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Chemistry LibreTexts

Elementary Steps

The mechanism of a reaction is a series of steps leading from the starting materials to the products. After each step, an intermediate is formed. The intermediate is short-lived, because it quickly undergoes another step to form the next intermediate. These simple steps are called elementary reactions. Because an overall reaction is composed of a series of elementary reaction, the overall rate of the reaction is somehow dependent on the rates of those smaller reactions. But how are the two related? Let's look at two cases. We'll keep it simple and both cases will be two-step reactions.

Case 1

The first case begins with a reaction having a pretty low activation barrier. Maybe there is a lot of bond-making involved at the beginning, so this reaction gets an easy start. However, after that, things get harder. The second step has a higher activation barrier.Both steps occur at particular rates. The first step has a low barrier, so it occurs quickly. The second step has a high barrier, so it happens only slowly.

In that case, the intermediate will be formed quickly and it will sit around for a while before the second step has a chance to happen. The second step is like a bottleneck or a traffic jam. No matter how quickly things were going in the first step, the reaction has to wait to get through the second step. The rate of the overall reaction really depends on the second step. We call this step the "rate-determining step".

Each step has its own rate constant associated with it. We'll call these constants \(k_1\) and \(k_2\). The lower the barrier, the faster the rate, and the larger the rate constant.We can predict rate laws for elementary reactions (although the same thing isn't true for overall reactions). The rate law for an elementary reaction is simply its rate constant, times the concentrations of any species involved in that step.

The rate of formation of product will be determined by the rate of that second step, the rate-determining step. The rate of that second step is the rate constant, \(k_2\), times the concentration of the intermediate.

\[ {d[Product] \over dt} = k_2[Intermediate] \]

But what is the concentration of the intermediate?  The intermediate is produced in the first step, so its concentration depends on how quickly it is produced from the reactants. However, if the activation barrier is low, that first step may be reversible; it may really be an equilibrium reaction. So there is a third step in this reaction, and it goes backwards from the intermediate to the reactant. We'll call the rate constant for this step k-1.

In terms of kinetics, an equilibrium is really just a ratio of forward and reverse steps. If the forward step is much faster, this ratio is bigger than one, and products are favored. If the reverse step is much faster, the ratio is smaller than one, and reactants are favored.

\[ K_{eq} = {k_1 \over \kappa _1} = {[Intermediate] \over [Reactant] } \]

Or, rearranging,

\[ {k_1\over \kappa _1} [Reactant] = [Intermediate]\]

Now we know the concentration of the intermediate. That means the rate of the overall reaction is

\[ {d[Product] \over dt} = {k _2k _1 \over \kappa _1[Reactant]} \]

The take-home lesson is this: the second step is the rate determining step, so that step tells us how quickly the product forms. However, that step depends on an intermediate formed in an earlier step, so that earlier step also influences how quickly the product forms.

  • The rate-determining step controls the rate of the reaction.
  • The steps prior to the rate determining step influence the rate of the reaction by supplying intermediates needed for the rate determining step.

Case 2

Now consider the opposite case with the first step having a very high barrier and the second step has a lower one. We have to wait and wait and wait for the intermediate to be produced, but once it's there, it reacts pretty quickly to give products. In this case, the first step is the rate-determining step. That is the step that controls the rate of the reaction and as soon as that step is over, the rest of the reaction can occur quickly. In this case,

\[ {d[Product] \over dt} = k_1[Reactant]\]

Note

Any steps after the rate-determining step do not influence the rate of the reaction.

Problems

RK7.1

Consider the borohydride reduction of a ketone in methanol.

  1. Draw a mechanism for this reaction. (If you know about the competing reaction between borohydride and methanol, ignore it. If you didn't know anything about that, forget I said anything.)
  2. Assign rate constants (k1, k2, k3...) for each elementary step in the reaction.
  3. One of the steps along this reaction is probably reversible. Which one?  Why?
  4. What do you think is the rate-determining step in this reaction?
  5. Draw a reaction progress diagram for this reaction.
  6. Predict a rate law for this reaction.

RK7.2

Consider the ammonolysis of acetyl chloride.

  1. Draw a mechanism for this reaction.
  2. Assign rate constants (k1, k2, k3...) for each elementary step in the reaction.
  3. Draw alternative reaction progress diagrams for this reaction. In each case, assume a different rate determining step.
  4. Predict the rate law corresponding to each of your possible reaction progress diagrams.
  5. Consider your alternative reaction progress diagrams and decide which one is most likely. Justify your choices for eliminating the other one(s).
  6. Predict a rate law for this reaction.