Edexcel International A Level Chemistry: Introduction to kinetics and equilibria (YCH11) – Revision Notes

For full explanations and worked examples, read the kinetics and equilibria study guide.

These notes cover Topic 9: Introduction to Kinetics and Equilibria of the Pearson Edexcel International Advanced Subsidiary/Advanced Level in Chemistry specification (YCH11), Issue 1, September 2017, outcomes 9.1 to 9.11. Topic 9 is found in Unit 2: Energetics, Group Chemistry, Halogenoalkanes and Alcohols (IAS level). Pair them with the practice questions, the printable checklist, the course hub and the free diagnostics.

Key definitions

Word	What it means
Rate of reaction	How fast a reactant’s (or product’s) amount or concentration changes, per second or minute
Activation energy, Ea	Minimum energy colliding particles need for reaction to occur
Successful collision	A collision with energy ≥ Ea that leads to reaction
Catalyst	Increases rate by giving an alternative route of lower Ea; not used up overall
Intermediate	Short-lived species formed on the catalysed route; the dip between the humps on a profile
Reversible reaction	Can go forwards and backwards under the same conditions
Dynamic equilibrium	Forward rate = backward rate; concentrations of reactants and products constant
Homogeneous system	All species in the same phase
Position of equilibrium	The relative amounts of products and reactants at equilibrium

9.1 and 9.2: collision theory in one table

Factor	What changes	Fraction with E ≥ Ea
Concentration up	Collision frequency up	Same
Gas pressure up	Collision frequency up	Same
Surface area up	Collision frequency up	Same
Temperature up	Collision frequency up slightly	Much larger
Catalyst added	Ea lowered	Larger

Write “more frequent collisions” or “more collisions per unit time”, never just “more collisions”.

9.3: rate calculations

Rate from time. rate ∝ 1/time (unit s⁻¹).

Method in steps:

1. Find 1/t for each run.
2. Compare runs by the ratio of their 1/t values.
3. To test proportionality, divide each 1/t by the variable (such as concentration). A constant ratio means rate ∝ that variable.

Worked reminder: run A takes 50 s, run B takes 20 s. Rate B ÷ rate A = (1/20) ÷ (1/50) = 2.5. Run B is 2.5 times faster.

Rate from a graph.

1. Draw a tangent touching the curve at the chosen time (t = 0 for the initial rate).
2. Make the tangent long and pick two points on it far apart.
3. Gradient = Δy ÷ Δx. Units = y-axis unit ÷ x-axis unit.
4. Quote the rate as positive, even for a falling reactant curve.

Worked reminder: a tangent to a reactant concentration curve runs from (0 s, 0.0080 mol dm⁻³) to (400 s, 0 mol dm⁻³). Gradient = −0.0080 ÷ 400 = −2.0 × 10⁻⁵. Rate = 2.0 × 10⁻⁵ mol dm⁻³ s⁻¹.

The gradient falls over time because reactant concentration falls, so collisions become less frequent. A flat line means the reaction has finished.

9.4: Maxwell-Boltzmann distribution

Axes: number of molecules (y) against energy (x).

Shape checklist:

• begins at zero (every molecule has some energy);
• one peak, not symmetrical, long tail to the right;
• tail never touches the energy axis;
• area under curve = total number of molecules;
• Ea marked to the right of the peak; the area to its right is the number of molecules that can react.

Higher temperature: peak lower and further right; same area; curves cross; much more area beyond Ea. The main reason rate increases is that a greater proportion of collisions have energy ≥ Ea.

9.5, 9.6 and 9.8: catalysts

Feature	Uncatalysed	Catalysed
Humps on profile	One	Usually two, with an intermediate between
Ea	Higher	Lower; measured to the highest hump
ΔH	Same	Same
Maxwell-Boltzmann curve	Unchanged	Unchanged (same temperature); only Ea mark moves left

Profile method in steps:

1. Draw reactant and product levels; the gap is ΔH (products lower for exothermic).
2. Draw one high hump for the uncatalysed route.
3. Draw two lower humps for the catalysed route, with a dip (intermediate) between them.
4. Label Ea(uncatalysed) and Ea(catalysed) from the reactant level to the top of each route’s highest hump.

9.7: catalysts and sustainability

• Lower Ea, so a lower temperature gives a useful rate: less fuel burnt, lower cost, less CO₂.
• New routes with fewer steps or by-products: higher atom economy, less waste.
• Atom economy = (molar mass of the desired product ÷ sum of the molar masses of all products) × 100%.
• A route with one product and no by-product has an atom economy of 100%.

9.9: dynamic equilibrium

Both of these must be stated:

1. Forward and backward rates are equal.
2. Reactant and product concentrations do not change.

Equilibrium needs a closed system, so that no reactant or product can escape. Both reactions keep going at the molecular level, even though nothing seems to change when you watch the mixture.

9.10: predicting shifts

Change	Position moves
Temperature up	Endothermic direction
Temperature down	Exothermic direction
Pressure up	Side with fewer gas molecules
Pressure down	Side with more gas molecules
More reactant / less product	Right
Catalyst	Nowhere; equilibrium reached sooner

Method in steps:

1. Write down ΔH for the forward reaction, or whether it is exothermic.
2. Count gas molecules on each side, using the coefficients.
3. State the direction of shift.
4. Justify it: “to absorb the heat”, “to reduce the pressure”, “to use up the added reactant”.
5. Say what happens to the yield of the product named in the question.

Suggested practicals: NO₂ (brown) ⇌ N₂O₄ (colourless), forward exothermic; ICl (dark brown liquid) + Cl₂ ⇌ ICl₃ (yellow solid).

9.11: the compromise

• Exothermic forward reaction: low temperature gives high yield but a slow rate.
• Fewer gas molecules on the product side: high pressure gives high yield and faster rate, but costs more and is less safe.
• Industry picks a compromise temperature and pressure, adds a catalyst, and recycles unreacted reactants.
• When you evaluate data, quote figures: yield and time at each condition.

Must-know distinctions

• Rate vs yield. Rate is how fast; yield is how much product at equilibrium. A catalyst changes rate, not yield.
• Frequency vs energy. Concentration, pressure and surface area raise collision frequency; temperature and catalysts change the fraction of collisions that succeed.
• Constant vs equal. At equilibrium, concentrations are constant, not equal.
• Intermediate vs transition state. An intermediate sits in a dip; the top of a hump is not the intermediate.

Quick self-test

1. A run takes 40 s. Calculate the rate as 1/time.
2. A tangent to a gas volume curve goes through (0 s, 4 cm³) and (50 s, 54 cm³). Calculate the rate.
3. Why does the Maxwell-Boltzmann curve begin at zero?
4. State what happens to the area under the curve when temperature rises.
5. A catalysed route has Ea = 84 kJ mol⁻¹; uncatalysed Ea = 132 kJ mol⁻¹. By how much has the catalyst lowered Ea?
6. Does a catalyst change ΔH?
7. Give two ways a catalyst can make an industrial process more sustainable.
8. State the two features of dynamic equilibrium.
9. For H₂(g) + Br₂(g) ⇌ 2HBr(g), what is the effect of raising pressure on the position of equilibrium?
10. Doubling the concentration of a reactant doubles the rate. A run took 84 s. How long would the run at double concentration take?
11. An equilibrium has an endothermic forward reaction. What happens to the yield when the temperature falls?

Answers

1. 1/40 = 0.025 s⁻¹.
2. (54 − 4) ÷ 50 = 1.0 cm³ s⁻¹.
3. No molecule has zero energy.
4. It stays the same, because the number of molecules is unchanged.
5. 132 − 84 = 48 kJ mol⁻¹.
6. No. Reactants and products are the same, so ΔH is the same.
7. Lower operating temperature, so less energy is used; a route with higher atom economy, so less waste.
8. Forward rate equals backward rate; concentrations of reactants and products constant.
9. No effect: 2 gas molecules on each side.
10. Rate doubles, so time halves: 42 s.
11. The yield falls, because the position moves in the exothermic (backward) direction.

Where marks are usually lost

• Writing “more collisions” with no reference to time.
• Explaining a temperature rise by collision frequency alone.
• Drawing a curve that crosses or touches the energy axis, or with a different area at the higher temperature.
• Adding a second Maxwell-Boltzmann curve for a catalyst instead of a second Ea mark.
• Labelling the intermediate at the top of a hump.
• Missing units on a gradient, or taking points from the curve instead of the tangent.
• Saying the forward and backward reactions stop at equilibrium.
• Claiming a catalyst raises the yield or shifts the equilibrium.
• Counting atoms instead of gas molecules when predicting a pressure change.
• Choosing conditions in a compromise question without quoting the data.

Official syllabus

Pearson Edexcel International Advanced Subsidiary/Advanced Level in Chemistry (XCH11/YCH11) specification, Issue 1, September 2017, Topic 9: Introduction to Kinetics and Equilibria (outcomes 9.1 to 9.11). Published by Pearson Education Limited.