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ChemistryChemistry993 views·Updated 20 Aug 2026·25 pages

Higher Chemistry Notes: Unit 1 - Chemical Changes & Structure

user profile picture
Amy Neill@amyneill

Ever wondered why some chemical reactions happen super fast whilst...

1
of 10
Higher Chemistry: Unit 1- Chemical Changes and Structure Notes  – page 1

Chemical Changes and Structure - Unit One

This unit covers everything you need to know about controlling chemical reactions and understanding the energy involved. You'll learn why reactions speed up or slow down, and how to calculate exactly what's happening.

The key is understanding that chemistry isn't just about mixing things together - it's about controlling when and how fast reactions happen. This knowledge is essential for everything from cooking to industrial processes.

Remember: Mastering these concepts will give you a solid foundation for understanding all chemical processes!

2
of 10
Higher Chemistry: Unit 1- Chemical Changes and Structure Notes  – page 2

Collision Theory and Reaction Factors

Think of chemical reactions like a dance party - particles need to bump into each other in just the right way! There are five main factors that affect reaction rates: temperature, concentration, particle size, pressure, and catalysts.

For a successful collision to happen, two things are essential. First, particles need enough kinetic energy to overcome the activation energy (Ek ≥ Ea). Second, they must collide with the correct geometry - imagine trying to high-five someone who's facing the wrong way!

The activated complex is like a wobbly bridge between reactants and products. It's a high-energy, unstable arrangement that forms briefly during the reaction before breaking down into the final products.

Think of it: Activation energy is like the minimum effort needed to start a friendship - you need enough energy to overcome that initial awkwardness!

3
of 10
Higher Chemistry: Unit 1- Chemical Changes and Structure Notes  – page 3

Temperature and Concentration Effects

Temperature is simply a measure of how fast particles are moving (their kinetic energy). When you heat things up, more particles have enough energy to react successfully, leading to faster reactions. It's like giving everyone at that dance party more energy to move around!

For concentration, think of it as having more people at the party. With more particles in the same space, there are simply more chances for successful collisions to occur. Higher concentration equals more frequent meetings between reactive particles.

A typical exam question might ask you to explain why increasing temperature speeds up reactions. Your answer should mention that higher temperature means higher kinetic energy, so more particles exceed the activation energy threshold.

Exam tip: Always mention "more successful collisions" in your explanations - it's the key phrase examiners look for!

4
of 10
Higher Chemistry: Unit 1- Chemical Changes and Structure Notes  – page 4

Particle Size and Pressure

Smaller particle size means a bigger surface area for reactions to occur. Think of breaking a big chocolate bar into tiny pieces - suddenly there's loads more surface for reactions to happen on. More surface area equals more successful collisions.

Higher pressure squashes particles closer together, making them bump into each other more frequently. It's like cramming more people into a smaller dance floor - they're bound to interact more often!

Here's a quick summary: high temperature gives particles more energy, high concentration provides more particles, small particle size increases surface area, and high pressure brings particles closer together. All of these lead to more successful collisions and faster reactions.

Memory trick: Remember TCPS - Temperature, Concentration, Particle size, and Pressure all increase reaction rates!

5
of 10
Higher Chemistry: Unit 1- Chemical Changes and Structure Notes  – page 5

Calculating Reaction Rates

Average rate calculations use the formula: Rate = ΔQ/ΔT (change in quantity over change in time). If Barry collects 30 cm³ of gas in 20 seconds, his reaction rate is 30/20 = 1.5 cm³s⁻¹. Simple maths, really!

For rate at a specific time, you use Rate = 1/T. So at 30 seconds, the rate would be 1/30 = 0.033 s⁻¹. You can flip this formula to find time if you know the rate.

The SQA loves asking you to calculate time from a given rate, so practice rearranging T = 1/rate. These calculations might seem tricky at first, but they follow the same pattern every time.

Pro tip: Always check your units match the question - seconds, minutes, cm³, or dm³. Getting units wrong loses easy marks!

6
of 10
Higher Chemistry: Unit 1- Chemical Changes and Structure Notes  – page 6

Practical Rate Calculations

When decomposing hydrogen peroxide H2O2H2O+O2H₂O₂ → H₂O + O₂, different catalysts work at different speeds. MnO₂ and PbO catalysts were too fast to measure, whilst liver catalyst gave measurable results over time.

Using real experimental data, you can calculate both average rates and instantaneous rates. For the liver catalyst experiment, with 55 cm³ collected over 80 seconds, the average rate is 0.7 cm³s⁻¹. At 20 seconds specifically, the rate is 0.05 s⁻¹.

This type of practical calculation appears frequently in exams. You'll be given a table of results and asked to work out rates at different points. The key is staying organised with your working and double-checking your arithmetic.

Lab insight: Different catalysts can dramatically change reaction speeds - some are so effective they make reactions too fast to measure safely!

7
of 10
Higher Chemistry: Unit 1- Chemical Changes and Structure Notes  – page 7

Energy Distribution and Catalysts

Energy distribution diagrams show how particle energies are spread out in a reaction mixture. At higher temperatures, the curve flattens and shifts right, meaning more particles have enough energy to react successfully.

The area under the curve always stays the same regardless of temperature - this represents the total number of particles. What changes is how that energy is distributed among them.

Catalysts are like chemical shortcuts - they speed up reactions without being used up by providing an alternative pathway with lower activation energy. Think of them as building a tunnel through a mountain instead of climbing over it.

Remember: Catalysts don't change the starting or ending points of a reaction, just the route taken to get there!

8
of 10
Higher Chemistry: Unit 1- Chemical Changes and Structure Notes  – page 8

Potential Energy Diagrams

Enthalpy (ΔH) represents the energy difference between reactants and products. Exothermic reactions release energy (negative ΔH), whilst endothermic reactions require energy input (positive ΔH).

The activated complex sits at the peak of the energy curve - it's that unstable, high-energy arrangement that briefly forms during the reaction. Think of it as the wobbly moment when you're switching from one monkey bar to the next.

For reverse reactions, you flip everything around. If the forward reaction has an activation energy of 300 kJ and ΔH of -200 kJ, then the reverse reaction has an activation energy of 500 kJ and ΔH of +200 kJ.

Visual tip: Draw these diagrams yourself - seeing the energy changes helps you understand what's actually happening during reactions!

9
of 10
Higher Chemistry: Unit 1- Chemical Changes and Structure Notes  – page 9

Enthalpy of Combustion Calculations

Enthalpy of combustion is the energy released when one mole of substance burns completely in oxygen. You calculate it using ΔH = cmΔT, where c is specific heat capacity (4.18 for water), m is mass in kg, and ΔT is temperature change.

For ethanol combustion, if 3g burns and heats 100 cm³ of water by 12.2°C, you first calculate the energy released (5.1 kJ), then work out moles of ethanol (0.065), and finally scale up to find energy per mole (-78.2 kJ mol⁻¹).

The key steps are: calculate energy released, find moles of fuel, then scale to one mole. Remember that combustion values are always negative because energy is released, not absorbed.

Calculation tip: Always convert cm³ to kg by dividing by 1000, and don't forget to make your final answer negative for combustion!

10
of 10
Higher Chemistry: Unit 1- Chemical Changes and Structure Notes  – page 10

Advanced Enthalpy Calculations

These calculations can work backwards too - if you know the enthalpy of combustion, you can find how much fuel you need. For heating 100 cm³ of water by 10°C, you need 4.18 kJ of energy.

With ethanol's enthalpy of combustion being -1367 kJ mol⁻¹, you can calculate that 0.00305 moles will provide 4.18 kJ. Converting to mass using the formula mass = moles × molar mass gives 0.14g of ethanol needed.

Practice these calculations step by step - energy needed, moles required, then mass of fuel. The maths isn't complicated, but the method needs to become automatic for exam success.

Success strategy: Work through past paper questions systematically - these calculation patterns repeat frequently in exams!

We thought you’d never ask...

Our AI Companion is a student-focused AI tool that offers more than just answers. Built on millions of Knowunity resources, it provides relevant information, personalised study plans, quizzes, and content directly in the chat, adapting to your individual learning journey.

You can download the app from Google Play Store and Apple App Store.

That's right! Enjoy free access to study content, connect with fellow students, and get instant help – all at your fingertips.

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ChemistryChemistry993 views·Updated 20 Aug 2026·25 pages

Higher Chemistry Notes: Unit 1 - Chemical Changes & Structure

user profile picture
Amy Neill@amyneill

Ever wondered why some chemical reactions happen super fast whilst others take ages? Understanding reaction rates and energy changes is crucial for GCSE Chemistry - and it's actually pretty straightforward once you get the hang of it!

1
of 10
Higher Chemistry: Unit 1- Chemical Changes and Structure Notes  – page 1

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Chemical Changes and Structure - Unit One

This unit covers everything you need to know about controlling chemical reactions and understanding the energy involved. You'll learn why reactions speed up or slow down, and how to calculate exactly what's happening.

The key is understanding that chemistry isn't just about mixing things together - it's about controlling when and how fast reactions happen. This knowledge is essential for everything from cooking to industrial processes.

Remember: Mastering these concepts will give you a solid foundation for understanding all chemical processes!

2
of 10
Higher Chemistry: Unit 1- Chemical Changes and Structure Notes  – page 2

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  • Access to all documents
  • Improve your grades
  • Join milions of students

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Collision Theory and Reaction Factors

Think of chemical reactions like a dance party - particles need to bump into each other in just the right way! There are five main factors that affect reaction rates: temperature, concentration, particle size, pressure, and catalysts.

For a successful collision to happen, two things are essential. First, particles need enough kinetic energy to overcome the activation energy (Ek ≥ Ea). Second, they must collide with the correct geometry - imagine trying to high-five someone who's facing the wrong way!

The activated complex is like a wobbly bridge between reactants and products. It's a high-energy, unstable arrangement that forms briefly during the reaction before breaking down into the final products.

Think of it: Activation energy is like the minimum effort needed to start a friendship - you need enough energy to overcome that initial awkwardness!

3
of 10
Higher Chemistry: Unit 1- Chemical Changes and Structure Notes  – page 3

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Temperature and Concentration Effects

Temperature is simply a measure of how fast particles are moving (their kinetic energy). When you heat things up, more particles have enough energy to react successfully, leading to faster reactions. It's like giving everyone at that dance party more energy to move around!

For concentration, think of it as having more people at the party. With more particles in the same space, there are simply more chances for successful collisions to occur. Higher concentration equals more frequent meetings between reactive particles.

A typical exam question might ask you to explain why increasing temperature speeds up reactions. Your answer should mention that higher temperature means higher kinetic energy, so more particles exceed the activation energy threshold.

Exam tip: Always mention "more successful collisions" in your explanations - it's the key phrase examiners look for!

4
of 10
Higher Chemistry: Unit 1- Chemical Changes and Structure Notes  – page 4

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Particle Size and Pressure

Smaller particle size means a bigger surface area for reactions to occur. Think of breaking a big chocolate bar into tiny pieces - suddenly there's loads more surface for reactions to happen on. More surface area equals more successful collisions.

Higher pressure squashes particles closer together, making them bump into each other more frequently. It's like cramming more people into a smaller dance floor - they're bound to interact more often!

Here's a quick summary: high temperature gives particles more energy, high concentration provides more particles, small particle size increases surface area, and high pressure brings particles closer together. All of these lead to more successful collisions and faster reactions.

Memory trick: Remember TCPS - Temperature, Concentration, Particle size, and Pressure all increase reaction rates!

5
of 10
Higher Chemistry: Unit 1- Chemical Changes and Structure Notes  – page 5

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Calculating Reaction Rates

Average rate calculations use the formula: Rate = ΔQ/ΔT (change in quantity over change in time). If Barry collects 30 cm³ of gas in 20 seconds, his reaction rate is 30/20 = 1.5 cm³s⁻¹. Simple maths, really!

For rate at a specific time, you use Rate = 1/T. So at 30 seconds, the rate would be 1/30 = 0.033 s⁻¹. You can flip this formula to find time if you know the rate.

The SQA loves asking you to calculate time from a given rate, so practice rearranging T = 1/rate. These calculations might seem tricky at first, but they follow the same pattern every time.

Pro tip: Always check your units match the question - seconds, minutes, cm³, or dm³. Getting units wrong loses easy marks!

6
of 10
Higher Chemistry: Unit 1- Chemical Changes and Structure Notes  – page 6

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Practical Rate Calculations

When decomposing hydrogen peroxide H2O2H2O+O2H₂O₂ → H₂O + O₂, different catalysts work at different speeds. MnO₂ and PbO catalysts were too fast to measure, whilst liver catalyst gave measurable results over time.

Using real experimental data, you can calculate both average rates and instantaneous rates. For the liver catalyst experiment, with 55 cm³ collected over 80 seconds, the average rate is 0.7 cm³s⁻¹. At 20 seconds specifically, the rate is 0.05 s⁻¹.

This type of practical calculation appears frequently in exams. You'll be given a table of results and asked to work out rates at different points. The key is staying organised with your working and double-checking your arithmetic.

Lab insight: Different catalysts can dramatically change reaction speeds - some are so effective they make reactions too fast to measure safely!

7
of 10
Higher Chemistry: Unit 1- Chemical Changes and Structure Notes  – page 7

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Energy Distribution and Catalysts

Energy distribution diagrams show how particle energies are spread out in a reaction mixture. At higher temperatures, the curve flattens and shifts right, meaning more particles have enough energy to react successfully.

The area under the curve always stays the same regardless of temperature - this represents the total number of particles. What changes is how that energy is distributed among them.

Catalysts are like chemical shortcuts - they speed up reactions without being used up by providing an alternative pathway with lower activation energy. Think of them as building a tunnel through a mountain instead of climbing over it.

Remember: Catalysts don't change the starting or ending points of a reaction, just the route taken to get there!

8
of 10
Higher Chemistry: Unit 1- Chemical Changes and Structure Notes  – page 8

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Potential Energy Diagrams

Enthalpy (ΔH) represents the energy difference between reactants and products. Exothermic reactions release energy (negative ΔH), whilst endothermic reactions require energy input (positive ΔH).

The activated complex sits at the peak of the energy curve - it's that unstable, high-energy arrangement that briefly forms during the reaction. Think of it as the wobbly moment when you're switching from one monkey bar to the next.

For reverse reactions, you flip everything around. If the forward reaction has an activation energy of 300 kJ and ΔH of -200 kJ, then the reverse reaction has an activation energy of 500 kJ and ΔH of +200 kJ.

Visual tip: Draw these diagrams yourself - seeing the energy changes helps you understand what's actually happening during reactions!

9
of 10
Higher Chemistry: Unit 1- Chemical Changes and Structure Notes  – page 9

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Enthalpy of Combustion Calculations

Enthalpy of combustion is the energy released when one mole of substance burns completely in oxygen. You calculate it using ΔH = cmΔT, where c is specific heat capacity (4.18 for water), m is mass in kg, and ΔT is temperature change.

For ethanol combustion, if 3g burns and heats 100 cm³ of water by 12.2°C, you first calculate the energy released (5.1 kJ), then work out moles of ethanol (0.065), and finally scale up to find energy per mole (-78.2 kJ mol⁻¹).

The key steps are: calculate energy released, find moles of fuel, then scale to one mole. Remember that combustion values are always negative because energy is released, not absorbed.

Calculation tip: Always convert cm³ to kg by dividing by 1000, and don't forget to make your final answer negative for combustion!

10
of 10
Higher Chemistry: Unit 1- Chemical Changes and Structure Notes  – page 10

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  • Improve your grades
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Advanced Enthalpy Calculations

These calculations can work backwards too - if you know the enthalpy of combustion, you can find how much fuel you need. For heating 100 cm³ of water by 10°C, you need 4.18 kJ of energy.

With ethanol's enthalpy of combustion being -1367 kJ mol⁻¹, you can calculate that 0.00305 moles will provide 4.18 kJ. Converting to mass using the formula mass = moles × molar mass gives 0.14g of ethanol needed.

Practice these calculations step by step - energy needed, moles required, then mass of fuel. The maths isn't complicated, but the method needs to become automatic for exam success.

Success strategy: Work through past paper questions systematically - these calculation patterns repeat frequently in exams!

We thought you’d never ask...

Our AI Companion is a student-focused AI tool that offers more than just answers. Built on millions of Knowunity resources, it provides relevant information, personalised study plans, quizzes, and content directly in the chat, adapting to your individual learning journey.

You can download the app from Google Play Store and Apple App Store.

That's right! Enjoy free access to study content, connect with fellow students, and get instant help – all at your fingertips.

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AQA Biology: Key Concepts

Explore essential AQA Biology topics including Photosynthesis, Respiration, Homeostasis, Genetics, and Ecology. This comprehensive knowledge organizer covers key concepts such as energy transfer, hormonal control, and genetic variation, providing a solid foundation for your studies. Ideal for exam preparation and understanding biological processes.

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BiologyBiology

Biology P2: Evolution & Adaptation

Explore key concepts in AQA GCSE Biology P2, focusing on evolution, natural selection, genetic engineering, and adaptations in organisms. This summary covers essential topics such as DNA structure, speciation, and the impact of environmental changes on biodiversity. Ideal for exam preparation and understanding complex biological processes.

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