Analysing; Enthalpy of Decomposition of Sodium Hydrogencarbonate
Skill 3: Analysing; Enthalpy of Decomposition of Sodium Hydrogencarbonate
Procedures and results
Sodium Carbonate experiments
Chemical
Mass of Cup
(grams)
Mass of solid + cup
(grams)
Mass of solid
(grams)
Na2CO3
.041
3.541
2.500
Na2CO3
.114
3.514
2.500
Na2CO3
0.935
3.435
2.500
Table of results
Time
Experiment 1
Experiment 2
Experiment 3
Temp. of solid
(0C)
Temp. of solution (0C)
Temp. of solid
(0C)
Temp. of solution (0C)
Temp. of solid
(0C)
Temp. of solution (0C)
0
21.2
21.0
21.2
20.0
21.2
20.0
-
21.0
-
20.0
-
20.0
2
-
21.0
-
20.0
-
20.0
3
-
21.0
-
20.0
-
20.0
4
-
-
-
-
-
-
5
-
25.5
-
25.0
-
25.0
6
-
25.0
-
24.4
-
24.0
7
-
24.8
-
24.2
-
24.0
8
-
24.6
-
24.0
-
23.8
9
-
24.5
-
24.0
-
23.8
0
-
24.5
-
23.9
-
23.6
Analysis
Mr. of Sodium Carbonate = 2 x Ar (Sodium) + Ar (Carbon) + 3xAr (Oxygen)
= 2 x 23.0 + 12.0 + 3 x 16.0
= 106
Moles of sodium carbonate used = mass of solid used
Mr
= 2.500/106
= 0.0235 mol
Mass of acid = volume x density
= 30.00 x 1
= 30.0 grams
Mass of Solution = mass of acid + mass of solid
= 30.0 + 2.500
= 32.5 grams
The same amount of acid and solution was used throughout the 3 experiments
Experiment 1
(T = T2 - T1
= 25.61 - 21.0
= 4.6 0C
Energy transferred to surrounding = m x c x (T
= 32.5 x 4.2 x 4.6
= 627.9 Joules
Internal energy change = - 627.9 Joules
Enthalpy change of neutralisation = Internal energy change
No. Of moles of solid used
= -627.9/0.0235
= - 26719.15 Joules/mol
= - 26.7 KJ/mol
Experiment 2
(T = T2 - T1
= 24.9 - 20.0
= 4.9 0C
Energy transferred to surrounding = m x c x (T
= 32.5 x 4.2 x 4.9
= 668.85 Joules
Internal energy change = - 668.85 Joules
Enthalpy change of neutralisation = Internal energy change
No. Of moles of solid used
= -688.85/0.0235
= - 29312.76 Joules/mol
= - 29.3 kJ/ mol
Experiment 3
(T = T2 - T1
= 24.8 - 20.0
= 4.8 0C
Energy transferred to surrounding = m x c x (T
= 32.5 x 4.2 x 4.8
= 655.2 Joules
Internal energy change = - 655.2
Enthalpy change of neutralisation = Internal energy change
No. Of moles of solid used
= - 655.2/0.0235
= - 27880.85 J/mol
= - 27.9 KJ/mol
Average results = (result 1 + result 2 + result 3)/3
= (- 26.7 + - 29.3 + - 27.9)/3
= - 28.0 KJ/mol
Therefore, for further use to find the enthalpy of decomposition of Sodium Hydrogencarbonate, we will use the average value of -28.0 KJ/mol as the value of enthalpy of neutralisation of Sodium Carbonate.
Sodium Hydrogencarbonate experiments
Chemical
Mass of Cup
(grams)
Mass of solid + cup
(grams)
Mass of solid
(grams)
NaHCO3
.240
4.740
3.500
NaHCO3
.070
4.570
3.500
NaHCO3
.129
3.629
3.500
Table of results
Time
Experiment 1
Experiment 2
Experiment 3
Temp. of solid
(0C)
Temp. of solution
(0C)
Temp. of solid
(0C)
Temp. of solution
(0C)
Temp. of solid
(0C)
Temp. of solution
(0C)
0
21.1
20.0
21.0
20.0
21.1
9.9
-
20.0
-
20.0
-
20.0
2
-
20.0
-
20.0
-
20.0
3
-
20.0
-
20.0
-
20.0
4
-
-
-
-
-
-
5
-
3.2
-
2.4
-
2.0
6
-
3.5
-
2.8
-
2.4
7
-
3.8
...
This is a preview of the whole essay
20.0
21.0
20.0
21.1
9.9
-
20.0
-
20.0
-
20.0
2
-
20.0
-
20.0
-
20.0
3
-
20.0
-
20.0
-
20.0
4
-
-
-
-
-
-
5
-
3.2
-
2.4
-
2.0
6
-
3.5
-
2.8
-
2.4
7
-
3.8
-
3.0
-
2.9
8
-
4.0
-
3.5
-
3.2
9
-
4.2
-
3.8
-
3.5
0
-
4.4
-
4.0
-
3.8
Analysis
Mr. of NaHCO3 = Ar (Sodium) + Ar (Hydrogen) +Ar (Carbon) + 3xAr (Oxygen)
= 23.0 + 1.0 + 12.0 + 3 x 16.0
= 84
Moles of NaHCO3 used = mass of solid used
Mr
= 3.500/84
= 0.042 mol
Mass of acid = volume x density
= 30.00 x 1
= 30.0 grams
Mass of Solution = mass of acid + mass of solid
= 30.0 + 3.500
= 33.5 grams
The same amount of acid and solution was used throughout the 3 experiments
Experiment 1
(T = T2 - T1
= 13.0 - 20.0
= -7.0 0C
Energy transferred to surrounding = m x c x (T
= 33.5 x 4.2 x -7.0
= - 984.9 J
Internal energy change = 984.9 J
Enthalpy change of neutralisation = Internal energy change
No. Of moles of solid used
= 984.9/0.042
= 23450 J/mol
= 23.5 kJ/mol
Experiment 2
(T = T2 - T1
= 12.1- 20.0
= -7.9 0C
Energy transferred to surrounding = m x c x (T
= 33.5 x 4.2 x -7.9
= -1111.53 J
Internal energy change = 1111.53 J
Enthalpy change of neutralisation = Internal energy change
No. Of moles of solid used
= 1111.53/0.042
= 26465 J/mol
= 26.5 kJ/mol
Experiment 3
(T = T2 - T1
= 11.7 - 20.0
= -8.3 0C
Energy transferred to surrounding = m x c x (T
= 33.5 x 4.2 x -8.3
= - 1167.81 J
Internal energy change = 1167.81 J
Enthalpy change of neutralisation = Internal energy change
No. Of moles of solid used
= 1167.81/0.042
= 27805 J/mol
= 27.9 kJ/mol
I decided to disregard the first experiment, because the results are not concordant, probably due to the factor that I used a different acid for both experiment 2 and 3, as the acid I used in experiment 1 was all used up.
Average results = (result 1 + result 2)/2
= (26.5 + 27.9)/2
= 27.2 kJ/mol
Thus, we use the value of 27.2 kJ/mol as the value of enthalpy of neutralisation of Sodium Hydrogencarbonate.
Calculations of enthalpy of decomposition
We can calculate the value of enthalpy of decomposition using the Hess's law, which states 'the enthalpy change of a reaction depends only on the initial and final states of the reaction and is independent of the route by which the reaction may occur'. Thus, it can be concluded, that we can choose other paths to reach the final product, through the following energy diagram:
2 NaHCO3 (s) Na2CO3 (s) + H2O(l) + CO2(g)
2 NaCl + 2 H2O + 2 CO2
By using both the enthalpy of neutralisation of Sodium carbonate and Sodium hydrogencarbonate, we can ultimately find the enthalpy of decomposition of Sodium hydrogencarbonate. We do not need to determine the neutralisation of water and carbon dioxide, because they are already neutral.
Thus we can determine the value of the enthalpy of decomposition by using the following equation:
Enthalpy decomposition of 2 moles = 2 x enthalpy of neutralisation of sodium hydrogencarbonate - enthalpy of neutralisation of sodium carbonate
= 2 x 27.2 - (-28.0)
= 82.4 kJ/mol
Therefore the enthalpy of decomposition of 1 mol = 82.4/2
= 41.2 kJ/mol
From my experiment, it can be concluded that the enthalpy of decomposition of sodium hydrogencarbonate to be 41.2 kJ/mol.
Maximum error of reading an apparatus = maximum error x 100
Amount read
Maximum error of balance = 0.1/(the "difference" of 2 measurements) x 100
= 0.1/(1.0) x 100
= 10.0 %
As per worksheet. However, in reality we used a more accurate balance that reads to 0.0001, thus, the reality maximum error of balance = 0.0001/1.0 x 100
= 0.01%
Maximum error of burette = 0.15/30 x 100
= 0.50 %
Maximum error of thermometer = 0.1/(the "difference" of 2 measurements)
= 0.1/(24.5-13.8)
= 0.93 %
Maximum error from apparatus reading = sum of maximum errors
= 10.0 % + 0.50% + 0.93%
= 11.43 %
There are other possible unquantified errors that could have interfered with the reaction; one main one would be heat loss. Heat loss cannot be quantified, as cannot be the methods in which the graphs were extrapolated, thus reduces the inaccuracies of this experiment. There could have also been substances, in particular solids that were left in the washing cup, despite the fact that I tried to use all the solid and use the "wash" the cup with some acid that I had previously extracted from the given amount of 30.00 ml.
Skill 4: Evaluation; Enthalpy of Decomposition of Sodium Hydrogencarbonate
Look at your graphs and comment on your results. Are you lines of best fit good enough for you to extrapolate the confidence? Are there any anomalous results? Suggest 1 possible reason for your anomalous result even if your results are within tolerance!
I have attempted my best to extrapolate the line of best fit, by not actually drawing the line of best fit, but calculating and predicting it statistically using the 'regression line' equation. Calculating the regression line includes plugging in all the obtained results in to the following equation:
Y = a +bx (where y and x are the axis's)
B = ((xI - (x)(yI - (y)
((xI - (x)2
a = (y - B(x
Where (y = average of all y values
(x = average of all x values
xI = x value
yI = y value
thus, we can find the equation of the regression line (the line of best fit) and not only accurately extrapolate, we can accurately estimate or predict the value at a certain point using the line equation, by plugging in different values of x. However, the regression line could only have been accurate for the given data, and anything outside the given data could have been out of the trend, thus, we cannot say that the estimation from the regression line is correct, we can only minimise the inaccuracies from extrapolating the regression line.
I have, as I did the experiment thrice, 1 anomalous result, which is the first experiment with Sodium hydrogencarbonate, which I decided not to use. The unconcordancies of the results may have been a consequence of losing some of the powdered solid, which I saw flying around, and thus not included in the reaction mixture. However, even within an experiment there were anomalous results, which could have been due to reading the temperature incorrectly. Or the fact that the solid given was not always fine, and some were lumpy. Despite the fact that I had attempted to grind the lumps into a finer powder, some remained lumpy, thus the anomalous results could have been the lumps reacting with the acids, giving a sudden increase in energy that is released to the surroundings, thus an anomalous temperature reading.
A data source is gives the enthalpy change of decomposition of sodium hydrogencarbonate as +91.6 kJ/mol. Calculate the difference between your value from part 7 of the analysis and the data source value. Express this as a percentage of the data source value!
Difference of my value with the data = 91.6 - 41.2 kJ/mol
= 50.4 kJ/mol
Percentage of data value = difference x 100
Data value
= 50.4/91.6 x 100
= 55.0%
Assuming that the data source value is correct comment on the magnitude of the difference between the data source value and your value
There is quite a considerable difference in the magnitude of my results with the data source. An inaccuracy of 55.0% is far more than the "allowed" error of 11. 43%, which would be due to reading error. Thus we can conclude that the main error in this experiment is not laid on the reading error.
The considerable difference of the values also can make the data, somewhat of less reliability, as the value is far off, but this could have been due to the fact that the given value of the enthalpy of decomposition was using the standard value, thus in the standard state of 298 K of temperature and 1 atm of pressure. Clearly, the conditions in the lab was different from these values, thus resulting in a value that is different, and somewhat incomparable to the data value.
Identify the main source of error in this experiment; suggest one improvement to minimise this main source of error
One main source of error, is undoubtedly, heat loss. Despite the fact that I had tried to minimise heat loss further by covering both the plastic cups and beaker and provided a lid with aluminium foil, heat loss would be inevitable in this reaction. Heat loss, unfortunately is one factor of the reaction that cannot be quantified, and thus we do not know or cannot predict how much of the inaccuracy was due to heat loss, and thus increases the unreliability of this experiment. One way that we could further the heat loss is by using a calorimeter, or a polystyrene cup and lag the calorimeter, to further prevent heat loss.
Identify one other source of error in this experiment. Suggest one improvement to minimise this other source of error.
The reading of the thermometer could have been misleading. The actual thermometer only measures to 0.5 0C, and we were supposed to estimate each measurements to 1 0C, which was not easy, thus leading to a large possibility that I misinterpreted the readings from the thermometer (the results we have are only estimates, not the actual figures). In order to improve and make the temperature more accurate we could have used a digital thermometer, which accurately measures to 0.10C. however, due to unfortunate circumstances, we only had 1 digital thermometer at school, therefore we couldn't have used it, because it would take up too much time to share it.
Another apparatus that could have interfered with the results of the experiment is the use of the balance. Although we used a balance that reads to 0.0001 g, on the calculations of maximum errors, we were asked to use the figure of error to 0.1 g, therefore the calculations of the maximum error could be misleading.
My high increase in temperature could have been due to the difference between the temperature of the acid and the base, and therefore, even before the reaction started, the initial temperature of the solution could be higher than its actual temperature calculated, the temperature of the acid. We could avoid this by trying to start the experiment with the same temperature of the acid and the base. Also, as I measured 2 things, different thermometer gives out different results, thus it was hard to determine whether the temperature difference would be acceptable. It could be avoided by using the same temperature, but not forgetting to wipe the acid or bases off (wash and clean the thermometer before measuring another chemical).
The timescale of that was involved could have also been misleading, we recorded the temperature every minute, while the reaction could have occurred for a scale of, probably 15-20 seconds, thus, it would probably be better if we had recorded the temperature at a shorter time scale difference, over a longer period. Because the timescale we used may have bee too short, and maybe by using a longer time span, for example, 15 minutes, we could have mead the patterns and trends of the cooling curve more apparent.
There could have also been substances, in particular solids that were left in the washing cup, despite the fact that I tried to use all the solid and use the "wash" the cup with some acid that I had previously extracted from the given amount of 30.00 ml. A way of avoiding this is to ensure that the weighing cup is properly rinsed with the acid, and that the entire solid is included in the reaction mixture.
Other ways of improving this experiment would include the use of a pipette instead of a burette to measure out the acid, as it gives off a more accurate reading. Using a higher concentration of both acids and bases, to ensure reaction to occur, and the use of more standard conditions in the lab.
Estimated using regression line equation- see appendix A
