x

Revision notes for OCR GCSE Chemistry Controlling reactions. Open the guide for explanations and worked examples. Written against the OCR GCSE Chemistry (J248) specification, so the content matches what's examinable rather than general Chemistry background.

Controlling reactions

How do you slow down food going bad, or speed up the manufacture of essential chemical products like fertilisers? It all comes down to controlling how fast chemical reactions happen.

In this topic, you will explore the science of reaction rates, learn how to measure them, and discover how we can use collision theory and catalysts to speed up or slow down chemical changes.

What you'll learn

  • How to measure and calculate the rate of chemical reactions using practical methods and graphs.
  • How to explain the effects of temperature, concentration, pressure, and surface area in terms of particle collisions.
  • How to calculate the rate of reaction at a specific point on a curve using a tangent (Higher Tier only).
  • How catalysts and biological enzymes speed up reactions by lowering the activation energy.

Measuring and Representing Reaction Rates

Before you can control a reaction, you need to know how fast it is going. The speed of a reaction is called its rate of reaction.

Definition

Rate of reaction

The rate of reaction is a measure of how quickly reactants are turned into products. It is calculated by dividing the quantity of reactant used, or product formed, by the time taken.

To measure the rate of reaction experimentally, you can monitor how a physical property changes over time.

Common Practical Methods

Depending on the reactants and products involved, you can choose from three main laboratory setups:

  1. Measuring mass loss (for reactions that produce a gas): Place the reaction vessel (like a conical flask) on a digital mass balance. As a gas (such as carbon dioxide) is produced and escapes, the mass of the flask decreases. You record this decrease in mass at regular time intervals.
  2. Measuring the volume of gas produced: Connect the reaction flask to a gas syringe or an upside-down measuring cylinder filled with water. As gas is produced, it is trapped, and you can read its volume directly at regular intervals (e.g. every 10 seconds).
  3. The disappearing cross method (turbidity): For reactions that form a solid precipitate (such as the reaction between sodium thiosulfate and hydrochloric acid), place the reaction flask over a piece of paper marked with a black cross. Look down through the solution and measure the time (ttt) it takes for the precipitate to cloud the liquid so much that the cross is completely obscured.
Tip

Proportionality and reaction time

The time taken (ttt) for a reaction to finish or for a cross to disappear is inversely proportional to the rate of reaction. This means that a shorter time indicates a faster rate. You can use the formula 1/t1/t1/t as a simple measure of the relative rate of reaction.

Interpreting Rate of Reaction Graphs

When you plot your results on a graph, you will usually put the volume of gas or mass loss on the vertical y-axis, and time on the horizontal x-axis.

Rate of reaction graph

Looking at the graph, you can see three distinct phases:

  • The start of the reaction (steepest gradient): The reaction rate is at its fastest. This is because there are plenty of reactant particles available to collide.
  • The middle of the reaction (curve flattens): The rate is slowing down. Reactant particles are being used up, so there are fewer collisions.
  • The end of the reaction (flat line): The rate is zero. One or more of the reactants have been completely used up, so the reaction has stopped.

Calculating Rates of Reaction

To find the average (mean) rate of a reaction over a certain time window, use this simple formula:

Mean rate of reaction=Quantity of reactant used or product formedTime taken \text{Mean rate of reaction} = \frac{\text{Quantity of reactant used or product formed}}{\text{Time taken}} Mean rate of reaction=Time takenQuantity of reactant used or product formed​

The units for the rate of reaction depend on what you measured:

  • If you measured mass loss in grams (g), the rate is in grams per second (g/s).
  • If you measured gas volume in cubic centimetres (cm³), the rate is in cubic centimetres per second (cm³/s).

Calculating the Rate at a Specific Time (Higher Tier Only)

A curve on a rate graph is constantly changing its steepness. To find the exact rate of reaction at one specific second, you cannot just use the mean rate formula. Instead, you must draw a tangent to the curve at that exact point.

Definition

Tangent

A tangent is a straight line that touches a curve at a single point, matching the slope of the curve at that exact point.

Example

Calculating reaction rate using a tangent

Question: Use the graph below to calculate the rate of reaction at 30 seconds.

Imagine a curve where at 30 seconds, a tangent line has been drawn touching the curve.

  1. Draw the tangent: Place a ruler on the curve at exactly 30 seconds. Adjust the angle of your ruler so it matches the slope of the curve at that point. Draw a long, straight line extending in both directions.
  2. Select two points on the tangent: Choose two widely separated points on your straight tangent line that are easy to read from the grid. Let's choose Point 1 at (10 s,15 cm3)(10\text{ s}, 15\text{ cm}^3)(10 s,15 cm3) and Point 2 at (50 s,45 cm3)(50\text{ s}, 45\text{ cm}^3)(50 s,45 cm3).
  3. Calculate the change in yyy (Δy\Delta yΔy): Subtract the initial volume from the final volume on your tangent line.
Δy=45 cm3−15 cm3=30 cm3 \Delta y = 45\text{ cm}^3 - 15\text{ cm}^3 = 30\text{ cm}^3 Δy=45 cm3−15 cm3=30 cm3
  1. Calculate the change in xxx (Δx\Delta xΔx): Subtract the initial time from the final time on your tangent line.
Δx=50 s−10 s=40 s \Delta x = 50\text{ s} - 10\text{ s} = 40\text{ s} Δx=50 s−10 s=40 s
  1. Calculate the gradient: Divide the change in yyy by the change in xxx.
Rate=ΔyΔx=30 cm340 s=0.75 cm3/s \text{Rate} = \frac{\Delta y}{\Delta x} = \frac{30\text{ cm}^3}{40\text{ s}} = 0.75\text{ cm}^3/\text{s} Rate=ΔxΔy​=40 s30 cm3​=0.75 cm3/s

Collision Theory

To understand how we can control reactions, we must look at the nanoscale. Collision theory explains why and how chemical reactions happen.

Definition

Collision theory

Collision theory states that for a chemical reaction to occur, reactant particles must collide with each other with a minimum amount of energy and in the correct orientation.

Not every collision leads to a reaction. Collisions that result in the formation of products are called successful collisions. There are two requirements for a collision to be successful:

  1. The particles must collide.
  2. The particles must have energy equal to or greater than the activation energy.
Definition

Activation energy

The activation energy (EaE_aEa​) is the minimum amount of energy that reacting particles must possess in order to break bonds and start a chemical reaction.


Explaining the Factors That Affect Rates

You can speed up a reaction by doing one of two things:

  • Increasing the frequency of collisions (making them happen more often).
  • Increasing the energy of the colliding particles (making more of them exceed the activation energy).

Here is how changing the conditions alters the rate:

1. Concentration

If you increase the concentration of a solution, you place more reactant particles into the same volume. Because the particles are crowded closer together, they will collide with each other more frequently. This increases the frequency of successful collisions, speeding up the reaction.

2. Pressure

For reacting gases, increasing the pressure squashes the gas molecules closer together, increasing the number of particles in a given volume. This has the exact same effect as increasing concentration: it increases the frequency of collisions, which in turn increases the frequency of successful collisions.

3. Temperature

Increasing the temperature speeds up reactions dramatically. This is because temperature affects the particles in two distinct ways:

  • Higher energy: The particles absorb thermal energy and convert it into kinetic energy. They move much faster, leading to more frequent collisions.
  • More successful collisions: Because the particles are moving with much more kinetic energy, a significantly higher proportion of the collisions have energy equal to or greater than the activation energy (EaE_aEa​).
Common Mistake

Frequency vs amount of collisions

Do not just say "there are more collisions" in the exam. You must write frequency of collisions (or "collisions per second"). A reaction does not speed up because it has more collisions in total, but because the collisions are happening more often.

4. Surface Area

If a reactant is a solid, only the particles on the very outside of the solid can collide with the other reactant. If you break the solid down into smaller pieces (or a fine powder), you expose many more inner particles to the surface.

This is Higher Tier only: We describe this using the surface-area-to-volume ratio (SA:V). Smaller pieces of solid have a much larger surface-area-to-volume ratio than larger pieces of the same mass.

Example

Calculating surface-area-to-volume ratio

Question: Compare the surface-area-to-volume ratio of a large solid cube of side length 2 cm to that of the same material broken down into eight smaller cubes of side length 1 cm.

  1. Calculate the volume and surface area of the large cube:
    • Volume=2 cm×2 cm×2 cm=8 cm3\text{Volume} = 2\text{ cm} \times 2\text{ cm} \times 2\text{ cm} = 8\text{ cm}^3Volume=2 cm×2 cm×2 cm=8 cm3
    • Surface Area of one face=2 cm×2 cm=4 cm2\text{Surface Area of one face} = 2\text{ cm} \times 2\text{ cm} = 4\text{ cm}^2Surface Area of one face=2 cm×2 cm=4 cm2
    • Total Surface Area (6 faces)=6×4 cm2=24 cm2\text{Total Surface Area (6 faces)} = 6 \times 4\text{ cm}^2 = 24\text{ cm}^2Total Surface Area (6 faces)=6×4 cm2=24 cm2
  2. Calculate the volume and surface area of the eight smaller cubes:
    • Volume of one small cube=1 cm×1 cm×1 cm=1 cm3\text{Volume of one small cube} = 1\text{ cm} \times 1\text{ cm} \times 1\text{ cm} = 1\text{ cm}^3Volume of one small cube=1 cm×1 cm×1 cm=1 cm3
    • Total Volume (8 cubes)=8×1 cm3=8 cm3\text{Total Volume (8 cubes)} = 8 \times 1\text{ cm}^3 = 8\text{ cm}^3Total Volume (8 cubes)=8×1 cm3=8 cm3 (Notice the total volume remains the same!)
    • Surface Area of one small cube (6 faces)=6×(1 cm×1 cm)=6 cm2\text{Surface Area of one small cube (6 faces)} = 6 \times (1\text{ cm} \times 1\text{ cm}) = 6\text{ cm}^2Surface Area of one small cube (6 faces)=6×(1 cm×1 cm)=6 cm2
    • Total Surface Area of 8 small cubes=8×6 cm2=48 cm2\text{Total Surface Area of 8 small cubes} = 8 \times 6\text{ cm}^2 = 48\text{ cm}^2Total Surface Area of 8 small cubes=8×6 cm2=48 cm2
  3. Calculate the SA:V ratio for both cases:
    • For the large cube:
SA:V=248=3 \text{SA:V} = \frac{24}{8} = 3 SA:V=824​=3
  • For the smaller cubes:
SA:V=488=6 \text{SA:V} = \frac{48}{8} = 6 SA:V=848​=6
  1. Compare the ratios: The smaller cubes have double the surface area of the single large cube, despite having the exact same volume. This means there are many more exposed particles available to collide, leading to a much higher frequency of successful collisions.

Catalysts: Lowering the Barrier

Sometimes, raising the temperature or concentration is too expensive or dangerous in an industrial setting. In these cases, we use a catalyst.

Definition

Catalyst

A catalyst is a substance that increases the rate of a chemical reaction without being chemically changed or used up at the end of the reaction.

Because catalysts are not consumed, they can be used over and over again. They are highly specific, meaning a catalyst that speeds up one reaction might have no effect on another.

How Catalysts Work

Instead of giving the particles more energy, a catalyst works by providing an alternative reaction pathway that has a lower activation energy.

Because the activation energy barrier is lower, a much higher proportion of reactant collisions have enough energy to react. This increases the frequency of successful collisions.

Reaction profile diagram showing the effect of a catalyst

As shown in the reaction profile above:

  • The reactants and products start and end at the exact same energy levels; the overall energy change (ΔH\Delta HΔH) of the reaction does not change.
  • The peak of the curve represents the activation energy. The catalysed pathway (light blue) has a significantly lower peak than the uncatalysed pathway (dark blue).
Key Idea

Catalysts do not change the yield

A catalyst speeds up how quickly a reaction reaches its end-point, but it does not increase the amount of product (the yield) you get. It simply helps you get that same amount of product in a shorter time!

Biological Catalysts: Enzymes

Catalysts are not just used in chemical factories; they are also vital to life. Enzymes are proteins that act as biological catalysts in living organisms. They speed up metabolic processes (such as respiration and digestion) keeping you alive at body temperature (37 ∘C37\text{ }^\circ\text{C}37 ∘C), which would otherwise be far too cold for these essential reactions to occur quickly enough.


Exam technique

In the exam

  1. Always link to collisions: If an exam question asks you to explain why a rate changed, you must mention both the frequency of collisions (or frequency of successful collisions) and the energy of the particles.
  2. Be specific about temperature: When explaining the effect of temperature, always state that the main reason the rate increases so much is because a greater proportion of particles have energy greater than or equal to the activation energy.
  3. Do not say catalysts take part: Students often lose marks by saying "catalysts do not take part in the reaction". They do take part by offering an alternative pathway, but they are not chemically changed or used up by the end of it.
Self review

Check yourself

  • A student measures the rate of a reaction by tracking the volume of gas produced. The reaction stops after 5 minutes. Why does the curve flatten out?
  • Describe the difference between how temperature increases reaction rate and how a catalyst increases reaction rate.
  • Explain why breaking a large lump of calcium carbonate into a fine powder increases the rate of its reaction with hydrochloric acid (use the term surface-area-to-volume ratio in your answer). [Higher Tier]

Recap questions

Test yourself with 5 quick questions on this guide. Answer them all correctly to complete it.

You've reached the end

Test yourself on this topic, or move on to the next guide.

Practice questionsTake a quick quiz on this topicFlashcardsSelf-test with active recall
EquilibriaUp next

How was this guide?

Controlling reactions Revision Guide

  1. GCSE
  2. /Chemistry
  3. /Controlling reactions