Measuring Enthalpy Changes by Calorimetry

AS · 14 min

Enthalpy changes are measured by letting a reaction heat or cool a known mass of water, measuring the temperature change, and working out the energy transferred. This note covers the two key relationships, $q = mc\Delta T$ and $\Delta H = -mc\Delta T/n$ , the standard experiments (reactions in solution in a polystyrene cup and combustion of a fuel), the temperature–time graph method for correcting heat loss, and the evaluation of errors and uncertainties. Calorimetry calculations are standard in Paper 2, and the experiments are among the most common in Paper 3.

The two equations

When a reaction happens in water (or heats water), the energy released or absorbed changes the temperature of the water. The energy needed to change the temperature of a substance depends on its mass and on its specific heat capacity, $c$ : the energy needed to raise the temperature of 1 g of the substance by 1 K.

Key result

Energy transferred to or from the water (or solution):

q = mc\Delta T

$q$ : heat energy transferred, in J
$m$ : mass of the water or solution being heated, in g
$c$ : specific heat capacity, $4.18\ \text{J g}^{-1}\ \text{K}^{-1}$ for water (data booklet)
$\Delta T$ : temperature change, in K (a change of 1 K is the same as a change of $1\ ^\circ\text{C}$ )

Enthalpy change per mole:

\Delta H = -\frac{mc\Delta T}{n}

$n$ : amount, in mol, of the substance the $\Delta H$ refers to (usually the limiting reactant, or water for neutralisation)
the minus sign: if the temperature rises ( $\Delta T$ positive), the reaction is exothermic and $\Delta H$ is negative

$q$ comes out in joules; $\Delta H$ is quoted in $\text{kJ mol}^{-1}$ , so divide by 1000.

For reactions in aqueous solution, assume:

the density of the solution is $1.00\ \text{g cm}^{-3}$ , so the mass of solution in grams equals its volume in $\text{cm}^3$ ;
the specific heat capacity of the solution equals that of water, $4.18\ \text{J g}^{-1}\ \text{K}^{-1}$ ;
all the energy goes into (or comes from) the solution: none is lost to the surroundings or absorbed by the container.

Watch out

The mass $m$ is the mass of the water or solution that changes temperature, not the mass of the reactant. When $25.0\ \text{cm}^3$ of acid is mixed with $25.0\ \text{cm}^3$ of alkali, $m = 50.0\ \text{g}$ . When a small mass of solid is added to a solution, it is normally ignored: use the mass of the solution only.

Method

Calculating ΔH from experimental results

Find $\Delta T$ (final minus initial temperature, or from a corrected graph).
Calculate $q = mc\Delta T$ using the total mass of solution or water heated.
Calculate $n$ for the substance named in the $\Delta H$ : usually the limiting reactant (the one not in excess), or the moles of water formed for neutralisation, or the moles of fuel burnt for combustion.
Calculate $\Delta H = -\dfrac{q}{n}$ , convert to $\text{kJ mol}^{-1}$ , and give the sign: negative if the temperature rose, positive if it fell.
Give the answer to 3 significant figures (or as justified by the data).

Experiments in solution

A simple calorimeter for reactions in solution is an expanded polystyrene cup with a lid, standing in a beaker for stability. Polystyrene is a good thermal insulator and has a very low heat capacity, so little energy is lost through the sides or used to warm the cup.

Typical reactions:

neutralisation: an acid added to an alkali;
displacement: excess zinc powder added to copper(II) sulfate solution, $\ce{Zn(s) + CuSO4(aq) -> ZnSO4(aq) + Cu(s)}$ ;
dissolving a solid: for example ammonium nitrate (endothermic) or sodium hydroxide (exothermic);
metal with acid: magnesium ribbon with excess hydrochloric acid.

Combustion experiments

To measure an enthalpy change of combustion, a spirit burner containing the liquid fuel heats a known mass of water in a metal calorimeter (often a copper can), clamped a fixed distance above the flame. The burner is weighed before and after to find the mass of fuel burnt.

Combustion experiments usually give values far less exothermic than data-book values, often by 30% or more. Reasons:

heat loss to the surroundings: much of the hot gas from the flame goes around the can rather than heating it;
energy used to heat the calorimeter itself and the thermometer;
incomplete combustion (soot on the can shows carbon is formed, releasing less energy than complete combustion to $\ce{CO2}$ );
evaporation of the fuel from the wick, so the mass loss is not all fuel burnt;
non-standard conditions: water formed is a gas, not a liquid.

Improvements include a draught shield, a lid on the calorimeter, placing the flame closer, and using a bomb calorimeter (combustion in pure oxygen in a sealed container immersed in water), which gives accurate values.

Correcting for heat loss: temperature–time graphs

In a real experiment the solution starts losing heat as soon as it warms up, so the highest temperature recorded is lower than it would be if the reaction were instantaneous. A graph corrects for this.

Method

The extrapolation method

Record the temperature of the first solution every minute for a few minutes to get a steady starting temperature.
Add the second reactant at a recorded time (for example at $t = 4.0$ min), stir, and do not read the temperature at that instant.
Continue recording every 30 s or minute until the temperature has been falling steadily for several minutes.
Plot temperature against time. Draw a best-fit line through the points before mixing, and a best-fit line through the cooling points after the maximum.
Extrapolate the cooling line back to the time of mixing. The vertical gap between the two lines at that time is the corrected $\Delta T$ .

(0, 20) -- (4, 20) (4, 20) -- (4, 46) (4, 46) -- (5.5, 45.4) (5.5, 45.4) -- (11, 43.2)

(Horizontal axis: time in minutes; vertical axis: temperature in °C. Before mixing the temperature is steady at $20.0\ ^\circ\text{C}$ . The cooling line after the reaction, extrapolated back to the time of mixing at $4.0$ min, reaches $46.0\ ^\circ\text{C}$ , so the corrected $\Delta T = 26.0\ \text{K}$ .)

Worked examples

Neutralisation

$50.0\ \text{cm}^3$ of $1.00\ \text{mol dm}^{-3}$ hydrochloric acid was mixed with $50.0\ \text{cm}^3$ of $1.00\ \text{mol dm}^{-3}$ sodium hydroxide in a polystyrene cup. The temperature rose by $6.8\ \text{K}$ . Calculate the enthalpy change of neutralisation.

Solution

Mass of solution $= 50.0 + 50.0 = 100.0\ \text{g}$ .

q = mc\Delta T = 100.0 \times 4.18 \times 6.8 = 2842\ \text{J}

$n(\ce{H2O}) = n(\ce{HCl}) = 0.0500 \times 1.00 = 0.0500\ \text{mol}$ .

\Delta H_{\text{neut}} = -\frac{2842}{0.0500} = -56\,840\ \text{J mol}^{-1} = -56.8\ \text{kJ mol}^{-1}

Endothermic dissolving

$5.00\ \text{g}$ of ammonium nitrate, $\ce{NH4NO3}$ ( $M_r = 80.0$ ), was dissolved in $50.0\ \text{cm}^3$ of water. The temperature fell by $7.0\ \text{K}$ . Calculate the enthalpy change of solution.

Solution

$q = 50.0 \times 4.18 \times 7.0 = 1463\ \text{J}$ (absorbed from the water).

$n(\ce{NH4NO3}) = \dfrac{5.00}{80.0} = 0.0625\ \text{mol}$ .

The temperature fell, so the process is endothermic:

\Delta H = +\frac{1463}{0.0625} = +23\,400\ \text{J mol}^{-1} = +23.4\ \text{kJ mol}^{-1}

Combustion of a fuel

A spirit burner containing ethanol ( $M_r = 46.0$ ) was used to heat $200\ \text{g}$ of water. The mass of the burner decreased by $0.460\ \text{g}$ and the temperature of the water rose by $12.5\ \text{K}$ .

(a) Calculate the enthalpy change of combustion of ethanol. (b) The data-book value is $-1367\ \text{kJ mol}^{-1}$ . Calculate the percentage difference and suggest two reasons for it.

Solution

(a) $q = 200 \times 4.18 \times 12.5 = 10\,450\ \text{J}$ .

$n(\ce{C2H5OH}) = \dfrac{0.460}{46.0} = 0.0100\ \text{mol}$ .

\Delta H_c = -\frac{10\,450}{0.0100} = -1\,045\,000\ \text{J mol}^{-1} = -1050\ \text{kJ mol}^{-1}\ (\text{3 s.f.})

(b) Percentage difference $= \dfrac{1367 - 1045}{1367} \times 100 = 23.6\%$ .

Reasons: heat lost to the surroundings (the air) rather than to the water; heat absorbed by the copper can; incomplete combustion of ethanol; some ethanol evaporated from the wick rather than burning.

Displacement with a temperature–time graph

$25.0\ \text{cm}^3$ of $0.500\ \text{mol dm}^{-3}$ copper(II) sulfate was placed in a polystyrene cup. Its temperature was steady at $20.0\ ^\circ\text{C}$ . At $4.0$ minutes, excess zinc powder was added. Extrapolating the cooling curve back to $4.0$ minutes gave a temperature of $46.0\ ^\circ\text{C}$ (the graph above). Calculate $\Delta H$ for $\ce{Zn(s) + CuSO4(aq) -> ZnSO4(aq) + Cu(s)}$ .

Solution

Corrected $\Delta T = 46.0 - 20.0 = 26.0\ \text{K}$ .

$q = 25.0 \times 4.18 \times 26.0 = 2717\ \text{J}$ .

Zinc is in excess, so copper(II) sulfate is limiting: $n = 0.0250 \times 0.500 = 0.0125\ \text{mol}$ .

\Delta H = -\frac{2717}{0.0125} = -217\,400\ \text{J mol}^{-1} = -217\ \text{kJ mol}^{-1}

Exam-hard: uncertainties and design

In the neutralisation experiment in the first example, each temperature was read from a thermometer graduated in $1\ ^\circ\text{C}$ intervals, and each volume was measured using a $50\ \text{cm}^3$ measuring cylinder with an uncertainty of $\pm 0.5\ \text{cm}^3$ .

(a) Calculate the percentage uncertainty in $\Delta T$ and in the volume of acid. (b) Identify the main source of uncertainty and suggest an improvement. (c) Explain why using $2.00\ \text{mol dm}^{-3}$ solutions of both acid and alkali (same volumes) would reduce the percentage uncertainty in $\Delta T$ , and predict the new temperature rise.

Solution

(a) Each temperature reading has an uncertainty of $\pm 0.5\ ^\circ\text{C}$ . $\Delta T$ is the difference of two readings, so its uncertainty is $\pm 1.0\ ^\circ\text{C}$ :

\%\ \text{uncertainty in}\ \Delta T = \frac{1.0}{6.8} \times 100 = 14.7\%

Volume: $\dfrac{0.5}{50.0} \times 100 = 1.0\%$ .

(b) The thermometer reading dominates. Use a thermometer graduated in $0.1\ ^\circ\text{C}$ (or a digital temperature probe), which reduces the uncertainty in $\Delta T$ to $\pm 0.2\ ^\circ\text{C}$ , about $3\%$ .

(c) Doubling both concentrations doubles the moles of water formed in the same total volume, so twice the energy is released into the same mass of solution: $\Delta T$ doubles to about $13.6\ \text{K}$ . The absolute uncertainty in $\Delta T$ is still $\pm 1.0\ ^\circ\text{C}$ , so the percentage uncertainty halves to about $7.4\%$ .

Enthalpy change of neutralisation or displacement

Apparatus: expanded polystyrene cup with lid, 250 cm³ beaker (to stand the cup in), thermometer reading to $0.1\ ^\circ\text{C}$ or $0.5\ ^\circ\text{C}$ , burette or pipette, measuring cylinders, stopwatch, stirrer, balance (for solids).

Method:

Measure a known volume of the first solution into the cup; record its temperature every minute for 3 to 4 minutes.
Measure the temperature of the second solution (if it is a solution) and use the mean starting temperature.
Add the second reactant at a recorded time, put the lid on, and stir continuously.
Record the temperature every 30 seconds until it has been falling steadily for several minutes.
Plot a temperature–time graph and extrapolate to find $\Delta T$ at the time of mixing.

Variables: independent: the reaction or quantity being changed; dependent: temperature change; controlled: volumes and concentrations of solutions, starting temperature, type of cup, stirring.

Sources of error and improvements:

error	effect	improvement
heat lost to surroundings	$\Delta T$ too small, $\Delta H$ not exothermic enough	lid, insulation, extrapolation of cooling curve
heat absorbed by cup and thermometer	$\Delta T$ slightly too small	use polystyrene (low heat capacity)
reaction slow so cooling occurs before maximum	$\Delta T$ too small	use powdered solid, stir continuously, extrapolate
assumption $c$ and density equal those of water	small systematic error	use measured values for the solution
thermometer resolution	large % uncertainty for small $\Delta T$	more precise thermometer; larger concentrations to increase $\Delta T$

How Paper 3 asks about it: plotting the temperature–time graph and drawing the two best-fit lines; reading $\Delta T$ from the extrapolation; calculating $\Delta H$ with correct sign and significant figures; percentage uncertainty in $\Delta T$ ; identifying the limiting reactant; suggesting why the value differs from the data-book value.

Watch out

Wrong mass. Use the mass of solution heated (total volume in $\text{cm}^3$ , as grams), never the mass of the solid reactant or fuel.
Wrong sign. Temperature up means $\Delta H$ negative. Students who forget the minus sign lose the final mark even with perfect working.
Wrong moles. Use the reactant that is not in excess. If zinc is in excess, the moles come from copper(II) sulfate.
Forgetting to convert J to kJ, or quoting $\Delta H$ in J with $\text{kJ mol}^{-1}$ units.
Treating ΔT uncertainty as one reading. A temperature change involves two readings, so the uncertainty doubles.

Exam tip

Set out the calculation in three labelled lines: $q = \ldots$ , $n = \ldots$ , $\Delta H = \ldots$ . Each line is usually a separate mark.
In "explain why the experimental value is less exothermic than the data-book value", the first mark is almost always heat loss to the surroundings; give a second, specific reason (incomplete combustion; heat absorbed by the calorimeter; evaporation of the fuel).
When asked to "suggest an improvement", match it to the error you identified (insulation and lid for heat loss; a more precise thermometer for reading uncertainty).
Data booklet value: $c = 4.18\ \text{J g}^{-1}\ \text{K}^{-1}$ (sometimes written $4.18\ \text{kJ kg}^{-1}\ \text{K}^{-1}$ ; the number is the same).

Summary

$q = mc\Delta T$ , with $m$ the mass of water or solution heated, $c = 4.18\ \text{J g}^{-1}\ \text{K}^{-1}$ .
$\Delta H = -mc\Delta T/n$ ; $n$ for the limiting reactant (or water for neutralisation, or fuel for combustion).
Temperature rise: exothermic, $\Delta H$ negative. Temperature fall: endothermic, $\Delta H$ positive.
Assumptions: solution density $1.00\ \text{g cm}^{-3}$ ; $c$ of solution equals that of water; no heat lost.
Polystyrene cup with lid for solutions; spirit burner and copper can for combustion (large heat losses, incomplete combustion).
Temperature–time graph: extrapolate the cooling line back to the time of mixing to correct for heat loss.
Uncertainty in $\Delta T$ is twice the reading uncertainty; a bigger $\Delta T$ gives a smaller percentage uncertainty.

Practice

Question

State the two assumptions made about a dilute aqueous solution in calorimetry calculations.
$25.0\ \text{cm}^3$ of $1.00\ \text{mol dm}^{-3}$ sodium hydroxide was added to $25.0\ \text{cm}^3$ of $1.00\ \text{mol dm}^{-3}$ nitric acid. The temperature rose by $6.6\ \text{K}$ . Calculate $\Delta H_{\text{neut}}$ .
$2.00\ \text{g}$ of sodium hydroxide ( $M_r = 40.0$ ) was dissolved in $50.0\ \text{g}$ of water, and the temperature rose by $10.2\ \text{K}$ . Calculate the enthalpy change of solution of sodium hydroxide.
Burning $0.600\ \text{g}$ of propan-1-ol ( $M_r = 60.0$ ) raised the temperature of $150\ \text{g}$ of water by $25.0\ \text{K}$ . Calculate $\Delta H_c$ of propan-1-ol from these results.
Explain why, in a displacement experiment, zinc powder is used rather than granules, and why it is used in excess.
Excess zinc is added to $50.0\ \text{cm}^3$ of $0.400\ \text{mol dm}^{-3}$ copper(II) sulfate. Given $\Delta H = -217\ \text{kJ mol}^{-1}$ , predict the temperature rise, assuming no heat loss.
A student measures the temperature change in a neutralisation as $5.0\ ^\circ\text{C}$ with a thermometer graduated in $1\ ^\circ\text{C}$ divisions. Calculate the percentage uncertainty in the temperature change, and suggest two ways to reduce it.
$30.0\ \text{cm}^3$ of $2.00\ \text{mol dm}^{-3}$ hydrochloric acid is mixed with $20.0\ \text{cm}^3$ of $2.00\ \text{mol dm}^{-3}$ sodium hydroxide. Using $\Delta H_{\text{neut}} = -57.1\ \text{kJ mol}^{-1}$ , predict the temperature rise.
$25.0\ \text{cm}^3$ of hydrochloric acid of unknown concentration was mixed with $25.0\ \text{cm}^3$ of $2.00\ \text{mol dm}^{-3}$ sodium hydroxide (an excess). The temperature rose by $5.50\ \text{K}$ . Using $\Delta H_{\text{neut}} = -57.1\ \text{kJ mol}^{-1}$ , calculate the concentration of the acid.
A spirit burner burnt $1.50\ \text{g}$ of ethanol ( $M_r = 46.0$ , $\Delta H_c = -1367\ \text{kJ mol}^{-1}$ ) and raised the temperature of $250\ \text{g}$ of water by $30.0\ \text{K}$ . Calculate the percentage of the energy released by the ethanol that was transferred to the water, and explain where the rest went.

Answers

The density of the solution is $1.00\ \text{g cm}^{-3}$ (so $1\ \text{cm}^3$ has a mass of $1\ \text{g}$ ); the specific heat capacity of the solution is the same as that of water, $4.18\ \text{J g}^{-1}\ \text{K}^{-1}$ .
$q = 50.0 \times 4.18 \times 6.6 = 1379\ \text{J}$ . $n(\ce{H2O}) = 0.0250\ \text{mol}$ . $\Delta H = -\dfrac{1379}{0.0250} = -55\,200\ \text{J mol}^{-1} = -55.2\ \text{kJ mol}^{-1}$ .
$q = 50.0 \times 4.18 \times 10.2 = 2132\ \text{J}$ . $n = \dfrac{2.00}{40.0} = 0.0500\ \text{mol}$ . $\Delta H = -\dfrac{2132}{0.0500} = -42\,600\ \text{J mol}^{-1} = -42.6\ \text{kJ mol}^{-1}$ .
$q = 150 \times 4.18 \times 25.0 = 15\,675\ \text{J}$ . $n = \dfrac{0.600}{60.0} = 0.0100\ \text{mol}$ . $\Delta H_c = -\dfrac{15\,675}{0.0100} = -1\,567\,500\ \text{J mol}^{-1} \approx -1570\ \text{kJ mol}^{-1}$ .
Powder has a much larger surface area, so the reaction is fast and the maximum temperature is reached before much heat is lost. Excess zinc makes sure all the copper(II) sulfate reacts, so the amount reacting is known exactly from the copper(II) sulfate solution (the limiting reactant).
$n(\ce{CuSO4}) = 0.0500 \times 0.400 = 0.0200\ \text{mol}$ . $q = 0.0200 \times 217\,000 = 4340\ \text{J}$ . $\Delta T = \dfrac{4340}{50.0 \times 4.18} = 20.8\ \text{K}$ .
Uncertainty in $\Delta T = 2 \times 0.5 = \pm 1.0\ ^\circ\text{C}$ ; percentage $= \dfrac{1.0}{5.0} \times 100 = 20\%$ . Use a thermometer with $0.1\ ^\circ\text{C}$ divisions (or a temperature probe); use more concentrated solutions to give a larger temperature change.
$n(\ce{HCl}) = 0.0600\ \text{mol}$ , $n(\ce{NaOH}) = 0.0400\ \text{mol}$ : NaOH limiting, so $0.0400\ \text{mol}$ of water forms. $q = 0.0400 \times 57\,100 = 2284\ \text{J}$ . Total mass $= 50.0\ \text{g}$ . $\Delta T = \dfrac{2284}{50.0 \times 4.18} = 10.9\ \text{K}$ .
$q = 50.0 \times 4.18 \times 5.50 = 1149.5\ \text{J}$ . $n(\ce{H2O}) = \dfrac{1149.5}{57\,100} = 0.02013\ \text{mol} = n(\ce{HCl})$ (acid limiting). Concentration $= \dfrac{0.02013}{0.0250} = 0.805\ \text{mol dm}^{-3}$ .
Energy to water: $q = 250 \times 4.18 \times 30.0 = 31\,350\ \text{J}$ . Energy released: $n = \dfrac{1.50}{46.0} = 0.03261\ \text{mol}$ ; $0.03261 \times 1\,367\,000 = 44\,580\ \text{J}$ . Percentage $= \dfrac{31\,350}{44\,580} \times 100 = 70.3\%$ . The rest was lost to the surroundings (heating the air around the flame and can), used to heat the calorimeter and thermometer, or not released because of incomplete combustion; some ethanol may have evaporated rather than burned.

The two equations

Experiments in solution

Combustion experiments

Correcting for heat loss: temperature–time graphs

Worked examples

Practice

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