Most chemical reactions do not happen in a single step where all reactant molecules collide simultaneously. Instead, they occur via a sequence of simpler, individual steps known as a reaction mechanism. The overall rate of the reaction is governed by the slowest step in this sequence.
🔑 Key Principle
The rate equation of a reaction only contains reactants that are involved in the rate-determining step (or reactants from which those species are formed). Reactants that only participate in fast steps after the rate-determining step do not affect the rate and are zero order.
What is the Rate-Determining Step?
Imagine a motorway where traffic flows freely until it reaches a toll booth. The toll booth slows down the flow, creating a bottleneck. The overall speed at which cars travel the entire length of the motorway is determined by how fast they can pass through the toll booth. The toll booth is the rate-determining step.
The slowest step in a multi-step reaction mechanism. It acts as a bottleneck and determines the maximum overall rate of the reaction.
Multi-Step Energy Profiles
For a multi-step reaction, the reaction profile diagram has multiple activation energy humps. The step with the highest activation energy barrier (relative to its starting state) is the slowest step, representing the rate-determining step:
How the RDS Relates to the Rate Equation
There are two key rules that connect the rate-determining step to the rate equation:
- Only species involved in the rate-determining step (or in fast equilibrium steps preceding the rate-determining step) appear in the rate equation.
- The orders of reaction in the rate equation match the stoichiometry of the reactants in the rate-determining step (and preceding fast steps).
A species formed in one step of a reaction mechanism and consumed in a subsequent step. Intermediates do not appear in the overall balanced equation and do not appear directly in the rate equation.
A species that is consumed in an early step of a mechanism and regenerated in a later step. Catalysts do not appear in the overall balanced equation, but they can appear in the rate equation because they participate in the rate-determining step.
Step 1: \( \text{NO}_2 + \text{NO}_2 \rightarrow \text{NO}_3 + \text{NO} \) (Slow)
Step 2: \( \text{NO}_3 + \text{CO} \rightarrow \text{NO}_2 + \text{CO}_2 \) (Fast)
Step 1: Why is it multi-step?
If the reaction occurred in a single step, the rate equation would be \( \text{Rate} = k[\text{NO}_2][\text{CO}] \) (first order with respect to both). Because the rate equation is \( \text{Rate} = k[\text{NO}_2]^2 \) (and carbon monoxide is zero order), the reaction must occur via a multi-step mechanism where CO does not participate in the rate-determining step.
Step 2: Validate the mechanism
To check if a proposed mechanism is valid, it must meet two criteria:
- Sum of steps equals overall equation:
Step 1 + Step 2: \( \text{NO}_2 + \text{NO}_2 + \text{NO}_3 + \text{CO} \rightarrow \text{NO}_3 + \text{NO} + \text{NO}_2 + \text{CO}_2 \)
Cancel species on both sides (\( \text{NO}_2 \) and \( \text{NO}_3 \)):
\( \text{NO}_2 + \text{CO} \rightarrow \text{NO} + \text{CO}_2 \)
This matches the overall equation. (Yes)
- Mechanism matches the rate equation:
The rate-determining step is Step 1 (Slow). The reactants in Step 1 are two molecules of \( \text{NO}_2 \). Therefore, the predicted rate equation is:
\( \text{Rate} = k[\text{NO}_2][\text{NO}_2] = k[\text{NO}_2]^2 \)
This matches the experimental rate equation. (Yes)
Conclusion: The proposed mechanism is valid. Note that \( \text{NO}_3 \) is an intermediate because it is produced in Step 1 and consumed in Step 2.
What happens when the first step is fast?
Sometimes, the rate-determining step is the second step, and the first step is a fast reversible step. Under these conditions, the reactants of the fast first step will appear in the rate equation.
Step 1: \( \text{NO} + \text{NO} \rightleftharpoons \text{N}_2\text{O}_2 \) (Fast equilibrium)
Step 2: \( \text{N}_2\text{O}_2 + \text{O}_2 \rightarrow 2\text{NO}_2 \) (Slow)
Deduce the rate equation predicted by this mechanism.
Step 1: Write the rate equation for the RDS (Step 2)
\( \text{Rate} = k_2[\text{N}_2\text{O}_2][\text{O}_2] \)
However, \( \text{N}_2\text{O}_2 \) is an intermediate and cannot be in the final rate equation.
Step 2: Use the fast equilibrium to replace the intermediate
Since Step 1 is in fast equilibrium, the rate of the forward reaction equals the rate of the reverse reaction:
\( k_f[\text{NO}]^2 = k_r[\text{N}_2\text{O}_2] \)
Rearranging this gives an expression for \( [\text{N}_2\text{O}_2] \):
\( [\text{N}_2\text{O}_2] = \frac{k_f}{k_r}[\text{NO}]^2 \)
Step 3: Substitute back into the RDS rate equation
\( \text{Rate} = k_2 \left(\frac{k_f}{k_r}[\text{NO}]^2\right) [\text{O}_2] \)
Combining all the constant terms (\( k_2, k_f, k_r \)) into a single rate constant \( k \):
\( \text{Rate} = k[\text{NO}]^2[\text{O}_2] \)
The rate equation contains \( \text{NO} \) to the power of 2 and \( \text{O}_2 \) to the power of 1, showing how preceding fast steps affect the overall rate expression.
In AQA A-Level exams, questions about preceding fast equilibria are less common but represent the highest difficulty level. Remember: the rate equation simply reflects the sum of all reactant particles involved up to and including the rate-determining step. In Example 2, the RDS involves \( \text{N}_2\text{O}_2 \) and \( \text{O}_2 \), and since \( \text{N}_2\text{O}_2 \) is made from \( 2\text{NO} \), the total reactant particles up to the RDS are \( 2\text{NO} + 1\text{O}_2 \), giving \( \text{Rate} = k[\text{NO}]^2[\text{O}_2] \).
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