R1.1 Measuring enthalpy changes · Educerie Chemistry

Educerie · IB Diploma · Chemistry

Reactivity 1 What drives chemical reactions? · R1.1 Measuring enthalpy changes

Level: SL and HL. Nothing here is HL only, so every section is examinable for both.
Themes (key concepts): energy is conserved; measurement; models. Every chemical change moves energy between the reacting chemicals and their surroundings, and a thermometer in the surroundings is how you measure it, with all the uncertainty a real measurement carries.
The question this unit answers: what can be deduced from the temperature change that accompanies chemical or physical change?
Where it is examined: Paper 1A, for the signs and the shape of energy profiles; Paper 1B, where a calorimetry experiment is described and you process its data, read a temperature–time graph and evaluate the method; Paper 2, where a Q = mcΔT calculation typically carries 3 or 4 marks and a sketched energy profile 2 or 3.

What you must be able to do

You must be able to	Level	What it looks like in the exam
Distinguish between heat and temperature	SL, HL	"Distinguish between heat and temperature" (2 marks)
Define system and surroundings, and state that total energy is conserved	SL, HL	Inside an explanation of an energy change
Say which way the temperature moves in an exothermic and an endothermic reaction	SL, HL	Paper 1A; "Deduce whether the reaction is exothermic" from data (1 mark)
Sketch and interpret energy profiles, with reaction coordinate and potential energy axes	SL, HL	"Sketch an energy profile for this reaction, labelling ΔH and Eₐ" (3 marks)
Relate the relative stability of reactants and products to the sign of ΔH	SL, HL	"Explain, in terms of stability, why the reaction is endothermic" (2 marks)
State what ΔH⦵ means: heat transferred at constant pressure, standard conditions and states, in kJ mol⁻¹	SL, HL	Paper 1A, or a definition inside a longer question
Use Q = mcΔT and ΔH = −Q/n to calculate an enthalpy change	SL, HL	Paper 2 or 1B: "Calculate the enthalpy change of neutralisation" (3–4 marks)
Correct a temperature rise for heat loss by extrapolation, and evaluate a calorimetry method	SL, HL	Paper 1B: a graph to extrapolate, then "Suggest why the value is less exothermic than the literature value" (2–3 marks)

Before you start

You need the idea from S1.1 that temperature measures the average kinetic energy of particles, and the mole from S1.4: amount = mass ÷ molar mass, and amount = concentration × volume in dm³. You need to identify a limiting reactant. No other maths is needed beyond rearranging an equation.

1The idea in one paragraph

When a reaction happens, bonds break and new bonds form, and the energy stored in the chemicals changes. That energy does not vanish or appear from nowhere: whatever the chemicals lose, their surroundings gain, and whatever the chemicals gain, their surroundings lose. So if you put a reaction in water and watch a thermometer, the water's temperature change tells you how much energy moved and in which direction. Divide by the amount of reaction that happened and you have the enthalpy change, ΔH, in kilojoules per mole. The rest is how to draw it, calculate it, and why school measurements come out too small.

2Heat and temperature are not the same thing

Temperature is a measure of the average kinetic energy of the particles in a substance. It tells you how hot something is, and it does not depend on how much of the substance there is. It is measured in °C or K; a change of 1 °C is the same size as a change of 1 K, so for a temperature change you can use either.

Heat is energy transferred from one place to another because of a difference in temperature. It flows from hotter to colder. It is measured in joules, and the amount depends on the mass as well as the temperature change.

A cup of tea at 80 °C is hotter than a bath at 40 °C: its particles have a higher average kinetic energy. But warming the 250 g of tea from 20 °C took about 63 kJ, while warming the 100 kg of bath water from 20 °C took about 8,400 kJ. The bath holds far more thermal energy at a lower temperature, simply because there is so much more of it. Temperature is a property of the particles; heat is a quantity of energy on the move.

3System, surroundings, and which way the energy goes

Chemists divide the world in two. The system is the reacting chemicals. The surroundings are everything else: the water they are dissolved in, the container, the thermometer, the air. Energy can pass between them, but the total is conserved.

An exothermic reaction transfers energy from the system to the surroundings. The chemicals end up with less stored energy, and the surroundings get warmer. An endothermic reaction transfers energy from the surroundings to the system, so the surroundings get cooler. Figure 1 shows both.

Figure 1 · System and surroundings

The trap is where the thermometer is. It measures the surroundings, not the reacting particles. So an exothermic reaction makes the thermometer rise, and an endothermic one makes it fall. Figure 2 shows what you see in each case: a steady temperature, then a jump up or down at the moment of mixing, then a slow drift back towards room temperature as heat is exchanged with the air.

Figure 2 · What the thermometer shows

Some familiar examples, and what you would observe:

Change	Type	What you notice
Burning a fuel; neutralising an acid with an alkali	exothermic	flame, or the solution warms
Magnesium ribbon in dilute acid	exothermic	fizzing, and the tube warms
Dissolving solid sodium hydroxide	exothermic	the solution gets hot
Dissolving ammonium nitrate	endothermic	the beaker feels cold
Citric acid with sodium hydrogencarbonate solution	endothermic	fizzing, and the temperature drops
Thermal decomposition of calcium carbonate	endothermic	only happens while heating continues

Physical changes count too, which is why the guide's question says "chemical or physical change". Melting and boiling are endothermic, because energy is needed to separate particles; freezing and condensing are exothermic.

4Energy profiles: why some reactions release energy

An energy profile is a sketch of how the potential energy of the system changes as reactants turn into products. The guide fixes the axes: the horizontal axis is the reaction coordinate, which means the progress of the reaction, and the vertical axis is potential energy. Figure 3 shows the two shapes.

Figure 3 · Energy profiles

Every profile has three features:

a level for the reactants and a level for the products;
the enthalpy change, ΔH, drawn as an arrow from the reactants' level to the products' level. It points down for exothermic and up for endothermic;
a hump between them. Its height above the reactants is the activation energy, Eₐ: the minimum energy the particles must have for a collision to lead to reaction. Even a strongly exothermic reaction needs this energy first, which is why paper does not burst into flames on its own. You meet Eₐ properly in R2.2.

The profile carries the guide's key idea: the relative stability of reactants and products decides the sign of ΔH. Lower potential energy means more stable. In an exothermic reaction the products are lower, so more stable, than the reactants, and the difference is released to the surroundings. In an endothermic reaction the products are less stable than the reactants, and the difference has to be taken in.

The guide asks one more question here: most combustions are exothermic, so why is the combustion of nitrogen endothermic? The answer is in the bonding. The two atoms in N₂ are held by a triple bond, one of the strongest bonds in chemistry, so N₂ is exceptionally stable. The bonds formed in nitrogen monoxide release less energy than it takes to break the N≡N and O=O bonds, so

N₂(g) + O₂(g) → 2NO(g) ΔH ≈ +180 kJ mol⁻¹products less stable than reactants

That is why air does not burn, and why nitrogen oxides, a cause of acid rain (S3.1), form only at very high temperatures such as inside an engine.

5Enthalpy change, and the standard version

Enthalpy, H, is the total heat content of a system. You cannot measure H itself, only the change in it. The enthalpy change, ΔH, is the heat transferred between the system and the surroundings at constant pressure, which is the situation in any open beaker. Its sign is from the system's point of view:

Exothermic: ΔH is negative, the surroundings warm up. Endothermic: ΔH is positive, the surroundings cool down.

ΔH depends on the conditions, so values are compared under the same ones. The standard enthalpy change, ΔH⦵, is measured under standard conditions: a pressure of 100 kPa and a stated temperature, usually 298 K, with solutions at 1 mol dm⁻³ and every substance in its standard state, its normal physical state under those conditions. The units of ΔH⦵ are kJ mol⁻¹: kilojoules per mole of the reaction as written in the equation.

That is why an enthalpy change is always given with an equation and state symbols. A thermochemical equation looks like this:

HCl(aq) + NaOH(aq) → NaCl(aq) + H₂O(l) ΔH = −57 kJ mol⁻¹

The state symbols matter. Forming liquid water releases more energy than forming steam, because the steam still has to condense to become liquid.

6Calorimetry: from a temperature change to ΔH

Calorimetry is measuring an enthalpy change from a temperature change. The guide's two equations do it in two steps, and both are in the data booklet or given to you.

Step 1: how much heat did the water gain or lose?

Q = mcΔT

Q is the heat transferred to the water, in J.
m is the mass of the water or solution being heated, in g. It is not the mass of the reactants.
c is the specific heat capacity: the energy needed to raise 1 g of the substance by 1 K. For water, c = 4.18 J g⁻¹ K⁻¹.
ΔT is the temperature change, final minus initial, in K or °C.

Step 2: turn it into an enthalpy change per mole.

ΔH = −Q ÷ n

n is the amount, in mol, of the reactant that limits the reaction, or of whichever substance the question asks about. The minus sign does the sign convention for you: when the water warms, ΔT and Q are positive, and ΔH comes out negative, which is exothermic. Divide Q by 1000 to go from J to kJ.

Figure 4 shows the standard apparatus for a reaction in solution. The polystyrene is a poor conductor, the lid stops heat escaping from the surface, and stirring makes sure the thermometer reads the temperature of the whole solution.

Figure 4 · A simple calorimeter for reactions in solution

Worked example 1: enthalpy change of neutralisation. 50.0 cm³ of 2.00 mol dm⁻³ hydrochloric acid and 50.0 cm³ of 2.00 mol dm⁻³ sodium hydroxide, both at 20.3 °C, are mixed in a polystyrene cup. The highest temperature reached is 33.1 °C. Calculate the enthalpy change of neutralisation.

ΔT = 33.1 − 20.3 = 12.8 K

m = 50.0 + 50.0 = 100.0 gtotal volume; density taken as 1.00 g cm⁻³

Q = mcΔT = 100.0 × 4.18 × 12.8 = 5350 J = 5.35 kJ

n(HCl) = 2.00 × 0.0500 = 0.100 mol = n(NaOH)neither is in excess; 0.100 mol of water forms

ΔH = −Q ÷ n = −5.35 ÷ 0.100 = −53.5 kJ mol⁻¹

Four assumptions sit behind that answer, and Paper 1B asks you to name them: the solution has the density and specific heat capacity of water; no heat is lost to the surroundings or absorbed by the cup and thermometer; the reaction is complete; and both solutions started at the same temperature. Data sources give about −57 kJ mol⁻¹. The measured value is less exothermic than that, and the first assumption to blame is the one about heat loss.

7Correcting for heat loss: the temperature–time graph

A calorimeter is never perfect. While the reaction is still releasing heat, some of that heat is already leaking out. The highest reading on the thermometer is therefore lower than the temperature the solution would have reached if the reaction had been instant. You can correct for this with a graph.

Take readings for a few minutes before mixing, add the reactant at a noted time, and keep taking readings for several minutes after the peak. Then draw a best-fit line through the cooling points and extrapolate it back to the moment of mixing. That extended line estimates the temperature the solution would have reached with no heat loss. Figure 5 is a real-shaped example.

Figure 5 · Correcting for heat loss by extrapolation

Worked example 2: a displacement reaction. Excess zinc powder is added at 3.5 minutes to 50.0 cm³ of 0.200 mol dm⁻³ copper(II) sulfate solution. The readings are plotted in Figure 5. Calculate ΔH for Zn(s) + CuSO₄(aq) → ZnSO₄(aq) + Cu(s).

initial temperature = 21.0 °C (steady before mixing)

extrapolated temperature at 3.5 min = 31.1 °Cfrom the cooling line

ΔT = 31.1 − 21.0 = 10.1 K

Q = mcΔT = 50.0 × 4.18 × 10.1 = 2110 J = 2.11 kJ

n(CuSO₄) = 0.200 × 0.0500 = 0.0100 molzinc is in excess, so CuSO₄ limits

ΔH = −2.11 ÷ 0.0100 = −211 kJ mol⁻¹

Had you used the highest reading, 30.5 °C, you would have got ΔT = 9.5 K and ΔH = −199 kJ mol⁻¹, further from the true value. The graph turned a systematic error into a smaller one. Notice also that the mass in Q is the 50.0 g of solution; the zinc is not included.

8Enthalpies of combustion, and why they come out too small

The enthalpy change of combustion is the enthalpy change when one mole of a substance burns completely in oxygen. Combustion is always exothermic. In school it is measured with a spirit burner heating water in a metal can, as Figure 6 shows. The burner is weighed before and after, so you know the mass of fuel burned.

Figure 6 · Measuring an enthalpy of combustion, and where the heat goes

Worked example 3: propan-1-ol. A spirit burner of propan-1-ol, C₃H₇OH (M = 60.11 g mol⁻¹), weighs 58.62 g before and 58.00 g after heating 150.0 g of water from 19.5 °C to 41.1 °C. Calculate the enthalpy change of combustion.

mass burned = 58.62 − 58.00 = 0.62 g

n = 0.62 ÷ 60.11 = 0.0103 mol

ΔT = 41.1 − 19.5 = 21.6 K

Q = 150.0 × 4.18 × 21.6 = 13 500 J = 13.5 kJ

ΔH = −13.5 ÷ 0.0103 = −1.3 × 10³ kJ mol⁻¹2 s.f., because 0.62 g has only 2

The data booklet value is −2021 kJ mol⁻¹, so this result is about 35% too small in size. That is typical, and the guide asks you to explain why calorimetry usually measures a smaller temperature change than theory predicts. Every clay arrow in Figure 6 is part of the answer:

Heat is lost to the surroundings: hot gases pass around the can instead of into it, and the can and water lose heat to the air.
Heat is absorbed by the apparatus: the can, the thermometer and the clamp are warmed too, and Q = mcΔT counts only the water.
Combustion is incomplete: in a limited oxygen supply some fuel forms carbon monoxide or soot (the black deposit on the can), releasing less energy than complete combustion.
Fuel evaporates from the wick, so some of the mass lost was never burned.
Conditions are not standard: the water in the products may leave as vapour, not liquid.

These are systematic errors: they push every repeat in the same direction, so repeating the experiment does not remove them. Improvements attack each one: a draught shield round the flame and can, a lid on the can, the can close above the flame, a cap on the burner when it is not lit, and, at the professional end, a sealed bomb calorimeter filled with oxygen. The same method measures the energy content of food: burn a known mass of a nut or crisp under a tube of water. Food is a mixture, so the result is given in kJ g⁻¹ rather than kJ mol⁻¹.

9Where marks are lost

Using the mass of the reactants in Q = mcΔT. m is the mass of the water or solution whose temperature changed. In worked example 2 it is 50.0 g of solution, not the zinc.

Dropping the sign of ΔH. A reaction that warmed its surroundings has a negative ΔH. "ΔH = 53.5 kJ mol⁻¹" for a neutralisation loses the final mark.

Forgetting to convert J to kJ. Q comes out in joules; ΔH is quoted in kJ mol⁻¹. An answer of −53 500 kJ mol⁻¹ is out by a factor of 1000.

Dividing by the wrong amount. n is the amount of the limiting reactant, or of the substance the question names. When one reactant is in excess, its amount is irrelevant.

Thinking the thermometer measures the system. In an endothermic reaction the thermometer falls because the surroundings lose energy to the chemicals.

Saying that heat and temperature are the same, or that temperature is energy. Temperature measures average kinetic energy; heat is energy transferred because of a temperature difference.

Drawing the energy profile against time. The guide specifies reaction coordinate on x and potential energy on y. Time is wrong.

Blaming "human error". In an evaluation, name the specific systematic error, such as heat lost to the air or incomplete combustion, and say which way it moves the result.

10Draw it right

Energy profile axes: x = reaction coordinate, y = potential energy. No numbers are needed on a sketch.
Reactants and products labelled on their levels, with formulas or names.
Products below reactants for exothermic, above for endothermic.
ΔH drawn as an arrow from the reactant level to the product level, pointing the right way, and labelled.
Eₐ drawn from the reactant level to the top of the hump, and labelled.
Temperature–time graphs: axes labelled with units, points plotted, a best-fit line through the cooling points extrapolated back to the time of mixing, and ΔT marked at that time.
On any apparatus sketch, label the insulation, the lid, the thermometer and what is being heated.

11Try it

Marks in brackets. Answers and marker's notes are at the end. Show your working in every calculation.

Q1. Distinguish between heat and temperature. 2 marks

Q2. Solid barium hydroxide and solid ammonium chloride are mixed in a flask. The flask becomes cold enough to freeze a film of water beneath it. Which is correct? 1 mark

A. The reaction is exothermic and ΔH is negative.

B. The reaction is exothermic and ΔH is positive.

C. The reaction is endothermic and ΔH is positive.

D. The reaction is endothermic and ΔH is negative.

Q3. Sketch an energy profile for an exothermic reaction. Label the axes, ΔH and the activation energy, and state which is more stable, the reactants or the products. 3 marks

Q4. 4.00 g of ammonium nitrate, NH₄NO₃, is dissolved in 50.0 cm³ of water. The temperature falls from 21.3 °C to 15.8 °C. Calculate the enthalpy change of solution of ammonium nitrate, in kJ mol⁻¹. Assume the solution has a mass of 50.0 g and the specific heat capacity of water. 3 marks

Q5. A student adds excess iron powder to 25.0 cm³ of 0.500 mol dm⁻³ copper(II) sulfate solution in a polystyrene cup and records the temperature every 30 seconds. The initial temperature is 20.0 °C. The highest reading is 35.2 °C. Extrapolating her cooling curve back to the time of mixing gives 37.1 °C.

(a) Explain why the student extrapolates the cooling curve instead of using the highest reading. 2 marks

(b) Calculate the enthalpy change for Fe(s) + CuSO₄(aq) → FeSO₄(aq) + Cu(s). 3 marks

(c) A data source gives −154 kJ mol⁻¹. Calculate the percentage error and state one reason the student's value is less exothermic. 2 marks

Q6. A student burns ethanol in a spirit burner under a copper can of water and obtains an enthalpy change of combustion of −890 kJ mol⁻¹. The data booklet value is −1367 kJ mol⁻¹. Explain two reasons for the difference and suggest an improvement for each. 4 marks

12In one breath

Temperature measures the average kinetic energy of particles; heat is energy transferred because of a temperature difference, and it depends on mass. The system is the reacting chemicals and the surroundings are everything else, and total energy is conserved between them. Exothermic reactions release energy, warm the surroundings and have a negative ΔH; endothermic reactions absorb energy, cool the surroundings and have a positive ΔH. An energy profile plots potential energy against reaction coordinate: products lower than reactants means more stable products and an exothermic reaction, ΔH is the arrow from reactants to products and Eₐ is the hump; N₂ burns endothermically because its triple bond makes it so stable. ΔH⦵ is the heat transferred at constant pressure under standard conditions and states, in kJ mol⁻¹. Calorimetry uses Q = mcΔT, with m the mass of water or solution and c = 4.18 J g⁻¹ K⁻¹, then ΔH = −Q/n with n for the limiting reactant, converting J to kJ. Heat loss makes measured values too small; extrapolating the cooling curve back to the time of mixing corrects part of it, and combustion results come out low because of heat lost to the air and apparatus, incomplete combustion and evaporation of fuel.

Answers

Q1. Temperature is a measure of the average kinetic energy of the particles in a substance; it does not depend on the amount of substance. Heat is energy transferred from a hotter to a colder body because of a difference in temperature; it is measured in joules and depends on the mass as well as the temperature change. 1 for temperature as average kinetic energy of particles, 1 for heat as energy transferred due to a temperature difference. "Heat is how hot something is" scores 0.

Q2. C. The flask gets cold, so energy has moved from the surroundings into the reacting system: the reaction is endothermic, and endothermic reactions have a positive ΔH. C only. B and D mix up the sign convention.

Q3. Axes: potential energy (y) against reaction coordinate (x). Reactant level higher than product level, with a hump between. ΔH: a downward arrow from reactants to products, labelled. Eₐ: an arrow from the reactant level up to the top of the hump, labelled. The products are more stable, because they have lower potential energy. 1 for both axes labelled correctly with products below reactants, 1 for ΔH and Eₐ both correctly drawn and labelled, 1 for products more stable because lower in energy. A graph with time on the x-axis loses the first mark; Eₐ drawn from the products loses the second.

Q4.

ΔT = 15.8 − 21.3 = −5.5 K

Q = mcΔT = 50.0 × 4.18 × (−5.5) = −1150 J = −1.15 kJ

n(NH₄NO₃) = 4.00 ÷ 80.06 = 0.0500 mol

ΔH = −Q ÷ n = −(−1.15) ÷ 0.0500 = +23 kJ mol⁻¹

M1 for Q = 50.0 × 4.18 × 5.5 = 1150 J, M1 for n = 0.0500 mol, A1 for +23 kJ mol⁻¹ with the positive sign. Using the mass of ammonium nitrate, 4.00 g, as m scores 0 for M1. A value of −23 loses the A1. Error carried forward applies to the final mark.

Q5. (a) Heat is lost to the surroundings while the reaction is still happening, so the highest reading is lower than the temperature the solution would have reached with no heat loss. Extending the cooling line back to the time of mixing estimates that temperature, which compensates for the heat lost. (b) and (c):

ΔT = 37.1 − 20.0 = 17.1 K

Q = 25.0 × 4.18 × 17.1 = 1790 J = 1.79 kJ

n(CuSO₄) = 0.500 × 0.0250 = 0.0125 moliron is in excess

ΔH = −1.79 ÷ 0.0125 = −143 kJ mol⁻¹

percentage error = (154 − 143) ÷ 154 × 100 = 7.1%

Reason: heat is still lost to the surroundings, or absorbed by the cup and thermometer, and the extrapolation corrects only part of it; or the solution's specific heat capacity and density are not exactly those of water. (a) 1 for heat lost during the reaction making the maximum reading too low, 1 for the extrapolated value estimating the no-loss temperature at the moment of mixing. (b) M1 for Q, M1 for n, A1 for −143 kJ mol⁻¹ with the sign. (c) 1 for 7.1% (7% accepted), 1 for a specific reason. Using the highest reading, 35.2 °C, in (b) gives −127 kJ mol⁻¹ and loses only the A1.

Q6. Any two, each with an improvement. Heat is lost to the surroundings, as hot gases pass around the can and the can loses heat to the air: use a draught shield and a lid, and put the can closer to the flame. Heat is absorbed by the can and thermometer, which Q = mcΔT ignores: include the can's heat capacity in the calculation, or use a thinner can. Combustion is incomplete, with soot seen on the can: improve the air supply to the flame. Ethanol evaporates from the wick, so less was burned than was weighed: cap the burner and weigh it as soon as the flame is out. 1 for each of two reasons with a mechanism, 1 for each matching improvement. "Human error", "the thermometer was inaccurate" or "repeat the experiment" score 0, because repeating does not remove a systematic error.

Educerie · written from the published IB Diploma Programme Chemistry guide, first assessment 2025, section R1.1 Measuring enthalpy changes. Original text, examples and questions. Diagrams drawn by Educerie. Last reviewed 25 September 2026.

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