C21: Rate of Reaction
Understanding how fast reactions happen, what controls their speed, and how to measure and interpret rate graphs using collision theory.
Collision Theory
For a reaction to occur, reactant particles must collide with sufficient energy (the activation energy) and with the correct orientation.
Not all collisions lead to a reaction. A collision that produces a chemical reaction is called a successful collision. The rate of a reaction depends on the frequency of successful collisions between reacting particles.
Increasing the frequency of successful collisions increases the rate of reaction.
Factors Affecting Rate of Reaction
Temperature
Increasing temperature increases the rate of reaction because particles have more kinetic energy, move faster, and collide more frequently. A greater proportion of particles have energy equal to or greater than the activation energy.
As a rough guide, increasing the temperature by 10 °C approximately doubles the rate of reaction.
Concentration
Increasing the concentration of a solution increases the rate because there are more reactant particles per unit volume, leading to more frequent collisions.
Surface Area
Increasing the surface area of a solid (e.g. by powdering) increases the rate because more particles are exposed and available for collisions.
Catalysts
A catalyst increases the rate of reaction by providing an alternative reaction pathway with a lower activation energy, without being used up in the reaction.
Catalysts do not change the amount of product formed, only how quickly it forms. They are specific to particular reactions.
| Factor | How it Increases Rate | Effect on Particles | Effect on Activation Energy |
|---|---|---|---|
| Temperature | More frequent and more energetic collisions | More kinetic energy, faster movement | Lower proportion of particles below Ea |
| Concentration | More particles per unit volume | More frequent collisions | No change |
| Surface area | More exposed particles | More collision sites available | No change |
| Catalyst | Alternative pathway with lower Ea | More particles have sufficient energy | Lowers activation energy |
Measuring Rate of Reaction
There are three main methods to measure the rate of a reaction experimentally:
Method 1: Measure the volume of gas produced — use a gas syringe or an inverted measuring cylinder over water. Record volume at regular time intervals.
Method 2: Measure the change in mass — if a gas is produced, the reaction mixture loses mass. Use a balance and record mass at regular intervals.
Method 3: Measure the time for a visual change — e.g. time for a precipitate to obscure a cross mark (disappearing cross method).
Worked Example: Measuring gas volume
Marble chips (calcium carbonate) are added to hydrochloric acid. The volume of carbon dioxide gas collected every 10 seconds is recorded:
| Time / s | Volume / cm³ |
|---|---|
| 0 | 0 |
| 10 | 35 |
| 20 | 60 |
| 30 | 78 |
| 40 | 88 |
| 50 | 93 |
| 60 | 95 |
| 70 | 95 |
The reaction is fastest in the first 10 seconds (3.5 cm³/s) and slows down as the acid is used up. The reaction finishes when the volume stops changing at 95 cm³.
Rate Graphs
On a rate graph: the gradient of the curve at any point gives the rate of reaction at that time. A steeper gradient means a faster rate. When the line becomes horizontal, the reaction has finished.
Worked Example: Interpreting a rate graph
A student investigates the reaction between magnesium ribbon and excess dilute sulfuric acid. The volume of hydrogen gas is recorded over time.
Key observations from the graph:
- The curve is steepest at the start — the rate is highest at the beginning because the concentration of acid is greatest.
- The curve gradually becomes less steep — the acid is being used up, concentration decreases, so the rate decreases.
- The curve levels off — the reaction has stopped because one reactant (magnesium) has been completely used up.
- The final horizontal section shows the total volume of gas produced.
Worked Example: Comparing rates on graphs
Two experiments are carried out with the same amounts of reactants. Experiment A uses small marble chips and Experiment B uses large marble chips.
Both graphs reach the same final volume of gas (same amount of product) but Experiment A reaches it faster because the smaller chips have a larger surface area, giving a steeper initial gradient.
The Tangent Method (Higher)
To find the rate of reaction at a specific time from a curved graph, draw a tangent to the curve at that point. The gradient of the tangent equals the rate at that time.
Rate = gradient of tangent = change in volume (or mass) / change in time
Worked Example: Tangent method
From a volume-time graph for a gas-producing reaction, a tangent is drawn at 20 seconds. The tangent passes through the points (10, 45) and (40, 100).
Gradient = (100 − 45) / (40 − 10) = 55 / 30 = 1.83 cm³/s
The rate of reaction at 20 seconds is 1.83 cm³/s.
Worked Example: Tangent at the start
From a mass-time graph where a gas is released, the tangent at time = 0 passes through (0, 100.0) and (25, 97.5).
Gradient = (97.5 − 100.0) / (25 − 0) = −2.5 / 25 = −0.10 g/s
The initial rate is 0.10 g/s (the negative sign shows mass is decreasing).
Calculating Mean Rate
Mean rate of reaction = quantity of reactant used / time taken
or
Mean rate of reaction = quantity of product formed / time taken
Worked Example: Mean rate calculation
In a reaction, 0.48 g of magnesium reacts completely with hydrochloric acid in 60 seconds.
Mean rate = 0.48 / 60 = 0.008 g/s
Catalysts in Industry
Catalysts are essential in industrial processes because they speed up reactions without needing higher temperatures, which saves energy and reduces costs.
Examples of industrial catalysts:
- Iron — used in the Haber process to make ammonia
- Vanadium(V) oxide — used in the Contact process to make sulfuric acid
- Platinum and rhodium — used in catalytic converters in cars
Remember: a catalyst does NOT increase the yield of a reaction. It only speeds up how quickly the products are formed. The same amount of product is made, just in less time.
Activation Energy and Energy Profile Diagrams
The activation energy is the minimum amount of energy that particles must have to collide successfully and react.
On an energy profile diagram:
- The peak of the curve represents the activation energy
- A catalysed reaction has a lower peak — the catalyst provides a pathway with lower activation energy
- The overall energy change (ΔH) is the same with or without a catalyst
When explaining why a factor affects rate, always link back to collision theory: state what changes (particle energy / frequency of collisions), then state the effect on successful collisions, then state the effect on rate.
Practice Questions
1. Explain, in terms of collision theory, why increasing the concentration of a reactant increases the rate of reaction. (3 marks)
Increasing the concentration means there are more reactant particles per unit volume. This means particles collide more frequently. More frequent collisions mean more successful collisions per unit time, so the rate increases.
2. A reaction produces 72 cm³ of gas in 30 seconds. Calculate the mean rate of reaction in cm³/s. (2 marks)
Mean rate = 72 / 30 = 2.4 cm³/s
3. A tangent is drawn on a rate graph at 15 seconds. The tangent passes through (5, 20) and (35, 80). Calculate the rate of reaction at 15 seconds. (3 marks)
Gradient = (80 − 20) / (35 − 5) = 60 / 30 = 2.0 cm³/s
4. Explain why a catalyst increases the rate of a reaction but does not increase the amount of product formed. (3 marks)
A catalyst provides an alternative reaction pathway with a lower activation energy. This means more particles have sufficient energy to react successfully, increasing the rate. The catalyst is not used up and does not change the position of equilibrium or the amounts of reactants, so the same total amount of product is formed.
5. Describe how you would investigate the effect of surface area on the rate of reaction between calcium carbonate and hydrochloric acid. (4 marks)
Add the same mass of calcium carbonate in two forms (e.g. small chips and large chips) to equal volumes and concentrations of hydrochloric acid at the same temperature. Measure the volume of carbon dioxide gas produced at regular time intervals using a gas syringe. Compare the initial rates by comparing the initial gradients of the volume-time graphs. The small chips produce gas faster due to greater surface area.
Required Practical
Investigating the Effect of Concentration on Rate of Reaction (Disappearing Cross Method)
Method: Place a conical flask on a printed cross (e.g. on paper). Add a fixed volume of sodium thiosulfate solution to the flask and warm to the required temperature. Add a fixed volume of hydrochloric acid and start the stopwatch immediately. Look down through the flask and stop the watch when the cross is no longer visible due to the sulfur precipitate clouding the solution. Repeat at different concentrations of sodium thiosulfate (keeping temperature and HCl volume constant).
Independent variable: Concentration of sodium thiosulfate
Dependent variable: Time for the cross to disappear
Control variables: Volume of HCl, volume of thiosulfate, temperature, same cross, same eye distance
Equation: Na₂S₂O₃(aq) + 2HCl(aq) → 2NaCl(aq) + S(s) + SO₂(g) + H₂O(l)
Safety: SO₂ is a toxic gas — carry out in a well-ventilated room or fume cupboard. Avoid inhaling fumes. Wear eye protection.
Tips: Plot 1/time (which is proportional to rate) against concentration. The graph should show a directly proportional relationship if rate is proportional to concentration.
Maths Skills
Calculating Rate from Graphs
The rate of reaction at a particular time can be found from a concentration-time or volume-time graph by calculating the gradient at that point.
Gradient = change in y ÷ change in x = Δy / Δx
Tangent method: To find the rate at a specific time on a curved graph, draw a tangent to the curve at that point. Then calculate the gradient of the tangent using two points on the line well apart.
Example: If a tangent at 30 s passes through (10, 45) and (50, 15):
Gradient = (15 − 45) / (50 − 10) = −30 / 40 = −0.75 cm³/s
The negative sign shows the quantity is decreasing. The rate = 0.75 cm³/s.
Initial rate is found by drawing a tangent at t = 0 (the steepest part of the curve).
Mean rate can also be calculated: total volume of gas produced ÷ total time.
Common Misconceptions
Misconceptions About Catalysts and Rate
Catalysts increase the yield of a reaction — they make more product.
Catalysts speed up the rate of a reaction by providing an alternative pathway with a lower activation energy. They do NOT increase the yield — the same amount of product is formed, just faster. Catalysts are not used up in the reaction.
A higher temperature always gives a better result in industry.
Higher temperature increases the rate but may shift equilibrium positions (for reversible reactions) and always increases energy costs. Compromise conditions are often used.
More collisions always means a faster reaction.
Only collisions with energy equal to or greater than the activation energy lead to a reaction. Increasing the frequency of collisions increases the rate, but the energy of those collisions also matters.
6-Mark Question
Explain how temperature affects reaction rate using collision theory.
Increasing the temperature increases the rate of reaction. According to collision theory, when temperature is increased, the particles gain more kinetic energy and move faster. This means they collide more frequently, increasing the frequency of collisions. More importantly, a greater proportion of the particles have energy equal to or greater than the activation energy, so a greater proportion of collisions are successful. Both the increased frequency of collisions and the increased proportion of successful collisions contribute to a faster rate. As a rule of thumb, the rate roughly doubles for every 10 °C rise in temperature. Conversely, decreasing the temperature reduces particle kinetic energy, meaning fewer particles have sufficient energy to react on collision, and collisions occur less often, so the rate decreases.
AO3: Analysis and Evaluation
Interpreting Rate Experiment Data
A student investigated the effect of concentration on the rate of reaction between marble chips and hydrochloric acid. They recorded the volume of CO₂ produced every 10 seconds for two concentrations (0.5 mol/dm³ and 1.0 mol/dm³).
Evaluate the data: The higher concentration produced gas faster (steeper initial gradient) but both curves levelled off at the same final volume. Explain why the final volumes are the same even though the rates differ.
Answer: The same mass of marble chips was used, so the same number of moles of CaCO₃ was available. The HCl was in excess in both cases. Therefore the same total amount of CO₂ is produced — the limiting reactant (CaCO₃) determines the total yield. The rate differs because higher concentration means more HCl particles per unit volume, increasing collision frequency, but the total amount of product is determined by the limiting reactant.
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