What factors determine the thermal effect of the dissolution process. The thermal effect of dissolution (ENTALPIA dissolution)

Solutions are single-phase variable composition systems consisting of several components, one of which is a solvent, and other - dissolved substances. The fact that solutions are single-phase systems relating to them with chemical compounds, and the fact that they are variable composition systems, bringing them closer with mechanical mixtures. Therefore, it is believed that the solutions have a dual nature: on the one hand, they are similar to chemical compounds, and on the other - with mechanical mixtures.

Dissolution is a physico-chemical process. In physical phenomenon, the crystal lattice is destroyed and the diffusion of solute molecules occurs. When a chemical appearance, in the process of dissolution, the dissolved substance molecule is reacting with solvent molecules.

The dissolution process is accompanied by the release or absorption of heat. This warmth, assigned to one praying of the substance, is called the thermal effect of dissolution, QP.

The total thermal effect of dissolution depends on thermal effects:
a) the destruction of the crystal lattice (the process always comes with consideration - Q 1);
b) diffusion of the dissolved substance in the solvent (energy costs - Q 2);
c) hydration (heat outlook, + q 3, since hydrates are formed by the emergence of fragile chemical bond, which is always accompanied by the release of energy).

The total thermal effect of dissolution Qp will be equal to the sum of the title thermal effects: Qp \u003d (-q 1) + (- Q 2) + (+ Q 3); If q 1\u003e q 3\u003e The dissolution goes with the absorption of heat, that is, the process of endothermic, if Q 1< Q 3 , то растворение идет с выделением теплоты, то есть процесс экзотермический. Например, растворение NaCl, KN0 3 , NH 4 CNS идет с поглощением теплоты, растворение NaOH, H 2 S0 4 - с выделением теплоты.

A task. Why, when dissolved in water, sodium chloride temperature decreases, and when sulfuric acid is dissolved - rises?

Answer. When dissolved sodium chloride, the crystal lattice is destroyed, which is accompanied by considerable energy. The diffusion process is spent a slight amount of energy. The hydration of ions is always accompanied by the release of energy. Therefore, if the temperature is reduced during the dissolution, the energy required to destroy the crystal lattice turns out to be greater than the energy released during hydration, and in general the solution is cooled.

The thermal effect of the dissolution of sulfuric acid consists mainly of the heat of hydration of ions, so the solution is heated.

Solubility of matter - It is its ability to be distributed in the solvent environment. The solubility (or solubility coefficient) is determined by the maximum amount of grams of a substance that can dissolve in 100 grams of the solvent at a given temperature.

The solubility of most solids increases with heating. There are exceptions, that is, such substances whose solubility with increasing temperature changes (NaCl) or even falls (sa (O) 2).

The solubility of gases in water drops with heating and increases with an increase in pressure.

The solubility of substances is associated with the nature of the dissolved substance. Polar and ionic compounds are usually solved well in polar solvents, and non-polar compounds in non-polar solvents. Thus, chloride and ammonia are well soluble in water, while hydrogen, chlorine, nitrogen dissolve in water is much worse.

Dissolution is a physicochemical process leading to the formation of a homogeneous system. The thermal effects accompanying it are a consequence of a wide variety of reasons. Consider a few examples:

A) The process of dissolution in water of liquids can be accompanied by such phenomena as the dissociation of polar molecules to form ions, the occurrence of hydrogen bonds between the polar water molecules and molecules of substances containing elements with high electronegitability, hydration of chemical particles, etc.

From 2N 5 ON - H 2

This system corresponds to the formation of ideal solutions in a wide range of concentrations. The dissolution process should be accompanied by the formation of hydrogen bonds, therefore, is energetically profitable, that is, it has a positive thermal effect.

CH 3 coxy - H 2 o

Acetic acid is a weak monosocondary acid to d \u003d 1.8 10 -5, therefore, when dissolved in water, some part of the energy will be spent on the dissociation of molecules (negative thermal effect), and part 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 values.

B) The process of dissolving solids in water depends on the type of crystal lattice of the latter. As a rule, dissolving ionic crystals is associated with two opposite effects: the positive value of the energy of hydration of ions and negative - the destruction of the crystal lattice. Molecular crystals the first component is practically absent. When pluming dilute solutions of solts of strong electrolytes, the thermal effect is not observed. If the precipitate is formed, the thermal deposition effect is observed.

Integral heat dissolution - This is the amount of heat absorbed or released during dissolution 1 mol of substance in a very large (300 mol / mol substance) of the solvent.

Example of the settlement task:

Calculate the integral heat dissolution of ammonium chloride, if when dissolved 1,473 g of salts in 528.5 g of water, the temperature dropped by 0.174 o C. Mass heat capacity of the solution of 4,109 J / g. K. The heat capacity of the calorimeter of 181.4 J / g to

Decision:Integral heat dissolution can be calculated by the formula:

Q \u003d (C calorim. + C p-ra. M) × ΔT / n,

where C is heat capacity, n - the amount of the dissolved substance: n \u003d m / m

m (p-ra) \u003d 528.5 +1,473 \u003d 530 g,

ΔT \u003d -0.174 o C,

Q \u003d (4.109 × 530 + 181.4) × (-0.174) × 53.5 / 1,473 × 1000 \u003d -15,11 kJ / mol from the course of chemical thermodynamics It is known that the measure of the thermal effect of the chemical process during the isobaric process (the constancy of pressure in the system) is the thermodynamic function of the state - enthalpy

ΔH \u003d n con. - n beginning. The thermal effect is equal in the absolute value of enthalpy, but the sign is opposite to it. The exothermic process, accompanied by heat release, corresponds to -Δn, and endothermal, accompanied by heat absorption, corresponds to + ΔH. In the way, the process of dissolving ammonium chloride - endothermic, ΔH \u003d 15,11 kJ / mol.

The solution is called a homogeneous system consisting of two or more components. In the transition of a substance into solution, the intemolecular and ionic bonds of the crystalline solid grid occurs and the transition to the solution in the form of individual molecules or ions, which are evenly distributed among the solvent molecules.

To destroy the crystal lattice of the substance, it is necessary to spend more energy. This energy is exempt as a result of hydration (solvation) of ions and molecules, i.e., the chemical interaction of the soluble substance with water (or in general with the solvent).

It means that the solubility of the substance depends on the difference in the amount of energy of hydration (solvation) and the energy of the crystal lattice of the substance.

Energy dissolution ΔH Rast - energy absorbed (or released) when dissolved 1 mol of substance in such a volume of solvent, the further addition of which does not cause changes in the thermal effect.

The total thermal effect of dissolution depends on thermal effects:

· A) the destruction of the crystal lattice (the process always comes with the cost of Δn 1\u003e 0);

· B) diffusion of the dissolved substance in the solvent (energy cost ΔH 2\u003e 0);

· C) solvation (hydration) (heat release, ΔH 3<0, так как между растворителем и растворенным веществом образуются непрочные химические связи, что всегда сопровождается выделением энергии).

The total thermal effect of dissolution ΔH p will be equal to the sum of the named thermal effects

The dissolution energy is determined by Formula 1.1:

ΔH Pac T \u003d ΔH to p. R. + ΔH c, (1.1)

where ΔH ras - the dissolution energy of the substance, KJ / mol;

ΔН C - solvent interaction energy with solvent

substance (solvation energy), KJ / mol;

ΔН to p .r. - the energy of the destruction of the crystal lattice,

kJ / mol.

If the energy of the destruction of the crystal lattice is greater than the energy of solvation, then the dissolution process will be an endothermic process, since the energy spent on the destruction of the crystal structure will not be compensated for by the energy released during solvation.

If the energy of the destruction of the crystal lattice is less than the energy of solvation, then the dissolution process will be an exothermic process, since the energy spent 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 the destruction of the crystal lattice of the dissolved substance and the interaction energy of the dissolved substance with the solvent (solvation), the dissolution energy can be both positive and negative.

Thus, when dissolved in water sodium chloride, the temperature practically does not change, when dissolving potassium nitrate or ammonium, the temperature is sharply reduced, and when potassium hydroxide is dissolved or sulfuric acid, the solution temperature rises sharply.

The dissolution of solids in water is more likely to endothermic process, since in many cases the heat is released in the hydration less than is spent on the destruction of the crystal lattice.

The energy of the crystal lattice can be calculated theoretically. However, there is still no reliable methods for the theoretical calculation of the energy of solvation.

There are some patterns that bind the solubility of substances with their composition.

For salts of the same anion with different cations (or vice versa), solubility will be the smallest in the case when the salt is formed by ions of the same charge and about the same size, because In this case, the energy of the ion crystal lattice is maximum.

For example, the solubility of sulfates of elements of the second group of the periodic system is reduced by a subgroup from top to bottom (from magnesium to the barium). This is explained by the fact that barium and sulfate ions are most suitable for each other. While calcium and magnesium cations are much less than Anions SO 4 2-.

The solubility of the 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 the bary cations are very different from small hydroxyl anions.

However, there are exceptions, for example, for oxalates and calcium carbonates, strontium, barium, etc.

1) Using the temperature change during dissolution.

The amount of energy released during heating or cooling of the body is calculated by equation (1.2):

, (1.2)

where Δn is suited. - energy dissolution of substance, KJ / mol;

c A is the specific heat capacity of the substance A, J / (g ∙ K);

m 1 - mass of substance A, r;

ΔТ - change in temperature, hail.

Example 1.1. When the ammonium chloride is dissolved, the temperature dropped to 2 0 by 2 0. Calculate the heat dissolution of NH 4 C1 in water, taking the specific heat capacity of the resulting solution of water heat capacity of 4,1870 J / (g * K).

Decision:

Using equation (1.2), we calculate the energy absorbed by 291 g of water when dissolved 8g NH 4 C1, because At the same time, the temperature decreases by 2 0 s, then: ΔH is satisfied. \u003d - (4,187 ∙ 291 ∙ (-2)) \u003d 2436.8 J.

To determine the enthalpy of dissolution NH 4 C1 we compile proportion, M (NH 4 C1) \u003d 53.49 g / mol:

8g NH 4 Cl - 2436.8 J

53,49g NH 4 C1 - X J

x \u003d 1629,3J \u003d 16,3KJ. Consequently, the dissolution of NH 4 C1 is accompanied by heat absorption.

2) Using the consequence of the GESSA; the thermal effect of the chemical reaction (ΔH 0 H.R.) is equal to the sum of heat (enthalpy) of the formation of reaction products (ΔH 0 O 6P. NPO d.) Minus the amount of heat (enthalpy) of the formation of starting materials (ΔH 0 arr. Ex.) C accounting of coefficients before the formulas of these substances in the reaction equation.

ΔH 0 x.r.= ΣΔH 0 Obr.Prod - Σ ΔH 0 Obr.Ish, (1.3)

Example 1.2. Calculate the thermal effect of the dissolution of aluminum in dilute hydrochloric acid, if the standard heat of the formation of reactants is equal to (KJ / mol): ΔH 0 (NS1) (AQ) \u003d - 167.5; ΔH 0 A1C1 3 (A Q) \u003d -672.3.

Decision: The dissolution response A1 in hydrochloric acid flows through equation 2a1 + 6ns1 (AQ) \u003d 2ALCl 3 (AQ) + 3H 2. Since aluminum and hydrogen are simple substances, for them ΔH 0 \u003d 0 kJ / mol, then the thermal effect of the dissolution reaction is:

ΔH 0 298 \u003d 2 ∙ ΔH 0 A1C1 3 (A Q) -6 ∙ ΔH 0 ns1 (AQ)

ΔН 0 298 \u003d 2 ∙ (-672.3) -6 ∙ (-167,56) \u003d - 339,2kj.

Using the consequence of the GESS law, it is possible to determine the possibility of a dissolution reaction. In this case, it is necessary to calculate the energy of Gibbs.

Example 1.3. Will the copper sulfide in dilute sulfuric acid be dissolved if the Gibbs energy of the reactant substances is equal to (KJ / mol): Δg 0 (cus (k)) \u003d -48.95; Δg 0 (H 2 SO 4 (AQ)) \u003d - 742.5; Δg 0 (Cuso 4 (AQ)) \u003d -677,5, Δg 0 (H 2 S (g)) \u003d -33.02.

Decision. To respond, it is necessary to calculate Δg 0 298 of the dissolution reaction. The possible response of CUS dissolution in the diluted H 2 SO 4 flows through the equation:

CUS (K) + H 2 SO 4 (AQ) \u003d Cuso 4 (AQ) + H 2 S (g)

Δg 0 298 \u003d Δg 0 (Cuso 4 (AQ)) + Δg 0 (H 2 S (g)) -Δg 0 (cus (k)) -Δg 0 (H 2 SO 4 (AQ))

Δg 0 298 \u003d -677,5-33.02 + 742.5 + 48.95 \u003d 80.93 kJ / mol.

Since Δg\u003e 0, the reaction is not possible, i.e. CUS will not be dissolved in dilute H 2 SO 4.

Heat hydration ΔH 0 hydrate. - Heat, isolated by interaction of 1 mol soluble substance with a solvent - water.

Example 1.4. When dissolved 52.06g you1 2 in 400 mol H 2 o 2,16 kJ of heat, and when dissolved 1 mol of you1 2 ∙ 2N 2 o 400 mol H 2 Oh, 18.49 kJ heat is absorbed. Calculate the warmth of the hydration of anhydrous you1 2,

Decision. The process of dissolving anhydrous you1 2 can be represented as follows:

a) hydration of anhydrous salt of you1 2

You1 2 + 2n 2 o \u003d you1 2 ∙ 2n 2 o; ΔH hydra.<0

b) dissolution of the resulting hydrate

BACL 2 ∙ 2H 2 O + AQ * → You1 2 ∙ 2N 2 O (AQ); ΔH Rast. \u003e 0.

The amount of heat ΔH 0, released when dissolving anhydrous you1 2, is equal to the algebraic sum of the thermal effects of these two processes:

ΔH 0 \u003d\u003d ΔH 0 hydr + ΔH 0 is removed; ΔH 0 hydr \u003d ΔH 0 - ΔH 0

To calculate the heat of the hydration of anhydrous bary chloride, it is necessary to determine the heat dissolution of you1 2 for the same conditions as for you1 2 ∙ 2N 2 O, i.e. for 1 mol of you1 2 (the solution in both cases should have the same concentration); M (BaCl 2) \u003d 208.25 g / mol

52.06g you1 2 - 2,16kJ

208.25g you1 2 - x kj

x \u003d 8.64 kJ / mol. Consequently, Δn is satisfied \u003d -8.64 kJ / mol.

Then ΔH hydr \u003d 18.49 + 8.64 \u003d 27.13 kJ / mol.

It means that the solubility of the substance depends on the difference in the amount of energy of hydration (solvation) and the energy of the crystal lattice of the substance.

The total thermal effect of dissolution depends on thermal effects:

· A) the destruction of the crystal lattice (the process always comes with the cost of Δn 1\u003e 0);

· B) diffusion of the dissolved substance in the solvent (energy cost ΔH 2\u003e 0);

The total thermal effect of dissolution ΔH p will be equal to the sum of the named thermal effects

The dissolution energy is determined by Formula 1.1:

ΔH Pac T \u003d ΔH to p. R. + ΔH c, (1.1)

where ΔH ras - the dissolution energy of the substance, KJ / mol;

ΔН C - solvent interaction energy with solvent

substance (solvation energy), KJ / mol;

ΔН to p .r. - the energy of the destruction of the crystal lattice,

kJ / mol.

The dissolution of solids in water is more likely to endothermic process, since in many cases the heat is released in the hydration less than is spent on the destruction of the crystal lattice.

The energy of the crystal lattice can be calculated theoretically. However, there is still no reliable methods for the theoretical calculation of the energy of solvation.

There are some patterns that bind the solubility of substances with their composition.

However, there are exceptions, for example, for oxalates and calcium carbonates, strontium, barium, etc.

1) Using the temperature change during dissolution.

The amount of energy released during heating or cooling of the body is calculated by equation (1.2):

, (1.2)

where Δn is suited. - energy dissolution of substance, KJ / mol;

c A is the specific heat capacity of the substance A, J / (g ∙ K);

m 1 - mass of substance A, r;

ΔТ - change in temperature, hail.

Decision:

To determine the enthalpy of dissolution NH 4 C1 we compile proportion, M (NH 4 C1) \u003d 53.49 g / mol:

8g NH 4 Cl - 2436.8 J

53,49g NH 4 C1 - X J

x \u003d 1629,3J \u003d 16,3KJ. Consequently, the dissolution of NH 4 C1 is accompanied by heat absorption.

ΔH 0 x.r.= ΣΔH 0 Obr.Prod - Σ ΔH 0 Obr.Ish, (1.3)

ΔH 0 298 \u003d 2 ∙ ΔH 0 A1C1 3 (A Q) -6 ∙ ΔH 0 ns1 (AQ)

ΔН 0 298 \u003d 2 ∙ (-672.3) -6 ∙ (-167,56) \u003d - 339,2kj.

Using the consequence of the GESS law, it is possible to determine the possibility of a dissolution reaction. In this case, it is necessary to calculate the energy of Gibbs.

Decision. To respond, it is necessary to calculate Δg 0 298 of the dissolution reaction. The possible response of CUS dissolution in the diluted H 2 SO 4 flows through the equation:

CUS (K) + H 2 SO 4 (AQ) \u003d Cuso 4 (AQ) + H 2 S (g)

Δg 0 298 \u003d Δg 0 (Cuso 4 (AQ)) + Δg 0 (H 2 S (g)) -Δg 0 (cus (k)) -Δg 0 (H 2 SO 4 (AQ))

Δg 0 298 \u003d -677,5-33.02 + 742.5 + 48.95 \u003d 80.93 kJ / mol.

Since Δg\u003e 0, the reaction is not possible, i.e. CUS will not be dissolved in dilute H 2 SO 4.

Heat hydration ΔH 0 hydrate. - Heat, isolated by interaction of 1 mol soluble substance with a solvent - water.

Decision. The process of dissolving anhydrous you1 2 can be represented as follows:

a) hydration of anhydrous salt of you1 2

You1 2 + 2n 2 o \u003d you1 2 ∙ 2n 2 o; ΔH hydra.<0

b) dissolution of the resulting hydrate

BACL 2 ∙ 2H 2 O + AQ * → You1 2 ∙ 2N 2 O (AQ); ΔH Rast. \u003e 0.

The amount of heat ΔH 0, released when dissolving anhydrous you1 2, is equal to the algebraic sum of the thermal effects of these two processes:

ΔH 0 \u003d\u003d ΔH 0 hydr + ΔH 0 is removed; ΔH 0 hydr \u003d ΔH 0 - ΔH 0

52.06g you1 2 - 2,16kJ

208.25g you1 2 - x kj

x \u003d 8.64 kJ / mol. Consequently, Δn is satisfied \u003d -8.64 kJ / mol.

Then ΔH hydr \u003d 18.49 + 8.64 \u003d 27.13 kJ / mol.

SOLUBILITY

The most common liquid solvent is water. It has the most dissolving and dissociating ability. For water, the dissolution temperature is limited to the interval of 0-100 0 C.

Most substances dissolving in water are solid.

The process of dissolving the substance is accompanied by diffusion, i.e. The movement of molecules from the regions of a concentrated solution in the region with a smaller concentration. In other words, the substance at dissolution is evenly distributed over the entire mass of the solvent.

The dissolution process occurs until the concentration of this substance in the solution does not reach a certain value at which the equilibrium state occurs:

solid phase solution

The ability of the solid to translate into the solution is not irreplaceable. When introduced into a glass with water (T \u003d const), the first portions of the substance are completely dissolved, and formed unsaturated solution. In this solution, the dissolution of the following portions is possible until the substance ceases to move into the solution and part of it will remain as a sediment at the bottom of the glass.

The dissolution is a bidirectional process: the solid is transferred to the solution, and the dissolved substance in turn goes into a solid phase. If the amount of substance transmitting to the solution per unit of time is equal to the amount of substance that is released during the same time into a solid phase, then this means that the solution is saturated. The solution formed in this case is called saturated solution . An increase in the concentration of the solution slows down the establishment of equilibrium.

A condition of heterogeneous equilibrium is established between substance in a saturated solution and substance in the sediment. The particles of the dissolved substance are moving through the surface of the separation of their liquid phase (solution) into the solid phase (precipitate) and back, therefore the composition of the saturated solution remains constant at a certain fixed temperature. Saturated solutions are stable systems, i.e., they can exist at this temperature without changing the concentration arbitrarily for a long time.

The concentration of a saturated solution changes with a change in temperature. When the temperature decreases, the solution can under certain conditions for some time to maintain this concentration of the substance, i.e. the concentration of the solution may be higher than in a saturated solution at a given temperature. Such solutions are called improved . Explosive solutions are unstable systems. It is enough to mix such a solution or throw the smallest crystalline of the dissolved substance (seed) to start the solid phase. This process continues until the concentration of the substance reaches the concentration of a saturated solution at a given temperature. The possibility of existence of an intersening solution is due to the difficulty of crystallization centers.

In solutions of electrolytes, the processes of ionization and association continuously occur. In this case, the equilibrium is maintained, the composition of the solution is maintained by constant, but the electrolytic dissociation process does not stop. If you enter some other substance into the solution, then its ions can react with the first substance and form a new substance that has not been introduced into the solution. For example, in separate solutions of barium chloride and sodium sulfate, an equilibrium is established:

in the first solution: you1 2 ↔ VA 2+ + 2C1 -,

in the second solution: Na 2 SO 4 ↔ 2NA + + SO 4 2-.

Both of these compounds are salts and relate to strong electrolytes, i.e., in dilute solutions, these substances are mainly in the form of ions. If these two solutions merge, then SO 4 ions will meet not only with sodium ions, but also with the barium ions and come to the reaction:

SO 4 2- + Ba 2+ ↔ Baso 4.

This reaction occurs, since barium sulfate is a low-soluble compound and falls into a precipitate. In the solution, sodium cations and chlorine anions will remain in the solution - but the sediment is not formed, because sodium chloride is well soluble in water.

The precipitation process occurs gradually. First, very small crystals are formed - embryos, which gradually grow into large crystals or group of crystals. Time from the moment of mixing the solutions before the formation of embryos - small crystals call induction period . The duration of this period depends on the individual properties of the sediment. So, in the case of the formation of silver chloride, this time is very small, in the case of the formation of barium sulfate, this period is significantly larger.

Chemical analysis should be carried out so that a smaller number of small crystals (embryos) are formed if possible, then with the gradual addiction, the existing crystallization centers will increase, i.e. large crystals will grow.

Solubility of matter - high-quality and quantitative ability of a substance to form a solution when mixing with another substance (solvent).

The solubility of substances is determined by the concentration of a solution saturated at this temperature.

The composition of the saturated solution can be expressed by any known method (mass fraction, molar concentration, etc.). More often than other values s solubility coefficient K s - The ratio of the mass of anhydrous solute substance to the mass of the solvent, for example, at 20 0 s solubility coefficient is 0.316 for KNO 3, which corresponds to a 24,012% or 2.759 m solution.

Solubility is often expressed by the amount of grams of soluble substance in 100g solvent.

Example 2.1 Calculate the solubility coefficient of you1 2 in water at 0 ° C, if at this temperature of 13.1g solution contains 3.1g you1 2.

Decision. The solubility coefficient is expressed by a mass of substance (g), which can be dissolved in 100g solvent at a given temperature. Mass solution of you1 2 13.1g. Therefore, in a 10g solvent at 0 ° C contains 3.1g you1 2. The solubility coefficient of you1 2 at 0 0 s is:

In the case of dissolving solid or liquid substances in liquids, solubility increases with an increase in temperature, and for gases - decreases. Pressure has a great influence on the solubility of gases.

By solubility at T \u003d const distinguishes:

1) well-soluble substances (form\u003e 0.1 m saturated solutions),

2) low-soluble substances (form 0.1 - 0.001m saturated solutions),

3) practically insoluble substances (form<0,001М насыщенные растворы).

For example, MgCl 2 is a well-soluble substance (at 20 0 s forms a saturated solution), MgCo 3 is a low-soluble substance (forms 0.02m solution) and Mg (OH) 2 - almost insoluble substance (forms 1.2 ∙ 10 -4 m Solution).

The solubility of the substance depends on its nature and the aggregate state to dissolution, as well as on the nature of the solvent and the temperature of the preparation of the solution, and for gases also on pressure.

The amount of heat that is released or is absorbed by dissolving 1 praying substance in such an amount of solvent, the further addition of which no longer causes changes in the thermal effect is called the heat of dissolution.

When dissolving salts in water, the sign and the magnitude of the thermal effect of dissolution δ N. Determined by two values: the energy spent on the destruction of the crystal lattice of the substance (δ H. 1) - endothermic process, and energy allocated in the physicochemical interaction of particles of the soluble substance with water molecules (hydration process) (δ 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 both positive and negative.

To practically determine the heat of dissolution, the amount of heat absorbed or allocated during the dissolution of an arbitrary amount of salt is determined. Then this magnitude is recalculated by 1 mol, since the amount of heat is directly proportional to the amount of solute substance.

For thermochemical measurements, a device called a calorimeter is used.

The determination of heat dissolution leads to a change in the temperature of the solution, so the accuracy of the definition depends on the price of dividing (accuracy) of the thermometer used. Typically, the range of measured temperatures lies in the range of 2-3 ° C, and the price of dividing the thermometer is not more than 0.05 ° C.

PROGRESS

To perform work, use a calorimeter consisting of a thermal insulation body, covers with a built-in electrical stirrer and a thermometer, as well as a hole with a plug.

Get a task from the teacher: the type of soluble substance.

Open the cork on the calorimeter lid and fill it into it 200 ml of water, close the cork and hold 10-15 minutes to establish a constant temperature ( t. nach. ). During this time, on scales using a traction or hour glass, get a sample of your substance (1.5 - 2.0 g) pre-gripped into a mortar. The resulting hitch, if possible, quickly, paste through the hole in the lid in the calorimeter when the stirrer is turned on. Follow the temperature change. After establishing thermal equilibrium (temperature stabilizes), write down the maximum solution temperature ( t. MAH) and calculate Δ t. = t. Max - t. nach According to the data obtained, calculate the heat dissolution of salt using the equation:

∆N. Sol \u003d q.M / m., J / Mol, (1)

where q. - heat released (or absorbed) in calorimeter (CJ); m. - Salt Side (D); M is the molar mass of the soluble substance (g / mol);

Heat q. Determined on the basis of experimental data from the ratio:

Q. = (m. Art C. st + m. R-R. C. p-ra) Δ t.,(2)

where m. st - weight of glass (g); m. p-ra - mass of solution, equal to the sum of mass of water and salt in a glass (g); FROM ST - specific glass heat capacity 0,753 J / g ∙ K;

FROM Р-р - specific heat capacity of the solution (water) 4,184 j / g ∙ to.

By comparing the result obtained with the data Table 2, calculate the relative experience error (in%).

Heat hydration salts and its definition

The physicochemical process of the interaction of particles of the dissolved substance with water molecules (solvent) is called hydration. In the process of hydration, complex spatial structures are formed, called hydrates, and at the same time the energy in the form of heat is distinguished into the environment.

The thermal effect of the formation reaction 1 mol of the hydrated salt from anhydrous salt is called heat of hydration.

When anhydrous salt dissolved in water capable of forming hydrates, two processes are consistently proceed: hydration and dissolution of the formal crystallohydrate. For example:

CUSO 4 (TV) + 5N 2 O (g) \u003d CUSO 4 × 5H 2 O (TV),

CUSO 4 × 5H 2 O (TV) + n.H 2 O (g) \u003d Cuso 4 (P),

CUSO 4 (P) + n.H 2 O (g) \u003d Cu 2+ (p) + SO 4 2- (P)

The dissolution of electrolytes is accompanied by the electrolytic dissociation process. The heat of the hydration of the molecule is equal to the sum of the heat of hydration formed at the same time, taking into account the heat of dissociation. The process of hydration-exothermic.

Approximately the heat of hydration of the substance can be determined as a difference between the heat dissolving the anhydrous salt and its crystallohydrate:

∆H. Hydra \u003d Δ. H. versions - Δ. H. CRIST, (3)

where Δ. H. Hydra - heat hydration of molecules;

∆H. versions - heat dissolving anhydrous salt;

∆H. Crystal is the warmth of the dissolution of the crystallohydrate.

Thus, to determine the heat of hydration of molecules, it is necessary to pre-determine the heat dissolving of anhydrous salt and the heat of dissolution of the crystallohydrate of this salt.

PROGRESS

The warmth of the dissolution of anhydrous cust0 4 copper sulfate and CUS0 4 × 5H 2 0 crystalline is needed using a laboratory calorimeter and a working method 1.

To more accurately determine the heat of hydration, it is necessary to obtain samples of 10-15 g of crystallohydrate and anhydrous salt of copper sulfate. It is necessary to know that anhydrous copper salt easily absorbs water from the air and goes into a hydrated state, so anhydrous salt must be weighed immediately before experience. According to the data obtained, it is necessary to calculate the heat dissolving of anhydrous salt and crystallohydrate, and then determine the heat of hydration from the ratio (3). Calculate the relative error of experience as a percentage using the data and data Table 2.