Goal of the work - determination of the thermal effect of the process of dissolving salt in water and the heat of the neutralization reaction using a calorimeter with an isothermal shell.
Regarding the processes being studied, the following must be kept in mind: chemical reactions, in contrast to phase transformations, are accompanied by a change in the composition of substances in the system. Dissolution processes occupy an intermediate position between them. These processes, if you do not know their nature, seem difficult to explain. For example, in order to destroy sodium chloride crystals into individual ions, it is necessary to expend quite significant energy (ΔE cr):
NaCl solid → Na + gas + Cl – gas; DН° destruction = +777.26 kJ/mol. (18)
According to the first law of thermochemistry, the reverse process of crystal formation from ions will be exothermic, that is, DН° image = – 777.26 kJ/mol.
At the same time, when sodium chloride interacts with water, the process of combining Na + and Cl – ions with polar water molecules occurs, which is considered as a process of ion hydration; it is accompanied by the release of a significant amount of heat.
Table 11 shows the values of bond energies Eb in some substances and enthalpies of hydration DН° of hydride ions under standard conditions.
As a result, the processes of dissolution of ionic compounds are considered as ordinary chemical reactions and are clearly characterized by thermal effects. To find them, it is necessary either to conduct an experimental study, for example, calorimetric, or to use tabulated values of the heats of formation of all hydrated ions and compounds involved in the dissolution process.
Typically, the heat of solution is referred to as the dissolution of one mole of a substance. It is assumed that an infinitely dilute solution is formed. As a result, the dissolution mechanism is presented as a process of destruction of the crystal lattice of a substance under the influence of a solvent (endothermic effect) and as a process of hydration of the resulting ions (exothermic effect). The total thermal effect is determined precisely by these processes.
Table 11.
Using the first corollary of the second law of thermochemistry, we can calculate according to those available in Table 11. These are the thermal effects of the dissolution of these substances, as well as the heat of neutralization of an acid with an alkali.
For example, the enthalpy of dissolution of crystalline sodium chloride in water is found by the equation:
NaCl TV aqua→ Na + aq + Сl – aq , (19)
DН° p ast. = DН° hydr (Na + aq) + DН° hydr (Cl – aq) – = (20)
420.1 - 353.7 - (- 777.3) = + 3.5 kJ/mol.
The positive sign of the thermal effect indicates that the dissolution process occurs with the absorption of heat and the temperature of the solution should decrease.
The heat of a neutralization reaction is the amount of heat released when 1 equivalent of a strong acid reacts with 1 equivalent of a strong base. This produces 1 equivalent of liquid water.
It has been found that in the case of dilute solutions, the heat of reactions of strong bases (such as NaOH and KOH) with strong acids (for example, HCl or H 2 SO 4) does not depend on the nature of the acid and base. This constancy of the neutralization heat is explained by the almost complete dissociation into ions of strong acids and bases, as well as salts formed as a result of the neutralization reaction. Therefore, when mixing dilute solutions of a strong acid and a strong base, only one chemical reaction actually occurs, namely, between the hydrated ions of hydronium H 3 O + aq and hydroxyl OH – a q:
1/2 H 3 O + aq + 1/2 OH – a q → H 2 O liquid, (21)
DН° neutr = DН° image (Н–ОН) – (1/2)
= – 459.8 – (1/2) · (– 477.8 –– 330.0) = – 55.9 kJ/mol. (22)
The negative sign of the thermal effect indicates that the neutralization reaction proceeds with the release of heat and the temperature of the solution should increase.
“Thermal effects when dissolving substances in water” Andronova Alina Petrosyan Anahit Shirmanova Alina Pupils of the 11th grade Leader: Shkurina Natalya Aleksandrovna, chemistry teacher.
Consider the thermal effects when substances are dissolved in water. Experimentally establish which substances dissolve in water is accompanied by the release of heat (+Q), and which by absorption (-Q). Share the research with your classmates.
Each substance stores a certain amount of energy. We encounter this property of substances already at breakfast, lunch and dinner, since food allows our body to use the energy of a wide variety of chemical compounds contained in food. In the body, this energy is converted into movement, work, and is used to maintain a constant (and quite high!) body temperature.
The energy of chemical compounds is concentrated mainly in chemical bonds. To break a bond between two atoms requires ENERGY. When a chemical bond is formed, energy is RELEASED. Any chemical reaction consists of breaking some chemical bonds and forming others.
When, as a result of a chemical reaction during the formation of new bonds, MORE energy is released than was required to destroy the “old” bonds in the starting substances, then the excess energy is released in the form of heat. An example is combustion reactions. For example, natural gas (methane CH 4) burns in oxygen in the air, releasing a large amount of heat. The reaction can even occur with an explosion - so much energy is contained in this transformation. Such reactions are called EXOTHERMAL from the Latin "exo" - outward (meaning the energy released).
In other cases, the destruction of bonds in the original substances requires more energy than can be released during the formation of new bonds. Such reactions occur only when energy is supplied from the outside and are called ENDOTHERMIC (from the Latin “endo” - inside). An example is the formation of carbon monoxide (II) CO and hydrogen H2 from coal and water, which occurs only when heated
Thus, any chemical reaction is accompanied by the release or absorption of energy. Most often, energy is released or absorbed in the form of heat (less often in the form of light or mechanical energy). This heat can be measured. The measurement result is expressed in kilojoules (kJ) for one MOLE of reactant or (less commonly) for one mole of reaction product. This quantity is called THE THERMAL EFFECT OF THE REACTION. For example, the thermal effect of the combustion reaction of hydrogen in oxygen can be expressed by any of two equations: 2 H 2 (g) + O 2 (g) = 2 H 2 O (l) + 572 kJ or H 2 (g) + 1/ 2 O 2 (g) = H 2 O (l) + 286 k. J
Equations of chemical reactions in which the thermal effect of the reaction is written along with the reagents and products are called THERMOCHEMICAL EQUATIONS
The thermal effects of chemical reactions are needed for many technical calculations. Imagine yourself for a moment as a designer of a powerful rocket capable of launching spaceships and other payloads into orbit. The world's most powerful Russian rocket, Energia, before launch at the Baikonur Cosmodrome. The engines of one of its stages operate on liquefied gases - hydrogen and oxygen. Let's say you know the work (in kilojoules) that will have to be spent to deliver a rocket with cargo from the surface of the Earth to orbit; you also know the work to overcome air resistance and other energy costs during the flight. How to calculate the required supply of hydrogen and oxygen, which (in a liquefied state) are used in this rocket as fuel and oxidizer? Without the help of the thermal effect of the reaction of the formation of water from hydrogen and oxygen, this is difficult to do. After all, the thermal effect is the very energy that should launch the rocket into orbit. In the combustion chambers of a rocket, this heat is converted into the kinetic energy of molecules of hot gas (steam), which escapes from the nozzles and creates jet thrust. In the chemical industry, thermal effects are needed to calculate the amount of heat to heat reactors in which endothermic reactions occur. In the energy sector, thermal energy production is calculated using the heat of combustion of fuel. Dietitians use the thermal effects of food oxidation in the body to create proper diets not only for patients, but also for healthy people - athletes, workers in various professions. Traditionally, calculations here use not joules, but other energy units - calories (1 cal = 4.1868 J). The energy content of food is referred to any mass of food products: 1 g, 100 g, or even standard packaging of the product. For example, on the label of a jar of condensed milk you can read the following inscription: “calorie content 320 kcal/100 g.”
The branch of chemistry that studies the transformation of energy in chemical reactions is called thermochemistry. There are two laws of thermochemistry: 1. Lavoisier-Laplace’s law (the thermal effect of a forward reaction is always equal to the thermal effect of a reverse reaction with the opposite sign.) 2. G. I. Hess’s law (thermal effect reactions depends only on the initial and final states of the substances and does not depend on the intermediate stages of the process.
Thus, dissolution is a physicochemical process. The dissolution of substances is accompanied by a thermal effect: release (+Q) or absorption (-Q) of heat - depending on the nature of the substances. The dissolution process itself is determined by the interaction of particles of the soluble substance and the solvent.
Experimentally determine which substances dissolve in water is accompanied by the release of heat (+Q), and which by absorption (-Q). Materials: acetone, sucrose, sodium chloride, sodium carbonate (anhydrous and/or crystalline hydrate), sodium bicarbonate, citric acid, glycerin, water, snow. Equipment: electronic medical thermometer or temperature sensor from sets of digital sensors in school chemistry, physics or biology classrooms.
1. Sucrose 2. Sodium chloride 3. Sodium carbonate (anhydrous) 4. Sodium bicarbonate 5. Citric acid 6. Glycerin 7. Snow 1 2 3 4 5 6 7
Conclusion Dissolution of sodium carbonate (anhydrous) and sodium bicarbonate occurs with the release of heat. Snow with water - with heat absorption, the rest are unchanged.
1. We collected half a cup of snow. 2. Put some snow on the board. Let it melt into a small puddle. 
Test 1. Under standard conditions, the heat of formation is 0 for: a) hydrogen b) water c) hydrogen peroxide d) aluminum. 2. The reaction, the equation of which N 2 + O 2 = 2 NO-Q refers to the reactions of: a) endothermic compound b) exothermic compound c) endothermic decomposition d) exothermic decomposition.
3. The reaction is endothermic: a) combustion of hydrogen b) decomposition of water c) combustion of carbon d) combustion of methane. 4. Which definition is incorrect for this reaction: 2 Na. NO 3(solid)=2 Na. NO 2(solid)+O 2(g)-Q a) homogeneous b) endothermic c) reaction of the compound d) redox. 5. The basic law of thermochemistry is the law of: a) Gay-Lussac b) Hess c) Avogadro d) Proust
Conclusion Results of pedagogical research: 1. Students understood the essence of thermal effects when substances are dissolved in water. 2. Exo- and endothermic reactions were determined. 3. Test results (83% of students completed the test tasks).
The amount of heat that is released or absorbed when 1 mole of a substance is dissolved in such an amount of solvent, the further addition of which no longer causes a change in the thermal effect, is called the heat of solution.
When dissolving salts in water, the sign and magnitude of the thermal effect of dissolution ∆ N is determined by two quantities: the energy spent on the destruction of the crystal lattice of the substance (∆ H 1) - an endothermic process, and the energy released during the physico-chemical interaction of particles of the dissolved substance with water molecules (hydration process) (∆ N 2) - exothermic process. The thermal effect of the dissolution process is determined by the algebraic sum of the thermal effects of these two processes:
∆N = ∆H 1 + ∆H 2 .
The thermal effect of the dissolution process can be either positive or negative.
For the practical determination of heats of solution, the amount of heat absorbed or released when an arbitrary amount of salt is dissolved is usually determined. This value is then recalculated per 1 mole, since the amount of heat is directly proportional to the amount of dissolved substance.
For thermochemical measurements, a device called a calorimeter is used.
The heat of solution is determined by the change in the temperature of the solution, so the accuracy of the determination depends on the division value (accuracy) of the thermometer used. Typically, the range of measured temperatures lies in the range of 2-3°C, and the temperature division of the thermometer is no more than 0.05°C.
PROGRESS
To do the work, use a calorimeter consisting of a heat-insulating body, a lid with a built-in electric stirrer and thermometer, and a hole with a plug.
Get an assignment from your teacher: type of solute.
Open the cap on the calorimeter lid and pour 200 ml of water into it, close the cap and wait for 10-15 minutes to establish a constant temperature ( t beginning ). During this time, on the scales, using tracing paper or a watch glass, obtain a sample of your substance (1.5 - 2.0 g), previously thoroughly ground in a mortar. Place the resulting sample as quickly as possible through the hole in the lid into the calorimeter with the stirrer turned on. Watch the temperature change. After thermal equilibrium has been established (the temperature stabilizes), record the maximum temperature of the solution ( t max) and calculate ∆ t = t max – t beginning Based on the data obtained, calculate the heat of solution of the salt using the equation:
∆N dist = q M/ m, J/mol, (1)
Where q- heat released (or absorbed) in the calorimeter (kJ); m- weight of salt (g); M is the molar mass of the solute (g/mol);
Heat q determined on the basis of experimental data from the relationship:
q = (m st C st + m solution C solution)∆ t,(2)
Where m st - mass of the glass (g); m solution - mass of solution equal to the sum of the masses of water and salt in a glass (g); WITH st - specific heat capacity of glass 0.753 J/g∙K;
WITH solution - specific heat capacity of solution (water) 4.184 J/g∙K.
Comparing the obtained result with the data in Table 2, calculate the relative error of the experiment (in%).
Heat of hydration of salt and its definition
The physicochemical process of interaction between particles of a dissolved substance and molecules of water (solvent) is called hydration. During the process of hydration, complex spatial structures called hydrates are formed, and at the same time energy is released into the environment in the form of heat.
The heat effect of the reaction of forming 1 mole of hydrated salt from anhydrous salt is called the heat of hydration.
When an anhydrous salt capable of forming hydrates is dissolved in water, two processes occur sequentially: hydration and dissolution of the resulting crystalline hydrate. For example:
CuSO 4 (tv) + 5H 2 O (l) = CuSO 4 × 5H 2 O (tv),
CuSO 4 ×5H 2 O (tv) + n H 2 O (l) = CuSO 4 (r),
CuSO 4(p) + n H 2 O (l) = Cu 2+ (r) + SO 4 2- (r)
The dissolution of electrolytes is accompanied by the process of electrolytic dissociation. The heat of hydration of a molecule is equal to the sum of the heats of hydration of the resulting ions, taking into account the heat of dissociation. The hydration process is exothermic.
Approximately, the heat of hydration of a substance can be defined as the difference between the heats of dissolution of an anhydrous salt and its crystalline hydrate:
∆H hydr = ∆ H without - ∆ H Krist, (3)
where ∆ H hydr - heat of hydration of molecules;
∆H anhydrous - heat of dissolution of anhydrous salt;
∆H krist - heat of dissolution of crystalline hydrate.
Thus, to determine the heat of hydration of molecules, it is necessary to first determine the heat of dissolution of an anhydrous salt and the heat of dissolution of the crystalline hydrate of this salt.
PROGRESS
The heat of dissolution of anhydrous copper sulfate CuS0 4 and crystalline hydrate CuS0 4 × 5H 2 0 must be determined using a laboratory calorimeter and work procedure 1.
To more accurately determine the heat of hydration, it is necessary to obtain 10-15 g samples of crystalline hydrate and anhydrous copper sulfate salt. You need to know that anhydrous copper salt easily absorbs water from the air and goes into a hydrated state, so the anhydrous salt must be weighed immediately before the experiment. Based on the data obtained, it is necessary to calculate the heats of dissolution of anhydrous salt and crystalline hydrate, and then from relation (3) determine the heat of hydration. Calculate the relative experimental error as a percentage using the data obtained and the data in Table 2.
Dissolution is a physicochemical process leading to the formation of a homogeneous system. The thermal effects that accompany it are the result of a wide variety of reasons. Let's look at a few examples:
A) The process of dissolving liquids in water can be accompanied by such phenomena as the dissociation of polar molecules with the formation of ions, the occurrence of hydrogen bonds between polar water molecules and molecules of substances containing elements with high electronegativity, hydration of chemical particles, etc.
C 2 H 5 OH - H 2 O
This system is responsible for the formation of ideal solutions over a wide range of concentrations. The dissolution process must be accompanied by the formation of hydrogen bonds, therefore, it is energetically favorable, that is, it has a positive thermal effect.
CH 3 COOH - H 2 O
Acetic acid is a weak monobasic acid K d = 1.8 10 -5, therefore, when dissolved in water, some of the energy will be spent on the dissociation of molecules (negative thermal effect), and some of the energy, on the contrary, will be released in the form of heat during hydration ions. The total effect will depend on the ratio of these quantities.
B) The process of dissolving solids in water depends on the type of crystal lattice of the latter. As a rule, the dissolution of ionic crystals is associated with two opposite effects: a positive energy of hydration of ions and a negative energy of destruction of the crystal lattice. In molecular crystals, the first component is practically absent. When draining dilute solutions of salts of strong electrolytes, no thermal effect is observed. If a precipitate is formed, the thermal effect of deposition is observed.
Integral heat of solution is the amount of heat absorbed or released when 1 mole of a substance is dissolved in a very large (300 mol/mol of substance) amount of solvent.
Example of a calculation problem:
Calculate the integral heat of dissolution of ammonium chloride if, when 1.473 g of salt is dissolved in 528.5 g of water, the temperature decreases by 0.174 o C. The mass heat capacity of the solution is 4.109 J/g. K. Heat capacity of the calorimeter 181.4 J/g. K
Solution: The integral heat of solution can be calculated using the formula:
Q = (C calor. + C solution m)× ΔТ/n,
where C is the heat capacity, n is the amount of dissolved substance: n = m/M
m (solution) = 528.5 +1.473 = 530 g,
ΔT = -0.174 o C,
Q = (4.109 × 530 + 181.4) × (-0.174) × 53.5/ 1.473 × 1000 = -15.11 kJ/mol From the course of chemical thermodynamics it is known that the measure of the thermal effect of a chemical process in an isobaric process (constant pressure in the system) is the thermodynamic function of the state - enthalpy
ΔН = Н con. - N beginning The thermal effect in this case is equal in absolute value to enthalpy, but opposite in sign. The exothermic process, accompanied by the release of heat, corresponds to –ΔH, and the endothermic process, accompanied by the absorption of heat, corresponds to +ΔH. Thus, in the problem considered above, the dissolution process of ammonium chloride is endothermic, ΔH = 15.11 kJ/mol.
Section 5. ROSCHIN. THEORY OF ELECTROLYTIC DISSOCIATION
§ 5.3. Thermal phenomena during dissolution
The dissolution of substances is accompanied by a thermal effect: the release or absorption of heat, depending on the nature of the substance. When, for example, potassium hydroxide or sulfuric acid is dissolved in water, a strong heating of the solution is observed, i.e. release of heat, and when ammonium nitrate dissolves, strong cooling of the solution, that is, absorption of heat. In the first case, an exothermic process occurs (∆H 0), in the second - an endothermic process (∆H > 0). The heat of solution ∆H is the amount of heat that is released or absorbed when 1 mole of a substance is dissolved. So, for potassium hydroxide ∆Н° = -55.65 kJ/mol, and for ammonium nitrate ∆ H = +26.48 kJ/mol.
As a result of the chemical interaction of a solute with a solvent, compounds are formed that are called solvates (or hydrates if the solvent is water). The formation of such compounds makes solutions similar to chemical compounds.
The great Russian chemist D.I. Mendeleev created the chemical theory of solutions, which he substantiated by numerous experimental data set out in his work “Study of aqueous solutions by their specific gravity,” published in 1887. “Solutions are chemical compounds determined by the forces acting between the solvent and the dissolved substance,” - he wrote. The nature of these forces is now known. Solvates (hydrates) are formed due to donor-acceptor, ion-dipole interactions, due to hydrogen bonds, as well as disperse interactions (in the case of solutions of related substances, for example benzene and toluene). They are especially prone to hydration (combination with water) ions. Ions attach polar water molecules, resulting in the formation of hydrated ions (see § 5.4); therefore, for example, in solution the cuprum(II) ion is blue, and in anhydrous cuprum sulfate it is colorless. Many of these compounds are fragile and easily decompose when isolated in a free state, but in some cases strong compounds are formed that can be easily isolated from solution by crystallization. In this case, crystals containing water molecules fall out.
Crystalline substances containing water molecules are called crystalline hydrates, and the water included in crystalline hydrates is called crystallization. Many natural minerals are crystal hydrates. A number of substances (including organic ones) are extracted in their pure form only in the form of crystalline hydrates. DI. Mendeleev proved the existence of sulfuric acid hydrates, as well as a number of other substances.
Therefore, dissolution is not only a physical, but also a chemical process. Solutions are formed by the interaction of solute particles with solvent particles. Student D.I. Mendeleeva D.P. Konovalov always emphasized that there are no boundaries between chemical compounds and solutions.
Liquid solutions occupy an intermediate position between chemical compounds of constant composition and mechanical mixtures. Like chemical compounds, they are homogeneous and characterized by thermal phenomena, as well as concentration, which is often observed - a decrease in volume when mixing liquids. On the other hand, unlike chemical compounds, solutions are not subject to the law of constancy of composition. they, like mixtures, can be easily divided into their component parts. The dissolution process is a physical-chemical process, and solutions are physical-chemical systems.
M.V. paid much attention to the study of solutions. Lomonosov. He conducted research to establish the dependence of the solubility of substances on temperature, studied the phenomena of heat release and absorption during dissolution, and discovered cooling mixtures. M.V. Lomonosov was the first to establish that solutions freeze (crystallize) at a lower temperature than the solvent. He also gave a molecular kinetic explanation of dissolution, close to the modern one, considering that the particles of the substance, dissolved, are evenly distributed among the particles of the solvent.
1 In chemical formulas and hydrates of crystalline hydrates, the formula of water is written separately (through a dot), for example H 2 SO 4 ∙ H 2 O, H 2 SO 4 2H 2 O, H 2 SO 4 ∙ 4H 2 O, H 2 C 2 O 4 ∙ 2H 2 O, N 2 SO 4 ∙ 10 H 2 O, Al 2 (S 0 4) 3 1 8H 2 O, etc.
D.I. devoted about 40 years of scientific work to the study of solutions. Mendeleev. His chemical theory of solutions turned out to be extremely fruitful. On its basis, new scientific disciplines were formed - such as physical and chemical analysis, chemistry of complex compounds, electrochemistry of non-aqueous solutions. Now this theory is generally accepted.
A significant contribution to the development of the chemical theory of solutions was made by famous Russian scientists D.P. Konovalov, 1.0. Kablukov, M.S. Kurnakov.
