Pre-Equilibrium Approximation
A reaction mechanism often has a slow step that uses an intermediate as a reactant. But you can't measure an intermediate in the lab, and a rate law must be written with reactants only. So how do you get rid of it? When the step that makes the intermediate is fast and reversible, there is a neat trick: the pre-equilibrium approximation.
Why intermediates are a problem
The rate of a mechanism is set by its slowest step, the rate-determining step. For an elementary step, the rate law comes straight from the coefficients: a step $\text{X} + \text{B} \rightarrow \text{P}$ has rate $= k[\text{X}][\text{B}]$.
If $\text{X}$ is an intermediate, that is a problem. An intermediate is made in one step and used up in a later step, so it never appears in the overall equation. The final rate law may contain only reactants (and sometimes catalysts).
There are two cases:
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The slow step is first. Its reactants are all real reactants, so you write the rate law directly from it.
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A fast, reversible step comes before the slow step. The slow step contains an intermediate. You must replace it. That is the job of the pre-equilibrium approximation.
For each mechanism, state whether the rate law can be written directly from the slow step or whether you must use a pre-equilibrium to eliminate an intermediate. (i)…
For each mechanism, state whether the rate law can be written directly from the slow step or whether you must use a pre-equilibrium to eliminate an intermediate.
(i) Step 1 slow: $\text{A} + \text{B} \rightarrow \text{X}$; Step 2 fast: $\text{X} + \text{B} \rightarrow \text{P}$.
(ii) Step 1 fast, reversible: $\text{A} + \text{B} \rightleftharpoons \text{X}$; Step 2 slow: $\text{X} + \text{A} \rightarrow \text{P}$.
(iii) Step 1 slow: $\text{A} \rightarrow \text{X}$; Step 2 fast: $\text{X} \rightarrow \text{P}$.
The pre-equilibrium idea
Picture a fast, reversible first step: $\text{A} + \text{B} \rightleftharpoons \text{C}$. The forward and reverse reactions are both much faster than the slow step that follows. So only a tiny fraction of $\text{C}$ is ever drained away by the slow step. The first step therefore keeps up with its own back-and-forth and sits at equilibrium.
The key condition for the approximation: the forward and reverse rates of the fast step are equal (and much larger than the rate of the slow step). Then Step 1 obeys its equilibrium constant:
Why does it work? The slow step is a bottleneck. Intermediate $\text{C}$ is made and unmade many times before one molecule finally moves on, so its concentration is set by the equilibrium, not by the slow step.
In a mechanism, the first step is a fast, reversible step: $\text{A} + \text{B} \rightleftharpoons \text{C}$, with equilibrium constant $K$. Write the expression for…
In a mechanism, the first step is a fast, reversible step: $\text{A} + \text{B} \rightleftharpoons \text{C}$, with equilibrium constant $K$. Write the expression for $K$, then solve it for $[\text{C}]$. Explain in one sentence what must be true about the forward and reverse rates of this step for the pre-equilibrium approximation to apply.
Step-by-step method
Use these steps every time:
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Find the slow step. Write its rate law from its coefficients.
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Check whether it contains an intermediate.
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If so, write the equilibrium expression for the fast step that makes the intermediate and solve for its concentration.
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Substitute, then combine the constants: $k_{\text{obs}} = k_2 K$.
Notice that the rate law matches the overall equation here: second order in NO and first in $\text{O}_2$. The mechanism is consistent with the data, but that does not prove it is correct.
A mechanism for $2\,\text{NO} + \text{O}2 \rightarrow 2\,\text{NO}2$ is: Step 1 (fast equilibrium): $2\,\text{NO} \rightleftharpoons \text{N}2\text{O}2$ Step 2 (slow):…
A mechanism for $2\,\text{NO} + \text{O}_2 \rightarrow 2\,\text{NO}_2$ is:
Step 1 (fast equilibrium): $2\,\text{NO} \rightleftharpoons \text{N}_2\text{O}_2$
Step 2 (slow): $\text{N}_2\text{O}_2 + \text{O}_2 \rightarrow 2\,\text{NO}_2$
Identify the intermediate and write an expression for $[\text{N}_2\text{O}_2]$ in terms of $K$ and $[\text{NO}]$.
Use the mechanism from the previous section, Step 1 (fast equilibrium) $2\,\text{NO} \rightleftharpoons \text{N}2\text{O}2$ with constant $K$, and Step 2 (slow)…
Use the mechanism from the previous section, Step 1 (fast equilibrium) $2\,\text{NO} \rightleftharpoons \text{N}_2\text{O}_2$ with constant $K$, and Step 2 (slow) $\text{N}_2\text{O}_2 + \text{O}_2 \rightarrow 2\,\text{NO}_2$ with rate constant $k_2$.
(a) Derive the rate law in terms of $[\text{NO}]$ and $[\text{O}_2]$ only. (b) Write the observed rate constant $k_{\text{obs}}$ in terms of $k_2$ and $K$. (c) If $K = 0.040\ \text{M}^{-1}$ and $k_2 = 5.0 \times 10^{3}\ \text{M}^{-1}\text{s}^{-1}$, calculate $k_{\text{obs}}$ with units.
Products in the equilibrium step
Sometimes the fast step makes more than one species, so the equilibrium expression has a product of concentrations or a power. Take the ozone mechanism in the upper atmosphere:
Step 1 (fast): $\text{O}_3 \rightleftharpoons \text{O}_2 + \text{O}$. Step 2 (slow): $\text{O} + \text{O}_3 \rightarrow 2\,\text{O}_2$.
The order in $\text{O}_2$ is $-1$. A negative order is a signature of a pre-equilibrium: $\text{O}_2$ is a product of Step 1, so extra $\text{O}_2$ pushes that equilibrium back toward $\text{O}_3$ (Le Châtelier). Less $\text{O}$ means a slower reaction.
For the $\text{H}_2 + \text{Br}_2$ mechanism, the fast step is $\text{Br}_2 \rightleftharpoons 2\,\text{Br}$, so $K = [\text{Br}]^2/[\text{Br}_2]$ and $[\text{Br}] = (K[\text{Br}_2])^{1/2}$. The slow step, rate $= k_2[\text{Br}][\text{H}_2]$, then gives rate $= k_2K^{1/2}[\text{H}_2][\text{Br}_2]^{1/2}$. Fractional orders can appear. The intermediate $\text{Br}$ is made in the fast equilibrium, so it must be replaced using $K$. The intermediate $\text{H}$ is made in the slow step and used up in a fast step after it, so it does not affect the rate law.
The decomposition of ozone in the upper atmosphere is proposed to follow this mechanism: Step 1 (fast equilibrium): $\text{O}3 \rightleftharpoons \text{O}2 + \text{O}$…
The decomposition of ozone in the upper atmosphere is proposed to follow this mechanism:
Step 1 (fast equilibrium): $\text{O}_3 \rightleftharpoons \text{O}_2 + \text{O}$
Step 2 (slow): $\text{O} + \text{O}_3 \rightarrow 2\,\text{O}_2$
(a) Derive the rate law. (b) State the order with respect to $\text{O}_3$ and with respect to $\text{O}_2$. (c) By what factor does the rate change if $[\text{O}_3]$ is tripled and $[\text{O}_2]$ is doubled? (d) Explain why adding $\text{O}_2$ slows the reaction, using the equilibrium in Step 1.
A mechanism for $\text{H}2 + \text{Br}2 \rightarrow 2\,\text{HBr}$ is: Step 1 (fast equilibrium): $\text{Br}2 \rightleftharpoons 2\,\text{Br}$ Step 2 (slow):…
A mechanism for $\text{H}_2 + \text{Br}_2 \rightarrow 2\,\text{HBr}$ is:
Step 1 (fast equilibrium): $\text{Br}_2 \rightleftharpoons 2\,\text{Br}$
Step 2 (slow): $\text{Br} + \text{H}_2 \rightarrow \text{HBr} + \text{H}$
Step 3 (fast): $\text{H} + \text{Br}_2 \rightarrow \text{HBr} + \text{Br}$
(a) Derive the rate law from the slow step and the pre-equilibrium. (b) State the order with respect to each reactant and the overall order. (c) Identify all intermediates.
Comparing mechanisms with data
Different mechanisms give different rate laws, so experiments can rule some out. Compare the initial-rate data to each predicted rate law. If the exponents don't match, that mechanism is wrong. If they match, the mechanism is only consistent.
Temperature adds a twist. Because $k_{\text{obs}} = k_2K$, it depends on both a rate constant and an equilibrium constant. $k_2$ rises with temperature. But if Step 1 is exothermic, $K$ falls as temperature rises. If $K$ drops faster than $k_2$ climbs, $k_{\text{obs}}$ decreases, and the reaction slows when heated. This explains an apparent negative activation energy.
The reaction $2\,\text{NO} + \text{Br}2 \rightarrow 2\,\text{NOBr}$ is proposed to proceed by: Step 1 (fast equilibrium):…
The reaction $2\,\text{NO} + \text{Br}_2 \rightarrow 2\,\text{NOBr}$ is proposed to proceed by:
Step 1 (fast equilibrium): $\text{NO} + \text{Br}_2 \rightleftharpoons \text{NOBr}_2$, constant $K$
Step 2 (slow): $\text{NOBr}_2 + \text{NO} \rightarrow 2\,\text{NOBr}$, rate constant $k_2$
(a) Derive the rate law. (b) With $K = 0.30\ \text{M}^{-1}$ and $k_2 = 2.0 \times 10^{2}\ \text{M}^{-1}\text{s}^{-1}$, calculate $k_{\text{obs}}$ and the rate when $[\text{NO}] = 0.10\ \text{M}$ and $[\text{Br}_2] = 0.050\ \text{M}$. (c) Experiments show that the rate decreases slightly as temperature increases. If the activation energy of Step 2 is small, explain how this can happen when Step 1 is exothermic.
For the reaction $2\,\text{A} + \text{B} \rightarrow \text{C}$, two mechanisms are proposed. Mechanism 1: $2\,\text{A} \rightleftharpoons \text{A}2$ (fast equilibrium,…
For the reaction $2\,\text{A} + \text{B} \rightarrow \text{C}$, two mechanisms are proposed.
Mechanism 1: $2\,\text{A} \rightleftharpoons \text{A}_2$ (fast equilibrium, constant $K$); $\text{A}_2 + \text{B} \rightarrow \text{C}$ (slow, $k_2$).
Mechanism 2: $2\,\text{A} \rightarrow \text{A}_2$ (slow); $\text{A}_2 + \text{B} \rightarrow \text{C}$ (fast).
Initial-rate data at a fixed temperature:
| Experiment | $[\text{A}]$ (M) | $[\text{B}]$ (M) | Initial rate (M/s) |
|---|---|---|---|
| 1 | 0.10 | 0.10 | $4.0 \times 10^{-4}$ |
| 2 | 0.20 | 0.10 | $1.6 \times 10^{-3}$ |
| 3 | 0.10 | 0.20 | $8.0 \times 10^{-4}$ |
(a) Derive the rate law predicted by each mechanism. (b) Determine the experimental rate law and $k_{\text{obs}}$ with units, and decide which mechanism is supported. (c) If $K = 0.50\ \text{M}^{-1}$, calculate $k_2$. (d) Explain what dependence on $[\text{B}]$ the other mechanism would have predicted.
Summary
Back to the start: the intermediate disappears because the fast step ties its concentration to the reactants through $K$.