Transition States and Reaction Rates

Figure 11.2 depicts a plot of the free energy of the system versus the reaction coordinate for a chemical reaction. This simplistic view illustrates that the standard state free energy of the products is lower than that of the reactants. It shows little of what happens in the transition from reactant to product.

Reactions do not occur all at once. A molecule in a first-order reaction must only occasionally reach an energy state in which the process can occur-otherwise all molecules would react at once. In reality, only a certain fraction of the molecules which are sufficiently energetic can undergo reactions. The same is true in second-order reactions.

Reactions have free energy "barriers" to them. An activated or "transition state" for a molecule occurs when it has reached an energy that is sufficient to react. Figure 11.2b more accurately depicts the free energy barrier for a reaction. Figure 11.2c shows the same conversion for an actual molecular alteration-a boat --> chair conversion of a pyranose ring. Here the transition state is drawn as a half-boat/half-chair structure.

As shown in Figures 11.2b and c, the transition state (symbolized by ) represents an intermediate molecular state having an increase in free energy of above that of the initial substrate. If the barrier to reaction is high, only a small fraction of the molecules will have enough energy to react.

If we let [A] represent the concentration of molecules having the activation energy, then we can write the equilibrium constant K as [A]/[A]. Substituting into equation 3.23, we obtain

> [A] = [A]e-/RT

Because only molecules at the transition state can proceed to react (in either direction), the rate constant can be then expressed as proportional to the population of the transition state:

where Q represents the frequency of forming the product (right side of reaction equation).

Substituting = H -TS, we obtain

> ,

which is known at the Arrhenius equation. Q' is a constant equal to

> Qe-(S0/R) ,

so we take the natural logarithms of both sides and simplify to obtain

Since we expect H to be positive, reactions should go faster at higher temperatures. From equation 11.11, a graph of ln k versus 1/T should be a straight line and its slope should give H. An example (the reaction of L-malate to yield fumarate and water) is shown in Figure 11.3.

According to Figure 11.2b, the activation energy opposes the reaction in both directions, with and representing the transition state free energy barrier in the forward and reverse directions, respectively. Incorporating this and K = k 1/k-1,

Because - = , equation 11.12 simplifies to equation 3.25.

Thus, the equilibrium constant K tells nothing about the rates of process. K depends only on the free energy difference between final and initial states and carries no information about the height of the barrier between these states.

Summary of important points:

> 1. Reactions should go faster at higher temperatures.

> 2. The activation energy opposes the reaction in both directions.

> 3. K reveals nothing about the rates of process-it depends only on the free energy difference between final and initial states and carries no information about the height of the barrier between these states.

See also: Free Energy Change and the Equilibrium Constant (from Chapter 3), How Catalysts Work

INTERNET LINKS:

> 1. Characteristics of Enzymes

> 2. Enzyme Outline