Laboration 2: Thermal analysis of boric acid (B(H) 3 ) Henrik Bergvall Berglund, William Sjöström, Uppsala 2016-09-21 Course: Solid State Chemistry Mentors: William Brant, Johan Cedervall, Dennis Karlsson Summary: The purpose of the laboration was to study TG- and DTA-plots for boric acid to conclude what happens to the structure and composition between 0 and 500 C. The program IGR was used to collect data from the plots. The conclusion was that for the pressure that the test was performed in, boric acid decomposes in three steps and releases water vapour, H 2 (g). B 6 is formed in the middle step and the final product becomes boron oxide B 2.
1 Introduction ne way to identify how the composition of a sample changes when the temperature varies is with thermogravimetry (TG). Thermograimetry means measuring how the sample s mass changes when the temperature is variated, the mass is usually measured as a function of temperature (but it could also be measured as a function of time). TG is often combined with differential thermal analysis (DTA), which means measuring how the sample s temperature changes compared to an inert substance (reference). Since the sample and reference is supplied with the same amount heat (energy), a diffrence between their temperatures must correspond to a change of structure or composition of the substance. By comparing if the temperature of the sample is cooler or warmer than the reference, it can also be determined if the change is endotherm or exotherm (change of enthalpy). 2 Purpose The purpose of this laboration was to examine a DT/TGA-plot of boric acid, B(H) 3. 3 Method In this laboratory, the TG-DTA-plot in figure 1 was given. It is known that the plot is for a sample of boric acid,b(h) 3, that was heated from 0 to 500 C, the pressure at which the test was performed is unknown. By studying the TG-curve, it is possible to determine at which temperatures the mass of the sample has changed and by how much. This gives the temperatures at which the sample changes structure. By comparing the changes of mass of the sample to the mass of possible products (substances with different structure and composition), the products can be determined. Since boiling is an endothermic process and melting is a exothemic process, the DTA-curve can verify these guessed products if their boiling/melting-points are known. The program IGR was used to analyse data from the TG-DTA-plot. 1
4 Result and discussion The given TG- and DTA-plot of boric acid, B(H) 3, can be seen in figure 1. A zoomed in version of the same plot can be seen in figure 2. Figure 1. The given TG- and DTA-plot for boric acid B(H) 3. Green line is TG-plot, red line is DTA plot and black line temperature. The numbers 1-5 are reference-numbers to important events; At 1, there is huge loss of mass combined with a very endothermic reaction. At 2, there is a small loss of mass combined with a endothermic reaction. At 3-4, there is loss of mass combined with a small endothermic reaction. At 5, there is no change in mass but a endothermic reaction. At 6 there is no change in mass by a exothermic reaction. Figure 2. A zoomed part of figure 1, please note that the values on the time axis is *10 3. 2
By studying figure 1 and 2, it was concluded that the TG-curve (green) have 3 areas of interest, marked in the figures as reference-points 1, 2 and 3. All these three peaks corresponds to reducing of the sample s mass together with an endotermic reaction according to the DTA-curve, and thus a change of the sample s composition. There are also two peaks on the DTA-curve that does not correspond to any peaks on the TG-curve, see reference points 5 and 6 in figure 1. Thus, at these two points, the sample has a change in enthalpy but no change in mass. Data from these points were collected using the program IGR and noted in table 1. Table 1. Data collected from the plot in figure 1. The program IGR were used to collect the data. Note the entaply change in point 5 and 6 is difficult to measure so these have high uncertanty. Temperature / Reference point in figure 1 and/or 2. Masschange, Δm / mg Enthalpy change / µvs 86 1 0 0 86-155 2-14.13-23183,1 155-190 3-2.51-6871.53 190-268 4-3.13-1714.94 419-419 5 ~0-1912.69 419-20 6 0 +7726.37 From the data in table 1, the individual steps of change of structure and/or composition can be studied together with figure 1 and 2: 1. The first step of composition starts at 86 and is linked to the endotherm DTA-peak at reference point 1, where the sample start to decompose. This decomposition is linked to the highest change of enthalpy and also the highest loss in mass. Later in this report it is concluded that this enthalpy change corresponds to a release of 3H 2 molecules. 2. The second step starts at 155 and is linked to the endotherm DTA-peak at reference point 3. This decomposition is linked to the lowest change of mass. And this is roughly 1 3 of the enthalpy at point 1. Later in the report, it is concluded that this change in enthalpy is due to the release of one H 2 molecule. This explains why this enthalpy change is roughly 1 3 of the enthalpy change at point 1 above, where 3H 2 molecules are released. 3. The third step starts at 190 and is linked to the endotherm DTA-peak at reference point 4. 4. The fourth step starts at 419, and is linked to the endothermic DTA-peak at reference point 5. There is no peak in the TG-curve, thus the mass does not change in this step. Since the mass does not change but there is a endothermic process, this 3
is probably where the sample starts to melt. Note that the endothermic value in table 1 is hard to measure accurately from figure 1, however, it can easily be seen that the value is negative (which is the only important thing for this discussion). 5. The final step starts when the temperature is lowered again. This step is linked to the exothermic DTA-peak at reference point 6, but there is no mass change. Since there is no change of mass but an exotherm process, this is probably where the sample starts to solidify again. 4
To identify what these changes of composition could have been, the masschange, noted in table 1, was compared to probable compositions and the best fitting were noted in table 2. Table 2.The compositions in this table has been normalised to one B. Δ m for HB 2 was calculated by: m n M n M Δ = HB2 HB 2 B(H)3 B(H) 3 Δ m for BH 11/6 4/6 and B 3/2 were calculated in the same way but with their values for n and M. It was concluded that the substance decomposites in three steps: 1. 3x(B(H) 3 ) H 3 6 + 3H 2 2. 2x(H 3 6 ) B 6 + H 2 3. B 6 3x(B 2 ) + 2H 2 The structure of each step is illustrated in figure 3. Figure 3. Illustation of the structure of each decomposition step. Table 3. Melting points for substances examined Composition/Substance Melting point [ C] Boiling point [ C] B(H) 3 d169 [1] (decomposes) - 5
H 2 0 100 HB 2 176 [3] - H 5 - - B 2 3 450 [1] 2065 [1] Table 3 shows melting points for boric acid and the decompose substances. Together with this and table 1, the first reaction happens at 86 where water is released and boric acid decomposed. But water has a boiling point at 100 which don t correlate. This is probably due to the pressure at the which the experiment was performed. All reactions happen before their standard decomposition temperatures, which could be explained by the pressure lower than 1 atm. There was no information about what pressure this experiment was performed in. The decomposing temperature for boric acid B(H) 3 and metaboric acid HB 2 are very close. This correlate to the two close peaks in enthalpy change at reference points 2 and 3 in figure 2. The melting point for B 2, 450 C, proves that this is a possible product, since the sample starts to melt at 419 C as discussed earlier. The small temperature difference here can be explained by the unknown pressure at which the experiment was performed. 5 Conclusions By using TG together with DTA this laboratory shows that boric acid decomposes in steps while releasing water in differerent amount corresponding to entalpy change. 1. 3x(B(H) 3 ) H 3 6 + 3H 2 ΔH=-23183,1 µvs 2. 2x(H 3 6 ) B 6 + H 2 ΔH=-6871.53 µvs 3. B 6 3x(B 2 ) + 2H 2 ΔH=-1714.94 µvs And thus our final solid product is boron oxide B 2. 6 References [1] G. Aylward, T. Findlay, SI Chemical Data, John Wiley and Sons, Milton (2014). [2] Zachariasen, W.H. The crystal structure of monoclinic metaboric acid, Acta Crystallographica (1963), 16, p385-p389 [3] Dale L. Perry, Handbook of Inorganic Compounds, Second Edition (2011), p72 6