Doc Brown's A Level Chemistry - Advanced Level Theoretical Physical Chemistry Revision Notes - Basic Thermodynamics

GCE A Level Thermodynamics–thermochemistry sub–index links below

Part 1 – ΔH Enthalpy Changes – The thermochemistry of enthalpies of reaction, formation, combustion and neutralisation

Part 1.3 Experimental methods to determine enthalpy changes and treatment of data–results–calculations

This page describes experimental methods for determining enthalpy changes of chemical reactions e.g. using a simple calorimeter and a bomb calorimeter. Treatment of experimental results is covered i.e. how to calculate the molar enthalpy for the reaction under investigation.

See also a set of enthalpy problems to solve, worked out answers given on a separate page!

Energetics-Thermochemistry-Thermodynamics Notes INDEX


Enthalpy determinations using a calorimeter system described and explained on this page

1. Determining the enthalpy of combustion of an alcohol

2. Determining the enthalpy of combustion of benzoic acid with a bomb calorimeter

3. Determining the enthalpy of dissolution of ammonium nitrate with a polystyrene calorimeter

4. Determining the enthalpy of dissolution of potassium chloride/sodium carbonate with a polystyrene calorimeter

5. Determining the enthalpy of neutralisation of hydrochloric acid and sodium hydroxide

6. Determining the enthalpy of reaction of zinc displacing copper from copper(II) sulfate solution

7. Using Hess's Law and experimental data to determine the enthalpy of hydration of anhydrous copper(II) sulfate

8. Using Hess's Law and experimental data to determine the enthalpy of hydration of anhydrous magnesium sulfate

9. Using Hess's Law and experimental data to determine the enthalpy of the decomposition of sodium hydrogencarbonate to sodium carbonate, water and carbon dioxide

These are preceded by sections describing and discussing types of calorimeter and procedures, the principles behind the enthalpy calculations, laboratory equipment needed, sources of error and the specific heat of water and salt solutions.


Energetics index: GCSE Notes on the basics of chemical energy changes – important to study and know before tackling any of the three Advanced Level Chemistry pages Parts 1–3 here * Part 1a–b ΔH Enthalpy Changes 1.1 Advanced Introduction to enthalpy changes – reaction, formation, combustion : 1.2a & 1.2b(i)–(iii) Thermochemistry – Hess's Law and Enthalpy Calculations – reaction, combustion, formation etc. : 1.2b(iv) Bond Enthalpy Calculations  : 1.3a–b Experimental methods for determining enthalpy changes and treatment of results : 1.4 Some enthalpy data patterns : 1.4a The combustion of linear alkanes and linear aliphatic alcohols : 1.4b Some patterns in Bond Enthalpies and Bond Length : 1.4c Enthalpies of Neutralisation : 1.4d Enthalpies of Hydrogenation of unsaturated hydrocarbons and evidence of aromatic ring structure in benzene : Extra Q page A set of practice enthalpy calculations with worked out answers ** Part 2 ΔH Enthalpies of ion hydration, solution, atomisation, lattice energy, electron affinity and the Born–Haber cycle : 2.1a–c What happens when a salt dissolves in water and why? : 2.1d–e Enthalpy cycles involving a salt dissolving : 2.2a–c The Born–Haber Cycle *** Part 3 ΔS Entropy and ΔG Free Energy Changes : 3.1a–g Introduction to Entropy : 3.2 Examples of entropy values and comments * 3.3a ΔS, Entropy and change of state : 3.3b ΔS, Entropy changes and the feasibility of a chemical change : 3.4a–d More on ΔG, Free energy changes, feasibility and applications : 3.5 Calculating Equilibrium Constants : 3.6 Kinetic stability versus thermodynamic feasibility * PLEASE note that delta H/S/G values vary slightly from source to source, so I apologise in advance for any inconsistencies that may arise as I've researched and developed each section.

 

1.3 Experimental methods for determining enthalpy changes and treatment of results

1.3a Experimental methods

All the methods described here rely on measuring a temperature change knowing the molar quantities of reactants and the mass of water used in a calorimeter system

(c) doc b

1.3a1 A simple polystyrene calorimeter of low heat capacity can be used for any non–combustion reaction that will happen spontaneously at room temperature involving solutions or solid reacting/dissolving with/in a liquid like water and it doesn't matter if the reaction is exothermic or endothermic.

Reasonably accurate results can be obtained for a school/college laboratory. The reactants are weighed in if solid and a known volume of any liquid (usually water or aqueous solution). The mixture could be a salt and water (heat change on dissolving) or an acid and an alkali solution (heat change of neutralisation). It doesn't matter whether the change is exothermic (heat released or given out, temperature increases) or endothermic (heat absorbed or taken in, temperature decreases).

Ideally a very accurate mercury thermometer with 0.1 or 0.2 oC graduations should be used or an equally accurate electronic digital thermometer (can read to 0.01oC via a thermistor, a solid–state electronic device).

A double polystyrene cup system provides good thermal insulation for the system.


Typical apparatus needed for this kind of calorimetric work with a simple polystyrene cup calorimeter

safety spectacles, pipette and suction pipette filler (or a burette, double polystyrene cup calorimeter with insulating lid plus hole for the thermometer or temperature probe, chemicals of appropriate mass/volume/concentration, weighing bottle, spatula, accurate electronic balance, mercury thermometer (preferably 0.1 or 0.2 oC graduations) or electronic thermistor temperature probe, accurate electronic stop watch or clock.


Note on graphical analysis

To get the best value for the temperature change (ΔT) you should take multiple readings before and after mixing the reactants and then plotting a graph versus time.

The initial readings give you a baseline, but the reaction may take a few seconds or a few minutes, and so you cannot get an immediate true ΔT. However by drawing a baseline for the initial temperature and extrapolating back to the start of the reaction (e.g. at 1.5 minutes) you can then estimate the real temperature change.

What happens is quite simple, but it leads to inaccuracy:

For exothermic reactions the system will continuously lose heat once the reaction has started, so the temperature starts to fall once the reaction is complete, so extrapolating back up gives the true temperature rise.

For endothermic reactions the system will continuously gain heat once the reaction has started, so the temperature starts to rise once the reaction is complete, so extrapolating back down gives the true temperature fall.

 Without allowing for these unavoidable experimental circumstances, you will always measure too low a temperature change.


Note on the specific heat of water

The specific heat of water is exceptionally high due to the energy required to break down the hydrogen bonding. When you heat water, some of the absorbed heat goes into increasing the vibration of the hydrogen bonds and weakening them and increasing disorder, rather than being absorbed directly to increase the kinetic energy of the water molecules i.e. not all the absorbed heat goes to raising the temperature of liquid water.

When salts (and anything else) are dissolved in water, the hydrogen bonding is disrupted as water molecules congregate around the ions in the process called hydration. This has the effect of lowering the specific heat capacity of water and the more concentrated the salt solution, the lower the heat capacity of the solution.

Therefore using the specific heat of pure water in calorimeter data calculations automatically incurs an error!

 


Examples of specific heats (SHC) of salt solutions

All of these compounds happen to consist of two ions and seem to show a similar pattern of specific heat capacity reduction with increasing salt concentration.
molarity of salt solution, mol dm–3 0.0 0.1 0.5 0.9 1.0
SHC sodium chloride, J g–1 oC–1 4.18 4.15 4.03   3.90
approximate g of NaCl per 100 ml (100 g) water 0.0 0.6 2.9   5.9
SHC ammonium chloride, J g–1 oC–1 4.18   4.04 3.93  
approximate g of NH4Cl per 100 ml (100 g) water 0.0   2.7 4.8  
SHC ammonium nitrate, J g–1 oC–1 4.18   4.01 3.98  
approximate g of NH4NO3 per 100 ml (100 g) water 0.0   4.0 7.2  
This is all the data I could find after several hours on the internet!

If anybody finds any useful SHC data for salt solutions in J g–1 oC–1 please send me the link!

         

 


Note on source of errors

Despite the use of a poorly conducting polystyrene container and lid, they still absorb/release (exothermic/endothermic) heat and a small amount of heat will still be lost/gained to/from the surroundings. The amount of heat involved is small, but not zero!

The glass mercury thermometer or thermistor temperature probe will absorb some heat (if exothermic) or release heat (if endothermic), therefore reducing or increasing the measured temperatures.

If a solid reactant is used or formed e.g. metals in a displacement reaction, there is a small error from unreacted metal or metal formed, but the specific heat of metals or other solids is usually quite small.

The specific heat of aqueous solutions is less than that of pure water (see examples of data above and the multiple calculation and discussion in experiment 3.).

A 'low resolution' thermometer, reading to the nearest 0.5oC is not accurate enough unless the temperature change is quite big e.g. ΔT >20o.

(c) doc b

1.3a2 A simple calorimeter for combustion is specifically for determining the heat energy released (given out) for burning fuels. The burner is weighed before and after combustion to get the mass of liquid fuel burned. The thermometer records the temperature rise of the known mass of water (1g ~ 1cm3 since density of water is ~1.0 gcm–1). You can use this system to compare the heat output from burning various fuels. The bigger the temperature rise, the more heat energy is released. You can take a series of measurements with time and from the graph extrapolate the maximum temperature rise.

This is a very inaccurate method because of huge losses of heat e.g. radiation from the flame and calorimeter, conduction through the copper calorimeter, convection from the flame gases passing by the calorimeter etc. BUT, at least using the same burner and set–up, you can do a reasonable comparison of the heat output of different fuels.

Other sources of error: Heat is also absorbed by the copper calorimeter. The flame may be smokey, indicating combustion is incomplete.

You can also do a calibration by burning a fixed amount of material whose enthalpy of combustion is accurately known.


Typical apparatus needed for this kind of calorimetric work with a copper calorimeter

safety spectacles, accurate measuring cylinder for water (or weigh calorimeter before and after adding the water), copper vessel calorimeter with insulating lid plus hole for the thermometer or temperature probe, draught shielding around the burner and calorimeter, a small burner/lamp with wick, suitable combustible material (alcohols like ethanol/propanol burn more cleanly than alkanes like hexane), accurate electronic balance, mercury thermometer (preferably 0.1 or 0.2 oC graduations) or an accurate electronic thermistor temperature probe, accurate electronic stop watch or clock.

1.3a3 The Adiabatic Bomb Calorimeter: Method 1.3a2 is quite a crude and inaccurate method for determining enthalpies of combustion. The bomb calorimeter method delivers very accurate results. The idea is execute the reaction adiabatically – meaning no heat loss from the system. The compound undergoing combustion is electrically ignited, and does so under pressure in an atmosphere of pure oxygen gas and this ensures complete and rapid combustion. There may be several water baths and thermometers to make sure every joule of energy released is absorbed by the calorimeter and measured.

The heat capacity of a bomb calorimeter can be very accurately determined by combustion of a standard substance like benzoic acid whose enthalpy of combustion is known from previous experiments.

The bomb calorimeter method involves measuring the heat released at constant volume and is strictly speaking called the internal energy change ΔE or ΔU. Methods 1.3a1 and 1.3a2 involve heat energy changes at constant pressure and directly measure ΔH. From 1.3a bomb calorimeter measurements you can calculate the enthalpy change from the equation

ΔH = ΔE + ΔnRT (n = the net change in moles of gas in the reaction) I don't think this equation is needed for any UK pre–university advanced level chemistry course these days?

If there are no gaseous reactants or products (i.e. only liquids/solids involved) OR if moles gaseous reactants = moles gaseous products, then Δn = 0 and ΔH = ΔE or ΔU).

ENERGETICS INDEX 


1.3b Treatment of experimental results

In any calorimeter the heat released or absorbed is given by

energy transferred = SHCH2O x m x ΔT    (sometimes expressed simply as q = m c ΔT)

SHCH2O = specific heat capacity of water (4.18 J g–1 K–1) i.e. it takes 4.18 J of heat energy to raise 1g/1cm3 of water by 1o.

This assumes the heat capacity of the water is the same as the solution in method 1.3a1

m = mass of the water absorbing the heat, usually grammes.

This ignores the mass of the calorimeter, thermometer, insulation etc.

ΔT = temperature change  (Tfinal – Tinitial)

This cannot take into account heat energy losses, which any experiment should be designed to minimise, so its only what you can actually measure directly.

You then have to relate this heat change to the mass/molar quantities used to get the ΔH enthalpy change in kJ mol–1.


Examples of calculations using data from various calorimetric methods

The calculations are based on experimental data alone OR a combination of standard data and experimental data.

Either way, many involve using Hess's Law, e.g. a 'simple' triangular arrangement (see Hess's Law Notes and below)

Application of Hess's Law (i)

ΔHθ1

  B

ΔHθ2

(c) doc b    

ΔHθ3

C
Clearly there are two pathways from A to B

direct and via C following the arrows direction

from Hess's Law: ΔHθ1  =  ΔHθ2  +  ΔHθ3

or    ΔHθ2  =  ΔHθ1  –  ΔHθ3

or    ΔHθ3  =  ΔHθ1  –  ΔHθ2

Application of Hess's Law (ii)

ΔHθ1

  B

ΔHθ2

(c) doc b     (c) doc b

ΔHθ3

C
In this case there are two pathways from A to C

direct and via B following the arrows direction

from Hess's Law: ΔHθ2  =  ΔHθ1  +  ΔHθ3

and   ΔHθ1  =  ΔHθ2  –  ΔHθ3

or    ΔHθ3  =  ΔHθ2  –  ΔHθ1

Always take care with the ΔH signs which ever way you set up the triangle to apply Hess's Law Although slightly more awkward, (ii) might be better suited to the way the experiment results are obtained

 


1. Determining the enthalpy of combustion of an alcohol (using method 1.3a2)

  • 100 cm3 of water (100g) was measured into the calorimeter.

  • The spirit burner contained the fuel ethanol CH3CH2OH ('alcohol') and weighed 18.62g at the start.

  • After burning it weighed 17.14g and the temperature of the water rose from 18 to 89oC.

  • The temperature rise = 89 – 18 = 71oC (exothermic, heat energy given out).

  • Mass of fuel burned = 18.62–17.14 = 1.48g.

  • Heat given out to the water = mass of water x SHCwater x temperature change

    • = 100 x 4.18 x 71 = 29678 J (for 1.48g)

  • Mr(ethanol) = 46 (H=1, C=12, O=16)

  • Therefore 1.48g ethanol = 1.48/46 = 0.03217 mol

  • So, scaling up to 1 mole of ethanol burned gives 29678 x 1 / 0.03217 = 922536 J

  • Enthalpy of combustion of ethanol = ΔHc(ethanol) = –923 kJmol–1  (only accurate to 3 sf at best)

  • for the reaction: CH3CH2OH(l) + 3O2(g) ===> 2CO2(g) + 3H2O(l)

  • The data book value for the heat of combustion of ethanol is –1367 kJmol–1, showing lots of heat loss in the experiment!

  • It is possible to get more accurate values by calibrating the calorimeter with a substance whose energy release on combustion is known.

 


2. Determining the enthalpy of combustion of benzoic acid with a bomb calorimeter (using method 1.3a3)

A bomb calorimeter had a heat capacity equivalent to 800.0 g of water.

After the complete combustion of 1.200 g of benzoic acid the temperature had risen by 9.500oC.

This an adiabatic reaction, no heat is lost to the surroundings beyond the calorimeter and the volume remains constant, so it is ΔE, the internal energy change, not ΔH.

Heat given out to the water = mass of water x SHCwater x temperature change

ΔEcalorimeter = 800 x 4.18 x 9.5 = 31768 J

Mr(C6H5COOH) = 122,  mol benzoic acid = 1.20 / 122 = 0.009836

Scaling up to 1 mole of benzoic acid: ΔE = 31768 x 1/0.009836 = 3229768 J

ΔH = ΔE + ΔnRT (n = the net change in moles of gas in the reaction)

The balanced equation under standard conditions is ...

C6H5COOH(s) +  7.5O2(g)  ===>  7CO2(g)  +  3H2O(l)

ΔnRT = (7 – 7.5) x 8.314 x 298 = –1239 J

ΔH = ΔE + ΔnRT = 3229768 – 1239 = 3228529 J

ΔHθc(C6H5COOH) = 3229 kJ mol–1  (accurate to 4 sf)

 


3. Determining the enthalpy of solution of ammonium nitrate with a simple 'coffee cup' calorimeter (method 1.3a1)

To measure the enthalpy of dissolution of ammonium nitrate.

Using a simple polystyrene calorimeter 7.2 g of ammonium nitrate was dissolved in 100 ml (100 g) of water.

Typical graph shape for an endothermic experiment

Graphical analysis (e.g. like above) showed the temperature fell by 5.45oC.

Calculate the enthalpy of dissolution (dissolving) of ammonium nitrate.

Heat absorbed by the water = mass of water x SHCwater x temperature

q = m c ΔT

I'm now going to do the calculation three times, explaining why, but only for example 3.

Mr(NH4NO3) = 14 + 4 + 14 + (3 x 16) = 80; mol NH4NO3 = 7.2/80 = 0.09 mol

(i) Using the real value of the heat capacity of ammonium nitrate for this particular concentration and total mass.

q = m c ΔT

q = 107.2 x 3.98 x 5.45 = 2325 J, 2.325 kJ

Scaling up to 1 mole = 2.325 x (1/0.09) = 25.8 kJ mol–1

Since the temperature has fallen, the change is endothermic

So ΔHdissolution(NH4NO3) = +25.8 kJ mol–1   (only accurate to 3 sf)

(ii) Using the SHC of pure water with just the mass of the water

q = m c ΔT

q = 100 x 4.18 x 5.45 = 2278 J, 2.278 kJ

Scaling up to 1 mole = 2.278 x (1/0.09) = 25.3 kJ mol–1

Since the temperature has fallen, the change is endothermic

So ΔHdissolution(NH4NO3) = +25.3 kJ mol–1   (only accurate to 3 sf)

(iii) Using the SHC of pure water and total mass of solution

q = m c ΔT

q = 107.2 x 4.18 x 5.45 = 2442 J, 2.442 kJ

Scaling up to 1 mole = 2.442 x (1/0.09) = 27.1 kJ mol–1

Since the temperature has fallen, the change is endothermic

So ΔHdissolution(NH4NO3) = +27.1 kJ mol–1   (only accurate to 3 sf)

Discussion of calculations (i), (ii) and (iii)

(i) This is the most accurate answer because it uses the true specific heat capacity of the salt solution.

–

(ii) This is the next most accurate answer by ignoring the mass of the salt!

It seems that neglecting the mass of the salt roughly off–sets the fall of the heat capacity of water, because all that students will be given will be the higher value of the SHC of pure water. Some textbooks/internet pages just use the mass of water and its probably the best compromise for the calculation.

(iii) This is the least accurate value calculated.

Although it does use the actual total mass of solution, the real heat capacity of the solution is 5% less. In calculation (ii) the mass is 7% less than (iii) but the SHC of water is 5% higher, so that's why (ii) is more accurate than (iii).

 


4. Determining the enthalpy of solution of potassium chloride and sodium carbonate with a simple 'coffee cup' calorimeter (method 1.3a1)

The determination of the enthalpy of dissolution of potassium chloride.

Using a simple polystyrene calorimeter 4.0 g of ammonium nitrate was dissolved in 50 ml (50 g) of water.

Typical graph shape for an endothermic experiment

Graphical analysis (e.g. like above) showed the temperature fell by 4.40oC.

Calculate the enthalpy of dissolution (dissolving) of potassium chloride.

Heat absorbed by the water = mass of water x SHCwater x temperature

q = m c ΔT

Mr(KCl) = 39 + 35.5 = 74.5; mol KCl = 4.0/74.5 = 0.05369 mol

(i) Using the value of the heat capacity of pure water and just the mass of the water.

q = m c ΔT

q = 50 x 4.18 x 4.40 = 919.6 J, 0.9196 kJ

Scaling up to 1 mole = 0.9196 x (1/0.05369) = 17.1 kJ mol–1

Since the temperature has fallen, the change is endothermic

So ΔHdissolution(KCl) = +17.1 kJ mol–1   (only accurate to 3 sf)

This value is not too dissimilar from the theoretically better calculation (i) below.

(ii) Using the total mass and the likely SHC of the potassium chloride solution, assuming it is similar to sodium chloride at around a 1 molar concentration.

q = m c ΔT

q = 54 x 3.90 x 4.40 = 926.6 J, 0.9266 kJ

Scaling up to 1 mole = 0.9266 x (1/0.05369) = 17.3 kJ mol–1

Since the temperature has fallen, the change is endothermic

So ΔHdissolution(KCl) = +17.3 kJ mol–1  (only accurate to 3 sf)

Both methods (i) and (ii) give similar results, again justifying the textbooks/internet pages that ignore the mass of the salt and using the SHC of pure water.

3. and 4. so far have been endothermic changes, so time for a change!

Typical graph shape for an exothermic experiment.

An example of an exothermic enthalpy of solution (dissolution) is dissolving anhydrous sodium carbonate in water.

On dissolving 8.5 g of anhydrous sodium carbonate in 50 cm3 of water, the temperature rose by a maximum of 8.1oC e.g. using an extrapolation graph method illustrated above.

Calculate the enthalpy of solution of anhydrous sodium carbonate. Formula mass of Na2CO3 = 106

q = m c ΔT,  q = 50 x 4.18 x 8.1 = 1692.9 J, 1.6929 kJ

mol Na2CO3 = 8.5/106 = 0.0802

Scaling up to 1 mol = 1.6929 x 1.0/0.0802 = 21.108

Since the temperature rose, exothermic, therefore

So ΔHdissolution(Na2CO3) = –21.1 kJ mol–1  (only accurate to 3 sf)

 


5. Determining the enthalpy of neutralisation of hydrochloric acid and sodium hydroxide (method 1.3a1)

You can do this experiment by mixing equal volumes of equimolar concentrations of dilute hydrochloric acid and dilute sodium hydroxide. e.g. 25 cm3 of each in the polystyrene calorimeter as previously described.

Read the temperature every 30 seconds before and after the mixing.

Typical graph shape for an exothermic experiment

A graph is drawn of temperature versus time and the true temperature rise estimated by extrapolation as shown on the group.

Suppose after mixing, via accurate pipettes, 25.0 cm3 of 1.0 mol dm–3 hydrochloric acid and 25.0 of 1.0 mol dm–3, sodium hydroxide solutions the extrapolated temperature rise from a graph like the one above, is 7.1oC, calculate the enthalpy of neutralisation for the reaction:

HCl(aq)  +  NaOH(aq)  ===>  NaCl(aq)  +  H2O(l)

Calculation.

(i) Using the SHC for pure water and the total mass is effectively 50 g (actually 50 cm3 of NaCl solution).

q = m c ΔT  =  50 x 4.18 x 7.1 = 1483.9 J, 1.4839 kJ

From the equation: mol HCl = mol NaOH = 1.0 x 25/1000 = 0.025 mol

Therefore scaling up to 1 mol gives a numerical enthalpy change of 1.4839 x 1/0.025 = 59.4 kJ

Since the temperature rose indicating an exothermic reaction, the enthalpy of neutralisation is ..

ΔHneutralisation(HCl + NaOH) = –59.4 kJ mol–1  (only accurate to 3 sf)

 

(ii) Using the SHC of 0.50 molar sodium chloride solution, which is 4.03 J g–1 oC–1

(see the discussion on the specific heat capacity of salt solutions)

By mixing 1 molar solutions of HCl and NaOH, you actually produce a 0.5 molar solution of sodium chloride, for which, we actually have the true specific heat capacity!

q = m c ΔT  =  50 x 4.03 x 7.1 = 1430.65 J, 1.43065 kJ

From the equation: mol HCl = mol NaOH = 1.0 x 25/1000 = 0.025 mol

Therefore scaling up to 1 mol gives a numerical enthalpy change of 1.43065 x 1/0.025 = 57.2 kJ

Since the temperature rose indicating an exothermic reaction, the enthalpy of neutralisation is ..

ΔHneutralisation(HCl + NaOH) = –57.2 kJ mol–1  (only accurate to 3 sf)

 

The data book value is –57.1 kJ mol–1

% error = 100 x  (experimental value – accepted value) / accepted value

Calculation (i) gives an error of 4.0% and calculation (ii) gives an error of 0.2%.

So you can see that by using the wrong specific heat capacity you incur a 4.0% error due to using the higher specific heat capacity of pure water.


6. Determining the enthalpy of reaction of zinc displacing copper from copper(II) sulfate solution (method 1.3a1)

The reaction equation is: Zn(s)  +  Cu2+(aq)  ===>  Zn2+(aq)  +  Cu(s)

About 3 g of fine zinc powder is added to exactly 25.0 cm3 of copper(II) sulfate solution of 1.0 mol dm–3 concentration, which can be pipetted in using a suction bulb. As long as its about 3 g, you be sure it is in excess, so the calculation can be based on the moles of copper(II) ion.

The temperature is plotted every 30 seconds, including a few minutes before adding the zinc powder. The reaction is very exothermic and the temperature should rise rapidly.

A graph is then plotted of temperature versus time and the maximum temperature rise deduced by interpolation (see exemplar graph below.

Typical graph shape for an exothermic experiment

If the temperature rise was 52oC calculate the enthalpy change for the zinc – copper sulfate displacement reaction

mol Cu2+(aq) = 1.0 x 25/1000 = 0.025 mol;  mol Zn = 3/65 = 0.046 mol (showing zinc to be well in excess)

q = m c ΔT, q = 25 x 4.18 x 52 = 5434 J,  5.434 kJ

Scaling up to 1 mole of copper ions: ΔH = 5.434 x 1/0.025 = 217 kJ mol–1

Since the temperature rose, this displacement reaction is exothermic

ΔHdisplacement(Zn + CuSO4) = –217 kJ mol–1  (only accurate to 3 sf)

  (only accurate to 3 sf)

 

Theoretical calculation from data book information

Zn(s)  +  Cu2+(aq)  ===>  Zn2+(aq)  +  Cu(s)

ΔHθreaction,298 = ∑ΔHθf,298(products) – ∑ΔHθf,298(reactants)

ΔHθdisplacement reaction,298 = ΔHθf,298(Zn2+(aq)) – ΔHθf,298(Cu2+(aq))

ΔHθdisplacement reaction,298 = –152.4 – (+64.4) = –216.8 kJ mol–1

Note: Enthalpies of formation of elements in the normal standard stable states at 298K are zero and so can be excluded from the calculation (though you can put them in formally if you want). The enthalpies of formation of the ions are measured with respect to the formation of the hydrogen ion, H+(aq) which is arbitrarily assigned an enthalpy of formation value of zero.


7. Using Hess's Law and experimental data to determine the enthalpy of hydration of anhydrous copper(II) sulfate (1.3a1)

By the calorimetric methods already described you can separately determine the enthalpy of dissolution (solution) of anhydrous copper sulfate (white) and hydrated copper sulfate crystals (blue) using the calculation method already described for experiments 4. to 6.

See example 9. for another case of combining experimental enthalpy values to obtain another one not obtainable by experiment.

From this data, using a Hess's Law cycle described below, you can then calculate the enthalpy of hydration of anhydrous copper(II) sulfate which you cannot .

Application of Hess's Law

ΔHθ1

  B

ΔHθ2

(c) doc b     (c) doc b

ΔHθ3

C
from Hess's Law:   ΔHθ2  =  ΔHθ1  +  ΔHθ3     or     ΔHθ1  =  ΔHθ2  –  ΔHθ3     or     ΔHθ3  =  ΔHθ2  –  ΔHθ1

 

ΔHθ1 ΔHθhydration[CuSO4(s)]
CuSO4(s) + 5H2O(l) CuSO4.5H2O(s)

ΔHθ2 = ΔHθdissolution[CuSO4(s)]

(c) doc b   +

aq

(c) doc b

ΔHθ3 = ΔHθdissolution[CuSO4.5H2O(s)]

CuSO4(aq) + 5H2O(l)
from Hess's Law: ΔHθ2  =  ΔHθ1  +  ΔHθ3      and     ΔHθ1  =  ΔHθ2  –  ΔHθ3

ΔHθhydration[MgSO4(s)] = ΔHθdissolution[MgSO4(s)] – ΔHθdissolution[MgSO4.7H2O(s)]

 

DATA and theoretical calculations for experiment examples 7. and 8.

The enthalpy of dissolution is sometimes just called the enthalpy of solution.

The best enthalpy data I can find from data books or computation from: ΔHθ1  =  ΔHθ2  –  ΔHθ3
salt and enthalpy value

in kJmol–1

ΔHθ2 =

ΔHθdissolution(anhydrous salt)

ΔHθ3 =

ΔHθdissolution(hydrated salt)

ΔHθ1 =

ΔHθhydration(anhydrous salt)

(a) copper(II) sulfate ΔHθ2 =

ΔHθdissolution(CuSO4)

= –73.3

ΔHθ3 =

ΔHθdissolution(CuSO4.5H2O)

= +4.7

ΔHθ1 =

ΔHθhydration(CuSO4)

= –73.3 – (+4.7) = –78.0

(b) magnesium sulfate ΔHθ2 =

ΔHθdissolution(MgSO4)

= –91.2

ΔHθ3 =

ΔHθdissolution(MgSO4.7H2O)

= +12.8

ΔHθ1 =

ΔHθhydration(MgSO4)

= –91.2 – (+12.8) = –104.0

       

Theoretical calculations for enthalpy of hydration of anhydrous salt to hydrated salt crystals (examples 7. and 8.)

ΔHθ1 = ΔHθhydration(anhydrous salt) cannot be obtained directly by experiment, but can be calculated from experimental data (experiments 7./8.) using a Hess's Law cycle, and can be theoretically calculated from enthalpy of formation data (see below). This is the object of experimental exercises 7. and 8.

ΔHθ2 = ΔHθdissolution(anhydrous salt)  can be obtained by experiment (e.g. like experiments 3. to 6.), and can also be obtained from standard data books!

Note that the values quoted are for infinite dilution, because the enthalpy values are dependent in a small way on the ratio of salt to water - and the specific heat capacity of solutions also depends on concentration!

ΔHθ3 = ΔHθdissolution(hydrated salt) can be obtained by experiment, couldn't find any reliable data in books or internet, but can, theoretically, be calculated from reliable book/internet data for ΔHθ1 and ΔHθ2.

 

Theoretical calculation (see methods of enthalpy calculations)

ΔHθreaction,298 = ∑ΔHθf,298(products) – ∑ΔHθf,298(reactants)

All enthalpy values used in the calculations below are given in kJ mol–1

 

(a) Copper(II) sulfate

CuSO4(s)  +  5H2O(l)  ===>  CuSO4.5H2O(s)

ΔHθf,298(CuSO4(s)) = –770, ΔHθf,298(H2O(l)) = –286, ΔHθf,298(CuSO4.5H2O(s)) = –2278

ΔHθreaction,298 = ∑ΔHθf,298(products) – ∑ΔHθf,298(reactants)

ΔHθhydration,298 = ΔHθf,298(CuSO4.5H2O(s)) – {ΔHθf,298(CuSO4(s)) + 5 x ΔHθf,298(H2O(l))}

ΔHθhydration,298(CuSO4(s)) = –2278 – {–770 – 5 x 286} = –78.0 kJ mol–1

 

(b) Magnesium sulfate

MgSO4(s)  +  7H2O(l)  ===>  MgSO4.7H2O(s)

ΔHθf,298(MgSO4(s)) = –1278, ΔHθf,298(H2O(l)) = –286, ΔHθf,298(MgSO4.7H2O(s)) = –3384

ΔHθreaction,298 = ∑ΔHθf,298(products) – ∑ΔHθf,298(reactants)

ΔHθhydration,298 = ΔHθf,298(MgSO4.7H2O(s)) – {ΔHθf,298(MgSO4(s)) + 7 x ΔHθf,298(H2O(l))}

ΔHθhydration,298(MgSO4(s)) = –3384 – {–1278 – 7 x 286} = –104.0 kJ mol–1

 


8. Using Hess's Law and experimental data to determine the enthalpy of hydration of anhydrous magnesium sulfate (1.3a1)

By the calorimetric methods already described you can separately determine the enthalpy of dissolution (solution) of anhydrous magnesium sulfate and hydrated magnesium sulfate crystals using the calculation method already described for experiments 4. to 6.

See example 9. for another case of combining experimental enthalpy values to obtain another one not obtainable by experiment.

From this data, using a Hess's Law cycle described below, you can then calculate the enthalpy of hydration of anhydrous magnesium sulfate.

 
ΔHθ1 = ΔHθhydration[MgSO4(s)]
MgSO4(s) + 7H2O(l) MgSO4.7H2O(s)

 ΔHθ2 = ΔHθdissolution[MgSO4(s)]

(c) doc b   +

aq

(c) doc b

ΔHθ3 = ΔHθdissolution[MgSO4.7H2O(s)]

MgSO4(aq) + 7H2O(l)
from Hess's Law: ΔHθ2  =  ΔHθ1  +  ΔHθ3       and      ΔHθ1  =  ΔHθ2  –  ΔHθ3

ΔHθhydration[MgSO4(s)] = ΔHθdissolution[MgSO4(s)] – ΔHθdissolution[MgSO4.7H2O(s)]

See data table and calculations in related experiment 7. above.


9. Using Hess's Law and experimental data to determine the enthalpy of the decomposition of sodium hydrogencarbonate to sodium carbonate, water and carbon dioxide (method 1.3a1)

You cannot directly obtain experimental results to determine (1) the enthalpy of thermal decomposition of sodium hydrogen carbonate. However, you can measure the enthalpy change for two other reactions and using Hess's Law calculate the 'unobtainable' by experiment.

The two other reactions involve (2) mixing sodium hydrogencarbonate and (3) sodium carbonate with hydrochloric acid.

The Hess's Law cycle connecting these three reactions is shown below.

ΔHθ1
2NaHCO3(s)     Na2CO3(s)  +  H2O(l)  +  CO2(g)

+ 2HCl(aq)

ΔHθ2

(c) doc b     (c) doc b

+ 2HCl(aq)

ΔHθ3

2NaCl(aq)  +  2H2O(l)  +  2CO2(g)
∴  ΔHθ2  =  ΔHθ1  +  ΔHθ3    and     ΔHθ1  =  ΔHθ2  –  ΔHθ3

By the calorimetric methods already described you can separately determine the enthalpy of reaction of firstly, (a) sodium hydrogencarbonate with hydrochloric acid and then (b) the reaction of anhydrous sodium carbonate with hydrochloric acid using the calculation methods already described e.g. experiments 4. to 6.

Examples of experimental results:

atomic masses: Na = 23, H = 1, C = 12, O = 16. formula masses: NaHCO3 = 84   and   Na2CO3 = 106

Enthalpy change for reaction 1.

The unknown!

Enthalpy change for reaction 2.

On adding 3.36 g of sodium hydrogen carbonate to 50 cm3 of 1 molar hydrochloric acid (an excess), the maximum temperature fall was found to be 5.20oC. Assume the density of hydrochloric acid is the same as pure water.

mol HCl = 1.0 x 50/1000 = 0.05 mol (in excess, check the next line and equation, 1:1 mole ratio)

mol NaHCO3 = 3.36/84 = 0.04 mol (the limiting reactant on which the calculation must be based)

NaHCO3(s)  +  HCl(aq)  ===>  NaCl(aq)  +  H2O(l)  +  CO2(g)

q = m c ΔT,  q = 50 x 4.18 x 5.2 = 10868.8 J,  1.08688 kJ

Scaling up to 1 mol of NaHCO3: 1.08688 x 1.0/0.04 = 27.17 kJ

Since the temperature fell, the reaction is endothermic, heat absorbed

Therefore ΔHθr,298 = ΔHθ2 = +27.2 kJ mol–1 per mole of NaHCO3  (equivalent to +54.4 for 2 mol of NaHCO3)

(b) On adding 2.12 g of anhydrous sodium carbonate to 50 cm3 of 1 molar hydrochloric acid (an excess), the maximum temperature rise was estimated to be 3.30oC.

mol HCl = 1.0 x 50/1000 = 0.05 mol (in excess, check the next few lines!)

mol Na2CO3 = 2.12/106 = 0.02 mol (the limiting reactant on which the calculation must be based)

Na2CO3(s)  +  2HCl(aq)  ===>  2NaCl(aq)  +  H2O(l)  +  CO2(g)

0.02 mol Na2CO3 will react with 0.04 mol HCl

q = m c ΔT,  q = 50 x 4.18 x 3.3 = 689.7 J,  0.6897 kJ

Scaling up to 1 mol of Na2CO3: 1.08688 x 1.0/0.02 = 34.485 kJ

Since the temperature rose, the reaction is exothermic, heat released

Therefore ΔHθr,298 = ΔHθ3 = –34.5 kJ mol–1 per mole of Na2CO3

The temperature rises are quite small, particularly for reaction 3., so I would suggest using double the masses and 2 molar hydrochloric acid. However, the real heat capacity of the resulting sodium chloride solution is likely to be <3.9 J g–1 oC–1.

From this data, using a Hess's Law cycle described below, you can then calculate the enthalpy of hydration of anhydrous copper(II) sulfate. Note that reaction 2 has double molar quantities to that the cycle as a whole is perfectly balanced.

ΔHθ1  =  ΔHθ2  –  ΔHθ3

ΔHθ1  =  +54.4  –  (–34.5)  =  +88.9 kJ mol–1 for the thermal decomposition of NaHCO3

After the Hess's Law cycle below (repeat!), for this experiment, I've worked everything out from scratch using accurate thermodynamic date i.e. all the enthalpies of formation to calculate all three enthalpies of reaction.

 
ΔHθ1
2NaHCO3(s)     Na2CO3(s)  +  H2O(l)  +  CO2(g)

+ 2HCl(aq)

ΔHθ2

(c) doc b     (c) doc b

+ 2HCl(aq)

ΔHθ3

2NaCl(aq)  +  2H2O(l)  +  2CO2(g)
∴  ΔHθ2  =  ΔHθ1  +  ΔHθ3    and     ΔHθ1  =  ΔHθ2  –  ΔHθ3

Theoretical calculations of ΔHθ1–3 from standard enthalpy of formation data.

ΔHθr,298 = standard enthalpy of reaction at 298K and 1 atm pressure i.e. the usual standard conditions

The thermodynamic data for NaCl(aq) and HCl(aq) apply to concentrations of about 1.0 mol dm–3 (it varies with concentration)

ΔHθf,298 data (all in kJ mol–1): ΔHθf,298(NaHCO3(s)) = –948,  ΔHθf,298(Na2CO3(s)) = –1131,  ΔHθf,298(H2O(l)) = –286,

ΔHθf,298(CO2(g)) = –394,  ΔHθf,298(NaCl(aq, ~1.0 mol dm–3)) = –407,  ΔHθf,298(HCl(aq, ~1.0 mol dm–3)) = –165

 

(ΔHθ1) 2NaHCO3(s)  ===>   Na2CO3(s)  +  H2O(l)  +  CO2(g)

ΔHθr,298 = ∑ΔHθf,298(products) – ∑ΔHθf,298(reactants)

ΔHθr,298 = {ΔHθf,298(Na2CO3(s))  +  ΔHθf,298(H2O(l))  +  ΔHθf,298(CO2(g))} – {2 x ΔHθf,298(NaHCO3(s))}

ΔHθr,298 = {–1131 +   –286  +  –394}  –  {2 x –948)

ΔHθr,298  = –1811 – (–1896)  = ΔHθ1 = +85 kJ mol–1

The thermal decomposition of sodium hydrogencarbonate is endothermic.

 

(ΔHθ2) 2NaHCO3(s)  +  2HCl(aq)  ===>  2NaCl(aq)  +  2H2O(l)  +  2CO2(g)

ΔHθr,298 = ∑ΔHθf,298(products) – ∑ΔHθf,298(reactants)

ΔHθr,298 = {2 x ΔHθf,298(NaCl(aq))  + 2 x ΔHθf,298(H2O(l))  +  2 x ΔHθf,298(CO2(g))}  –  {2 x ΔHθf,298(NaHCO3(s))  +  2 x ΔHθf,298(HCl(aq))}

ΔHθr,298 = {(2 x –407) + (2 x –286) + (2 x –394)} – {(2 x –948) + (2 x –165)}

= –2174 – (–2226) = ΔHθ2 = +52 kJ mol–1    (equivalent to +26 kJ mol–1 per mole of NaHCO3)

Therefore the reaction of sodium hydrogen carbonate with hydrochloric acid is endothermic, so you should see a temperature fall in the calorimeter solution.

 

(ΔHθ3) Na2CO3(s)  +  2HCl(aq)  ===>  2NaCl(aq)  +  H2O(l)  +  CO2(g)

ΔHθr,298 = ∑ΔHθf,298(products) – ∑ΔHθf,298(reactants)

ΔHθr,298 =  {2 x ΔHθf,298(NaCl(aq))  + ΔHθf,298(H2O(l))  +  ΔHθf,298(CO2(g))} – {ΔHθf,298(Na2CO3(s))  +  2 x ΔHθf,298(HCl(aq))}

ΔHθr,298 = {(2 x –407)  +  –286  +  –394} – {–1131  +  (2 x –165)} = –1494 – (–1461) = ΔHθ3 = –33 kJ mol–1

Therefore the reaction of sodium carbonate with hydrochloric acid is exothermic, so you should see a temperature rise in the calorimeter solution.

 

Using the theoretical data from calculations 2 and 3 to simulate 'perfect' experimental results:

ΔHθ1  =  ΔHθ2  –  ΔHθ3  =  +52 – (–33)  = +85 kJ mol–1

which was a great relief after dealing with all those numbers and brackets!


For the next problem to designed!

If anybody has a good suggestion for a pre–university practical, I'd be happy to look at it.

ΔHθ1
  B

ΔHθ2

(c) doc b     (c) doc b

ΔHθ3

C

 


A set of enthalpy calculation problems with worked out answers – based on enthalpies of reaction, formation, combustion

Energetics-Thermochemistry-Thermodynamics Notes INDEX


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See also a set of enthalpy problems to solve, worked out answers given on a separate page!

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