Reaction mechanisms, intermediates and the rate determining step

There are many reactions which happen as the stoichiometric (chemical) equation would suggest …

e.g. the reaction between chlorine radicals and ozone in the stratosphere

O₃ + Cl^• ⇾ ClO^• + O₂

rate = k [O₃][Cl^•]

e.g. the nucleophilic substitution of a primary halogenoalkane

CH₃Br + OH^– ⇾ CH₃OH + Br^–

rate = k [CH₃Br] [OH^–]

In both of these examples the reaction proceeds as a result of a successful collision between the two reactants and the rate equation is written directly from the chemical equation. These are often called elementary reactions.

However many reactions are more complex, such as the reaction between bromate(V) ions and bromide ions in acidic solution

BrO₃^– + 6H⁺ + 5Br^– ⇾ 3Br₂ + 3H₂O

Complex reactions take place via a series of steps (a set of elementary reactions) as the likelihood of 12 ions colliding successfully at the same time to make 6 product molecules seems virtually impossible!

In a complex reaction each step produces an intermediate molecule or ion which is used up in the subsequent step (and so doesn’t feature in the stoichiometric equation). Put all these steps together and we have reaction mechanism.

Intermediates are real molecules although they are neither reactants nor products. They tend to be reactive and short lived but it is really important not to confuse them with transition states. A transition state is best thought of as an arrangement of atoms the reaction passes through on route from reactant to product (such as a change in geometry).

The transition state is the highest energy point on an energy profile for a chemical reaction. We can see this illustrated on the energy profile for the reaction between bromomethane and hydroxide ions below:

Intermediates are found at the minima on the energy profile. Consider the nucleophilic substitution of a tertiary halogenoalkane, as in the reaction between 2-bromo-2-methylpropane and hydroxide ions …

(CH₃)₃CBr + OH^– ⇾ (CH₃)₃OH + Br^–

rate = k [(CH₃)₃CBr]

The reaction is first order with respect to the halogenoalkane and zero order with respect to the hydroxide ions. It proceeds via two steps, the first step results in an intermediate which is used up in the second step when it reacts with a hydroxide ion.

The activation energy for step 1 is significantly greater than the activation energy for step 2 which means that all things being equal, the rate of step 2 is greater than the rate of step 1.

The moment the intermediate cation is formed it immediately reacts to form the products (hence the concentration of the intermediate will always be very low). The rate at which the products form is determined by the rate of step 1, the step with the greatest activation energy, because the following steps cannot proceed until the intermediate is formed.

Step 1 is the rate determining step and it depends only on the concentration of (CH₃)₃CBr. We can think of it as the bottleneck in the reaction mechanism.

the rate equation describes the rate determining step
rate = k [(CH₃)₃CBr]

We can use the rate equation for a complex reaction to propose a mechanism in which one of the steps must describe the rate determining step exactly. We could propose two different mechanisms for the reaction between molecules A and B in the following reaction:

A + 2B ⇾ AB₂

rate = k[B]²

1. In step 1, B + B ⇾ B₂ and in step 2, A + B₂ ⇾ AB₂
2. In step 1, A + B ⇾ AB and in step 2, AB + B ⇾ AB₂

The rate equation describes the rate determining step – in this example that means a step in the mechanism in which two molecules of B react to form an intermediate, since rate = k[B]². The first mechanism is therefore possible as in step 1 B + B ⇾ B₂. The second proposed mechanism is not possible.

Finally, let’s consider a two step complex reaction in which the activation energy for the second step is greater than the activation energy for the first step.

Essentially, the reactants and the intermediate are in equilibrium, rapidly interconverting, and only a very small fraction of intermediates will form products. Step 2 is the rate determining step here.

Practice questions

1. Suggest a possible mechanism for the reaction between hydrogen and iodine monochloride given that the chemical equation and the rate equation are as follows:

H_2(g) + 2ICl_(g) ⇾ 2HCl_(g) + I_2(g)

rate = k [ICl][H₂]

2. Given that the decomposition of dinitrogen pentoxide is first order with respect to N₂O₅, suggest a possible mechanism that is consistent with the rate equation.

2N₂O_5(g) ⇾ 4NO_2(g) + O_2(g)

3. The rate equation for the reaction between hydrogen and nitrogen(II) oxide is rate = k [H₂][NO]². Propose a mechanism that is consistent with the rate equation.

2H_2(g) + 2NO_(g) ⇾ 2H₂O_(g) + N_2(g)

Answers

1. The rate equation tells us that the rate determining step involves one molecule of ICl and one of H₂. If these two molecules react we can propose the products of this step to be HCl and HI.

ICl + H₂ ⇾ HCl + HI

The chemical equation tells us that there is a step involving a second molecule of ICl, and also that HI is not final product so must be used up in the second step.

ICl + HI ⇾ HCl + I₂

If we combine these two steps we end up with the chemical equation as the HI cancels out.

2. The decomposition reaction is first order so the rate equation must be rate = k [N₂O₅]. A possible mechanism could be a step in which one molecule of N₂O₅ decomposes: N₂O₅ ⇾ NO₂ + NO₃ (rate determining step), followed by: NO₃ + N₂O₅ ⇾ 3NO₂ + O₂. In the second step the second molecule of N₂O₅ reacts, satisfying the chemical equation, and the intermediate NO₃ created in the first step is used up.
3. The rate determine step involves a reaction between one molecule of hydrogen and two of nitrogen(II) oxide, as described by the rate equation: H₂ + 2NO ⇾ N₂ + H₂O + O. A second step would then be: O + H₂ ⇾ H₂O, using up the oxygen radical intermediate formed in the first step as well as the second molecule of hydrogen, satisfying the chemical equation.

These are all possible mechanisms – there may well be others that include the rate determining step and a subsequent step or steps. The trick is to make at least one of the products in your first step (as per the chemical equation), not to worry too much about whether your intermediate is a common molecule / ion / radical, and be sure that all your intermediates cancel out from one step to the next.

For example, another solution to question three might look like this:

H₂ + 2NO ⇾ 2NOH (rate determining step)

NOH + H₂ ⇾ H₂O + NH

NOH + NH ⇾ H₂O + N₂