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D. Energy Changes and Rates of Reaction · Energy Changes and Rates of Reaction

Reaction Mechanisms, Rate-Determining Steps and Rate Laws

D3.7 D3.5 D3.6 A1.12 A2.2About 70 minutes
Learning goals
  • I can explain what elementary steps, reaction mechanisms and molecularity are.
  • I can identify intermediates and catalysts in a mechanism and check that the steps add up to the overall equation.
  • I can explain how the rate-determining step controls the overall rate and relate it to a potential energy diagram.
  • I can determine a rate law and rate constant from initial-rate data and use it to test a proposed mechanism.

Words to know

elementary step
A single collision or breakdown event in a reaction mechanism.
reaction mechanism
The series of elementary steps that add up to an overall reaction.
molecularity
The number of particles reacting in an elementary step: unimolecular, bimolecular or termolecular.
intermediate
A substance formed in one step of a mechanism and used up in a later step.
rate-determining step
The slowest step in a mechanism, which limits the overall rate.
rate law
An equation, rate = k[A]ᵐ[B]ⁿ, that shows how the rate depends on reactant concentrations; found by experiment.
rate constant (k)
The proportionality constant in a rate law; it changes with temperature and with a catalyst.
order of reaction
The exponent of a reactant's concentration in the rate law.

What actually happens, one collision at a time?

The balanced equation for a reaction tells you the starting materials and the final products. It does not usually tell you how the reaction happens. Consider the combustion of propane:

C₃H₈(g) + 5O₂(g) → 3CO₂(g) + 4H₂O(g)

If this happened in one step, one propane molecule and five oxygen molecules would all have to collide at the same instant, with the right energy and orientation. That is extremely unlikely. Even a collision of three particles at once is rare. Instead, most reactions happen through a series of simple steps, called a reaction mechanism. (Combustion of hydrocarbons involves many steps with reactive fragments called radicals.)

Elementary steps and mechanisms

An elementary step is a single collision event (or the breakdown of one particle). A reaction mechanism is the series of elementary steps that add up to the overall reaction.

  • The molecularity of a step is the number of particles that react in it: unimolecular (1), bimolecular (2) or termolecular (3, rare).
  • An intermediate is formed in one step and used up in a later step. It does not appear in the overall equation.
  • A catalyst is used up in an early step and regenerated in a later step. It also does not appear in the overall equation.

Example: at temperatures below about 225 °C, nitrogen dioxide and carbon monoxide react through two steps:

  • Step 1 (slow): NO₂(g) + NO₂(g) → NO₃(g) + NO(g)
  • Step 2 (fast): NO₃(g) + CO(g) → NO₂(g) + CO₂(g)
  • Overall: NO₂(g) + CO(g) → NO(g) + CO₂(g)

Adding the steps: one NO₃ appears on both sides and one NO₂ appears on both sides, so they cancel. NO₃ is an intermediate.

reaction progress →Eₚreactants2NO₂ + COproductsNO + NO₂ + CO₂intermediateNO₃ + NO + COstep 1 (slow)high Eastep 2 (fast)low EaEa₁Ea₂
Potential energy diagram for the two-step NO₂ + CO mechanism (not to scale). Each step has its own activated complex (peak). The intermediate sits in the valley between the peaks. Step 1 has the higher activation energy, so it is the slow step.

The rate-determining step

The overall reaction cannot go faster than its slowest step, just as a line of people cannot move through a door faster than one at a time. The slowest step is called the rate-determining step. Speeding up a fast step makes little difference; speeding up the slow step speeds up the whole reaction.

On a potential energy diagram, the rate-determining step is usually the step with the highest activation energy (the biggest climb from the valley or level before it).

wash30 s per dishdry4 min per dishput away20 s per dishslowest station = bottleneckDishes leave the line about one every 4 min, no matter how fast washing is.In a mechanism, the slowest step is the rate-determining step.
An assembly-line analogy: the overall rate is set by the slowest station, just as the overall rate of a reaction is set by its rate-determining step.

Rate laws

Experiments show how the rate depends on concentration. The result is written as a rate law:

rate = k[A]ᵐ[B]ⁿ

  • k is the rate constant. It is constant at a given temperature and gets larger when temperature increases or a catalyst is added.
  • m and n are the orders of the reaction with respect to A and B. They are usually 0, 1 or 2, and they must be found by experiment. They are not simply the coefficients of the overall balanced equation.

What orders mean: if a reactant is first order, doubling its concentration doubles the rate. If it is second order, doubling its concentration multiplies the rate by 2² = 4. If it is zero order, changing its concentration has no effect on the rate.

The link to mechanisms: for a single elementary step, the rate law can be written from the step itself, because the rate depends on how often those particles collide. For the step A + B → products, rate = k[A][B]; for 2A → products, rate = k[A]². So the rate law of the rate-determining step usually gives the rate law of the whole reaction. For NO₂ + CO, step 1 (NO₂ + NO₂) is slow, so:

rate = k[NO₂]²

This is exactly what experiments find: the rate does not depend on [CO] at all (CO is zero order), because CO only takes part in the fast step.

Worked example

Example 1: Finding orders from initial rates

Sample data for 2NO(g) + O₂(g) → 2NO₂(g) at a fixed temperature:

  • Trial 1: [NO] = 0.0100 mol/L, [O₂] = 0.0100 mol/L, initial rate = 7.0 × 10⁻³ mol/(L·s)
  • Trial 2: [NO] = 0.0200 mol/L, [O₂] = 0.0100 mol/L, initial rate = 2.8 × 10⁻² mol/(L·s)
  • Trial 3: [NO] = 0.0100 mol/L, [O₂] = 0.0200 mol/L, initial rate = 1.4 × 10⁻² mol/(L·s)

Find the rate law and the value of k.

  1. Trials 1 and 2: [NO] doubles, [O₂] stays the same. Rate increases by 2.8 × 10⁻² ÷ 7.0 × 10⁻³ = 4 = 2². So the reaction is second order in NO.
  2. Trials 1 and 3: [O₂] doubles, [NO] stays the same. Rate increases by 1.4 × 10⁻² ÷ 7.0 × 10⁻³ = 2 = 2¹. So the reaction is first order in O₂.
  3. Rate law: rate = k[NO]²[O₂]. The overall order is 2 + 1 = 3.
  4. Using trial 1: k = rate ÷ ([NO]²[O₂]) = 7.0 × 10⁻³ ÷ ((0.0100)² × 0.0100) = 7.0 × 10⁻³ ÷ 1.0 × 10⁻⁶ = 7.0 × 10³ L²/(mol²·s).

rate = k[NO]²[O₂], with k = 7.0 × 10³ L²/(mol²·s).

Worked example

Example 2: Analysing a mechanism

Hydrogen peroxide decomposes faster when iodide ions are added. A proposed mechanism is:

Step 1 (slow): H₂O₂(aq) + I⁻(aq) → H₂O(l) + IO⁻(aq)

Step 2 (fast): H₂O₂(aq) + IO⁻(aq) → H₂O(l) + O₂(g) + I⁻(aq)

(a) Write the overall equation. (b) Identify the intermediate and the catalyst. (c) Predict the rate law.

  1. (a) Add the steps and cancel what appears on both sides (IO⁻ and I⁻): 2H₂O₂(aq) → 2H₂O(l) + O₂(g).
  2. (b) IO⁻ is formed in step 1 and used in step 2: it is the intermediate. I⁻ is used in step 1 and regenerated in step 2: it is the catalyst.
  3. (c) Step 1 is the rate-determining step and is bimolecular, so rate = k[H₂O₂][I⁻]. This matches the experimental rate law, which supports (but cannot prove) the mechanism.

2H₂O₂ → 2H₂O + O₂; IO⁻ is the intermediate; I⁻ is the catalyst; rate = k[H₂O₂][I⁻].

A mechanism must pass two tests

A proposed mechanism is acceptable only if (1) its steps add up to the overall balanced equation, and (2) it predicts the experimental rate law. Passing both tests makes a mechanism reasonable; it does not prove it. Chemists look for further evidence, such as detecting the intermediate.

Mechanisms in the atmosphere

In the stratosphere, chlorine atoms released from chlorofluorocarbons (CFCs) by ultraviolet light destroy ozone through a two-step mechanism:

Step 1: Cl(g) + O₃(g) → ClO(g) + O₂(g)

Step 2: ClO(g) + O(g) → Cl(g) + O₂(g)

Overall: O₃(g) + O(g) → 2O₂(g)

The chlorine atom is a catalyst (regenerated in step 2), and ClO is an intermediate. Because each chlorine atom is regenerated, a single atom can destroy many ozone molecules before it is removed by other reactions. Understanding this mechanism led to the Montreal Protocol (1987), an international agreement signed in Montréal to phase out CFCs. The ozone layer is slowly recovering.

A Canadian contribution

John Polanyi, a chemist at the University of Toronto, studied the energy of molecules formed in individual elementary reactions by detecting the faint infrared light they give off. He shared the 1986 Nobel Prize in Chemistry with Dudley Herschbach and Yuan T. Lee for their contributions to understanding the dynamics of elementary chemical processes. Work like this tests the ideas in this lesson, such as how energy and orientation determine whether a collision leads to a reaction.

Just for fun

Termolecular steps are rare because three particles must arrive at the same place at the same instant. Anyone who has tried to get three busy friends to the same coffee shop at the same time knows the feeling.

Hands-on activity

Build a family assembly line

You need: 20 sheets of scrap paper, a stapler (or paper clips), a pen, a timer, and two or three family members.

  1. Set up three stations: fold a sheet in half, write a three-word message on it, and staple (or clip) it shut.
  2. Run the line for 3 minutes. Count how many finished notes come out. Watch where sheets pile up.
  3. Now make the slowest station faster (give it an extra person, or make the message one word). Run again. Then make a fast station even faster instead, and run again.
  4. Compare the three counts.

What to look for: the pile-up shows the rate-determining step. Speeding up the slowest station increases the output a lot; speeding up a fast station barely changes it. Write two sentences connecting this to the NO₂ + CO mechanism.

Your turn

1

Define an elementary step and a reaction mechanism.

2

State the molecularity of the step 2NO₂(g) → NO₃(g) + NO(g).

3

Mechanism: Step 1: A + B → C (slow); Step 2: C + B → D (fast). Write the overall equation and identify the intermediate.

4

Write the rate law for the elementary step A + 2B → products.

5

A reaction is first order in X. If [X] is tripled, what happens to the rate?

6

A reaction is second order in Y. If [Y] is halved, what happens to the rate?

7

For the ozone mechanism, explain why Cl is a catalyst and not an intermediate.

More practice: warm-up

1

What is the rate-determining step?

2

Why are termolecular steps rare?

3

In rate = k[A]²[B], what is the overall order?

4

Can the orders in a rate law be read from the coefficients of the overall balanced equation?

5

Does an intermediate appear in the overall equation?

More practice: on your own

1

For 2NO + O₂ → 2NO₂ with rate = 7.0 × 10³[NO]²[O₂], find the rate when [NO] = 0.0300 mol/L and [O₂] = 0.0100 mol/L. Explain your method.

2

Sample data for A + B → C: trial 1 [A] = 0.10, [B] = 0.10, rate 2.0 × 10⁻⁴; trial 2 [A] = 0.20, [B] = 0.10, rate 4.0 × 10⁻⁴; trial 3 [A] = 0.10, [B] = 0.20, rate 2.0 × 10⁻⁴. Find the rate law and explain.

3

Explain why adding more CO does not speed up the NO₂ + CO reaction below 225 °C.

4

A proposed one-step mechanism for NO₂ + CO → NO + CO₂ would predict rate = k[NO₂][CO]. Does it agree with experiment? What does that tell you?

5

Explain, using the potential energy diagram, why step 1 of the NO₂ + CO mechanism is slower than step 2.

Challenge

1

A proposed mechanism for 2NO(g) + Br₂(g) → 2NOBr(g) is: Step 1: NO + Br₂ → NOBr₂ (slow); Step 2: NOBr₂ + NO → 2NOBr (fast). (a) Check that it adds up. (b) Predict the rate law. (c) Experiments give rate = k[NO]²[Br₂]. Does this mechanism pass both tests?

2

Sample data for 2A + B → products: [A] doubled (B constant) gives 4 times the rate; [B] tripled (A constant) gives 3 times the rate. If the rate is 1.2 × 10⁻³ mol/(L·s) at [A] = [B] = 0.050 mol/L, find k and its units.

3

The rate constant k for a reaction increases from 0.010 to 0.040 when the temperature rises by 20 °C, but the rate law stays rate = k[A]. Explain, using collision theory, why k changes but the orders do not.

Check your understanding

Quiz

1.

An intermediate in a reaction mechanism is

2.

The rate-determining step of a mechanism is

3.

For rate = k[A]²[B], doubling [A] while keeping [B] constant

4.

Below 225 °C, NO₂ + CO → NO + CO₂ has rate = k[NO₂]². Which statement explains why [CO] does not appear?

5.

In the mechanism Cl + O₃ → ClO + O₂ followed by ClO + O → Cl + O₂, the catalyst is

6.

Which is the best reason to reject a proposed mechanism?

For parents and teachers

Big ideas (D3.7): most reactions happen through a series of simple elementary steps called a mechanism. The slowest step (the rate-determining step) controls the overall rate. A rate law, rate = k[A]ᵐ[B]ⁿ, is found by experiment, and a good mechanism must both add up to the overall equation and predict the experimental rate law. Intermediates are made and then used up; catalysts are used and then regenerated.

Ontario context: the depth of rate-law calculations varies between SCH4U classrooms. The curriculum focuses on elementary steps and the rate-determining step; rate laws and initial-rate calculations are included here because many SCH4U courses and textbooks teach them, and they help explain why the slow step matters.

How to help: The assembly-line activity makes the rate-determining step concrete. Ask your teen to point to where the pile-up was and explain why speeding up a fast station did not help.

Common difficulties:

  • Using the coefficients of the overall equation as the orders in the rate law.
  • Mixing up intermediates (made, then used) and catalysts (used, then remade).
  • Forgetting to square the factor for second order (doubling gives 4 times, not 2 times).
  • Thinking that the first step is always the slow step.