Reactivity How fast? The rate of chemical change :R2.2.8 Molecularity IB DP Chemistry Study Notes - New Syllabus 2025
Reactivity How fast? The rate of chemical change – IB DP Chemistry- Study Notes
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Reactivity 2.2.8 – Molecularity of an Elementary Step
Reactivity 2.2.8 – Molecularity of an Elementary Step
Definition of Molecularity:
The molecularity of an elementary step refers to the number of reactant particles (atoms, molecules, or ions) that simultaneously collide and take part in a single step of a chemical reaction mechanism. It is a theoretical value assigned to individual steps in a reaction mechanism — not the overall reaction.
Key Characteristics of Molecularity:
- Applies only to elementary reactions — steps that occur in a single collision event.
- Molecularity must be a whole number: 1 (unimolecular), 2 (bimolecular), or 3 (termolecular).
- Not determined from experiments; rather, it is proposed based on the mechanism.
- Each elementary step in a reaction mechanism can have a different molecularity.
Molecularity vs Reaction Order:
- Molecularity is theoretical and applies only to elementary steps.
- Reaction order is determined experimentally and applies to the overall rate law of the reaction.
Why Molecularity Matters:
Understanding the molecularity of a step helps:
- Predict the rate law for that step (if it is elementary).
- Understand the reaction mechanism and how complex a collision event is.
- Identify the rate-determining step (usually with higher molecularity or slower kinetics).
Limitations of Molecularity:
- Steps with molecularity >3 are considered extremely unlikely due to the low probability of simultaneous collisions involving more than three species.
- Overall reactions do not have molecularity — only individual elementary steps do.
Example:
Consider the following elementary reaction:
\( \text{O}_3 + \text{NO} \rightarrow \text{O}_2 + \text{NO}_2 \)
Determine molecularity of the elementary Step.
▶️Answer/Explanation
- This reaction involves the direct collision of two reactant species: ozone (\( \text{O}_3 \)) and nitric oxide (\( \text{NO} \)).
- Since two particles are involved in this single-step reaction, the molecularity is bimolecular.
- Being an elementary step, its molecularity can be used to predict the rate law: \( \text{rate} \propto [\text{O}_3][\text{NO}] \)
Interpretation of “Unimolecular”, “Bimolecular” and “Termolecular” Elementary Steps:
1. Unimolecular Reactions
A unimolecular reaction involves the transformation of a single reactant particle. This typically includes isomerization or decomposition.
General form: \( \text{A} \rightarrow \text{Products} \)
Example: \( \text{N}_2\text{O}_5 \rightarrow \text{NO}_2 + \text{NO}_3 \)
- Only one molecule is involved in the rate-determining step.
- Rate law: \( \text{rate} = k[\text{N}_2\text{O}_5] \)
- Molecularity = 1
2. Bimolecular Reactions
A bimolecular reaction involves the collision between two particles, either of the same or different species.
General form: \( \text{A} + \text{B} \rightarrow \text{Products} \)
Example: \( \text{NO} + \text{O}_3 \rightarrow \text{NO}_2 + \text{O}_2 \)
- Two molecules collide in one step.
- Rate law: \( \text{rate} = k[\text{NO}][\text{O}_3] \)
- Molecularity = 2
3. Termolecular Reactions
A termolecular reaction involves the simultaneous collision of three reactant particles in a single elementary step.
General form: \( \text{A} + \text{B} + \text{C} \rightarrow \text{Products} \)
Example: \( 2\text{NO} + \text{O}_2 \rightarrow 2\text{NO}_2 \)
- All three particles must collide with proper orientation and sufficient energy at the same time.
- Such reactions are rare due to the low probability of three-body collisions.
- Rate law: \( \text{rate} = k[\text{NO}]^2[\text{O}_2] \)
- Molecularity = 3
Summary Table:
| Type | Number of Particles | Example | Rate Law |
|---|---|---|---|
| Unimolecular | 1 | \( \text{N}_2\text{O}_5 \rightarrow \text{NO}_2 + \text{NO}_3 \) | \( \text{rate} = k[\text{N}_2\text{O}_5] \) |
| Bimolecular | 2 | \( \text{NO} + \text{O}_3 \rightarrow \text{NO}_2 + \text{O}_2 \) | \( \text{rate} = k[\text{NO}][\text{O}_3] \) |
| Termolecular | 3 | \( 2\text{NO} + \text{O}_2 \rightarrow 2\text{NO}_2 \) | \( \text{rate} = k[\text{NO}]^2[\text{O}_2] \) |
Example:
A reaction has the following proposed mechanism:
- \( \text{A}_2 \rightarrow 2\text{A} \) (slow)
- \( \text{A} + \text{B} \rightarrow \text{C} \) (fast)
- \( \text{C} + \text{A} \rightarrow \text{D} \) (fast)
For each elementary step above:
- Identify the molecularity.
- Classify the step as unimolecular, bimolecular, or termolecular.
- Identify any intermediates formed.
▶️Answer/Explanation
- Step 1: \( \text{A}_2 \rightarrow 2\text{A} \)
– Only one molecule of \( \text{A}_2 \) is involved → Unimolecular
– No collision with other species is required.
– Product: 2 atoms of A. - Step 2: \( \text{A} + \text{B} \rightarrow \text{C} \)
– Two species react → Bimolecular
– C is formed → likely an intermediate. - Step 3: \( \text{C} + \text{A} \rightarrow \text{D} \)
– Two species involved again → Bimolecular
– C is consumed → confirms it’s an intermediate. - Overall Intermediates: C (produced in step 2, consumed in step 3)
Example:
The reaction A → products appears first-order, but it involves an intermediate A*. The proposed mechanism is:
- \( \text{A} + \text{M} \rightleftharpoons[k_{-1}]{k_1} \text{A}^* + \text{M} \) (fast equilibrium)
- \( \text{A}^* \xrightarrow{k_2} \text{Products} \) (slow)
Tasks:
- Identify the molecularity of Step 1.
- Identify the molecularity of Step 2.
- What is the role of A* in the mechanism?
- Why does the overall reaction appear unimolecular even though it begins with a bimolecular step?
▶️Answer/Explanation
- Step 1 molecularity: A + M collide → bimolecular.
- Step 2 molecularity: Decomposition of A* → unimolecular.
- Intermediate A*: A transient energetic form of A, formed in Step 1 and consumed in Step 2; consistent with a steady-state intermediate assumption.
- Apparent kinetics: Because Step 2 is slow and Step 1 is fast equilibrium, rate depends primarily on [A*], which in turn depends on [A]. Thus the reaction behaves as unimolecular overall despite the bimolecular initiation.