Chemical Kinetics

Multistep Reactions

Some chemical reactions occur in more than one step. The slowest step, known as the rate-determining step, determines the rate of the reaction.
Chemical reactions often occur in a number of steps that are individual chemical reactions. The rate of the overall reaction is affected by the rates of the reactions in these steps. The chemical reaction between nitrogen dioxide (NO2) and carbon monoxide (CO) occurs in two steps at low temperatures.
In the first step, two nitrogen dioxide (NO2) molecules collide to form a nitrate radical (NO3\rm{NO_3\cdot}) and nitrogen oxide (NO).
The nitrate radical (NO3\rm{NO_3\cdot}) then collides with carbon monoxide (CO), forming nitrogen dioxide (NO2) and carbon dioxide (CO2).
The nitrate radical (NO3\rm{NO_3\cdot}) is produced in the first step and then consumed in the second step. It is not part of the overall reaction. A species that is produced in one step and consumed in another step of a chemical reaction is an intermediate. Multistep reactions always involve intermediates. If a potential energy graph of this reaction is plotted, it shows one potential energy peak for each step. The graph therefore has two peaks.

Potential Energy-Reaction Progress Graph for a Two-Step Reaction

A potential energy graph for a chemical reaction has as many peaks as there are reaction steps.
The two steps of this reaction and the overall reaction are:
Elementary reactions occur at different rates. The rate of reaction 1 is different than that of reaction 2. The rates of reactions 1 and 2 affect the overall reaction rate.

Multistep reactions occur in series. One reaction occurs after the other. If one of the reactions in a multistep reaction is slow, then it will slow down all the other reactions. The slowest reaction in a multistep reaction that determines the overall rate is the rate-determining step. One way to think of this is like a highway with a traffic jam. If one part of the highway has heavy traffic, cars have to slow down in that part. It may take a long time to get through the heavy traffic area, but on the rest of the highway, traffic can proceed at a faster rate.

For the reaction in this example, experiments reveal that the rate law for the reaction is k[NO2]2. The concentration of carbon dioxide (CO) does not have an effect on the rate. Since nitrogen dioxide (NO2) is the reactant for the first step, the first step determines the rate and is the slowest.
NO2(g)+NO2(g)NO3(g)+NO(g)slow step: determines overall rateNO3(g)+CO(g)NO2(g)+CO2(g)fast step\begin{alignedat}{3}{\rm{NO_2}}(g)+{\rm{NO_2}}(g)&\rightarrow{\rm{NO_3\cdot}}(g)+{\rm{NO}}(g)\hspace{25px}&&\text{slow step: determines overall rate}\\{\rm{NO_3\cdot}}(g)+{\rm{CO}}(g)&\rightarrow{\rm{NO_2}}(g)+{\rm{CO}}_2(g)&&\text{fast step}\end{alignedat}
Step-By-Step Example
Multistep Reaction with a Slow Initial Step
A multistep reaction occurs in two steps.
What is the overall reaction and the rate expression for this reaction?
Step 1
To determine the overall reaction, add the two steps of the reaction together:
Note that the Cs cancel each other out. C is produced at the first step and is consumed in the second step, so C is an intermediate.
The overall reaction can be simplified.
2A2B+D2\rm A\rightarrow2\rm B+\rm D
To determine the rate of the overall reaction, determine the slowest step. Step 1 is slow in this example and determines the rate. The first step is a unimolecular reaction, and its rate is k[A].
If the first step in a multistep reaction is fast, enough product can build up while the slower second step is occurring that the first step begins to occur in reverse, producing more reactant. Eventually the first step can reach equilibrium while the slow step is still occurring.
Step-By-Step Example
Multistep Reaction with a Fast Initial Step
The oxidation of nitric oxide (NO) produces nitrogen dioxide (NO2).
The rate of the reaction is experimentally found to be k[NO]2[O2]. Two steps are proposed for the reaction:
Show that the reaction steps are possible, and calculate the rate expression.
Step 1

Possible reaction steps must have three properties:

1. The proposed steps must equal the overall reaction.

2. Each step should be unimolecular or bimolecular.

3. The rate equation calculated from the proposed steps must equal the experimentally determined rate equation.

Step 2
Check the first property by adding the proposed steps to determine whether they equal the overall reaction.
NO(g)+O2(g)+NO3(g)+NO(g)NO3(g)+2NO2(g){\rm{NO}}(g)+{\rm{O}}_2(g)+{\rm{NO}}_3(g)+{\rm{NO}}(g)\rightarrow{\rm{NO}}_3(g) + 2{\rm{NO}}_2(g)
The only intermediate is NO3(g). Canceling this from each side gives the overall reaction.
Step 3
Check the second property. Both proposed steps are bimolecular, so the proposed steps have the second property.
Step 4
To check the third property, first write rate expressions for both steps. The first step is reversible, so a rate expression for both forward and reverse reactions must be written.
rate1,forward=k1,forward[NO][O2]rate1,reverse=k1,reverse[NO3]rate2=k2[NO3][NO]\begin{aligned}&\rm{rate}_{1, \rm{forward}}=k_{1,\rm{forward}}[\rm{NO}][\rm{O}_2]\\\\&\rm{rate}_{1, \rm{reverse}}=k_{1,\rm{reverse}}[\rm{NO}_3]\\\\&\rm{rate}_{2}=k_{2}[\rm{NO}_3][\rm{NO}]\end{aligned}
Step 5
The third rate equation describes the rate of the (slow) rate-determining step, but it cannot be used as the rate equation for the reaction because it contains the intermediate. Use the first two equations to obtain an expression for the intermediate. Use the fact that the forward and reverse reaction rates are equal at equilibrium, and solve for the intermediate.
Step 6
Substitute the expression for the intermediate into the rate equation for the slow step.
rate2=k2[NO3][NO]=k2(k1,forwardk1,reverse[NO][O2])[NO]=k2k1,forwardk1,reverse[NO]2[O2]\begin{aligned}\rm{rate}_{2}&=k_{2}[\rm{NO}_3][\rm{NO}]\\&=k_{2}\left(\frac{k_{1,\rm{forward}}}{k_{1,\rm{reverse}}}[\rm{NO}][\rm{O}_2]\right)[\rm{NO}]\\&=\frac{k_{2}k_{1,\rm{forward}}}{ k_{1,\rm{reverse}}}[\rm{NO}]^2[\rm{O}_2]\end{aligned}
Combining all of the constants into a single constant k shows that the rate for the slow step is equivalent to the experimentally observed rate.
Since the proposed steps have all three required properties, the proposed reaction steps are possible.