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First teaching 2023

First exams 2025

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Reaction Mechanisms (HL) (HL IB Chemistry)

Revision Note

Caroline

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Caroline

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Physics Lead

Reaction Mechanisms

What is a reaction mechanism?

  • Most reactions do not occur in one step but in a series of simple steps
  • Each step is called an elementary step and involves a small number of particles
  • Some of the products of an elementary step exist as intermediates and react in subsequent steps
  • The sequence of elementary steps is called the reaction mechanism
  • The sum of the elementary steps must equal the overall reaction equation
    • Intermediates that are produced in one elementary step and react in another step cancel out
  • Chemical kinetics can only suggest a reaction mechanism, they cannot prove it
    • However, they can be used to disprove a proposed mechanism
  • Elementary steps are the steps involved in a reaction mechanism
    • For example, in the following general reaction:

A + B → C + D

    • The elementary steps could involve the formation of an intermediate:

Elementary step 1: A → R + D

Elementary step 2: R + B → C

  • It is important that the elementary steps for a proposed mechanism agree with the overall stoichiometric equation
    • For example, combining the 2 elementary steps above gives the overall stoichiometric equation

A + R + B → R + C + D

A + B → C + D

Worked example

Sulfur dioxide reacts with oxygen to form sulfur trioxide

  1. Propose a one-step mechanism for the above reaction
  2. The above reaction is catalysed by the formation of nitrogen dioxide from nitrogen monoxide. Propose a two-step mechanism for this reaction.

 

Answer 1:

  • A one-step reaction mechanism is simply the overall stoichiometric equation
  • Therefore, the correct answer is 2SO2 + O2 → 2SO3

Answer 2:

  • One of the two elementary steps for this two-step mechanism can be taken from the question:
    • Elementary step 1: 2NO + O2 → 2NO2
  • The second elementary step must involve the reaction of the nitrogen dioxide formed with sulfur dioxide:
    • Elementary step 2: NO2 + SO2 → NO + SO3 (or 2NO2 + 2SO2 → 2NO + 2SO3)

Exam Tip

  • It is important that you check the equations you are proposing for a reaction mechanism
  • They must add up to the overall stoichiometric equation, otherwise the proposed mechanism is wrong.

What is the rate-determining step?

  • A chemical reaction can only go as fast as the slowest part of the reaction
    • So, the rate-determining step is the slowest step in the reaction

What does the rate equation tell us about the rate-determining step?

  • If a reactant appears in the rate-determining step, then the concentration of that reactant will also appear in the rate equation
  • The order with respect to a reactant describes the number of particles of that reactant that take part in the rate-determining step

Predicting the reaction mechanism

  • The overall reaction equation and rate equation can be used to predict a possible reaction mechanism of a reaction
  • For example, nitrogen dioxide (NO2) and carbon monoxide (CO) react to form nitrogen monoxide (NO) and carbon dioxide (CO2)
    • The overall reaction equation is:

NO2 (g) + CO (g) → NO (g) + CO2 (g)

    • The rate equation is:

Rate = k [NO2]2

  • From the rate equation, it can be concluded that the reaction is zero-order with respect to CO (g) and second-order with respect to NO2 (g)
  • This means that there are two molecules of NO2 (g) involved in the rate-determining step and zero molecules of CO (g)
  • A possible reaction mechanism could therefore be:

   Step 1:

2NO2 (g) → NO (g) + NO3 (g)                   slow (rate-determining step)

   Step 2:

NO3 (g) + CO (g) → NO2 (g) + CO2 (g)                             fast

   Overall:

2NO2 (g) + NO3 (g) + CO (g) → NO (g) + NO3 (g) + NO2 (g) + CO2 (g)

   Simplify (remove species on both sides of the equation):

NO2 (g) + CO (g) → NO (g) + CO2 (g)

Exam Tip

  • It is important that the elementary steps for a proposed mechanism also agree with the experimentally determined rate equation
    • The rate equation and the overall reaction must be related, i.e. the correct chemical species involved 
  • Remember: There is no direct link between the orders in the rate equation and the stoichiometry of the overall equation
    • However, the rate equation can be derived directly from the rate-determining step and its stoichiometry

Predicting the reaction order & deducing the rate equation

  • The order of a reactant and thus the rate equation can be deduced from a reaction mechanism if the rate-determining step is known
  • For example, the reaction of nitrogen oxide (NO) with hydrogen (H2) to form nitrogen (N2) and water

2NO (g) + 2H2 (g) → N2 (g) + 2H2O (l)

  • The reaction mechanism for this reaction is:
Step 1:

NO (g) + NO (g) → N2O2 (g)  fast

Step 2:
 
N2O2 (g) + H2 (g) → H2O (l) + N2O (g)  slow (rate-determining step)

Step 3:

 N2O (g) + H2 (g) → N2 (g) + H2O (l)  fast

  • The second step in this reaction mechanism is the rate-determining step
  • The rate-determining step consists of:
    • N2O2 which is formed from the reaction of two NO molecules
    • One H2 molecule
  • The reaction is, therefore, second order with respect to NO and first order with respect to H2
  • So, the rate equation becomes:

Rate = k [NO]2 [H2]

  • The reaction is, therefore, third order overall

Exam Tip

  • Intermediates in the mechanism cannot appear as substances in the rate equation
  • Instead, the chemicals required to make the intermediate feature in the rate equation
  • This is why you substitute the N2O2 in the above example
    • Step 1 shows that 2NO molecules are required to form the necessary N2O2

Energy Profiles in Multistep Reactions

Single-step reactions

  • When any reacting molecules collide with bond breaking and bond formation occurring, the interacting molecules will be in an unstable, high-energy state temporarily
    • This transition state will be of a higher energy than either the reactants or products and corresponds to the activation energy
  • The exothermic reaction of hydrogen and iodine to form hydrogen iodide will be used to discuss how energy level diagrams relate to the rate-determining step

Energy profile of an exothermic reaction, showing the transition state

Energy profile of a single step reaction has one peak corresponding to a transition state

The energy profile for the exothermic reaction of hydrogen and iodine

  • As the reaction proceeds, covalent bonds start to form between the hydrogen and iodine atoms from the hydrogen and iodine molecules
  • At the same time, the covalent bonds within the hydrogen and iodine molecules grow longer and become weaker
  • This results in the transition state complex shown

Transition state

The transition state complex contains 2 hydrogen atoms and 2 iodine atoms

The transition state complex for the reaction of hydrogen and iodine

  • From this transition state, the bonds between the hydrogen and iodine atoms can continue to grow shorter and stronger resulting in the formation of hydrogen iodide
  • Alternatively, the bonds within the hydrogen and iodine molecules can grow shorter and stronger which would result in the formation of the reactants
  • Hydrogen iodide will only form if the hydrogen and iodine molecules collide with kinetic energy greater than or equal to the activation energy
    • The molecules will also need to collide in the correct orientations
  • The reaction, or elementary, step with the greatest activation energy will be the rate-determining step and can be used to determine the rate equation

Multi-step reactions

  • The exothermic reaction of nitrogen dioxide and fluorine to form nitryl fluoride (NO2F) will be used to relate rate equations and rate-determining steps to the energy level diagram of a multi-step reaction:

2NO2 (g) + F2 (g) → 2NO2F (g)

  • This reaction is unlikely to occur in a single step as that would require three molecules to collide in the correct orientation and with sufficient kinetic energy
    • This is even less likely to occur as all three molecules are gaseous
  • Experimental data shows that the rate equation for this reaction is:

Rate = k[NO2][F2]

  • One proposed reaction mechanism for this reaction involves the following elementary steps:
    • Step 1: NO2 + F2 → NO2F + F
    • Step 2: NO2 + F → NO2F
  • Step 1 must be the rate-determining step as it is the only step that has reactants matching the rate equation
    • Therefore, on an energy profile, the activation energy for step 1 will be greater than the activation energy for step 2
  • The species present after the first step must be:

NO2 + NO2F + F

  • This can be deduced using the equation for elementary step 1 and the overall equation
    • The overall equation states that two NO2 react with one F2
    • Elementary step 1 uses one NO2 to form the intermediates, NO2F + F
    • This leaves one NO2 along with NO2F + F
  • This leads to the following energy profile:

Energy profile of a multistep reaction

The energy profile of this multistep reaction has two peaks corresponding to two different transition states

Energy profile for the formation of nitryl fluoride

  • Key points from the energy profile are:
    • The overall reaction is exothermic, as stated
    • The rate-determining step is the step that has the greatest activation energy
    • There is a labelled energy level for the species present after the first step which include the intermediates and unreacted reactant

Exam Tip

  • Multi-step reactions where the first elementary step is not the rate-determining step prove more challenging in terms of energy profiles
  • Remember: The rate-determining step must have the greatest activation energy
    • Do not confuse the greatest with the highest!
    • When people talk about the highest they are normally referring to where the activation energy peak is, not the actual value it represents

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Caroline

Author: Caroline

Caroline graduated from the University of Nottingham with a degree in Chemistry and Molecular Physics. She spent several years working as an Industrial Chemist in the automotive industry before retraining to teach. Caroline has over 12 years of experience teaching GCSE and A-level chemistry and physics. She is passionate about creating high-quality resources to help students achieve their full potential.