16.1.4 Rate Determining Step

Rate-determining step & intermediates

• 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

• If a reactant appears in the rate-determining step, then the concentration of that reactant will also appear in the rate equation
• For example, the rate equation for the reaction below is rate = k[CH₃Br][OH^-]

CH₃Br + OH^- → CH₃OH + Br^-

• • This suggests that both CH₃Br and OH^- take part in the slow rate-determining step
  • Molecularity is the number of reactant particles that participate in the rate-determining step
  • This reaction is a bimolecular reaction
    • Unimolecular: one species involved in the rate-determining step
    • Bimolecular: two species involved in the rate-determining step

• The intermediate is derived from substances that react together to form it in the rate-determining step
  • For example, for the reaction above the intermediate would consist of CH₃Br and OH^-

The intermediate formed from the species that are involved in the rate-determining step (and thus appear in the rate equation)

Identifying the rate-determining step

• The rate-determining step can be identified from a rate equation given that the reaction mechanism is known
• For example, propane (CH₃CH₂CH₃) undergoes bromination under alkaline solutions
• The overall reaction is:

CH₃CH₂CH₃ + Br₂ + OH^- → CH₃CH₂CH₂Br + H₂O + Br^-

• The reaction mechanism is:

Reaction mechanism for the bromination of propane under alkaline conditions

• The rate equation is:

Rate = k[CH₃CH₂CH₃][OH^-]

• From the rate equation, it can be deduced that only CH₃CH₂CH₃ and OH^- are involved in the rate-determining step and not bromine (Br₂)
• CH₃CH₂CH₃ and OH^- are only involved in the first step of the reaction mechanism, therefore the rate-determining step is:
  • CH₃CH₂CH₃ + OH^- → CH₃CH₂CH₂^- + H₂O

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

The potential energy level diagram 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

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 endothermic 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 a potential energy level diagram the activation energy for step 1 will be greater than the activation energy for step 2

• The transition state for this multi-step reaction must be:

NO₂ + NO₂F + F

• • This can be deduced using the 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 NO₂F + F
    • This leaves one NO₂ along with NO₂F + F

• This leads to the following potential energy level diagram:

Potential energy level diagram for the formation of nitryl fluoride

• Key points from the potential energy level diagram are:
  • The overall reaction is endothermic, as stated
  • The rate-determining step is the step that has the greatest activation energy
  • There is a labelled energy level for the transition state

Identifying intermediates & catalyst

• When a rate equation includes a species that is not part of the chemical reaction equation then this species is a [popover id="mT5UZwSwYE9Vt6Fl" label=catalyst]
• For example, the halogenation of butanone under acidic conditions
• The overall equation for this reaction is:

• One commonly accepted reaction mechanism is:

Reaction mechanism for the halogenation of butanone under acidic conditions

• The rate-determining step is the slowest step, identified as step 2
• However, the CH₃CH₂C(OH⁺)CH₃ cannot be used in the rate equation as it is an intermediate, not a reactant

Reaction to form the intermediate found in the rate-determining step

• Therefore, the CH₃CH₂COCH₃ and H⁺ that form the intermediate feature in the rate equation:

Rate = k[CH₃CH₂COCH₃][H⁺]

• The H⁺ is not a reactant in the overall chemical equation but it does appear in the rate equation
  • H⁺ must, therefore, be a catalyst
  • This is also supported by the H⁺ being used in step 1 and forming as a product in step 2