R1.1 Measuring enthalpy changes | IB DP Chemistry

Home › Learning Hub › IB DP Chemistry › R1.1 Measuring enthalpy changes

🎯What you need to be able to do

Understand the difference between heat and temperature, and describe energy transfer between the system and the surroundings.
Understand the temperature change that accompanies an endothermic and an exothermic reaction.
Sketch and interpret energy profiles for endothermic and exothermic reactions, with correctly labelled axes.
Explain how the relative stability of reactants and products determines whether a reaction is endothermic or exothermic.
Apply \( Q = mc\Delta T \) and the relationship between \( Q \) and \( \Delta H \) to calculate the enthalpy change of a reaction from calorimetric data.
Explain why calorimetry experiments typically give a smaller temperature change than theory predicts.

📚The chemistry

System, surroundings, heat and temperature

The system is the reaction under study; the surroundings are everything else — in practice the solvent, the container and the air. Chemical reactions transfer energy between the two, while the total energy is conserved.

Heat and temperature are not the same thing. Heat is the energy transferred because of a temperature difference, measured in joules; it depends on how much substance there is. Temperature is a measure of the average kinetic energy of the particles, measured in kelvin; it does not depend on quantity. A bathful of warm water contains far more energy than a spoonful of boiling water, but it is at the lower temperature.

Exothermic and endothermic

Exothermic
energy transferred from the system to the surroundings
the surroundings get warmer
\( \Delta H \) is negative
products more stable (lower in energy) than reactants

Endothermic
energy transferred from the surroundings to the system
the surroundings get cooler
\( \Delta H \) is positive
products less stable (higher in energy) than reactants

Note what is actually observed in the laboratory: you measure the temperature of the surroundings — the solution in the cup. An exothermic reaction releases energy into that solution, so the thermometer reading rises. Students often write that an exothermic reaction “gets hotter”, which inverts the logic; it is the surroundings that get hotter, because the system has given energy away.

Energy profiles

An energy profile plots potential energy on the y-axis against reaction coordinate on the x-axis — the guide is explicit about the labels, and marks are given for them.

Both profiles rise to a peak: the energy barrier that must be climbed, whose height is the activation energy \( E_{\mathrm{a}} \) (R2.2).
In an exothermic profile the products end below the reactants, and \( \Delta H \) is drawn as a downward arrow between the two levels.
In an endothermic profile the products end above the reactants, and \( \Delta H \) points upward.

Two things follow that questions like to test. First, \( \Delta H \) is the difference between reactants and products only — the height of the peak has no effect on it. Second, an endothermic reaction always has \( E_{\mathrm{a}} \) at least as large as \( \Delta H \), because the system must climb at least to the product level.

Standard enthalpy change

The standard enthalpy change \( \Delta H^{\ominus} \) is the heat transferred at constant pressure under standard conditions and states: 100 kPa, substances in their standard states, and a stated temperature (usually 298 K). Its units are kJ mol⁻¹ — per mole, which is why the value must always be divided by an amount at the end of a calorimetry calculation.

Calorimetry

Two equations do all the work, and both are in the data booklet:

\[ Q = mc\Delta T \qquad\qquad \Delta H = -\frac{Q}{n} \]

In \( Q = mc\Delta T \):

\( m \) is the mass of the solution absorbing the heat, in grams — not the mass of the reactant. For dilute aqueous solutions take the density as 1.00 g cm⁻³, so 50.0 cm³ has a mass of 50.0 g. If two solutions are mixed, add both volumes.
\( c \) is the specific heat capacity of water, 4.18 J g⁻¹ K⁻¹ (data booklet).
\( \Delta T \) is the temperature change. Because a kelvin and a degree Celsius are the same size, a change in temperature has the same numerical value in either — this is the one place in the course where you may safely leave Celsius alone.
\( Q \) comes out in joules, so divide by 1000 before quoting \( \Delta H \) in kJ mol⁻¹.

Then \( n \) is the amount of the limiting reactant, and the minus sign converts the perspective: heat gained by the surroundings is heat lost by the system. If the temperature rose, \( Q \) is positive and \( \Delta H \) must come out negative.

Using the mass of the reactant as \( m \). The energy is absorbed by the solution, not by the solid you dropped into it, and \( c \) is the specific heat capacity of water, so \( m \) must be the mass of water present. Dissolve 2.5 g of a salt in 100 cm³ of water and \( m \) is 100 g (or 102.5 g if the question tells you to include the solute); it is never 2.5 g. The companion error is dividing by the amount of the excess reactant instead of the limiting one.

Why the experiment always under-reads

Measured enthalpy changes from a school calorimeter are almost always less exothermic than the accepted value, and the syllabus asks you to explain why. The reasons are worth learning as a set, because Paper 1B asks for them repeatedly:

Heat loss to the surroundings — through the walls of the container, and by evaporation from an open top. This is much the largest effect for combustion in an open calorimeter.
Heat absorbed by the apparatus — the thermometer, the stirrer and the container itself warm up, and that energy is not counted in \( mc\Delta T \).
Incomplete reaction or incomplete combustion — less energy is released than the equation implies; the soot on the bottom of a beaker is the visible evidence.
The assumptions that the solution has the density and specific heat capacity of pure water are only approximate.

These are systematic errors, all acting in the same direction, so repeating the experiment does not remove them. Improvements: insulate and lid the calorimeter, use a draught shield, minimise the distance between flame and vessel, and — the best answer — take temperature readings at regular intervals before and after mixing and extrapolate the cooling curve back to the moment of mixing to recover the temperature change that would have occurred with no heat loss.

✏️Worked example

50.0 cm³ of 1.00 mol dm⁻³ hydrochloric acid at 21.0 °C is added to 50.0 cm³ of 1.00 mol dm⁻³ sodium hydroxide at 21.0 °C in a polystyrene cup. The maximum temperature reached is 27.8 °C.
(a) Calculate the enthalpy change of neutralization, in kJ mol⁻¹.
(b) The accepted value is −57.3 kJ mol⁻¹. Calculate the percentage error and suggest two reasons for the discrepancy.
(c) Sketch the energy profile for this reaction, labelling the axes, \( \Delta H \) and \( E_{\mathrm{a}} \).
Take the density of the solutions as 1.00 g cm⁻³ and \( c \) as 4.18 J g⁻¹ K⁻¹.

(a) Find \( Q \) first, then the amount, then divide.

The mass absorbing the heat is the total solution: \( 50.0 + 50.0 = 100.0\ \mathrm{cm^3} \), so \( m = 100.0\ \mathrm{g} \). \( \Delta T = 27.8 - 21.0 = 6.8\ \mathrm{K} \).

\[ Q = mc\Delta T = (100.0)(4.18)(6.8) = 2842\ \mathrm{J} = 2.842\ \mathrm{kJ} \]

Amount of each reactant: \( n = cV = 1.00 \times 0.0500 = 0.0500\ \mathrm{mol} \). They react in a 1 : 1 ratio and there are equal amounts, so neither is in excess and \( n = 0.0500\ \mathrm{mol} \) of water is formed.

\[ \Delta H = -\frac{Q}{n} = -\frac{2.842}{0.0500} = -56.8\ \mathrm{kJ\,mol^{-1}} \]

The temperature rose, so the reaction is exothermic and the negative sign is correct.

(b)

\[ \text{percentage error} = \left|\frac{-56.8 - (-57.3)}{-57.3}\right| \times 100 = \frac{0.5}{57.3} \times 100 = 0.87\% \]

Reasons (any two): heat lost to the surroundings through the walls of the cup and the open top, so the measured temperature rise is smaller than the true one; heat absorbed by the thermometer, stirrer and cup, which is not included in \( mc\Delta T \); the assumption that the solution has the same density and specific heat capacity as pure water; and the practical difficulty of reading the true maximum temperature, since the mixture begins cooling as soon as it starts warming.

(c) Axes: potential energy vertically, reaction coordinate horizontally. Draw the reactants at a level on the left, a curve rising to a peak, and the products at a lower level on the right. Mark \( E_{\mathrm{a}} \) as the vertical distance from the reactant level up to the peak, and \( \Delta H \) as a downward arrow from the reactant level to the product level, labelled negative.

Check it. Three checks, all quick. Sign: the temperature rose, so \( \Delta H \) must be negative — if your answer is positive, you have dropped the minus sign in \( \Delta H = -Q/n \). Magnitude: the enthalpy of neutralization of a strong acid with a strong base is always close to −57 kJ mol⁻¹, because the reaction occurring is the same in every case, \( \mathrm{H^{+}}(aq) + \mathrm{OH^{-}}(aq) \rightarrow \mathrm{H_2O}(l) \). An answer of −2.8 or −5680 means the division by \( n \) or by 1000 went wrong. Direction of error: heat loss makes the measured value less exothermic, so a correct experiment should give a value slightly above −57.3 — and −56.8 is, which is reassuring.

Dividing by the total amount of both reactants. Here the acid and the base each supply 0.0500 mol, and it is tempting to write \( n = 0.100 \). It is not: they react 1 : 1, so 0.0500 mol of acid reacts with 0.0500 mol of base to produce 0.0500 mol of water, and the enthalpy change is quoted per mole of water formed. Halving \( n \) doubles \( \Delta H \) to −114 kJ mol⁻¹, an answer twice the accepted value — which the magnitude check above catches immediately, if you do it.

📝Practise

Work through these on paper, then reveal the answer.

1. Distinguish between heat and temperature, and explain why a reaction described as exothermic causes the thermometer in the reaction mixture to rise.

Heat is energy transferred between a system and its surroundings because of a temperature difference; it is measured in joules and depends on the quantity of substance present. Temperature is a measure of the average kinetic energy of the particles in a sample; it is measured in kelvin and is independent of quantity. A litre of water at 30 °C contains far more energy than a millilitre at 80 °C, but is at the lower temperature. In an exothermic reaction, energy is transferred from the system to the surroundings. The thermometer is measuring the surroundings — the solution in which the reaction is taking place — so as the system releases energy into it, the average kinetic energy of the solution particles increases and the reading rises. The system itself is losing energy, which is why \( \Delta H \) is negative even though the thermometer goes up.

2. 2.50 g of ammonium nitrate is dissolved in 100.0 cm³ of water. The temperature falls from 20.4 °C to 18.5 °C. Calculate the enthalpy change of solution, in kJ mol⁻¹ (\( M(\mathrm{NH_4NO_3}) = 80.06 \)).

Heat absorbed from the water: \( m = 100.0\ \mathrm{g} \) (the water is the surroundings), \( \Delta T = 20.4 - 18.5 = 1.9\ \mathrm{K} \). \( Q = mc\Delta T = (100.0)(4.18)(1.9) = 794\ \mathrm{J} = 0.794\ \mathrm{kJ} \). Amount of ammonium nitrate: \( n = \dfrac{2.50}{80.06} = 0.0312\ \mathrm{mol} \). \( \Delta H = -\dfrac{Q}{n} \), but here the temperature fell, meaning the surroundings lost energy to the system, so \( Q \) as calculated is heat taken from the water and \( \Delta H \) is positive: \( \Delta H = +\dfrac{0.794}{0.0312} = \mathbf{+25.4\ kJ\,mol^{-1}} \). The process is endothermic, which is exactly why ammonium nitrate is used in instant cold packs. The safest way to get the sign right is not to memorise where the minus goes, but to ask whether the surroundings warmed (exothermic, negative) or cooled (endothermic, positive).

3. Sketch, on the same axes, energy profiles for an exothermic and an endothermic reaction. Label the axes, \( \Delta H \) and \( E_{\mathrm{a}} \), and state whether a large activation energy makes a reaction more exothermic.

Axes: potential energy on the y-axis, reaction coordinate on the x-axis — not "time" and not "energy" alone; the guide specifies these labels. Exothermic: reactants on the left at some level, the curve rising to a peak, products ending below the reactant level. \( E_{\mathrm{a}} \) is the vertical distance from the reactants up to the peak; \( \Delta H \) is a downward arrow from reactant level to product level, and is negative. Endothermic: the same shape, but the products end above the reactant level, and \( \Delta H \) is an upward arrow and positive. No — a large activation energy does not make a reaction more exothermic. \( \Delta H \) depends only on the difference in energy between reactants and products; the height of the barrier between them affects only the rate. A reaction can be strongly exothermic and immeasurably slow, which is exactly the case for the combustion of a hydrocarbon at room temperature.

4. In a combustion calorimetry experiment, 0.740 g of ethanol (\( M = 46.08 \)) is burned and heats 200.0 cm³ of water from 19.0 °C to 39.6 °C. Calculate the enthalpy of combustion, and explain why it is much less exothermic than the accepted −1367 kJ mol⁻¹.

\( m = 200.0\ \mathrm{g} \) (the water absorbing the heat), \( \Delta T = 39.6 - 19.0 = 20.6\ \mathrm{K} \). \( Q = (200.0)(4.18)(20.6) = 17\,222\ \mathrm{J} = 17.22\ \mathrm{kJ} \). \( n(\mathrm{ethanol}) = \dfrac{0.740}{46.08} = 0.01606\ \mathrm{mol} \). \( \Delta H_{\mathrm{c}} = -\dfrac{17.22}{0.01606} = \mathbf{-1072\ kJ\,mol^{-1}} \) — about 22% less exothermic than the accepted value. Reasons, all systematic and all in the same direction: (i) heat loss to the surrounding air, which is very large in an open spirit-burner set-up, and the largest single contributor; (ii) heat absorbed by the apparatus — the beaker, the thermometer, the stand — which is not included in \( mc\Delta T \); (iii) incomplete combustion, producing carbon monoxide and soot rather than carbon dioxide and so releasing less energy per mole, visible as blackening on the base of the beaker; and (iv) evaporation of ethanol from the wick without burning. Improvements: use a draught shield and a lid, reduce the distance from flame to vessel, use a copper calorimeter with a known heat capacity, or burn the sample in a sealed bomb calorimeter in excess oxygen.

5. A student mixes 25.0 cm³ of 2.00 mol dm⁻³ HCl with 25.0 cm³ of 1.00 mol dm⁻³ NaOH and records a temperature rise of 6.7 °C. Calculate \( \Delta H \) of neutralization and identify the error the student would make by dividing by the amount of acid.

\( m = 25.0 + 25.0 = 50.0\ \mathrm{g} \); \( Q = (50.0)(4.18)(6.7) = 1400\ \mathrm{J} = 1.400\ \mathrm{kJ} \). Amounts: \( n(\mathrm{HCl}) = 2.00 \times 0.0250 = 0.0500\ \mathrm{mol} \); \( n(\mathrm{NaOH}) = 1.00 \times 0.0250 = 0.0250\ \mathrm{mol} \). They react 1 : 1, so the sodium hydroxide is limiting and only 0.0250 mol of water is formed; the acid is in twofold excess and half of it does not react. \( \Delta H = -\dfrac{1.400}{0.0250} = \mathbf{-56.0\ kJ\,mol^{-1}} \), close to the accepted −57.3. Dividing by the amount of acid instead would give \( -\dfrac{1.400}{0.0500} = -28.0\ \mathrm{kJ\,mol^{-1}} \), half the correct value. The general rule: always divide by the amount of the limiting reactant, because only that amount actually reacted and released energy.

6. Explain why the enthalpy change of neutralization of a strong acid with a strong base is essentially the same (about −57 kJ mol⁻¹) whichever acid and base are used, and predict qualitatively what happens if a weak acid such as ethanoic acid is used instead.

A strong acid and a strong base are both essentially fully dissociated in aqueous solution, so the solutions already contain free \( \mathrm{H^{+}}(aq) \) and \( \mathrm{OH^{-}}(aq) \) ions together with spectator ions that take no part. Whatever the acid and base, the only reaction actually occurring is the same one: \( \mathrm{H^{+}}(aq) + \mathrm{OH^{-}}(aq) \rightarrow \mathrm{H_2O}(l) \). The same bonds form in every case, so the same energy is released per mole of water, and the value is essentially constant. With a weak acid such as ethanoic acid, the enthalpy of neutralization is less exothermic — typically around −55 kJ mol⁻¹ or less. A weak acid is only partially dissociated, so before neutralization can proceed the remaining molecules must ionise, and that ionisation is an endothermic step which absorbs some of the energy released by the neutralization itself. The measured net value is therefore smaller in magnitude.

🔗Go deeper — other people’s work

These are external resources, not mine. If one stops working, tell me and everything above it on this page still stands.

RSC Learn Chemistry — the standard enthalpy of neutralization and enthalpy of combustion practicals, both with a full error analysis, and the graphical extrapolation method for correcting heat loss.
PhET — Energy Forms and Changes, useful for making the system/surroundings distinction concrete before you meet the sign convention.
Your data booklet — \( Q = mc\Delta T \), the value of \( c \) for water, and the standard conditions. All three are given, so none should be memorised.