📘 GCSE Chemistry · Notes & worksheet
Rates, Equilibrium and Energy Changes.
Notes
Rate of reaction
- Mean rate = amount of reactant used or product formed ÷ time.
- The rate at a point on a graph = gradient of the tangent at that point.
- A graph steepest at the start then levels off: reactants are used up, concentration falls, and the reaction slows and stops.
Collision theory
A reaction needs particles to collide with at least the activation energy. The rate depends on the frequency of collisions and the proportion that are successful.
| Change | Effect | Why |
|---|---|---|
| Higher temperature | faster | particles move faster, so more frequent collisions and more with energy above activation energy |
| Higher concentration or pressure | faster | more particles per unit volume, so more frequent collisions |
| Larger surface area (powder) | faster | more exposed particles, so more frequent collisions |
| Catalyst | faster | provides an alternative pathway with a lower activation energy; not used up |
Reversible reactions and equilibrium
A ⇌ B. At dynamic equilibrium in a closed system the forward and reverse rates are equal and concentrations stay constant.
Le Chatelier's principle: if a condition changes, the system shifts to oppose the change.
- Increase concentration of a reactant: shifts towards the products.
- Increase pressure: shifts towards the side with fewer gas molecules.
- Increase temperature: shifts in the endothermic direction.
- A catalyst changes the rate only, not the position of equilibrium.
The Haber process
N₂ + 3H₂ ⇌ 2NH₃, ΔH = −92 kJ/mol (exothermic)
- Conditions: about 450 °C, 200 atmospheres, iron catalyst.
- High pressure favours ammonia (4 gas molecules → 2) but is expensive.
- Low temperature favours ammonia (exothermic) but is too slow, so a compromise is used.
Energy changes
- Exothermic: releases energy (temperature rises); products have less energy than reactants.
- Endothermic: takes in energy (temperature falls).
- Bond energy calculation: ΔH = energy to break bonds − energy released making bonds. A negative answer means exothermic.
Common mistakes
- Saying a catalyst "gives particles more energy". It lowers the activation energy.
- Saying the equilibrium "stops". It is dynamic: both reactions continue.
- Mixing up the sign in bond-energy calculations.
Exam tips
- For "explain" questions about rate use: frequency of collisions, energy of collisions, or both.
- In a compromise question, say what each condition does to yield and rate.
Worksheet
Prints the questions only, without answers. Choose “Save as PDF” in the print window to download.
Total: 27 marks.
- 48 cm³ of gas was produced in the first 80 seconds of a reaction. Calculate the mean rate in cm³/s. (2)
- Using collision theory, explain why increasing the concentration of an acid increases the rate of reaction. (3)
- Explain why magnesium powder reacts faster with acid than a magnesium ribbon of the same mass. (2)
- Explain how a catalyst increases the rate of reaction. (2)
- A rate graph for a reaction between marble and acid is steep at the start and then levels off. Explain both features. (3)
- The Haber process: N₂(g) + 3H₂(g) ⇌ 2NH₃(g), ΔH = −92 kJ/mol.
(a) State what happens to the position of equilibrium when pressure is increased, and explain why. (2)
(b) State what happens to the yield of ammonia when the temperature is increased, and explain why. (2)
(c) Explain why the industrial process uses a compromise temperature of about 450 °C. (2)
(d) State what effect the iron catalyst has on the position of equilibrium. (1) - Calculate the energy change for CH₄ + 2O₂ → CO₂ + 2H₂O using bond energies (kJ/mol): C–H = 413, O=O = 498, C=O = 805, O–H = 464. State whether the reaction is exothermic or endothermic. (5)
- Explain what is meant by "dynamic equilibrium". (3)
Show mark scheme and answers (tutor copy)
Try the worksheet first, then check your work here.
- 48 ÷ 80 (M1) = 0.6 cm³/s. (A1)
- More particles in the same volume (B1), so more frequent collisions (B1), so more successful collisions per second and a higher rate. (B1)
- Powder has a larger surface area (B1), so more particles are exposed and collisions are more frequent. (B1)
- A catalyst provides an alternative pathway with a lower activation energy (B1), so a greater proportion of collisions are successful. (B1)
- At the start reactants are at their highest concentration, so the rate is greatest (B1). The rate falls as reactants are used up (B1), and the graph levels off when one reactant has been completely used up and the reaction stops. (B1)
- (a) The equilibrium shifts to the right (B1) because there are fewer gas molecules on the right (2 instead of 4). (B1)
(b) The yield decreases (B1) because the forward reaction is exothermic, so higher temperature favours the reverse (endothermic) direction. (B1)
(c) A lower temperature gives a higher yield but the rate is too slow (B1). 450 °C gives an acceptable yield at an acceptable rate. (B1)
(d) None. A catalyst speeds up both directions equally and only helps reach equilibrium faster. (B1) - Energy to break: 4 × 413 + 2 × 498 = 1652 + 996 = 2648 (M1, A1). Energy released making bonds: 2 × 805 + 4 × 464 = 1610 + 1856 = 3466 (M1, A1). ΔH = 2648 − 3466 = −818 kJ/mol and the reaction is exothermic. (A1)
- The forward and reverse reactions both continue (B1) at the same rate (B1), so the concentrations of reactants and products stay constant in a closed system. (B1)