1.1.1 · The system and the surroundings SL core
The syllabus asks for one thing here: “Understand the difference between heat and temperature.” That sounds like a definition question. It is almost never assessed as one.
The system is the reaction — the chemicals you are interested in. The surroundings are everything else: the water, the beaker, the air, the thermometer.
Heat is energy in transit between the two. It is measured in joules, and it depends on how much substance there is. Temperature is a measure of the average kinetic energy of the particles. It does not depend on how much there is.
A bath and a cup of tea can be at the same temperature; the bath holds far more heat. This is why the mass appears in every calculation in this topic and the temperature appears only as a change.
You never measure the system. A thermometer in a beaker measures the surroundings. The whole of this topic is the inference from what the surroundings did to what the system did, and the inference runs backwards: if the surroundings gained energy, the system lost it.
This is the most persistent single observation in the evidence behind this document, and it is about practical work rather than calculation. Published examiner feedback records, of one session, A worrying number of candidates focus on the temperature of the surroundings ignoring that of the system — and then repeats the same sentence, almost word for word, in the following session. Two years later the wording has shifted but the fault has not: Temperature was often not controlled or poorly done. Reports usually focus on the room temperature ignoring the system's temperature.
The point is not that room temperature does not matter. It is that recording the room temperature is not the same as knowing what happened inside the flask, and an investigation that reports only the former has not measured its own reaction.
Examiner feedback from the most recent session in this evidence base is unusually specific: Few students correctly identified the direction of systematic error. A common misconception was reference to temperature errors in the surroundings rather than those of the system itself. Naming an error earns little. Saying which way it pushes the answer is what is being asked for, and section 1.1.4 below sets out the one direction you are expected to be able to argue.
1.1.2 · Endothermic and exothermic SL core
Two words, one convention, and a sign that carries a mark on its own.
| Reaction | Energy transfer | The surroundings | ΔH |
|---|---|---|---|
| Exothermic | system → surroundings | warm up | negative |
| Endothermic | surroundings → system | cool down | positive |
Exo = out of the system. The thermometer goes up because the energy left the reaction and went into the water.
A published report on a multiple-choice item: Mediocre performance in a question which did not discriminate well. Many more candidates chose answer B, rather than the correct answer D. The question states that the dissolving process increases the temperature of the solution; it is an exothermic, thus eliminating answer B.
On another item in the same session, feedback records that Just over 40% chose the correct response for this endothermic reaction as B, however a similar number of students incorrectly believed the reaction to be exothermic and chose D. As many students chose the exact reverse as chose the right answer.
Both are the same fault: the temperature of the surroundings was read as though it were the energy of the system. Say the sentence out loud — “the water got hotter, so the reaction gave energy out” — before choosing.
Examiner feedback on a structured calculation: Many however did not remember that, as it is endothermic, ΔH must be positive. The arithmetic in those answers was already right. The sign was the whole of what was lost. Feedback from an earlier session names the same thing alongside the substitution errors: candidates used incorrect masses or temperatures for the mcT formula or forgot the sign for enthalpy change.
Examiners defended one item explicitly: There were some concerns expressed about this question since students identified the endothermic reaction by eliminating the known exothermic reactions. The endothermic reaction given is not one that students are expected to know. Students should be reading all answers given and selecting the best option. Knowing the others are exothermic allowed them to identify the endothermic process. You are not expected to recognise every reaction. You are expected to recognise combustion, neutralisation and most oxidations as exothermic — and then to take what is left.
1.1.3 · Energy profiles SL core
The guide is specific about the axes: Axes for energy profiles should be labelled as reaction coordinate x, potential energy y. Label them that way even when the sketch is otherwise perfect.
A multiple-choice item asked for the interpretation of a profile with the products drawn above the reactants. Each option paired a relative stability with a temperature of the surroundings. Published feedback gives the whole distribution:
70 % of students selected the option that correctly described the energy profile of an endothermic reaction, which was A. 13 % chose B, which gave the correct relative stability of reactants and products, but incorrect change in temperature of the surroundings. 11 % made the opposite mistake and chose option C as a result.
Thirteen per cent failed on one half and eleven per cent on the other, in almost equal numbers. That is the shape of a genuinely two-part question: answer both parts deliberately, and never let one follow from the other by feel. Feedback on a related item makes the same point from the other side — Weaker students correctly identified the temperature of surroundings decreases but failed to realize that reactants are more stable than products.
A structured question asked candidates to Sketch an energy profile for this reaction and label the “Reactants”, “Products” and “ΔH” for two marks. The published scheme awards them like this:
| M1 | exothermic reaction profile |
| M2 | ΔH correct AND labelled reactants and products |
The second mark is a conjunction. A correct ΔH arrow with unlabelled levels earns nothing for M2, and so do labelled levels with no arrow. Draw all three.
Examiner feedback: candidates did not show the energy levels of reactants and products clearly, or lacked arrows to indicate directionality — suggesting unfamiliarity with conventions for reaction coordinate diagrams. On another item, some did not note the equation was shown as exothermic and drew the profile the wrong way up. Read the equation for its sign before you draw anything.
A profile shows the products below the reactants. (a) Is the reaction endothermic or exothermic? (b) Which is the more stable? (c) Does the temperature of the surroundings rise or fall? (d) What is the sign of ΔH? Answers at the end.
Measuring enthalpy changes — the energy profile
Set the activation energy and the enthalpy change. Everything else on the diagram, including the activation energy of the reverse reaction, follows from those two.
A catalyst moves the peak down and nothing else. Reactants and products keep the same energies, so ΔH is untouched — and both activation energies fall by the same amount.
Enthalpy against reaction coordinate. The axes are labelled the way the guide asks for them: reaction coordinate on x, potential energy on y.
1.1.4 · Standard enthalpy change, and how it is measured SL core
The standard enthalpy change ΔH is the heat transferred at constant pressure, under standard conditions and states, per mole of reaction. Its units are kJ mol–1. Everything below is the route from a thermometer reading to that number.
Both equations, and the value of c for water, are given in the data booklet. What is not given is which mass, which temperature change and which n to use — and that is where the marks are.
A multiple-choice question: burning 0.321 g of methanol raises the temperature of 200 cm3 of water by 5.00 K. The question supplies c = 4.18 J g–1 K–1, Mr(methanol) = 32.05 g mol–1 and the equation Q = mcΔT. The four printed options were –417, –579, –23 200 and –417 000 kJ mol–1.
| Step 1 — which mass? | The water, because the water is what changed temperature. 200 cm3 of water has a mass of 200 g. Not 0.321 g. |
| Step 2 — Q | Q = 200 × 4.18 × 5.00 = 4180 J |
| Step 3 — kJ | 4180 J = 4.18 kJ |
| Step 4 — n | n(methanol) = 0.321 / 32.05 = 0.01002 mol |
| Step 5 — sign | The water warmed, so the reaction is exothermic and ΔH is negative. |
Where the distractors come from. –417 000 is the identical calculation with step 3 skipped — joules never turned into kilojoules. That single omission is a factor of one thousand, and it is printed on the paper as an option because enough candidates make it.
A multiple-choice question: 0.050 mol of sodium hydrogencarbonate is added to 50 cm3 of 2.0 mol dm–3 citric acid.
The temperature falls from 298 K to 297 K. The printed options were +4.18, –4.18, +12.5 and –12.5 kJ mol–1.
| Sign | The temperature fell. The reaction absorbed energy from the solution: endothermic, ΔH positive. |
| Q | 50 × 4.18 × 1 = 209 J = 0.209 kJ |
| Which n? | n(acid) = 0.0500 × 2.0 = 0.100 mol; n(NaHCO3) = 0.050 mol. The equation needs three NaHCO3 per acid, so the hydrogencarbonate runs out first and the amount of reaction is 0.050 / 3 = 0.0167 mol. |
Both traps are printed as options. Dividing by 0.050 mol instead of 0.0167 gives +4.18 — the sign right, the ratio wrong, and the answer exactly three times too small. Getting the sign wrong instead gives –12.5. The two errors are independent, which is how all four options become reachable.
A structured, data-based question. A 1.28 g metal sample is heated in boiling water to 100.00 °C, then transferred into 5.65 g of water at 25.00 °C. The final temperature of both is 26.77 °C. The question states the assumption explicitly: the heat absorbed by the water in beaker B in (a) is equal to the heat released by the metal.
| Energy gained by the water | 5.65 × 4.18 × (26.77 – 25.00) = 41.8 J |
| Temperature change of the metal | 100.00 – 26.77 = 73.23 K — the metal falls all the way to the final temperature, not to 25.00 °C. |
| Specific heat capacity | 41.8 / (1.28 × 73.23) = 0.446 J g–1 K–1 |
Published feedback on this item names the errors precisely: many candidates used mass and/or the temperature change of the water for beaker A and did not score M1, and Some candidates did not apply math correctly and got a negative answer and scored 1 mark only. Using the water's temperature change (1.77 K) for the metal gives 18.5 J g–1 K–1 — more than four times the value for water itself, and a number no metal has. A result that absurd is telling you the substitution is wrong.
Calorimetry questions routinely ask for a source of error and an improvement. Published feedback from the most recent session in this evidence base is worth reading twice:
Few students working with calorimetry recognized or explicitly stated that assuming all heat is transferred solely to or from the solution introduces systematic errors. This simplification ignoring heat losses to the surroundings or the heat absorbed by the calorimeter commonly leads to an underestimation of ΔH.
Learn the direction, because it is the part that is missing. Heat that leaks away never reaches the water, so the measured temperature rise is too small, so the measured ΔH comes out smaller in magnitude than the true value — a less negative number for an exothermic reaction. The error is systematic, not random: it always pushes the same way, which is exactly what makes it a systematic error rather than a random one.
And an improvement has to be something the apparatus does not already have. Feedback records that the accepted answers were insulating reaction apparatus or put a lid on the beaker, while Some incorrectly suggested use of Styrofoam cup or performance of the combustion inside the calorimeter — describing equipment that was already in use.
From published feedback: Many students gained the mark by proposing ways in which the system could be better thermally insulated. Quite a number incorrectly suggested using a water bath which, as temperature rise is being measured, achieves exactly the opposite of what is required. A bath holds the temperature constant. In an experiment whose entire measurement is a temperature change, that destroys the reading. Insulate; do not thermostat.
(1) What changed temperature? That is your m and your ΔT. (2) What is one mole of? That is your n, and it is one mole of the reaction as written, not one mole of whatever was weighed out. If those two answers are right, the rest is arithmetic and a sign.
Putting it together
Four outcomes, one chain of reasoning. The thermometer measures the surroundings (1.1.1). Which way it moved tells you whether the system released or absorbed energy, and therefore the sign of ΔH (1.1.2). The same information drawn as a picture is an energy profile, where the vertical positions give relative stability and the arrow gives the sign (1.1.3). How far it moved, with a mass and an amount, gives the number (1.1.4).
| Outcome | What is assessed | The error published feedback keeps recording |
|---|---|---|
| 1.1.1 | System against surroundings; heat against temperature | Monitoring the room instead of the reaction, in practical work |
| 1.1.2 | Direction of transfer, and the sign of ΔH | Reading the thermometer backwards; omitting the sign entirely |
| 1.1.3 | Sketching and interpreting profiles | Getting one of the two readings right and the other wrong; unlabelled axes and undirected arrows |
| 1.1.4 | Q = mcΔT and ΔH = –Q/n | The wrong mass, the wrong temperature change, the wrong n, or no unit conversion |
- A reaction in a beaker makes the beaker feel cold. Is it endothermic or exothermic, and what is the sign of ΔH?
- Why does the mass of water appear in Q = mcΔT, but never the mass of the fuel?
- 25.0 cm3 of solution rises by 4.00 K during a reaction of 0.0100 mol. Take c = 4.18 J g–1 K–1. What is ΔH in kJ mol–1?
- A sketch shows correctly placed reactant and product levels and a correct ΔH arrow, but no labels on the levels. On the two-mark scheme quoted above, how many marks?
- Heat escapes from an open calorimeter. Is the measured ΔH of an exothermic reaction too negative or not negative enough?
- A reaction consumes 0.060 mol of a reagent that appears with a coefficient of 3. How many moles of reaction is that?
- Which two axes labels does the guide require on an energy profile?
Profiles quick check: (a) exothermic; (b) the products; (c) rise;
(d) negative.
1. The beaker is cold because the reaction took energy from it — endothermic, ΔH positive.
2. Because the water is what changed temperature; Q = mcΔT measures the energy the water gained.
The fuel's mass is used only to find n.
3. Q = 25.0 × 4.18 × 4.00 = 418 J = 0.418 kJ; ΔH = –0.418 / 0.0100 =
–41.8 kJ mol–1.
4. One. M1 is earned for the correct profile; M2 requires the arrow and the labels together.
5. Not negative enough — too small in magnitude.
6. 0.060 / 3 = 0.020 mol of reaction.
7. Reaction coordinate on x, potential energy on y.