Simon Johnson
Wednesday, 3 June 2009
Year 13 Chemistry Practical
DCP, CE
Hess’ Law
Aim
MgSO4 (s)
+
7H2O (l) ®
MgSO4 . 7H2O (s)
Background
anhydrous MgSO4 in water and hydrated MgSO4 . 7H2O in water, and
Hess’s Law.
aq
MgSO4 (s)
then use
MgSO4 (aq)
1
7H2
2
3
aq
MgSO4 . 7H2O (s)
Equipment
Plastic Coffee Mug
Hydrated MgSO4.7H2O
100 mL Measuring Cylinder
Data Logger with Temperature Probe
Anhydrous MgSO4 (in dessicator)
Method
1.
Use the 100 mL measuring cylinder to measure 100.0 mL of water into
a plastic coffee mug and record the temperature.
2.
Accurately weigh about 7.5 g of anhydrous MgSO4.
3.
Add this to the water and stir with the temperature probe.
4.
Record the temperature change using the data logger.
5.
Repeat the whole procedure using 15.5 g of hydrated MgSO4.7H2O
Theoretical Enthalpy Change = -100 kJ mol-1
Enthalpy of Hydration for anhydrous magnesium sulfate = -84 kJ mol-1
Enthalpy of Hydration for hydrated magnesium sulfate = +16 kJ mol-1
(Experiments in Seventm Form Chemistry – a Laboratory Manual. Dept
Education New Zealand 1978)
Data
of
Simon Johnson
Wednesday, 3 June 2009
A Data Table Showing the Relationship between Time Elapsed
and the Temperature of the Reaction of both Hydrous and Anhydrous Magnesium Sulphate
Time
Sec, ±0.05sec
Anhydrous MgSO4
Temperature
°C, ±0.2°C
0
0.5
1
1.5
2
2.5
3
3.5
4
4.5
5
5.5
6
6.5
7
7.5
8
8.5
9
9.5
10
10.5
11
11.5
12
12.5
13
13.5
14
14.5
15
15.5
16
16.5
17
17.5
18
18.5
19
19.5
20
13.30
13.30
13.30
13.27
13.20
13.15
13.10
13.00
12.95
12.87
12.80
12.75
12.67
12.59
12.54
12.47
12.39
12.34
12.27
12.27
12.29
12.29
12.34
12.34
12.34
12.32
12.29
12.32
12.37
12.44
12.49
12.54
12.57
12.62
12.72
12.77
12.80
12.80
12.85
12.87
12.90
16.49
16.51
16.63
16.90
17.31
18.24
19.00
19.24
19.19
19.71
20.82
21.61
22.78
24.87
26.31
26.84
27.72
28.16
27.58
26.33
25.45
24.71
24.22
23.89
23.87
23.78
23.57
23.41
23.27
23.15
23.01
22.85
22.71
22.69
22.62
22.55
22.52
22.57
22.64
22.71
22.78
Change in Temperature =12.27-13.30
(±0.2°C + ±0.2°C) =-1.03
±0.4°C
Initial Temperature,
°C, ±0.2ml
Initial Weight, g, ±0.001g
Initial Volume, ml, ±0.5ml
Hydrated MgSO4
Temperature
°C, ±0.2°C
=28.16-16.49
=11.67
13.30
16.67
7.500
100.
15.523
100.
Simon Johnson
Wednesday, 3 June 2009
Observations: There was no visible colour change in either reaction: the water remained colourless. In the
anhydrous reaction, a white precipitate was formed. No precipitate was formed in the other reaction.
Note that data has been rounded to 4 significant figures. Whilst the data was supplied by the data logger
was to 8 significant figures, such a level of accuracy is futile, considering the random error on the
thermometer is ±0.2. 4 significant figures was therefore chosen for tabulation.
Calculations
The enthalpy change between MgSO4 (s) + 7H20 and MgSO4 . 7H2O (s) (hereafter ΔH2), is impossible to
determine directly. Hence an intermediate step must be applied according to the diagram below
aq
MgSO4 (s)
MgSO4 (aq)
ΔH1
7H2O
ΔH2
ΔH3
aq
MgSO4 . 7H2O (s)
For this calculation, ΔH1 = ΔH2 + ΔH3 according to Hess’ law. As we can determine ΔH1 and ΔH3, we will be
able to determine ΔH2 by rearranging.
Reaction One: hydrating Anhydrous MgSO4
For this, and the next calculation, it is required that we assume that the mass of one millimetre of water to
be one gram. That is, that the density of water is 1 g/ml-1. I have therefore converted into kilograms below
without further comment.
ΔH1
= mass of water x specific heat capacity x change in temperature
= 0.1 x 4.18 x 11. 56
= 4.43…. kJ
= ± 0.5% + 3.46….%
= ±3.96…..%
The next step is to determine the amount of moles that reacted.
Moles (MgSO4) = mass(MgSO4) / molar mass (MgSO4)
Moles (MgSO4) = 7.500 / 120
=0.00625
From this, we can say that
ΔH1
= 4.43…. kJ per 0.00625 mol
=-77.7… kJ per 1 mol
=±0.0133….%
=±3.96…%+ 0.013…%
=±3.977…%
Note that we can say that it is exothermic due to the gain in temperature and hence has a negative state
symbol.
Simon Johnson
Wednesday, 3 June 2009
Reaction Two: aquifying Hydrated MgSO4
ΔH3
= mass of water x specific heat capacity x change in temperature
= 0.1 x 4.18 x1.03
= 0.43054 kJ
= ± 0.5% + 38.8%
= ±39.3…%
The next step is to determine the amount of moles that reacted
Moles (MgSO4.7H20)
= mass(MgSO4.7H20)) / molar mass (MgSO4.7H20))
Moles (MgSO4)
= 15.523 / 246.51
=0.006…mol
=±0.006….%
From this, we can say that
ΔH3
= 0.43…. kJ per 0.006… mol
=±39.3…%+ 0.006…%
=±39.3…%
=+6.83… kJ per 1 mol
Note that we can say that it is endothermic due to the loss in temperature and hence has a positive state
symbol.
The next thing is to calculate the value of ΔH2. Remembering that ΔH2 + ΔH3 = ΔH1 and so, ΔH1 - ΔH3 = ΔH2
ΔH2
= ΔH1 - ΔH3
=-77.7 kJ - 6.83…
=±39.3…% + ±3.977…%
=±43.2…%
-1
=-84.5 kJ mol (3sf) ±40kJ (1sf)
Percentage Deviation
The accepted literature value for reaction ΔH2 is -100kJ mol-1. 1Accordingly, we can calculate the percentage
deviation of our final answer:
Further, given the literature value of ΔH1 of -84 kJ mol-1 and an accepted literature value of of ΔH3 of +16 kJ
mol-1, we can calculate the percentage deviation of the other two reactions.2
ΔH1
ΔH3
Data Comparison
The final table shows a comparison of results, attained by individual groups. The methodology and
execution of the experiment was essentially identical. Hence, this data can be used for a comment on the
precision of this methodology in calculating the enthalpy change. Note that the random error will be
identical, as the method was identical.
1
2
3
4
5
1
2
ΔH1
-55.0
-77.7
-54.3
-53.9
-54.2
ΔH3
+8.07
+6.83
+7.29
+9.26
+5.30
ΔH2
-63.1
-84.5
-61.5
-63.2
-64.3
Experiments in Seventh Form Chemistry- a Laboratory Manual. Dept of Education New Zealand, 1978
Both ibid.
Simon Johnson
Wednesday, 3 June 2009
Conclusion
The aim of this experiment was to determine experimentally the enthalpy change for the following reaction:
MgSO4 (s)
+
7H2O (l)  MgSO4 . 7H2O (s) and hence to compare this to the literature value. As it
is not possible to do this directly, Hess’ law was used to calculate the reaction indirectly. We were able to
calculate the enthalpy change for this reaction to be -83.4kJ mol-1. This compared to a literature value of 100kJ mol-1, a percentage deviation of 15.5%. There are two conclusions that can be drawn from this: firstly,
that this reaction is an exothermic reaction, secondly, that our experiment has problems and errors in it that
caused the deviation from the literature value.
How can this be explained scientifically? Breaking a bond is an endothermic reaction: it requires energy
from the surroundings. Conversely, forming a bond is an exothermic reaction. The enthalpy change is the
measure of this energy change, which is calculated by the mass of the substance multiplied by the specific
heat capacity of the substance multiplied by the specific heat capacity of water. Here, a negative enthalpy
change can explain the fact that the reaction which we are investigating is exothermic.
Some comment is necessary on our data.. Our result was reasonably consistent with the literature value,
giving a percentage deviation of 15.5%, but the total error was considerable: 43.2%. This suggests that the
methodology should be reviewed to reduce the total percentage error to ensure a result closer to the
literature value.
The other comment is required on the precision of the data. This experiment, owing to logistical problems,
could not be repeated several times. Instead, the values can be compared to other groups who undertook
experiments under identical conditions. In theory, these results should be the same. As can be seen from
the final table above, the results are reasonably precise, 4 of the 5 groups achieved roughly similar results.
The differences in this can be explained by random error in this experiment: a considerable source of error
for our results. A change in the experiment should initially concentrate on reducing the random error, a
factor that was considerably more important to this experiment than the systematic error of the
experiment.
Evaluation
There were two principle sources of error in this experiment: random error that was carried through by error
propagation and from systemic error, most notably due to heat loss.
One key source of systemic error was the heat lost in this experiment. The experiment was conducted using a plastic
coffee mug. This is considerably more insulated than other forms of container that have been used. However, it is not
perfect and some heat will have been lost, most noticeably through the process of opening and closing the beaker’s
lid, and through the stirring rod. This would have been seen in the final result: if the reaction in ΔH 3 had lost less heat,
the change in temperature would have been greater, and hence the enthalpy change would have been bigger. This
would equally apply to ΔH2 .
Such a problem is impractical to attempt to deal with under experimental conditions. An insulated beaker is as
insulated as is practically possible. The process becomes one of control. If we undertook the experiment in an object of
known heat capacity, we would be able to calculate the energy lost. This could be added to the enthalpy change
already calculated, giving a more accurate result.
The other key aspect of systemic error is the problem in stirring. This was a key aspect of the methodology: by stirring
the reaction, we ensured that the reaction reacted faster. This had the effect of reducing the duration of the reaction,
and hence the amount of heat lost to the surroundings. This had a noticeable effect on the data in a similar vein to as
been already described: if reaction ΔH2 had lost less heat to the surroundings, the change in temperature would have
been greater, and hence the enthalpy change bigger overall. This would especially help to explain the disparities
between each group. Stirring is a subjective descriptor, and one group could have stirred much more vigorously than
the other. This would have had the effect described previously of decreasing the final enthalpy change.
Simon Johnson
Wednesday, 3 June 2009
Fortunately, this can easily be rectified. The process of stirring can be replaced by a magnetic stirrer. If this was done,
the magnet would stir the solution an equal amount of times in every experiment. This would help to create a much
more precise set of results for our data.
The key cause of random error can be seen clearly in this experiment. In order to calculate the total error percentage,
the percentage error of each step of the reaction was required. The percentage error of the change in temperature
was ±0.4°C. Such a value is reasonably low. However, it can generate a very high percentage error when small raw
numbers are used. The calculation of the percentage error of the change in temperature in ΔH 3 produced such an
error. Because the raw value is so low, the percentage error was considerable at nearly 40% and was then propagated
throughout the experiment. The effect of this was to artificially increase the random error in the experiment
considerably.
A solution is very hard to determine for this. The experiment requires the change in temperature value determined,
and error propagation requires a percentage error. The most effective solution is to suggest that this problem casts
doubts upon the usefulness of the total percentage error, and that it should be used with other methods of statistical
analysis, for example, percentage deviation and the precision of the results, when we assess the accuracy of the
experiment.
A comment on the equipment used is important. For this experiment, we used a measuring cylinder, scales and a data
probe to gain our results. The scales and data probe have a very low percentage error on them naturally. The
measuring cylinder has a much greater one. The random error on a measuring cylinder is ±0.5ml. This is a reasonably
large error, and will have the effect of increasing the total percentage error considerably. Because of this, the
confidence that we can have in our final answer is decreased.
Again, a simple step can be taken to reduce the percentage error in this respect. There is no reason that a measuring
cylinder has to be used. The water could be measured using a pipette which has a significantly lower percentage error.
However, as a pipette can only hold 25ml, four pipettes would be required; quadrupling the percentage error. More
effectively the water could be weighed. Assuming that one millilitre of water weighs one gram, the gravimetrical scales
could be effectively used to reduce the extent of the error. The scales have a much lower random error, and so we
would reduce the total percentage error significantly. Whilst the assumption that one millimetre of water weighs a
gram requires standard temperature and conditions for the density to remain accurate, that should not be a practical
problem for our experiment.
One key error came in this experiment during the calculation of individual enthalpy changes. For both ΔH 1
and ΔH3 the literature value was greater than the value which we found. This can be explained by a key assumption.
We assumed in the process of calculation that the mass of the water reacting was 0.1kJ. This is correct; however, it
does not take into account the fact that both masses were greater than 0.1kJ when the reacting MgSO4 is added. Both
of the enthalpy values were smaller than expected because of this problem: had it been included, the enthalpy
changes would have been greater. The percentage deviation in ΔH3 was greater than the percentage deviation for ΔH1
because of the difference in masses. The mass of MgSO4 in ΔH3 was double that of ΔH1. This helps to explain the
greater percentage deviation: the greater mass in ΔH3 had a greater effect on the enthalpy change than the smaller
mass in ΔH1.
The solution to this is reasonably simple. Instead of assuming 0.1kJ of water, we could use the combined weights of
water and the MgSO4. This could be done without further work: the weight of the MgSO4 is already gained in the
experiment to calculate the moles. Whilst this would increase the total percentage error slightly, such a drawback
would be worth it for the benefit to the experiment of a much more accurate set of enthalpy changes.
However, one further error came during the process of calculation. We assumed the specific heat capacity of the
solution to be 4.18 kJ, the same to that of distilled water. There are two errors in this assumption: firstly that the water
was distilled when it wasn’t. The specific heat capacity of impure water is greater than that of distilled water. As we
used impure water but the heat capacity of distilled water, a smaller enthalpy change than expected was gained. To
solve the problem of impure water is simple. The method should be simply changed to the use of distilled water;
making the specific heat accurate and meaning that the enthalpy change is more consistent with the literature value.