Unit 5 · Topic 5.9 Beta

Pre-Equilibrium Approximation

When a fast reversible step comes before the slow step, its forward and reverse rates are equal.

Practice 5: Mathematical RoutinesPractice 6: Argumentation

Question set for this topic

Part 1 · Hook

Why this matters

Picture a busy doorway between two rooms, with people walking in and out all the time, and a slow ticket booth at the far end of the second room. The number of people waiting in the second room stays steady, because as many come in as go back out. The ticket booth sets the pace, but how many are in line depends on that busy doorway.

Part 2 · Before you start

What this builds on

Part 3 · Prerequisite check

Quick check before you start

1. If the first step of a mechanism is slow, the overall rate law is:

  1. the rate law of that first step
  2. from the overall coefficients
  3. the rate law of the last step
  4. first order in each reactant
Show the answer

The slow step limits the rate; if it comes first, its rate law is the overall one.

  • Correct: the rate law of that first step:
  • from the overall coefficients:
  • the rate law of the last step:
  • first order in each reactant:

2. What is a dynamic equilibrium?

  1. Two opposite processes going on at equal rates
  2. A process that has stopped completely
  3. A process that goes in one direction only
  4. Two processes at different rates
Show the answer

In a dynamic equilibrium both directions continue at equal rates, so nothing changes overall.

  • Correct: Two opposite processes going on at equal rates:
  • A process that has stopped completely:
  • A process that goes in one direction only:
  • Two processes at different rates:

Part 4 · See it

See it first

Step 1 is drawn with a pair of arrows: A plus B forms the intermediate I at rate k1 times [A] times [B], and I falls back to A plus B at rate k-1 times [I]. Both are fast and equal. Step 2, slow: I plus C forms products at rate k2 times [I] times [C]. Below, setting the two step 1 rates equal gives [I] equal to k1 over k-1 times [A][B]; substituting into the slow step gives rate equals k2 k1 over k-1 times [A][B][C].
Forward and reverse rates of the fast step are equal, which gives the intermediate's concentration in terms of reactants. LevlPrep original diagram.

Part 5 · Step by step

How it works, step by step

  1. A fast reversible step comes before a slow stepthe fast step runs back and forth many times and stays balanced
  2. Its forward and reverse rates are equalk₁[reactants] = k₋₁[intermediate] gives the intermediate in terms of reactants
  3. The slow step's rate law contains that intermediatesubstituting removes it, leaving only reactant concentrations
  4. The constants combine into one kthe predicted rate law can be compared with the measured one

Part 6 · Key ideas

Key ideas

  • A reversible reaction step runs forward and in reverse (⇌), each with its own rate constant.
  • When a fast reversible step comes before a slow step, set its forward rate equal to its reverse rate (pre-equilibrium).
  • Solve for the intermediate, substitute into the slow step's rate law, and combine constants: k = k₂k₁/k₋₁.
  • A final rate law contains no intermediates.

Part 7 · Misconception

A common mistake

The wrong idea: The rate law of a mechanism is the slow step's rate law, even if it contains an intermediate.

What actually happens: A rate law cannot contain an intermediate. If the slow step has one, replace it using the fast reversible step before it.

Part 8 · Check yourself

Check yourself

Exam-style questions. Anything you miss goes into your review queue.

Model

A fast reversible first step

A proposed mechanism for 2 NO(g) + O₂(g) → 2 NO₂(g):

  1. NO + NO ⇌ N₂O₂  (fast, reversible; k₁ = 2.0 × 10³ M⁻¹ s⁻¹ forward, k₋₁ = 5.0 × 10⁴ s⁻¹ back)
  2. N₂O₂ + O₂ → 2 NO₂  (slow; k₂ = 1.5 × 10² M⁻¹ s⁻¹)

1. What is the rate law of the slow step, written as it stands?

  1. rate = k₂[N₂O₂][O₂]
  2. rate = k₂[NO]²[O₂]
  3. rate = k₂[NO₂]²
  4. rate = k₂[NO][O₂]
Show the answer

Step 2 is elementary: one N₂O₂ and one O₂ collide, so rate = k₂[N₂O₂][O₂]. It contains N₂O₂, an intermediate, so it is not yet the final rate law.

  • Correct: rate = k₂[N₂O₂][O₂]: Right: written from the step's coefficients.
  • rate = k₂[NO]²[O₂]: That is the final rate law after substitution, not the slow step as written.
  • rate = k₂[NO₂]²: NO₂ is the product of step 2.
  • rate = k₂[NO][O₂]: NO is not a reactant in step 2.

2. Setting the forward and back rates of step 1 equal, what is [N₂O₂]?

  1. (k₁/k₋₁)[NO]²
  2. (k₋₁/k₁)[NO]²
  3. (k₁/k₋₁)[NO]
  4. k₁k₋₁[NO]²
Show the answer

Forward rate k₁[NO]² equals back rate k₋₁[N₂O₂], so [N₂O₂] = (k₁/k₋₁)[NO]².

  • Correct: (k₁/k₋₁)[NO]²: Right: k₁[NO]² = k₋₁[N₂O₂].
  • (k₋₁/k₁)[NO]²: The ratio is upside down: solve k₁[NO]² = k₋₁[N₂O₂] for [N₂O₂].
  • (k₁/k₋₁)[NO]: Two NO molecules collide in step 1, so [NO] is squared.
  • k₁k₋₁[NO]²: The constants divide, they do not multiply.

3. What overall rate law does the mechanism predict?

  1. rate = k[NO]²[O₂]
  2. rate = k[NO][O₂]
  3. rate = k[N₂O₂][O₂]
  4. rate = k[NO]²
Show the answer

Substitute [N₂O₂] = (k₁/k₋₁)[NO]² into rate = k₂[N₂O₂][O₂]: rate = (k₂k₁/k₋₁)[NO]²[O₂] = k[NO]²[O₂].

  • Correct: rate = k[NO]²[O₂]: Right: the intermediate is replaced by reactants.
  • rate = k[NO][O₂]: Step 1 involves two NO, so NO is second order.
  • rate = k[N₂O₂][O₂]: A final rate law cannot contain an intermediate.
  • rate = k[NO]²: O₂ is a reactant in the slow step, so it stays.

4. Using the constants given, calculate the overall rate constant k = k₂k₁/k₋₁. Include units.

Type a number and its unit.

Show the answer

k = k₂k₁/k₋₁ = (1.5 × 10² M⁻¹ s⁻¹)(2.0 × 10³ M⁻¹ s⁻¹)/(5.0 × 10⁴ s⁻¹) = 6.0 M⁻² s⁻¹. Units: (M⁻¹ s⁻¹)(M⁻¹ s⁻¹)/s⁻¹ = M⁻² s⁻¹, as a third-order rate law needs.

  • Answer: 6.0 M^-2 s^-1

5. Mechanism: (1) NO + Br₂ ⇌ NOBr₂ (fast, reversible); (2) NOBr₂ + NO → 2 NOBr (slow). What rate law does it predict?

  1. rate = k[NO]²[Br₂]
  2. rate = k[NO][Br₂]
  3. rate = k[NOBr₂][NO]
  4. rate = k[NO]²
Show the answer

Slow step: rate = k₂[NOBr₂][NO]. Fast step balanced: k₁[NO][Br₂] = k₋₁[NOBr₂], so [NOBr₂] = (k₁/k₋₁)[NO][Br₂]. Rate = (k₂k₁/k₋₁)[NO]²[Br₂].

  • Correct: rate = k[NO]²[Br₂]: Right: one NO from each step.
  • rate = k[NO][Br₂]: That would be the rate law if step 1 were slow.
  • rate = k[NOBr₂][NO]: This contains the intermediate NOBr₂.
  • rate = k[NO]²: Br₂ is a reactant in step 1, so it carries into the rate law.

6. In the fast reversible step NO + NO ⇌ N₂O₂, k₁ = 2.0 × 10³ M⁻¹ s⁻¹ and k₋₁ = 5.0 × 10⁴ s⁻¹. What is [N₂O₂], in M, when [NO] = 0.010 M?

Type a number in M.

Show the answer

[N₂O₂] = (k₁/k₋₁)[NO]² = (2.0 × 10³ / 5.0 × 10⁴ M⁻¹)(0.010 M)² = (0.040 M⁻¹)(1.0 × 10⁻⁴ M²) = 4.0 × 10⁻⁶ M. The intermediate is present at a tiny concentration, as expected.

  • Answer: 4.0 × 10-6 M

Part 9 · Summary

Summary

When a fast reversible step comes before the slow step, its forward and reverse rates are equal. That equality gives the intermediate's concentration in terms of reactants, which is substituted into the slow step's rate law. The result contains only reactants, with a combined rate constant, and can be compared with the measured rate law.

Part 10 · Up next

What comes next

Part 11 · Connections

Connections