1.3 Experimental methods
for determining enthalpy changes and treatment of results
Page introduction
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!
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
1.3a1 A simple
polystyrene calorimeter of low heat capacity can be used for any
noncombustion 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 solidstate
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.
 Above are
two graphs from simple calorimeter experiments (picture on right).
On the left are typical results from an exothermic
reaction e.g. metal plus acid or metal plus metal salt displacement
reaction.
On the right are typical results from an endothermic
change e.g. when certain salts dissolve in water.
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.
In this case
ΔT (corrected) = 28.4 -
20.4 = 8.0oC
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.
In this case
ΔT (corrected) = 19.5 -
11.2 = 8.3oC
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 dm3 |
0.0 |
0.1 |
0.5 |
0.9 |
1.0 |
| SHC
sodium chloride, J g1 oC1 |
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 g1 oC1 |
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 g1 oC1 |
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 g1 oC1
please send me the link! |
|
|
|
|
|
Note on
other 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.
|
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 gcm1).
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 setup, 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
You need 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 preuniversity 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).
|
|
1.3b Treatment of
experimental results
(general method for any calorimeter)
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 g1 K1) i.e. it takes
4.18 J of heat energy to raise 1g or 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 mol1.
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 |
A  B |
|
ΔHθ2 |
 |
|
 |
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 |
|
A
B |
|
ΔHθ2 |
|
|
 |
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 =
18.6217.14 = 1.48g.
-
Heat given out to the water =
mass of water x SHCwater x temperature change
-
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 kJmol1 (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 kJmol1, 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.
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|
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 mol1 (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 mol1
Since the temperature has fallen, the
change is endothermic
So
ΔHdissolution(NH4NO3)
= +25.8 kJ mol1
(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 mol1
Since the temperature has fallen, the
change is endothermic
So
ΔHdissolution(NH4NO3)
= +25.3 kJ mol1
(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 mol1
Since the temperature has fallen, the
change is endothermic
So
ΔHdissolution(NH4NO3)
= +27.1 kJ mol1
(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 offsets 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).
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|
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 mol1
Since the temperature has fallen, the
change is endothermic
So
ΔHdissolution(KCl)
= +17.1
kJ mol1
(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 mol1
Since the temperature has fallen, the
change is endothermic
So
ΔHdissolution(KCl)
= +17.3
kJ mol1
(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 mol1
(only accurate to 3 sf)
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|
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 dm3 hydrochloric acid and 25.0
of 1.0 mol dm3, 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 mol1
(only accurate to 3 sf)
(ii) Using the SHC of 0.50 molar sodium
chloride solution, which is 4.03 J g1 oC1
(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 mol1 (only accurate to 3 sf)
The data book value is 57.1 kJ mol1
% 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.
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|
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 dm3 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 mol1
Since the temperature rose, this
displacement reaction is exothermic
ΔHdisplacement(Zn + CuSO4)
= 217
kJ mol1
(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 mol1
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 (method 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 |
|
A
B |
|
ΔHθ2 |
|
|
|
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)] |
|
+ aq |
|
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
kJmol1 |
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 mol1
(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 mol1
(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 mol1
|
|
8. Using Hess's Law and
experimental data to determine the enthalpy of hydration of
anhydrous magnesium sulfate (method 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)] |
|
+ aq |
|
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 |
|
|
|
+ 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 mol1
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 mol1 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 g1 oC1.
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 mol1
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 |
|
|
|
+ 2HCl(aq)
ΔHθ3 |
|
2NaCl(aq) + 2H2O(l)
+ 2CO2(g) |
|
∴ ΔHθ2
= ΔHθ1 + ΔHθ3 and ΔHθ1
= ΔHθ2 ΔHθ3 |
Theoretical calculations of ΔHθ13
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 dm3
(it varies with concentration)
ΔHθf,298
data (all in kJ mol1): 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 dm3)) = 407, ΔHθf,298(HCl(aq,
~1.0 mol dm3)) = 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 mol1
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 mol1 (equivalent to
+26
kJ mol1 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 mol1
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 mol1
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
preuniversity practical, I'd be happy to look at it.
Enthalpy calculation problems with worked out answers based on
enthalpies of reaction,
formation, combustion
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QUICK INDEX for
Energetics:
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 13 here
* Part 1ab
ΔH Enthalpy Changes
1.1 Advanced Introduction to enthalpy changes
of reaction,
formation, combustion
etc. : 1.2a & 1.2b(i)(iii)
Thermochemistry Hess's Law and Enthalpy
Calculations reaction, combustion, formation etc. : 1.2b(iv)
Enthalpy of reaction from bond enthalpy
calculations : 1.3ab
Experimental methods
for determining enthalpy changes and treatment of results and
calculations :
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 BornHaber cycle : 2.1ac What happens when a
salt dissolves in water and why? :
2.1de Enthalpy
cycles involving a salt dissolving : 2.2ac
The
BornHaber Cycle *** Part 3
ΔS Entropy and ΔG Free Energy Changes
: 3.1ag 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 : 3.4ad
More on ΔG,
free energy changes, feasibility and
applications : 3.5
Calculating Equilibrium
Constants from ΔG the free energy change : 3.6
Kinetic stability versus thermodynamic
feasibility - can a chemical reaction happen? and will it happen?
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