C17: Rate of Reaction
Factors affecting rate and measuring rate
📋 Key Definitions
Rate of reaction: How fast a chemical reaction takes place. It measures how quickly reactants are used up or how quickly products are formed.
Collision theory: Chemical reactions can only happen when reactant particles collide with each other with enough energy. The minimum energy needed for a reaction is called the activation energy.
Activation energy: The minimum amount of energy that colliding particles must have for a reaction to occur. Not all collisions lead to a reaction — only successful collisions with enough energy.
Catalyst: A substance that speeds up a chemical reaction without being used up. It works by lowering the activation energy, providing an alternative reaction pathway.
Rate of reaction formulae:
Rate = amount of reactant used ÷ time
Rate = amount of product formed ÷ time
Units: g/s, cm³/s, or mol/s
📝 Factors Affecting Rate
| Factor | Effect on Rate | Explanation (Collision Theory) |
|---|---|---|
| Temperature | Higher temperature → faster reaction | Particles have more kinetic energy, move faster, collide more frequently and with more energy. More collisions exceed the activation energy. |
| Concentration | Higher concentration → faster reaction | More particles per unit volume, so collisions are more frequent. More successful collisions per second. |
| Surface area | Larger surface area → faster reaction | More particles exposed to the other reactant, so more frequent collisions. Powder reacts faster than lumps. |
| Catalyst | Adding a catalyst → faster reaction | Lowers the activation energy by providing an alternative pathway. More collisions have enough energy to be successful. Catalyst is not used up. |
Important distinction: Increasing temperature increases both the frequency of collisions AND the energy of each collision. Increasing concentration or surface area only increases the frequency of collisions — the energy of each collision stays the same.
💥 Collision Theory Explained
For a reaction to happen, particles must:
- Collide with each other
- Collide with enough energy (at least the activation energy)
If particles collide but do not have enough energy, they simply bounce apart — this is an unsuccessful collision. Only collisions with enough energy lead to a reaction — these are successful collisions.
Worked Example 1: Explaining temperature effect
Explain, in terms of particles, why increasing temperature increases the rate of reaction.
At a higher temperature, particles have more kinetic energy and move faster. This means:
1. Particles collide more frequently (more collisions per second)
2. A greater proportion of collisions have energy equal to or greater than the activation energy
Therefore, there are more successful collisions per second, so the rate of reaction increases.
Worked Example 2: Explaining surface area effect
Explain why calcium carbonate powder reacts faster with acid than the same mass of calcium carbonate lumps.
The powder has a much larger surface area than the lumps. More particles of CaCO₃ are exposed and available to collide with the acid particles. This means more frequent collisions, so more successful collisions per second, and a faster rate of reaction.
📏 Measuring Rate of Reaction
There are three common methods for measuring the rate of a reaction:
| Method | Measures | Example Reaction |
|---|---|---|
| Collecting gas volume over time | Volume of gas produced | Mg + HCl → MgCl₂ + H₂ |
| Mass lost over time | Mass decreases as gas escapes | CaCO₃ + HCl → CaCl₂ + H₂O + CO₂ |
| Colour change / disappearance | Time for solution to become opaque | Na₂S₂O₃ + HCl → sulfur precipitate |
Worked Example 3: Calculating rate from data
In a reaction, 60 cm³ of hydrogen gas is produced in 120 seconds. Calculate the mean rate of reaction.
Rate = volume of gas ÷ time = 60 ÷ 120 = 0.5 cm³/s
📈 Rate Graphs
Rate graphs show how the quantity of product (or reactant) changes over time. The gradient of the graph at any point gives the rate of reaction at that moment.
Reading rate graphs:
- The steeper the gradient, the faster the reaction
- The gradient is steepest at the start (fastest rate) and becomes shallower as the reaction slows down
- The graph levels off (plateau) when the reaction is complete — one reactant has been used up
- To calculate the rate at a specific time, draw a tangent to the curve at that point and find its gradient
Worked Example 4: Calculating rate from gradient
A graph of volume of gas produced against time shows a straight line from the origin to (40 s, 80 cm³). Calculate the rate of reaction.
Gradient = change in volume ÷ change in time = 80 ÷ 40 = 2 cm³/s
🔬 Required Practical: Effect of Concentration on Rate
Method 1: Magnesium and hydrochloric acid
Add magnesium ribbon to different concentrations of HCl and measure the volume of hydrogen gas produced over time.
Method 2: Sodium thiosulfate and hydrochloric acid
Add HCl to sodium thiosulfate solution and time how long it takes for a cross underneath the flask to disappear (as sulfur precipitate forms).
Worked Example 5: Sodium thiosulfate practical
In the sodium thiosulfate practical, the following results were obtained:
0.2 mol/dm³ HCl: time = 60 s
0.4 mol/dm³ HCl: time = 30 s
0.6 mol/dm³ HCl: time = 20 s
Calculate the rate for each concentration (rate = 1 ÷ time).
0.2 mol/dm³: rate = 1/60 = 0.0167 s⁻¹
0.4 mol/dm³: rate = 1/30 = 0.0333 s⁻¹
0.6 mol/dm³: rate = 1/20 = 0.0500 s⁻¹
Conclusion: higher concentration gives a faster rate.
❓ Practice Questions
Q1: Foundation State four factors that affect the rate of a chemical reaction.
Q2: Foundation Explain, using collision theory, why increasing the concentration of a reactant increases the rate of reaction.
Q3: Higher A reaction produces 45 cm³ of gas in 90 seconds. Calculate the mean rate. Give the unit.
Q4: Higher Explain why a catalyst increases the rate of a reaction but is not used up.
Q5: Foundation In the sodium thiosulfate practical, why is it important to use the same volume of solution and the same size cross each time?
✅ Answers
- Temperature, concentration, surface area, and the presence of a catalyst.
- Increasing concentration means more particles per unit volume. This means the particles collide more frequently. More frequent collisions means more successful collisions per second, so the rate of reaction increases.
- Rate = 45 ÷ 90 = 0.5 cm³/s
- A catalyst provides an alternative reaction pathway with a lower activation energy. This means a greater proportion of collisions have enough energy to be successful, so there are more successful collisions per second. The catalyst is not used up because it is regenerated at the end of the reaction.
- These must be kept the same to ensure a fair test — only the concentration should be changed. Different volumes or cross sizes would introduce other variables that could affect the results.
🎯 Exam Tips
- When explaining rate changes using collision theory, always mention both the frequency of collisions AND whether they have enough energy (activation energy)
- Temperature increases BOTH collision frequency AND collision energy — this is why it has the biggest effect on rate
- Concentration and surface area only increase collision frequency
- A catalyst does NOT change the amount of product formed — it only makes the reaction happen faster
- In rate calculations, rate = 1/time is only valid when comparing reactions that produce the same amount of product
- For graph questions, remember: gradient = rate, steeper = faster, plateau = reaction finished
- In required practical questions, always mention controlling variables for a fair test
🔬 Required Practical
Required Practical: Effect of Concentration on Rate
Aim: Investigate how the concentration of hydrochloric acid affects the rate of reaction with sodium thiosulfate.
Method: 1. Add 10 cm³ of sodium thiosulfate solution to a conical flask. 2. Add 10 cm³ of HCl at a specific concentration. 3. Swirl the flask and start the timer. 4. Look down through the flask at a cross drawn on a piece of paper underneath. 5. Stop the timer when the cross is no longer visible (sulfur precipitate makes the solution opaque). 6. Repeat with different concentrations of HCl. 7. Repeat each concentration three times and calculate a mean time.
Variables: IV: concentration of HCl, DV: time for cross to disappear (or rate = 1/time), Control: volume of sodium thiosulfate, volume of HCl, temperature, same cross, same flask
🔢 Maths Skills
Mathematical Skills
You need to calculate rates from data (rate = amount ÷ time or rate = 1/time), calculate gradients of rate graphs, and interpret data from required practical experiments.
Maths Example
Calculate the mean rate from this data: 48 cm³ of gas produced in 120 seconds.
Rate = volume ÷ time = 48 ÷ 120 = 0.4 cm³/s
Maths Example 2
The following times were recorded for the sodium thiosulfate experiment at 0.2 mol/dm³: 45 s, 50 s, 47 s. Calculate the mean rate.
Mean time = (45 + 50 + 47) ÷ 3 = 47.3 s. Rate = 1 ÷ 47.3 = 0.0211 s⁻¹
⚠️ Common Misconceptions
Watch Out!
1. Wrong: Increasing concentration increases the energy of collisions Correct: Increasing concentration only increases the FREQUENCY of collisions — particles do not gain more kinetic energy. Only increasing temperature increases collision energy
2. Wrong: A catalyst increases the yield of a reaction Correct: A catalyst increases the RATE but does NOT change the amount of product formed — the same amount of product is made, just faster
3. Wrong: All collisions between particles lead to reactions Correct: Only collisions with energy equal to or greater than the activation energy are successful — most collisions are unsuccessful
4. Wrong: Rate = 1/time can be used to compare any reactions Correct: Rate = 1/time only works when comparing reactions that produce the SAME amount of product — if the total product differs, you must use rate = amount ÷ time
✍️ 6-Mark Question
Extended Answer
6 marks: Explain, using collision theory, why increasing the temperature increases the rate of a chemical reaction. Include ideas about both collision frequency and collision energy.
When the temperature is increased, the reactant particles gain more kinetic energy. This means the particles move faster, so they collide more frequently with each other — there are more collisions per second. However, the key effect is that a greater proportion of the particles now have energy equal to or greater than the activation energy. This means that a higher percentage of the collisions are successful — they have enough energy to react on impact. The combination of more frequent collisions AND a greater proportion of successful collisions means that the number of successful collisions per second increases significantly, so the rate of reaction increases. This is why temperature has a greater effect on rate than concentration or surface area — those factors only increase collision frequency, not collision energy.
Mark scheme: 1 mark for particles gain kinetic energy; 1 mark for increased collision frequency; 1 mark for greater proportion exceed activation energy; 1 mark for more successful collisions per second; 1 mark for linking to increased rate; 1 mark for comparison with concentration/surface area (only affect frequency)
📊 AO3: Analyse & Evaluate
Analysis and Evaluation
A student investigated the effect of surface area on the rate of reaction between calcium carbonate and hydrochloric acid. They used the same mass (2.0 g) of CaCO₃ in two forms: large lumps and fine powder. The volume of CO₂ produced was measured over time.
- Large lumps: 50 cm³ CO₂ produced in 300 s (total 100 cm³ produced overall)
- Fine powder: 100 cm³ CO₂ produced in 50 s (total 100 cm³ produced overall)
(a) Calculate the initial rate for each experiment.
(b) Explain the difference in rate using collision theory.
(c) What can you conclude about the effect of surface area on the TOTAL amount of product formed?
Answers: (a) Large lumps rate = 50 ÷ 300 = 0.167 cm³/s. Fine powder rate = 100 ÷ 50 = 2.0 cm³/s. (b) The powder has a much larger surface area than the lumps. More CaCO₃ particles are exposed and available to collide with HCl particles. This means more frequent collisions and more successful collisions per second, so the rate is much faster. (c) Both experiments produced the same total volume (100 cm³) of CO₂. Surface area affects the RATE of reaction but NOT the total amount of product — this is determined by the amount of limiting reactant.
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