Rate-Determining Step

Learning Objectives (OCR Spec 5.1.1)

By the end of this lesson you should be able to:

• Define the rate-determining step (RDS) as the slowest step in a reaction mechanism
• Use the rate equation to deduce information about the mechanism
• Understand that species appearing in the rate equation are involved in the RDS
• Interpret simple two-step mechanisms and check them against a rate equation

The Idea of a Rate-Determining Step

Most reactions proceed via a series of elementary steps, not in a single act. The rate of the overall reaction is determined by the slowest step — the bottleneck. This is known as the rate-determining step (RDS).

Analogy: Imagine a production line where one stage takes much longer than all the others. The rate at which finished products emerge from the line is set by that slow stage, regardless of how fast the other stages could work. Speeding up a fast step makes little or no difference; only speeding up the slow step changes the overall rate.

Linking the Rate Equation to the Mechanism

A crucial insight: only species that appear in the rate equation are involved in (or before) the rate-determining step. Species not in the rate equation either:

• Appear only in fast steps after the RDS, or
• Are catalysts, intermediates or spectators that do not appear in the RDS.

This gives us a powerful way to narrow down possible mechanisms:

1. Identify species in the rate equation. These must be involved up to and including the RDS.
2. Count how many of each species appears in the rate equation (the orders tell you). If rate = k[A]^2[B], then two A and one B are involved in the RDS.
3. Check candidate mechanisms by looking at which species appear in the slow step.

Worked Example 1 — SN1 and SN2

Two mechanisms for the hydrolysis of haloalkanes:

SN1 (unimolecular)

Slow: R-X -> R+ + X-
Fast: R+ + OH- -> R-OH

Only R-X in the slow step. Predicted rate equation: rate = k[RX].

SN2 (bimolecular)

Slow: R-X + OH- -> R-OH + X-

Both species in the slow step. Predicted rate equation: rate = k[RX][OH-].

Experimentally, (CH3)3C-Br hydrolyses by SN1 (rate = k[(CH3)3CBr]) because the tertiary carbocation is stable. CH3-Br hydrolyses by SN2 (rate = k[CH3Br][OH-]) because a primary cation would be very unstable.

So the rate equation itself distinguishes between mechanisms.

Worked Example 2 — Multi-Step Mechanism

For the reaction NO2(g) + CO(g) -> NO(g) + CO2(g), the experimentally determined rate equation is rate = k[NO2]^2. CO does not appear.

A proposed two-step mechanism:

Slow: 2 NO2 -> NO + NO3
Fast: NO3 + CO -> NO2 + CO2

• Slow step involves 2 x NO2 and gives rate ∝ [NO2]^2 — matches the experimental rate equation.
• CO only appears in the fast step — consistent with it NOT being in the rate equation.
• Adding the two steps gives the overall equation NO2 + CO -> NO + CO2. Correct.

This is a valid mechanism.

Worked Example 3 — NO and O2

For 2NO(g) + O2(g) -> 2NO2(g), experimental rate = k[NO]^2[O2]. Overall order = 3.

A single three-molecule collision is extremely unlikely, so the mechanism is probably:

Step 1 (fast equilibrium): 2NO <-> N2O2
Step 2 (slow):             N2O2 + O2 -> 2NO2