I have always wondered about how stuff works and what is the science behind it. Observing incidents and phenomenon in daily life and understanding the science behind it is what am always been passionate about it. Being a sportsman, using ice-packs is very common for me. I have always wondered, how ‘ice-packs’ works and what is the chemical composition of it. After some research, I came to Know that ice packs use endothermic reactions. These ice-packs contains salts like ammonium nitrates and other Group-I nitrates. These salts when dissolved in water is an endothermic reaction. After discussion with my teacher, I understood how the concepts of enthalpy of solution and enthalpy of hydration is related to it. These topics, when taught during the classes for Topic-5 did not seem so interesting but when I got to know about a real life application for these topics, I got more interested towards it. Thus, when these salts are dissolved they cause reduction of temperature through absorption of heat. The composition of ice packs are not same everywhere. Most of the commercially available ice packs contains nitrate salts – ammonium nitrate, calcium nitrate, urea. Nowadays, reusable ice packs are also available which mainly contains water along with some chemicals that reduces the freezing point of water (Shaaban Khalil). The commonly used reagents added along with water are silica gel, propylene glycol and other polyacrylate crystals. Choice of chemicals while making an ice pack is thus an important factor to be considered. This brought me to the question- if the type of nitrates inside the ice packs are varied, how will the effectiveness of the ice pack change. Thus, I arrived at the research question given below.
Is there a trend in the magnitude of heat absorbed (measured in J) on dissolution of Group-I nitrates (LiN03, NaNO3 and KNO3) down the group from Li to K, determined using calorimetry?
Enthalpy change when 1 mole of an ionic compound is dissolved in water is known as the enthalpy of solution (Bazhin).
For example, NaCl (s) + excess H2O → NaCl (aq) ∆Hsolution of NaCl
The enthalpy of solution is related to two other enthalpy changes - ∆Hlattice and ∆Hhydration. The energy cycle shown below will establish the relationship:
Applying Hess’s law, we have,
1 = 2 + 3 + 4
Thus, ∆ Hsolution of MNO3 = ∆Hlattice of MNO3 + ∆Hhydration of M+ + ∆Hhydration of NO3-
The amount of heat absorbed can be calculated using the equation given below:
Q = m.c.∆t
where, m = mass of the solution
∆t = change of temperature = initial temperature – final temperature
As per assumptions made that the density of solution = density of water (1 g/cm3 ) and heat capacity of the
Mass of the solution = Volume of the solution × Density of the solution = V (in cm3 ) X 1 g cm-3 = V g
Therefore, Q = V.4.184. ∆t
The magnitude of ∆t will be determined graphically. 50 cm3 of water will be taken and salt will be added to it after a certain time period. The temperature of the solution will be monitored against time and the temperature versus time graph will be used to calculate ∆t graphically as illustrated below: The red line represents the temperature of water before the salt was added. The blue line represents the temperature of the solution after the salt was added. So, a perpendicular is drawn from the point when the salt was added and the perpendicular intersects the red and the blue line. The intersection point shows the initial and the final temperature and thus the value of ∆t can be determined.
The enthalpy of solution is the heat change when 1 mole of the salt is dissolved in water. Thus, the heat absorbed would be related to the enthalpy of solution in the following way:
Q = n × ∆Hsolution of MNO3
Q = heat absorbed
n = number of moles
∆Hsolution of MNO3 = enthalpy of solution of MNO3
As stated earlier, the enthalpy of solution and the heat absorbed is given as per the equation – Q = n × ∆Hsolution of MNO3
Thus, if the number of moles is kept constant, the heat absorbed increases as the ∆Hsolution of MNO3 increases. The values of ∆Hsolution are - 12.51 kJ mol-1 , 35.00 kJ mol-1 and 49.00 kJ mol-1 for LiNO3, NaNO3 and KNO3 respectively (Kinchin et al.). Hence, if the number of moles of the salt taken is 0.1,
Q for LiNO3 = 0.1 × ∆Hsolution of LiNO3 = 0.1 × 12.5 kJ = 1.25 kJ = 1250 J
Q for NaNO3 = 0.1 × ∆Hsolution of NaNO3 = 0.1 × 3500 kJ = 3.50 kJ = 3500 J
Q for KNO3 = 0.1 × ∆Hsolution of KNO3 = 0.1 × 49.00 kJ = 4.90 kJ = 4900 J
Thus, the values of heat absorbed increases as we go down the group from Li to K.
The three Group-I nitrates are chosen – LiNO3, NaNO3 and KNO3. As the nitrates chosen are of the first three alkali metals, study of the amount of heat absorbed will give us the trend down the group.
50.00 cm3 of distilled water was taken. A burette was used to control this volume.
As indicated in the equation between enthalpy of solution is related to the heat absorbed according to the equation Q = n × ∆H, the number of moles must be kept constant.
The salts used are LiNO3, NaNO3 and KNO3. So, the type of anion is used nitrate.
0.10 cm3
± 0.10 cm3
The considerations are made in reference to the MSDS (Material Safety Data Sheet):
The experiment has been done in a way so that minimum amount of chemicals are used. For example- only 0.1 moles of the salt were added. To get a proper reading thus, 50 cm3 of water was used instead of 100 cm3 . The experiment was performed under the supervision of a laboratory personnel.
The nitrate salts used are toxic in nature. They will cause serious dangers to the ecosystem. So, the unused solutions must be disposed carefully. Initially, large amount of tap water was added to dilute the solutions and then they were flushed off the drain.
Figure 9 - Showing Variation Of Temperature Against Time For LiNO3
In Figure-8, the temperature for the dissolution of Li(NO3)3 in ± 0.01°C was plotted against time in ± 0.01 s. Two trend lines were plotted as indicated by the red dotted lines. A perpendicular (a black dotted line) has been drawn from x = 180.00 ± 0.01 s which intersects these two trend lines. The co-ordinates of these two points has been also displayed in the graph in the form of (x,y). The standard errors of each data point have been also shown. The intersection of the black dotted line and the lower red dotted line which is (180.00,29.20) here is used to find the final temperature while the intersection of the black dotted line and the upper red dotted line which is (180.00, 35.00) is used to determine the final temperature.
Initial temperature (t1) = 35.00 ± 0.01°C
Final temperature (t2) = 29.20 ± 0.01 °C
Change in temperature (∆t) = t2 – t1 = (35.00 ± 0.01°C) – (29.20 ± 0.010 °C) = 5.80 ± 0.02 C
Similar calculations were done for NaNO3 and KNO3.
Initial temperature (t1) = 35.50 ± 0.01 °C
Final temperature (t2) = 19.20 ± 0.01 °C
Change in temperature (∆t) = t2 – t1 = (35.50 ± 0.01°C) – (19.20 ± 0.010 °C) = 16.30 ± 0.02 °C
Initial temperature (t1) = 35.50 ± 0.01 °C
Final temperature (t2) = 12.90 ± 0.01 °C
Change in temperature (∆t) = t2 – t1 = (35.50 ± 0.01 °C) – (12.90 ± 0.010 °C) = 22.60 ± 0.02 °C
LiNO3
NaNO3
KNO3
Water was added using a burette.
So, volume of solution added
= Final burette reading – Initial burette reading
= (50.00 ± 0.05 cm3) – (0.00 ± 0.05 cm3)
= 50.00 ± 0.10 cm3
Mass of solution (m) = (Volume of solution added × Density of solution ) = (50.00 ± 0.10 cm3) × ( 1.00 g cm-3) = 50.00 g
[It has been assumed that the density of the solution is equal to the density of water and is thus considered to be 1.00 g cm-3]
Heat change (Q) = (mass × heat capacity × change in temperature) = (m × C × ∆t) = (50.00 × 4.18 × 5.80) = 1212.20 J
[It has been assumed that the heat capacity of the solution is equal to the heat capacity of the water which is 4.18 J g-1 °C-1]
Q = mC ∆t
Mass of solution (m) = Volume × Density
As the density is taken from the literature value, the absolute error is taken only for volume.
Absolute uncertainty in mass (∆m) = ± 0.10
Fractional uncertainty in mass = \(\frac{∆m}{m}=\frac{0.10}{50.00}\)
Change of temperature (∆t) = 5.80 ± 0.02 °C
Fractional uncertainty in change of temperature = \(\frac{∆(∆t)}{∆t}=\frac{0.02}{5.80}\)
Fractional uncertainty in heat change
\(\frac{∆q}{q}=\frac{∆m}{m}+\frac{∆(∆t)}{∆t}=\frac{0.10}{50.00}+\frac{0.02}{5.80}\)
Percentage uncertainty in heat change = \(\frac{∆q}{q}\) × 100 = \((\frac{0.10}{50.00}+\frac{0.02}{5.80})\) × 100 = 0.54
The bar graph above compares the values of heat change in J against the type of Group-I nitrates. The values obtained for LiNO3, NaNO3 and KNO3 are 1212.20 J, 3406.70 J and 4723.40 J respectively. The data shows that as we go down the group from Li to K, the values are increasing. The values are positive which clearly shows that the process is endothermic. Thus, the process of dissolution of alkali metal nitrates becomes more endothermic as we go down the group from Li to K. This means that as we go down the group from Li to K, more heat is absorbed when the salt is dissolved. The sign of the heat change is positive in all cases which confirms that heat is absorbed by the system and the process is endothermic in all cases.
LiNO3
NaNO3
KNO3
Sample calculation:
Percentage error = \(\frac{literature\ value-experimental\ value}{literature\ value}×100=\frac{1250.00-1212.00}{1250.00}×100=3.02\)
In all cases, the error is a positive error which means that the experimental value is lower than the calculated value. This indicates that the decrease in temperature on dissolving the salt is lower than expected. This may have happened because of heat exchange between the surroundings and the system (beaker with the solution) which is an inevitable methodological limitation in this investigation. To be more clear, if the temperature of water is expected to decrease to 25.000 °C from 35.00 °C, it has decreased to 26.00 °C instead (random values has been chosen to ease the discussion). This may have happened as the solution in the beaker may have absorbed some heat from the surroundings even after the nitrate is dissolved which has finally elevated the temperature from the desired value. This discussion makes it evident that there is a methodological limitation involved in this investigation which is the exchange of heat between the system and surrounding.
As we move down the group from Li to K, the magnitude of heat change increases; the process of dissolution becomes more endothermic. As we descend down the group, the electrons are added to new shells; the outermost electrons move away from the nucleus; the ionic size of the cation increases. This increase in ionic size interferes with the solubility of the ions. When an ionic lattice is dissolved in the solution, the lattice dissociates into its respective cations and anions. These ions are then encapsulated (surrounded by) the water molecules. The dissolution of nitrate salts is endothermic as the stability of the system increases as the salts are dissolved.
MNO3 (s) → M+ (aq) +. NO3 - (aq)
As the reaction is endothermic, it is clear that the products are less stable than the reactants. It means that the dissolved ions (M+ ion and NO3 - ions surrounded by water) are less stable than the ionic lattice. The amount of heat absorbed or technically the heat change / energy change depends on how much the stability of the system is decreasing as we move from the reactant side to the product side. As the stability of the system decreases more on moving from the ionic lattice to the dissolved ions, the amount of heat change is more and thus the energy change is more and thus the heat absorbed will be of higher value. The stability of the product side which is the dissolved ions depends on the interaction of the water molecules with the cations and anions. As we go down the group from Li to K, the size of the ions is increasing; dissolving the cations becomes more difficult and thus the interaction of the water molecules with the ions is getting weaker and this in turn decreases the stability of the dissolved ions. Due to this, there is more decrease in stability going from the reactant to the product side as we move from Li to K.
As we move from Li to K, the size of the ions increases, the interaction of the ions and the water molecules becomes weaker, the stability on the product side (dissolved ions) decreases, there is more decrease of stability on moving from the reactant side (ionic lattice) to the product side (dissolved ions) ; energy difference is more between the reactant side and the product side; more heat is absorbed. This explanation is perfectly in coherence with the fact that the magnitude of enthalpy of dissolution of the Group-I nitrates increases (becomes more endothermic) as we are moving down the group from Li to K. (Refer to Hypothesis for the values).
Is there a trend in the magnitude of heat absorbed (measured in J) on dissolution of Group-I nitrates (LiN03, NaNO3 and KNO3) down the group from Li to K, determined using calorimetry?
The values obtained for heat change for dissolution of LiNO3, NaNO3 and KNO3 are 1212.20 J, 3406.70 J and 4723.40 J respectively. The data shows that as we go down the group from Li to K, the values are increasing. This means that as we go down the group from Li to K, more heat is absorbed when the salt is dissolved. As we move down the group from Li to K, the magnitude of heat change increases; the process of dissolution becomes more endothermic. As we move from Li to K, the size of the ions increases, the interaction of the ions and the water molecules becomes weaker, the stability on the product side (dissolved ions) decreases, there is more decrease of stability on moving from the reactant side (ionic lattice) to the product side (dissolved ions) ; energy difference is more between the reactant side and the product side; more heat is absorbed. The hypothesis is thus considered to be valid. The qualitative observations are in agreement with the conclusion made. The beaker appeared colder on touching in case of NaNO3 in comparison to that in case of LiNO3 and coldest in case of KNO3
As already discussed in the rationale, ice packs also contain water mixed with cooling reagents apart from aqueous solutions of nitrate salts. One of the most common cooling agent is ethylene glycol which is a dihydric alcohol. I would like to investigate if the mass of ethylene glycol added has an effect on the temperature depression or not. I would like to take specific volume of water, add various mass of ethylene glycol and measure the decrease of temperature in each case. The decrease of temperature can be estimated using a temperature versus time graph as done in this case. Thus, we can understand if there is a relationship between the mass of ethylene glycol we are adding and the decrease of temperature of water.
Bazhin, Nikolai. “The Born Formula Describes Enthalpy of Ions Solvation.” ISRN Thermodynamics, vol. 2012, 2012, pp. 1–3. DOI.org (Crossref), doi:10.5402/2012/204104.
Kinchin, A. N., et al. “Pitzer Equation for the Enthalpy of Solution of Electrolytes in N,N- Dimethylformamide in a Wide Temperature Range.” Theoretical Foundations of Chemical Engineering, vol. 34, no. 4, July 2000, pp. 351–55. DOI.org (Crossref), doi:10.1007/BF02758683.
Shaaban Khalil, Nahla. “Effect of Application of Ice Pack on Reducing Pain during the Arterial Puncture.” Clinical Practice, vol. 14, no. 4, 2017. DOI.org (Crossref), doi:10.4172/clinical- practice.1000115.