How To Remove An Intermediate In A Rate Law

Understanding reaction mechanisms is a cornerstone of chemical kinetics. Consider this: since intermediates are typically unstable and their concentrations are unknown or difficult to measure experimentally, they cannot appear in the final, experimentally verifiable rate law. When a reaction proceeds through multiple elementary steps, the overall rate law is often dictated by the slowest step, known as the rate-determining step (RDS). Even so, a frequent complication arises when the rate law for this slow step includes a reaction intermediate—a species that appears in the mechanism but not in the overall balanced equation. Learning how to remove an intermediate in a rate law is therefore an essential skill for deriving a rate expression that depends only on reactant (and sometimes product) concentrations And it works..

People argue about this. Here's where I land on it.

The Role of Intermediates in Reaction Mechanisms

Before diving into the elimination techniques, it is the kind of thing that makes a real difference. In a multi-step mechanism, an intermediate is formed in one elementary step and consumed in a subsequent step. It does not appear in the overall stoichiometric equation. Take this: in the reaction $2NO_2 + F_2 \rightarrow 2NO_2F$, a proposed mechanism might involve the formation of an $NO_2F$ intermediate or a fluorine atom radical Turns out it matters..

The rate law for an elementary step is written directly from its molecularity (stoichiometry). Think about it: if the RDS is the second step: $Intermediate + Reactant \rightarrow Products$, the rate law would be $Rate = k[Intermediate][Reactant]$. Because $[Intermediate]$ is not a measurable starting condition, we must substitute it with an expression involving known concentrations. There are two primary approved methods for achieving this substitution: the Steady-State Approximation and the Pre-equilibrium (Fast Equilibrium) Approximation.

Method 1: The Steady-State Approximation (SSA)

The Steady-State Approximation is the more general and widely applicable method. It relies on the physical reality that reactive intermediates are highly reactive and exist in very low concentrations. After a brief induction period, the concentration of the intermediate reaches a near-constant, low level where its rate of formation equals its rate of consumption And that's really what it comes down to. That's the whole idea..

Applying the Steady-State Approximation: Step-by-Step

To apply SSA and remove the intermediate, follow this systematic procedure:

1. Identify the Intermediate: Pinpoint the species that appears in the mechanism but not in the overall reaction.
2. Write Formation and Consumption Rates: Write the rate expressions for every elementary step that produces the intermediate (formation) and every step that consumes it (consumption). Remember to use the stoichiometric coefficients as reaction orders for elementary steps.
3. Set Net Rate to Zero: Apply the approximation $\frac{d[Intermediate]}{dt} \approx 0$. This means: $Rate_{formation} = Rate_{consumption}$ $\sum k_{formation}[Reactants] = \sum k_{consumption}[Intermediate][Other\ Species]$
4. Solve for [Intermediate]: Algebraically isolate the concentration of the intermediate ($[I]$) on one side of the equation.
5. Substitute into the RDS Rate Law: Take the rate law for the Rate-Determining Step (or the overall rate expression if defined by product formation) and replace $[I]$ with the expression derived in step 4.

Worked Example: Ozone Decomposition

Consider the gas-phase decomposition of ozone: $2O_3 \rightarrow 3O_2$

Proposed Mechanism:

1. $O_3 \xrightleftharpoons[k_{-1}]{k_1} O_2 + O$ (Fast, reversible)
2. $O + O_3 \xrightarrow{k_2} 2O_2$ (Slow)

Here, atomic oxygen ($O$) is the intermediate. The RDS is Step 2, giving a preliminary rate law: $Rate = k_2[O][O_3]$

Applying SSA to Intermediate O:

• Formation: Step 1 (forward) $\rightarrow Rate = k_1[O_3]$
• Consumption: Step 1 (reverse) $\rightarrow Rate = k_{-1}[O_2][O]$ AND Step 2 $\rightarrow Rate = k_2[O][O_3]$

Set $\frac{d[O]}{dt} = 0$: $k_1[O_3] = k_{-1}[O_2][O] + k_2[O][O_3]$

Solve for $[O]$: $[O] = \frac{k_1[O_3]}{k_{-1}[O_2] + k_2[O_3]}$

Substitute back into the RDS rate law: $Rate = k_2[O_3] \left( \frac{k_1[O_3]}{k_{-1}[O_2] + k_2[O_3]} \right)$ $Rate = \frac{k_1 k_2 [O_3]^2}{k_{-1}[O_2] + k_2[O_3]}$

The intermediate $[O]$ has been successfully removed. The final rate law depends only on reactants ($O_3$) and products ($O_2$).

Method 2: The Pre-equilibrium (Fast Equilibrium) Approximation

The Pre-equilibrium Approximation is a specific, simplified case of the Steady-State Approximation. It applies when the step forming the intermediate is fast and reversible, and the step consuming the intermediate (the RDS) is slow. Because the first step is fast, it reaches equilibrium long before the slow step significantly depletes the intermediate. Because of this, the equilibrium constant ($K_{eq}$ or $K_c$) for that fast step can be used to express the intermediate's concentration Less friction, more output..

When to Use Pre-equilibrium vs. Steady-State

• Use Pre-equilibrium if: The mechanism explicitly states a "fast equilibrium" step preceding the RDS, or if $k_{-1} \gg k_2$ (the reverse of the first step is much faster than the second step).
• Use Steady-State if: No equilibrium is established (e.g., the first step is irreversible), the RDS is not clearly the last step, or the problem asks for the general derivation without assuming relative rate constant magnitudes.

Applying the Pre-equilibrium Approximation: Step-by-Step

1. Identify the Fast Equilibrium Step: Locate the reversible step preceding the RDS that produces the intermediate.
2. Write the Equilibrium Constant ($K_c$): For a generic fast step $A + B \rightleftharpoons I$, the equilibrium constant is $K_c = \frac{[I]}{[A][B]}$.
3. Solve for [Intermediate]: Rearrange to $[I] = K_c[A][B]$. Recall that $K_c = \frac{k_{forward}}{k_{reverse}}$.
4. Substitute into RDS Rate Law: Replace $[I]$ in the slow step's rate law.

Worked Example: NO Oxidation

Consider the reaction: $2NO + O_2 \rightarrow 2NO_2$

Proposed Mechanism:

1. $NO + NO \xrightleftharpoons[k_{-1}]{k_1} N_2O_2$ (Fast Equilibrium)
2. $N_2O_2 + O_2 \xrightarrow{k_2} 2NO_2$ (Slow)

Intermediate: $N_2O_2$ (Dinitrogen dioxide). RDS Rate Law: $Rate = k_2[N_2O_2][O_2]$

Applying Pre-equilibrium: For Step 1 at equilibrium: $K_c = \frac{[N_2O_2]}{[NO]^2} = \frac{k_1}{k_{-1}}$

Solve for intermediate: $[N