Topic 5.9 Notes – Pre-Equilibrium Approximation

1. What the Pre-Equilibrium Approximation Is

In a multi-step mechanism, the rate-determining step (RDS) controls how fast the overall reaction happens.

Sometimes the first step is fast and reversible, and the second step is slow. That’s when pre-equilibrium applies.

Here’s the situation:

• Step 1: Fast and reversible
• Step 2: Slow
• Step 1 is not rate-limiting

Because Step 1 is fast and reversible, it reaches equilibrium quickly compared to the slow step. So we can treat Step 1 as an equilibrium and use an equilibrium expression to describe the concentration of any intermediate it forms.

Big picture:

• The slow step determines the rate law form
• The fast step determines how much intermediate exists
• We use equilibrium math to rewrite the intermediate in terms of reactants

If the first step is fast and reversible and not the RDS, this is your signal to use pre-equilibrium.

2. When to Use It and What to Look For in a Mechanism

When you’re given a mechanism, read it carefully before writing anything.

Fast Reversible First Step

You’ll see:

• Labeled fast
• Double arrow ($\ce{<=>}$)
• Formation of an intermediate

Because it’s reversible and fast, it establishes equilibrium quickly.

Slow Step

You’ll see:

• Labeled slow
• This is the rate-determining step

The rate law always starts from this step.

Intermediates

An intermediate:

• Is formed in one step
• Is consumed in a later step
• Does not appear in the overall balanced reaction

Intermediates cannot appear in the final rate law.

Here’s the difference students mix up all the time:

Overall Reaction

Add all steps together and cancel intermediates.

Your final rate law must contain only reactants from this overall equation.

If you see an intermediate in your final answer, something is wrong.

3. How to Derive the Rate Law Using Pre-Equilibrium

Let’s walk through a sample mechanism:

$$\ce{A + B <=> C} \quad \text{(fast)}$$

$$\ce{C + D -> E} \quad \text{(slow)}$$

Step 1: Write the Rate Law from the Slow Step

Because it’s elementary:

$$\text{rate} = k[C][D]$$

But $C$ is an intermediate. We must eliminate it.

Step 2: Use Equilibrium from the Fast Step

At equilibrium:

$$k_f[A][B] = k_r[C]$$

Solve for $C$:

$$[C] = \frac{k_f}{k_r}[A][B]$$

Let $\frac{k_f}{k_r} = k'$:

$$[C] = k'[A][B]$$

Step 3: Substitute into the Slow Step

$$\text{rate} = k(k'[A][B])[D]$$

Combine constants:

$$\text{rate} = k''[A][B][D]$$

That’s your final rate law.

✔️ Only overall reactants
✔️ No intermediates
✔️ Based on slow step

This exact type of derivation shows up often in free-response questions.

4. Why This Works and How It Differs from Other Approximations

Because the first step is fast, it reaches equilibrium quickly.

When the slow step consumes the intermediate, the fast equilibrium shifts to replace it. So the intermediate concentration depends directly on reactant concentrations through the equilibrium expression.

The slow step still controls the speed. The equilibrium step just controls how much intermediate is available.

If the first step were slow, you wouldn’t need any of this. You’d just write the rate law directly from that first step.

Pre-equilibrium is only needed when:

• First step is fast and reversible
• A later step is slow
• The slow step includes an intermediate

The steady-state approximation exists in higher-level chemistry, but it is outside AP scope.

5. Common AP Exam Traps

• Writing the rate law from the overall equation
• Leaving an intermediate in the final expression
• Forgetting that equilibrium means forward rate = reverse rate
• Using pre-equilibrium when the first step is slow

One subtle trap. Sometimes the first step is fast but not reversible. If it’s not reversible, you cannot use pre-equilibrium because no equilibrium is established.