Inspection of the trend in the 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