Welcome back...continuing from earlier with (equilibrium) constant calculations and how to define.So now let's move ahead to some reactional stages.
A situation that is only slightly more complicated than the sequential reaction scheme described in last part-3
A + B ---k1 → C---k2 → D
← k1---
d[A] / dt = d[B] / dt = -k1[A][B] + k1[C]
d[C] / dt = k1[A][B] - k2[C]
d[D] / dt = k2[C]
These cannot be solved analytically, and in general would have to be integrated numerically to
obtain an accurate solution. However, the situation simplifies considerably if k-1 >> k2. In this case,
an equilibrium is reached between the reactants A and B and the intermediate C, and the
equlibrium is only perturbed very slightly by C ‘leaking away’ very slowly to form the product D.
If we assume that we can neglect this perturbation of the equilibrium, then once equilibrium is
reached, the rates of the forward and reverse reactions must be equal. i.e.
k1[A][B] = k-1[C]
Rearranging this equation, we find
k1 / k-1 = [C] / [A][B] = K
The equilibrium constant K is therefore given by the ratio of the rate constants k1 and k-1 for the
forward and reverse reactions. The rate of the overall reaction is simply the rate of formation of the
product D, so
ν = d[D] / dt = k2[C] = k2K[A][B]
The reaction therefore follows second order kinetics, with an effective rate constant keff = k2K. Note
that this rate law will not hold until the equilibrium between A, B and C has been established, and
so is unlikely to be accurate in the very early stages of the reaction.
Apart from the two simple examples described above, the rate equations for virtually all complex reaction mechanisms generally comprise a complicated system of coupled differential equations that cannot be solved analytically. In state-of-the-art kinetic modelling studies, fairly sophisticated software is generally used to obtain numerical solutions to the rate equations in order to determine the time-varying concentrations of all species involved in a reaction sequence. However, very good approximate solutions may often be obtained by making simple assumptions about the nature of reactive intermediates.
Almost by definition, a reactive intermediate R will be used up virtually as soon as it is formed, and therefore its concentration will remain very low and essentially constant throughout the course of the reaction. This is true at all times apart from at the very start of the reaction, when [R] must necessarily build up from zero to some small non-zero value, and at the very end of the reaction in the case of a reaction that goes to completion, when [R] must return to zero. During the period of time when [R] is essentially constant, because d[R]/dt is so much less than the rates of change of the reactant and product concentrations, it is a good approximation to set d[R]/dt = 0. This is known as the steady state approximation.
Steady state approximation: if a reactive intermediate R is present at low and constant concentration throughout (most of) the course of the reaction, then we can set d[R]/dt = 0 in the rate equations.
As we shall see, applying the steady state approximation has the effect of converting a mathematically intractable set of coupled differential equations into a system of simultaneous algebraic equations, one for each species involved in the reaction. The algebraic equations may be solved to find the concentrations of the reactive intermediates, and these may then be substituted back into the more general equations to obtain an expression for the overall rate law.
A number of gas phase reactions follow first order kinetics and apparently only involve one chemical species. Examples include the structural isomerisation of cyclopropane to propene, and the decomposition of azomethane
10.Pre-equilibria
A situation that is only slightly more complicated than the sequential reaction scheme described in last part-3
A + B ---k1 → C---k2 → D
← k1---
d[A] / dt = d[B] / dt = -k1[A][B] + k1[C]
d[C] / dt = k1[A][B] - k2[C]
d[D] / dt = k2[C]
These cannot be solved analytically, and in general would have to be integrated numerically to
obtain an accurate solution. However, the situation simplifies considerably if k-1 >> k2. In this case,
an equilibrium is reached between the reactants A and B and the intermediate C, and the
equlibrium is only perturbed very slightly by C ‘leaking away’ very slowly to form the product D.
If we assume that we can neglect this perturbation of the equilibrium, then once equilibrium is
reached, the rates of the forward and reverse reactions must be equal. i.e.
k1[A][B] = k-1[C]
Rearranging this equation, we find
k1 / k-1 = [C] / [A][B] = K
The equilibrium constant K is therefore given by the ratio of the rate constants k1 and k-1 for the
forward and reverse reactions. The rate of the overall reaction is simply the rate of formation of the
product D, so
ν = d[D] / dt = k2[C] = k2K[A][B]
The reaction therefore follows second order kinetics, with an effective rate constant keff = k2K. Note
that this rate law will not hold until the equilibrium between A, B and C has been established, and
so is unlikely to be accurate in the very early stages of the reaction.
11.The steady state approximation
Apart from the two simple examples described above, the rate equations for virtually all complex reaction mechanisms generally comprise a complicated system of coupled differential equations that cannot be solved analytically. In state-of-the-art kinetic modelling studies, fairly sophisticated software is generally used to obtain numerical solutions to the rate equations in order to determine the time-varying concentrations of all species involved in a reaction sequence. However, very good approximate solutions may often be obtained by making simple assumptions about the nature of reactive intermediates.
Almost by definition, a reactive intermediate R will be used up virtually as soon as it is formed, and therefore its concentration will remain very low and essentially constant throughout the course of the reaction. This is true at all times apart from at the very start of the reaction, when [R] must necessarily build up from zero to some small non-zero value, and at the very end of the reaction in the case of a reaction that goes to completion, when [R] must return to zero. During the period of time when [R] is essentially constant, because d[R]/dt is so much less than the rates of change of the reactant and product concentrations, it is a good approximation to set d[R]/dt = 0. This is known as the steady state approximation.
Steady state approximation: if a reactive intermediate R is present at low and constant concentration throughout (most of) the course of the reaction, then we can set d[R]/dt = 0 in the rate equations.
As we shall see, applying the steady state approximation has the effect of converting a mathematically intractable set of coupled differential equations into a system of simultaneous algebraic equations, one for each species involved in the reaction. The algebraic equations may be solved to find the concentrations of the reactive intermediates, and these may then be substituted back into the more general equations to obtain an expression for the overall rate law.
12.‘Unimolecular’ reactions – the Lindemann-Hinshelwood mechanism
A number of gas phase reactions follow first order kinetics and apparently only involve one chemical species. Examples include the structural isomerisation of cyclopropane to propene, and the decomposition of azomethane
(CH2N2CH3 → C2H6 + N2,
with experimentally determined rate law
ν = k[CH3N2CH3])
The mechanism by which these molecules acquire enough energy to react remained a puzzle for some time, particularly since the rate law seemed to rule out a bimolecular step. The puzzle was solved by Lindemann in 1922, when he proposed the following mechanism for ‘thermal’ unimolecular reactions..
....continued
