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 ( N A ) = κ ( N A + 1) P ( N A + 1) − κ N A P ( N A ) dt ◮ The average molecules of A at time t is A 0 � � N A � = N A P ( N A ) N A =0 A 0 ∴ d � N A � dP ( N A ) � = N A dt dt N A =0

Case 1: First-order reaction (continued) A 0 ∴ d � N A � � = N A [ κ ( N A + 1) P ( N A + 1) − κ N A P ( N A )] dt N A =0   A 0 − 1 A 0   � � N 2 = κ N A ( N A + 1) P ( N A + 1) − A P ( N A )   N A =0 N A =0   A 0 A 0   � � N 2 = κ ( N A − 1) N A P ( N A ) − A P ( N A )   N A =1 N A =0   A 0 A 0   � � N 2 = κ ( N A − 1) N A P ( N A ) − A P ( N A )   N A =0 N A =0

Case 1: First-order reaction (continued) A 0 ∴ d � N A � � = − κ N A P ( N A ) dt N A =0 = − κ � N A � ◮ If we associate [A] = � N A � / LV , we get d [A] = − κ [A] dt which is the mass-action rate law. ◮ k = κ

Case 1: First-order reaction Variance ◮ The variance (square of the standard deviation) is A 0 A � − � N A � 2 = � σ 2 A = � N 2 N 2 A P ( N A ) − � N A � 2 N A =0 A 0 ∴ d σ 2 dP ( N A ) − 2 � N A � d � N A � � N 2 A = A dt dt dt N A =0

Case 1: First-order reaction Variance (continued) A 0 ∴ d σ 2 � A N 2 A [ κ ( N A + 1) P ( N A + 1) − κ N A P ( N A )] + 2 κ � N A � 2 = dt N A =0   A 0 − 1 A 0   � � N 2 N 3 A P ( N A ) + 2 � N A � 2 = κ A ( N A + 1) P ( N A + 1) −   N A =0 N A =0   A 0 A 0   � � ( N A − 1) 2 N A P ( N A ) − N 3 A P ( N A ) + 2 � N A � 2 = κ   N A =1 N A =0

Case 1: First-order reaction Variance (continued)  A 0 ∴ d σ 2  � ( N 2 A = κ A − 2 N A + 1) N A P ( N A ) dt  N A =0  A 0  � N 3 A P ( N A ) + 2 � N A � 2 −  N A =0   A 0   � N A (1 − 2 N A ) P ( N A ) + 2 � N A � 2 = κ   N A =0 � � N A � − 2 � N 2 A � + 2 � N A � 2 � = κ � N A � − 2 σ 2 � � = κ A

Case 1: First-order reaction Variance (continued) ◮ The variance reaches a maximum when d σ 2 A / dt = 0, i.e. A , max = 1 σ 2 2 � N A � or a standard deviation of � � N A � σ A , max = 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 = σ A , max � N A � = [2 � N A � ] − 1 / 2 ◮ Relatively speaking, fluctuations go to zero when the number of molecules is large.

Case 2: A + B reaction ◮ Here, we will use the fact that for a reaction A + B → , N A − N B = A 0 − B 0 = ∆ is constant. ◮ Master equation: dP ( N A ) = κ ( N A +1)( N A +1 − ∆) P ( N A +1) − κ N A ( N A − ∆) P ( N A ) dt ◮ The average number of A molecules obeys A 0 A 0 d � N A � dP ( N A ) = d � � N A P ( N A ) = N A dt dt dt N A =0 N A =0  A 0  � = κ N A ( N A + 1)( N A + 1 − ∆) P ( N A + 1)  N A =0  A 0  � N 2 − A ( N A − ∆) P ( N A )  N A =0

Case 2: A + B reaction (continued)  A 0 d � N A �  � = κ ( N A − 1) N A ( N A − ∆) P ( N A ) dt  N A =0  A 0  � N 2 − A ( N A − ∆) P ( N A )  N A =0 A 0 � = − κ N A ( N A − ∆) P ( N A ) N A =0 � N 2 � � = − κ A � − ∆ � N A �

Case 2: A + B reaction (continued) d � N A � � N 2 � � = − κ A � − ∆ � N A � dt ◮ Compare the mass-action rate equation d [A] = − k [A][B] = − k [A]([A] − δ ) dt where δ = [A] 0 − [B] 0 . ◮ � N 2 A � � = � N A � 2 ◮ To finish this we need to ◮ show that � N 2 A � ≈ � N A � 2 , and ◮ figure out the relationship between k and κ .

Case 2: A + B reaction Variance   A 0 d σ 2 = d � A N 2 A P ( N A ) − � N A � 2   dt dt N A =0 A 0 − d � N A � 2 dP ( N A ) � N 2 = A dt dt N A =0  A 0  � N 2 = κ A ( N A + 1)( N A + 1 − ∆) P ( N A + 1)  N A =0  A 0  − 2 � N A � d � N A �  � N 3 − A ( N A − ∆) P ( N A ) dt N A =0

Case 2: A + B reaction Variance (continued)  A 0 ∴ d σ 2  � A ( N A − 1) 2 N A ( N A − ∆) P ( N A ) = κ dt  N A =0  A 0  � N 3 − A ( N A − ∆) P ( N A )  N A =0 � � N 2 � + 2 κ � N A � A � − ∆ � N A � � A 0 � = κ (1 − 2 N A ) N A ( N A − ∆) P ( N A ) N A =0 � � N 2 � � + 2 � N A � A � − ∆ � N A �

Case 2: A + B reaction Variance (continued) ∴ d σ 2 A � � N 2 A � (1 + 2∆) − ∆ � N A � − 2 � N 3 = κ A � dt � N 2 � �� + 2 � N A � A � − ∆ � N A � ◮ Using � N 2 A � = σ 2 A + � N A � 2 , we get (after some algebra) d σ 2 1 A (2∆ + 1 + 2 � N A � ) + � N A � 2 − ∆ � N A � A = σ 2 κ dt � N 3 A � − � N A � 3 � � − 2

Case 2: A + B reaction Variance (continued) ◮ The term � N 3 A � − � N A � 3 � � is related to the skewness of the distribution. Assuming this is small, and setting d σ 2 A / dt = 0 to find the point of maximum variance, we get A , max = � N A � 2 − ∆ � N A � σ 2 2∆ + 1 + 2 � N A � ◮ The CV is then � 1 ∆ � N A � − � CV = σ A , max � N A � 2 � � N A � = � 2∆+1 � N A � + 2

Case 2: A + B reaction Variance (continued) � 1 ∆ � N A � − � CV = σ A , max � N A � 2 � � N A � = � 2∆+1 � N A � + 2 ◮ For large � N A � , CV → 0. ◮ Relatively speaking, σ A is not significant in this limit, which proves that � N 2 A � ≈ � N A � 2 .

Case 2: A + B reaction Rate constant ◮ Using the result of the variance calculation, we have d � N A � ≈ − κ � N A � {� N A � − ∆ } dt ◮ To convert to concentrations in molar units, divide both sides by V and by L : d [A] = − κ [A] {� N A � − ∆ } dt = − 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 = 2 k 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.