Heat of Solution
by Jiun-Yiing Hu, Thamy Giritharan, Nicole Pun, Charissa Poon
Heat of Solution (2007) Powerpoint Show













Introduction



The heat of solution (more commonly referred to as the enthalpy of solution) is the enthalpy change, or simply the heat change, in a solution with respect to the dissolution of a substance in another at an invariable pressure. It is calculated by the sum of the endothermic energy and the exothermic energy in kJ/mol. The endothermic energy is classified as all the energy absorbed into the solution whereas the exothermic energy is energy that is released. Some compounds undergo this heat change when dissolved in water, either decreasing or increasing in temperature. However, when referring to gases, the dissolution of gases is always exothermic, since a gas’ solubility is decreased when heated. This can be explained using kinetic molecular theory, which states that particles in gas move at very high speeds compared to those in solids and even liquids. If one took a solid, such as sucrose, and placed it in hot water, it would be evident that the sucrose dissolves faster than if it were to be placed in cold water. This is because the water molecules of hot water are moving faster (they have more energy), which means that the crystal structure of sucrose is broken down faster, and the sucrose dissolves at a higher rate. However, gas particles are already moving very quickly. When the water molecules with added heat (which are moving at a rapid speed) collide with these gas particles, the gas particles may be ejected out of the solution, which means that it can no longer dissolve in the solution. Therefore, a gas’ solubility is decreased when heated.


Dissolution itself can be further broken down into three general phases – breaking of solute-solute attractions (endothermic), breaking of solvent-solvent attractions (endothermic), and forming of solvent-solute attractions (exothermic). The sum of the heat changes at each of the three steps is the overall enthalpy change or the heat of solution. It is the amount of heat given off or absorbed during the process of solution and is equal to the difference between the energy that must be supplied to break up the crystals of the solute and the energy that is released when the solute particles are taken into solution by the solvent. From this, it is inferred that solutions with negative heats of solution—decreased temperature—form stronger bonds.


To study the heat changes in the solution, one requires the knowledge of calorimetry. Calorimetry is the study of measuring the heat change produced through a reaction. It can be executed through two points of focus, either with constant volume or constant pressure as the basis. The use of the calorimeter is required to study this heat change. Substances under investigation must be within a calorimeter, where the heat of the reaction is accounted for. Therefore, both the initial and final temperatures are recorded through the calorimeter.

In calorimetry, an assumption is made that the change in heat of the substance (∆Hsolute) is equal to the change in heat of the water (∆Hwater).


The change in heat of the water is calculated using the formula:

total_heat_formula.jpg
where Q is the amount of heat (in J) absorbed/lost by the water, m is the mass of water (in g), c is the specific heat of water (in
J/g °C), and ΔT is the change in temperature (in °C). The Q value obtained through this formula gives us the amount of heat absorbed/lost by the water; to obtain the amount of heat absorbed/lost by the solute, simply switch the sign (positive or negative) in front of this value, because the amount of heat lost by water is the amount of heat absorbed by the solute, and vice versa.

The molar heat of solution is how much energy is gained or lost for one mole of the substance, and is calculated using the following formula,


molar_heat_forumla.jpg
In this experiment, a simple calorimeter will be constructed using every day items to measure the heat of solution of two different compounds.

Materials



  • For every compound, 100mL of water is used
  • 9.5g of each compound used:
    - Sodium Hydroxide (NaOH)
    - Potassium Nitrate (KNO3)
    - Ammonium Chloride (NH4Cl)
    - Sodium Chloride (NaCl)
  • 2 Styrofoam bowls (put one in the other to reduce heat loss and potential sources of error; 1 has a hole large enough to fit a thermometer inside)
  • 1 Thermometer
  • Safety goggles
  • Scoopula


Procedure



  1. The thermometer was slid into the hole of the Styrofoam bowl with the hole to ensure a secure fit.
  2. 100 mL of water was measured using a graduated cylinder and then poured the Styrofoam bowl without the hole.
  3. 10g of the compound was measured and then gently poured into the water.
  4. The second Styrofoam bowl (the one with the hole in the bottom) was placed on top of the other Styrofoam bowl (containing the water and the compound) snugly (refer to Fig. 1 for a diagram).
  5. The thermometer was placed inside the hole in the Styrofoam bowl on top; it was ensured that the bulb of the thermometer was inside the water.
  6. The entire apparatus was gently shaken to swirl the liquid inside the Styrofoam bowl, taking care not to tip the thermometer over; the temperature of the mixture was noted after it had stabilized.
  7. The apparatus, including both Styrofoam bowls and the thermometer, was washed and dried thoroughly.
  8. Steps 2 to 7 were repeated for the remaining compounds.
  9. The results were compiled and the respective heats of solution were calculated.


diagram2.jpg


Note: Take note of the experimental errors that take place during the experiment, and any outside factors that might have influenced the results.

Safety Precautions



  • Always wear safety goggles (if a chemical is splashed into your eye, immediately wash it out with water)
  • Do not consume any food or beverages in the lab
  • Perform the experiment under teacher supervision
  • Care must be exercised when using the thermometer to prevent unnecessary injury or spillage— if mercury is spilled, report it immediately to the teacher and do not try to clean it up yourself
  • Do not consume solute, solutions, or any experiment materials
  • Wash your hands after handling chemicals
  • Avoid transportation of chemicals through congested areas
  • Sink drains should be thoroughly flushed after disposing of chemicals
  • Sodium hydroxide (NaOH) is a highly corrosive substance and should be treated with extra caution

Observations



Compound



Water temperature without compound (°C)



Water temperature after adding compound (°C)



Δ Temperature (°C)



NaOH



22



47



25



NH4Cl



22



16



-6



NaCl



22



21



-1



KNO3



22



15



-7



Of the four reactions that took place, sodium hydroxide (NaOH) and water (H2O) was the only exothermic reaction. The latter three - ammonium chloride (NH4Cl), sodium chloride (NaCl), and potassium nitrate (KNO3) were all endothermic. The temperature change of sodium hydroxide (NaOH) was the most dramatic, with an absolute value of 25 degrees, and sodium chloride (NaCl) was the least dramatic, with an absolute temperature change of 1 degree. It is interesting to note that both these compounds have the element sodium (Na).


Calculations



The total heat of solution is how much heat was lost or absorbed in the experiment using the actual masses of the substances, and is calculated through the following formula,

total_heat_formula.jpg
where Q is the amount of heat (in J) absorbed/lost by the water, m is the mass of water (in g), c is the specific heat of water (4.18 J/g °C), and ΔT is the change in temperature (in °C). The Q value obtained through this formula gives us the amount of heat absorbed/lost by the water; to obtain the amount of heat absorbed/lost by the solute, simply switch the sign (positive or negative) in front of this value, because the amount of heat lost by water is the amount of heat absorbed by the solute, and vice versa.

The molar heat of solution is how much energy is gained or lost for one mole of the substance, and is calculated using the following formula,

molar_heat_forumla.jpg
Using the above two formulae, the total heat of solution, as well as the molar heat of solution, is calculated for each of the compounds that were tested.

Compound



Total heat of solution



Molar heat of solution



NaOH



total_heat_formula.jpg

Q = ?

m = 100g

c = 4.18 J/g °C

ΔT = 21°C


Q = 100g * 4.18 J/g °C *25°C

Q =
10450J

Amount of heat gained by water = 10450J

Amount of heat lost by solute = -10450J


Total heat of solution = -10450J




Amount of heat lost by solute = -10450J

molar_heat_forumla.jpg

NaOH.jpg



Molar heat of solution = -44.00kJ/mol





NH4Cl




total_heat_formula.jpg

Q = ?

m = 100g

c = 4.18 J/g °C

ΔT = 6°C


Q = 100g * 4.18 J/g °C *-6°C

Q = -2508J


Amount of heat gained by water = -2508J

Amount of heat lost by solute = 2508J


Total heat of solution = 2508J




Amount of heat lost by solute = 2508J

molar_heat_forumla.jpg

ammonium.jpg


Molar heat of solution = 14.12kJ/mol





NaCl



total_heat_formula.jpg

Q = ?

m = 100g

c = 4.18 J/g °C

ΔT = -1°C


Q = 100g * 4.18 J/g °C *-1°C

Q = -418J


Amount of heat gained by water = -418J

Amount of heat lost by solute = 418J


Total heat of solution = 418J





Amount of heat lost by solute = 418J

molar_heat_forumla.jpg

NaCl_molar_heat.jpg



Molar heat of solution = 2.57kJ/mol





KNO3



total_heat_formula.jpg


Q = ?

m = 100g

c = 4.18 J/g °C

ΔT = -7°C


Q = 100g * 4.18 J/g °C *-7°C

Q = -2926J


Amount of heat gained by water = -2926J

Amount of heat lost by solute = 2926J


Total heat of solution = 2926J




Amount of heat lost by solute = 2926J

molar_heat_forumla.jpg

potassium.jpg



Molar heat of solution = 31.14kJ/mol





Conclusion



Compound




Molar heat of solution obtained from lab



Official molar heat of solution



NaOH



-44.00kJ/mol




-44.51kJ/mol



NH4Cl




14.12kJ/mol



14.78kJ/mol



NaCl



2.57kJ/mol




3.88kJ/mol



KNO3



31.14kJ/mol




34.89 kJ/mol



Based on our observations, we found that out of the four compounds, only sodium hydroxide (NaOH) was exothermic, yielding a heat of solution of -44.00kJ/mol. The latter three compounds - ammonium chloride (NH4Cl), sodium chloride (NaCl), and potassium nitrate (KNO3), were all endothermic, and yielded heats of solution of 13.42kJ/mol, 2.44kJ/mol, and 31.14kJ/mol respectively.

The following calculations provide the percentages by which the results obtained differed from the officials results:

NaOH

100% - (-44.00kJ/mol / -44.51kJ/mol * 100) = 1.14%


NH4Cl

100% - (14.12kJ/mol / 14.78kJ/mol * 100) = 4.47%


NaCl

100% - (2.57kJ/mol / 3.88kJ/mol * 100) = 33.72%


KNO3

100% - (31.14kJ/mol / 34.89kJ/mol * 100) = 10.74%


As can be seen, all the results were within a 33.72% range to the theoretical values.



Discussion



Lattice Energy and Hydration energy


Dissolution can occur in three steps: breaking of solute-solute attractions (endothermic), breaking of solvent-solvent attractions (endothermic), and forming of solvent-solute attractions (exothermic). It is the sum of the heat from these three steps that prove to be the enthalpy change, or heat of solution. To attain this number with theoretical values, without experimentation, the use of only two variables, lattice energy and hydration energy, are required.


Picture1.jpg
Fig. 2 - Diagram outlining different bonds that must be broken in order for dissolution to take place



By breaking the solute-solute attractions (see Fig. 2), the ions of the compound must be separated by overcoming the lattice energy. The lattice energy of ionic compounds (salts) is essentially the measure of how strong the bonds are in the compound. Lattice energy is defined as the amount of energy consumed or required for the gaseous ions to form one mole of an ionic solid, in which it is seen as an endothermic process (symbolized as U). Instead of using the lattice energy as a variable, the reverse process, energy of crystallization, may be used instead. The energy of crystallization (Ecryst) is defined as the energy released as gaseous ions form a mole of an ionic solid, which is always an exothermic process. Therefore, U = - Ecryst. Lattice energies in ionic compounds are usually quite large since ionic bonds are strong (more energy is needed to break or form bonds).

After breaking solvent-solvent attractions, which involves the bonding between molecules within the solvent (in this case it is likely water and the bonding is greatly hydrogen bonding), the solute molecules would then combine with the solvent molecules. Since water is the most common solvent for ionic compounds, both the ion-dipole forces of attraction (∆Hion-dipole) and the force of hydrogen bonding (∆HH-Bond) must be overcome. The sum of these two variables is the hydration energy. The hydration energy (∆Hhydration) or heat of hydration is the amount of energy required as a mole of the substance is dissolved into water, an endothermic process. The dissolving process of the substance into the universal solvent of water is therefore referred to as hydration. Forming solute-solvent attractions (referring to the process of dissolution, which involves the association and the attraction of molecules from both the solvent and the solute) is essential in determining whether the compound is soluble or not. Similar to lattice energy, the reverse process to hydration is an exothermic process and is referred to as the enthalpy of hydration. The enthalpy of hydration of an ion is the amount of energy that is released when a mole of the ion is dissolved into water. Therefore, hydration energy and the enthalpy of hydration would have the same values for a certain substance, but hydration energy would always be a positive value, where the latter would always be a negative one.

These two viewpoints of hydration energy and lattice energy are comparable to the viewpoint of whether the glass of water is half full or half empty; in both cases, they result in the same basic inference but through different methods of analysis. It is important to note that different sources (such as textbooks) will define the lattice energy and hydration energy using either of the two viewpoints.

A general pattern that is noticed for lattice energy, is that as the ionic charges increase, so does the lattice energy. However, as the ionic radii increase, the lattice energy will decrease. For hydration energy, as the ionic radii increase, there is a decrease in hydration energy.

How Substances are Determined to be Endothermic and Exothermic


The relationship between the heat of solution or enthalpy change (Hsoln), the hydration energy and the lattice energy can be generated through a simple mathematical equation:
Hsoln = – U + ∆Hhydration

It is important to note that -U
is also the Ecryst. This equation can be altered depending on how the variables are defined (for example, if the lattice energy was seen as an exothermic process instead of an endothermic one, the negative sign would consequently be removed). With this equation, the heat of solution can be evaluated without calorimetry through the sum of these two numbers, lattice energy and hydration energy. From this equation, one can deduce that if the lattice energy is greater than the hydration energy, the heat of solution would be negative (and vice versa). By determining the heat of solution, one can easily construe whether an endothermic or exothermic change has occurred. The heat of solution is negative for exothermic changes, and positive for endothermic changes. In fact, the experiment conducted produced positive results in showing which compounds were exothermic and endothermic.

Compound


Δ Temperature (°C)

Official molar heat of solution

NaOH
Sodium hydroxide

+25

-44.51kJ/mol

NH4Cl
Ammonium chloride

-6

+14.78kJ/mol

NaCl
Sodium chloride

-1

+3.88kJ/mol

KNO3
Potassium nitrate

-7

+34.89 kJ/mol


NaOH in water was the only exothermic reaction, with an increase in temperature of 25 degrees Celsius, which agrees with the accepted heat of solution being negative. These observations are thus proved valid.
With the results from the dissolution of NaOH, an exothermic process seemed to have occurred. This finding agrees with the accepted value of the ∆Hsoln of NaOH, which is -44.2° kJ/mol. 2 This means that NaOH loses -44.2°kJ/mol when dissolved in water and producing an exothermic reaction. We can assume that NaOH has lattice energy (U) that is greater than its hydration energy Hhydration. If the lattice energy given is 900 kJ/mol, the hydration energy should be:
-44.2 kJ/mol = - 900 kJ/mol + Hhydration
855.9 kJ/mol = Hhydration
This should be the hydration energy (the amount of energy absorbed during hydration) of NaOH.

If one were to compare NaCl and NaOH, one would find that NaCl is an endothermic process whereas the latter is an exothermic process (in water). Both compounds have similar hydration energies, but NaOH has a weaker crystal structure. Its lattice structure is easier to pull apart, meaning it should have lower lattice energy. However, NaCl in water produces an endothermic change whereas NaOH produces an exothermic one. If NaCl has a stronger crystal structure than NaOH, it would be reasonable to assume that NaCl would at least have a negative heat of solution as well. Although its crystal structure is tecnically stronger, NaOH would require more energy to separate due to the additional hydrogen bonding present in NaOH. Since the bonding forces are greater in NaOH, a greater amount of lattice energy needs to be overcome, making it possible that NaCl is endothermic and NaOH is exothermic, in water. This explicates the fact that solutions with negative heats have stronger bonds since more energy is needed to break these bonds.

Fundamentally, the balance between the energy needed to break apart the crystal structure (lattice energy) and the energy given off when ions are dissolved in water (hydration energy) will establish whether the reaction is endothermic or exothermic.


Applications in Real Life

p2.jpg
Fig. 3 - A salt crystal disassociates to form new bonds with water molecules
Science is a way of life— exothermic and endothermic reactions are everywhere, taking place in many commonplace situations out of the science laboratory. Used to treat minor injuries, such as headaches, muscle spasms or strains, hot or cold packs offer relief through the simple exothermic and endothermic reactions taking place inside them. These packs contain two chambers: one containing a salt and the other containing water. To initiate the reaction, one must break the seal betw een the compartments an d shake the pack robustly in order to combine the salt with the water.

Inside the hot or cold packs, the salts would disassociate and form new connections with the water molecules (see Fig. 3). For example, if one were to place NaCl in water (this would produce an endothermic reaction), it would separate into Na+ and Cl-. The electrostatic attraction (attraction between opposite charges) between the ions would be broken and left to bond with the water molecules. According to the specific compound properties, either an endothermic or exothermic change will take place.

Hot packs are created with salts that will dissolve into water and release heat (energy), making it an exothermic process. Some hot packs contain magnesium sulphate (MgSO4) since it generates heat when dissolved in water. Its enthalpy change is -1278.2°kJ/mol. Cold packs are produced with salts that will dissolve into water and absorb heat, making it an endothermic process. A substance that is often used in instant cold packs is ammonium nitrate (NH4NO3), which has an enthalpy change of 25.59°kJ/mol. The amount of heat released or absorbed relies on the concentration of the ionic compounds and the water.


Modifications to Improve the Experiment


This experiment involves few parameters, therefore minimizing the margin of possible error. This is clearly shown from the results obtained as the percentage of error is below 40%. Although some sources of experimental error are inevitable, there are certain aspects of the experiment that can be improved to reduce the margin of error. One aspect is the precautions that should be taken to ensure that the experiment will not be affected by changes in the surrounding air temperature and pressure. Proper insulation and sealing of the Styrofoam cups will ensure that the environment in which the experiment takes place will have little effect on the experiment; it will also prevent unnecessary heat loss. Other modifications may include thorough stirring to ensure that the solution is completely homogeneous in all parts. Apart from inaccurate results caused by human error and technical problems, such as a defective thermometer in monitoring temperature changes, if done correctly, this experiment should have little to no margin of error and requires no modification in process.


Sources of Experimental Error



In this lab, as with any other scientific procedure, there is some degree of experimental error. In this particular experiment, the most significant source of experiment was the electronic balance that was used to compute the masses of the compounds. This electronic balance was hyper-sensitive, causing the values to change even with a slight draft. Furthermore, the site of the experiment was located in a corner, which was bound by a door and a window. Students mistakenly thought that this area would be protected from such drafts such it was a corner. Unfortunately, the aforementioned door had a tendency to rebound a slight current off the corner, and the window – although as fully closed as possible – still leaked air. This caused the numbers displayed on the balance to consistently fluctuate dramatically without stop, meaning we had to guess at which values were the closest to the real mass of the compounds. This would have skewed the results.


In addition, the calorimeter used was made of Styrofoam. Although a very good insulator, Styrofoam is still liable to transfer heat, however little. Since temperature was paramount to the experiment conducted, any slight bit of heat lost would skew the results. Also, the lids used for this experiment were actually smaller, inverted Styrofoam cups. Although these cups seemed to fit sufficiently at the time of the experiment, it would not be irrational to reason that there were holes in between the two cups, and thus heat was also lost this way.