K inetics M echanism S imulation E xamples

Consecutive Reactions

The concentration of reactive intermediates is small and relatively constant during the course of a reaction. A reactive intermediate is a species that is typically formed slowly, but after being formed reacts very rapidly. The steady state approximation, d[B]/dt = 0, for B the reactive intermediate, is based on this idea. This example explores the formation of a reactive intermediate using simple first order irreversible reactions.

Set up the following reactions and rate constants:
 A = X kf = 0.02 kr = 0 X = P kf = 0.2 kr = 0
Set the Initial conditions: Ao=1. Plot A, X, and P for a 150 sec maximum time. The concentration of X will remain remarkably constant during the course of the reaction.

Kinetic vs. Thermodynamic Control

The production of products in chemical reactions can be under kinetic or thermodynamic control. Kinetic control results from product sampling at short time intervals. At long times, reactions are under thermodynamic control. This example explores kinetic verses thermodynamic control for simple first order reversible reactions. Remember that the equilibrium constant for a reaction is Keq = kf/ kr.

Set up the following reactions and rate constants:
 A = X kf = 0.02 kr = 0.0005 A = Y kf = 0.5 kr = 1.5
Set the Initial conditions: Ao=1. Plot A, X, and Y for a 150 sec maximum time. The equilibrium constant for the first reaction is 40, and for the second is 0.333. For short times, Y will be the major product. For long times, X will be the major product. In other words, Y is under kinetic control and X is under thermodynamic control.

Michaelis-Menton Enzyme Kinetics

Many, but not all, enzyme catalyzed reactions follow a pre-equilibrium mechanism. A simple example is the Michaelis-Menton mechanism:
S + E = [ES]
[ES] -> P + E
where E is the enzyme, S is the substrate, [ES] is the enzyme substrate complex, and P is the product. The enzyme-substrate complex is formed in an initial pre-equilibrium step and the enzyme-substrate complex then reacts to form products. The decomposition of the enzyme-substrate complex is assumed to be irreversible. Enzyme kinetics studies often use intial rates. Initial rates are particularly useful, because for this mechanism the initial time course for the formation of products is linear over a long period of time. The linear region is established quickly after an induction lag.

In our Michaelis-Menton simulation, we will use the symbols:
A + B = [X]
[X] -> P + B
where B is the enzyme, A is the substrate, X is the enzyme substrate complex, and P is the product. Many reactions follow a pre-equilibrium mechanism, so this simulation is useful for more than enzyme kinetics.

Set up the following reactions and rate constants:
 A + B = X kf = 0.4 kr = 0.1 X = P + B kf = 0.1 kr = 0
Set the Initial conditions: Ao=1, Bo=0.1. Plot A, X, and P for a 150 sec maximum time. You will see the product concentration increasing linearly. Change to a 300 sec maximum time. You will see that eventually the product line begins to curve as the reaction goes to completion. To see the induction lag, change the plot variables to just X and P and set 15 sec maximum time. Use a ruler (or the edge of a piece of paper) to extrapolate the product curve backwards to [P]=0. Notice that the extrapolated product curve doesn't go through t=0. This time delay is the induction lag. The induction lag is caused by the initial time that it takes for the enzyme- substrate complex, X, to build up in concentration through the pre-equilibrium step. Once the enzyme-substrate complex is formed in sufficient concentration, then the second step can proceed.

In this above example, the second step is slow. But, the second step does not need to be slow for the mechanism to be valid. In addition, the enzyme was 1/10 th the concentration of the initial substrate, which is large for many realistic processes in cells. Try the following alternative set of reaction condidtions, and then compare with the previous example. In the new conditions the decomposition of the enzyme-substrate complex, X, is fast and the substrate to enzyme ratio is 100.
 A + B = X kf = 2 kr = 0.05 X = P + B kf = 2 kr = 0
Set the Initial conditions: Ao=1, Bo=0.01. Plot A, X, and P for a 75 sec maximum time. You will still see the product concentration increasing linearly. The induction lag is harder to see in this example, however.

Lotka-Volterra Mechanism

Most chemical reactions progess uniformly to equilibrium, often exponentially. However, some reactions show the surprising ability to oscillate. See P. W. Atkins, "Physical Chemistry," section 26.8 for a detailed discussion. Reactions far from equilibrium that have an autocatalytic step and two stable states can oscillate. An auto-catalytic step is a step in a mechanism where the product of a step catalyses the production of more of that same product. A simple mechanism for an oscillating reaction is the Lotka-Volterra mechanism. The reactions occur in a constant flow reactor, where one of the reactants is supplied continuously. We will call this "reservoir" species M (Atkins calls the reservoir species A). The two intermediates that oscillate will be called A and B.

Set up the following reactions and rate constants:
 A + M = 2A kf = 0.1 kr = 0 A + B = 2B kf = 0.1 kr = 0 B = P kf = 0.05 kr = 0
Set the Initial conditions: Ao=1, Bo=1, and Mo=1. Plot A and B for a 300 sec maximum time. The concentrations of A and B will oscillate. The animation (click Initialize and then click repeadedly on Step) for this mechanism is fun. Question: how sensitive is the oscillation period to the initial concentrations and the rate constants?

Notes

To rescale the plot, after entering a new axis limit in one of the dialog boxes, click on the background.

Colby Chemistry, T. W. Shattuck, 11/29/01