• First-order reactionsFirst-order reactions – –

Remember that for a 1st order reaction: ln[A]t = ln[A]0 - kt

At time t = 0, [A] = [A]0

Then at time t = t½ (half-life), [A]t½ = [A]0/2

Substituting into above equation,

ln([A]0/2) = ln[A]o – kt½

ln([A]0/2) – ln[A]0 = -kt½

2/10

0

][

2/][ln kt

A

A

2/12

1ln kt

ln 1 – ln 2 = -kt½, where ln 1 = 0Therefore, ln 2 = kt ½

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Hence,

kt

2ln2/1 or

kt

693.02/1

What is/are the main point(s) to note from this expression??

For a 1st order reaction, the half-life is independent of reactant concentration but dependent on k.

The half-life is constant for a 1st order reaction

time

concentration

[A]0

[A]0/2

[A]0/4

[A]0/8

Recall: [A]t = [A]0e-kt

t1/2

t1/2t1/2

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• Second-order reactionsSecond-order reactions – –

ktAA t

2][

1

][

1

0

At time t = 0, [A] = [A]0

And when t = t½, [A]t½ = [A]0/2

2/100

2][

1

2

][1

ktAA

2/100

2][

1

][

2kt

AA

2/10

2][

1kt

A

02/1 ][2

1

Akt

So t1/2 for 2nd order reactions

depends on initial concentration

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Therefore, larger initial concentrations imply shorter half-lives

(so faster the reaction).

concentration

[A]0

[A]0/2

[A]0/4

[A]0/8

time

t1/2

t1/2

t1/2

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Determining Rate LawsDetermining Rate Laws

Rate laws have to be determined experimentally.

Techniques for monitoring the progress of a reaction include:

Absorption measurements (using a spectrophotometer)

Conductivity (reaction between ions in solution)

Polarimetry (if reactants/products are optically active, e.g. glucose)

Aliquot method (employing titration technique)

RecallA + B P, r = k[A]m[B]n

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(A) Isolation Method:This technique simplifies the rate law by making all the reactants except one, in large excess.

Therefore,

The dependence of the rate on each reactant can be found by isolating each reactant in turn and keeping all other substances (reactants) in large excess.

Using as example: r = k[A]tm [B]t

n

Make B in excess, so [B]>>[A].

Hence, by the end of the reaction [B] would not have

changed that much, although all of A has been used up

And we can say, [B] [B]0

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r = k’[A]tm , where k’ = k[B]0

n

Since A is the reactant that changes, then the rate

becomes dependent on A, and we can say

Created a ‘false’ first-order (imitating first-order) PSEUDO-FIRST-ORDER,

Logging both sides gives:

log r = log k’ + m log [A]t

y = c + m x

A plot of log r vs log [A]t gives a straight line with slope = m,and intercept log k’

where k’ is the pseudo-first-order rate constant

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If m = 1, the reaction is said to be pseudo-first-order

With the roles of A and B reversed, n can be found in a similar manner

k can then be evaluated using any data set along with the

known values of m and n

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(B) Initial Rate Method: - often used in conjunction with the isolation method,

-The rate is measured at the beginning of the reaction

for several different initial concentrations of reactants.

[A]t

t / s

Initial rate

Follow reaction to ~ 10%

completion

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RecallA + B P, Rate0 = k[A]0

a[B]0b

Taking ‘logs’

log Rate0 = log k + a log [A]0 + b log[B]0

y m xc

** Keep [A]0 constant for varying values of [B]0 to find b

Log Ro

log[B]0

slope = b

Intercept = log k + a log[A]0

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** Keep [B]0 constant for varying values of [A]0 to find a from the slope

of the graph, log R0 vs log [A]0

** Substitute values of a, b, [A]0, [B]0 to find k.

However, in some cases, there may be no need to use

the plots as shown previously.

EXAMPLEEXAMPLE

R1 = k[A]a[B]b

R2 = k[nA]a[B]b

Dividing R2 by R1

For these experiments, B is kept constant

while A is varied and R1 and R2 are known.

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ba

ba

BAk

BnAk

R

R

][][

][][

1

2 a

a

A

nA

][

][

a

aa

A

An

][

][ an

naR

Rloglog

1

2

n

R

R

alog

log1

2

(a) If R2 = 2R1, and n=2, then a = 1, so 1st order with respect to A(b) If R2 = 4R1, and n=2, then a = 2, so 2nd order with respect to A

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Concluding: if n=2,

and Rate doubles 1st order

Rate increases by a factor of 4 2nd order

Rate increases by a factor of 9 3rd order

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COLLISION THEORY COLLISION THEORY & ARRHENIUS EQUATION& ARRHENIUS EQUATION

According to the Collision Theory Model: a bimolecular reaction

occurs when two properly orientedproperly oriented reactant molecules come

together in a sufficiently energetic collisionsufficiently energetic collision.

i.e. for a reaction to occur, molecules, atoms or ions must first collide.

Consider the hypothetical reaction: A + BC AB + C

A + BC A----B----C AB + C

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Potential Energy Profile

Ea

A---B---C

A + BC

AB + CReactants

Products

Potential Energy

Reaction Progress

The height of the barrier is called the

activation energy, Ea.

The configuration of atoms at the

maximum in the P.E. profile is called

the transition state.

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**If the collision energy < Ea, the reactant molecules cannot

surmount the barrier and they simply bounce apart.

**If the collision energy is Ea, the reactants will be able to

surmount the barrier and be converted to products.

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Very few collisions are productive because very few occur with a collision energy as large as the activation energy. Also, proper orientation is necessary for product formation.

There must be some effect by Temperature on reaction systems.

Temperature can result in an increase in energy.

This leads us to say: The average kinetic energy of a

collection of molecules is proportional to the absolute temperature.

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At a temperature T1, a certain fraction of the reactant

molecules have sufficient K.E., i.e. K.E. > Ea.

At a higher temperature T2, a greater fraction of the

molecules possess the necessary activation energy,

and the reaction proceeds at a faster rate.

**In fact it has been found that reaction rates tend to double

when the temperature is increased by 10 oC.

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Fraction of molecules

Kinetic Energy

Ea

T1

T2

T2 > T1

(i) The total area under the curve is proportional to

the total # molecules present.

(ii) Total area is the same at T1 and T2.

(iii) The shaded areas represent the number of particles

that exceed the energy of activation, Ea.

Maxwell-Boltzmann distribution curve

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It was observed by Svante Arrhenius that almost all of the

reaction rates (obtained from experiments)

accumulated over a period showed similar

dependence on temperature.

This observation led to the development of the Arrhenius EquationArrhenius Equation:

k = Ae-Ea/RT

Collectively, AA and EEaa are called the Arrhenius parametersArrhenius parameters

of the reaction.

EEaa = activation energy (kJ mol-1), and is the minimum kinetic energy required to allow reaction to occur

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This fraction goes up when T is increased because of the

negative sign in the exponential term.

However, most of the collisions calculated by ee-Ea/RT-Ea/RT do not lead to

products, and so

The exponential term ee-Ea/RT-Ea/RT is simply the fraction of collisions that

have sufficient energy to react.

AA = the frequency factor or pre-exponential factor (same units as k),

is the fraction of sufficiently energetic collisions that actually

lead to reaction.

TT = Kelvin temperature RR = ideal gas constant (8.314 J mol-1 K-1)

kk is the rate constant

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Logarithmic form of the Arrhenius equation:

RT

EAk alnln

y mxc

A plot of ln k versus 1/T gives slope= –Ea/R and intercept= ln A

ln k

1/T

x

y

Cannot extrapolate for intercept. Obtain AA

by substituting one of the data values

along with value of Ea into equation.

Recall : k = Ae-Ea/RT

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High activation energy corresponds to a reaction rate that is

very sensitive to temperature (the Arrhenius plot has a steep slope).

Converse also applies.

ln k

1/T

High

activation

energy

Low activation energy

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Manipulation of Arrhenius equation:

(i) Once the activation energy of a reaction is known, it is a simple

matter to predict the value of a rate constant k’ at a temperature,

T’ from another value of k at another temperature, T.

ln k’ = ln A – Ea/RT’

ln k = ln A – Ea/RTSubtract these equations

ln k’ – ln k = ln A – ln A – Ea/RT’ – (-Ea/RT)

(ii) Can also find Ea if k’, k, T’ and T are known.

'

11'ln

TTR

E

k

k a