O-Level Chemistry · Secondary 3 and 4
Energy Changes
Every chemical reaction involves an energy change, because bonds are broken and new bonds are made. In O-Level Chemistry you learn to classify reactions as exothermic or endothermic, read and draw energy profile diagrams, and calculate the enthalpy change from bond energies. These ideas explain why fuels release heat, why some reactions need heating to start, and how catalysts work, which links this topic closely to rate of reaction.
What the syllabus expects
- Describe exothermic and endothermic reactions in terms of heat transfer to or from the surroundings
- State that ΔH is negative for exothermic reactions and positive for endothermic reactions
- Explain that bond breaking absorbs energy and bond making releases energy
- Calculate the enthalpy change of a reaction from bond energies
- Draw and interpret energy profile diagrams showing ΔH and the activation energy
- Describe the effect of a catalyst on the activation energy shown in an energy profile diagram
Exothermic and endothermic reactions
An exothermic reaction transfers heat energy to the surroundings, so the temperature of the surroundings, such as the solution or the test tube, rises. Combustion, neutralisation, respiration and the reaction of metals with acids are exothermic. An endothermic reaction takes in heat energy from the surroundings, so the temperature of the surroundings falls. Thermal decomposition of calcium carbonate, photosynthesis and dissolving ammonium nitrate in water are endothermic. The enthalpy change, ΔH, is the heat energy change of a reaction, usually given in kJ/mol. For exothermic reactions ΔH is negative, because the chemicals lose energy. For endothermic reactions ΔH is positive, because the chemicals gain energy.
Bond breaking and bond making
Breaking a bond always needs energy, so it is endothermic. Making a bond always releases energy, so it is exothermic. Whether a whole reaction is exothermic or endothermic depends on the balance between these two. If more energy is released making the new bonds in the products than is absorbed breaking the bonds in the reactants, the reaction is exothermic. If less is released than absorbed, it is endothermic. Bond energy is the energy needed to break one mole of a particular covalent bond. So ΔH = total energy absorbed to break bonds − total energy released when bonds form. Always count how many of each bond there are, including the numbers in front of each formula in the balanced equation.
Energy profile diagrams
An energy profile diagram plots energy on the vertical axis against the progress of the reaction. The reactants are drawn as a level on the left and the products as a level on the right. For an exothermic reaction the products are lower than the reactants, and for an endothermic reaction they are higher. The difference between the two levels is ΔH, drawn as an arrow from the reactants to the products, pointing down for exothermic and up for endothermic. Between them is a hump. The activation energy, Ea, is the minimum energy the colliding particles need for the reaction to happen, shown as the height from the reactant level to the top of the hump. Even exothermic reactions such as burning methane need this energy to start, which is why a flame or spark is needed.
Catalysts and energy
A catalyst provides an alternative pathway for the reaction with a lower activation energy. On an energy profile diagram this is drawn as a lower hump between the same reactant and product levels. The value of ΔH is unchanged, because the reactants and products are the same. With a lower activation energy, a greater proportion of colliding particles have enough energy to react, so the reaction is faster. The catalyst is chemically unchanged at the end. Iron in the Haber process and manganese(IV) oxide in the decomposition of hydrogen peroxide are examples you may meet.
Worked example
Hydrogen reacts with chlorine: H₂ + Cl₂ gives 2HCl. Use the bond energies H-H = 436 kJ/mol, Cl-Cl = 242 kJ/mol and H-Cl = 431 kJ/mol to calculate ΔH, and state whether the reaction is exothermic or endothermic.
- Bonds broken: one H-H and one Cl-Cl, so energy absorbed = 436 + 242 = 678 kJ.
- Bonds formed: two H-Cl, because the equation makes 2HCl, so energy released = 2 × 431 = 862 kJ.
- ΔH = energy absorbed − energy released = 678 − 862 = −184 kJ/mol.
- ΔH is negative because more energy is released making bonds than is absorbed breaking them.
Answer: ΔH = −184 kJ/mol, so the reaction is exothermic.
Common mistakes
- Subtracting the wrong way round. ΔH = bonds broken − bonds formed, so a reaction that releases more energy than it absorbs has a negative ΔH.
- Forgetting to multiply by the number of bonds. Methane has four C-H bonds, and 2O₂ means two O=O bonds.
- Saying the temperature falls in an exothermic reaction. Heat is given out, so the surroundings get warmer.
- Writing that breaking bonds releases energy. Bond breaking always absorbs energy.
- Claiming that a catalyst changes ΔH. It lowers the activation energy only, and the reactant and product levels stay the same.
- Measuring the activation energy from the product level or from the bottom of the graph. It is drawn from the reactant level to the top of the hump.
Exam tips
- Set out bond energy answers in two labelled lines, 'bonds broken' and 'bonds formed', before subtracting. Markers can then award method marks even if one value slips.
- Always write the sign in front of ΔH, including the plus sign for endothermic values, and include the unit kJ/mol.
- On energy profile diagrams, label the axes, both levels with their formulae, ΔH and Ea, and draw the arrows in the correct direction.
Practice questions
1. Easy: When ammonium nitrate dissolves in water, the temperature falls from 25.0 °C to 19.0 °C. Is the process exothermic or endothermic, and what is the sign of ΔH?
Show answer
Endothermic, with a positive ΔH. The temperature of the water falls, which shows heat energy is taken in from the surroundings by the process.
2. Medium: Calculate ΔH for CH₄ + 2O₂ gives CO₂ + 2H₂O using the bond energies C-H = 410, O=O = 496, C=O = 805 and O-H = 460 kJ/mol.
Show answer
ΔH = −818 kJ/mol. Bonds broken: 4 × 410 + 2 × 496 = 1640 + 992 = 2632 kJ. Bonds formed: 2 × 805 + 4 × 460 = 1610 + 1840 = 3450 kJ. ΔH = 2632 − 3450 = −818 kJ/mol.
3. Harder: For H₂ + Br₂ gives 2HBr, ΔH = −103 kJ/mol. Given H-H = 436 kJ/mol and Br-Br = 193 kJ/mol, calculate the H-Br bond energy.
Show answer
366 kJ/mol. Bonds broken = 436 + 193 = 629 kJ. Let the H-Br bond energy be x, so bonds formed = 2x. Then 629 − 2x = −103, so 2x = 732 and x = 366 kJ/mol.
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