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Reaction mechanism and rate law

AP.Chem:
TRA‑5 (EU)
,
TRA‑5.A (LO)
,
TRA‑5.A.1 (EK)
,
TRA‑5.A.2 (EK)
,
TRA‑5.A.3 (EK)
,
TRA‑5.B (LO)
,
TRA‑5.B.1 (EK)

Key points

  • A reaction mechanism is the sequence of elementary steps by which a chemical reaction occurs.
  • A reaction that occurs in two or more elementary steps is called a multistep or complex reaction.
  • A reaction intermediate is a chemical species that is formed in one elementary step and consumed in a subsequent step.
  • The slowest step in a reaction mechanism is known as the rate-determining step.
  • The rate-determining step limits the overall rate and therefore determines the rate law for the overall reaction.

Introduction: Multistep reactions

One of the most important applications of kinetics is to the study of reaction mechanisms, or the sequences of steps by which chemical reactions occur. For example, consider the reaction of nitrogen dioxide with carbon monoxide:
NOX2(g)+CO(g)NO(g)+CO2(g)\ce{NO2}(g) + \ce{CO}(g) \rightarrow \ce{NO}(g) + \ce{CO}_2(g)
Based on the balanced equation, we might hypothesize that this reaction occurs by a single collision between a molecule of nitrogen dioxide and a molecule of carbon monoxide. In other words, we might hypothesize that this reaction is an elementary reaction.
If that were the case, then the rate law would be based on the reactant coefficients in the balanced chemical equation:
rate=k[NOX2][CO]\text{rate} = k[\ce{NO2}][\text{CO}]
However, when this reaction is studied experimentally, the rate law is in fact observed to be
rate=k[NOX2]2\text{rate} = k[\ce{NO2}]^2
Since the experimental rate law does not match the one derived by assuming an elementary reaction, we know immediately that the reaction must involve more than one step. Reactions that occur in two or more elementary steps are called multistep or complex reactions. As we'll see in the next section, we can use the experimental rate law to help deduce the individual steps that might be involved in a multistep reaction mechanism.

Multistep reaction mechanisms

Once the experimental rate law for a reaction is known, chemists can begin to devise and investigate possible reaction mechanisms. At minimum, a possible reaction mechanism must meet the following two conditions:
  1. The equations for the elementary steps in the mechanism must add up to the overall equation for the reaction.
  2. The mechanism must be consistent with the experimental rate law.
Let's use these conditions to evaluate a proposed mechanism for the reaction between NOX2\ce{NO2} and C, O. It is generally believed that this reaction occurs through two elementary steps:
Step 1:NOX2(g)+NOX2(g)NO(g)+NOX3(g)Step 2:NOX3(g)+CO(g)NOX2(g)+COX2(g)\begin{aligned}\text{Step 1:}&\quad \ce{NO2}(g) + \ce{NO2}(g) \xrightarrow{} \ce{NO}(g) + \ce{NO3}(g) \\\\[-0.50em] \text{Step 2:}&\kern1.5em \ce{NO3}(g) + \ce{CO}(g) \xrightarrow{} \ce{NO2}(g) + \ce{CO2}(g)\end{aligned}
First, let's check that the equations for these two steps add up to the overall reaction equation:
Step 1:NOX2(g)+NOX2(g)NO(g)+NOX3(g)Step 2:NOX3(g)+CO(g)NOX2(g)+COX2(g)NOX2(g)+CO(g)NO(g)+COX2(g)Overall:NOX2(g)+CO(g)NO(g)+COX2(g)\begin{aligned} \text{Step 1:}&\quad \ce{NO2}(g) + \blueD{\cancel{\ce{NO2}(g)}} \xrightarrow{} \ce{NO}(g) + \maroonD{\cancel{\ce{NO3}(g)}} \\\\[-0.50em] \text{Step 2:}&\kern1.5em \maroonD{\cancel{\ce{NO3}(g)}} + \ce{CO}(g) \xrightarrow{} \blueD{\cancel{\ce{NO2}(g)}} + \ce{CO2}(g) \\\\[-0.75em] &\kern1.4em \overline{\phantom{\ce{NO2}(g) + \ce{CO}(g) \xrightarrow{} \ce{NO}(g) + \ce{CO2}(g)}} \\\\[-2em] \text{Overall:}&\kern1.5em \ce{NO2}(g) + \ce{CO}(g) \xrightarrow{} \ce{NO}(g) + \ce{CO2}(g) \end{aligned}
Check! Note that one molecule, NOX3\ce{NO3}, appears in both steps of the mechanism but doesn't show up in the overall equation. In this case, NOX3\ce{NO3} is a reaction intermediate, a species that is formed in one step and consumed in a subsequent step.
Next, let's determine if the two-step mechanism is consistent with the experimental rate law. To do so, we need to know which of the two steps is the rate-determining step, or the slowest step in the mechanism. Because a reaction can occur no faster than its slowest step, the rate-determining step effectively limits the overall rate of a reaction. This is analogous to how a traffic jam limits the overall rate at which cars can move along a highway, even if other parts of the highway are clear.
In our proposed mechanism, the rate-determining step is believed to be step 1:
Step 1:NOX2(g)+NOX2(g)slowNO(g)+NOX3(g)Step 2:NOX3(g)+CO(g)fastNOX2(g)+COX2(g)\begin{aligned} \text{Step 1:}&\quad \ce{NO2}(g) + \ce{NO2}(g) \xrightarrow{slow} \ce{NO}(g) + \ce{NO3}(g) \\\\[-0.50em] \text{Step 2:}&\kern1.5em \ce{NO3}(g) + \ce{CO}(g) \xrightarrow{fast} \ce{NO2}(g) + \ce{CO2}(g) \end{aligned}
Since step 1 limits the overall rate of the reaction, the rate law for this step will be the same as the overall rate law. The predicted rate law for the overall reaction is therefore
rate=k[NOX2]2\text{rate} = k[\ce{NO2}]^2
This rate law is in agreement with the experimentally-determined rate law we saw earlier, so the mechanism also meets the second condition (check!). Since the reaction mechanism meets both conditions, we can safely say that it is a valid mechanism for the reaction.

Practice: Analyzing a reaction mechanism

The elementary steps of a proposed reaction mechanism are represented below.
Step 1:2NO(g)fastNX2OX2(g)Step 2:NX2OX2(g)+HX2(g)slowNX2O(g)+HX2O(g)Step 3:NX2O(g)+HX2(g)fastNX2(g)+HX2O(g)\begin{aligned} \text{Step 1:}&\kern5.0em \ce{2NO}(g) \xrightleftharpoons{fast} \ce{N2O2}(g) \\\\[-0.50em] \text{Step 2:}&\quad \ce{N2O2}(g) + \ce{H2}(g) \xrightarrow{slow} \ce{N2O}(g) + \ce{H2O}(g) \\\\[-0.50em] \text{Step 3:}&\kern1.5em \ce{N2O}(g) + \ce{H2}(g) \xrightarrow{fast} \ce{N2}(g) + \ce{H2O}(g) \end{aligned}
Based on this information, try to answer the following questions:
1. What is the overall equation for the reaction?
Choose 1 answer:
Choose 1 answer:

2. What are the reaction intermediates?
Choose all answers that apply:
Choose all answers that apply:

3. What is the rate-determining step?
Choose 1 answer:
Choose 1 answer:

Summary

  • A reaction mechanism is the sequence of elementary steps by which a chemical reaction occurs.
  • A reaction that occurs in two or more elementary steps is called a multistep or complex reaction.
  • A reaction intermediate is a chemical species that is formed in one elementary step and consumed in a subsequent step.
  • The slowest step in a reaction mechanism is known as the rate-determining step.
  • The rate-determining step limits the overall rate and therefore determines the rate law for the overall reaction.

Want to join the conversation?

  • leafers sapling style avatar for user Nishant
    How do we determine the slow or fast step?
    (30 votes)
    Default Khan Academy avatar avatar for user
    • aqualine ultimate style avatar for user Talos
      There are multiple ways to find which step is the slow/fast step without it being given to you.

      1) Look at k. The smaller k is probably the slower step.

      2) Look at the activation energy for each step. The step with larger activation energy is the slower step (as the fraction of collisions w/ enough energy to react will be smaller - activation energy requirement is higher) This derives from the Arrhenius equation:

      k = A * e^(-Ea/RT)

      where A and R are constants, T = temp, and Ea = activation energy. As you can see, as Ea increases, k decreases. Thus, the higher the Ea, the slower the reaction rate.
      (52 votes)
  • blobby green style avatar for user pabaaaa
    How can you determine which step is fast or slow if you are told reaction order. For example, if you are told the reaction is a first order, how do you know which step is fast or which is slow?
    (9 votes)
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    • hopper cool style avatar for user mrhappy1104
      If you are provided the complete mechanisms and the rate law of the overall reaction, it may be possible to infer the bottleneck reaction. If a curve that illustrates the change in energy throughout the reaction is provided, then the RDS is the step with the highest summit.
      (1 vote)
  • blobby green style avatar for user Ann Laubstein
    If A + 2B > C is a third order reaction does it have to be Rate = k {A} [B] squared? If B is mor influential on rate than A?
    (3 votes)
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  • blobby green style avatar for user karthik.subramanian
    Are we always going to be given whether or not the elementary steps are slow or fast
    (5 votes)
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  • leaf green style avatar for user Devina Thasia Wijaya
    How do we decide whether a reaction is slow or fast?
    (3 votes)
    Default Khan Academy avatar avatar for user
  • blobby green style avatar for user MJ
    Hi,
    I have some questions:
    1- How we can determine the intermediate (elementary reactions) for our overall reaction? Assume that we are doing a leaching experiments and we have derived a reaction about our process. Now, how we can derive intermediate reactions? Should we assume them and use the trial and error method?
    2- Is there any way, for example fixing some parameters and varying the other one to derive the leaching mechanism?
    Thanks for veryyyyyyy helpful courses :)
    (2 votes)
    Default Khan Academy avatar avatar for user
    • hopper jumping style avatar for user Ribhu Saha
      In all problems of equilibrium systems and reaction rates, we mainly deal with the gaseous substances. Liquids or solids( in small amount) don't affect the reaction. The leaching experiment doesn't conclude any gaseous substance. So, we can't derive intermediate reaction for a leaching experiment.
      For your second question, the same logic can be applied. Leaching experiments are not associated with any gaseous substance. So, if you will change the phases, by varying parameters, it'll not be a leaching experiment any more.
      (2 votes)
  • male robot hal style avatar for user Yash
    The article says, "Intermediates are produced in one step and consumed in later step, so they do not appear in the overall reaction equation or overall rate law." What if the rate determining step involves a reaction intermediate? Then shouldn't the overall rate law contain the intermediate as well? Please tell what am I missing.
    (2 votes)
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    • blobby green style avatar for user Jasper N
      They don't go over this, but from what I've learned from my textbook, it is possible to have a rate-limiting step containing an intermediate. However, algebraic maneuvering is required to substitute an expression for the concentration of the intermediate so that it's removed from the overall rate law. These reactions will often have a fast equilibrium reaction as their first step.
      (11 votes)
  • mr pants teal style avatar for user Shruthi K
    does the overall reaction rate always have to be equal to the slow step? What if there are intermediates in the slower step?
    (2 votes)
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  • blobby green style avatar for user owebster181
    Hello friends,
    Just to be sure, mechanisms are given, correct? Or you have to perform the experiment. You don't have to just make one up?
    (1 vote)
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    • blobby green style avatar for user Abijit.Jay
      Mechanisms don't necessarily have to be given. For example, in a test you may be asked to suggest a possible reaction mechanism considering a certain rate expression. This could mean that there could be multiple possible "correct" answers for the reaction mechanism. So yeah, you may "have to just make one up." I hope this answers your question.
      (2 votes)
  • starky sapling style avatar for user Archana
    How do we find what the catalyst is without having to do the experiment?
    (1 vote)
    Default Khan Academy avatar avatar for user