1. Chemical kinetics Mrs. Khadijah Hanim bt Abdul Rahman Sem II 2011/2012
Week 12 & 13: 7-18 May 2012

2. Course Outcome
Ability to evaluate problems concerning chemical kinetics, phase diagrams and electrochemistry

3. Topics Covered Experimental Chemical and Kinetics Reactions
First Order Reactions
Second Order Reactions
Reaction Rates and Reaction Mechanisms
Enzyme Catalysis

4. CHEMICAL KINETICS Also called reaction kinetics
Study of the rates & mechanisms of chemical reactions
2 types of reaction;
a)homogeneous – reaction occurs
in 1 phase phase)
b)heterogeneous – reaction occurs
in > phase

5. Experimental Chemical and Kinetics Reactions
Rates of chemical Reactions:
Reaction rates depend on the composition and the temperature of the reaction mixture.
the rate of speed with which a reactant disappears or a product appears.
the rate at which the concentration of one of the reactants decreases or of one of the products increases with time.
The instantaneous rate of reaction is the slope of a tangent to the graph of conc against time.
mol L-1 s-1.

6. Definition of rate
Consider A + 2B  3C + D, where the molar conc of participant J is [J] and the vol of the system is constant.
The rate of consumption of one of the reactants at a given time is d[R]/dt, where R is A or B. rate is positive quantity.
The rate of formation of one of the products (C or D, denote as P) is d[P]/dt. This rate is +ve quantity.

7. From the stoichiometry for reaction
A + 2B  3C + D:
Rate of reaction is related to the rates of change of conc of products and reactants.
For a homogenous reaction, the rate of reaction, v:
Reaction rates of homogenous reactions :
mol dm-3 s-1.

8. Example 1:
The rate of change of molar concentration of CH3 radicals in the reaction
2CH3(g)  CH3CH3(g) was reported as d(CH3)/dt = -1.2 mol dm-3 s-1 under particular conditions.
Calculate the rate of reaction
Calculate the rate of formation CH3CH3.

9. Example 2:
From the figure , determine the rate of decomposition of N2O5 at 1900s.
Note: the rate of reaction can be expressed as the slope of a tangent line.

10. Based on the graph of concentration of reactant vs time,
the slope of a tangent line at t=1900s,

11. The Rate Law for Chemical Reactions
Rate of reaction is proportional to the conc of the reactants raised to a power.
For example, the rate of a reaction may be proportional to the molar conc of 2 reactants A and B:
Each conc raised to the 1st power.
Coefficient kr, rate constant for the reaction.
kr is independent of the conc but depends on the temp.
Experimentally determined equation like this is called the rate law of reaction.

12. Normally, a rate law is an equation that expresses the rate of reaction as a function of the conc
For homogeneous gas-phase reactions, it is often more convenient to express the rate law in terms of partial pressures.

13. Many reactions are found to have rate laws of the form:
The power to which the conc of a product/reactant is raised in rate law of this kind is the order of the reaction.
A reaction , A is first-order and B is first-order.
The overall order of a reaction is the sum of individual orders. Therefore, second-order overall

14. Example 3:
The data of three reactions involving S2O82- and I- were given in the below table.
(i) Use the data to establish the order of reaction with respect to S2O82-, the order with respect to I- & the overall order.

15. (ii) Determine the value of k for the above reaction.
(iii) What is the initial rate of disappearance of S2O82- reaction in which the initial concentrations are
[S2O82] =0.050M & [I-]=0.025M?
(iv) What is the rate of formation of SO42- in Experiment 1?

16. First-order reactions
The integrated form of the first-order rate law: is
Where [A]o is the initial conc of A (at t=0)
If ln([A]/[A]0) is plotted against t, then first- order reaction will give a straight line of slope –kr.

17. Half-lives and time constants
Useful indication of the rate of a first-order chemical reaction is the half-life, t1/2, of a substance.
t1/2 , the time taken for the conc of a reactant to fall to half its initial value.
The time for [A] to decrease from [A]o to ½ [A]o in first order reaction is given:
hence

18. Second-order reactions
Integrated form of the second-order rate law
Is either of the following 2 forms or
For a second-order reaction, 1/[A] against expect to get straight line. The slope of the graph is kr.

19. Summary

21. second order
Zero order
First order

22. Example 4
(a) When [N2O5] =0.44M, the rate of decomposition of N2O5 is 2.6 x 10-4 mol L-1 s-1.
what is the value of k for this first-order reaction?
(b) N2O5 initially at a concentration of 1.0 mol/L in CCl4, is allowed to decompose at 450C. At what time will [N2O5] be reduced to 0.50M?

23. Example 5
The rate constant for the first-order decomposition of a compound A in the reaction 2A  P is kr=3.56 x 10-7 s-1 at 25oC.
What is the half-life of A?
What will be the pressure, initially 33.0 kPa at:
(i) 50 s
(ii) 20 min after initiation of reaction?

24. The temperature dependence of reaction rates
Effects of temperature
Chemical reactions tend to go faster at higher temperature.
slow down some reactions by lowering the temperature.
Increasing the temperature increases the fraction of the molecules that have energies in excess of the activation energy.
The Ea is the minimum kinetic energy required for reaction during a molecular encounter.
this factor is so important that for many chemical reactions it can lead to a doubling or tripling of the reaction rate for a temperature increase of only 100C.

25. The temperature dependence of reaction rates
The rate constants of most reactions increase as the temp is raised.
Experimentally, for many reactions the plot of ln kr against 1/T gives a straight line.
It normally expressed mathematically by introducing 2 parameters, one representing the intercept and other is the slope.
Arrhenius equation or

26. The parameter A correspond to the intercept of the line at 1/T=0 (at infinite temp) is called the pre-exponential factor
Parameter Ea, which is obtained from the slope of the line (Ea/R) is called the activation energy.

27. Ea = slope of the plot of ln kr against 1/T means that: the higher the activation energy, the stronger the temp dependence of the rate constant, kr.
If reaction is Ea=0, its rate is independent of temp.
In some cases, the Ea is –ve, indicates that the rate decreases as the temp is raised.

28. Example 6
The rate constant for the decomposition of a certain substance is 1.7x10-2 dm3 mol-1 s-1 at 24oC and 2.01x10-2 dm3 mol-1s-1 at 37oC. Evaluate the Arrhenius parameters of the reaction.

29. Reaction Mechanisms
Most reactions occur in a sequence of steps called elementary reactions which involves only a small no of molecules or ion. A typical elementary reaction is
H + Br2  HBr + Br
The mechanism of a reaction:
the sequence of elementary reactions that add up to give the overall reaction.
A mechanism is a hypothesis about the elementary steps through which chemical change occurs.

30. The molecularity of an elementary reaction is the no of molecules coming together to react in an elementary reaction.
In unimolecular reaction, a single molecules shakes itself apart or its atoms into new arrangements
In bimolecular reaction, a pair of molecules collide and exchange energy, atoms or groups of atoms or undergo some kind of change.

31. Elementary processes in which a single molecule dissociates (unimolecular) or two molecules collide (bimolecular) much more probable than a process requiring the simultaneous collision of three bodies (termolecular).
All elementary processes are reversible and may reach a steady-state condition. In the steady state the rates of the forward & reverse processes become equal.
One elementary process may occur much more slower than all the others. In this case, it determines the rate at which the overall reaction proceeds & is called the rate-determining/ limiting step.

32. The observed rate law provides information on the mechanism of a reaction, in that any proposed mechanism must yield the observed rate law.
Usually an exact deduction of rate law from differential rate equations is not possible
Therefore, one of two approximation methods is generally used, the rate- determining step or the steady-state approximation.

33. Rate-determining step approximation
The reaction mechanism is assumed to consist of one or more reversible reactions that stay close to equilibrium during most of the reaction, followed by relatively slow-rate determining step, which in turn is followed by one or more rapid reactions.
As an example, consider the following mechanism composed of unimolecular reactions k1 k2 k3
k-1
k-2
k-3

34. Step 2, B ⇌ C is assumed to be the rate- limiting step.
For this assumption to be valid, we must have .
The slow rate of B  C compared to B A ensures that most B molecules go back to A rather than to C, thereby ensuring that A ⇌ B remain close to equilibrium.
To ensure that step 2 acts as bottleneck and that product D is rapidly form from C, we must have and
the overall rate is controlled by the rate- determining step B  C.

35. The Hydrogen-Iodine reaction
H2 (g) + I2 (g) → 2HI (g)
Rate of formation of HI = k [H2][I2]
The hydrogen-iodine reaction is proposed to be a two-step mechanism [Sullivan J. (1967). J.Chem.Phys.46:73].
1st step: iodine molecules are believed to dissociate into iodine atoms.
2nd step: simultaneous collision of two iodine atoms and a hydrogen molecule.
(this termolecular step is expected to occur
much more slowly – the rate-determining step).

36. The first step: rapid equilibrium 2nd step: slow k1
The first step: rapid equilibrium
2nd step: slow
In this reaction the intermediate is atom I and we want to express r in term of reactant and products
Since step 1 is in equilibrium, the forward and reverse reaction is nearly equal.
k-1 k2

37. =
Second step is the rate determining step. So, the rate of formation of HI
Substituting [I]2
where

38. Example 7 The thermal decomposition of ozone to oxygen:
2O3 (g) → 3O2 (g)
The observed rate law: r =
Show the following mechanism is consistent with the experimental rate law.
1st step:
2nd step:

39. Exercise 1 The rate law for the Br- catalyzed aqueous reaction
H+ + HNO2 + C6H5NH2 C6H5N2+ + 2H2O
is observed to be
r =k[H+][HNO2][Br-]
A proposed mechanism is
H+ + HNO2 ⇌ H2NO rapid equilibrium
H2NO2+ + Br-  ONBr + H2O slow
ONBr + C6H5NH2  C6H5N2+ + H2O + Br- fast
Deduce the rate law for this mechanism and relate the observed rate constant k to the rate constants in assumed mechanism k1
k-1 k2 k3

40. Steady-state approximation
Multistep reaction mechanisms usually involved 1 or more intermediate species that do not appear in the overall equation.
For example the species H2NO+ in the Ex. 1 is a reaction intermediate.
Frequently, these intermediates are very reactive therefore do not accumulate to any significant extent during reaction:
and during most of the reaction.

41. we can assume that [I] will start at 0, rise to [I]max and then fall back to 0.
If [I] remains small during the reaction, [I]max will be small compared to [R]max and [P]max.
It is therefore frequently a good approx. to take d[I]/dt=0 for each reactive intermediate. This is the steady-state approximation.
The steady-state approx. assumes that the rate of formation of a reaction intermediate essentially equals its rate of destruction, so as to keep it at a near-constant steady-state approx.

42. Example 8
Apply steady state approximation to Ex. 1, dropping the assumption that step 1 and -1 are in equilib and that step 2 is slow.
Step 1: H+ + HNO2 ⇌ H2NO2+
Step 2: H2NO2+ + Br-  ONBr + H2O
Step 3: ONBr + C6H5NH2  C6H5N2+ + H2O + Br-

43. a) Take the reaction rate, r equal to the rate of formation of a product in the last step of the mechanism

44. The intermediates are ONBr and H2NO2+
b) Eliminate the conc of any reaction intermediates that occur in the rate expression obtained in (a) using d[I]/dt=0 to find the conc of each such intermediate, I
The intermediates are ONBr and H2NO2+
To eliminate the intermediate ONBr from the rate expression, we apply the steady state approx d[ONBr]/dt=0.
The species ONBr is formed by elementary step 2 at rate:
And is consumed by step 3 with:

45. The net rate of change of [ONBr] equals (d[ONBr]2/dt + d[ONBr]3/dt).

46. Substitute this expression for [ONBr] in the above equation for r gives:

47. The species H2NO+ is formed by elementary step 1 at rate:
c) If step (b) introduces conc of other intermediates, apply d[I]/dt=0 to these intermediates to eliminate their conc.
The species H2NO+ is formed by elementary step 1 at rate:
And is consumed by step -1 and 2.
The net rate of change of [H2NO+] equals

48. Substitute to:
To obtain agreement with the observed rate law , r =k[H+][HNO2][Br-] we must further assume that
The assumption means that the rate k1[H2NO2+] of reversion of H2NO2+ back to H+ and HNO2 is much greater than the rate k2 [H2NO2+][Br-] of reaction of H2NO2+ with Br-

49. In summary: To apply rate-determining step:
Take the reaction rate r as equal to the rate determining step
Eliminate the conc of any reaction intermediates that occur in the rate expression obtained in (a) using equilibrium constant expressions.
To apply steady-state approximation:
Take the reaction rate, r equal to the rate of formation of a product in the last step of the mechanism
Eliminate the conc of any reaction intermediates that occur in the rate expression obtained in (a) using d[I]/dt=0 to find the conc of each such intermediate, I
) If step (b) introduces conc of other intermediates, apply d[I]/dt=0 to these intermediates to eliminate their conc.

50. Enzyme Catalysis
Most of the reactions that occur in living organisms are catalyzed by molecules called enzyme.
Enzyme is very specific in its action
Enzymes speed up reaction rates very substantially
In the absence of enzyme most biochemical reactions occur at negligible rates.
Enzyme acts on substrate
The substrate binds to a specific active site on the enzyme to form an enzyme-substrate complex.
While bound to the enzyme, the substrate is converted to product, which is then released from the enzyme.

51. Consider the following mechanism:
E = enzyme, S = substrate, ES=enzyme- substrate complex, P= product.
In most experimental studies on enzyme kinetics, the enzyme conc is much less than the substrate conc
Hence, the conc of the intermediate ES is much less than that S and steady-state approx d[ES]/dt=0 can be used. k1 k2
k-1
k-2

52. Usually, the reaction is followed to only a few percent completion and the initial rate determined.
Therefore, [P] will be very small and we shall neglect step-2.
If [E]o is the initial enzyme conc, then [E]o=[E]+[ES].
Since the enzyme conc [E] during the reaction is unknown while [E]o is known, replace [E] by [E]o-[ES].

53. The reaction rate is r=d[P]/dt=k2[ES] (since step -2 is being neglected, gives:
Setting [S] equal to its initial conc [S]o, we get the initial rate ro,
, Michaelis-Menten equation

54. KM, Michaelis-Menten constant difined as KM ≡ (k-1+k2)/k1.
The reciprocal of Michaelis-Menten equation is
, Lineweaver-Burk equation.
The constant k2 and KM are found from the intercept and slope of a plot of 1/ro vs 1/[S]o, since [E]o is known.