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Pre-Equilibrium Approximation (College Board AP® Chemistry) : Study Guide

Oluwapelumi Kolawole

Written by: Oluwapelumi Kolawole

Reviewed by: Stewart Hird

Updated on

Pre-equilibrium Approximation

Mechanisms with an Initial Fast Step

  • When the first step of a reaction mechanism is the rate-limiting step, the rate law can be easily determined

  • However, it is less straightforward to derive rate laws for a mechanism where an intermediate is a reactant in the rate-limiting step

    • This happens when the rate-limiting step follows an initial fast step

  • For example, consider the gas-phase reaction between nitric oxide, NO and bromine gas, Br2

2NO (g) + Br2 (g) → 2NOBr (g)

  • The experimental rate law for this reaction is given as:

Rate = k[NO]2[Br2]

  • A possible reaction mechanism consistent with this rate law could be an elementary reaction, involving three reactant molecules - a termolecular reaction

NO (g) + NO (g) + Br2 (g)  → 2NOBr (g)                        Rate = k[NO]2[Br2]

  • However, termolecular reactions are rare

  • So let’s consider an alternative mechanism which does not involve termolecular reactions:

NO (g) + Br2 (g) ⇋ NOBr2 (g)                    (fast step)

NOBr2 (g) + NO (g) → 2NOBr (g)                        (slow step)  

  • From the proposed elementary reactions above:

    • The first step is fast and reversible

    • The rate-determining step involves an intermediate, NOBr2

  • Based on the equation of the rate-limiting step, the rate law would be:

Rate = k[NOBr2][NO]

  • This is not consistent with the experimental rate law because it involves an NOBr2 intermediate

  • To eliminate the intermediate and express the rate law in terms of reactants, we use the pre-equilibrium approximation

  • The pre-equilibrium approximation assumes that the fast initial step establishes an equilibrium

NO (g) + Br2 (g) ⇋ NOBr2 (g)                    (fast step)

  • The rate laws for the forward and backward reverse reactions are:

Rateforward = kf[NO][Br2]

Ratebackward = kb[NOBr]

  • Since the rate of forward and backward reactions are equal:

kf[NO][Br2] = kb[NOBr2]

Examiner Tips and Tricks

As the first step is not the rate-limiting step, we apply the pre-equilibrium approximation to connect intermediates back to reactants.

  • By rearranging the above expression, we can express the concentration of the NOBr2 intermediate, NOBr2 in terms of the concentration of the reactants:

[NOBr2] = kf/kb[NO][Br2]

  • We then rewrite the rate law equation for the rate-limiting step by substituting the above expression for the intermediate:

Rate = k[NOBr2][NO]

Rate = kopen parentheses k subscript f over k subscript b close parentheses[NO][Br2][NO]

  • All the k terms are constants, which means that they can be combined to give a rate law that is consistent with the experimental rate law:

Rate = k'[NO]2[Br2]

  • In general, whenever a fast step precedes a slow one and an equilibrium is established in the fast step, we can eliminate intermediates by applying the pre-equilibrium approximation

Worked Example

The experimental rate law for the reaction between hydrogen gas and iodine gas to produce hydrogen iodide is first order with respect to both hydrogen and iodine gas.

The overall balanced chemical equation is:

H2 (g) + I2 (g) → 2HI (g)

A chemistry student was asked to propose a possible reaction mechanism and provided the following elementary reactions:

                        I2(g) ⇋ 2I(g)                                       (fast)

                        H2(g) + I(g) +I(g) → 2HI(g)             (slow)

Show that the rate law is consistent with the proposed mechanism.

Answer:

  • Step 1: Deduce the experimental rate equation.

  • Since the reaction is first order with respect to each reactant:

Rate = k[H2][I2]

  • Step 2: Deduce the rate law based on the rate-limiting step.

  • Use the reactant coefficients in the rate-limiting step to write the rate law equation:

Rate = k[H2][I]2 

  • Step 3: Express [I]2 in terms of [I2] using the pre-equilibrium step.

  • At equilibrium for the fast step:

kf[I2] = kb[I]2

  • This can be rearranged to:

[I]2 = k subscript f over k subscript b[I2]

  • Step 4: Substitute into the rate law.

  • Substitute for [I]2:

Rate = kopen parentheses k subscript f over k subscript b close parentheses[I2][H2]

  • On simplifying:

Rate = k[I2][H2]

Examiner Tips and Tricks

Use the pre-equilibrium approximation when the first step is fast and reversible.

Substitute intermediates out to express the rate law in terms of reactants only.

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    Author
    Reviewer
    Oluwapelumi Kolawole

    Author: Oluwapelumi Kolawole

    Expertise: Chemistry Content Creator

    Oluwapelumi is a Pharmacist with over 15000+ hours of AP , IB, IGCSE, GCSE and A-Level chemistry tutoring experience. His love for chemistry education has seen him work with various Edtech platforms and schools across the world. He’s able to bring his communication skills as a healthcare professional in breaking down seemingly complex chemistry concepts into easily understood concepts for students.

    Stewart Hird

    Reviewer: Stewart Hird

    Expertise: Chemistry Lead

    Stewart has been an enthusiastic GCSE, IGCSE, A Level and IB teacher for more than 30 years in the UK as well as overseas, and has also been an examiner for IB and A Level. As a long-standing Head of Science, Stewart brings a wealth of experience to creating Topic Questions and revision materials for Save My Exams. Stewart specialises in Chemistry, but has also taught Physics and Environmental Systems and Societies.