DP Chemistry HL R2.2 How Fast? The Rate of Chemical Change

DP Chemistry HL R2.2 How Fast? The Rate of Chemical Change - Full notes

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DP Chemistry HL | R2.2 How Fast? The Rate of Chemical Change

DP CHEMISTRY · REACTIVITY

R2.2

How Fast? The Rate of Chemical Change

Full notes for HL · IB-style

Contents

1	Rate of reaction and how to measure it
2	Collision theory and activation energy
3	The Maxwell-Boltzmann distribution
4	Factors affecting rate
5	Catalysts
6	Rate equations and orders of reaction (HL only)
7	Reaction mechanisms (HL only)
8	The Arrhenius equation (HL only)
9	Measuring rates: methods
10	Rates from graphs
11	Collision theory in detail
12	The Maxwell–Boltzmann distribution
13	Effect of concentration and pressure
14	Effect of surface area
15	Effect of temperature
16	Catalysts
17	Designing rate experiments
18	Rate and 1/time
19	Rate equations and orders (HL)
20	Mixed review: worked problems
21	Additional exam-style practice
22	Rates in industry and biology
23	Units of rate
24	Rate graphs: concentration against time
25	Common misconceptions and links
26	Exam-style question with markscheme
27	Check your understanding
28	Summary and topic map

These are full teaching notes. Read each section in order, try the Think about it question before you read on, and do every Quick check before moving to the next section.

Before you start

Rate yourself on each statement now, and again after you've worked through the notes. Shade R (not yet), A (getting there) or G (confident).

I can…	R	A	G
define rate of reaction and describe methods of measuring it
calculate rates from graphs using gradients and tangents
explain collision theory and the role of activation energy
sketch and interpret Maxwell-Boltzmann distributions
explain the effects of concentration, pressure, surface area, temperature and catalysts on rate
explain how catalysts work using energy profiles and Maxwell-Boltzmann curves
deduce rate equations and orders from experimental data
relate rate equations to reaction mechanisms and the rate-determining step
use the Arrhenius equation to determine activation energy

Sections marked HL only (Sections 6 to 8) are Additional Higher Level content. Every other section is common to SL and HL.

Guiding questions

How can the rate of a reaction be measured and controlled? Why do some factors increase rate so much more than others?

How this topic is examined

R2.2 is examined through rate graphs, explanations using collision theory and Maxwell-Boltzmann curves, and experimental design. HL questions include determining orders, rate constants and activation energy from data.

Explain every factor in terms of the frequency of collisions and the proportion of collisions with energy ≥ Ea.

Command term	What the examiner wants
State / Identify	a brief answer, no explanation
Outline / Describe	a brief or detailed account, without reasons
Explain	a detailed account with causes, reasons or mechanisms
Calculate / Determine	a numerical answer with working, correct significant figures and units
Deduce / Predict	reach a conclusion from given information
Draw / Sketch	a clear labelled diagram, structure or graph
Evaluate / Discuss	weigh strengths and limitations; reach a judgement

How to use these notes

Each section teaches one part of the syllabus in full. You'll see these boxes throughout; each has one job:

Box	What it's for
Think about it	a question to get you thinking before a section; the answer comes in the section
Key point	the one idea you must not forget
Common mistake	an error examiners see every year, and how to avoid it
In the real world	where the chemistry shows up outside the classroom
Try this	something quick you can do yourself to test the idea
Worked example	full solutions with IB-style marks (M1 method, A1 answer, R1 reasoning)
Quick check	short questions to test yourself at the end of each section

1 Rate of reaction and how to measure it

How quickly reactants are used or products form.

Think about it

Fireworks explode in a fraction of a second; iron takes months to rust. What makes some reactions so much faster than others?

The rate of reaction is the change in concentration (or amount, mass, volume) of a reactant or product per unit time. Units are usually mol dm−3 s−1.

Rate can be followed by measuring any property that changes during the reaction: volume of gas produced (gas syringe or inverted measuring cylinder); loss of mass as gas escapes (balance); colour change (colorimeter measuring absorbance); change in pH or conductivity; or the time for a visible change (such as a precipitate obscuring a cross in the sodium thiosulfate reaction).

On a graph of product against time, the rate at any instant is the gradient of the tangent at that time. The rate is fastest at the start (highest concentration of reactants) and decreases as reactants are used up, reaching zero when the reaction is complete.

Figure 1. Measuring the volume of gas produced with a gas syringe.

Figure 2. Volume of gas produced against time: the gradient (rate) decreases.

WORKED EXAMPLE 1 Rates from a graph

Using the graph, calculate the mean rate over the first 20 s, and explain why the rate decreases.

Solution

Mean rate = 38 ÷ 20 = 1.9 cm3 s−1A1

Reactant concentration decreasesR1

so collisions become less frequentR1

Quick check

1. Define rate of reaction.

2. How is the instantaneous rate found from a graph?

Answers: 1. change in concentration per unit time 2. gradient of the tangent at that time

2 Collision theory and activation energy

Particles must collide with enough energy and the right orientation.

Collision theory states that for a reaction to occur, particles must collide with each other, with sufficient energy (at least the activation energy, Ea) and with the correct orientation (geometry). Most collisions are not successful.

The activation energy is the minimum energy that colliding particles need to react: energy is needed to begin breaking bonds. It appears as the energy barrier on an energy profile (R1.1). Reactions with high Ea are slow at room temperature.

The rate depends on the frequency of collisions and the proportion of collisions that have energy ≥ Ea and correct orientation.

Figure 3. Activation energy is the energy barrier between reactants and products.

3 The Maxwell-Boltzmann distribution

Not all particles have the same energy.

In a gas (or solution), particles have a range of kinetic energies. The Maxwell-Boltzmann distribution shows the number of particles with each energy. Features: it starts at the origin (no particles have zero energy); it rises to a peak (the most probable energy); it has a long tail at high energies and never touches the x-axis; the area under the curve equals the total number of particles.

Only particles with energy ≥ Ea (the area under the curve to the right of Ea) can react on collision. At a higher temperature, the curve shifts to the right and becomes flatter, with the peak lower; the area under the curve stays the same (same number of particles), but a much larger proportion of particles have energy ≥ Ea.

Figure 4. Maxwell-Boltzmann distributions at two temperatures: more particles exceed Eₐ at the higher temperature (shaded).

Quick check

1. What does the area under a Maxwell-Boltzmann curve represent?

2. How does the curve change at higher temperature?

Answers: 1. the total number of particles 2. shifts right, flatter, lower peak; more particles above Ea

4 Factors affecting rate

Frequency of collisions and fraction with enough energy.

Concentration (solutions) or pressure (gases): more particles per unit volume, so more frequent collisions, so faster rate.

Surface area (solids): smaller pieces or powder expose more particles at the surface, giving more frequent collisions. Powdered calcium carbonate reacts faster with acid than marble chips.

Temperature: particles move faster, so collisions are slightly more frequent; but much more importantly, a greater proportion of collisions have energy ≥ Ea (Maxwell-Boltzmann). A 10 °C rise often roughly doubles the rate.

Catalyst: provides an alternative pathway with lower Ea (Section 5).

Factor	Effect on collisions	Effect on rate
Higher concentration / pressure	more frequent collisions	increases
Larger surface area	more frequent collisions at surface	increases
Higher temperature	more frequent; much greater fraction ≥ Ea	increases greatly
Catalyst	greater fraction ≥ lower Ea	increases

Common mistake

Explaining the effect of temperature only by 'more collisions'. The main reason is that a larger proportion of collisions have energy equal to or greater than Ea.

5 Catalysts

A different route with a lower barrier.

A catalyst increases the rate of a reaction without being permanently changed (it is regenerated at the end). It provides an alternative reaction pathway with a lower activation energy. On a Maxwell-Boltzmann diagram, the Ea line moves to the left, so a greater proportion of particles have enough energy to react; the curve itself does not change.

A catalyst does not change ΔH, and does not change the position of equilibrium (R2.3): it speeds up forward and reverse reactions equally.

Homogeneous catalysts are in the same phase as the reactants; heterogeneous catalysts are in a different phase, often a solid with gaseous reactants adsorbing onto its surface (iron in the Haber process; platinum and rhodium in catalytic converters). Enzymes are biological catalysts. Catalysts are important in industry and green chemistry because they reduce energy use.

Figure 5. A catalyst lowers the activation energy; ΔH is unchanged.

Quick check

1. How does a catalyst increase rate?

2. Does a catalyst change ΔH?

Answers: 1. provides an alternative pathway with lower Ea 2. no

6 Rate equations and orders of reaction (HL only)

Rate = k[A]m[B]n.

HL only

This section is Additional Higher Level content. SL students do not need it.

The rate equation (rate law) shows how rate depends on concentration: rate = k[A]m[B]n. The order with respect to each reactant (m, n) is determined experimentally, not from the balanced equation. The overall order is m + n. k is the rate constant; its units depend on the overall order, and its value changes only with temperature (and catalyst).

Zero order: rate independent of [A]; concentration-time graph is a straight line; rate-concentration graph is horizontal. First order: rate ∝ [A]; doubling [A] doubles the rate; concentration-time graph has a constant half-life. Second order: rate ∝ [A]2; doubling [A] quadruples the rate.

Orders are found by the initial rates method: change the concentration of one reactant while keeping others constant, and see how the initial rate changes.

Experiment	[A] / mol dm−3	[B] / mol dm−3	Initial rate / mol dm−3 s−1
1	0.10	0.10	2.0 × 10−4
2	0.20	0.10	4.0 × 10−4
3	0.10	0.20	8.0 × 10−4

WORKED EXAMPLE 2 Determining a rate equation

Use the table to deduce the rate equation and calculate k.

Solution

Exp 1 → 2: [A] doubles, rate doubles: first order in AM1

Exp 1 → 3: [B] doubles, rate quadruples: second order in BM1

rate = k[A][B]2; k = 2.0 × 10−4 ÷ (0.10 × 0.102) = 0.20 dm6 mol−2 s−1A1

Overall order	Units of k
0	mol dm−3 s−1
1	s−1
2	dm3 mol−1 s−1
3	dm6 mol−2 s−1

7 Reaction mechanisms (HL only)

The slowest step controls the rate.

HL only

This section is Additional Higher Level content. SL students do not need it.

Most reactions occur in a series of elementary steps called the mechanism. The molecularity of a step is the number of particles colliding in it (unimolecular, bimolecular); termolecular steps are very rare. The rate-determining step is the slowest step; the rate equation includes only species involved in (or before) the rate-determining step. Species that are formed in one step and used in another are intermediates; they do not appear in the overall equation.

Example: for 2NO2 + F2 → 2NO2F, the rate equation is rate = k[NO2][F2]. A consistent mechanism: step 1 (slow): NO2 + F2 → NO2F + F; step 2 (fast): NO2 + F → NO2F. The slow step involves one NO2 and one F2, matching the rate equation.

8 The Arrhenius equation (HL only)

Rate constants rise exponentially with temperature.

HL only

This section is Additional Higher Level content. SL students do not need it.

The Arrhenius equation relates the rate constant to temperature: k = Ae−Ea/RT, where A is the frequency factor (related to collision frequency and orientation), Ea is the activation energy (J mol−1), R = 8.31 J K−1 mol−1 and T is in kelvin. Taking logarithms: ln k = −Ea/R × (1/T) + ln A.

A graph of ln k against 1/T is a straight line with gradient = −Ea/R and intercept ln A. So Ea can be found experimentally: Ea = −gradient × R.

Figure 6. Arrhenius plot: gradient = −Ea/R.

WORKED EXAMPLE 3 Activation energy from an Arrhenius plot

The gradient of the Arrhenius plot is −6000 K. Calculate Ea.

Solution

Ea = −gradient × R = 6000 × 8.31 = 49 860 J mol−1M1

= 49.9 kJ mol−1A1

Quick check

1. What is the gradient of a ln k against 1/T plot?

2. Which step determines the rate equation?

Answers: 1. −Ea/R 2. the slowest (rate-determining) step

9 Measuring rates: methods

Following a reaction.

Rates are measured by following a change over time: gas volume (gas syringe), mass loss (balance, as gas escapes), colour (colorimeter), conductivity, pH, or the time for a precipitate to obscure a mark. Choose a method that matches a property that changes during the reaction.

Reaction	Suitable method
CaCO3 + HCl	gas volume or mass loss
Na2S2O3 + HCl	time for cross to disappear (sulfur precipitate)
I2 reactions	colorimetry
ester hydrolysis	titration of acid formed, or pH

WORKED EXAMPLE 4 Choosing a method (standard)

Suggest a method to follow the reaction of magnesium with hydrochloric acid.

Solution

Collect H2 in a gas syringe and record volume against timeR1

Quick check

1. Why is mass loss not suitable for reactions producing hydrogen in small amounts?

2. Suggest a method for a reaction that changes colour.

Answers: 1. hydrogen has a very small mass: changes are hard to detect 2. colorimetry

10 Rates from graphs

Gradients of tangents.

The average rate over an interval is the change in quantity divided by time. The instantaneous rate at a time t is the gradient of the tangent to the curve at t. The initial rate is the gradient at t = 0.

Figure 7. Gas volume against time: the rate decreases as reactants are used up.

WORKED EXAMPLE 5 Average rate (standard)

Using the graph, find the average rate over the first 20 s.

Solution

V(20) ≈ 33.0 cm3A1

33.0/20 = 1.65 cm3 s−1A1

WORKED EXAMPLE 6 Initial rate (exam level)

The tangent at t = 0 passes through (0, 0) and (25, 60). Find the initial rate.

Solution

60/25 = 2.4 cm3 s−1A1

Quick check

1. Explain why the gradient decreases over time.

2. Why does the curve level off?

Answers: 1. reactant concentration falls, fewer collisions per second 2. a reactant is used up

11 Collision theory in detail

Energy and orientation.

For a reaction to occur, particles must collide with energy at least equal to the activation energy Ea and with the correct orientation. Rate depends on the frequency of collisions and the fraction of collisions that are successful.

WORKED EXAMPLE 7 Orientation (standard)

Explain why not all collisions with sufficient energy lead to reaction.

Solution

Particles must collide with the correct orientationR1

so that the reacting groups come into contactR1

Quick check

1. Define activation energy.

Answers: 1. the minimum energy that colliding particles need to react

12 The Maxwell–Boltzmann distribution

Energy of particles.

The distribution shows how many particles have each kinetic energy. The area under the curve is the total number of particles; the area beyond Ea is the number able to react. At a higher temperature, the curve flattens and shifts right, so a much larger fraction exceeds Ea.

Figure 8. Maxwell–Boltzmann distributions at two temperatures.

WORKED EXAMPLE 8 Temperature effect (exam level)

Explain, using the distribution, why a 10 K rise can roughly double the rate.

Solution

Higher T: more particles have energy ≥ EaR1

Fraction of successful collisions increases greatly; collision frequency increases slightlyR1

Common mistake

Drawing the higher-temperature curve with a higher peak. The area (number of particles) is constant, so the peak is lower and moves to the right.

Quick check

1. What does the area under the curve represent?

Answers: 1. the total number of particles

13 Effect of concentration and pressure

More frequent collisions.

Increasing concentration (solutions) or pressure (gases) increases the number of particles per unit volume, so collisions are more frequent and the rate increases. The fraction of successful collisions does not change.

WORKED EXAMPLE 9 Concentration (standard)

Explain why doubling the concentration of HCl increases the rate of reaction with marble chips.

Solution

More acid particles per unit volumeR1

More frequent collisions with the marble surfaceR1

Quick check

1. Explain why increasing pressure speeds up a gas reaction.

Answers: 1. more particles per unit volume; more frequent collisions

14 Effect of surface area

Smaller pieces react faster.

Breaking a solid into smaller pieces increases its surface area, exposing more particles to collisions. Powders react much faster than lumps, which explains dust explosions in flour mills and coal mines.

WORKED EXAMPLE 10 Surface area (standard)

Compare the initial rates of 1 g marble chips and 1 g marble powder with excess acid, and the final volumes of gas.

Solution

Powder reacts faster (greater surface area)R1

Same final volume (same mass of CaCO3)R1

In the real world

Flour and coal dust can explode because the enormous surface area of fine particles allows extremely rapid combustion.

Quick check

1. Why are catalysts often used as fine powders?

Answers: 1. larger surface area for reactions

15 Effect of temperature

Energy and frequency.

Raising temperature increases particle kinetic energy: collisions are more frequent and, more importantly, a greater proportion have energy ≥ Ea. As a rule of thumb, a 10 °C rise roughly doubles the rate of many reactions.

WORKED EXAMPLE 11 Rule of thumb (standard)

A reaction takes 120 s at 20 °C. Estimate the time at 50 °C.

Solution

Three 10 °C rises: rate × 8M1

Time ≈ 15 sA1

Quick check

1. Why is food stored in a refrigerator?

Answers: 1. lower temperature slows the reactions that cause spoilage

16 Catalysts

An alternative pathway.

A catalyst increases the rate by providing an alternative reaction pathway with a lower activation energy. It is not used up. On a Maxwell–Boltzmann diagram, the lower Ea line shows more particles able to react. Catalysts do not change ΔH.

WORKED EXAMPLE 12 Energy profile (standard)

Describe how an energy profile changes when a catalyst is added.

Solution

Lower peak (smaller Ea)R1

Reactants and products at the same levels (ΔH unchanged)R1

In the real world

Enzymes are biological catalysts. Catalase decomposes hydrogen peroxide about 1012 times faster than without a catalyst.

Quick check

1. Name the catalyst in the Haber process.

2. Does a catalyst change the yield at equilibrium?

Answers: 1. iron 2. no; it reaches equilibrium faster

17 Designing rate experiments

Variables and control.

Change one variable at a time (independent variable), measure the rate or time (dependent variable), and keep everything else constant (control variables: temperature, volumes, concentrations, particle size, mixing).

WORKED EXAMPLE 13 Thiosulfate experiment (exam level)

Design an experiment to investigate the effect of temperature on the rate of reaction between sodium thiosulfate and hydrochloric acid.

Solution

Warm thiosulfate solution to different temperatures; same volumes and concentrationsR1

Time how long until the cross disappears; rate ∝ 1/timeR1

Repeat for reliabilityR1

18 Rate and 1/time

Clock reactions.

When the time is measured for a fixed amount of reaction (for example, a cross disappearing), the average rate is proportional to 1/time. Plotting 1/t against concentration or temperature shows the effect of the variable.

WORKED EXAMPLE 14 Processing data (standard)

At 0.10, 0.20 and 0.40 mol dm−3, times were 80, 40 and 20 s. Find 1/t and describe the relationship.

Solution

1/t = 0.0125, 0.025, 0.050 s−1A1

Rate is directly proportional to concentrationR1

19 Rate equations and orders (HL)

rate = k[A]m[B]n.

The rate equation shows how the rate depends on concentrations. Orders (m, n) are found experimentally, not from the equation. The overall order is m + n. k is the rate constant; its units depend on the overall order.

WORKED EXAMPLE 15 Order from data (exam level)

Doubling [A] quadruples the rate; doubling [B] has no effect. Write the rate equation.

Solution

Second order in A, zero order in BA1

rate = k[A]2A1

20 Mixed review: worked problems

Bringing the ideas together.

WORKED EXAMPLE 16 Mixed problem (exam level)

Magnesium ribbon reacts with 1.0 mol dm−3 HCl at 25 °C. Predict and explain the effect on the initial rate of (a) using 2.0 mol dm−3 HCl; (b) using magnesium powder; (c) cooling to 15 °C.

Solution

(a) faster: more frequent collisionsR1

(b) faster: larger surface areaR1

Quick check

1. Why does the rate decrease as the reaction proceeds?

Answers: 1. reactant concentrations fall

21 Additional exam-style practice

Paper 1 and Paper 2.

WORKED EXAMPLE 17 Paper 1 question (standard)

Which change increases the fraction of collisions with energy above Ea? A. increasing concentration B. increasing surface area C. increasing temperature D. increasing pressure

Solution

CA1

WORKED EXAMPLE 18 Paper 2 question (exam level)

(a) Sketch Maxwell–Boltzmann distributions at T1 and a higher T2. (b) Mark Ea. (c) Explain the effect of a catalyst using the diagram.

Solution

(a) T2 curve lower peak, shifted right, same areaA1

(b) vertical line at EaA1

22 Rates in industry and biology

Fast enough, safe enough.

Industry needs reactions fast enough to be economical but controllable. Catalysts reduce energy costs. In biology, enzymes allow reactions at body temperature; enzymes denature at high temperatures, so rates fall above an optimum.

WORKED EXAMPLE 19 Enzyme optimum (standard)

Explain why the rate of an enzyme reaction falls above about 40 °C.

Solution

Enzyme denatures: active site changes shapeR1

Substrate no longer fitsR1

23 Units of rate

Concentration per unit time.

Rate is usually expressed in mol dm−3 s−1, but cm3 s−1 (gas volume) or g s−1 (mass loss) are also used. Always include units and state whether the rate is average or instantaneous.

WORKED EXAMPLE 20 Converting (standard)

A rate is 0.024 mol dm−3 min−1. Convert to mol dm−3 s−1.

Solution

0.024/60M1

= 4.0 × 10−4 mol dm−3 s−1A1

24 Rate graphs: concentration against time

Curves for reactants and products.

Reactant concentration falls steeply at first and then more slowly; product concentration rises in a mirror image. Changing conditions changes the steepness; changing the amount of limiting reactant changes the final value.

WORKED EXAMPLE 21 Sketching (standard)

Describe the graph of [product] against time when the temperature is raised (same amounts).

Solution

Steeper initial gradientR1

Same final concentration, reached soonerR1

25 Common misconceptions and links

Students often think…	Actually…
temperature increases rate only by more collisions	mainly by more collisions with energy ≥ Ea
catalysts are used up	they are regenerated
catalysts change ΔH	ΔH is unchanged
orders come from the balanced equation	they are determined experimentally

How R2.2 connects to the rest of the course

Topic	Connection
R1.1 Enthalpy	energy profiles
R2.3 How far?	catalysts and equilibrium
S1.5 Ideal gases	kinetic theory
R1.4 Entropy (HL)	kinetics vs thermodynamics

26 Exam-style question with markscheme

Try this under exam conditions (about 15 minutes) before reading the markscheme.

Figure 9. Gas produced when marble chips react with excess acid.

(a) Calculate the mean rate of gas production in the first 30 s. [1]

(b) Sketch the curve if the experiment were repeated with powdered marble of the same mass. [2]

(d) Explain how a catalyst increases the rate of a reaction. [2]

(e) Doubling [X] doubles the rate; doubling [Y] has no effect. Write the rate equation and state the units of k. [2]

Markscheme

Part	Answer	Marks
(a)	50 ÷ 30 = 1.7 cm3 s−1	A1
(b)	steeper initial gradient	A1
	same final volume	A1
(c)	curve shifts right and flattens	R1
	more particles with E ≥ Ea	R1
	particles move faster: more frequent collisions	R1
	more successful collisions per unit time	R1
(d)	alternative pathway	R1
	lower activation energy	R1
(e)	rate = k[X]	A1
	s−1	A1

Examiner's comments. In (c), weaker answers mentioned only more frequent collisions.

27 Check your understanding

Section A: multiple choice

1. Rate of reaction is measured in

A. mol dm−3 s−1 B. mol dm−3 C. kJ mol−1 D. s

2. Activation energy is

A. energy released B. ΔH C. minimum energy for a successful collision D. average kinetic energy

3. At higher temperature the Maxwell-Boltzmann peak

A. is higher B. is lower and shifted right C. shifts left D. does not change

4. Increasing surface area increases rate because

A. Ea decreases B. ΔH changes C. particles move faster D. collisions are more frequent

5. A catalyst

A. increases ΔH B. lowers Ea C. is used up D. shifts equilibrium

6. The area under a Maxwell-Boltzmann curve

A. stays the same at higher temperature B. increases C. decreases D. doubles

7. Units of k for a second-order reaction

A. s−1 B. mol dm−3 s−1 C. dm6 mol−2 s−1 D. dm3 mol−1 s−1

8. A first-order reaction has

A. a straight concentration-time graph B. rate independent of concentration C. a constant half-life D. no rate constant

Section B: structured

9. State the conditions for a successful collision. [3]

10. Describe an experiment to measure the effect of concentration on the rate of Mg with HCl. [4]

11. Explain why a catalyst does not change the yield at equilibrium. [2]

12. Explain the relationship between a mechanism and the rate equation. [3]

Answers

1. A 2. C 3. B 4. D 5. B 6. A 7. D 8. C

9. Collide; energy ≥ Ea; correct orientation.

10. Vary [HCl], same Mg mass and temperature; measure gas volume over time or time to dissolve; repeat.

11. Speeds forward and reverse equally; equilibrium reached faster, same position.

12. Rate equation includes species in the rate-determining step; intermediates; molecularity.

28 Summary and topic map

Idea	Remember
Rate	change in concentration per unit time; gradient
Collision theory	collisions with E ≥ Ea and correct orientation
Maxwell-Boltzmann	higher T: more particles above Ea
Factors	concentration, pressure, surface area, temperature, catalyst
Catalysts	alternative pathway, lower Ea; ΔH unchanged
HL	rate equations, orders, k; mechanisms; Arrhenius

Before the exam, make sure you…

•use collision frequency and energy in explanations
•draw Maxwell-Boltzmann curves correctly
•find rates from tangents
•distinguish catalyst effects on rate and equilibrium

How Fast? on one page

Cover this page with a sheet of paper and try to rebuild it from memory. Whatever you can't recall is what to revise next.

Figure 10. Topic map for How Fast?.

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