Inspection of the trend in amount of heat absorbed on dissolution of Group-I nitrates down the group

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) ∆H_solution of NaCl

The enthalpy of solution is related to two other enthalpy changes - ∆H_lattice and ∆H_hydration. The energy cycle shown below will establish the relationship: 

Applying Hess’s law, we have,

1 =  2 + 3 + 4

Thus,

∆H_solution of MNO₃ = ∆H_lattice of MNO₃ + ∆H_hydration of M⁺ + ∆H_hydration 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 × ∆H_solution of MNO₃

Q = heat absorbed

n = number of moles

∆H_solution 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 × ∆H_solution of MNO₃

Thus, if the number of moles is kept constant, the heat absorbed increases as the ∆H_solution of MNO₃ increases. The values of ∆H_solution 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 × ∆H_solution of LiNO₃ = 0.1 × 12.5 kJ = 1.25 kJ = 1250 J

Q for NaNO₃ = 0.1 × ∆H_solution of NaNO₃ = 0.1 × 3500 kJ = 3.50 kJ = 3500 J

Q for KNO₃ = 0.1 × ∆H_solution 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 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.

• A burette was taken and filled with distilled water. It must be taken care of that there are no bubbles in the water column inside the beaker.
• A 100 cm³ glass beaker was taken and a polystyrene cup was placed inside it.
• The glass beaker was kept on a magnetic stirrer.
• A temperature probe was connected to a Vernier Logger Pro and calibrated using boiling water and ice.
• 6.89 (0.10 moles) ± 0.01 g of LiNO₃ was weighed on a watch glass using a spatula.
• The water in the burette was emptied into the polystyrene cup inside the glass beaker.
• The stop-watch was started and the temperature of the water in the beaker was recorded using the temperature probe. Temperature was recorded at an interval 0f 30 seconds till 2 minutes 30 seconds (150 seconds).
• At 3.00 minutes, the weighed solid will be added to water.
• The temperature was recorded again at an interval of 30 seconds till 10 minutes.
• Repeat the same process with 8.49 g (0.10 moles) of NaNO₃ and 10.11 g (0.10 moles) of KNO₃.

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

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.

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

• If Figure - 9,10 and 11 are observed than two striking features can be noted. Firstly, the data points for the temperature of the water is same and there is no much variation as indicated by the fact that the trend line (red dotted line at the top) is almost perpendicular to the x axes. Secondly, the data points obtained after the salt was added at 180.00 s, are also obeying a trend line (red dotted line in the lower part) which again shows that there is not much temperature fluctuation after the loss of heat from the beaker to the surrounding on dissolving the salt. The existence of trend lines (one red dotted line in the top and one in the bottom) being almost perpendicular to the x axes clearly denotes that there is enough precision in the data collected.
• Figure - 14 shows the value of the percentage error obtained in reference to the calculated values obtained. The values are significantly less and thus claim a high level of accuracy in the data calculated for heat change. This definitely shows that the design of this investigation is reliable.
• The procedure to perform the data collection is pretty simple and does not involve any chemicals which is extremely hazardous or poisonous and not allowed to be used.

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.