Topic 16 Kinetics• Rate expressions• Reaction mechanism• Activation energy

A + B → C + D

16.1 Rate expression

• Change in concentration usually affects the rate of reaction

• The change in rate isn’t the same for all reactants (A and B)

• Must be determined by experiment. (Change the concentrations of one reactant and hold the others constant)

A + B → C + D rate =k*[A]a*[B]b

If a reaction involves the reactants A, B etc =>

The Rate expressionRate of reaction = - d[A]/dt = k[A]a[B]b k = rate constant

• Order of reaction: “a” in substance A and “b” in substance B

• Overall order of reaction: a+b

What happens to the rate in the reactionA + B C + …

[A] [B] Reaction rate

Order

Double the concentration

Keep constant No change

20 = 1 Zero order

Double the concentration

Keep constant X 2 21 = 2 First order

Double the concentration

Keep constant X 4 22 = 4 Second order

Reactants A, B and C in four experiments with altering concentrations:

A + B + C => ……

Experiment [A]mol*dm-3

[B]mol*dm-3

[C]mol*dm-3

Initial ratemol*dm-3*s-1

1 0.400 1.600 0.0600 4.86

2 0.800 1.600 0.0600 9.72

3 0.400 0.800 0.0600 4.86

4 0.800 1.600 0.1800 87.5

Compare Experiment 1 and 2:

Initial rate: [A]: 2x[B] and [C] : constant Þ 2X rate => [A]1

The reaction is first order in [A]

Experiment [A]mol*dm-3

[B]mol*dm-3

[C]mol*dm-3

Initial ratemol*dm-3*s-1

1 0.400 1.600 0.0600 4.86

2 0.800 1.600 0.0600 9.72

Compare Experiment 1 and 3:

Initial rate: [A]: constant [B]: ½ [C] : constant Þ same rate => [B]0

The reaction is zero order in [B]

Experiment [A]mol*dm-3

[B]mol*dm-3

[C]mol*dm-3

Initial ratemol*dm-3*s-1

1 0.400 1.600 0.0600 4.86

3 0.400 0.800 0.0600 4.86

Compare Experiment 2 and 4:

Initial rate: [A]: constant [B]: constant [C] : 3XÞ 32 = 9X=> [C]2

The reaction is second order in [C]

Experiment [A]mol*dm-3

[B]mol*dm-3

[C]mol*dm-3

Initial ratemol*dm-3*s-1

2 0.800 1.600 0.0600 9.72

4 0.800 1.600 0.1800 87.5

Conclusion

• Rate = k*[A]1’*[B]0*[C]2 = k*[A]1*[C]2

• Overall order 1+2 = 3• k can be calculated using the data from one of

the experiments above

Exercises

• 1-2 on page 120

The order can also be found in a graph where initial concentration is set against initial rate.

The gradient of the graph => rate of the reaction.

First order reactionsThey show an exponential decrease:

the time to half the concentration is equal to go from ½ to 1/4

Half life, t½ = 0.693/k

16.2 Reaction mechanism

Types of reactions: Molecularity• A Products

Unimolecular• A + B Products Bimolecular

• In a Bimolecular reaction the reactants collide and form an activated complex

Nucleophilic Substitution bimolecular, SN2- topic 10

2 molecules

If we study the reaction: CH3COCH3 + I2 CH3COCH2I + HI

• It could be a bimolecular process with a rate expression rate = k*[CH3COCH3] *[I2]

• The rate is independent of [I2], but first-order in propanone and acid => rate = k*[CH3COCH3]*[H+]

• The reaction must proceed through a series of steps, a mechanism must be found:

H+

CH3

C

OH

H2C CH3

CH2C

O

I

CH3

C

O

H3C

H+

CH3

C

OH

H3C

+

Propanone

Fast equilibrium

CH3

C

OH

H3CCH3

C

OH

H2C

++ H+

Slow-Rate DeterminingStep

+ I2 + HIFast

• CH3C(OH+)CH3 is known as a intermediate, not an activated complex, though it occur at an energy minimum. In the mechanism there will be several activated complexes

CH3

C

OH

H3C

+

Intermediate

Activated complex = Transition state, T

CH3

C

OH

H3C

+

CH3

C

O

H3C

CH3

C

OH

H2C

CH3

CH2C

O

I

Exercises

• 1 and 2 on page 122

16.3 Activation energyRecall: Maxwell-Boltzmann energy distribution curve.

Temperature Average speed

Higher temperature =>More particles with higher speed => Greater proportion of particles with energy enough

to react

The Arrhenius equation

• The rate constant, k, can be given if collision rate and orientation is given

• Ea = activation energy• T = temperature, K• R = Gas constant

The equation can also be given in a logarithmic form:

Exersize: Consider the following graph of ln k against (temperature in Kelvin) for the second order decomposition of N2O into N2 and O

N2O → N2 + O

(a) State how the rate constant, k varies with temperature, T(b) Determine the activation energy, Ea, for this reaction.(c) The rate expression for this reaction is rate = k [N2O]2 and the rate constant is0.244 dm3 mol–1 s–1 at 750 °C.A sample of N2O of concentration 0.200 mol dm–3 is allowed to decompose. Calculate the rate when 10 % of the N2O has reacted.

Solution:(a) State how the rate constant, k varies with temperature, TThe Arrhenius equation (it’s in the Data booklet): k=Ae(-Ea/RT)

Logaritming the equation on both sides:lnk=lnA –Ea/RT In the graph we see that the gradient is negativeAnswer: when k increases T decreases

(b) Determine the activation energy, Ea, for this reactionThe gradient (-Ea/R) can be calculated from the graph:DY/DX= -3/0,1*10-3= -3*104= -30000

Therefore:Ea=gradient*R=-30000*8.31= 2,49*105

Answer: Ea= 2,49*105

Solution:(c) The rate expression for this reaction is rate = k [N2O]2 and the rate constant is0.244 dm3 mol–1 s–1 at 750 °C.

A sample of N2O of concentration 0.200 mol dm–3 is allowed to decompose. Calculate the rate when 10 % of the N2O has reacted.

10 % has reacted → 90 % left → 0.200*0.9= 0.180 mol/dm3

Rate= 0.244 * [0.180]2 = 7.91*10-3