Welcome to Energy Changes in Chemistry!
Have you ever held a hand warmer in winter and felt it heat up, or used a cold pack on a sprained ankle? In chemistry, chemical reactions don't just rearrange atoms—they also transfer energy to or from their surroundings. Understanding these energy changes helps us predict how reactions behave and allows us to harness chemical energy for useful everyday items.
Don't worry if this topic seems a bit math-heavy or confusing at first. We will break every concept down into clear, bite-sized steps so you feel fully confident for your CCEA GCSE Chemistry Unit 2 exam!
---1. Exothermic and Endothermic Reactions
Understanding System vs. Surroundings
To understand energy changes, imagine two parts to every experiment:
1. The System: The chemical reaction itself (the atoms and bonds).
2. The Surroundings: Everything outside the reaction (the water in the beaker, the beaker itself, the thermometer, and the air around it).
Exothermic Reactions (Energy Out)
An exothermic reaction is a chemical reaction that gives out heat energy to the surroundings.
What you observe: The temperature of the surroundings increases (a thermometer placed in the reaction mixture will show a rise in temperature).
Sign of energy change: The overall energy change is represented by the symbol \(\Delta H\) (pronounced "delta H"). For exothermic reactions, \(\Delta H\) is always negative (\(-\Delta H\)) because the chemical system loses energy to the surroundings.
Key examples required for CCEA:
• Combustion: Burning fuels such as methane: \(\text{CH}_4 + 2\text{O}_2 \rightarrow \text{CO}_2 + 2\text{H}_2\text{O}\)
• Neutralisation: The reaction between an acid and an alkali/base
• Respiration: \(\text{C}_6\text{H}_{12}\text{O}_6 + 6\text{O}_2 \rightarrow 6\text{CO}_2 + 6\text{H}_2\text{O}\)
• Everyday uses: Disposable hand warmers and self-heating beverage cans
Endothermic Reactions (Energy In)
An endothermic reaction is a chemical reaction that takes in heat energy from the surroundings.
What you observe: The temperature of the surroundings decreases (a thermometer placed in the reaction mixture will show a drop in temperature).
Sign of energy change: \(\Delta H\) is always positive (\(+\Delta H\)) because the chemical system gains energy from the surroundings.
Key examples required for CCEA:
• Thermal decomposition: Heating and breaking down substances like calcium carbonate (\(\text{CaCO}_3\)) or hydrated salts
• Citric acid and sodium hydrogencarbonate: Mixing these causes a noticeable temperature drop
• Everyday uses: Instant sports injury cold packs
Quick Tip: The Thermometer Trap
Examiner Warning: Students often get confused when a thermometer reads a higher temperature, thinking that because the temperature went up, the reaction must have "taken in heat". Remember: the thermometer is part of the surroundings. If the thermometer gets hotter, energy was given out to the surroundings, meaning the reaction is exothermic!
Section 1 Key Takeaway
• Exothermic: Heat energy given out \(\rightarrow\) Temperature goes up \(\rightarrow -\Delta H\)
• Endothermic: Heat energy taken in \(\rightarrow\) Temperature goes down \(\rightarrow +\Delta H\)
2. Activation Energy and Reaction Profile Diagrams
What is Activation Energy (\(E_a\))?
Chemical reactions do not just happen automatically when particles touch. Particles must collide with enough energy to break bonds and start the reaction.
Activation energy (\(E_a\)) is defined as the minimum energy needed for a reaction to occur. You can think of it like rolling a boulder up a hill: you have to push it all the way to the top of the hill before it can roll down the other side.
Interpreting Reaction Profile Diagrams
A reaction profile diagram (or energy level diagram) shows how energy changes as reactants turn into products.
Standard Axes:
• y-axis: Energy (or Potential Energy)
• x-axis: Progress of reaction (or Reaction pathway)
A. Exothermic Reaction Profile
In an exothermic reaction, the reactants start with more chemical energy than the products because energy is given out during the reaction.
• Reactants level: Drawn higher on the y-axis than the products.
• Products level: Drawn lower on the y-axis.
• Activation Energy (\(E_a\)): Drawn as a vertical arrow starting strictly from the reactant level up to the peak of the curve.
• Overall Energy Change (\(\Delta H\)): Drawn as a vertical arrow starting from the reactant level down to the product level (pointing downwards because \(\Delta H\) is negative).
B. Endothermic Reaction Profile
In an endothermic reaction, the products have more chemical energy than the reactants because energy has been absorbed from the surroundings.
• Reactants level: Drawn lower on the y-axis than the products.
• Products level: Drawn higher on the y-axis.
• Activation Energy (\(E_a\)): Drawn as a tall vertical arrow starting from the reactant level up to the peak of the curve.
• Overall Energy Change (\(\Delta H\)): Drawn as a vertical arrow starting from the reactant level up to the product level (pointing upwards because \(\Delta H\) is positive).
Common Exam Pitfalls to Avoid:
1. Drawing \(E_a\) from the bottom axis: Never draw the activation energy arrow starting from the x-axis or the product level. It must always start directly at the reactant line and go to the top peak.
2. Forgetting arrowheads: Always clearly show the direction of your arrows. For \(\Delta H\), exothermic points down and endothermic points up.
Section 2 Key Takeaway
• \(E_a\) = Reactants to the peak (always positive).
• \(\Delta H\) = Reactants to products.
• Exothermic = Products are lower than reactants (\(-\Delta H\)).
• Endothermic = Products are higher than reactants (\(+\Delta H\)).
3. Bond Breaking, Bond Making, and Energy Calculations
The Chemistry of Bonds: Why Does Energy Change?
Every chemical reaction involves two distinct stages:
1. Breaking bonds in the reactants.
2. Making new bonds in the products.
• Bond breaking takes in energy: Think of snapping a strong plastic ruler in half—you have to put effort (energy) into breaking it. Bond breaking is an endothermic process.
• Bond making releases energy: When new chemical bonds form, energy is given out to the surroundings. Bond making is an exothermic process.
Memory Aid: MEXO BENDO
To remember which process is which in the exam, use the classic mnemonic:
• BENDO: Breaking bonds is ENDOthermic (energy in).
• MEXO: Making bonds is EXOthermic (energy out).
Why Reactions are Overall Exothermic or Endothermic
The overall energy change (\(\Delta H\)) is simply the balance between the energy taken in to break bonds and the energy released when new bonds are formed:
• If more energy is released when making bonds than is taken in when breaking bonds \(\rightarrow\) Exothermic reaction (\(-\Delta H\)).
• If more energy is taken in to break bonds than is released when making bonds \(\rightarrow\) Endothermic reaction (\(+\Delta H\)).
Calculating Energy Changes Using Bond Energies
A bond energy is the amount of energy (measured in \(\text{kJ}\) or \(\text{kJ/mol}\)) required to break one mole of a particular chemical bond.
The Master Formula:
\(\Delta H = \text{Total Energy In (Bonds Broken)} - \text{Total Energy Out (Bonds Formed)}\)
or
\(\Delta H = \sum (\text{Bonds Broken}) - \sum (\text{Bonds Formed})\)
4. Step-by-Step Worked Examples
Worked Example 1: Reaction of Hydrogen and Chlorine
Equation: \(\text{H}_2 + \text{Cl}_2 \rightarrow 2\text{HCl}\)
Given Bond Energies:
• \(\text{H}-\text{H} = 436\text{ kJ/mol}\)
• \(\text{Cl}-\text{Cl} = 242\text{ kJ/mol}\)
• \(\text{H}-\text{Cl} = 431\text{ kJ/mol}\)
Step 1: Calculate energy taken in to break all bonds in the reactants
• \(1 \times (\text{H}-\text{H}) = 436\text{ kJ}\)
• \(1 \times (\text{Cl}-\text{Cl}) = 242\text{ kJ}\)
• Total energy broken = \(436 + 242 = 678\text{ kJ}\)
Step 2: Calculate energy released when making all bonds in the products
• \(2 \times (\text{H}-\text{Cl}) = 2 \times 431 = 862\text{ kJ}\)
• Total energy formed = \(862\text{ kJ}\)
Step 3: Calculate \(\Delta H\)
\(\Delta H = \text{Bonds Broken} - \text{Bonds Formed}\)
\(\Delta H = 678 - 862 = -184\text{ kJ/mol}\)
Conclusion: Since \(\Delta H\) is negative (\(-184\text{ kJ/mol}\)), the reaction is exothermic.
---Worked Example 2: Combustion of Methane
Equation: \(\text{CH}_4 + 2\text{O}_2 \rightarrow \text{CO}_2 + 2\text{H}_2\text{O}\)
Given Bond Energies:
• \(\text{C}-\text{H} = 413\text{ kJ/mol}\)
• \(\text{O}=\text{O} = 498\text{ kJ/mol}\)
• \(\text{C}=\text{O} = 805\text{ kJ/mol}\)
• \(\text{O}-\text{H} = 464\text{ kJ/mol}\)
Step 1: Energy taken in to break bonds (Reactants)
• In \(\text{CH}_4\), there are four \(\text{C}-\text{H}\) bonds: \(4 \times 413 = 1652\text{ kJ}\)
• In \(2\text{O}_2\), there are two \(\text{O}=\text{O}\) double bonds: \(2 \times 498 = 996\text{ kJ}\)
• Total energy in = \(1652 + 996 = 2648\text{ kJ}\)
Step 2: Energy released when making bonds (Products)
• In \(\text{CO}_2\), there are two \(\text{C}=\text{O}\) double bonds: \(2 \times 805 = 1610\text{ kJ}\)
• In \(2\text{H}_2\text{O}\), each water molecule has two \(\text{O}-\text{H}\) bonds, so two water molecules have four \(\text{O}-\text{H}\) bonds in total: \(4 \times 464 = 1856\text{ kJ}\)
• Total energy out = \(1610 + 1856 = 3466\text{ kJ}\)
Step 3: Calculate \(\Delta H\)
\(\Delta H = 2648 - 3466 = -818\text{ kJ/mol}\)
Conclusion: The reaction is exothermic with \(\Delta H = -818\text{ kJ/mol}\).
---Worked Example 3: Formation of Ammonia
Equation: \(\text{N}_2 + 3\text{H}_2 \rightarrow 2\text{NH}_3\)
Given Bond Energies:
• \(\text{N}\equiv\text{N} = 945\text{ kJ/mol}\)
• \(\text{H}-\text{H} = 436\text{ kJ/mol}\)
• \(\text{N}-\text{H} = 391\text{ kJ/mol}\)
Step 1: Energy taken in to break bonds (Reactants)
• In \(\text{N}_2\), there is one \(\text{N}\equiv\text{N}\) triple bond: \(1 \times 945 = 945\text{ kJ}\)
• In \(3\text{H}_2\), there are three \(\text{H}-\text{H}\) single bonds: \(3 \times 436 = 1308\text{ kJ}\)
• Total energy in = \(945 + 1308 = 2253\text{ kJ}\)
Step 2: Energy released when making bonds (Products)
• In each \(\text{NH}_3\) molecule, there are three \(\text{N}-\text{H}\) bonds. For \(2\text{NH}_3\), there are \(2 \times 3 = 6\) bonds in total:
• \(6 \times (\text{N}-\text{H}) = 6 \times 391 = 2346\text{ kJ}\)
• Total energy out = \(2346\text{ kJ}\)
Step 3: Calculate \(\Delta H\)
\(\Delta H = 2253 - 2346 = -93\text{ kJ/mol}\)
Conclusion: The reaction is exothermic with \(\Delta H = -93\text{ kJ/mol}\).
---5. Exam Checklist & Common Calculation Traps
Before submitting your answers in the Unit 2 exam, run through this quick checklist:
• Did you count every bond inside a molecule? Remember that \(\text{H}_2\text{O}\) has two \(\text{O}-\text{H}\) bonds, \(\text{CO}_2\) has two \(\text{C}=\text{O}\) bonds, and \(\text{CH}_4\) has four \(\text{C}-\text{H}\) bonds.
• Did you multiply by the balancing numbers (stoichiometry)? For \(2\text{H}_2\text{O}\), multiply \(2 \times 2 = 4\) \(\text{O}-\text{H}\) bonds.
• Did you apply the formula in the correct order? It is always Reactants minus Products (Broken minus Formed).
• Did you include the correct sign? If your answer is negative, keep the minus sign clearly written (\(-\)). If asked whether it is exothermic or endothermic, state that a negative \(\Delta H\) means exothermic.