Topic 5.9 Pre-Equilibrium Approximation: derive the rate law of a mechanism with a fast initial equilibrium followed by a slow step by expressing the intermediate concentration in terms of reactant concentrations.

What this topic is asking

The College Board (Topic 5.9) wants you to derive the rate law for a mechanism in which a fast initial equilibrium precedes a slow step, by expressing the intermediate concentration in terms of reactant concentrations. This handles the case left open in Topic 5.8: when the slow step's rate law contains an intermediate, you must eliminate it before the rate law is acceptable.

The problem: an intermediate in the rate law

The rate-determining step gives the rate law (Topic 5.8), but if that step's reactants include an intermediate, the expression is not yet usable. The fix is to relate the intermediate back to the actual reactants using the fast step that made it.

Using the fast equilibrium

The reasoning is that a fast reversible step reaches equilibrium quickly compared with the slow step that drains the intermediate, so to a good approximation the forward and reverse rates of the fast step balance. This lets you write the intermediate's concentration as a ratio of rate constants times powers of the reactant concentrations.

Building the overall rate law

After substitution, the individual rate constants merge into a single composite constant $k = \dfrac{k_1 k_2}{k_{-1}}$ (for the common two-step case), and the orders come out of the powers of the reactant concentrations. The resulting overall order can be unusual, for example third order, which the overall stoichiometry alone would never reveal. Matching this derived rate law to the experimental one is the confirmation that the mechanism is viable.

Try this

Q1. For a fast equilibrium $2\text{X} \rightleftharpoons \text{Y}$, write $[\text{Y}]$ in terms of $[\text{X}]$ and the forward and reverse rate constants. [2 points]

• Cue. $k_1[\text{X}]^2 = k_{-1}[\text{Y}]$, so $[\text{Y}] = \dfrac{k_1}{k_{-1}}[\text{X}]^2$.

Q2. Explain why an intermediate cannot remain in the final rate law. [1 point]

• Cue. The rate law must be expressed in terms of measurable reactant concentrations; an intermediate is a transient species, not a starting material.