22.6 Elementary reactions

• Elementary reactions: reactions which involve only a small number of molecules or ions.

A typical example:

H + Br2 → HBr + Br

• Molecularity: the number of molecules coming together to react in an elementary reaction.

• Molecularity and the reaction order are different !!! Reaction order is an empirical quantity, and is obtained from

the experimental rate law; Molecularity refers to an elementary reaction proposed as an

individual step in a mechanism. It must be an integral.

• An elementary bimolecular reaction has a second-order rate law:

A + B → P

• If a reaction is an elementary bimolecular process, then it has second-order reaction kinetics; However, if the kinetics are second-order, then the reaction might be complex.

]][[][

BAkdt

Ad

22.7 Consecutive elementary reactions

• An example:

239U → 239Np →239Pu

• Consecutive unimolecular reaction

A → B → C

The rate of decomposition of A is:

• The intermediate B is formed from A, but also decays to C. The net formation rate of B is therefore:

• The substance C is produced from the unimolecular decay of B:

][][

Akdt

Ad1

][][][

BkAkdt

Bd21

][][

Bkdt

Cd2

• Integrated solution for the first order reaction (A) is:

• Then one gets a new expression for the reactant B:

the integrated solution for the above equation is:

when assuming [B]0 = 0.

• Based on the conservation law [A] + [B] + [C] = [A]0

tkeAA 10

][][

][ e[A]][

1k-0 Bkk

dt

Bd t21

012

1 21 ])[(][ Aeekk

kB tktk

012

2112

1 ][][ Akk

ekekC

tktk

Example. In an industrial batch process a substance A produces the desired compound B that goes on to decay to a worthless product C, each step of the reaction being first-order. At what time will B be present in the greatest concentration?

Solution: At the maximum value of B

Using the integrated solution of B and taking derivatives with respect to t:

In order to satisfy

= 0

tmax =

The maximum concentration of B can be calculated by plugging the tmax into the equation.

0dt

Bd ][

12

210121

kk

ekekAk

dt

Bd tktk

)(][][

0dt

Bd ][

tktk ekek 2121

2

1

21

1

k

k

kkln

Steady State Approximation• Assuming that after an initial induction period, the rates of change of

concentrations of all reaction intermediates are negligibly small. • Substitute the above expression back to the rate law of B 0 ≈ [B] = (k1/ k2)[A]

• Then

• The integrated solution of the above equation is

0dt

Bd ][

][][ BkAk 21

][][][

AkBkdt

Cd12

tkeAkdt

Cd1

01 ][

][

0)1( 1 AeC tk

In a batch reactor, steady state approximation can only be applied to intermediate products, not the final product!

Self-test 22.8 Derive the rate law for the decomposition of ozone in the reaction 2O3(g) → 3O2(g) on the basis of the following mechanism

O3 → O2 + O k1O2 + O → O3 k1’O + O3 → O2 + O2 k2

Solution: First, write the rate law for the reactant O3 and the intermediate product O

Applying the steady state approximation to [O]

Plug the above relationship back to the rate law of [O3]

]][[]][[][][ '

3221313 OOkOOkOk

dt

Od

]][[]][[][][ '

322131 OOkOOkOkdt

Od

]][[]][[][ '3221310 OOkOOkOk

])[][(][][ '

3221313 OkOkOk

dt

Od

Rate determining step

Simplifications with the rate determining step

• Suppose that k2 >> k1,

then k2 – k1 ≈ k2

therefore concentration C

can be reorganized as

[C] ≈ (1 - )[A]0

• The above result is the same as obtained with steady state approximation

tke 1

tktk ee 12

012

2112

1 ][][ Akk

ekekC

tktk

Kinetic and thermodynamic control of reactions

• Consider the following two reactionsA + B → P1 rate of formation of P1 = k1[A][B]

A + B → P2 rate of formation of P2 = k2[A][B]

• [P1]/[P2] = k1/k2 represents the kinetic control over the proportions of products.

• Problems 22.6 The gas phase decomposition of acetic acid at 1189 K proceeds by way of two parallel reactions:

(1) CH3COOH → CH4 + CO2, k1 = 3.74 s-1

(2) CH3COOH → H2C=C=O + H2O, k2 = 4.65 s-1

What is the maximum percentage yield of the ketene CH2CO obtainable at this temperature.

Pre-equilibrium

• Consider the reaction:

A + B ↔ I → P

when k1’ >> k2, the intermediate product, I, could reach an equilibrium with the reactants A and B.

• Knowing that A, B, and I are in equilibrium, one gets:

and

• When expressing the rate of formation of the product P in terms of the reactants, we get

]][[

][

BA

IK '

1

1

k

kK

]][[][][

BAKkIkdt

Pd22

Self-test 22.9: Show that the pre-equilibrium mechanism in which 2A ↔ I followed by I + B → P results in an overall third-order reaction.

Solution: A + A ↔ I k1, k1’

I + B → P k2

write the rate law for the product P

Because I, and A are in pre-equilibrium

so [I] = K [A]2

Therefore, the overall reaction order is 3.

]][[][

BIkdt

Pd2

2][

][

A

IK

][][][

BAKkdt

Pd 22

Kinetic isotope effect

• Kinetic isotope effect: the decrease in the rate of a chemical reaction upon replacement of one atom in a reactant by a heavier isotope.

• Primary kinetic isotope effect: the kinetic isotope effect observed when the rate-determining step requires the scission of a bond involving the isotope:

with

• Secondary kinetic isotope effect: the variation in reaction rate even though the bond involving the isotope is not broken to form product:

with

eHCk

DCk

)(

)(

2112

/

~

)()(

CD

CH

kT

HCvhc

eHCk

DCk

)(

)(

2112

/

~~

)()}()({

CD

CH

kT

HCvHCvhc

Kinetic isotope effect

22.8 Unimolecular reactions• The Lindemann-Hinshelwood mechanism A reactant molecule becomes energetically excited by collision with

another reactant molecule:A + A → A* + A

The energized molecule, A*, may lose its excess energy by colliding with another molecule:

A + A* → A + A

The excited molecule may shake itself apart to form product PA* → P

The net rate of the formation of A* is

21 ][

*][Ak

dt

Ad

*]][[*][ ' AAk

dt

Ad1

*][*][

Akdt

Ad2

*][*]][[][*][ ' AkAAkAk

dt

Ad21

21

• If the reaction step, A + A → A* + A, is slow enough to be the rate determining step, one can apply the steady-state approximation to A*, so [A*] can be calculated as

Then

The rate law for the formation of P could be reformulated as

Further simplification could be obtained if the deactivation of A* is much faster than A* P, i.e., then

in case

0212

1 *][*]][[][*][ ' AkAAkAk

dt

Ad

21

21

kAk

AkA

][

][*][

'

2'1

221

2 ][

][*][

][

kAk

AkkAk

dt

Pd

*][*]][[' AkAAk 21 21 kAk ]['

][][

'A

k

kk

dt

Pd

1

21

21 kAk ]['

21 ][

][Ak

dt

Pd

• The equation can be reorganized into

• Using the effective rate constant k to represent

• Then one has

])[][

][(

][

2'1

21 AkAk

Akk

dt

Pd

2'1

221

2 ][

][*][

][

kAk

AkkAk

dt

Pd

2'1

21

][

][

kAk

Akkk

][

11

121

'1

Akkk

k

k

The Rice-Ramsperger-Kassel (RRK) model

• Reactions will occur only when enough of required energy has migrated into a particular location in the molecule.

• s is the number of modes of motion over which the energy may be dissipated, kb corresponds to k2

1*

1

s

E

EP

b

s

b kE

EEk

1*

1)(

The activation energy of combined reactions

• Consider that each of the rate constants of the following reactions A + A → A* + AA + A* → A + AA* → P

has an Arrhenius-like temperature dependence, one gets

Thus the composite rate constant also has an Arrhenius-like form with activation energy,

E = E1 + E2 – E1’

Whether or not the composite rate constant will increase with temperature depends on the value of E,

if E > 0, k will increase with the increase of temperature

RTE

RTERTE

eA

eAeA

k

kk/'

//

' '1

21

1

21

1

21

RTEEEeA

AA /)('

'121

1

21

Combined activation energy

• Theoretical problem 22.20

The reaction mechanism

A2 ↔ A + A (fast)

A + B → P (slow)

Involves an intermediate A. Deduce the rate law for the reaction.

• Solution: