Unit 5 · Topic 5.11 Beta

Catalysis

A catalyst speeds a reaction by providing a different mechanism with lower activation energies.

Practice 4: Model AnalysisPractice 6: Argumentation

Question set for this topic

Part 1 · Hook

Why this matters

A bottle of hydrogen peroxide can sit in a cupboard for months, slowly turning into water and oxygen. Pour it on a cut and it foams within seconds, because an enzyme in your blood speeds the same reaction up enormously. The enzyme is not used up: the same molecule does the job again and again.

Part 2 · Before you start

What this builds on

Part 3 · Prerequisite check

Quick check before you start

1. In a mechanism, an intermediate is:

  1. made in one step and used in a later step
  2. used in one step and made in a later step
  3. a reactant in the overall equation
  4. the top of an energy barrier
Show the answer

An intermediate is formed, then consumed.

  • Correct: made in one step and used in a later step:
  • used in one step and made in a later step:
  • a reactant in the overall equation:
  • the top of an energy barrier:

2. A lower activation energy at the same temperature means:

  1. more collisions succeed, so the reaction is faster
  2. fewer collisions happen
  3. the products are lower in energy
  4. the molecules move faster
Show the answer

More molecules have at least Ea, so more collisions are effective.

  • Correct: more collisions succeed, so the reaction is faster:
  • fewer collisions happen:
  • the products are lower in energy:
  • the molecules move faster:

Part 4 · See it

See it first

Potential energy against reaction progress. Without a catalyst the curve climbs one tall hump. With a catalyst the reaction follows a different path with two smaller humps and a valley between them, where an intermediate sits. Both paths start at the same reactant level and end at the same product level, so the energy change of the reaction is unchanged; only the highest barrier is lower.
A catalyst opens a new pathway with lower barriers; the start and end levels do not change. LevlPrep original diagram.

Part 5 · Step by step

How it works, step by step

  1. A catalyst takes part in an early step of a new mechanismthe reaction goes by a different set of elementary steps
  2. The new steps have lower barriers than the uncatalyzed stepmore collisions succeed at the same temperature, so the rate rises
  3. The catalyst is remade in a later stepit is not used up and cancels from the overall equation
  4. The reactant and product levels are unchangedthe overall energy change and the final amount of product stay the same

Part 6 · Key ideas

Key ideas

  • A catalyst speeds a reaction by providing a new mechanism with lower barriers. It is used in one step and remade in a later step.
  • Catalyst: used, then remade. Intermediate: made, then used.
  • A catalyst does not change the overall energy change or the final amount of product, and it speeds the forward and reverse reactions alike.
  • Kinds: homogeneous, heterogeneous (surface), acid-base and enzymes.

Part 7 · Misconception

A common mistake

The wrong idea: A catalyst and an intermediate are the same thing, since both cancel out of the overall equation.

What actually happens: A catalyst is there at the start, is used in an early step and is remade later. An intermediate is not there at the start: it is made in an early step and used later.

Part 8 · Check yourself

Check yourself

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

Model

Hydrogen peroxide with iodide ions

Hydrogen peroxide breaks down slowly on its own: 2 H₂O₂(aq) → 2 H₂O(l) + O₂(g). Adding a little potassium iodide makes it fizz quickly. A proposed mechanism:

  1. H₂O₂ + I⁻ → H₂O + IO⁻ (slow)
  2. H₂O₂ + IO⁻ → H₂O + O₂ + I⁻ (fast)

1. In this mechanism, what is the role of I⁻?

  1. A catalyst: used in step 1 and remade in step 2
  2. An intermediate: made in step 1 and used in step 2
  3. A reactant: used up as the reaction goes
  4. A product: made in step 2
Show the answer

I⁻ is a reactant in step 1 and a product in step 2, so it is consumed and then regenerated, and it does not appear in the overall equation: a catalyst.

  • Correct: A catalyst: used in step 1 and remade in step 2: Right: consumed first, regenerated later.
  • An intermediate: made in step 1 and used in step 2: This is the classic mix-up. An intermediate is made first and then used; I⁻ is used first and then made.
  • A reactant: used up as the reaction goes: I⁻ is regenerated in step 2, so it is not used up.
  • A product: made in step 2: I⁻ is also used in step 1, so it is not just a product; it was there at the start.

2. Which species is an intermediate?

  1. IO⁻
  2. I⁻
  3. H₂O₂
  4. O₂
Show the answer

IO⁻ is made in step 1 and used in step 2, and it was not present at the start.

  • Correct: IO⁻: Right: made, then used.
  • I⁻: I⁻ is used, then remade: the catalyst.
  • H₂O₂: H₂O₂ is the reactant, used in both steps.
  • O₂: O₂ is a product.

3. What rate law does the mechanism predict?

  1. rate = k[H₂O₂][I⁻]
  2. rate = k[H₂O₂]²
  3. rate = k[H₂O₂][IO⁻]
  4. rate = k[H₂O₂]
Show the answer

The slow first step sets the rate: rate = k[H₂O₂][I⁻]. A catalyst can appear in a rate law, because its concentration is set by the experimenter.

  • Correct: rate = k[H₂O₂][I⁻]: Right: from the slow step, catalyst included.
  • rate = k[H₂O₂]²: That would match the overall coefficients, which is not how rate laws are found.
  • rate = k[H₂O₂][IO⁻]: IO⁻ is an intermediate and is in the fast step.
  • rate = k[H₂O₂]: I⁻ is a reactant in the slow step, so it must appear.

4. A student says, "I⁻ is in the rate law, so it must be a reactant that is used up." Which evidence refutes this?

  1. The two steps sum to an equation without I⁻; it is remade
  2. IO⁻ does not appear in the rate law
  3. The reaction fizzes more when KI is added
  4. The rate law is first order in H₂O₂
Show the answer

Adding the steps gives 2 H₂O₂ → 2 H₂O + O₂; I⁻ cancels because it is used in step 1 and remade in step 2. Being in the rate law only means the slow step involves it.

  • Correct: The two steps sum to an equation without I⁻; it is remade: Right: I⁻ cancels from the sum, so it is not consumed.
  • IO⁻ does not appear in the rate law: That is about the intermediate, not about I⁻.
  • The reaction fizzes more when KI is added: That shows I⁻ speeds the reaction, not whether it is used up.
  • The rate law is first order in H₂O₂: The order in H₂O₂ says nothing about whether I⁻ is consumed.

Graph

With and without a catalyst

Energy profiles for the same reaction by two pathways at the same temperature. Reaction progress has no units.

02550751001251501752000246810121416Reaction progressPotential energy (kJ/mol)

UncatalyzedCatalyzed

Data table
Reaction progressUncatalyzedCatalyzed
06060
16060
26060
368.476.3
491.3108.8
5122.5125
6153.8117.5
7176.6102.5
818595
9174.3105
10145125
11105135
1265107.5
1335.752.5
142525
152525
162525

5. Why is the catalyzed reaction faster at the same temperature?

  1. Its highest barrier is lower, so more collisions succeed
  2. The catalyst gives the molecules extra kinetic energy
  3. The catalyst makes the products more stable
  4. The catalyzed path has more steps, and more steps are faster
Show the answer

The catalyst opens a new mechanism whose barriers are lower than the single tall barrier. At the same temperature a larger fraction of collisions has enough energy, so the rate constant is larger.

  • Correct: Its highest barrier is lower, so more collisions succeed: Right: a lower-barrier pathway.
  • The catalyst gives the molecules extra kinetic energy: A catalyst does not add energy; the energy distribution depends on temperature.
  • The catalyst makes the products more stable: The product level is unchanged.
  • The catalyzed path has more steps, and more steps are faster: More steps do not make a reaction faster; lower barriers do.

6. In the mechanism (1) A + Q → AQ; (2) AQ + B → AB + Q, which is the catalyst and which is the intermediate?

  1. Catalyst Q; intermediate AQ
  2. Catalyst AQ; intermediate Q
  3. Catalyst B; intermediate AB
  4. Catalyst A; intermediate Q
Show the answer

Q is used in step 1 and remade in step 2 (catalyst). AQ is made in step 1 and used in step 2 (intermediate). Overall: A + B → AB.

  • Correct: Catalyst Q; intermediate AQ: Right: Q first used then remade; AQ first made then used.
  • Catalyst AQ; intermediate Q: Reversed: the catalyst is present at the start, the intermediate is not.
  • Catalyst B; intermediate AB: B is a reactant and AB the product.
  • Catalyst A; intermediate Q: A is a reactant, used up and never remade.

Part 9 · Summary

Summary

A catalyst speeds a reaction by providing a different mechanism with lower activation energies. It is consumed in one step and regenerated in a later one, unlike an intermediate, which is made and then used. A catalyst leaves the overall energy change and the final amount of product unchanged.

Part 10 · Up next

What comes next

Part 11 · Connections

Connections