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

This guide teaches Topic 9: Introduction to Kinetics and Equilibria from the Pearson Edexcel International Advanced Subsidiary/Advanced Level in Chemistry specification (YCH11), Issue 1, September 2017. It covers learning outcomes 9.1 to 9.11, in two parts: 9A Kinetics and 9B Equilibria. Its home in the specification is Unit 2: Energetics, Group Chemistry, Halogenoalkanes and Alcohols, an International AS (IAS) unit. Topics 11, 13 and 14 later build on it.

Links: the course hub, the printable checklist, the kinetics and equilibria revision notes, the kinetics and equilibria practice questions and the free diagnostics. For mole work, see the amount of substance guide.

Outcomes covered

Outcome	Skill
9.1	Explain rate changes with collision theory
9.2	Link successful collisions to activation energy
9.3	Find rate from 1/time and from tangents
9.4	Explain temperature effects with the Maxwell-Boltzmann distribution
9.5	Explain catalysts as a lower-Ea route
9.6	Draw catalysed and uncatalysed profiles, with the intermediate
9.7	Explain how catalysts make industry more sustainable
9.8	Explain catalysts with the Maxwell-Boltzmann distribution
9.9	Describe dynamic equilibrium
9.10	Predict and justify equilibrium shifts
9.11	Evaluate the yield versus rate compromise

9A: Kinetics

Collision theory and activation energy (9.1, 9.2)

Particles must collide to react, but most collisions do nothing. A collision leads to reaction only if its energy is at least the activation energy, Ea: the minimum energy a collision needs for reaction to happen.

So rate depends on how often particles collide and what fraction of those collisions have at least Ea.

Change	Effect on collisions
Higher concentration of a solution	More particles per unit volume, so more frequent collisions
Gas compressed to a higher pressure	More molecules per unit volume, so more frequent collisions
Larger surface area of a solid	More particles exposed, so more frequent collisions
Higher temperature	Collisions a little more frequent and, far more important, a larger fraction have energy ≥ Ea

Concentration, pressure and surface area change only the frequency of collisions. Temperature also changes the energy of the particles. Write “more frequent collisions” or “more collisions per second”, not just “more collisions”.

Calculating rates (9.3)

Rate tells you how quickly a reactant is used up or a product forms: the change in its amount or concentration divided by the time taken.

Method 1: rate = 1/time. If you time how long a reaction takes to reach a fixed point (a cross vanishing, a ribbon dissolving), rate is proportional to 1/time, in s⁻¹. It is an average rate, good for comparing runs.

Worked example 1. Sodium thiosulfate reacts with dilute hydrochloric acid to form a yellow precipitate of sulfur. A student times how long a cross under the flask takes to vanish, keeping total volume and acid concentration fixed.

[S₂O₃²⁻] / mol dm⁻³	0.040	0.032	0.024	0.016
Time / s	24	30	40	60

Find each rate and describe the relationship.

1/24 = 0.0417 s⁻¹; 1/30 = 0.0333 s⁻¹; 1/40 = 0.0250 s⁻¹; 1/60 = 0.0167 s⁻¹. Rate ÷ concentration is 1.04 for every run, so rate is directly proportional to thiosulfate concentration. Halving the concentration (0.032 to 0.016) doubles the time.

Method 2: gradient of a tangent. Plot amount of product or reactant against time. Draw a straight tangent touching the curve at the time you want, choose two points far apart on the tangent, and find change in y ÷ change in x. The tangent at t = 0 gives the initial rate; later tangents give the rate at time t.

Worked example 2. Excess marble chips react with dilute hydrochloric acid in a flask on a balance. Mass of carbon dioxide lost is plotted against time. The tangent at t = 0 runs from (0 s, 0 g) to (40 s, 1.20 g). The tangent at t = 150 s runs from (0 s, 1.62 g) to (200 s, 2.42 g). Find both rates and explain the difference.

Initial rate = 1.20 ÷ 40 = 0.030 g s⁻¹. Rate at 150 s = (2.42 − 1.62) ÷ 200 = 0.0040 g s⁻¹. The initial rate is 7.5 times larger. As acid is used up, its concentration falls, so collisions with the marble surface are less frequent and the rate falls. The curve levels off when the acid runs out.

Use a long tangent to cut reading errors, and build units from the axes (g s⁻¹, cm³ s⁻¹, mol dm⁻³ s⁻¹).

The Maxwell-Boltzmann distribution and temperature (9.4)

Gas molecules have a spread of energies. The Maxwell-Boltzmann distribution plots number of molecules against energy.

• The curve begins at zero, as every molecule has some energy.
• It rises to one peak (the most probable energy), then falls with a long tail.
• The tail approaches the energy axis but never meets it.
• Total area = number of molecules in the sample.
• Ea sits to the right of the peak; the area beyond it is the number of molecules that can react on collision.

At a higher temperature, with the same number of molecules, the peak is lower and moves to higher energy. The area is unchanged, so the curves cross. The area beyond Ea becomes much larger, so a much larger fraction of collisions succeed. This energy effect, not the small rise in collision frequency, explains why a modest temperature rise can increase the rate a lot.

Catalysts and reaction profiles (9.5, 9.6, 9.8)

A catalyst increases the rate of a reaction without being used up overall. It offers a different reaction route whose activation energy is lower, often by forming a short-lived intermediate that reacts on to give the product and release the catalyst.

A reaction profile plots energy against progress of reaction.

• Uncatalysed: one hump. Ea runs from reactants to the top of the hump; ΔH runs from reactants to products.
• Catalysed: usually two lower humps with a dip between them. The dip is the intermediate. The catalysed Ea runs from the reactants to the highest point on that route.
• ΔH is the same for both routes.

Worked example 3. An uncatalysed gas reaction has Ea = +125 kJ mol⁻¹ and ΔH = −46 kJ mol⁻¹. With a catalyst, the first step has a barrier 58 kJ mol⁻¹ above the reactants and forms an intermediate 18 kJ mol⁻¹ above the reactants. The second step has a barrier 31 kJ mol⁻¹ above the intermediate. Describe the catalysed profile and find its activation energy.

The second hump is 18 + 31 = 49 kJ mol⁻¹ above the reactants. The first hump (58) is higher, so the catalysed Ea is 58 kJ mol⁻¹, which is 67 kJ mol⁻¹ lower than uncatalysed. Sketch: reactants at 0; hump at +58; intermediate dip at +18; hump at +49; products at −46, the same level on both curves.

On the Maxwell-Boltzmann curve (9.8). At a fixed temperature a catalyst does not change the distribution. It moves the Ea mark to a lower energy. The area beyond the new Ea is larger, so a bigger share of the molecules can react when they collide, and the rate rises. Draw one curve with two Ea marks, labelled.

Catalysts and sustainability (9.7)

• Less energy. With a lower Ea, a useful rate is reached at a lower temperature. Less fuel is burnt to heat reactors, cutting cost and carbon dioxide emissions.
• Higher atom economy. A catalyst can make a route with fewer steps or fewer by-products possible. Ethanoic acid can be made by carbonylation of methanol using a rhodium or iridium catalyst: CH₃OH + CO → CH₃COOH. Using the Topic 1 definition, atom economy = (molar mass of the desired product ÷ sum of the molar masses of all products) × 100%. The only product here is the one wanted, so the atom economy is 100%.

9B: Equilibria

Dynamic equilibrium (9.9)

Many reactions are readily reversible. In a closed system the forward reaction slows and the backward reaction speeds up until the system reaches dynamic equilibrium:

• the forward and backward reactions run at the same rate;
• reactant and product concentrations no longer change.

Both reactions carry on; that is what “dynamic” means. “Constant” does not mean “equal”.

Predicting shifts (9.10)

For a homogeneous system (all species in one phase), use Le Chatelier’s principle: when a condition changes, the position of equilibrium moves to oppose the change.

Change	Shift	Why
Raise temperature	Endothermic direction	Absorbs the added heat
Lower temperature	Exothermic direction	Releases heat
Raise pressure (gases)	Side with fewer gas molecules	Lowers the pressure
Add a reactant, or remove a product	To the right	Uses up the reactant, or replaces the product
Add a catalyst	No shift	Speeds forward and backward reactions equally; equilibrium is reached sooner

With equal numbers of gas molecules on each side, pressure does not move the position.

Worked example 4. In the first stage of nitric acid manufacture: 4NH₃(g) + 5O₂(g) ⇌ 4NO(g) + 6H₂O(g); the forward reaction is exothermic. How does each change affect the equilibrium yield of NO: (a) raising the temperature, (b) raising the pressure, (c) adding oxygen?

(a) The backward reaction is endothermic, so the position moves left to absorb heat. Yield of NO decreases. (b) 9 gas molecules on the left, 10 on the right. Higher pressure favours the side with fewer, the left. Yield decreases. (c) The position moves right to use up the added oxygen. Yield increases.

Suggested practicals. The specification suggests two demonstrations.

• 2NO₂(g) ⇌ N₂O₄(g): NO₂ is brown, N₂O₄ colourless, and the forward reaction is exothermic. Cooling a sealed tube makes the gas paler. Compressing a syringe makes it darker at once (more molecules per unit volume), then slightly paler as the position moves towards N₂O₄.
• Chlorine with iodine gives dark brown liquid iodine(I) chloride. More chlorine turns it into yellow crystals of iodine(III) chloride: ICl + Cl₂ ⇌ ICl₃. Warming the solid turns it back towards brown ICl. This system includes a solid, so it is not homogeneous, but it shows the effect of adding a reactant.

Yield against rate: the industrial compromise (9.11)

Conditions that give the best yield often give a slow rate, so industry has to balance the two.

Worked example 5. An exothermic gas reaction, 2A(g) + B(g) ⇌ 2C(g), runs at a fixed pressure. The data are illustrative.

Temperature / °C	300	400	500	600
Equilibrium yield of C / %	45	32	19	9
Time to reach equilibrium / min	300	60	12	3

Evaluate the choice of temperature.

Lower temperature favours the exothermic forward reaction, so 300 °C gives the best yield, 45%. But reaching it takes 300 minutes, so little C is made per hour. At 600 °C equilibrium comes in 3 minutes, but the yield is only 9%. A temperature of 400–500 °C is a compromise: a moderate yield reached quickly enough for good output per hour. A catalyst lets the plant work towards the cooler end at an acceptable rate, and unreacted A and B can be recycled.

Higher pressure favours C (3 gas molecules become 2) and speeds the reaction, but very high pressure needs costly plant and compression energy, and is riskier, so a moderate pressure is chosen.

Common errors

• Explaining a temperature rise only by “more collisions”, without the larger fraction with energy ≥ Ea.
• Drawing a Maxwell-Boltzmann curve that starts above zero, touches the axis, or changes area.
• Moving the curve for a catalyst; only the Ea mark moves.
• Measuring catalysed Ea to the intermediate, not to the highest hump.
• Using points from the curve, not the tangent, or omitting units.
• Saying that at equilibrium the reaction has stopped or the concentrations are equal.
• Saying a catalyst increases the yield.
• Stating a shift with no reason given.

Next steps

Go on to the revision notes and practice questions.

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.