Rate-determining step Knowledge-Based Study

Study notes

1. Most chemical reactions do not occur in a single event where all reactant molecules collide simultaneously. Instead, they proceed through a sequence of simpler, discrete molecular events known as elementary steps. The detailed, step-by-step pathway leading from reactants to products is called the reaction mechanism.

   Unlike an overall balanced chemical equation, the stoichiometric coefficients in an elementary step directly reflect the number of colliding species, referred to as molecularity:

   • Unimolecular: A single molecule dissociates or isomerizes, e.g., A → products, with rate = $k[\text{A}]$.
   • Bimolecular: Two species collide, e.g., A + B → products, with rate = $k[\text{A}][\text{B}]$.
   • Termolecular: Three species collide simultaneously, which is statistically rare.

   Species that are produced in an early step and consumed in a subsequent step are called intermediates. Intermediates do not appear in the overall balanced equation.

   In a multistep mechanism, different elementary steps proceed at different intrinsic rates because each step has its own activation energy ($E_a$).

   The slowest elementary step in the reaction pathway is called the rate-determining step (RDS) or rate-limiting step. Because the overall transformation cannot proceed faster than its slowest stage, the rate-determining step acts as a kinetic bottleneck that governs the rate of the entire reaction.

   An everyday analogy is pouring sand through a funnel: no matter how wide the mouth of the funnel is, the rate at which sand exits depends solely on the diameter of the narrowest constriction.

   When the first elementary step of a mechanism is the rate-determining step, determining the overall rate law is straightforward. The overall rate law is identical to the rate law of this slow first step.

   Consider the reaction between nitrogen dioxide and carbon monoxide:

   $\text{NO}_2(g) + \text{CO}(g) \rightarrow \text{NO}(g) + \text{CO}_2(g)$

   A proposed two-step mechanism is:

   1. Step 1 (slow): $\text{NO}_2 + \text{NO}_2 \rightarrow \text{NO}_3 + \text{NO}$
   2. Step 2 (fast): $\text{NO}_3 + \text{CO} \rightarrow \text{NO}_2 + \text{CO}_2$

   Since Step 1 is the rate-determining step, the rate of the reaction depends only on the collision between two $\text{NO}_2$ molecules:

   Rate = $k_1[\text{NO}_2]^2$

   Notice that $\text{CO}$ does not appear in the rate law (it is zero-order with respect to $\text{CO}$) because it participates only in a step after the rate-determining step.

   When the rate-determining step is preceded by a fast reversible step, the slow step involves one or more reactive intermediates. Because rate laws must be expressed exclusively in terms of measurable reactant (or product) concentrations, the concentration of the intermediate must be eliminated.

   Consider the reaction:

   $2\text{NO}(g) + \text{O}_2(g) \rightarrow 2\text{NO}_2(g)$

   Proposed mechanism:

   1. Step 1 (fast equilibrium): $\text{NO} + \text{NO} \rightleftharpoons \text{N}_2\text{O}_2$ (forward rate constant $k_1$, reverse rate constant $k_{-1}$)
   2. Step 2 (slow, RDS): $\text{N}_2\text{O}_2 + \text{O}_2 \rightarrow 2\text{NO}_2$ (rate constant $k_2$)

   The rate is dictated by Step 2:

   $\text{Rate} = k_2[\text{N}_2\text{O}_2][\text{O}_2]$

   Assuming Step 1 establishes dynamic equilibrium, the forward and reverse rates are equal:

   $k_1[\text{NO}]^2 = k_{-1}[\text{N}_2\text{O}_2] \implies [\text{N}_2\text{O}_2] = \frac{k_1}{k_{-1}}[\text{NO}]^2$

   Substituting this expression into the rate equation gives:

   $\text{Rate} = k_2 \left(\frac{k_1}{k_{-1}}[\text{NO}]^2\right)[\text{O}_2] = k_{\text{obs}}[\text{NO}]^2[\text{O}_2]$

   where $k_{\text{obs}} = \frac{k_1 k_2}{k_{-1}}$.

   To master the rate-determining step concept, keep these principles in mind:

   • Molecularity vs. Reaction Order: Reaction orders in an overall rate law cannot be predicted from overall stoichiometry; they are determined experimentally. However, for an elementary step, the reaction orders equal its stoichiometric coefficients.
   • Intermediates in Rate Laws: An experimentally validated rate law never contains reactive intermediates. They must always be expressed in terms of reactants using prior equilibrium steps or steady-state approximations.
   • Steps After the RDS: Reactants involved only in steps occurring after the rate-determining step do not affect the rate of reaction under normal conditions and have an order of zero.
   • Catalysts: Catalysts speed up reactions by introducing an alternative reaction mechanism with a lower activation energy for the rate-determining step.