4.5: Reaction Mechanisms

It's often the case that one step in a multistep reaction mechanism is significantly slower than the others. Because a reaction cannot proceed faster than its slowest step, this step will limit the rate at which the overall reaction occurs. The slowest step is therefore called the rate-limiting step (or rate-determining step) of the reaction Figure \(\PageIndex{2}\).

Figure \(\PageIndex{2}\): A cattle chute is a nonchemical example of a rate-determining step. Cattle can only be moved from one holding pen to another as quickly as one animal can make its way through the chute. (credit: Loren Kerns)

As described earlier, rate laws may be derived directly from the chemical equations for elementary reactions. This is not the case, however, for ordinary chemical reactions. The balanced equations most often encountered represent the overall change for some chemical system, and very often this is the result of some multistep reaction mechanisms. In every case, we must determine the overall rate law from experimental data and deduce the mechanism from the rate law (and sometimes from other data). The reaction of NO2 and CO provides an illustrative example:

For temperatures above 225 °C, the rate law has been found to be:

The reaction is first order with respect to NO2 and first-order with respect to CO. This is consistent with a single-step bimolecular mechanism and it is possible that this is the mechanism for this reaction at high temperatures.

At temperatures below 225 °C, the reaction is described by a rate law that is second order with respect to NO2:

This is consistent with a mechanism that involves the following two elementary reactions, the first of which is slower and is therefore the rate-determining step:

\[\ce{NO2}(g)+\ce{NO2}(g)⟶\ce{NO3}(g)+\ce{NO}(g)\:\ce{(slow)}\\ \ce{NO3}(g)+\ce{CO}(g)⟶\ce{NO2}(g)+\ce{CO2}(g)\:\ce{(fast)}\]

The rate-determining step gives a rate law showing second-order dependence on the NO2 concentration, and the sum of the two equations gives the net overall reaction.

In general, when the rate-determining (slower) step is the first step in a mechanism, the rate law for the overall reaction is the same as the rate law for this step. However, when the rate-determining step is preceded by a step involving an equilibrium reaction, the rate law for the overall reaction may be more difficult to derive.

An elementary reaction is at equilibrium when it proceeds in both the forward and reverse directions at equal rates. Consider the dimerization of NO to N2O2, with k1 used to represent the rate constant of the forward reaction and k-1 used to represent the rate constant of the reverse reaction:

\[\ce{NO + NO ⇌ N2O2}\]

\[\ce{rate_{forward}=rate_{reverse}}\]

If N2O2 was an intermediate in a mechanism, this expression could be rearranged to represent the concentration of N2O2 in the overall rate law expression using algebraic manipulation:

However, once again, intermediates cannot be listed as part of the overall rate law expression, though they can be included in an individual elementary reaction of a mechanism. Example \(\PageIndex{1}\) will illustrate how to derive overall rate laws from mechanisms involving equilibrium steps preceding the rate-determining step.

Example \(\PageIndex{1}\): Deriving the Overall Rate Law Expression for a Multistep Reaction

Mechanism Nitryl chloride (NO2Cl) decomposes to nitrogen dioxide (NO2) and chlorine gas (Cl2) according to the following mechanism:

Determine the overall reaction, write the rate law expression for each elementary reaction, identify any intermediates, and determine the overall rate law expression.

Solution

For the overall reaction, simply sum the three steps, cancel intermediates, and combine like formulas:

Next, write the rate law expression for each elementary reaction. Remember that for elementary reactions that are part of a mechanism, the rate law expression can be derived directly from the stoichiometry:

The third step, which is the slow step, is the rate-determining step. Therefore, the overall rate law expression could be written as Rate = k3 [NOCl][ClO]. However, both NOCl and ClO are intermediates. Algebraic expressions must be used to represent [NOCl] and [ClO] such that no intermediates remain in the overall rate law expression.

Now substitute these algebraic expressions into the overall rate law expression and simplify:

Notice that this rate law shows an inverse dependence on the concentration of one of the product species, consistent with the presence of an equilibrium step in the reaction mechanism.

Exercise \(\PageIndex{1}\)

Atomic chlorine in the atmosphere reacts with ozone in the following pair of elementary reactions:

Determine the overall reaction, write the rate law expression for each elementary reaction, identify any intermediates, and determine the overall rate law expression.