Mechanisms and Rate Law
Car exhaust from highways in Houston and Los Angeles contains NO₂, and chemists study how it reacts with other gases in the air, such as CO and F₂ in the lab. Like most reactions, these don't happen in one jump. The balanced equation on paper shows only where you start and where you end up. Why can the equation $2\,\text{NO}_2 + \text{F}_2 \rightarrow 2\,\text{NO}_2\text{F}$ have a rate law that doesn't match its coefficients? To answer that, you have to look at the hidden path: the mechanism.
Elementary steps and molecularity
An elementary step is a single collision or breakup event that really happens at the particle level. The equation of an elementary step shows exactly which particles collide.
The number of particles that collide in a step is its molecularity:
| Molecularity | Step | Rate law |
|---|---|---|
| Unimolecular | $\text{A} \rightarrow \text{products}$ | $k[\text{A}]$ |
| Bimolecular | $\text{A} + \text{B} \rightarrow \text{products}$ | $k[\text{A}][\text{B}]$ |
| Bimolecular | $2\,\text{A} \rightarrow \text{products}$ | $k[\text{A}]^2$ |
| Termolecular (rare) | $2\,\text{A} + \text{B} \rightarrow \text{products}$ | $k[\text{A}]^2[\text{B}]$ |
Why? The rate depends on how often the particles meet. Doubling the concentration of A doubles the chance that A is in the right place. If two A particles must meet, the chance goes up with $[\text{A}]^2$. Termolecular steps are rare because three particles almost never collide at once.
An elementary step has the equation $2\,\text{A} + \text{B} \rightarrow \text{C}$. Write its rate law and state its molecularity. Then explain in one or two sentences…
An elementary step has the equation $2\,\text{A} + \text{B} \rightarrow \text{C}$. Write its rate law and state its molecularity. Then explain in one or two sentences why the rate law of an elementary step can be written directly from its equation, while the rate law of an overall reaction cannot.
Building a mechanism: overall reaction, intermediates, catalysts
A reaction mechanism is a set of elementary steps that add up to the overall reaction. Three things to find:
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Overall reaction: add the steps and cancel species that appear on both sides.
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Intermediate: produced in one step and used up in a later step. It never appears in the overall reaction.
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Catalyst: used in an early step and regenerated in a later step. It is a reactant first, then a product.
The way to tell them apart is the order of appearance. An intermediate is made first, then consumed. A catalyst is consumed first, then made again.
A proposed mechanism for the reaction of nitrogen dioxide with carbon monoxide is: Step 1 (slow): $\text{NO}2 + \text{NO}2 \rightarrow \text{NO}3 + \text{NO}$ Step 2…
A proposed mechanism for the reaction of nitrogen dioxide with carbon monoxide is:
Step 1 (slow): $\text{NO}_2 + \text{NO}_2 \rightarrow \text{NO}_3 + \text{NO}$
Step 2 (fast): $\text{NO}_3 + \text{CO} \rightarrow \text{NO}_2 + \text{CO}_2$
(a) Write the balanced overall reaction. (b) Identify the intermediate. (c) Write the rate law predicted by this mechanism.
The decomposition of hydrogen peroxide in the presence of iodide ion is proposed to occur as follows: Step 1 (slow):…
The decomposition of hydrogen peroxide in the presence of iodide ion is proposed to occur as follows:
Step 1 (slow): $\text{H}_2\text{O}_2 + \text{I}^- \rightarrow \text{H}_2\text{O} + \text{IO}^-$
Step 2 (fast): $\text{H}_2\text{O}_2 + \text{IO}^- \rightarrow \text{H}_2\text{O} + \text{O}_2 + \text{I}^-$
Write the overall reaction, identify the catalyst and the intermediate, and write the rate law.
The rate-determining step and the rate law
Think of a drive-thru with a fast cashier and a slow cook. How many cars leave per minute? It depends on the cook. In a mechanism, the rate-determining step (the slow step) sets the overall rate. Fast steps afterward only wait for its products.
Here, only a slow first step is treated. The case of a fast step before the slow one is in the next page (Pre-Equilibrium Approximation).
Notice what this means: the species and steps after the slow step do not affect the rate. A reactant that only enters in a fast step after the slow step does not appear in the rate law, even though it is in the overall equation.
The reaction $2\,\text{ICl} + \text{H}2 \rightarrow \text{I}2 + 2\,\text{HCl}$ is proposed to occur by this mechanism: Step 1 (slow):…
The reaction $2\,\text{ICl} + \text{H}_2 \rightarrow \text{I}_2 + 2\,\text{HCl}$ is proposed to occur by this mechanism:
Step 1 (slow): $\text{ICl} + \text{H}_2 \rightarrow \text{HI} + \text{HCl}$
Step 2 (fast): $\text{HI} + \text{ICl} \rightarrow \text{I}_2 + \text{HCl}$
(a) Write the rate law predicted by the mechanism. (b) Give the units of $k$ if the rate is measured in $\text{M/s}$. (c) By what factor does the rate change if $[\text{ICl}]$ is tripled and $[\text{H}_2]$ is doubled?
The substitution reaction $(\text{CH}3)3\text{CBr} + \text{OH}^- \rightarrow (\text{CH}3)3\text{COH} + \text{Br}^-$ is proposed to occur in two steps: Step 1 (slow):…
The substitution reaction $(\text{CH}_3)_3\text{CBr} + \text{OH}^- \rightarrow (\text{CH}_3)_3\text{COH} + \text{Br}^-$ is proposed to occur in two steps:
Step 1 (slow): $(\text{CH}_3)_3\text{CBr} \rightarrow (\text{CH}_3)_3\text{C}^+ + \text{Br}^-$
Step 2 (fast): $(\text{CH}_3)_3\text{C}^+ + \text{OH}^- \rightarrow (\text{CH}_3)_3\text{COH}$
(a) Write the rate law and identify the intermediate. (b) Ethan doubles the concentration of $\text{OH}^-$ and keeps everything else the same. Predict the effect on the rate and explain it using the mechanism.
Testing mechanisms against data
You can never prove a mechanism. You can only show it is consistent with the experimental rate law, or rule it out. The steps:
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Determine the rate law from experiments (the method of initial rates).
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Predict the rate law for each proposed mechanism.
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Keep the mechanisms that match. Reject those that don't.
A valid mechanism must also (a) add up to the overall reaction and (b) use reasonable elementary steps (usually unimolecular or bimolecular).
Sofia measures the initial rate of the reaction $\text{A} + \text{B} \rightarrow \text{P}$ at a fixed temperature: | Experiment | $[\text{A}]$ (M) | $[\text{B}]$ (M) |…
Sofia measures the initial rate of the reaction $\text{A} + \text{B} \rightarrow \text{P}$ at a fixed temperature:
| Experiment | $[\text{A}]$ (M) | $[\text{B}]$ (M) | Initial rate (M/s) |
|---|---|---|---|
| 1 | 0.10 | 0.10 | $2.0 \times 10^{-3}$ |
| 2 | 0.20 | 0.10 | $4.0 \times 10^{-3}$ |
| 3 | 0.10 | 0.30 | $2.0 \times 10^{-3}$ |
Two mechanisms are proposed. Mechanism I: $\text{A} \rightarrow \text{X}$ (slow), then $\text{X} + \text{B} \rightarrow \text{P}$ (fast). Mechanism II: a single elementary step $\text{A} + \text{B} \rightarrow \text{P}$.
(a) Determine the rate law and the value of $k$ with units. (b) Decide which mechanism is consistent with the data and justify your answer.
A mechanism for the overall reaction of species B to give E has three steps: Step 1 (slow): $\text{A} + \text{B} \rightarrow \text{C}$ Step 2 (fast):…
A mechanism for the overall reaction of species B to give E has three steps:
Step 1 (slow): $\text{A} + \text{B} \rightarrow \text{C}$
Step 2 (fast): $\text{C} + \text{B} \rightarrow \text{D}$
Step 3 (fast): $\text{D} \rightarrow \text{E} + \text{A}$
(a) Write the overall reaction and identify every catalyst and intermediate. (b) Write the rate law. (c) Malik says, "The overall equation has $2\,\text{B}$, so the rate must be $k[\text{B}]^2$." Evaluate his claim. (d) Describe, in words, how the potential energy diagram would look, and say which step has the highest activation energy.
The reaction $2\,\text{NO}2 + \text{F}2 \rightarrow 2\,\text{NO}2\text{F}$ has the experimental rate law $\text{rate} = k[\text{NO}2][\text{F}2]$ with…
The reaction $2\,\text{NO}_2 + \text{F}_2 \rightarrow 2\,\text{NO}_2\text{F}$ has the experimental rate law $\text{rate} = k[\text{NO}_2][\text{F}_2]$ with $k = 0.38\ \text{M}^{-1}\text{s}^{-1}$ at a given temperature.
Mechanism A: Step 1 (slow) $\text{NO}_2 + \text{F}_2 \rightarrow \text{NO}_2\text{F} + \text{F}$; Step 2 (fast) $\text{F} + \text{NO}_2 \rightarrow \text{NO}_2\text{F}$.
Mechanism B: one termolecular elementary step, $2\,\text{NO}_2 + \text{F}_2 \rightarrow 2\,\text{NO}_2\text{F}$.
(a) Show that both mechanisms add up to the overall reaction. (b) Predict the rate law of each mechanism and decide which one is consistent with experiment. (c) Give a reason, based on collisions, why mechanism B is unlikely even if it were consistent. (d) Calculate the rate when $[\text{NO}_2] = 0.020\ \text{M}$ and $[\text{F}_2] = 0.050\ \text{M}$.
Energy diagram of a multistep mechanism
Each elementary step has its own activation energy. In the energy profile, each step is a hill (a transition state), and each intermediate sits in a valley between hills. The slow step has the largest activation energy, measured from the species just before it to its transition state (the biggest climb).
Because the climb for step 1 is the biggest, step 1 is the rate-determining step. A very reactive intermediate sits in a shallow valley and is used up as soon as it forms.
Summary
Back to the start: the coefficients of the overall equation do not set the exponents, because the rate depends only on the slow step. That is why $2\,\text{NO}_2 + \text{F}_2 \rightarrow 2\,\text{NO}_2\text{F}$ has a rate that is first order in each reactant.