Complex Chemical Kinetics

Complex chemical reactions may incorporate several features

  1. Parallel Reactions
    A ® B
    A ® C
  2. Competitive Reactions
    A + B ® X
    A + C ® Y
  3. Opposing Reactions

  4. Consecutive Reactions
    A ® B ® C
  5. Catalysis

  6. Feedback


Sequential (Consecutive) Reactions

Let us write down the rate laws for the sequential reaction


 


The equation for [A] is a simple first-order equation whose solution is

We can solve this equation with a simple change of variables:

 

The easiest way to solve the equation for [C] is to use the conservation law instead of integrating (it is still necessary to know the initial conditions):

 


The plots of [A], [B], and [C] appear as follows (when k1 and k2 are similar in magnitude)

Observe the following:

(1)   [A](t) follows first-order kinetics

(2)   [B](t) and [C](t) each consist of two exponentials: one with a positive coefficient and one with a negative coefficient

(3)   Thus, [B](t) exhibits a maximum

(4)   [C](t) is sigmoidal in shape and exhibits a short “lag” or “induction period” before rising like a product of a 1st order reaction

Limiting Behavior of Consecutive Reactions

The solutions to the consecutive reactions scheme can be used to illustrate two very useful approximations that may be applied to more complicated reaction schemes.:

(1)   

k2 >> k1. B is consumed as quickly as it is formed. It is only present at a low, nearly constant concentration during the reaction. In this case, B is said to be present in the steady state. A disappears and C appears with an effective 1st order rate constant k » k1.

(2)    k2 >> k1. Almost immediately, A disappears and [B] approachs [A]0., then B ® C proceeds as a 1st order reaction with rate constant k2.

Note that when one rate constant is significantly slower than the other, the product is formed with that effective rate constant. The slow step is rate determining (or rate limiting).

The Steady-State Approximation

The steady-state approximation applies to reaction intermediates that are  present at low and near constant concentrations. (e.g.,  [B] in the consecutive reaction example when k2 >> k1)

In the SS-approximation, set the overall rate of appearance and disappearance of the intermediate to 0. In above example,

At t=0, [B] = k1[A]0/k2, which is the result obtained by neglecting k1 w.r.t. k2 in the expression for [B](t).

Detailed Balance

Stated by R.H. Fowler (1884-1950)

Each molecular collision has its exact counterpart in the reverse direction. In other words, the existence of a forward reaction implies that the reverse reaction also occurs.

Microscopic Reversibility

Stated by R.C. Tolman (1881-1948)

At equilibrium, any molecular process and its reverse occur at the same average rate.

Rate-Determining Step Approximation

Under this approximation, a reaction consists of one or more reversible reactions that stay close to equilibrium during the reaction, followed by a relatively slow rate-determining step, which may in turn be followed by one or more rapid steps.

Application of RDS approximation


The rate law for the Br- -catalyzed aqueous reaction

is

v = k[H+][HNO2][Br-]

 


A proposed mechanism is:

In this mechanism, reaction 2 is assumed to be the RDS. Since k3 >> k2, then

The step preceding the RDS is in equilibrium, so that

Substitution of the expression for [H2NO2+] into the expression for v gives the observed rate law.

Application of SS Approximation

We now apply the SS approximation to the above reaction and examine how the result differs from that given by the RDS approximation. We may this drop the assumption that reaction 2 is slow relative to reaction 3 and the reverse of reaction 1.


 


Again we start with the rate law for the appearance of product:

Since ONBr is an intermediate, we must eliminate it from the rate law. Making the steady-state approximation for ONBr,

But H2NO2+ is also an intermediate. We repeat the steady-state assumption for this intermediate to eliminate it from the rate expression:

Note that if we neglect k2 w.r.t. k-1 in this expression, we recover the expression given by the RDS approximation. Thus, the SS approximation is a less restrictive assumption than the RDS approximation. Our result tells us that the Br- catalyst may exhibit saturation behavior (or under some circumstances even be an inhibitor!) if reaction 2 is not rate-determining.

Michaelis-Menten Equation


 


This mechanism obeys a conservation law:

[E] + [ES] = [E]0

 

The rate of appearance of product is

We want a rate law with measurable quantities in it. (We can generally measure [E]0, but not [E]).

 

Strategy:
(1) Make steady state approximation for [ES]
(2) Solve steady-state expression for [E] and substitute into conservation law
(3) Solve result for[ES] and substitute into final rate expression

 

KM is called the Michaelis constant. Vmax represents the fastest the reaction can go, in the limit that all the enzyme is in the form [ES] (i.e., saturated  with substrate).

 

A typical experiment to determine these quantities is to measure the initial v as a function of initial substrate concentration:


Plot of

Note the following from the above plot and the rate law:

(1)   When [S] >> KM, the enzyme is saturated. If we neglect KM w.r.t. [S] in the denominator, [S] cancels and the rate expression becomes v » Vmax.

(2)   When [S] << KM, we can neglect [S] in the denominator, and the reaction is pseudo 1st order in [S]: v » [S](Vmax/KM)

(3)   When [S] = KM, v = Vmax/2

Linearized Plot Representations

It is easier to derive the parameters from the MM equation by applying a linearizing transformation and using a linear regression. There are three such transformations. The most commonly used is the Lineweaver-Burke representation because it is least susceptible to errors (the other two involve taking the ratio of two measurements)

 

Lineweaver-Burke

Take inverse of M-M equation, plot 1/v vs 1/[S]:


Eadie-Hofstee


Transform the MM equation as follows; plot v vs. v/[S]:

Hanes-Wolf


Multiply Lineweaver-Burke equation by [S] and plot [S]/v vs. [S]:

 


Inhibition

Many enzyme reactions (and some of the analogous surface reactions) can undergo inhibition. An example of inhibition


 


The conservation law for this mechanism is

[E] + [ES] + [EI] = [E]0

The rate of appearance of product is

Since we cannot generally measure [E], [ES] or [EI], we need to eliminate these from the rate expression. Strategy is similar to straight M-M equation:

 

Strategy:
(1) Make steady state approximations for [ES] and [EI]
(2) Solve steady-state expression for [E] and [EI] and substitute into conservation law
(3) Solve result for[ES] and substitute into final rate expression

 

The result for [E] is the same as for the M-M equation (using d[ES]/dt »0):

The SS approximation for [EI] gives

 

Substituting these expressions into the conservation law gives

which yields the final rate expression

Types of Reaction Inhibition

There are other types of inhibitory action, as listed below.

The various types of inhibition can be identified from a Lineweaver-Burke plot in the presence of different inhibitor concentrations, as summarized below:

 

Type of Inhibition

Slope

Intercept

None

Competitive
(I binds to E in place of S)

Uncompetitive

(I binds to ES)

Non-competitive
(I binds to E or ES)

 


 


 


Chain Reactions

Chain reactions have a number of identifying features:

(1)   Propagation step(s)
The essential feature of a chain reaction is one or more propagation steps that both consume and produce a reactive intermediate called a propagator, which is often a radical species. Typically, the propagators are present in very low concentrations, and reaction products are produced in at least one propagation step. Successive propagation steps form the “chain”.

(2)   Initiation step(s)
One or more initiation step is required to form a propagator for the chain reaction.

(3)   Termination step(s)
The chain reaction may be terminated by one or more reactions that consume a propagator.

 

A classic example of a chain reaction is the gas-phase reaction

H2 + Br2 ® 2HBr

The observed rate law for this reaction includes both a half-integer order and a sum in the denominator:


This expression can be derived from the following chain reaction mechanism

Strategy: Write steady-state approximations for the propagators. They may often be used to simplify the rate expression for appearance of product

 

The steady-state approximations for the propagators are

The rate of appearance of product is

This may be simplified by adding one of the SS expressions to it (in this case, the expression for d[H·]/dt leads to the simpler result):

Adding the two steady-state expressions gives an expression for the intermediate [Br·]:

Substitution into the steady-state expression for [H·] gives

Finally, substitution into the rate expression gives the observed expression

Rules of Thumb for Devising Mechanisms from Rate Laws

1.     Molecularity Elementary reactions are usually unimolecular or bimolecular, very rarely trimolecular, and never of molecularity greater than 3 (this rule includes the reverse of each elementary reaction according to the principle of microscopic reversibility).

2.     Composition of reactants in rate-determining step (r.d.s.)
(a) Rate law r = k[A]a[B]b¼[L]l Ž “total composition” of reactants in r.d.s. is aA + bB + ¼ + lL. (“total composition” means number of each type of reactant atom and total charge)

(b) Rate law  Ž “total composition” of reactants in r.d.s. is aA + bB + ¼ + lL - mM - nN - rR. Also, the species mM, nN, and rR appear as products in preceding equilibria and do not enter into the r.d.s.

3.     Solvent molecules in r.d.s. If order with respect to solvent S is unknown, total reactant composition of r.d.s. may be
 aA + bB + ¼ + lL - mM - nN - rR + xS, where x=0, ±1, ±2,¼

4.     Square root dependence If the rate law has the factor [B]1/2, the mechanism probably involves splitting a B molecule into two species before the r.d.s. (common in chain reactions)

5.     Sum of terms A rate law with a sum of terms in the denominator indicates a mechanism with one or more reactive intermediates to which the steady-state approximation applies.

Example 1

The gas-phase reaction 2NO + O2 ® 2NO2 has the observed rate law v=k[NO]2[O2]. Propose a consistent mechanism.

 

Solution:

By Rule 2(a), the total reactant composition of the R.D.S. is N2O4.

First work out all of the plausible molecules giving this composition as reactants in the RDS. Possible combinations are:

(a)   2NO + O2 ® products

(b)  N2O2 + O2 ® products

(c)   NO3 + NO ® products

(d)  N2 + 2O2 ® products

 

The first possibility is that the net reaction is in fact an elementary reaction, and the rate law reflects a molecularity of three. This is an unlikely candidate. (rule 1)

 

If rxn (b) is the RDS, it must be preceded by a rxn that produces N2O2 from the reactants.

2NO ® N2O2 (fast)

N2O2 + O2 ® 2NO2 (slow)

 

If rxn (c) is the RDS, it must be preceded by a rxn that produces NO3 from the reactants

NO + O2 ® NO3 (fast)

NO3 + NO ® 2NO2 (slow)

 

If rxn (d) is the RDS, it must be preceded by a rxn that produces N2 from the reactants

2NO ® N2 + O2 (fast)

N2 + 2O2 ® 2NO2 (slow)

This is technically a possibility, but is unlikely because the first step has a large positive DG. Examining the thermodynamics of a reaction is sometimes useful for evaluating mechanism candidates.

 

The actual mechanism for this reaction remains unknown.

Example 2

The reaction

 

2Fe(CN)63- + 2I- ® 2Fe(CN)64- + I2

 

has the rate law

 

v = k[Fe(CN)63-]2[I-]2[Fe(CN)64-]-1 .

 

Propose a consistent mechanism.

 

Solution:

Since the Fe(CN)6 ion appears to remain intact in the reaction and only change its oxidation state, a reasonable trial assumption is that it remains as a unit throughout the reaction.

 

By rule 2b, the species Fe(CN)64-, which enters to the -1 power in the rate law must appears as the product of an equilibrium preceding the RDS. The starting reactants Fe(CN)63- and I- are most likely to be the reactants in this pre-equilibrium. Then charge balance leaves 1 electron to be shared with 2 I atoms on the product side of the equilibrium. Since it is unlikely that I radicals will be found in solution, a plausible pre-equilibrium is

 

Fe(CN)63- + 2I- « Fe(CN)64- + I2-     (fast equilibrium)

 

Also by Rule 2b, the composition of the RDS reactants is Fe(CN)6 + 2I. The 2I may be accounted for by the intermediate I2-. Since Fe(CN)64-  appears as a product in the pre-equilibrium but not in the RDS reactants (rule 2b), the other reactant in the RDS must be Fe(CN)63-. The appearance of this species as reactant in both the fast equilibrium and the RDS is also consistent with the 2nd order dependence on [Fe(CN)63-]. These considerations suggest for the RDS:

 

Fe(CN)63- + I2- ® Fe(CN)64- + I2          (slow)

 

Consistency with rate law

Rate of formation of product I2: v = k2[I2-][Fe(CN)63-]

From the pre-equilibrium,

Substitution for [I2-] gives the observed rate law.

 

Example 3

The reaction CH3CHO ® CH4 + CO has the observed rate law v = k[CH3CHO]3/2. Propose a consistent mechanism.

 

Solution:

Since CH3CHO appears to a half-integral power, it is likely that this species is involved in a fast step in which it splits into two species, and that the reaction proceeds via a chain mechanism. Since [CH3CHO] appears at a power greater than ½, it may also be a reactant in a subsequent slow step.

For the fast equilibrium, we look for the weakest bond to break

CH3CHO « CH3 + CHO       fast

One of these radical intermediates must react with another CH3CHO. The only such reaction that can a product is

CH3 + CH3CHO ® CH4 + CH3CO

Note that this reaction produces another reactive species, CH3CO, and therefore could be a propagation step. This would require a second step to re-form CH3 and the other product, CO, from CH3CO.

CH3CO ® CH3 + CO

CH3 is most likely to form a stable compound in the termination step:

CH3 + CH3 ® C2H6

This set of equations does lead to the observed rate law after making the steady-state approximation for [CH3] and [CH3CO].