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Chemistry

Energetics is a direction question: which way did the heat flow? Exothermic reactions give heat OUT to the surroundings - the temperature of the mixture rises. Endothermic reactions take heat IN - the temperature falls. The thermometer is the detector; the sign of the enthalpy change is the verdict.

Sort & matchblock 2

Sort the reactions - exothermic or endothermic (Sections: Exothermic / Endothermic reactions).

choose a match

Page-true figure region in the book pane — Showing an exothermic change on an energy level diagram - products below reactants, delta H negative (chemistry only).

Examiner note - the sign IS the answer

On the energy level diagram the products sit BELOW the reactants for an exothermic change and delta H is negative; endothermic is the mirror image, products above, delta H positive. Write the sign even when the number is 'obvious' - '-4817 kJ/mol' and '4817 kJ/mol' are different answers, and the arrow labels (delta H, activation energy) are separately markable.

Check yourselfblock 5

Step throughblock 6

Activity 4 of record: 25.0 cm3 of 2.00 mol/dm3 KOH neutralised by HCl; the graph cross-over lands at 28.00 cm3 of acid. Find the acid concentration (Calculations for Activity 4).

1•Moles of KOH measured out (n = c x V, V in dm3)
(mol)
2•Moles of HCl at complete neutralisation (OH- + H+ is 1:1)
3•Concentration of the HCl (c = n / V)

Watchblock 7

Exothermic & Endothermic Reactions - Reaction Profiles | Activation Energy

Exothermic vs endothermic — energy profiles and ΔH signs.

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Video © Cognito (cognito.org) — embedded via YouTube

Watchblock 8

GCSE Chemistry - Bond Energies - Determining if Reactions are Exothermic or Endothermic

Bond energy calculations — bonds broken minus bonds made, sign conventions sorted.

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Video © Cognito (cognito.org) — embedded via YouTube

Watchblock 9

Temperature Changes - GCSE Science Required Practical

The required practical: temperature changes in reacting solutions — method and error control.

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Video © Cognito (cognito.org) — embedded via YouTube

Exothermic reactions

SB p.207

A reaction that gives out heat energy to the surroundings is called an exothermic reaction. Hold a test-tube in which an exothermic reaction is happening and the tube gets warmer: heat is flowing out of the reaction mixture into the glass, your fingers and the air. That is the definition and the observation in one — heat leaves the reaction for the surroundings, and the temperature of everything around the mixture goes up.

The clearest way to see what is happening energetically is to compare the chemical energy stored in the bonds of the chemicals before and after. In an exothermic reaction the products have less chemical energy than the reactants. The difference does not vanish — it is converted to heat energy and released to the surroundings. Chemical energy stored in bonds becomes heat energy in the environment, so the reaction mixture and its surroundings warm up.

The classic demonstration is adding water to solid calcium oxide. The heat produced is enough to boil the water and raise steam, and calcium hydroxide forms:

CaO(s) + H₂O(l) → Ca(OH)₂(s)

The reaction releases so much heat that the water can be seen steaming as the solid crumbles — a dramatic confirmation that chemical energy has been converted to heat on a large scale.

Did you know?

Calcium oxide is known as quicklime. Adding water to it is described as slaking it, and the calcium hydroxide produced is known as slaked lime. The same chemistry underlies traditional lime mortars, which harden as the quicklime slakes and then slowly absorbs carbon dioxide from the air.

Safety

It is less hazardous to use a lump of calcium oxide rather than powdered calcium oxide, and skin contact must be avoided — the solid is caustic and the slaking reaction scalds. Wear eye protection.

Exam hint

A one-mark "know" question wants exactly this chain: exothermic → heat given out to the surroundings → temperature of the surroundings rises. If the question then asks you to explain in terms of energy, add the second chain: products have less chemical energy than reactants → the difference is released as heat. Saying "it gets hot" without naming the surroundings earns nothing; the heat flow direction is the mark.

Common misconception

"An exothermic reaction must produce a flame." Flames belong to the small subset of exothermic reactions that burn — adding water to quicklime, or dilute acid neutralising dilute alkali, is fiercely exothermic with no flame at all. The test is the temperature of the surroundings, not light.

Quick check

Question-bank join: spec 3.1 exothermic-definition items. Write the definition of an exothermic reaction without looking, then explain the calcium oxide + water reaction in terms of the chemical energy of products versus reactants. Finally, name the product and its common name.

Key terms in this section

exothermic: — A reaction that gives out heat energy to the surroundings. Products have less chemical energy than reactants; the temperature of the mixture and its surroundings rises; ∆H is negative.
quicklime: — Calcium oxide, CaO. Adding water slakes it — a strongly exothermic reaction producing steam — and the calcium hydroxide formed is called slaked lime.

Keep goingExothermic & Endothermic 1-5-chemical-formulae-equations-calculations 2-4-reactivity-series 3-1-energetics 3-3-reversible-reactions-and-equilibria

Exothermic reactions in action

SB p.208

You have already met dozens of exothermic reactions — this section gathers the families the examiners expect you to recognise on sight.

Any reaction that produces a flame is exothermic. Burning things releases heat energy. The clearest example is hydrogen burning in oxygen, producing water and a large amount of heat:

2H₂(g) + O₂(g) → 2H₂O(l)

The burning of hydrogen is used in oxy-hydrogen cutting equipment underwater, because the flame is hot enough to cut metal even beneath the surface. You met the same reaction as the test for hydrogen (the squeaky pop) and in the reactivity-series work.

Metals reacting with acids are exothermic. When magnesium reacts with dilute sulfuric acid, the mixture gets very warm:

Mg(s) + H₂SO₄(aq) → MgSO₄(aq) + H₂(g)

Neutralisation reactions are exothermic. About the only interesting thing you can observe when sodium hydroxide solution reacts with dilute hydrochloric acid is that the temperature rises:

NaOH(aq) + HCl(aq) → NaCl(aq) + H₂O(l)

Nothing changes colour, no gas bubbles out — but the mixture is measurably warmer than the two starting solutions. You will measure exactly how much heat this family releases later in the chapter.

displacement-reactions are exothermic. The most spectacular example is the thermite reaction between powdered aluminium and iron(III) oxide, a displacement (competition) reaction that releases so much heat that the iron formed is molten — the reaction is used to weld railway lines together:

2Al(s) + Fe₂O₃(s) → 2Fe(l) + Al₂O₃(s)

Exam hint

"Give two examples of exothermic reactions" is a gift question if you keep a mental list: combustion (hydrogen, or any fuel), metal + acid, neutralisation, displacement. Name the chemicals, not just the family — "magnesium with dilute sulfuric acid" scores; "a reaction that gets hot" does not. If asked for equations, any balanced equation from this section is safe.

Common misconception

"Neutralisation cannot be exothermic because salt solution does not feel warm." With dilute solutions the temperature rise is only a few degrees — real, measurable with a thermometer, but easy to miss with a finger. The thermometer, not your hand, decides whether a reaction is exothermic; later in this chapter the measured rise for KOH + HCl works out at ΔH = −55.4 kJ/mol.

Quick check

Question-bank join: spec 3.1 exothermic-examples items. List the four exothermic families from memory, write a balanced equation for one member of each family, and state what you would observe (other than a flame) for the metal + acid and neutralisation examples.

Key terms in this section

exothermic: — A reaction that gives out heat energy to the surroundings. Products have less chemical energy than reactants; the temperature of the mixture and its surroundings rises; ∆H is negative.
displacement reaction: — A reaction in which a more reactive element takes the place of a less reactive one in a compound (a competition reaction). Zinc displacing copper from copper(II) sulfate is the calorimetry example here.

Keep goingExothermic & Endothermic 1-5-chemical-formulae-equations-calculations 2-4-reactivity-series 3-1-energetics 3-3-reversible-reactions-and-equilibria

Enthalpy change (∆H)

SB p.209

You can measure the amount of heat energy taken in or released in a chemical reaction. It is called the enthalpy-change of the reaction and is given the symbol ∆H. ∆H is pronounced "delta H" — the Greek letter ∆ is used to mean "change in", so ∆H means "change in heat". One subtlety matters: it is not possible to measure how much enthalpy (H) a substance has — you can only measure the change in enthalpy when it reacts. The enthalpy change is the amount of heat energy taken in or given out in a chemical reaction; it is the difference between the energy of the products and the energy of the reactants.

∆H is given a minus or a plus sign to show whether heat is being given out or absorbed, and you always read the sign from the point of view of the reactants. For an exothermic reaction, ∆H is negative, because the reactants are losing energy as heat — heat is transferred to the surroundings, which then get warmer. ∆H is measured in units of kJ/mol (kilojoules per mole); you may also see the units written kJ mol⁻¹.

Written next to an equation, the enthalpy change looks like this:

Mg(s) + H₂SO₄(aq) → MgSO₄(aq) + H₂(g) ∆H = −466.9 kJ/mol

The ∆H here represents the enthalpy change of the reaction: 466.9 kJ of heat is given out when one mole of magnesium reacts in this way. You know heat has been given out because ∆H carries a negative sign.

The word stability describes the relative energies of reactants and products: the more energy a chemical has, the less stable it is. So when an exothermic reaction runs, the products (lower energy) are more stable than the reactants — energy has drained out of the chemicals into the surroundings. And remember the vocabulary: the reactants are the chemicals you start with. The per-mole basis is why the mole matters here — every ∆H you quote is for the amounts shown in the balanced equation.

Key point

∆H is the enthalpy change: the heat energy taken in or given out by a reaction, measured from the reactants' point of view. Negative ∆H = heat given out (exothermic). Positive ∆H = heat taken in (endothermic). Units: kJ/mol (equivalently kJ mol⁻¹).

Exam hint

Two habits earn the sign-and-units marks every time. First, never quote a ∆H without its sign — an answer of "466.9 kJ/mol" for the magnesium reaction above loses the mark that "−466.9 kJ/mol" scores. Second, keep the units per mole: if a question gives you heat in joules for 0.78 g of fuel, dividing by moles before quoting kJ/mol is part of the calculation, not decoration (the full method arrives in the worked example later in this chapter).

Common misconception

"A negative ∆H means energy is missing." The minus sign records a loss from the reactants, not a loss from the universe: the heat has been transferred to the surroundings, where it warms them up. Energy is conserved — the sign just says which way it flowed relative to the chemicals.

Quick check

Question-bank join: spec 3.4 enthalpy-change items. Define enthalpy change, state why exothermic reactions carry a negative ∆H, and write the two acceptable forms of the units. Then explain why a chemical with more energy is described as less stable.

Key terms in this section

enthalpy change: — The heat energy taken in or given out during a chemical reaction — the difference between the energy of the products and the reactants. Only the change (∆H) is measurable, never absolute enthalpy.
stability: — Describes the relative energies of reactants and products: the more energy a chemical has, the less stable it is. Exothermic products are more stable than the reactants that formed them.
mole: — The unit for the amount of a substance. Molar enthalpy changes and bond energies are both quoted per mole, so mole arithmetic from u1:c05 feeds every calculation in this chapter.

Keep goingEnergetics Calculations 2-1-group-1-alkali-metals 3-1-energetics

Showing an exothermic change on an energy level diagram

SB p.209

An energy-level-diagram is the standard picture for the energy story of a reaction. The vertical axis is energy (label it!), the horizontal axis is the progress of the reaction, and the reactants and products each sit on their own horizontal level.

For an exothermic reaction the reactants sit higher than the products — they have more chemical energy — so the arrow from reactants to products points down. The drop between the two levels is the energy that has been released to the surroundings as heat. Because the products have less energy, we say the products are more stable than the reactants. As the reaction runs, the temperature of both the reaction mixture and its surroundings goes up — that warming is the released energy showing itself.

Draw the diagram in this order and you cannot lose the labels:

Two horizontal levels — reactants on the upper level, products on the lower level.
Vertical axis labelled "energy", horizontal axis labelled "progress of the reaction".
An arrow pointing down from reactants to products.
∆H written next to the arrow with its negative sign — the drop is the magnitude, the sign records the direction.

Exam hint

Paper 2C loves three details on this diagram. Label both axes — an unlabelled energy axis costs the mark even if the shape is perfect. Put the reactants above the products for an exothermic change, with a downward arrow. And write ∆H beside the arrow with its minus sign; "∆H" alone, or a positive value on a downward diagram, contradicts itself and loses the mark. If the question asks you to explain the shape, one sentence does it: the products have less chemical energy than the reactants, so the difference is released as heat.

Common misconception

"The products are drawn lower because the reaction mixture ended up colder." The levels show chemical energy stored in the chemicals, not the temperature at the end. An exothermic reaction leaves the surroundings hotter precisely because the chemicals ended with less energy — the two observations point in opposite directions, and the diagram records the chemicals.

Quick check

Question-bank join: spec 3.5C energy-level-diagram items. Sketch the exothermic energy level diagram from memory: axes, levels, arrow, sign. Then answer cold — for the reaction Mg + H₂SO₄ (ΔH = −466.9 kJ/mol), which level is higher and what does the drop represent?

Key terms in this section

energy level diagram: — A chart with energy on the vertical axis and progress of the reaction on the horizontal axis. Reactants and products sit on levels; the arrow and ∆H sign show the direction of the energy change.

Keep goingEnergy Level Diagrams 1-9-electrolysis 3-1-energetics 3-2-rates-of-reaction 3-3-reversible-reactions-and-equilibria

Endothermic reactions

SB p.209

A reaction that absorbs heat from the surroundings is called an endothermic reaction. Hold a test-tube in which an endothermic reaction is happening and the tube gets colder — heat is flowing the other way, out of your fingers and the air and into the reaction mixture.

In energy terms the picture mirrors the exothermic case exactly. In an endothermic reaction the products have more chemical energy than the reactants. To supply the extra energy needed to convert low-energy reactants into high-energy products, heat energy is absorbed from the surroundings and converted to chemical energy stored in the new bonds. The temperature of the reaction mixture and the surroundings goes down because heat energy has been drained and converted into a different form.

Chemical cold packs work on exactly this principle: an endothermic reaction inside the pack absorbs heat from the surroundings, lowering the temperature of the injured arm it rests against.

The everyday laboratory example is the thermal-decomposition of metal carbonates, which you met with Group 2 chemistry. These are endothermic — you have to heat a carbonate constantly to make it decompose, because the reaction keeps drawing heat from the flame. Copper(II) carbonate (green) decomposes on heating to produce copper(II) oxide (black):

CuCO₃(s) → CuO(s) + CO₂(g)

Zinc carbonate decomposes similarly to form zinc oxide when heated.

One warning that examiners turn into questions: not every reaction that needs heating is endothermic. Sometimes we heat a mixture simply to make the reaction go faster — the burning of magnesium needs a hot flame to start it, yet the reaction itself is fiercely exothermic. The direction of the temperature change during the reaction tells you which kind you have; the presence of a Bunsen burner does not.

Exam hint

The examiner's favourite endothermic definition question adds a twist: "explain why the decomposition of copper(II) carbonate is endothermic." The full-mark chain is: the products have more chemical energy than the reactants → the extra energy must be absorbed from the surroundings as heat → so the temperature of the surroundings falls and heating must continue throughout the reaction. Keep the heating-continuously point — it is what separates an endothermic decomposition from a merely slow reaction.

Common misconception

"If you have to light a Bunsen burner, the reaction must be endothermic." Heat applied to start or speed up a reaction tells you nothing about the direction of the overall energy change. Magnesium burning needs a flame to get going and then releases far more heat than it absorbed. Classify by the temperature change of the mixture itself: drops during the change → endothermic; rises → exothermic.

Quick check

Question-bank join: spec 3.1 endothermic-definition items. Define an endothermic reaction, state the energy relationship between products and reactants, write the equation for the thermal decomposition of copper(II) carbonate, and explain why a reaction needing a flame to start is not necessarily endothermic.

Key terms in this section

endothermic: — A reaction that absorbs heat from the surroundings. Products have more chemical energy than reactants; the temperature of the mixture and its surroundings falls; ∆H is positive.
thermal decomposition: — Breaking a compound down by heating — an endothermic change that needs the heat supplied continuously. Metal carbonates decompose to the metal oxide and carbon dioxide on heating.

Keep goingExothermic & Endothermic 1-5-chemical-formulae-equations-calculations 2-4-reactivity-series 3-1-energetics 3-3-reversible-reactions-and-equilibria

Showing an endothermic change on an energy level diagram

SB p.210

The endothermic energy-level-diagram is the mirror image of the exothermic one. In an endothermic change the products have more energy than the reactants, so we say the products are less stable than the reactants. That extra energy has to come from somewhere, and it is taken from the surroundings — for the thermal decomposition of carbonates in the laboratory, it comes from the Bunsen burner.

Because the reactants are gaining energy, the enthalpy change of the reaction is given a positive sign. The thermal decomposition of calcium carbonate carries the value you should be able to quote and explain:

CaCO₃(s) → CaO(s) + CO₂(g) ∆H = +178 kJ/mol

This means that 178 kJ of heat energy must be absorbed to convert one mole of calcium carbonate into calcium oxide and carbon dioxide. On the diagram, the arrow from reactants to products points up, the products level sits above the reactants level, and ∆H is written beside the arrow with its plus sign. Label the vertical axis "energy" and the horizontal axis "progress of the reaction" exactly as before.

Exam hint

Paper 2C asks for both diagrams in a single question so the comparison marks are easy to bank: exothermic = products lower, ∆H negative, arrow down; endothermic = products higher, ∆H positive, arrow up. For the calcium carbonate example, quote the value with its plus sign and attach the meaning — "178 kJ absorbed per mole of calcium carbonate decomposed." Omitting "per mole" leaves the answer incomplete; the value is defined for the equation's amounts.

Common misconception

"The +178 kJ/mol is the total heat the Bunsen burner supplied." The value is the minimum absorbed by the reaction per mole of calcium carbonate decomposed — the burner also heats the tube, the air and everything else, which is why a real flame delivers far more than 178 kJ in practice. The quoted ∆H is the reaction's share, not the fuel bill.

Quick check

Question-bank join: spec 3.5C endothermic-diagram items. Draw the endothermic energy level diagram for CaCO₃ → CaO + CO₂ with every label, then state the meaning of ∆H = +178 kJ/mol in one sentence. Finally, name where the energy comes from in the laboratory and what that does to the surroundings.

Key terms in this section

energy level diagram: — A chart with energy on the vertical axis and progress of the reaction on the horizontal axis. Reactants and products sit on levels; the arrow and ∆H sign show the direction of the energy change.

Keep goingEnergy Level Diagrams 1-9-electrolysis 3-1-energetics 3-2-rates-of-reaction 3-3-reversible-reactions-and-equilibria

Specific heat capacity and the Q = mc∆T equation

SB p.211

When we heat something up, it gets hotter. The specific-heat-capacity tells us how much energy has to be put in to increase the temperature of something. It is defined as the amount of heat needed to raise the temperature of 1 gram of a substance by 1 °C. The unit is J/g/°C (or J g⁻¹ °C⁻¹).

For water the value is 4.18 J/g/°C — 4.18 J of heat energy raises the temperature of 1 g of water by 1 °C. The idea scales in both directions, and the multiplication is easy to reason through:

1 g of water by 2 °C needs 4.18 × 2 = 8.36 J.
2 g of water by 2 °C needs 2 × 8.36 = 16.72 J.

The heat energy required is directly proportional to the mass (m) and to the temperature change (∆T), which compresses the whole idea into one equation — the workhorse of this chapter:

heat energy change = mass × specific heat capacity × temperature change

Q = m × c × ∆T

Each symbol carries a discipline. Q is the heat energy change in joules. m is the mass of the substance being heated (in grams — usually water). c is that substance's specific heat capacity (4.18 J/g/°C for water). ∆T is its temperature change in °C. The same equation runs backwards for cooling: to cool 1 g of water by 1 °C you must remove 4.18 J, so an endothermic process sitting in water still hands you a usable Q — you just read the temperature drop.

A calorimeter is simply the container you run a calorimetry experiment in: in one experiment it is a copper can, in others a polystyrene cup. What matters is that the reaction's heat ends up changing the temperature of a known mass of water inside it.

Key point

The unit of specific heat capacity is J/g/°C (J g⁻¹ °C⁻¹). Water's value is 4.18 J/g/°C — memorise it, every calculation in this chapter uses it. The ∆ in ∆T is the same "change in" symbol as in ∆H, this time applied to temperature, and the equation applies to cooling as well as heating.

Exam hint

Q = mc∆T questions are won and lost on which mass and which ∆T you substitute. m is the mass of the water (or solution) being heated, never the mass of the solid burned or dissolved. ∆T is final minus initial (or the magnitude of the drop for cooling processes). Carry an extra significant figure through the calculator and round at the end; divide by 1000 to convert J to kJ before computing a ∆H in kJ/mol.

Common misconception

"Specific heat capacity is a property of heating only." The quantity describes energy needed per degree of temperature change in either direction — removing 4.18 J cools 1 g of water by 1 °C just as supplying 4.18 J warms it. Endothermic experiments in this chapter depend on the cooling direction: the reaction takes heat from the water, ∆T is a drop, and Q still comes out of the same equation.

Quick check

Question-bank join: spec 3.3 Q=mc∆T items. State the definition and unit of specific heat capacity, calculate the energy needed to warm 50 g of water from 20.0 °C to 35.0 °C, and name the container used in a calorimetry experiment with the two common laboratory examples.

Key terms in this section

specific heat capacity: — The heat needed to raise the temperature of 1 gram of a substance by 1 °C. Water's value is 4.18 J/g/°C; the same quantity governs cooling in the reverse direction.
calorimeter: — The container a calorimetry experiment runs in — a copper can in the combustion experiment, a polystyrene cup for reactions in solution. Its contents are the known mass that absorbs or releases the heat.

Keep goingEnergetics Calculations 2-1-group-1-alkali-metals 3-1-energetics

Calorimetry and the combustion-of-an-alcohol experiment

SB p.211

Measuring the heat absorbed or given out by a reaction is calorimetry, and it is more straightforward than the name suggests. The idea: let the reaction's heat change the temperature of a known mass of water, then use Q = mc∆T to calculate the heat released. The mass, specific heat capacity and temperature change all belong to the substance heated — the water. If you also know how many moles of reactant were used up, you can divide Q by n to reach the molar enthalpy change ∆H in kJ/mol.

The classic International GCSE experiment measures the heat given off when small alcohols burn — methanol, ethanol, propan-1-ol and butan-1-ol. The alcohol burns in a small spirit burner and the heat produced warms water in a copper can (the calorimeter). The molar enthalpy change of combustion is how much heat energy is released when one mole of the alcohol burns.

The procedure, in the order that earns method marks:

Measure 100 cm³ of cold water with a measuring cylinder and transfer it to the copper can.
Take the initial temperature of the water.
Weigh a spirit burner containing the alcohol with its lid on — the lid stays on whenever the wick is not lit, to stop the alcohol evaporating.
Arrange the apparatus so the burner can heat the water in the can, and shield it as far as possible to prevent draughts.
Light the wick. Stop heating when the water has a reasonable temperature rise (about 40.0 °C); extinguish the flame by putting the lid back on the wick.
Stir the water thoroughly and measure the maximum temperature.
Weigh the spirit burner again with its lid on.
Repeat with the same alcohol to check reliability, then repeat with other alcohols.

How big a temperature rise should you aim for? There is no ideal answer — it is a deliberate compromise. The higher the temperature the water reaches, the greater the heat losses to the surroundings during the experiment; but if you only warm the water a little, errors in reading the thermometer and in finding the small mass change of the alcohol become too significant. A rise of around 40 °C balances the two.

Safety

Wear eye protection. Never carry a lit spirit burner, and never fill or re-fill one when there is a naked flame nearby. Keep the lid on the burner whenever the wick is not lit.

Key point

The enthalpy change when 1 mole of a substance is burned is called the enthalpy-change-of-combustion. Every combustion result in this chapter is quoted per mole of fuel burned.

Exam hint

Method questions on this experiment reward the control details: lid on when not lit (stops evaporation changing the mass), shielding from draughts (cuts heat loss), stir before reading (distributes the heat so the maximum is true), weigh the burner with the lid both times. If asked for an improvement, "use a lid or shield around the can to reduce heat loss" or "a wind shield to prevent draughts" are the standard reliable answers.

Common misconception

"The mass to use in Q = mc∆T is the mass of alcohol that burned." The Q equation describes the water: m is the 100 g of water in the can (100 cm³ × density 1 g/cm³). The mass of alcohol burned — the change in the burner's reading — enters later, when you convert Q into a per-mole value. Mixing the two masses up is the single most common arithmetic error on this experiment.

Quick check

Question-bank join: spec 3.2 + 3.8 combustion-calorimetry items. List the procedure in order from memory, then answer: why does the burner lid go back on between weighings, why shield the flame, and what temperature rise do you aim for — with the reason it is a compromise?

Key terms in this section

calorimetry: — Measuring heat changes by letting a reaction warm or cool a known mass of water, then calculating with Q = mc∆T and dividing by moles for the molar enthalpy change.
enthalpy change of combustion: — The enthalpy change when 1 mole of a substance is burned completely. Always negative for burning; measured for alcohols with the spirit-burner-and-copper-can experiment.

Keep goingCalorimetry Practical: Investigating Temperature Changes 2-1-group-1-alkali-metals 2-4-reactivity-series 2-6-acids-alkalis-and-titrations

Worked example: the combustion calculation

SB p.212

Combustion is exothermic, so the temperature of the water goes up; as the ethanol burns, the total mass of burner plus ethanol goes down. Here is a complete data set and the full route from it to a molar enthalpy change of combustion — the calculation the examiner asks you to reproduce with fresh numbers.

Measurement	Value
Volume of water / cm³	100
Mass of burner + ethanol before / g	137.36
Mass of burner + ethanol after / g	136.58
Original temperature of water / °C	21.5
Final temperature of water / °C	62.8

Worked example

Find the molar enthalpy change of combustion of ethanol from the data above.

Step 1 — temperature change. ∆T = 62.8 − 21.5 = 41.3 °C.

Step 2 — mass of water. 100 cm³ of water has mass 100 g (density of water ≈ 1 g/cm³ at room temperature).

Step 3 — heat gained by the water. c = 4.18 J/g/°C (it is the water that is being heated):

Q = mc∆T = 100 × 4.18 × 41.3 = 17 260 J

Divide by 1000 to convert to kJ: Q = 17.26 kJ. This is the heat released by the ethanol burned in this experiment.

Step 4 — moles of ethanol burned. Mass burned = 137.36 − 136.58 = 0.78 g. Ethanol, C₂H₅OH, has relative molecular mass (2 × 12) + (5 × 1) + 16 + 1 = 46:

n = mass ÷ Mr = 0.78 ÷ 46 = 0.01696 mol

Step 5 — molar enthalpy change. The molar-enthalpy-change is the heat energy change divided by the number of moles burned:

∆H = Q ÷ n = −17.26 ÷ 0.01696 = −1017.7 ≈ −1020 kJ/mol (3 s.f.)

Written on the equation:

CH₃CH₂OH + 3O₂ → 2CO₂ + 3H₂O ∆H = −1020 kJ/mol

The negative sign shows heat is released — the combustion is exothermic, as the rising temperature already told you.

Key point

The enthalpy-change-of-combustion is defined for one mole of substance burned — that is why Step 4 exists. Quoting −17.26 kJ as a final answer confuses the heat released in the experiment with the per-mole quantity the question asked for.

Exam hint

Two technique rules decide whether the arithmetic marks survive to the end. First, significant figures: quote the final answer to 3 significant figures, carry more figures through the calculator, and round only at the end — the least precise piece of experimental data fixes how many figures your answer can honestly carry. Second, the sign: add it at the ∆H stage and sanity-check it against the temperature change. Temperature went up → exothermic → ∆H negative. Temperature down → endothermic → ∆H positive. A sign-and-trend mismatch means a step has gone wrong.

Common misconception

"Rounding at every step keeps the working tidy." Rounding early compounds: 0.01696 mol rounded to 0.02 mol turns −1017.7 kJ/mol into −863 kJ/mol — off by more than 150 kJ and outside the mark scheme's tolerance. Keep the calculator's full display until the final line, then round once.

Quick check

Question-bank join: spec 3.3 + 3.4 combustion-calculation items. Redo the five steps with the burner masses 200.00 g and 199.30 g, the same 100 cm³ of water warming from 20.0 °C to 54.0 °C, and butan-1-ol (Mr = 74). Check: your Q = 14 212 J, your n = 0.00946 mol, and your answer near −1500 kJ/mol with a minus sign.

Key terms in this section

molar enthalpy change: — The heat energy change divided by the number of moles of reactant consumed (∆H = Q ÷ n), in kJ/mol. The moles used must be the limiting reactant's, and the sign follows the temperature trend.
enthalpy change of combustion: — The enthalpy change when 1 mole of a substance is burned completely. Always negative for burning; measured for alcohols with the spirit-burner-and-copper-can experiment.

Keep goingEnergetics Calculations 2-1-group-1-alkali-metals 3-1-energetics

Evaluating the combustion experiment

SB p.213

How accurate was the −1020 kJ/mol obtained for ethanol? Data booklets quote an accepted value of −1370 kJ/mol for the enthalpy change of combustion of ethanol — 1370 kJ of heat should be given out when one mole burns. The experimental value is less exothermic than expected: the experiment seemed to give out less heat than it should have. Explaining that gap is a guaranteed exam question, and the answer has two parts.

Heat loss dominates. The warm water gives out heat to the air; heat from the flame escapes straight into the surroundings rather than into the water; and heat is used to raise the temperature of the copper can and the thermometer as well as the water. Every joule that takes those side-routes is a joule the Q = mc∆T calculation never sees, so the measured temperature rise is too small and the calculated ∆H is too little exothermic.

The combustion is incomplete. incomplete-combustion happens when there is not enough oxygen present, and it releases less heat than complete combustion. You can see the evidence directly: the flame of the wick is often yellow-orange rather than blue, and soot (carbon) collects on the bottom of the copper can. In complete combustion the flame is blue and carbon dioxide is produced instead of carbon. Both observations — flame colour and soot — are acceptable evidence in the exam.

When the same experiment is repeated across the alcohol family, a pattern emerges. Plotting the molar enthalpy change of combustion against the number of carbon atoms in the alcohol shows the combustion reaction getting more exothermic as the alcohol chain becomes longer — longer alcohols give out more heat energy per mole when they burn than shorter ones. The increase is regular: each step from one alcohol to the next adds one extra CH₂ unit, so the number of extra bonds broken and made increases in a regular way, and so does the heat evolved.

One plotting discipline belongs with that graph. The points should be joined as discrete points, not a smooth curve: a smooth curve is only correct for a continuous independent variable — one that can take any value. There is no such thing as an alcohol with 0.5 or 1.64 carbon atoms; the number of carbon atoms is non-continuous, so a line through the points can mislead. The comparison remains valid: repeat with other alcohols under conditions as similar as possible, and the trend in heat evolved with chain length is reliable.

Key point

Evaluation formula for this experiment: measured ∆H less exothermic than the data booklet → (1) heat lost to surroundings (air, can, thermometer), (2) incomplete combustion (yellow-orange flame, soot) because less heat is released when oxygen is short.

Exam hint

"Suggest why the student's value differs from the data book value" wants the two-part answer above, with evidence. State the heat-loss routes specifically (to the air, from the flame, into the can and thermometer), then quote an observation for incomplete combustion — soot on the can or a yellow-orange flame instead of blue. For the improvement question, "insulate the can / add a lid / shield from draughts" attacks heat loss; "burn the alcohol in plenty of oxygen" attacks incompleteness. Match each improvement to the error it fixes.

Common misconception

"A value of −1020 kJ/mol instead of −1370 kJ/mol means the arithmetic was wrong." The arithmetic is fine; the method is systematically imperfect. Heat escaping to the surroundings and incomplete combustion both push the measured temperature rise down, and a smaller ∆T produces a smaller — less exothermic — ∆H. The gap's direction is predictable, which is exactly what makes it a favourite evaluation question.

Quick check

Question-bank join: spec 3.2 + 3.8 experiment-evaluation items. List the three heat-loss routes, give two observations that prove the combustion was incomplete, and state the trend in ∆H of combustion as alcohol chain length increases — with the reason the increase is regular.

Key terms in this section

incomplete combustion: — Burning with insufficient oxygen; releases less heat than complete combustion and produces carbon (soot) instead of only carbon dioxide. Seen as a yellow-orange rather than blue flame.

Keep goingCalorimetry Practical: Investigating Temperature Changes 2-1-group-1-alkali-metals 2-4-reactivity-series 2-6-acids-alkalis-and-titrations

Calorimetry for reactions in solution

SB p.214

Very similar methods to the alcohol experiment measure molar enthalpy changes for three more reaction families:

displacement reactions — for example zinc metal in copper(II) sulfate solution;
dissolving — for example ammonium chloride dissolving in water to form a solution;
neutralisation reactions — for example potassium hydroxide solution with dilute hydrochloric acid.

These experiments also involve "heating water" — but this time the water is part of the solutions themselves. The reaction happens in (or into) a polystyrene cup, the solution warms or cools, and the same equation Q = mc∆T converts the temperature change into heat energy. Two assumptions do the converting, and you must be able to state both:

the density of the solution is the same as that of water, so 1 cm³ of solution has a mass of 1 g;
the specific heat capacity of the mixture is the same as that of water — reasonable, because the reaction mixture is mostly water.

From there the route to the molar-enthalpy-change is identical to the combustion example: Q = mc∆T for the heat, moles of the limiting reactant for n, then ∆H = Q ÷ n with the sign fixed by whether the temperature rose (exothermic, negative) or fell (endothermic, positive). The next three sections run the full procedure and calculation for each family in turn.

Exam hint

"Describe how to measure the enthalpy change of neutralisation" and its siblings for displacement and dissolving are describe-the-method questions: polystyrene cup in a beaker (insulation), measured volumes, initial temperature, add and stir, record maximum (or minimum) temperature. Then the assumption pair — density of solution = density of water, and specific heat capacity of mixture = 4.18 J/g/°C because the mixture is mostly water — usually sits on the mark scheme as its own line. Learn it verbatim once and reuse it for all three families.

Common misconception

"Each reaction family needs a different equation." The equation never changes: Q = mc∆T, then ∆H = Q ÷ n. What changes is what reacts (a metal and a salt solution, a salt and water, an acid and an alkali) and therefore which moles you divide by. The constant structure is why one worked example per family is enough preparation for all of them.

Quick check

Question-bank join: spec 3.2 solution-calorimetry items. Name the three solution-reaction families, state the two standard assumptions of the method, and write the two-step calculation route from temperature change to molar enthalpy change.

Key terms in this section

molar enthalpy change: — The heat energy change divided by the number of moles of reactant consumed (∆H = Q ÷ n), in kJ/mol. The moles used must be the limiting reactant's, and the sign follows the temperature trend.

Keep goingCalorimetry 3-1-energetics

Practical: measuring the enthalpy change of a displacement reaction

SB p.215

To determine the enthalpy change for the displacement-reaction between zinc and copper(II) sulfate, the procedure is:

Place a polystyrene cup in a 250 cm³ glass beaker.
Transfer 50 cm³ of 0.200 mol/dm³ copper(II) sulfate solution into the cup using a measuring cylinder.
Weigh 1.20 g of zinc using a weighing boat on a balance.
Record the initial temperature of the copper(II) sulfate solution.
Add the zinc.
Stir the solution as quickly as possible.
Record the maximum temperature reached.

Measurement	Value
Initial temperature of copper(II) sulfate solution / °C	17.0
Maximum temperature of the reaction mixture / °C	27.3

The calculation uses the same assumptions as every solution experiment: the solution's density is that of water (1 cm³ has mass 1 g) and its specific heat capacity is 4.18 J/g/°C, because the mixture is mostly water.

Worked example

Calculate the enthalpy change for this displacement reaction when 1 mole of copper(II) sulfate reacts with zinc.

Step 1 — heat given out. ∆T = 27.3 − 17.0 = 10.3 °C:

Q = mc∆T = 50 × 4.18 × 10.3 = 2152.7 J = 2.1527 kJ

Step 2 — which reactant runs out first? Moles of CuSO₄ = (50 ÷ 1000) × 0.200 = 0.0100 mol. Moles of Zn = 1.20 ÷ 65 = 0.0185 mol. One mole of Zn reacts with one mole of CuSO₄, so 0.0100 mol of CuSO₄ reacts with 0.0100 mol of Zn — we added 0.0185 mol of zinc, so the zinc is in excess and the copper(II) sulfate is the limiting reactant. (Excess means more than enough is present to ensure all the copper(II) sulfate reacts.)

Step 3 — per mole. The heat released when 1 mole of CuSO₄ reacts is:

∆H = Q ÷ n = 2.1527 ÷ 0.0100 = 215.3 ≈ −215 kJ/mol

Zn(s) + CuSO₄(aq) → ZnSO₄(aq) + Cu(s) ∆H = −215 kJ/mol

The negative sign is added because the temperature of the reaction mixture went up: the displacement is exothermic and heat is released. (Remember the ÷1000 when converting the 50 cm³ to dm³ — the concentration is given in mol/dm³.)

You could repeat this experiment with metals of different reactivities: the more reactive the metal, the more heat should be released in the displacement. Keep everything else the same — the number of moles of metal, the size of the solid particles, and the volume and concentration of the copper(II) sulfate solution. Do not use metals more reactive than magnesium: those react with the water itself, and you would be measuring the heat released by the metal–water reaction instead.

Key point

The moles you divide Q by are the moles of the limiting reactant — the chemical that is not in excess. In this experiment the zinc was deliberately in excess so that all 0.0100 mol of copper(II) sulfate reacted, making the per-mole division clean.

Exam hint

The examiner builds this question in three layers: the Q = mc∆T arithmetic (2 marks), the moles comparison that identifies the excess reactant (1–2 marks), and the division plus sign (2 marks). The layer candidates drop is the middle one — say explicitly that 0.0185 mol of zinc exceeds the 0.0100 mol of copper(II) sulfate, so copper(II) sulfate is limiting and its moles go in the denominator of ∆H = Q ÷ n. A final answer without the negative sign, or divided by the zinc's moles, loses the last two marks.

Common misconception

"Use the moles of the reactant you weighed, because its mass was measured." The per-mole definition attaches to the balanced equation's reacting ratio, not to whichever mass was easier to weigh. With zinc in excess, dividing by 0.0185 mol would understate the per-mole value; the reaction's heat belongs to the 0.0100 mol of copper(II) sulfate that actually reacted.

Safety

Wear eye protection and avoid skin contact with the salts and their solutions.

Quick check

Question-bank join: spec 3.2 + 3.4 + 3.8 displacement-calorimetry items. Run the three-layer calculation with 50 cm³ of 0.100 mol/dm³ CuSO₄ and 1.30 g of zinc (Ar = 65): identify the limiting reactant, show which moles divide Q, and justify the sign of ∆H. Check: n(CuSO₄) = 0.00500 mol, n(Zn) = 0.0200 mol, zinc in excess.

Key terms in this section

displacement reaction: — A reaction in which a more reactive element takes the place of a less reactive one in a compound (a competition reaction). Zinc displacing copper from copper(II) sulfate is the calorimetry example here.
excess: — More than enough of a reactant to ensure the other reactant is fully used up. The excess reactant's moles never enter ∆H = Q ÷ n — the limiting reactant's do.

Keep goingCalorimetry Energetics Calculations Practical: Investigating Temperature Changes 2-1-group-1-alkali-metals 2-4-reactivity-series

Practical: measuring the enthalpy change when a salt dissolves

SB p.216

Calorimetry also measures the heat given out or taken in when salts dissolve in water. The set-up is the same polystyrene cup in a 250 cm³ beaker used for the displacement experiment. For ammonium chloride:

Place a polystyrene cup in a 250 cm³ glass beaker.
Transfer 100 cm³ of water into the cup using a measuring cylinder.
Record the initial temperature of the water.
Weigh 5.20 g of ammonium chloride using a weighing boat on a balance.
Add the ammonium chloride to the water and stir vigorously until it has all dissolved.
Record the minimum temperature.

Measurement	Value
Initial temperature of water / °C	18.3
Minimum temperature of salt solution / °C	15.1

The temperature decreases, so the process is endothermic — heat is absorbed from the surroundings for the dissolving to occur. Note one difference of kind, not just of sign: dissolving is a physical process rather than a chemical reaction. When some salts dissolve heat is given out (exothermic); when others dissolve the process is endothermic — ammonium chloride belongs to the second group.

The calculation follows the standard assumptions: the diluted solution has the specific heat capacity of water (it is mostly water), and the mass of the solution is 100 g — the small mass of ammonium chloride is ignored. Heat absorbed from the surrounding air is a further source of error, one that matters much more here than in the hotter experiments.

Worked example

Calculate the molar enthalpy change for dissolving ammonium chloride in water.

Step 1 — heat absorbed.

Q = mc∆T = 100 × 4.18 × (18.3 − 15.1) = 100 × 4.18 × 3.2 = 1337.6 J = 1.3376 kJ

Step 2 — moles dissolved. The relative formula mass of ammonium chloride, NH₄Cl, is 14 + (4 × 1) + 35.5 = 53.5:

n = mass ÷ Mr = 5.20 ÷ 53.5 = 0.0972 mol

Step 3 — per mole. The molar-enthalpy-change of solution:

∆H = Q ÷ n = 1.3376 ÷ 0.0972 = +13.8 kJ/mol

The positive sign shows that heat is absorbed — dissolving ammonium chloride in water is an endothermic process, which is why the temperature went down.

Key point

Dissolving is a physical process. The direction of its temperature change depends on the salt: some dissolvings warm the water (exothermic), others cool it (endothermic). Read the thermometer — do not assume.

Exam hint

Two marks hide in the details here. The thermometer reads a minimum for this experiment — "maximum temperature" is copy-paste from the exothermic experiments and drops the observation mark. And the sign discipline inverts: temperature down → endothermic → ∆H positive. If asked why the assumptions are reasonable, the answer is that the solution is mostly water, so it inherits water's specific heat capacity; if asked for a major error, heat absorbed from the surrounding air during the (slow) dissolving is the honest one.

Common misconception

"Endothermic means the thermometer reads a lower number because the thermometer is losing heat." The drop is real heat transfer: the dissolving salt takes energy from the water to pull its particles apart and into solution, so the water's temperature falls. The energy has not disappeared — it has become chemical energy in the solution, exactly the reverse of the exothermic bookkeeping.

Safety

Wear eye protection and avoid skin contact with the salts and their solutions.

Quick check

Question-bank join: spec 3.2 + 3.4 + 3.8 dissolving-calorimetry items. Recalculate with 6.40 g of ammonium chloride in 100 cm³ of water falling from 19.6 °C to 15.6 °C: show Q, n and the signed ∆H. Check: Q = 1672 J, n = 0.1196 mol, ∆H ≈ +14.0 kJ/mol. Then state whether potassium chloride dissolving exothermically would give a maximum or minimum reading.

Key terms in this section

dissolving: — A physical process in which a salt's particles separate into water. Depending on the salt it releases heat (exothermic) or absorbs heat (endothermic) — the thermometer decides.
molar enthalpy change: — The heat energy change divided by the number of moles of reactant consumed (∆H = Q ÷ n), in kJ/mol. The moles used must be the limiting reactant's, and the sign follows the temperature trend.

Keep goingCalorimetry Energetics Calculations Practical: Investigating Temperature Changes 2-1-group-1-alkali-metals 2-4-reactivity-series

Practical: measuring the enthalpy change of neutralisation

SB p.217

The reaction between an alkali and an acid is essentially the reaction between hydroxide ions and hydrogen ions to form water:

OH⁻(aq) + H⁺(aq) → H₂O(l)

For the potassium hydroxide + hydrochloric acid example, suppose you know the concentration of the potassium hydroxide but are not so sure of the dilute hydrochloric acid — the label has gone missing. The following method finds the unknown concentration and the heat released, in one run:

Place a polystyrene cup in a 250 cm³ glass beaker.
Transfer 25 cm³ of 2.00 mol/dm³ potassium hydroxide into the cup.
Record the initial temperature.
Fill a burette with 50.00 cm³ of dilute hydrochloric acid.
Use the burette to add 5.00 cm³ of the acid to the potassium hydroxide.
Stir vigorously and record the maximum temperature reached.
Keep adding further 5.00 cm³ portions, stirring and recording the maximum temperature each time, until 50.00 cm³ in total has been added.

Plot the temperature of the mixture against the volume of acid added. The shape tells the story: the temperature increases at first because the acid reacts with the alkali and gives out heat; once all the alkali has been used up, further additions are just cold acid poured into warm solution — nothing left to react with — so the temperature goes down.

Draw two lines of best fit, one through the rising points and one through the falling points: the point where the lines cross represents complete neutralisation. Reading from the graph, the maximum temperature is 31.8 °C, reached when 28.00 cm³ of acid has been added (the initial temperature for the ∆T is 19.3 °C).

Worked example

Calculate the heat released per mole and the unknown acid concentration.

Step 1 — heat given out. Total volume at neutralisation = 25 + 28 = 53 cm³:

Q = mc∆T = 53 × 4.18 × (31.8 − 19.3) = 2769.3 J = 2.7693 kJ

(Assumptions as always: density of the mixture = 1 g/cm³, and specific heat capacity = that of water because the neutralised solution is mostly water.)

Step 2 — moles. Moles of KOH = (25 ÷ 1000) × 2.00 = 0.0500 mol. At the neutralisation point the moles of HCl are exactly the same — that is what the crossing point guarantees.

Step 3 — the molar-enthalpy-change. The heat released when 1 mole of KOH reacts with 1 mole of HCl:

∆H = 2.7693 ÷ 0.0500 = −55.4 kJ/mol

KOH(aq) + HCl(aq) → KCl(aq) + H₂O(l) ∆H = −55.4 kJ/mol

The negative sign shows heat is released in this exothermic neutralisation.

Step 4 — the unknown concentration. 28.00 cm³ of acid contained the 0.0500 mol of HCl:

concentration = moles ÷ volume in dm³ = 0.0500 ÷ (28.00 ÷ 1000) = 0.0500 ÷ 0.02800 = 1.79 mol/dm³

One accuracy footnote: strictly, the water formed would first appear as steam, and extra heat is given out when steam condenses to liquid water — the attraction forces between molecules form exothermically. At this level the value is taken for the liquid product.

Key point

The two-lines-of-best-fit construction exists to fix the true neutralisation point: the crossing volume is where moles of acid = moles of alkali, which is why the HCl moles can be read off the KOH moles without knowing the acid concentration in advance.

Exam hint

This experiment is the richest mark-scheme target in the chapter. Know the graph shape and its explanation (rises — exothermic neutralisation; falls — cold acid diluting warm mixture with nothing left to react), the two-best-fit-lines crossing construction, and the neat two-in-one finish: ∆H = −55.4 kJ/mol and the acid concentration 1.79 mol/dm³ from the same run. When asked why the temperature falls after neutralisation, the phrase "no further reaction — only dilution with cold acid" is the mark.

Common misconception

"The maximum temperature must be at the largest volume of acid added." After the crossing point you are adding cold acid to a warm, spent mixture — each 5.00 cm³ portion drags the temperature down further. The peak sits at the neutralisation volume (28.00 cm³ here), which is smaller than the 50.00 cm³ delivered in total; that is the entire logic of finding the crossing point.

Quick check

Question-bank join: spec 3.2 + 3.4 + 3.8 neutralisation-calorimetry items. Sketch the temperature-versus-volume graph from memory with both best-fit lines, calculate n(KOH), and show how the unknown HCl concentration follows from the crossing volume. Check: n(KOH) = 0.0500 mol; if the crossing came at 25.00 cm³, the acid would be 2.00 mol/dm³.

Key terms in this section

neutralisation: — The reaction between an alkali and an acid — essentially OH⁻(aq) + H⁺(aq) → H₂O(l). Exothermic; measured here with the two-best-fit-lines graph that also yields an unknown acid concentration.
molar enthalpy change: — The heat energy change divided by the number of moles of reactant consumed (∆H = Q ÷ n), in kJ/mol. The moles used must be the limiting reactant's, and the sign follows the temperature trend.

Keep goingCalorimetry Energetics Calculations Practical: Investigating Temperature Changes 2-1-group-1-alkali-metals 2-4-reactivity-series

Why do reactions give out or absorb heat?

SB p.219

During chemical reactions, bonds in the reactants have to be broken and new ones formed to make the products. The whole energy story of every reaction rests on two facts:

Breaking bonds needs energy — endothermic. Making bonds releases energy — exothermic.

Think about hydrogen burning in oxygen to make water:

2H₂(g) + O₂(g) → 2H₂O(l)

Energy has to be supplied to break the bonds in the hydrogen molecules and in the oxygen molecule. Energy is released when new bonds form between the hydrogen and oxygen atoms inside the water molecules. In this particular reaction the products are more stable than the reactants: the total energy released when the covalent bonds form in the water molecules is more than the energy required to break the covalent bonds in the hydrogen and oxygen molecules. Released > absorbed, so the reaction is exothermic overall.

Heating calcium carbonate runs the ledger the other way. Breaking up the original bonds in the compound uses more energy than is obtained when the new ones are made: the reactants are more stable than the products, so when the reaction finishes more energy has been absorbed than released — an endothermic change.

This is also where the small starting push many reactions need comes from: energy is required to begin breaking the reactant bonds even in a reaction that releases heat overall (the quantity is called the activation-energy and this chapter's bond-energy arithmetic is what lets you account for it). But the classification of a reaction depends only on the balance: absorbed minus released.

released > absorbed → exothermic, ∆H negative, products more stable.
absorbed > released → endothermic, ∆H positive, products less stable.

Key point

The sign of ∆H is decided by competition: energy in to break reactant bonds versus energy out from making product bonds. Whichever term is bigger wins the sign — and the winner also decides which side of the energy level diagram sits higher.

Exam hint

"Explain, in terms of breaking and making bonds, why this reaction is exothermic" has a three-part mark scheme: (1) bond breaking is endothermic / needs energy, (2) bond making is exothermic / releases energy, (3) more energy is released making the new bonds than is needed to break the original ones. All three sentences, in that order. Omitting the comparison sentence is the most common way to drop the third mark — naming both processes is not the same as comparing them.

Common misconception

"Bonds are like springs — breaking them must release the stored energy." Everyday intuition says breaking things releases energy (snapping a stick, smashing a plate), but chemical bonds work the opposite way: the bond is the low-energy arrangement, so pulling it apart costs energy, and forming it pays energy back. That inversion is the single most examined idea in the chemistry-only half of this chapter.

Quick check

Question-bank join: spec 3.6C bond-breaking/making items. Write the two law-statements (breaking/making) from memory, apply them to explain why hydrogen burning is exothermic, and why calcium carbonate decomposition is endothermic — one sentence each with the comparison word more or less included.

Key terms in this section

endothermic: — A reaction that absorbs heat from the surroundings. Products have more chemical energy than reactants; the temperature of the mixture and its surroundings falls; ∆H is positive.
exothermic: — A reaction that gives out heat energy to the surroundings. Products have less chemical energy than reactants; the temperature of the mixture and its surroundings rises; ∆H is negative.
activation energy: — The small amount of energy needed to get a reaction started — beginning the bond-breaking in the reactants. Even strongly exothermic reactions need it; a flame is often its source.

Keep goingBond Energy 3-1-energetics 4-3-alkanes

Bond energies

SB p.220

Breaking chemical bonds requires energy, and the stronger the bond, the more energy is needed to break it. The bond-energy is defined as the amount of energy needed to break 1 mole of covalent bonds in gaseous molecules, measured in kJ/mol. The same number works in reverse: the bond energy is also the amount of energy released when 1 mole of those bonds forms.

Two examples anchor the idea. The Cl–Cl bond energy is 243 kJ/mol: it takes 243 kJ to break all the Cl–Cl bonds in 1 mole of chlorine gas, splitting each molecule into two chlorine atoms — and if those atoms recombine into their original molecules, exactly 243 kJ is released again. The H–Cl bond energy is 432 kJ/mol: if you see 2HCl(g) in an equation, it takes 2 × 432 kJ to break all the H–Cl bonds in the two moles of hydrogen chloride — or 2 × 432 kJ will be released when 2HCl(g) is formed from H and Cl atoms. The coefficient in front multiplies the bond energy; that one sentence is the whole counting rule of the next section.

Bond	C–H	C–Cl	I–Cl	Cl–Cl	I–I	H–Cl	H–I
Bond energy / kJ/mol	413	346	234	243	151	432	298

Some bonds are much stronger than others. A striking comparison from the table: the bond between iodine and hydrogen (298 kJ/mol) is about twice as strong as the bond between two iodine atoms (151 kJ/mol).

Bond energies can be used to calculate how much heat will be absorbed or released during reactions of covalent molecules. If a reaction involves ionic compounds, different energy terms are needed — beyond International GCSE. The method is an estimate by dismantling and rebuilding: calculate how much energy would be needed to break the reactants up into individual atoms, then how much would be given out when those atoms recombine into the new arrangements of the products.

Exam hint

Bond-energy questions always supply the table — you never memorise values. What you must bring is the counting discipline: multiply each bond energy by how many moles of that bond appear in the balanced equation. The 1-mole definition is also a mark: writing "energy to break a bond" instead of "energy to break 1 mole of bonds in gaseous molecules" loses the definition mark. Note the word gaseous — it constrains where the method is valid (see the next section's worked example).

Common misconception

"A bond energy of 432 kJ/mol for H–Cl means one HCl molecule needs 432 kJ to break." The value is per mole of bonds — one molecule needs 432 kJ divided by the Avogadro number, an utterly tiny amount. All bond-energy arithmetic runs in per-mole quantities multiplied by the equation's coefficients.

Quick check

Question-bank join: spec 3.7C bond-energy-definition items. State the definition of bond energy with its unit, interpret Cl–Cl = 243 kJ/mol in both directions (breaking and forming), and calculate the energy to break all the bonds in 2 moles of HCl gas.

Key terms in this section

bond energy: — The energy needed to break 1 mole of covalent bonds in gaseous molecules, in kJ/mol; the same quantity is released when the bonds form. Quoted values are averages for estimates.

Keep goingBond Energy 3-1-energetics 3-2-rates-of-reaction 4-3-alkanes 4-4-alkenes

Bond-energy calculations

SB p.221

The calculation recipe has one sign rule and two sums. When energy is given out you show it with a negative sign; when energy is taken in, positive. Then:

energy needed to break all the bonds in the reactants — every term positive;
energy released when the new bonds are made in the products — every term negative;
overall change = (energy in to break) + (energy out to make) — a positive total means the reaction is endothermic, a negative total exothermic.

A quick illustration of the rule: if breaking all the reactant bonds costs +1000 kJ and making the product bonds releases −1200 kJ, the overall change is +1000 + (−1200) = −200 kJ — the reaction releases 200 kJ of energy overall, so it is exothermic.

Worked example

Methane reacts with chlorine in the presence of ultraviolet light to produce chloromethane and hydrogen chloride. You can imagine all the bonds being broken in the methane and chlorine, and then being reformed in new ways in the products. Calculate the overall energy change.

Bonds that need to be broken (endothermic):

Bond	Count	Sum
4 C–H	4 × (+413)	+1652 kJ
1 Cl–Cl	1 × (+243)	+243 kJ
	total in	+1895 kJ

New bonds made (exothermic):

Bond	Count	Sum
3 C–H	3 × (−413)	−1239 kJ
1 C–Cl	1 × (−346)	−346 kJ
1 H–Cl	1 × (−432)	−432 kJ
	total out	−2017 kJ

More energy is released when bonds are formed than is required to break them, so the reaction releases energy overall: it is exothermic.

Overall energy change = +1895 + (−2017) = −122 kJ

The negative sign of the answer shows that, overall, heat is given out as the bonds rearrange.

Two honesty notes belong beside every bond-energy answer. First, bond-energy calculations only ever give estimates: the strength of a bond varies slightly depending on what else is around it in the molecule, so quoted bond energies are average values — good, but not perfect. Enthalpy changes measured by the calorimetry experiments of this chapter are more specific for particular compounds. Second, the "break everything, rebuild everything" route is an imaginary one — the real reaction does not actually happen by all the bonds being broken this way. That does not matter: the overall amount of heat released or absorbed is the same however you do the reaction, as long as the reactants and products are the same. That fact is hesss-law, which you will meet again if you take chemistry further.

Worked example

Combustion of methane. The equation is:

CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(g)

Bond	C–H	O=O	C=O	O–H
Bond energy / kJ/mol	413	498	743	464

Start by drawing out the structures showing all the bonds, so the bond counts are right: CH₄ holds 4 C–H; the two O₂ molecules hold 2 O=O double bonds (498 kJ/mol is the energy to break the whole double bond); CO₂ holds 2 C=O; the two H₂O molecules hold 4 O–H.

Bonds that need to be broken (endothermic):

4 C–H = 4 × (+413) = +1652 kJ 2 O=O = 2 × (+498) = +996 kJ total in = +2648 kJ

New bonds made (exothermic):

2 C=O = 2 × (−743) = −1486 kJ 4 O–H = 4 × (−464) = −1856 kJ total out = −3342 kJ

Enthalpy change = +2648 + (−3342) = −694 kJ/mol

More energy is released when bonds are formed than is required to break them: the reaction is exothermic. Energy is given out as the reactants are converted to the products, so the products must have less energy than the reactants — the products are more stable, and an energy level diagram for this reaction shows the products level sitting below the reactants level, with a downward arrow and ∆H = −694 kJ/mol.

The reaction is written with H₂O(g) rather than H₂O(l) deliberately: bond energies can only be used for gaseous molecules. Finding the enthalpy change for the same reaction forming liquid water would need the energy released by H₂O(g) → H₂O(l) as well — beyond International GCSE.

Exam hint

Paper 2C marks these in four slots: correct bond counts from the drawn-out structures (the balancing coefficients multiply the bonds — 2O₂ means 2 O=O, and 2H₂O means 4 O–H), correct signs (+ in, − out), the addition to a total, and the exothermic/endothermic verdict tied to the sign. Draw the molecules first: every miscounted bond in this question family traces back to skipping the structures. If an energy level diagram is required, the sign of your total dictates which level the products occupy.

Common misconception

"O=O at 498 kJ/mol breaks into two O– bonds worth 249 kJ each, so count the double bond as two single bonds." The 498 kJ/mol is the energy to break the whole double bond at once — count one O=O per oxygen molecule. Bond tables list whole bonds (single or double as printed); inventing half-values or counting double bonds twice is the standard arithmetic corruption in this question.

Quick check

Question-bank join: spec 3.7C bond-energy-calculation items. Recalculate the methane + chlorine example with the C–H value 410 kJ/mol: your totals are +1883 in and −2008 out, giving −125 kJ — confirm the sign does not change. Then state the two honesty notes (average values; imaginary route) in one sentence each.

Key terms in this section

bond energy: — The energy needed to break 1 mole of covalent bonds in gaseous molecules, in kJ/mol; the same quantity is released when the bonds form. Quoted values are averages for estimates.
Hess's law: — The overall heat released or absorbed by a reaction is the same however the reaction is carried out, provided reactants and products are identical — even if the calculated route is imaginary.

Keep goingBond Energy 3-1-energetics 3-2-rates-of-reaction 4-3-alkanes 4-4-alkenes

Looking ahead: entropy

SB p.223

Many reactions you meet at International GCSE are exothermic — combustion and neutralisation, for instance. Despite needing a small amount of energy to get started (the activation-energy), these reactions release heat energy overall because the products are more stable than the reactants. Enthalpy is clearly a driving force: reactions that give out heat happen readily.

But endothermic processes happen too. Dissolving ammonium nitrate in water lowers the temperature of the solution — yet it dissolves anyway, pulling heat out of its surroundings and still going. How? A second factor is at work: entropy.

Entropy is a measure of the number of ways of distributing the energy in a system. It is greater in a more disordered system, because the energy can be dispersed more flexibly. The ordering runs solid → liquid → gas: solids have lower entropy than liquids, and liquids lower entropy than gases — a gas's particles roam freely, so its energy has the most ways to be shared out.

If the entropy of a system increases during a change, the change is more likely to happen. When ammonium nitrate dissolves, although the process is endothermic, the NH₄⁺ and NO₃⁻ ions in the giant ionic lattice become separated from each other and gain freedom of movement in the water — the entropy of the system increases enough to outweigh the enthalpy unfavourability. The same logic explains a kitchen classic: the reaction between baking soda and vinegar (an endothermic reaction that proceeds spontaneously, fizzing out carbon dioxide gas) — producing a gas increases the entropy of the system, and the increase carries the change.

You do not need to calculate entropy at this level. What the examiners want is the concept: enthalpy is not the only driving force, and an increase in disorder can power changes that absorb heat.

Exam hint

This is enrichment, not a numbered specification point — but it earns extension credit and it future-proofs the explanation questions. If asked "why does ammonium nitrate dissolve even though the process is endothermic?", the mature two-part answer is: the dissolving is endothermic, but the ions leaving the lattice for the solution increases the entropy of the system, and that increase outweighs the enthalpy change. The same pattern answers for any spontaneous endothermic change that produces a gas.

Common misconception

"Endothermic changes cannot happen by themselves because they need heat pumped in." They cannot happen if enthalpy were the only driving force — but disorder counts too. A change that absorbs heat can still run on its own when it increases the number of ways the system's energy can be distributed: ammonium nitrate dissolving in cool water is the standard proof.

Quick check

Question-bank join: extension entropy items. Define entropy in one sentence, order the three states of matter by entropy, and explain in two sentences why ammonium nitrate dissolves in water even though the temperature drops.

Key terms in this section

entropy: — A measure of the number of ways the energy in a system can be distributed; greater in more disordered systems. Solids have lower entropy than liquids, liquids lower than gases.
activation energy: — The small amount of energy needed to get a reaction started — beginning the bond-breaking in the reactants. Even strongly exothermic reactions need it; a flame is often its source.

Keep goingExothermic & Endothermic 1-5-chemical-formulae-equations-calculations 2-4-reactivity-series 3-1-energetics 3-3-reversible-reactions-and-equilibria

Chapter 19 questions

SB p.224

You have the complete toolkit now: the exothermic/endothermic split, the ∆H sign convention, the Q = mc∆T and ∆H = Q ÷ n pair, four calorimetry setups with their assumptions and errors, and — for Paper 2C — energy level diagrams and bond-energy arithmetic. The question bank for this chapter joins the platform by the specification points below, so every practice item you are served targets exactly the skills these sections taught.

Before you open the question bank, audit yourself against the chapter's core moves. Can you define both reaction types with the heat-flow direction? Can you run a five-step combustion calculation from a burner data table? Can you state the two solution-calorimetry assumptions and the two errors that make combustion results less exothermic? On the Paper 2C track: can you draw both energy level diagrams with every label, and run a bond-energy sum with correct counts and signs? Anything that wobbled is a section to revisit, not a reason to start over.

Summary

exothermic: heat given out to the surroundings, temperature rises, products have less chemical energy (more stable), ∆H negative. Families: combustion, metal + acid, neutralisation, displacement.
endothermic: heat taken in from the surroundings, temperature falls, products have more chemical energy (less stable), ∆H positive. Example: thermal decomposition of carbonates — heating must continue throughout. Not everything that needs a flame to start is endothermic.
enthalpy-change ∆H: heat taken in or given out per the equation's amounts, read from the reactants' point of view, units kJ/mol; you can only measure the change, never the absolute enthalpy.
Q = mc∆T converts a temperature change of a known mass (usually water, c = 4.18 J/g/°C) into heat energy; ∆H = Q ÷ n converts it to a per-mole value with the sign from the temperature trend.
Calorimetry families: combustion (spirit burner + copper can; lid, shielding, stir), displacement (polystyrene cup; excess metal; divide by the limiting reactant), dissolving (physical process; minimum temperature; read the sign carefully), neutralisation (two best-fit lines; crossing point = complete neutralisation; doubles as an unknown concentration method).
Evaluation: combustion values come out less exothermic than data books because of heat loss and incomplete combustion (yellow-orange flame, soot); longer alcohol chains burn more exothermically, rising regularly with each extra CH₂.
Paper 2C: energy level diagrams (label both axes; products' level follows the sign of ∆H); bond-breaking endothermic, bond-making exothermic; bond energies are per-mole averages for gaseous molecules — count bonds from drawn-out structures, add +in and −out, and the total's sign classifies the reaction.

Can you…?

Define exothermic and endothermic reactions and give two balanced equations for each?
Draw and label energy level diagrams for both types (Paper 2C)?
Calculate Q from mc∆T and quote it with the right sign and units?
Convert Q into a molar enthalpy change, dividing by the limiting reactant's moles?
Describe all four calorimetry procedures with their key controls and assumptions?
Explain why a combustion experiment under-reads, with evidence?
Calculate an overall enthalpy change from bond energies, including double bonds and balancing coefficients (Paper 2C)?

Exam hint

Past-paper items for this chapter cluster into three question shapes: a calculate-from-data Q = mc∆T / ∆H = Q ÷ n chain (always with a sign mark), a describe-the-method-and-its-errors practical question (assumptions, improvements, evidence of incomplete combustion), and the Paper 2C diagram-plus-bond-energy pair. Practise all three shapes against the clock — and remember the chapter-wide discipline: sign first, then size.

Common misconception

"Revising the formulas is enough for this chapter." The formula questions are the predictable half. The unpredictable half is the practical understanding — which mass goes into m, why the lid stays on the burner, why two lines of best fit, why the zinc is in excess. The examiner tests the why behind every step; the sections in this chapter were written so each step's reason is one sentence you can quote.

Quick check

Question-bank join: spec 3.1–3.8 chapter assessment items. Work through the can-you list honestly, marking each item secure or shaky, then take the platform's spec-joined question set for 3.2–3.4 first (the calculation chain earns the most marks per minute of revision).

Key terms in this section

exothermic: — A reaction that gives out heat energy to the surroundings. Products have less chemical energy than reactants; the temperature of the mixture and its surroundings rises; ∆H is negative.
endothermic: — A reaction that absorbs heat from the surroundings. Products have more chemical energy than reactants; the temperature of the mixture and its surroundings falls; ∆H is positive.
enthalpy change: — The heat energy taken in or given out during a chemical reaction — the difference between the energy of the products and the reactants. Only the change (∆H) is measurable, never absolute enthalpy.

Keep goingExothermic & Endothermic Calorimetry Energetics Calculations Energy Level Diagrams Bond Energy