These revision notes cover every content standard of the Form 5 Thermochemistry chapter, exothermic and endothermic reactions, energy level diagrams, and the four heats of reaction, and show the step-by-step heat calculation using Q = mcθ then dividing by the number of moles, with units and sign.
These notes work through the Thermochemistry chapter one content standard at a time. Thermochemistry is calculation-heavy, so alongside the definitions we set out the standard calculation route that every heat-of-reaction question follows. Keep these notes beside past-paper questions and use them to confirm both your definitions and your working.
11.1 Exothermic and endothermic reactions
An exothermic reaction releases heat to the surroundings, so the temperature of the mixture rises and the heat change is given a negative sign. Examples include combustion, neutralisation and the reaction of a reactive metal with acid. An endothermic reaction absorbs heat from the surroundings, so the temperature falls and the heat change is given a positive sign. Examples include the thermal decomposition of many carbonates and dissolving ammonium salts such as ammonium chloride in water.
Bonds and energy. Breaking bonds absorbs energy; forming bonds releases energy. If more energy is released forming new bonds than is absorbed breaking old bonds, the reaction is exothermic; if less is released, it is endothermic. This is the reasoning the exam wants behind the sign.
11.2 Energy level diagrams
An energy level diagram shows the total energy of the reactants and of the products, with the difference marked as the heat change, ΔH. For an exothermic reaction the products sit lower than the reactants, and ΔH points downward and is negative. For an endothermic reaction the products sit higher than the reactants, and ΔH points upward and is positive. The “hump” between them represents the activation energy, the minimum energy needed to start the reaction.
Drawing tip. Always label the axis “Energy”, label reactants and products, and draw the arrow for ΔH from the reactant level to the product level. An exothermic diagram drawn the endothermic way round (products higher) loses the mark even if your words are correct.
The standard heat calculation
Almost every numerical question follows the same two steps.
- Step 1, heat change:
Q = mcθ, where m is the mass of solution or water in grams, c is the specific heat capacity (for aqueous solutions this is taken as 4.2 J g⁻¹ °C⁻¹, the value the question gives you), and θ is the temperature change in °C. Because the density of the solution is taken as 1 g cm⁻³, a volume in cm³ gives the same number in grams. - Step 2, per mole: divide Q by the number of moles that the definition refers to, then attach the sign. Heat of reaction is always quoted per mole, so
heat of reaction = Q ÷ number of moles, converted to kJ mol⁻¹.
Units and sign. Q comes out in joules; divide by 1000 to get kilojoules. A temperature rise means exothermic, so the answer takes a negative sign; a temperature fall means endothermic, so it takes a positive sign.
11.3 Heat of neutralisation
The heat of neutralisation is the heat released when one mole of water is formed from the neutralisation of an acid by an alkali. You mix known volumes of acid and alkali, record the highest temperature reached, find Q = mcθ using the total volume of the mixture, then divide by the moles of water formed.
Strong vs weak. A strong acid and a strong alkali are fully ionised, so their heat of neutralisation is higher (more exothermic). With a weak acid or weak alkali, some energy is absorbed to complete the ionisation of the acid or alkali, so the measured heat of neutralisation is lower. This comparison is a very common structured question.
11.4 Heat of displacement and precipitation
The heat of displacement is the heat released when one mole of a metal is displaced from its salt solution by a more electropositive metal, for example zinc displacing copper from copper(II) sulfate solution. The heat of precipitation is the heat change when one mole of a precipitate is formed from its ions in solution. Both use the same route: Q = mcθ on the solution, then divide by the moles of metal displaced or precipitate formed.
Which mass? In Q = mcθ use the mass of the solution, not the mass of the metal added or the precipitate formed. Using the metal’s mass is a frequent and costly slip.
11.5 Heat of combustion
The heat of combustion is the heat released when one mole of a substance is completely burnt in excess oxygen. In the school experiment a measured mass of fuel, often an alcohol, heats a known mass of water in a copper can. You find Q = mcθ using the mass of water, then divide by the moles of fuel burnt (mass burnt ÷ molar mass) to get the heat of combustion in kJ mol⁻¹.
Heat loss. The experimental value is usually lower than the accepted value because heat is lost to the surroundings and to the can, and combustion may be incomplete. Being able to state these reasons is worth easy marks.
How to use these notes
Revise one content standard at a time and say each definition aloud, making sure the phrase “per mole” is there. For 11.1 and 11.2, practise pairing the sign of ΔH with the direction of the temperature change and the shape of the energy level diagram. For 11.3 to 11.5, rehearse the two-step calculation until it is automatic: Q = mcθ first, then divide by the correct number of moles, then units and sign. A one-to-one teacher can check that your working shows every step and that your final answer always carries a unit and a sign, the two places students most often drop marks in this chapter. Because the same method underlies all four heats of reaction, mastering it once secures a large, predictable block of SPM Chemistry marks across both the written and practical papers. Finish by attempting the worked examples and practice questions for this chapter, marking yourself against the definitions and the calculation route set out above.
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