5.9 Pre-Equilibrium Approximation

AP Chem 5.9 Pre-Equilibrium Approximation Summary

Use the pre-equilibrium approximation when a reaction mechanism has a fast, reversible first step followed by a slower rate-limiting step. You set the forward and reverse rates of the fast step equal, solve for the intermediate's concentration, and substitute that into the slow step's rate law so the final rate law only uses real reactants. This gives you a rate law you can match against experimental data.

The pre-equilibrium approximation lets you write a rate law for a mechanism with a fast reversible first step and a slower later step. AP Chemistry expects you to move between experimental data, a rate law, and a proposed mechanism, and to justify whether they agree.

Replacing the Intermediate in the Slow-Step Rate Law

• Use the pre-equilibrium approximation when the first step is fast and reversible and a later step is slow (rate-limiting).
• A fast reversible step reaches equilibrium quickly, so its forward rate equals its reverse rate.
• An intermediate is produced in an earlier step and consumed later, so it must not appear in the final rate law.
• Solve the fast-step equilibrium for the intermediate concentration, then substitute it into the slow-step rate law.
• The final rate law should contain only species from the overall balanced reaction, with a combined rate constant.
• Always check that your elementary steps add up to the overall balanced equation.

Deriving a Rate Law From a Fast-Then-Slow Mechanism

Problem Solving: Follow these steps when a mechanism gives you a fast reversible first step and a slow second step:

1. Identify the slow step and write its rate law directly from its molecularity (the elementary-step stoichiometry).
2. Spot the intermediate, a species that shows up as a product in one step and a reactant in another. It cannot stay in your final rate law.
3. Set the forward and reverse rates of the fast equilibrium step equal to each other.
4. Solve that equality for the concentration of the intermediate.
5. Substitute that expression into the slow-step rate law.
6. Combine the constants into a single observed rate constant.

Worked example:

Step 1: 2NO ⇌ N₂O₂ (Fast)

Step 2: N₂O₂ + H₂ → H₂O + N₂O (Slow)

The overall reaction is:

2NO + H₂ → H₂O + N₂O

Here N₂O₂ is the intermediate because it is produced in Step 1 and consumed in Step 2.

The slow step gives:

rate = k[N₂O₂][H₂]

But N₂O₂ is an intermediate, so you cannot leave it in the rate law. Use the fast equilibrium:

forward rate = kₐ[NO]²
reverse rate = kₑ[N₂O₂]

Setting them equal:

kₐ[NO]² = kₑ[N₂O₂]

Solve for the intermediate:

[N₂O₂] = (kₐ/kₑ)[NO]²

Substitute into the slow-step rate law:

rate = k(kₐ/kₑ)[NO]²[H₂]

Combine the constants into one observed constant k":

rate = k"[NO]²[H₂]

Common Trap: Watch the difference between an intermediate and a catalyst. An intermediate is produced first, then consumed, so it appears as a product before it appears as a reactant. A catalyst is present at the start, gets consumed in one step, and is regenerated later, so its net concentration stays constant. Both can disappear from the final rate law, but they are not the same thing, and the AP exam may ask you to identify which is which.

Common Misconceptions

• Misconception: The slow step is always the first step.
  Actually: The slow step is not always the first step. When the slow step comes later, you cannot just read the rate law off the first step; you need the pre-equilibrium approach.
• Misconception: An intermediate can stay in a final rate law.
  Actually: An intermediate cannot stay in a final rate law. The rate law should only include species that appear in the overall balanced reaction.
• Misconception: You ignore a fast and reversible step.
  Actually: "Fast and reversible" does not mean you ignore that step. You use it to express the intermediate concentration, which is what makes the substitution possible.
• Misconception: Setting forward rate equal to reverse rate means the whole reaction is at equilibrium.
  Actually: Setting forward rate equal to reverse rate is an equilibrium condition for the fast step, not a claim that the whole reaction is at equilibrium.
• Misconception: Combining rate constants makes them disappear.
  Actually: Combining rate constants does not make them disappear. The final observed constant is built from the individual step constants, so it still depends on temperature.
• Misconception: The elementary steps do not have to add up to the overall equation.
  Actually: The elementary steps must add up to the overall equation. If they do not, the mechanism is not valid for that reaction.

Related AP Chemistry Guides

• Unit 5 Review: Kinetics
• 5.1 Reaction Rates
• 5.2 Introduction to Rate Law
• 5.3 Concentration Changes Over Time
• 5.4 Elementary Reactions
• 5.5 Collision Model