Kinetics Notes: Steady-State Approximation, Pre-Equilibrium, and Collision Theory

Steady State Approximation

  • When the mechanism involves intermediates but there is no information that the first equilibrium is fast (no fast forward/reverse balance) or no rate information is given, use the steady state approximation instead of the pre-equilibrium method.

  • Core idea: focus on the intermediate (I). It is produced by some steps and consumed by others. Over the course of the reaction, the concentration of I tends to stay relatively low and change only slowly compared to reactants and products.

  • Intuition from the lecture:

    • The intermediate is generated by a step (or steps) and consumed by multiple subsequent steps.

    • If production roughly balances consumption, the intermediate’s concentration profile looks like a shallow, flat line over time.

    • The rate of the overall reaction is governed by how fast the steps that change reactants to products proceed, but the intermediate’s concentration often remains small and near steady state.

  • General takeaway: In steady-state problems you attempt to express the rate of change of the intermediate as zero and solve for the intermediate’s concentration in terms of the reactants, then substitute into the observable rate law.

When to use steady state vs pre-equilibrium

  • Use pre-equilibrium if you have a fast equilibrium at the start followed by a slow rate-determining step. In that case you can replace the intermediate by its equilibrium expression in terms of reactants.

  • Use steady state if there is no fast pre-equilibrium (no reliable rate information for an initial equilibrium) or if the first step is not reversible or not fast enough to establish equilibrium.

  • The instructor emphasized: in problems with no rate information, or where the first equilibrium is not fast, apply the steady state approximation; do not force a pre-equilibrium solution.

Steady-state framework (general)

  • Consider a mechanism with an intermediate I formed by a reaction A + B → I (and possibly reversible) and consumed by subsequent steps I → products.

  • Production rate of I:
    r{ ext{prod}} = k1 [A]^m [B]^n

  • Consumption rates of I:
    r{ ext{cons}} = k{-1} [I] + k_2 [I] \,( ext{and possibly other terms if multiple paths})

  • Steady-state condition:
    rac{d[I]}{dt} = r{ ext{prod}} - r{ ext{cons}} \n \approx 0

  • Solve for [I] and substitute into the rate law of the step that consumes I to obtain an overall rate law in terms of reactants only.

  • Important sign note: when you write production versus consumption, keep signs consistent; the net rate of change is production minus all consumption terms.

A concrete steady-state example (NO/NO₂ system)

  • Overall reaction (example from the lecture):
    2 ext{NO} + ext{H}2 ightarrow ext{N}2 ext{O} + ext{H}_2 ext{O}

  • Proposed mechanism:

    • Step 1 (reversible): 2 ext{NO}
      ightleftharpoons ext{N}2 ext{O}2 ag{I}

    • Step 2 (irreversible): ext{N}2 ext{O}2 + ext{H}2 ightarrow ext{N}2 ext{O} + ext{H}_2 ext{O} ag{II}

  • Intermediates: ext{N}2 ext{O}2 is the intermediate.

  • Reason for steady-state: There is no rate information that makes Step 1 fast enough to justify pre-equilibrium; use steady-state for I.

  • Step-by-step for applying steady state: 1) Write a rate law for the step you will use as the basis (the one that is simplest and, ideally, irreversible). Here Step II depends on [I] and [H₂]: r2 = k2 [ ext{N}2 ext{O}2][ ext{H}_2] 2) Recognize the intermediate appears in the rate law; so apply steady state to I. 3) Write rate laws for the production and consumption of I:

    • Production of I from Step I (forward): r{ ext{prod}} = k1 [ ext{NO}]^2

    • Consumption of I via reverse of Step I: r{-1} = k{-1} [ ext{N}2 ext{O}2]

    • Consumption of I via Step II: r2 = k2 [ ext{N}2 ext{O}2][ ext{H}2] 4) Steady-state condition for [N₂O₂]: rac{d [ ext{N}2 ext{O}2]}{dt} = k1 [ ext{NO}]^2 - k{-1} [ ext{N}2 ext{O}2] - k2 [ ext{N}2 ext{O}2][ ext{H}2] = 0 5) Solve for the intermediate concentration: [ ext{N}2 ext{O}2] = rac{ k1 [ ext{NO}]^2 }{ k{-1} + k2 [ ext{H}2] } 6) Substitute [N₂O₂] into the overall rate (the rate-determining step is Step II): r = r2 = k2 [ ext{N}2 ext{O}2][ ext{H}2] = rac{ k1 k2 [ ext{NO}]^2 [ ext{H}2] }{ k{-1} + k2 [ ext{H}2] }
      7) Limiting cases (isolation test concepts):

    • If [H₂] is large, so that $k{-1} lesssim k2[ ext{H}2]$, the denominator ≈ $k2[ ext{H}2]$ and r oughly rac{ k1 k2 [ ext{NO}]^2 [ ext{H}2] }{ k2 [ ext{H}2] } = k_1 [ ext{NO}]^2
      This is pseudo-second-order in NO (rate ∝ [NO]^2 when [H₂] is in large excess).
      8) Isolation test predictions (practice):

    • With [NO] very small and [H₂] very large, the rate should appear to be second-order in NO (pseudo-second-order in NO).

    • If you reverse: [NO] very large and [H₂] very small, the observed rate law tends to be pseudo-first-order with respect to the limiting reactant (as discussed in the lecture).

  • Takeaway: This example demonstrates how steady-state yields a rate law in terms of reactants only and how approximations based on relative rates (large excess of one reagent) give simpler, testable pseudo-order forms.

Isolating the mechanism: the isolation method

  • Purpose: Test whether the derived rate law matches experimental data by varying one reactant while keeping the other in excess.

  • Strategy:

    • Make one reactant very large in concentration (to keep its concentration effectively constant) and the other very small (to amplify its effect on the rate).

    • See whether the observed rate depends on the limiting reactant as predicted by the derived rate law.

  • Example from the lecture:

    • Predicted under NO-low, H₂-high conditions: rate ∝ [NO]^2 (pseudo-second-order in NO).

    • Predicted under reversed conditions (NO-high, H₂-low): rate becomes pseudo-first-order with respect to the other reactant (as discussed, to be tested numerically in follow-up problems).

  • The instructor noted: this is a standard way to validate a proposed mechanism and its rate law; the method is used repeatedly in lab and exams to test understanding of the steady-state approach.

Step-by-step workflow (recap)

  • Identify the mechanism and the intermediate(s).

  • Decide whether steady state or pre-equilibrium is appropriate (based on information about rate constants and whether a fast equilibrium is stated).

  • Write the rate law for the simplest step (preferably irreversible) to anchor the analysis.

  • Express the rate law in terms of the intermediate; write separate rate laws for production and consumption of the intermediate.

  • Apply the steady-state condition: set the net rate of change of the intermediate to zero.

  • Solve for the intermediate concentration in terms of reactants.

  • Substitute back into the overall rate law to obtain a rate law depending only on reactants.

  • If appropriate, apply limiting-case approximations (e.g., large excess of one reactant) to obtain a simpler, testable expression (pseudo-order).

  • Use isolation method to propose experimental tests and validate the mechanism.

Collision theory and activation energy

  • Chemistry at the molecular level is governed by collisions between reactant molecules.

    • Reactions occur when collisions have enough energy and proper orientation to break bonds and form products.

  • Key energy concepts:

    • Kinetic energy: energy due to the motion of molecules; directly influenced by temperature.

    • Potential energy: stored energy within bonds and molecular configurations.

  • Activation energy ($E_a$): the minimum energy required to reach the transition state and convert reactants to products.

  • The energy barrier concept:

    • The energy profile from reactants to products shows a peak (the transition state) at height $E_a$ above the reactants.

    • The height of this barrier governs how often collisions lead to a reaction.

  • Temperature and kinetic energy distribution:

    • Molecules in a gas have a distribution of speeds/kinetic energies (not all the same).

    • At higher temperature, more molecules have kinetic energy exceeding $E_a$; this increases the fraction of successful collisions.

    • Conceptually, this shifts the distribution to the right, increasing the area under the curve beyond $E_a$.

  • Connection to rate constants:

    • A larger fraction of collisions with sufficient energy yields a higher rate constant $k$.

    • Therefore, increasing temperature generally increases reaction rates by increasing the likelihood of overcoming $E_a$.

  • Energy diagrams and terminology:

    • Activation energy $E_a$ is the barrier height separating reactants from the transition state.

    • The transition state is a high-energy, unstable arrangement of atoms at the top of the barrier.

    • In exothermic vs endothermic reactions, the thermodynamics determine product/reactant energies, but the rate is governed by the barrier height, not the overall energy difference.

  • Bond energies (illustrative example):

    • Bond dissociation energy (e.g., O=O double bond) is one factor that contributes to $E_a$ for reactions involving bond-breaking steps. A typical value mentioned was about
      364\;\text{kJ/mol} for the O=O bond (as an example memory aid in the lecture).

  • Practical takeaway:

    • Temperature control is a practical lever for changing reaction rates because it shifts the energy distribution of molecules relative to the activation barrier.

    • Collision geometry and orientation also matter; even when energy is sufficient, proper alignment is needed for reaction to proceed.

Energy profile intuition: transition state and barriers

  • The reaction coordinate moves from reactants on the left to products on the right, crossing a peak at the transition state.

  • The peak height corresponds to the activation energy $E_a$; the higher the barrier, the slower the reaction at a given temperature.

  • Lowering $E_a$ (catalysis, alternative pathways) increases the rate by allowing more collisions to surpass the barrier.

Quick contrasts to remember

  • Pre-equilibrium vs Steady State:

    • Pre-equilibrium: fast initial equilibrium, then a slow step; allows expressing an intermediate by an equilibrium constant and leads to a simple rate law in terms of reactants.

    • Steady state: no assumption of fast initial equilibrium; assume intermediate concentration stays nearly constant; solve for [I] from production and consumption rates.

  • Pseudo-order concepts:

    • Pseudo-second-order in NO: rate appears second order with respect to NO when another reactant is in large excess.

    • Pseudo-first-order in a limiting reactant: rate appears first order with respect to that limiting reactant when other reactants remain effectively constant.

Exam strategy reminders (in practice)

  • Read the mechanism carefully to decide which method applies (pre-equilibrium vs steady-state).

  • If there is no explicit fast equilibrium information, lean toward steady-state.

  • When formulating rate laws, keep intermediates out of the final expression; substitute their steady-state concentrations in terms of reactants.

  • Use the isolation method to validate the proposed mechanism by predicting how the rate depends on each reactant and testing experimentally.

Summary of key formulas (quick reference)

  • General steady-state for an intermediate I with production and two consumption paths:
    rac{d[I]}{dt} = k1 [A]^m [B]^n - k{-1}[I] - k2[I] = 0 \ ightarrow [I] = \frac{ k1 [A]^m [B]^n }{ k{-1} + k2 }

  • For a two-step NO + NO with a single intermediate I = N₂O₂:

    • Production rate: r{ ext{prod}} = k1 [ ext{NO}]^2

    • Consumption rates: r{-1} = k{-1} [ ext{N}2 ext{O}2], \ r2 = k2 [ ext{N}2 ext{O}2][ ext{H}_2]

    • Steady-state for I:
      0 = k1 [ ext{NO}]^2 - k{-1} [ ext{N}2 ext{O}2] - k2 [ ext{N}2 ext{O}2][ ext{H}2]

    • Solve for [N₂O₂]:
      [ ext{N}2 ext{O}2] = rac{ k1 [ ext{NO}]^2 }{ k{-1} + k2 [ ext{H}2] }

    • Overall rate (Step II rate):
      r = k2 [ ext{N}2 ext{O}2][ ext{H}2]
      = rac{ k1 k2 [ ext{NO}]^2 [ ext{H}2] }{ k{-1} + k2 [ ext{H}2] }

    • Limiting-case (large [H₂], pseudo-second-order in NO):
      r \,\approx\, k_1 [ ext{NO}]^2

  • Pre-equilibrium alternative (for contrast):

    • If A + B ⇌ I is fast and I → products is slow, then
      [I] = K{eq} [A][B], \ r = k2 [I] = k2 K{eq} [A][B]

These notes summarize the main ideas from the transcript: the steady-state approximation, how to apply it with a concrete NO/H₂ example, the use of the isolation method to test mechanisms, and the basic collision-energy view of why reactions run faster at higher temperatures due to activation energy barriers and the kinetic-energy distribution of molecules. The mathematical steps are kept explicit so you can reproduce the derivations during an exam or in tutorials.