Thermal effect of dissolution (enthalpy of dissolution). A textbook on chemistry for those entering higher education Thermal effects during dissolution
The dissolution of substances is accompanied by various thermal effects 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 (heat is released), and when ammonium nitrate is dissolved, a strong cooling of the solution occurs (heat is absorbed). In the first case it occurs exothermic process (?N < 0), in the second - endothermic process(?H > 0).
Heat of solution?H growth V -This is the amount of heat that is released or absorbed when 1 mole of a substance is dissolved. So, for example, under standard conditions for potassium hydroxide?H o dist = - 55.65 kJ/mol, and for ammonium nitrate?H o dist = +26.48 kJ/mol.
The heat of solution is the algebraic sum of the thermal effects of all endo- and exothermic stages of the process.
Let's consider the mechanism of dissolution of sodium chloride, a substance with an ionic crystal lattice (Fig. 2).
- 1 stage. Water molecules are dipoles, so due to electrostatic attraction get their bearings corresponding poles to the positively and negatively charged sodium and chlorine ions located on the surface of the crystal, ?Н orien? 0.
- 2 stage. Chemical bonds are formed between water molecules and sodium and chlorine ions due to ion-dipole interaction, therefore, this process is accompanied by the release of energy, ?H hydr
- 3 stage. The emergence of such bonds and the release of energy leads to the fact that the bonds in the crystal lattice weaken, and the ions in hydrated form go into solution, leaving the surface of the crystal. The process of abstraction of ions from the crystal is endothermic, ?H of abstraction > 0.
- 4 stage. Diffusion of hydrated ions throughout the entire volume of the solution, ?Н diff? 0.
If the energy of destruction of the crystal lattice is less than the energy of hydration of the dissolved substance, then dissolution occurs with the release of heat. If the energy of destruction of the crystal lattice is greater than the energy of hydration, then dissolution occurs with the absorption of heat.
When ideal solutions There are no thermal and volumetric effects: i.e. ?H dist = 0, ?V = 0, no chemical bonds are formed, but entropy increases.
The process of interaction between a solvent and a solute, as mentioned earlier, is called solvation, and if the solvent is water - hydration. 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.
Solvates (hydrates) are formed due to donor-acceptor, ion-dipole interactions, due to hydrogen bonds, as well as dispersion interactions (during the dissolution of related substances, such as benzene and toluene).
Particularly prone to hydration, i.e. connection with water molecules, ions. The ions attach to polar water molecules, resulting in the formation of hydrated ions. Therefore, for example, in solution, copper (II) ion is blue, but in anhydrous copper sulfate it is colorless. Many solvates (hydrates) are fragile and easily decompose when isolated in free form, 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 calledcrystal hydrates ,and the water included in the crystalline hydrates is calledcrystallization . Many natural minerals are crystalline hydrates. A number of substances (including organic ones) are obtained in pure form only in the form of crystalline hydrates.
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.
Sign of the change in dissolution entropy (D S o dissolution) depends on the degree of change in order in the system before and after dissolution. When gases dissolve in a liquid, the entropy of the system decreases and the enthalpy increases, so the dissolution of gases decreases as the temperature increases.
The sign of the change in enthalpy of the system during dissolution (D N o dissolution) is determined by the sum of the thermal effects of all processes accompanying dissolution. When a solid substance dissolves, its crystal lattice is destroyed and the particles of the substance are evenly distributed throughout the entire volume of the solution. This process requires energy, therefore, D N o cr > 0. At the same time, the process of interaction of particles of a dissolved substance with water occurs with the formation of hydrates, accompanied by the release of heat (D N about hydr< 0).
The overall thermal effect of solid dissolution (D N o dissolve) is determined by the ratio of the thermal effects of these processes and can be either positive or negative, or equal to zero, as when sugar is dissolved in water.
The dissolution of liquids and gases in most cases is accompanied by the release of a small amount of heat and, according to Le Chatelier's principle, their solubility decreases with decreasing temperature.
Solubility
When preparing a solution of any substance, the molecules of the solute continuously pass into the solution and, thanks to diffusion, are evenly distributed throughout the entire volume of the solvent. The molecules of the dissolved substance that have passed into the solution, hitting the surface of the not yet dissolved substance, again enter into its composition. As the solution concentration increases, the rate of solid formation increases. When the rates of these processes are equal, equilibrium is established in the system (D G o growth =0):
undissolved substance "substance in solution"
in this case, the number of molecules of a dissolved substance entering and leaving the solution per unit time becomes equal.
A solution of maximum concentration, which at a given temperature can remain in equilibrium indefinitely with an excess of solute, is called rich.
The concentration of a saturated solution is called solubility.
Solubility is expressed by the number of grams of solute contained in 100 grams of solvent, or the number of moles of solute contained in 1 liter of solution.
A solution whose concentration at a given temperature is less than saturated is called unsaturated .
The solubility of solids (for example, salts), as a rule, decreases with decreasing temperature. If you slowly cool a saturated solution, you can get oversaturated solution, i.e. a solution whose concentration is greater than the solubility of the substance at a given temperature. Supersaturated solutions are unstable (D G o rise >0) and spontaneously or under external influence (shaking, adding crystals) go into a state of equilibrium (D G o growth =0). In this case, the excess solute precipitates.
Concentration of solutions
Solution concentration is the amount of solute contained in a certain quantity or volume of a solution or solvent.
In chemistry, the following methods of expressing concentration are most commonly used.
Percentage concentration. Shows the number of grams of solute contained in 100 g of solution. For example, a 15% aqueous solution of salt is a solution whose 100 g contains 15 g of salt and 85 g of water.
Molar concentration (molarity). Shows the number of moles of a dissolved substance contained in 1 liter of solution, denoted by mol/l or the formula of the substance enclosed in square brackets. For example, =2 mol/L is a solution containing 2 moles (or 80 g) of sodium hydroxide in one liter of solution.
Molar concentration of equivalents. Shows the number of moles of solute equivalents contained in 1 liter of solution, designated WITH eq. For example, WITH eq H 2 SO 4 =0.1 mol eq/l - this is a solution of H 2 SO 4 containing 0.1 mol equivalents of sulfuric acid (or 4.9 g) in 1 liter of solution.
Equivalent(denoted by the letter E) is a real or fictitious particle of a substance that can replace, add, release, or be otherwise equivalent to one hydrogen ion in acid-base or ion-exchange reactions or one electron in redox reactions.
Acid equivalent equal to the molar mass of the acid divided by its basicity, i.e. on the number of hydrogen atoms in an acid molecule that can be replaced by a metal.
Base equivalent equal to the molar mass of the base divided by the valency of the metal.
Oxide equivalent is equal to the molar mass of the oxide divided by the product of the number of atoms of the element that make up the molecule and the valency of this element.
Salt equivalent equal to the molar mass of the salt divided by the product of the valence of the metal and the number of metal atoms in its molecule.
For example:
mole eq. H 2 SO 4 (M=98 g/mol) is equal to
mole eq. Ca(OH) 2 (M=74 g/mol) is equal to
mole eq. Al 2 O 3 (M=102 g/mol) is equal to
mole eq. Al 2 (SO 4) 3 (M=342 g/mol) is equal to
Solutions with molar concentration equivalents are widely used when carrying out reactions between solutes. Using this concentration, it is easy to calculate in advance in what volume ratios the dissolved substances must be mixed in order for them to react without leaving a residue. According to the law of equivalents the quantities of substances that react are proportional to their equivalents :
Consequently, for a reaction it is always necessary to take such volumes of solutions that would contain the same number of moles of equivalents of dissolved substances. At the same molar concentration of solutions, the volumes of reacting substances are proportional to their WITH eq. If the volumes of solutions spent on the reaction are denoted by V 1 and V 2, and their molar concentrations of equivalents are WITH eq.1 and WITH eq.2, then the relationship between these quantities will be expressed by the relation:
those. the volumes of reactants are inversely proportional to the molar concentrations of their equivalents.
Based on these dependencies, it is possible not only to calculate the volumes of solutions required for reactions, but also to find their concentrations based on the volumes of solutions spent on the reaction.
Titer. Shows the number of grams of solute contained in 1 ml of solution. Denoted by the letter T.
Knowing the titer of a solution, it is easy to calculate its molar concentration of the equivalent, and vice versa:
Molal concentration (molality). Shows the number of moles of solute contained in 1000 g of solvent, denoted WITH m:
 ,
| (5.3) |
Where m– amount of dissolved substance, – amount of solvent, g; M– molar mass of the dissolved substance, g/mol.
Raoult's laws
Each liquid at a given temperature corresponds to a certain saturated vapor pressure r 0 . As the temperature rises p 0 increases. When any non-volatile substance is dissolved in a liquid, the saturated vapor pressure of the solvent above the solution becomes lower than above the pure solvent at the same temperature. Moreover, the decrease in pressure is proportional to the concentration of the solution.
The relative decrease in the saturated vapor pressure of the solvent above the solution is equal to the mole fraction of the solute (Raoult’s law):
 | (5.4) |
Where p 0 – saturated vapor pressure over a pure solvent;
p– saturated vapor pressure of the solvent above the solution; N– mole fraction of the dissolved substance; n 1– number of moles of dissolved substance; n 2– number of moles of solvent.
Mole fraction (N i) is equal to the ratio of the number of moles of a given substance ( n i) to the sum of the number of moles of all substances (including the solvent) in solution:
A decrease in the saturated vapor pressure of a solvent over a solution of a nonvolatile substance leads to an increase in the boiling point and a decrease in the freezing point of the solution compared to a pure solvent.
According to Raoult's law, the pressure of water vapor above an aqueous solution is lower than above water.
Liquid boiling point T kip is the temperature at which the saturated vapor pressure reaches atmospheric pressure; for water it is 100°C (at a pressure of 101.3 kPa or 1.013∙10 5 N/m 2). Since the saturated vapor pressure of the solvent above the solution is lower, in order for the solution to boil, it must be heated to a higher temperature than the pure solvent.
Consequences of Raoult's law
1. The decrease in freezing point DT and the increase in boiling point DT of a non-electrolyte solution are directly proportional to the amount of substance dissolved in a given amount of solvent.
2. Equimolar (i.e. containing the same number of moles of substance equivalents) amounts of dissolved substances, being dissolved in the same amount of a given solvent, equally lower its freezing point and equally increase its boiling point.
The decrease in freezing point caused by the dissolution of one mole of a substance in 1000 g of solvent is a constant value for a given solvent. It is called cryoscopic constant K To solvent. In the same way, the increase in boiling point caused by the dissolution of one mole of a substance in 1000 g of solvent is called ebullioscopic constant K uh solvent. Cryoscopic and ebullioscopic constants depend only on the nature of the solvent.
A solution is a homogeneous system consisting of two or more components. When a substance passes into a solution, the intermolecular and ionic bonds of the crystal lattice of the solid substance are broken and it passes into the solution in the form of individual molecules or ions, which are evenly distributed among the solvent molecules.
To destroy the crystal lattice of a substance, it is necessary to expend a lot of energy. This energy is released as a result of hydration (solvation) of ions and molecules, i.e., the chemical interaction of a solute with water (or with a solvent in general).
This means that the solubility of a substance depends on the difference between the hydration (solvation) energy and the energy of the crystal lattice of the substance.
Energy of dissolution ∆H dist is the energy absorbed (or released) when 1 mole of a substance is dissolved in such a volume of solvent, the further addition of which does not cause a change in the thermal effect.
The overall thermal effect of dissolution depends on the thermal effects:
· a) destruction of the crystal lattice (the process always occurs with energy consumption ∆Н 1 >0);
· b) diffusion of a dissolved substance in a solvent (energy consumption ∆H 2 >0);
c) solvation (hydration) (heat release, ∆H 3<0, так как между растворителем и растворенным веществом образуются непрочные химические связи, что всегда сопровождается выделением энергии).
The total thermal effect of dissolution ∆H p will be equal to the sum of the above thermal effects
The energy of dissolution is determined by formula 1.1:
∆Н pac t =∆Н to p. R. + ∆Н c , (1.1)
where ∆H dist is the energy of dissolution of the substance, kJ/mol;
∆H c - energy of interaction of the solvent with the soluble
substance (solvation energy), kJ/mol;
∆H to p.r. - energy of destruction of the crystal lattice,
kJ/mol.
If the energy of destruction of the crystal lattice is greater than the energy of solvation, then the dissolution process will be an endothermic process, since the energy expended on the destruction of the crystal structure will not be compensated by the energy released during solvation.
If the energy of destruction of the crystal lattice is less than the energy of solvation, then the dissolution process will be an exothermic process, since the energy expended on the destruction of the crystal structure is completely compensated by the energy released during solvation. Consequently, depending on the relationship between the energy of destruction of the crystal lattice of the solute and the energy of interaction of the solute with the solvent (solvation), the energy of dissolution can be either positive or negative.
Thus, when sodium chloride is dissolved in water, the temperature practically does not change, when potassium or ammonium nitrate is dissolved, the temperature decreases sharply, and when potassium hydroxide or sulfuric acid is dissolved, the temperature of the solution rises sharply.
The dissolution of solids in water is often an endothermic process, since in many cases less heat is released during hydration than is spent on the destruction of the crystal lattice.
The energy of the crystal lattice can be calculated theoretically. However, there are still no reliable methods for theoretical calculation of solvation energy.
There are some regularities that relate the solubility of substances to their composition.
For salts of the same anion with different cations (or vice versa), the solubility will be lowest in the case when the salt is formed by ions of the same charge and approximately the same size, because in this case, the energy of the ionic crystal lattice is maximum.
For example, the solubility of sulfates of elements of the second group of the periodic table decreases by subgroup from top to bottom (from magnesium to barium). This is explained by the fact that barium and sulfate ions are most similar in size to each other. While calcium and magnesium cations are much smaller than SO 4 2- anions.
The solubility of hydroxides of these elements, on the contrary, increases from magnesium to barium, because the radii of magnesium cations and hydroxide anions are almost the same, and barium cations are very different in size from small hydroxyl anions.
However, there are exceptions, for example, for oxalates and carbonates of calcium, strontium, barium, etc.
1) using the temperature change during dissolution.
The amount of energy released when a body is heated or cooled is calculated using equation (1.2):
, (1.2)
where ∆Н sol. – energy of dissolution of the substance, kJ/mol;
c A - specific heat capacity of substance A, J/(g∙K);
m 1 - mass of substance A, g;
∆T – temperature change, degrees.
EXAMPLE 1.1 When 8 g of ammonium chloride was dissolved in 291 g of water, the temperature decreased by 2 0 . Calculate the heat of dissolution of NH 4 C1 in water, taking the specific heat of the resulting solution equal to the heat capacity of water 4.1870 J/(g * K).
Solution:
Using equation (1.2), we calculate the energy absorbed by 291 g of water when dissolving 8 g of NH 4 C1, because in this case, the temperature decreases by 2 0 C, then: ∆Н sol. = -(4.187∙291∙(-2)) = 2436.8 J.
To determine the enthalpy of dissolution of NH 4 C1, we compose the proportion, M (NH 4 C1) = 53.49 g/mol:
8g NH 4 Cl - 2436.8 J
53.49g NH 4 C1 - x J
x = 1629.3 J = 16.3 kJ. Consequently, the dissolution of NH 4 C1 is accompanied by heat absorption.
2) using a corollary from Hess’s law: the thermal effect of a chemical reaction (ΔH 0 c.r.) is equal to the sum of the heats (enthalpies) of formation of the reaction products (ΔH 0 o 6р. . npo d.) minus the sum of the heats (enthalpies) of formation of the starting substances (ΔH 0 arr. ref.) with taking into account the coefficients in front of the formulas of these substances in the reaction equation.
ΔН 0 h.r.= ΣΔН 0 return prod - Σ ΔН 0 return out, (1.3)
EXAMPLE 1.2 Calculate the thermal effect of the reaction of dissolving aluminum in dilute hydrochloric acid if the standard heats of formation of the reacting substances are equal (kJ/mol): ∆H 0 (HC1) ( aq ) = - 167.5; ∆Н 0 А1С1 3 (а q) = -672.3.
Solution: The dissolution reaction of A1 in hydrochloric acid proceeds according to the equation 2A1 + 6HC1 (aq) = 2AlCl 3 (aq) + 3H 2. Since aluminum and hydrogen are simple substances, then for them ΔН 0 =0 kJ/mol, the thermal effect of the dissolution reaction is equal to:
∆Н 0 298 =2∙∆Н 0 А1С1 3 (а q) -6∙∆Н 0 НС1 (aq)
∆Н 0 298 =2∙(-672.3)-6∙(-167.56)=-339.2 kJ.
Using a consequence of Hess's law, one can determine the possibility of a dissolution reaction occurring. In this case, it is necessary to calculate the Gibbs energy.
EXAMPLE 1.3 Will copper sulfide dissolve in dilute sulfuric acid if the Gibbs energy of the reactants is equal (kJ/mol): ∆G 0 (CuS (k)) = -48.95; ∆G 0 (H 2 SO 4 (aq)) = -742.5; ∆G 0 (CuSO 4 (aq)) = -677.5, ∆G 0 (H 2 S (g)) = -33.02.
Solution. To answer, you need to calculate ∆G 0 298 dissolution reactions. A possible reaction of dissolving CuS in dilute H 2 SO 4 proceeds according to the equation:
CuS (k) + H 2 SO 4 (aq) = CuSO 4 (aq) + H 2 S (g)
∆G 0 298 =∆G 0 (CuSO 4(aq)) + ∆G 0 (H 2 S (g)) -∆G 0 (CuS (K)) -∆G 0 (H 2 SO 4(aq))
∆G 0 298 = -677.5-33.02 + 742.5 + 48.95 =80.93 kJ/mol.
Since ∆G>0, the reaction is impossible, i.e. CuS will not dissolve in dilute H 2 SO 4 .
Heat of hydration ∆Н 0 hydrate. - heat released during the interaction of 1 mole of a solute with a solvent - water.
EXAMPLE 1.4. When 52.06 g of BaCl 2 is dissolved in 400 mol of H 2 O, 2.16 kJ of heat is released, and when 1 mol of BaC1 2 ∙2H 2 O is dissolved in 400 mol of H 2 O, 18.49 kJ of heat is absorbed. Calculate the heat of hydration of anhydrous BaCl 2,
Solution. The dissolution process of anhydrous BaCl 2 can be represented as follows:
a) hydration of anhydrous salt BaCl 2
BaC1 2 +2H 2 O = BaC1 2 ∙2H 2 O; ∆H hydr.<0
b) dissolution of the formed hydrate
BaCl 2 ∙2H 2 O+aq* → BaCl 2 ∙2H 2 O (aq); ∆Н rast. >0
The amount of heat ∆H 0 released during the dissolution of anhydrous BaCl 2 is equal to the algebraic sum of the thermal effects of these two processes:
∆Н 0 == ∆Н 0 hydr +∆Н 0 sol; ∆H 0 hydr = ∆H 0 - ∆H 0 solution
To calculate the heat of hydration of anhydrous barium chloride, it is necessary to determine the heat of solution of BaCl 2 for the same conditions as for BaCl 2 ∙2H 2 O, i.e. for 1 mol of BaCl 2 (the solution in both cases must have the same concentration); M(BaCl 2) = 208.25 g/mol
52.06 g BaCl 2 - 2.16 kJ
208.25 g BaCl 2 - x kJ
x=8.64 kJ/mol. Therefore, ∆Н sol = -8.64 kJ/mol.
Then ∆H hydr =18.49+8.64 =27.13 kJ/mol.