# Rate of Reaction

> Chemistry · CIE IGCSE 0620 2026-2028
> Source: https://www.owlsprep.com/study/cie-0620-u6-rate-of-reaction/

This guide covers all Core and Extended rate of reaction content for CIE IGCSE Chemistry 0620, including calculations, measurement methods, factor effects, and Extended-only collision theory explanations for exam success.

**Prerequisites:** [Knowledge of basic chemical reactions and word equations](https://www.owlsprep.com/study/cie-0620-u3-chemical-equations/); Ability to interpret line graphs and calculate gradients

## Learning objectives

- Define rate of reaction and calculate average rate using reactant use or product formation over time
- Describe 3 common practical methods to measure reaction rate for different reaction types
- Explain the effect of concentration, pressure, surface area, temperature and catalysts on reaction rate (Core content)
- Apply collision theory and activation energy to explain rate changes (Extended only)
- Interpret rate vs time graphs to compare reaction speeds and identify reaction completion

## What is Rate of Reaction?

**Rate of Reaction** — The change in amount of reactant used up or product formed per unit time. Units vary based on measurement type, common units include g/s, cm³/s and mol/s.

*Example:* If 50cm³ of hydrogen gas is produced in 10 seconds, the average rate is 5cm³/s.

Rate can be calculated using either the decrease in mass/volume of a reactant, or the increase in mass/volume of a product, over the time taken for the change. The formula for average rate is:

$$rate = \frac{\text{change in amount of reactant or product}}{\text{time taken}}$$

**Worked example:** A reaction between magnesium and hydrochloric acid produces 24g of magnesium chloride in 2 minutes. Calculate the average rate of reaction in g/s.

1. Convert time to the required unit (seconds):

   $$2 \times 60 = 120 s$$
2. Substitute values into the rate formula:

   $$rate = \frac{24 g}{120 s} = 0.2 g/s$$

## Measuring Rate of Reaction

Three standard practical methods are used to measure rate, chosen based on the products of the reaction:

- **Gas collection**: Use a gas syringe or inverted measuring cylinder over water to collect gaseous products, record volume at regular time intervals
- **Mass loss**: Place the reaction flask on a digital balance, record mass decrease at intervals as gas escapes the container
- **Disappearing cross**: Place the reaction flask over a marked paper cross, time how long until the cross is no longer visible through the cloudy precipitate formed

**Worked example:** A student uses the disappearing cross method for the reaction between sodium thiosulfate and hydrochloric acid. The cross disappears after 40 seconds. Calculate the relative rate of reaction for this test.

1. Relative rate for the disappearing cross method is calculated as 1 divided by time taken for the cross to vanish:

   $$relative\thinspace rate = \frac{1}{40 s} = 0.025 s^{-1}$$
2. Higher relative rate values correspond to faster reactions: if the cross disappears in 20 seconds instead, the relative rate doubles to 0.05 s⁻¹.

> **Exam tip:** When evaluating the disappearing cross method, note that it relies on subjective observation, so use the same observer for all tests to get reliable results.

## Factors Affecting Rate of Reaction (Core)

Five key factors change the speed of a chemical reaction. Core candidates only need to link these factors to collision frequency, no reference to activation energy is required:

- **Higher concentration (solution)**: More reactant particles per unit volume → more frequent collisions between reactant particles → faster rate
- **Higher pressure (gaseous reactants)**: Particles are squeezed into a smaller volume → more frequent collisions → faster rate
- **Larger surface area (solid reactants)**: Smaller solid pieces have more exposed surface particles → more frequent collisions → faster rate
- **Higher temperature**: Particles have more kinetic energy, move faster → more frequent collisions → faster rate
- **Catalyst added**: Speeds up reaction without being used up → faster rate, no change to product yield

**Worked example:** A student reacts large marble chips (calcium carbonate) with dilute hydrochloric acid. State and explain the effect of using powdered marble instead of large chips on the reaction rate.

1. State the effect: The rate of reaction increases.
2. Explain the cause: Powdered marble has a much larger total surface area than the same mass of large marble chips.
3. Link to collisions: More calcium carbonate particles are exposed to the acid, so more frequent collisions between reactant particles occur per second.

## Collision Theory Explanations (Extended Only)

> **warning**
>
> This section contains content exclusive to Extended (Paper 2/4) candidates. Core (Paper 1/3) candidates may skip this section.

**Collision Theory** — Theory stating that two conditions must be met for a reaction to occur: 1) Reactant particles collide with the correct orientation, 2) Collisions have energy equal to or greater than the activation energy (minimum energy needed for a reaction to happen).

*Notation:* $E_a$ = activation energy

*Example:* A temperature increase of 10°C roughly doubles reaction rate because far more collisions meet the activation energy requirement.

For Extended answers, you must add detail to core factor explanations, linking changes to the *proportion of successful collisions* (collisions that meet both orientation and energy requirements) rather than just collision frequency.

**Worked example:** Explain, using collision theory, why increasing the temperature of a reaction mixture increases the rate of reaction.

1. Increasing temperature gives all reactant particles more kinetic energy, so they move faster, leading to more frequent collisions.
2. A significantly higher proportion of colliding particles have energy equal to or greater than the activation energy, so the number of successful collisions per second increases.
3. Both effects combine to increase the reaction rate.

## Interpreting Rate Graphs

Rate graphs plot amount of product (or remaining reactant) on the y-axis against time on the x-axis. The gradient of the line at any point equals the rate of reaction at that time:

- Steeper gradient = faster reaction rate
- Gradient decreases over time as reactants are used up
- Flat horizontal line = reaction is complete, no more product is being formed

**Worked example:** A graph plotting volume of carbon dioxide produced against time for a marble-acid reaction has a gradient of 10 cm³/s at 10 seconds, and 2 cm³/s at 30 seconds. Explain what these values show about the reaction progress.

1. At 10 seconds, the reaction is fast: 10 cm³ of CO₂ is being produced every second.
2. At 30 seconds, the rate has slowed to 2 cm³/s, as the concentration of hydrochloric acid and mass of marble have decreased as they are used up.
3. When the gradient becomes 0, the limiting reactant has been fully consumed and the reaction is finished.

## Common pitfalls

- **Wrong:** Stating that catalysts are used up in reactions
  - Why it fails: Catalysts only lower activation energy, they are not reactants and remain chemically unchanged at the end of the reaction
  - Correct: Always specify that catalysts speed up reactions without being used up, so they can be reused indefinitely
- **Wrong:** Forgetting to convert time units when calculating rate
  - Why it fails: Exam questions often give time in minutes, so failing to convert to seconds will produce incorrect rate values in g/s or cm³/s
  - Correct: Check the required rate units first, convert time to the matching unit before substituting into the rate formula
- **Wrong:** Only referencing more frequent collisions for temperature increase in Extended answers
  - Why it fails: Extended marking schemes require reference to a higher proportion of collisions meeting activation energy requirements, not just more frequent collisions
  - Correct: For Extended collision theory answers, always link temperature increases to both more frequent collisions and more successful collisions due to higher particle energy
- **Wrong:** Claiming larger solid pieces have larger surface area
  - Why it fails: Smaller solid pieces have more total exposed surface area than the same mass of larger pieces, e.g. 100g of powder has far more surface area than 100g of single solid block
  - Correct: Remember: smaller particle size = larger surface area = faster reaction rate
- **Wrong:** Calculating rate as time divided by amount of product/reactant
  - Why it fails: Rate is defined as amount per unit time, not time per amount, so inverted calculations give incorrect values
  - Correct: Use the formula $rate = \frac{\text{change in amount}}{\text{time taken}}$ and check units after calculation to confirm they match g/s, cm³/s or other required units

## Cheatsheet

| Factor | Core Effect on Rate | Extended Collision Theory Explanation |
| --- | --- | --- |
| Higher solution concentration | Faster rate | More particles per unit volume → more frequent successful collisions |
| Higher gas pressure | Faster rate | Particles closer together → more frequent successful collisions |
| Larger solid surface area | Faster rate | More exposed reactant particles → more frequent successful collisions |
| Higher temperature | Faster rate | Particles move faster, more have energy ≥ $E_a$ → more successful collisions per second |
| Catalyst added | Faster rate | Lowers activation energy → higher proportion of collisions are successful |

## What's next

Now that you have mastered rate of reaction content for CIE IGCSE Chemistry 0620, you can move on to reversible reactions and equilibrium, which build on the kinetic concepts covered here. Be sure to practice structured past paper questions on rate calculation, graph interpretation, and factor explanations to consolidate your learning, paying close attention to tier-specific requirements for Core and Extended answers. Extended candidates should prioritize collision theory questions, as these are often high-mark extended response items in Paper 4.

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