Foundations of Chemical Kinetics Lecture 24: Relationship between - - PowerPoint PPT Presentation

Foundations of Chemical Kinetics Lecture 24: Relationship between the chemical master equation and mass-action kinetics

Marc R. Roussel Department of Chemistry and Biochemistry

The chemical master equation vs mass-action kinetics

◮ At least some aspects of the chemical master equation must

be right: Chemical reactions occur randomly and we therefore expect fluctuations around the equilibrium composition.

◮ On the other hand, we know from experience that mass-action

kinetics gives a very good account of the rates of chemical reactions for larger (lab-scale) systems.

◮ In physics, theories that are special cases are contained in

more general theories. For example, Newtonian mechanics arises from quantum mechanics in the limit of large mass.

◮ Does the chemical master equation reduce to mass-action

kinetics in the limit of a large number of molecules?

Case 1: First-order reaction

◮ The master equation for the first-order reaction A → is

dP(NA) dt = κ(NA + 1)P(NA + 1) − κNAP(NA)

◮ The average molecules of A at time t is

NA =

A0

• NA=0

NAP(NA) ∴ dNA dt =

A0

• NA=0

NA dP(NA) dt

Case 1: First-order reaction (continued)

∴ dNA dt =

A0

• NA=0

NA [κ(NA + 1)P(NA + 1) − κNAP(NA)] = κ   

A0−1

• NA=0

NA(NA + 1)P(NA + 1) −

A0

• NA=0

N2

AP(NA)

   = κ   

A0

• NA=1

(NA − 1)NAP(NA) −

A0

• NA=0

N2

AP(NA)

   = κ   

A0

• NA=0

(NA − 1)NAP(NA) −

A0

• NA=0

N2

AP(NA)

  

Case 1: First-order reaction (continued)

∴ dNA dt = −κ

A0

• NA=0

NAP(NA) = −κNA

◮ If we associate [A] = NA/LV , we get

d[A] dt = −κ[A] which is the mass-action rate law.

◮ k = κ

Case 1: First-order reaction

Variance

◮ The variance (square of the standard deviation) is

σ2

A = N2 A − NA2 = A0

• NA=0

N2

AP(NA) − NA2

∴ dσ2

A

dt =

A0

• NA=0

N2

A

dP(NA) dt − 2NAdNA dt

Case 1: First-order reaction

Variance (continued) ∴ dσ2

A

dt =

A0

• NA=0

N2

A [κ(NA + 1)P(NA + 1) − κNAP(NA)] + 2κNA2

= κ   

A0−1

• NA=0

N2

A(NA + 1)P(NA + 1) − A0

• NA=0

N3

AP(NA) + 2NA2

   = κ   

A0

• NA=1

(NA − 1)2NAP(NA) −

A0

• NA=0

N3

AP(NA) + 2NA2

  

Case 1: First-order reaction

Variance (continued) ∴ dσ2

A

dt = κ   

A0

• NA=0

(N2

A − 2NA + 1)NAP(NA)

−

A0

• NA=0

N3

AP(NA) + 2NA2

   = κ   

A0

• NA=0

NA(1 − 2NA)P(NA) + 2NA2    = κ

• NA − 2N2

A + 2NA2

= κ

• NA − 2σ2

A

Case 1: First-order reaction

Variance (continued)

◮ The variance reaches a maximum when dσ2

A/dt = 0, i.e.

σ2

A,max = 1

2NA

• r a standard deviation of

σA,max =

• NA

2

◮ The coefficient of variation (CV) is the standard deviation

divided by the average. The coefficient of variation at the maximum standard deviation is therefore CV = σA,max NA = [2NA]−1/2

◮ Relatively speaking, fluctuations go to zero when the number

• f molecules is large.

Case 2: A + B reaction

◮ Here, we will use the fact that for a reaction A + B →,

NA − NB = A0 − B0 = ∆ is constant.

◮ Master equation:

dP(NA) dt = κ(NA+1)(NA+1−∆)P(NA+1)−κNA(NA−∆)P(NA)

◮ The average number of A molecules obeys

dNA dt = d dt

A0

• NA=0

NAP(NA) =

A0

• NA=0

NA dP(NA) dt = κ   

A0

• NA=0

NA(NA + 1)(NA + 1 − ∆)P(NA + 1) −

A0

• NA=0

N2

A(NA − ∆)P(NA)

  

Case 2: A + B reaction (continued)

dNA dt = κ   

A0

• NA=0

(NA − 1)NA(NA − ∆)P(NA) −

A0

• NA=0

N2

A(NA − ∆)P(NA)

   = −κ

A0

• NA=0

NA(NA − ∆)P(NA) = −κ

• N2

A − ∆NA

Case 2: A + B reaction (continued)

dNA dt = −κ

• N2

A − ∆NA

• ◮ Compare the mass-action rate equation

d[A] dt = −k[A][B] = −k[A]([A] − δ) where δ = [A]0 − [B]0.

◮ N2

A = NA2

◮ To finish this we need to ◮ show that N2

A ≈ NA2, and

◮ figure out the relationship between k and κ.

Case 2: A + B reaction

Variance dσ2

A

dt = d dt  

A0

• NA=0

N2

AP(NA) − NA2

  =

A0

• NA=0

N2

A

dP(NA) dt − dNA2 dt = κ   

A0

• NA=0

N2

A(NA + 1)(NA + 1 − ∆)P(NA + 1)

−

A0

• NA=0

N3

A(NA − ∆)P(NA)

   − 2NAdNA dt

Case 2: A + B reaction

Variance (continued) ∴ dσ2

A

dt = κ   

A0

• NA=0

(NA − 1)2NA(NA − ∆)P(NA) −

A0

• NA=0

N3

A(NA − ∆)P(NA)

   + 2κNA

• N2

A − ∆NA

• = κ

A0

• NA=0

(1 − 2NA)NA(NA − ∆)P(NA) + 2NA

• N2

A − ∆NA

Case 2: A + B reaction

Variance (continued) ∴ dσ2

A

dt = κ

• N2

A(1 + 2∆) − ∆NA − 2N3 A

+ 2NA

• N2

A − ∆NA

• ◮ Using N2

A = σ2 A + NA2, we get (after some algebra)

1 κ dσ2

A

dt = σ2

A (2∆ + 1 + 2NA) + NA2 − ∆NA

− 2

• N3

A − NA3

Case 2: A + B reaction

Variance (continued)

◮ The term

• N3

A − NA3

is related to the skewness of the distribution. Assuming this is small, and setting dσ2

A/dt = 0 to find the

point of maximum variance, we get σ2

A,max = NA2 − ∆NA

2∆ + 1 + 2NA

◮ The CV is then

CV = σA,max NA =

• 1

NA − ∆ NA2 2∆+1 NA + 2

Case 2: A + B reaction

Variance (continued) CV = σA,max NA =

• 1

NA − ∆ NA2 2∆+1 NA + 2

◮ For large NA, CV → 0. ◮ Relatively speaking, σA is not significant in this limit, which

proves that N2

A ≈ NA2.

Case 2: A + B reaction

Rate constant

◮ Using the result of the variance calculation, we have

dNA dt ≈ −κNA {NA − ∆}

◮ To convert to concentrations in molar units, divide both sides

by V and by L: d[A] dt = −κ[A] {NA − ∆} = −LV κ[A] {[A] − δ} where δ = ∆/LV .

◮ Comparing to the mass-action form, this implies that

κ = k LV

◮ The stochastic rate constant κ depends on the system volume.

Conclusions

◮ If we do this work again for a reaction of the type A + A →,

we find κ = 2k LV

◮ As a rule, stochastic kinetics gives mass-action kinetics in the

limit of a large system.

◮ Except for first-order reactions, we have to make some

assumptions about the distribution to get this result (e.g. negligible skewness).

◮ This relationship only holds for systems with relatively simple

behavior. In particular, in open systems, we can sometimes see qualitative differences between the behavior of a stochastic system and of its mass-action counterpart.