However, can also react with NO if the first step occurs in the reverse direction (NO + → 2 ) with rate r−1, where the minus sign indicates the rate of a reverse reaction. The concentration of a reactive intermediate such as [] remains low and almost constant. It may therefore be estimated by the steady-state approximation, which specifies that the rate at which it is formed equals the (total) rate at which it is consumed. In this example is formed in one step and reacts in two, so that The statement that the first step is the slow step actually means that the first step in the reverse direction is slower than the second step in the forward direction, so that almost all is consumed by reaction with CO and not with NO. That is, r−1 ≪ r2, so that r1 − r2 ≈ 0. But the overall rate of reaction is the rate of formation of final product (here ), so that r = r2 ≈ r1. That is, the overall rate is determined by the rate of the first step, and (almost) all molecules that react at the first step continue to the fast second step. Pre-equilibrium: if the second step were rate-determining The other possible case would be that the second step is slow and rate-determining, meaning that it is slower than the first step in the reverse direction: r2 ≪ r−1. In this hypothesis, r1 − r−1 ≈ 0, so that the first step is (almost) at equilibrium. The overall rate is determined by the second step: r = r2 ≪ r1, as very few molecules that react at the first step continue to the second step, which is much slower. Such a situation in which an intermediate (here ) forms an equilibrium with reactants prior to the rate-determining step is described as a pre-equilibrium For the reaction of and CO, this hypothesis can be rejected, since it implies a rate equation that disagrees with experiment. + → NO + (fast step) + CO → + (slow step, rate-determining) If the first step were at equilibrium, then its equilibrium constant expression permits calculation of the concentration of the intermediate in terms of more stable (and more easily measured) reactant and product species: The overall reaction rate would then be which disagrees with the experimental rate law given above, and so disproves the hypothesis that the