Chemistry- the science of the composition, structure, properties and transformations of substances.

Atomic-molecular doctrine. Substances are made up of chemical particles (molecules, atoms, ions) that have complex structure and consist of elementary particles(protons, neutrons, electrons).

Atom- a neutral particle consisting of a positive nucleus and electrons.

Molecule- a stable group of atoms linked by chemical bonds.

Chemical element A type of atom with the same nuclear charge. Element designate

where X is the symbol of the element, Z is the ordinal number of the element in Periodic system elements D.I. Mendeleev, A- mass number. Serial number Z equal to the charge of the atomic nucleus, the number of protons in the atomic nucleus and the number of electrons in the atom. Mass number A is equal to the sum of the numbers of protons and neutrons in an atom. The number of neutrons is equal to the difference A-Z

isotopes Atoms of the same element with different mass numbers.

Relative atomic mass(A r) is the ratio of the average mass of an atom of an element of natural isotopic composition to 1/12 of the mass of an atom of the carbon isotope 12 C.

Relative molecular weight(M r) - the ratio of the average mass of a molecule of a substance of natural isotopic composition to 1/12 of the mass of an atom of the carbon isotope 12 C.

Atomic mass unit(a.u.m) - 1/12 part of the mass of an atom of the carbon isotope 12 C. 1 a.u. m = 1.66? 10 -24 years

mole- the amount of a substance containing as many structural units (atoms, molecules, ions) as there are atoms in 0.012 kg of the carbon isotope 12 C. mole- the amount of a substance containing 6.02 10 23 structural units (atoms, molecules, ions).

n = N/N A, Where n- amount of substance (mol), N is the number of particles, a N A is the Avogadro constant. The amount of a substance can also be denoted by the symbol v.

Avogadro constant N A = 6.02 10 23 particles/mol.

Molar massM(g / mol) - the ratio of the mass of a substance m(d) to the amount of substance n(mol):

M = m/n, where: m = M n And n = m/M.

Molar volume of gasV M(l/mol) – ratio of gas volume V(l) to the amount of substance of this gas n(mol). Under normal conditions V M = 22.4 l/mol.

Normal conditions: temperature t = 0°C or T = 273 K, pressure p = 1 atm = 760 mm. rt. Art. = 101 325 Pa = 101.325 kPa.

V M = V/n, where: V = V M n And n = V/V M .

The result is a general formula:

n = m/M = V/V M = N/N A .

Equivalent- a real or conditional particle interacting with one hydrogen atom, or replacing it, or equivalent to it in some other way.

Molar mass equivalents M e- the ratio of the mass of a substance to the number of equivalents of this substance: M e = m/n (eq) .

In charge exchange reactions, the molar mass of substance equivalents

with molar mass M equal to: M e = М/(n ? m).

In redox reactions, the molar mass equivalents of a substance with a molar mass M equal to: M e = M/n(e), Where n(e) is the number of electrons transferred.

Law of Equivalents– the masses of reactants 1 and 2 are proportional to the molar masses of their equivalents. m1/m2= M E1 / M E2, or m 1 / M E1 \u003d m 2 / M E2, or n 1 \u003d n 2, Where m 1 And m2 are the masses of two substances, M E1 And M E2 are the molar masses of equivalents, n 1 And n 2- the number of equivalents of these substances.

For solutions, the law of equivalents can be written in the following form:

c E1 V 1 = c E2 V 2, Where with E1, with E2, V 1 And V 2- molar concentrations of equivalents and volumes of solutions of these two substances.

Combined gas law: pV = nRT, Where p– pressure (Pa, kPa), V- volume (m 3, l), n- the amount of gas substance (mol), T- temperature (K), T(K) = t(°C) + 273, R- constant, R= 8.314 J / (K? mol), while J \u003d Pa m 3 \u003d kPa l.

2. The structure of the atom and the Periodic Law

Wave-particle duality matter - the idea that each object can have both wave and corpuscular properties. Louis de Broglie proposed a formula linking the wave and particle properties of objects: ? = h/(mV), Where h is Planck's constant, ? is the wavelength that corresponds to each body with a mass m and speed v. Although wave properties exist for all objects, but they can be observed only for micro-objects having masses of the order of the mass of an atom and an electron.

Heisenberg Uncertainty Principle: ?(mV x) ?x > h/2n or ?V x ?x > h/(2?m), Where m is the mass of the particle, x is its coordinate Vx- speed in direction x, ?– uncertainty, determination error. The uncertainty principle means that it is impossible to simultaneously specify the position (coordinate) of x) and speed (Vx) particles.

Particles with small masses (atoms, nuclei, electrons, molecules) are not particles in the understanding of this by Newtonian mechanics and cannot be studied classical physics. They are being studied quantum physics.

Principal quantum numbern takes the values ​​1, 2, 3, 4, 5, 6 and 7 corresponding to the electronic levels (layers) K, L, M, N, O, P and Q.

Level- space where electrons with the same number are located n. Electrons of different levels are spatially and energetically separated from each other, since the number n determines the energy of electrons E(the more n, the more E) and distance R between electrons and the nucleus (the more n, the more R).

Orbital (side, azimuthal) quantum numberl takes values ​​depending on the number n:l= 0, 1,…(n- 1). For example, if n= 2, then l = 0.1; If n= 3, then l = 0, 1, 2. Number l characterizes the sublevel (sublayer).

sublevel- the space where the electrons are located with certain n And l. Sublevels of this level are designated depending on the number l:s- If l = 0, p- If l = 1, d- If l = 2, f- If l = 3. The sublevels of a given atom are designated depending on the numbers n And l, ex: 2s (n = 2, l = 0), 3d(n= 3, l = 2), etc. The sublevels of a given level have different energies (the more l, the more E): E s< E < Е А < … and different shapes of the orbitals that make up these sublevels: the s-orbital has the shape of a ball, p-orbital has the shape of a dumbbell, etc.

Magnetic quantum numberm 1 characterizes the orientation of the orbital magnetic moment equal to l, in space relative to the outer magnetic field and takes the values: – l,…-1, 0, 1,…l, i.e. total (2l + 1) value. For example, if l = 2, then m 1 =-2, -1, 0, 1, 2.

Orbital(part of a sublevel) - the space where electrons are located (no more than two) with certain n, l, m 1 . Sublevel contains 2l+1 orbital. For example, d– the sublevel contains five d-orbitals. Orbitals of the same sublevel having different numbers m 1 , have the same energy.

Magnetic spin numberm s characterizes the orientation of the intrinsic magnetic moment of the electron s, equal to?, relative to the external magnetic field and takes two values: +? And _ ?.

Electrons in an atom occupy levels, sublevels and orbitals according to the following rules.

Pauli's rule: Two electrons in one atom cannot have four identical quantum numbers. They must differ by at least one quantum number.

It follows from the Pauli rule that an orbital can contain no more than two electrons, a sublevel can contain no more than 2(2l + 1) electrons, a level can contain no more than 2n 2 electrons.

Klechkovsky's rule: the filling of electronic sublevels is carried out in ascending order of the amount (n+l), and in the case of the same amount (n+l)- in ascending order of number n.

Graphic form of the Klechkovsky rule.

According to the Klechkovsky rule, the filling of sublevels is carried out in the following order: 1s, 2s, 2p, 3s, Zp, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s, 5f, 6d, 7p, 8s,…

Although the filling of sublevels occurs according to the Klechkovsky rule, in the electronic formula, sublevels are written sequentially by levels: 1s, 2s, 2p, 3s, 3p, 3d, 4s, 4p, 4d, 4f etc. Thus, the electronic formula of the bromine atom is written as follows: Br (35e) 1s 2 2s 2 2p 6 3s 2 3p 6 3d 10 4s 2 4p 5 .

The electronic configurations of a number of atoms differ from those predicted by the Klechkovsky rule. So, for Cr and Cu:

Cr(24e) 1s 2 2s 2 2p 6 3s 2 3p 6 3d 5 4s 1 and Cu(29e) 1s 2 2s 2 2p 6 3s 2 3p 6 3d 10 4s 1.

Hund's (Gund's) rule: the filling of the orbitals of a given sublevel is carried out so that the total spin is maximum. The orbitals of a given sublevel are first filled by one electron.

Electronic configurations of atoms can be written down by levels, sublevels, orbitals. For example, the electronic formula P(15e) can be written:

a) by levels)2)8)5;

b) by sublevels 1s 2 2s 2 2p 6 3s 2 3p 3;

c) by orbitals

Examples of electronic formulas of some atoms and ions:

V(23e) 1s 2 2s 2 2p 6 3s 2 3p 6 3d 3 4s 2;

V 3+ (20e) 1s 2 2s 2 2p 6 3s 2 3p 6 3d 2 4s 0.

3. Chemical bond

3.1. Valence bond method

According to the method of valence bonds, the bond between atoms A and B is formed using a common pair of electrons.

covalent bond. Donor-acceptor connection.

Valency characterizes the ability of atoms to form chemical bonds and is equal to the number chemical bonds formed by an atom. According to the method of valence bonds, valence is equal to the number of common pairs of electrons, and in the case covalent bond valence is equal to the number of unpaired electrons per external level an atom in its ground or excited states.

Valence of atoms

For example, for carbon and sulfur:

Saturability covalent bond: atoms form a limited number of bonds equal to their valency.

Hybridization of atomic orbitals– mixing of atomic orbitals (AO) of different sublevels of the atom, the electrons of which are involved in the formation of equivalent?-bonds. The equivalence of hybrid orbitals (HO) explains the equivalence of the formed chemical bonds. For example, in the case of a tetravalent carbon atom, there is one 2s– and three 2p-electron. To explain the equivalence of the four?-bonds formed by carbon in the CH 4, CF 4, etc. molecules, the atomic one s- and three R- orbitals are replaced by four equivalent hybrid sp 3-orbitals:

Orientation covalent bond is that it is formed in the direction of maximum overlap of the orbitals that form a common pair of electrons.

Depending on the type of hybridization, hybrid orbitals have a certain spatial arrangement:

sp– linear, the angle between the axes of the orbitals is 180°;

sp 2– triangular, the angles between the axes of the orbitals are 120°;

sp 3– tetrahedral, the angles between the axes of the orbitals are 109°;

sp 3 d 1– trigonal-bipyramidal, angles 90° and 120°;

sp2d1– square, the angles between the axes of the orbitals are 90°;

sp 3 d 2– octahedral, the angles between the axes of the orbitals are 90°.

3.2. Theory of molecular orbitals

According to the theory of molecular orbitals, a molecule consists of nuclei and electrons. In molecules, electrons are in molecular orbitals (MOs). The MO of outer electrons have a complex structure and are considered as a linear combination of the outer orbitals of the atoms that make up the molecule. The number of formed MOs is equal to the number of AOs participating in their formation. The energies of MOs can be lower (bonding MOs), equal (non-bonding MOs), or higher (loosening, anti-bonding MOs) than the energies of the AOs that form them.

JSC interaction conditions

1. AO interact if they have similar energies.

2. AOs interact if they overlap.

3. AO interact if they have the appropriate symmetry.

For a diatomic AB molecule (or any linear molecule), the MO symmetry can be:

If a given MO has an axis of symmetry,

If a given MO has a plane of symmetry,

If MO has two perpendicular planes of symmetry.

The presence of electrons on bonding MOs stabilizes the system, since it reduces the energy of the molecule compared to the energy of atoms. The stability of a molecule is characterized connection order n, equal to: n \u003d (n sv - n res) / 2, Where n sv and n res - the number of electrons in bonding and loosening orbitals.

The filling of an MO with electrons occurs according to the same rules as the filling of an AO in an atom, namely: the Pauli rule (there cannot be more than two electrons on an MO), the Hund rule (the total spin must be maximum), etc.

The interaction of 1s-AO atoms of the first period (H and He) leads to the formation of a bonding?-MO and a loosening?*-MO:

Electronic formulas of molecules, bond orders n, experimental bond energies E and intermolecular distances R for diatomic molecules from atoms of the first period are given in the following table:

Other atoms of the second period contain, in addition to 2s-AO, also 2p x -, 2p y - and 2p z -AO, which can form ?- and ?-MO upon interaction. For the O, F, and Ne atoms, the energies of 2s– and 2p-AO are significantly different, and the interaction between the 2s-AO of one atom and the 2p-AO of another atom can be neglected, considering the interaction between the 2s-AO of two atoms separately from the interaction of their 2p-AO. The MO scheme for O 2 , F 2 , Ne 2 molecules has the following form:

For B, C, N atoms, the energies of 2s– and 2p-AO are close in their energies, and the 2s-AO of one atom interacts with the 2p z-AO of another atom. Therefore, the order of MO in B 2 , C 2 and N 2 molecules differs from the order of MO in O 2 , F 2 and Ne 2 molecules. Below is the MO scheme for B 2 , C 2 and N 2 molecules:

Based on the above schemes of MO, one can, for example, write down the electronic formulas of the molecules O 2 , O 2 + and O 2 ?:

O 2 + (11e)? s2? s *2 ? z 2 (? x 2 ? y 2)(? x *1 ? y *0)

n = 2 R = 0.121 nm;

O 2 (12e)? s2? s *2 ? z 2 (? x 2 ? y 2)(? x *1 ? y *1)

n = 2.5 R = 0.112 nm;

O2?(13e)? s2? s *2 ? z 2 (? x 2 ? y 2)(? x *2 ? y *1)

n = 1.5 R = 0.126 nm.

In the case of the O 2 molecule, the MO theory makes it possible to foresee the greater strength of this molecule, since n = 2, the nature of the change in binding energies and internuclear distances in the O 2 + – O 2 – O 2 ? series, as well as the paramagnetism of the O 2 molecule, on the upper MO of which there are two unpaired electrons.

3.3. Some types of connections

Ionic bond – electrostatic bond between ions of opposite charges. The ionic bond can be considered as the limiting case of the covalent bond. polar bond. An ionic bond is formed if the difference in the electronegativity of atoms? X is greater than 1.5–2.0.

Ionic bond is non-directional non-saturable connection. In a NaCl crystal, the Na + ion is attracted by all Cl ions? and is repelled by all other Na + ions, regardless of the direction of interaction and the number of ions. This predetermines the greater stability of ionic crystals in comparison with ionic molecules.

hydrogen bond- the bond between the hydrogen atom of one molecule and the electronegative atom (F, CI, N) of another molecule.

The existence of a hydrogen bond explains the anomalous properties of water: the boiling point of water is much higher than that of its chemical counterparts: t bale (H 2 O) = 100 ° C, and t bale (H 2 S) = -61 ° C. Hydrogen bonds do not form between H 2 S molecules.

4. Patterns of the course of chemical processes

4.1. Thermochemistry

Energy(E)- the ability to do work. mechanical work(A) is performed, for example, by a gas during its expansion: A \u003d p? V.

Reactions that go with the absorption of energy - endothermic.

Reactions that take place with the release of energy exothermic.

Types of energy: heat, light, electrical, chemical, nuclear energy, etc.

Energy types: kinetic and potential.

Kinetic energy- the energy of a moving body, this is the work that a body can do before it reaches rest.

Heat (Q)- a type of kinetic energy - associated with the movement of atoms and molecules. When imparting a mass to the body (m) and specific heat capacity (c) of heat? Q its temperature rises by an amount? t: ?Q = m with ?t, where? t = ?Q/(c t).

Potential energy- the energy acquired by the body as a result of a change in it or its constituent parts positions in space. The energy of chemical bonds is a kind of potential energy.

First law of thermodynamics: energy can pass from one form to another, but cannot disappear or arise.

Internal energy (U) is the sum of the kinetic and potential energies particles that make up the body. The heat absorbed in the reaction is equal to the difference between the internal energy of the reaction products and reactants (Q \u003d? U \u003d U 2 - U 1), provided that the system has not done work on environment. If the reaction proceeds at constant pressure, then the released gases do work against the forces of external pressure, and the heat absorbed during the reaction is equal to the sum of the changes in internal energy ?U and work A \u003d p? V. This heat absorbed at constant pressure is called the enthalpy change: H = ?U + p?V, defining enthalpy How H \u003d U + pV. Reactions of liquid and solids flow without a significant change in volume (?V= 0), so what is for these reactions? H close to ?U (?H = ?U). For reactions with a change in volume, we have ?H > ?U if expansion is in progress, and ?H< ?U if compression is in progress.

The change in enthalpy is usually attributed to the standard state of matter: i.e., for a pure substance in a certain (solid, liquid or gaseous) state, at a pressure of 1 atm = 101 325 Pa, a temperature of 298 K and a concentration of substances 1 mol / l.

Standard enthalpy of formation? H arr- heat released or absorbed during the formation of 1 mol of a substance from simple substances, its constituents, under standard conditions. For example, ?N arr(NaCl) = -411 kJ/mol. This means that in the reaction Na(tv) + ?Cl 2 (g) = NaCl(tv), 411 kJ of energy is released during the formation of 1 mol of NaCl.

Standard enthalpy of reaction?- enthalpy change during a chemical reaction, is determined by the formula: ?H = ?N arr(products) - ?N arr(reagents).

So for the reaction NH 3 (g) + HCl (g) \u003d NH 4 Cl (tv), knowing? H o 6 p (NH 3) \u003d -46 kJ / mol,? H o 6 p (HCl) \u003d -92 kJ / mol and? H o 6 p (NH 4 Cl) = -315 kJ / mol we have:

H \u003d? H o 6 p (NH 4 Cl) -? H o 6 p (NH 3) -? H o 6 p (HCl) \u003d -315 - (-46) - (-92) \u003d -177 kJ.

If? H< 0, the reaction is exothermic. If? H > 0, the reaction is endothermic.

Law Hess: the standard enthalpy of reaction depends on the standard enthalpies of the reactants and products and does not depend on the reaction path.

Spontaneous processes can be not only exothermic, i.e., processes with a decrease in energy (?H< 0), but can also be endothermic processes, i.e. processes with an increase in energy (?H > 0). In all these processes, the "disorder" of the system increases.

EntropyS – physical quantity characterizing the degree of disorder of the system. S is the standard entropy, ?S is the change in the standard entropy. If?S > 0, disorder grows if AS< 0, то беспорядок системы уменьшается. Для процессов в которых растет число частиц, ?S >0. For processes in which the number of particles decreases, ?S< 0. Например, энтропия меняется в ходе реакций:

CaO (tv) + H 2 O (l) \u003d Ca (OH) 2 (tv),? S< 0;

CaCO 3 (tv) \u003d CaO (tv) + CO 2 (g), ?S\u003e 0.

Processes proceed spontaneously with the release of energy, i.e. for which? H< 0, and with an increase in entropy, i.e., for which?S > 0. Accounting for both factors leads to an expression for Gibbs energy: G = H - TS or? G \u003d? H - T? S. Reactions in which the Gibbs energy decreases, i.e. ?G< 0, могут идти самопроизвольно. Реакции, в ходе которых энергия Гиббса увеличивается, т. е. ?G >0, spontaneously do not go. The condition? G = 0 means that an equilibrium has been established between the products and the reactants.

At low temperature, when the value T is close to zero, only exothermic reactions take place, since T?S– few and? G = ? H< 0. At high temperatures, the values T?S large, and, neglecting the magnitude? H, we have? G = – T?S, i.e., processes with an increase in entropy will spontaneously occur, for which? S > 0, and ?G< 0. При этом чем больше по absolute value value? G, the more completely passes this process.

The value of AG for a particular reaction can be determined by the formula:

G = ?С arr (products) – ?G o b p (reagents).

In this case, the values? G o br, as well as? H arr and?s o br for a large number substances are given in special tables.

4.2. Chemical kinetics

The rate of a chemical reaction(v) is determined by the change in the molar concentration of the reactants per unit time:

Where v is the reaction rate, s is the molar concentration of the reagent, t- time.

The rate of a chemical reaction depends on the nature of the reactants and the reaction conditions (temperature, concentration, presence of a catalyst, etc.)

Influence of concentration. IN In the case of simple reactions, the reaction rate is proportional to the product of the concentrations of the reactants, taken in powers equal to their stoichiometric coefficients.

For reaction

where 1 and 2 are respectively the direction of the forward and backward reactions:

v 1 \u003d k 1? [A]m? [B]n and

v 2 \u003d k 2? [C]p? [D]q

Where v- speed reaction, k is the rate constant, [A] is the molar concentration of substance A.

Reaction molecularity is the number of molecules involved in the elementary act of the reaction. For simple reactions, for example: mA + nB> pC + qD, molecularity is equal to the sum of the coefficients (m + n). Reactions can be one-molecular, two-molecular and rarely three-molecular. Higher molecular reactions do not occur.

Reaction order is equal to the sum of the indicators of the degrees of concentration in the experimental expression of the rate of a chemical reaction. So, for a complex reaction

mA + nB > рС + qD the experimental expression for the reaction rate has the form

v 1 = k1? [A] ? ? [IN] ? and the reaction order is (? + ?). Wherein? And? are experimental and may not coincide with m And n respectively, since the equation of a complex reaction is the result of several simple reactions.

The effect of temperature. The reaction rate depends on the number of effective collisions of molecules. An increase in temperature increases the number of active molecules, giving them the necessary for the reaction to proceed. activation energy E act and increases the rate of a chemical reaction.

Van't Hoff's rule. With an increase in temperature by 10°, the reaction rate increases by a factor of 2–4. Mathematically, this is written as:

v2 = v1? ?(t 2 - t 1) / 10

where v 1 and v 2 are the reaction rates at the initial (t 1) and final (t 2) temperatures, ? - temperature coefficient reaction rate, which shows how many times the reaction rate increases with an increase in temperature by 10 °.

More precisely, the dependence of the reaction rate on temperature is expressed as Arrhenius equation:

k = A? e - E/(RT) ,

Where k is the rate constant, A- constant, independent of temperature, e = 2.71828, E is the activation energy, R= 8.314 J/(K? mol) – gas constant; T– temperature (K). It can be seen that the rate constant increases with increasing temperature and decreasing activation energy.

4.3. Chemical equilibrium

A system is in equilibrium if its state does not change with time. The equality of the rates of the direct and reverse reactions is a condition for maintaining the equilibrium of the system.

An example of a reversible reaction is the reaction

N 2 + 3H 2 - 2NH 3.

Mass action law: the ratio of the product of the concentrations of the reaction products to the product of the concentrations of the starting substances (all concentrations are indicated in powers equal to their stoichiometric coefficients) is a constant called equilibrium constant.

The equilibrium constant is a measure of the progress of a direct reaction.

K = O - no direct reaction;

K =? - the direct reaction goes to the end;

K > 1 - the balance is shifted to the right;

TO< 1 - the balance is shifted to the left.

Reaction equilibrium constant TO is related to the change in standard Gibbs energy?G for the same reaction:

G= – RT ln K, or ?g= -2.3RT lg K, or K= 10 -0.435?G/RT

If K > 1, then lg K> 0 and?G< 0, т. е. если равновесие сдвинуто вправо, то реакция – переход от исходного состояния к равновесному – идет самопроизвольно.

If TO< 1, then lg K < 0 и?G >0, i.e. if the equilibrium is shifted to the left, then the reaction does not spontaneously go to the right.

Equilibrium displacement law: If an external influence is exerted on a system in equilibrium, a process arises in the system that counteracts the external influence.

5. Redox reactions

Redox reactions- reactions that go with a change in the oxidation states of elements.

Oxidation is the process of giving up electrons.

Recovery is the process of adding electrons.

Oxidizer An atom, molecule, or ion that accepts electrons.

Reducing agent An atom, molecule, or ion that donates electrons.

Oxidizing agents, accepting electrons, go into the reduced form:

F2 [ca. ] + 2e > 2F? [rest.].

Reducing agents, donating electrons, pass into the oxidized form:

Na 0 [restore ] – 1e > Na + [approx.].

The equilibrium between the oxidized and reduced forms is characterized by Nernst equations for redox potential:

Where E 0 is the standard value of the redox potential; n is the number of transferred electrons; [rest. ] and [ca. ] are the molar concentrations of the compound in the reduced and oxidized forms, respectively.

Standard values electrode potentials E 0 are given in tables and characterize the oxidizing and reducing properties of the compounds: the more positive the value E 0, the stronger the oxidizing properties, and the negative value E 0, the stronger the restorative properties.

For example, for F 2 + 2e - 2F? E 0 = 2.87 volts, and for Na + + 1e - Na 0 E 0 =-2.71 volts (the process is always recorded for reduction reactions).

A redox reaction is a combination of two half-reactions, oxidation and reduction, and is characterized by electromotive force(emf) ? E 0:?E 0= ?E 0 ok – ?E 0 restore, Where E 0 ok And? E 0 restore– standard potentials oxidizing agent and reducing agent for this reaction.

emf reactions? E 0 is related to the change in the Gibbs free energy?G and the equilibrium constant of the reaction TO:

?G = –nF?E 0 or? E = (RT/nF) ln K.

emf reactions at non-standard concentrations? E is equal to: ? E =?E 0 - (RT / nF)? Ig K or? E =?E 0 -(0,059/n)lg K.

In the case of equilibrium? G \u003d 0 and? E \u003d 0, where? E =(0.059/n)lg K And K = 10n?E/0.059.

For the spontaneous occurrence of the reaction, the following relations must be satisfied: ?G< 0 или K >> 1 that the condition matches? E 0> 0. Therefore, to determine the possibility of a given redox reaction, it is necessary to calculate the value? E 0 . If? E 0 > 0, the reaction is on. If? E 0< 0, there is no reaction.

Chemical current sources

Galvanic cells devices that convert the energy of a chemical reaction into electrical energy.

Daniel's galvanic cell consists of zinc and copper electrodes immersed in ZnSO 4 and CuSO 4 solutions, respectively. Electrolyte solutions communicate through a porous partition. At the same time, oxidation occurs on the zinc electrode: Zn > Zn 2+ + 2e, and reduction occurs on the copper electrode: Cu 2+ + 2e > Cu. In general, the reaction is going on: Zn + CuSO 4 = ZnSO 4 + Cu.

Anode- the electrode at which oxidation takes place. Cathode- the electrode on which the reduction is taking place. In galvanic cells, the anode is negatively charged and the cathode is positively charged. In the element diagrams, the metal and solution are separated by a vertical line, and two solutions by a double vertical line.

So, for the reaction Zn + CuSO 4 \u003d ZnSO 4 + Cu, the galvanic cell circuit is written: (-) Zn | ZnSO 4 || CuSO4 | Cu(+).

The electromotive force (emf) of the reaction is? E 0 \u003d E 0 ok - E 0 restore= E 0(Cu 2+ /Cu) - E 0(Zn 2+ / Zn) \u003d 0.34 - (-0.76) \u003d 1.10 V. Due to losses, the voltage created by the element will be slightly less than? E 0 . If the concentrations of solutions differ from the standard ones, equal to 1 mol/l, then E 0 ok And E 0 restore are calculated according to the Nernst equation, and then the emf is calculated. corresponding galvanic cell.

dry element consists of a zinc body, NH 4 Cl paste with starch or flour, a mixture of MnO 2 with graphite and a graphite electrode. In the course of its work, the following reaction takes place: Zn + 2NH 4 Cl + 2MnO 2 = Cl + 2MnOOH.

Element diagram: (-)Zn | NH4Cl | MnO 2 , C(+). emf element - 1.5 V.

Batteries. A lead battery consists of two lead plates immersed in a 30% sulfuric acid solution and covered with a layer of insoluble PbSO 4 . When the battery is charged, the following processes take place on the electrodes:

PbSO 4 (tv) + 2e > Pb (tv) + SO 4 2-

PbSO 4 (tv) + 2H 2 O > РbO 2 (tv) + 4H + + SO 4 2- + 2e

When the battery is discharged, the following processes take place on the electrodes:

Pb(tv) + SO 4 2-> PbSO 4 (tv) + 2e

РbO 2 (tv) + 4H + + SO 4 2- + 2e> PbSO 4 (tv) + 2Н 2 O

The overall reaction can be written as:

To work, the battery needs regular charging and control of the concentration of sulfuric acid, which may decrease slightly during battery operation.

6. Solutions

6.1. Solution concentration

Mass fraction of a substance in solution w is equal to the ratio of the mass of the solute to the mass of the solution: w \u003d m in-va / m solution or w = m in-va / (V ? ?), because m p-ra \u003d V p-pa? ?r-ra.

Molar concentration With is equal to the ratio of the number of moles of the solute to the volume of the solution: c = n(mol)/ V(l) or c = m/(M? V( l )).

Molar concentration of equivalents (normal or equivalent concentration) with e is equal to the ratio of the number of equivalents of the solute to the volume of the solution: with e = n(mol equiv.)/ V(l) or with e \u003d m / (M e? V (l)).

6.2. Electrolytic dissociation

Electrolytic dissociation – decomposition of the electrolyte into cations and anions under the action of polar solvent molecules.

Degree of dissociation? is the ratio of the concentration of dissociated molecules (c diss) to the total concentration of dissolved molecules (c vol): ? = s diss / s rev.

Electrolytes can be divided into strong(?~1) and weak.

Strong electrolytes(for them? ~ 1) - salts and bases soluble in water, as well as some acids: HNO 3, HCl, H 2 SO 4, HI, HBr, HClO 4 and others.

Weak electrolytes(for them?<< 1) – Н 2 O, NH 4 OH, малорастворимые основания и соли и многие кислоты: HF, H 2 SO 3 , H 2 CO 3 , H 2 S, CH 3 COOH и другие.

Ionic reaction equations. IN In ionic reaction equations, strong electrolytes are written as ions, and weak electrolytes, poorly soluble substances and gases are written as molecules. For example:

CaCO 3 v + 2HCl \u003d CaCl 2 + H 2 O + CO 2 ^

CaCO 3 v + 2H + + 2Cl? \u003d Ca 2+ + 2Cl? + H 2 O + CO 2 ^

CaCO 3 v + 2H + = Ca 2+ + H 2 O + CO 2 ^

Reactions between ions go in the direction of the formation of a substance that gives fewer ions, i.e., in the direction of a weaker electrolyte or less soluble substance.

6.3. Dissociation of weak electrolytes

Let us apply the mass action law to the equilibrium between ions and molecules in a weak electrolyte solution, for example acetic acid:

CH 3 COOH - CH 3 COО? + H +

The equilibrium constants of dissociation reactions are called dissociation constants. Dissociation constants characterize the dissociation of weak electrolytes: the smaller the constant, the less the weak electrolyte dissociates, the weaker it is.

Polybasic acids dissociate in steps:

H 3 PO 4 - H + + H 2 PO 4?

The equilibrium constant of the total dissociation reaction is equal to the product of the constants of the individual stages of dissociation:

H 3 PO 4 - ZN + + PO 4 3-

Ostwald's dilution law: the degree of dissociation of a weak electrolyte (a) increases with a decrease in its concentration, i.e., upon dilution:

Effect of a common ion on the dissociation of a weak electrolyte: the addition of a common ion reduces the dissociation of a weak electrolyte. So, when adding a weak electrolyte solution CH 3 COOH

CH 3 COOH - CH 3 COО? + H + ?<< 1

a strong electrolyte containing an ion common with CH 3 COOH, i.e. an acetate ion, for example CH 3 COONa

CH 3 COONa - CH 3 COO? +Na+? = 1

the concentration of the acetate ion increases, and the equilibrium of the dissociation of CH 3 COOH shifts to the left, i.e., the dissociation of the acid decreases.

6.4. Dissociation of strong electrolytes

Ion activity A is the concentration of an ion, which manifests itself in its properties.

Activity factorf is the ratio of ion activity A to concentration with: f= a/c or A = f.c.

If f = 1, then the ions are free and do not interact with each other. This occurs in very dilute solutions, in solutions of weak electrolytes, etc.

If f< 1, то ионы взаимодействуют между собой. Чем меньше f, тем больше взаимодействие между ионами.

The activity coefficient depends on the ionic strength of solution I: the greater the ionic strength, the lower the activity coefficient.

Ionic strength of solution I depends on charges z and concentrations from ions:

I= 0.52?s z2.

The activity coefficient depends on the charge of the ion: the greater the charge of the ion, the lower the activity coefficient. Mathematically, the dependence of the activity coefficient f from ionic strength I and ion charge z is written using the Debye-Hückel formula:

Ion activity coefficients can be determined using the following table:

6.5 Ionic product of water. Hydrogen indicator

Water, a weak electrolyte, dissociates to form H+ and OH? ions. These ions are hydrated, i.e., connected to several water molecules, but for simplicity they are written in non-hydrated form

H 2 O - H + + OH?.

Based on the law of mass action, for this equilibrium:

The concentration of water molecules [H 2 O], i.e., the number of moles in 1 liter of water, can be considered constant and equal to [H 2 O] \u003d 1000 g / l: 18 g / mol \u003d 55.6 mol / l. From here:

TO[H 2 O] = TO(H 2 O ) = [H + ] = 10 -14 (22°C).

Ionic product of water– the product of concentrations [H + ] and – is a constant value at a constant temperature and equal to 10 -14 at 22°C.

The ionic product of water increases with increasing temperature.

pH value is the negative logarithm of the concentration of hydrogen ions: pH = – lg. Similarly: pOH = – lg.

The logarithm of the ionic product of water gives: pH + pOH = 14.

The pH value characterizes the reaction of the medium.

If pH = 7, then [H + ] = is a neutral medium.

If pH< 7, то [Н + ] >- acid environment.

If pH > 7, then [H + ]< – щелочная среда.

6.6. buffer solutions

buffer solutions- solutions having a certain concentration of hydrogen ions. The pH of these solutions does not change when diluted and changes little when small amounts of acids and alkalis are added.

I. A solution of a weak acid HA, concentration - from acid, and its salts with a strong base BA, concentration - from salt. For example, an acetate buffer is a solution of acetic acid and sodium acetate: CH 3 COOH + CHgCOONa.

pH \u003d pK acidic + lg (salt /s acidic).

II. A solution of a weak base BOH, concentration - with basic, and its salts with a strong acid BA, concentration - with salt. For example, an ammonia buffer is a solution of ammonium hydroxide and ammonium chloride NH 4 OH + NH 4 Cl.

pH = 14 - рК basic - lg (from salt / from basic).

6.7. Salt hydrolysis

Salt hydrolysis- the interaction of salt ions with water with the formation of a weak electrolyte.

Examples of hydrolysis reaction equations.

I. Salt is formed by a strong base and a weak acid:

Na 2 CO 3 + H 2 O - NaHCO 3 + NaOH

2Na + + CO 3 2- + H 2 O - 2Na + + HCO 3? +OH?

CO 3 2- + H 2 O - HCO 3? + OH?, pH > 7, alkaline.

In the second stage, hydrolysis practically does not occur.

II. A salt is formed from a weak base and a strong acid:

AlCl 3 + H 2 O - (AlOH)Cl 2 + HCl

Al 3+ + 3Cl? + H 2 O - AlOH 2+ + 2Cl? + H + + Cl?

Al 3+ + H 2 O - AlOH 2+ + H +, pH< 7.

In the second stage, hydrolysis occurs less, and in the third stage it practically does not occur.

III. Salt is formed by a strong base and a strong acid:

K + + NO 3 ? + H 2 O? no hydrolysis, pH? 7.

IV. A salt is formed from a weak base and a weak acid:

CH 3 COONH 4 + H 2 O - CH 3 COOH + NH 4 OH

CH 3 COO? + NH 4 + + H 2 O - CH 3 COOH + NH 4 OH, pH = 7.

In some cases, when the salt is formed by very weak bases and acids, complete hydrolysis occurs. In the solubility table for such salts, the symbol is “decomposed by water”:

Al 2 S 3 + 6H 2 O \u003d 2Al (OH) 3 v + 3H 2 S ^

The possibility of complete hydrolysis should be taken into account in exchange reactions:

Al 2 (SO 4) 3 + 3Na 2 CO 3 + 3H 2 O \u003d 2Al (OH) 3 v + 3Na 2 SO 4 + 3CO 2 ^

Degree of hydrolysish is the ratio of the concentration of hydrolyzed molecules to the total concentration of dissolved molecules.

For salts formed by a strong base and a weak acid:

= ch, pOH = -lg, pH = 14 - pOH.

It follows from the expression that the degree of hydrolysis h(i.e. hydrolysis) increases:

a) with increasing temperature, since K(H 2 O) increases;

b) with a decrease in the dissociation of the acid that forms the salt: the weaker the acid, the greater the hydrolysis;

c) with dilution: the lower c, the greater the hydrolysis.

For salts formed from a weak base and a strong acid

[H + ] = ch, pH = – lg.

For salts formed by a weak base and a weak acid

6.8. Protolytic theory of acids and bases

Protolysis is the proton transfer process.

Protoliths acids and bases that donate and accept protons.

Acid A molecule or ion capable of donating a proton. Each acid has its conjugate base. The strength of acids is characterized by the acid constant To k.

H 2 CO 3 + H 2 O - H 3 O + + HCO 3?

K k = 4 ? 10 -7

3+ + H 2 O - 2+ + H 3 O +

K k = 9 ? 10 -6

Base A molecule or ion that can accept a proton. Each base has its conjugate acid. The strength of the bases is characterized by the base constant K 0 .

NH3? H 2 O (H 2 O) - NH 4 + + OH?

K 0 = 1,8 ?10 -5

Ampholytes- protoliths capable of recoil and proton attachment.

HCO3? + H 2 O - H 3 O + + CO 3 2-

HCO3? - acid.

HCO3? + H 2 O - H 2 CO 3 + OH?

HCO3? - base.

For water: H 2 O + H 2 O - H 3 O + + OH?

K (H 2 O) \u003d [H 3 O +] \u003d 10 -14 and pH \u003d - lg.

Constants K to And K 0 for conjugated acids and bases are linked.

ON + H 2 O - H 3 O + + A ?,

A? + H 2 O - ON + OH?,

7. Solubility constant. Solubility

In a system consisting of a solution and a precipitate, two processes take place - the dissolution of the precipitate and precipitation. The equality of the rates of these two processes is the equilibrium condition.

saturated solution A solution that is in equilibrium with the precipitate.

The law of mass action applied to the equilibrium between precipitate and solution gives:

Since = const,

TO = K s (AgCl) = .

In general, we have:

A m B n(TV) - m A +n+n B -m

K s ( A m B n)= [A +n ] m[IN -m ] n .

Solubility constantKs(or solubility product PR) - the product of ion concentrations in a saturated solution of a sparingly soluble electrolyte - is a constant value and depends only on temperature.

Solubility of an insoluble substance s can be expressed in moles per litre. Depending on the size s substances can be divided into poorly soluble - s< 10 -4 моль/л, среднерастворимые – 10 -4 моль/л? s? 10 -2 mol/l and highly soluble s>10 -2 mol/l.

The solubility of compounds is related to their solubility product.

Precipitation and dissolution condition

In the case of AgCl: AgCl - Ag + + Cl?

Ks= :

a) the equilibrium condition between the precipitate and the solution: = K s .

b) settling condition: > K s ; during precipitation, the ion concentrations decrease until equilibrium is established;

c) the condition for the dissolution of the precipitate or the existence of a saturated solution:< K s ; during the dissolution of the precipitate, the concentration of ions increases until equilibrium is established.

8. Coordination compounds

Coordination (complex) compounds are compounds with a donor-acceptor bond.

For K3:

ions of the outer sphere - 3K +,

ion of the inner sphere - 3-,

complexing agent - Fe 3+,

ligands - 6CN?, their denticity - 1,

coordination number - 6.

Examples of complexing agents: Ag +, Cu 2+, Hg 2+, Zn 2+, Ni 2+, Fe 3+, Pt 4+, etc.

Examples of ligands: polar molecules H 2 O, NH 3 , CO and anions CN?, Cl?, OH? and etc.

Coordination numbers: usually 4 or 6, rarely 2, 3, etc.

Nomenclature. They first name the anion (in the nominative case), then the cation (in genitive case). The names of some ligands: NH 3 - ammine, H 2 O - aqua, CN? – cyano, Cl? – chloro, OH? - hydroxo. Names of coordination numbers: 2 - di, 3 - three, 4 - tetra, 5 - penta, 6 - hexa. Indicate the degree of oxidation of the complexing agent:

Cl is diamminesilver(I) chloride;

SO 4 - tetramminecopper(II) sulfate;

K 3 is potassium hexacyanoferrate(III).

Chemical connection.

The theory of valence bonds assumes hybridization of the orbitals of the central atom. The location of the resulting hybrid orbitals determines the geometry of the complexes.

Diamagnetic complex ion Fe(CN) 6 4- .

Cyanide ion - donor

Iron ion Fe 2+ - acceptor - has the formula 3d 6 4s 0 4p 0. Taking into account the diamagnetism of the complex (all electrons are paired) and the coordination number (6 free orbitals are needed), we have d2sp3- hybridization:

The complex is diamagnetic, low-spin, intra-orbital, stable (no external electrons are used), octahedral ( d2sp3-hybridization).

Paramagnetic complex ion FeF 6 3- .

Fluoride ion is a donor.

Iron ion Fe 3+ - acceptor - has the formula 3d 5 4s 0 4p 0 . Taking into account the paramagnetism of the complex (electrons are steamed) and the coordination number (6 free orbitals are needed), we have sp 3 d 2- hybridization:

The complex is paramagnetic, high-spin, outer-orbital, unstable (outer 4d-orbitals are used), octahedral ( sp 3 d 2-hybridization).

Dissociation of coordination compounds.

Coordination compounds in solution completely dissociate into ions of the inner and outer spheres.

NO 3 > Ag(NH 3) 2 + + NO 3 ?, ? = 1.

Ions of the inner sphere, i.e., complex ions, dissociate into metal ions and ligands, like weak electrolytes, in steps.

Where K 1 , TO 2 , TO 1 _ 2 are called instability constants and characterize the dissociation of complexes: the smaller the instability constant, the less the complex dissociates, the more stable it is.

The decision on the need to maintain such a notebook did not come immediately, but gradually, with the accumulation of work experience.

At the beginning it was the place at the end workbook– several pages to record the most important definitions. Then the most important tables were placed there. Then came the realization that in order to learn how to solve problems, most students need strict algorithmic prescriptions, which they, first of all, must understand and remember.

It was then that the decision came to maintain, in addition to the workbook, another obligatory chemistry notebook - a chemical dictionary. Unlike workbooks, which can even be two in one school year, the dictionary is a single notebook for the entire chemistry course. It is best if this notebook has 48 sheets and a strong cover.

We arrange the material in this notebook as follows: at the beginning - the most important definitions that the guys write out from the textbook or write down under the dictation of the teacher. For example, in the first lesson in the 8th grade, this is the definition of the subject “chemistry”, the concept of “chemical reactions”. During the school year in the 8th grade, they accumulate more than thirty. According to these definitions, I conduct surveys in some lessons. For example, an oral question in a chain, when one student asks a question to another, if he answered correctly, then he already asks the next question; or, when one student is asked questions by other students, if he does not cope with the answer, then they answer themselves. In organic chemistry, these are mainly class definitions organic matter and the main concepts, for example, “homologues”, “isomers”, etc.

At the end of our reference book, material is presented in the form of tables and diagrams. On the last page is the very first table “Chemical elements. chemical signs". Then the tables “Valence”, “Acids”, “Indicators”, “Electrochemical series of voltages of metals”, “Series of electronegativity”.

I especially want to dwell on the contents of the table “Correspondence of acids to acid oxides”:

Correspondence of acids to acid oxides
acid oxide		Acid
Name	Formula	Name	Formula	Acid residue, valence
carbon monoxide (II)	CO2	coal	H2CO3	CO 3 (II)
sulfur(IV) oxide	SO2	sulphurous	H2SO3	SO3(II)
sulfur(VI) oxide	SO 3	sulfuric	H2SO4	SO4(II)
silicon(IV) oxide	SiO2	silicon	H2SiO3	SiO 3 (II)
nitric oxide (V)	N 2 O 5	nitric	HNO3	NO 3 (I)
phosphorus(V) oxide	P2O5	phosphoric	H3PO4	PO 4 (III)

Without understanding and memorizing this table, it is difficult for students of the 8th grade to compile equations for the reactions of acid oxides with alkalis.

When studying the theory of electrolytic dissociation, at the end of the notebook we write down schemes and rules.

Rules for compiling ionic equations:

1. In the form of ions, write down the formulas of strong electrolytes that are soluble in water.

2. In molecular form, write down the formulas of simple substances, oxides, weak electrolytes and all insoluble substances.

3. The formulas of poorly soluble substances on the left side of the equation are written in ionic form, on the right - in molecular form.

When studying organic chemistry we write in the dictionary summarizing tables for hydrocarbons, classes of oxygen- and nitrogen-containing substances, schemes for genetic relationships.

Physical quantities
Designation	Name	Units	Formulas
	amount of substance	mole	= N / N A ; = m / M; V / V m (for gases)
N A	Avogadro's constant	molecules, atoms and other particles	N A = 6.02 10 23
N	number of particles	molecules, atoms and other particles	N = N A
M	molar mass	g/mol, kg/kmol	M = m / ; / M/ = M r
m	weight	g, kg	m = M ; m = V
Vm	molar volume of gas	l / mol, m 3 / kmol	Vm \u003d 22.4 l / mol \u003d 22.4 m 3 / kmol
V	volume	l, m 3	V = V m (for gases) ;
	density	g/ml;	= m/V; M / V m (for gases)

During the 25 years of teaching chemistry at school, I had to work on different programs and textbooks. At the same time, it was always surprising that practically no textbook teaches how to solve problems. At the beginning of the study of chemistry, in order to systematize and consolidate knowledge in the dictionary, the students and I compile a table “Physical quantities” with new quantities:

When teaching students how to solve computational problems, great importance I give algorithms. I believe that the strict prescription of the sequence of actions allows a weak student to understand the solution of problems of a certain type. For strong students, this is an opportunity to reach creative level their further chemical education and self-education, since for a start you need to confidently master a relatively small number of standard techniques. On the basis of this, the ability to apply them correctly at different stages of solving more problems will develop. challenging tasks. Therefore, I have compiled algorithms for solving calculation problems for all types of problems school course and for extracurricular activities.

I will give examples of some of them.

Algorithm for solving problems by chemical equations.

1. Briefly write down the condition of the problem and make a chemical equation.

2. Above the formulas in the chemical equation, write the data of the problem, write the number of moles under the formulas (determined by the coefficient).

3. Find the amount of a substance, the mass or volume of which is given in the condition of the problem, using the formulas:

M/M; \u003d V / V m (for gases V m \u003d 22.4 l / mol).

Write the resulting number above the formula in the equation.

4. Find the amount of a substance whose mass or volume is unknown. To do this, reason according to the equation: compare the number of moles according to the condition with the number of moles according to the equation. Proportion if necessary.

5. Find the mass or volume using the formulas: m = M ; V = V m .

This algorithm is the basis that the student must master so that in the future he can solve problems using equations with various complications.

Tasks for excess and deficiency.

If in the condition of the problem the quantities, masses or volumes of two reacting substances are known at once, then this is a problem for excess and deficiency.

When solving it:

1. It is necessary to find the amounts of two reacting substances according to the formulas:

M/M; = V/V m .

2. The resulting numbers of moles are inscribed above the equation. Comparing them with the number of moles according to the equation, draw a conclusion about which substance is given in deficiency.

3. By deficiency, make further calculations.

Tasks for the share of the yield of the reaction product, practically obtained from the theoretically possible.

According to the reaction equations, theoretical calculations are carried out and theoretical data are found for the reaction product: theor. , m theor. or V theor. . When carrying out reactions in the laboratory or in industry, losses occur, so the practical data obtained are practical. ,

m practical or V practical. is always less than theoretically calculated data. The yield fraction is denoted by the letter (eta) and is calculated by the formulas:

(this) = pract. / theor. = m practical. / m theor. = V practical. / V theor.

It is expressed as a fraction of a unit or as a percentage. There are three types of tasks:

If the data for the starting substance and the share of the yield of the reaction product are known in the condition of the problem, then you need to find the practical. , m practical or V practical. reaction product.

Solution order:

1. Calculate according to the equation, based on the data for the original substance, find the theory. , m theor. or V theor. reaction product;

2. Find the mass or volume of the reaction product, practically obtained, according to the formulas:

m practical = m theor. ; V pract. = V theor. ; practical = theor. .

If in the condition of the problem the data for the starting substance and practice are known. , m practical or V practical. of the obtained product, while it is necessary to find the share of the yield of the reaction product.

Solution order:

1. Calculate according to the equation, based on the data for the starting substance, find

Theor. , m theor. or V theor. reaction product.

2. Find the share of the yield of the reaction product using the formulas:

Prakt. / theor. = m practical. / m theor. = V practical. /V theor.

If in the condition of the problem are known pract. , m practical or V practical. of the resulting reaction product and the share of its yield, in this case, you need to find data for the starting substance.

Solution order:

1. Find theor., m theor. or V theor. reaction product according to the formulas:

Theor. = practical / ; m theor. = m practical. / ; V theor. = V practical. / .

2. Calculate according to the equation, based on theor. , m theor. or V theor. reaction product and find data for the starting material.

Of course, we consider these three types of problems gradually, we work out the skills of solving each of them using the example of a number of problems.

Problems on mixtures and impurities.

A pure substance is that which is more in the mixture, the rest is impurities. Designations: the mass of the mixture - m cm, the mass of the pure substance - m q.v., the mass of impurities - m approx. , mass fraction of a pure substance - h.v.

The mass fraction of a pure substance is found by the formula: h.v. = m q.v. / m see, express it in fractions of a unit or as a percentage. We distinguish 2 types of tasks.

If in the condition of the problem the mass fraction of a pure substance or the mass fraction of impurities is given, then the mass of the mixture is given. The word "technical" also means the presence of a mixture.

Solution order:

1. Find the mass of a pure substance using the formula: m p.m. = q.v. m see.

If the mass fraction of impurities is given, then first you need to find mass fraction pure substance: h.v. = 1 - approx.

2. Based on the mass of a pure substance, make further calculations according to the equation.

If the condition of the problem gives the mass of the initial mixture and n, m or V of the reaction product, then you need to find the mass fraction of the pure substance in the initial mixture or the mass fraction of impurities in it.

Solution order:

1. Calculate according to the equation, based on the data for the reaction product, and find n hours. and m h.v.

2. Find the mass fraction of a pure substance in a mixture using the formula: q.v. = m q.v. / m see and mass fraction of impurities: approx. = 1 - h.c.

The law of volumetric ratios of gases.

The volumes of gases are related in the same way as their quantities of substances:

V 1 / V 2 = 1 / 2

This law is used in solving problems by equations in which the volume of a gas is given and it is necessary to find the volume of another gas.

The volume fraction of gas in the mixture.

Vg / Vcm, where (phi) is the volume fraction of gas.

Vg is the volume of gas, Vcm is the volume of the mixture of gases.

If the volume fraction of the gas and the volume of the mixture are given in the condition of the problem, then, first of all, you need to find the volume of the gas: Vg = Vcm.

The volume of the mixture of gases is found by the formula: Vcm \u003d Vg /.

The volume of air spent on burning a substance is found through the volume of oxygen found by the equation:

Vair \u003d V (O 2) / 0.21

Derivation of formulas of organic substances by general formulas.

Organic substances form homologous series that have common formulas. This allows:

1. Express the relative molecular weight in terms of the number n.

M r (C n H 2n + 2) = 12n + 1 (2n + 2) = 14n + 2.

2. Equate M r expressed in terms of n to the true M r and find n.

3. Compose reaction equations in general form and perform calculations on them.

Derivation of formulas of substances by combustion products.

1. Analyze the composition of combustion products and draw a conclusion about qualitative composition burnt substance: H 2 O -> H, CO 2 -> C, SO 2 -> S, P 2 O 5 -> P, Na 2 CO 3 -> Na, C.

The presence of oxygen in the substance requires verification. Designate the indices in the formula as x, y, z. For example, CxHyOz (?).

2. Find the amount of substances of combustion products using the formulas:

n = m / M and n = V / Vm.

3. Find the amounts of elements contained in the burnt substance. For example:

n (C) \u003d n (CO 2), n (H) \u003d 2 ћ n (H 2 O), n (Na) \u003d 2 ћ n (Na 2 CO 3), n (C) \u003d n (Na 2 CO 3) etc.

4. If a substance of unknown composition burned out, then it is imperative to check whether it contained oxygen. For example, СxНyОz (?), m (O) \u003d m in-va - (m (C) + m (H)).

b) if relative density is known: M 1 = D 2 M 2 , M = D H2 2, M = D O2 32,

M = D air. 29, M = D N2 28, etc.

1 way: find the simplest formula substances (see the previous algorithm) and the simplest molar mass. Then compare the true molar mass with the simplest and increase the indices in the formula by the required number of times.

2 way: find the indices using the formula n = (e) Mr / Ar (e).

If the mass fraction of one of the elements is unknown, then it must be found. To do this, subtract the mass fraction of another element from 100% or from unity.

Gradually, in the course of studying chemistry in the chemical dictionary, algorithms for solving problems are accumulated different types. And the student always knows where to find the right formula or the right information to solve the problem.

Many students like to keep such a notebook, they themselves supplement it with various reference materials.

As for extracurricular activities, the students and I also start a separate notebook for writing algorithms for solving problems that go beyond school curriculum. In the same notebook, for each type of task, we write down 1-2 examples, they solve the rest of the tasks in another notebook. And, if you think about it, among the thousands of different tasks encountered in the exam in chemistry in all universities, one can distinguish tasks of 25 - 30 different types. Of course, there are many variations among them.

In developing algorithms for solving problems in optional classes, A.A. Kushnarev. (Learning to solve problems in chemistry, - M., School - press, 1996).

The ability to solve problems in chemistry is the main criterion for the creative assimilation of the subject. It is through solving problems of various levels of complexity that a chemistry course can be effectively mastered.

If a student has a clear idea of ​​all possible types of tasks, has solved a large number of tasks of each type, then he is able to cope with passing the exam in chemistry in the form of the Unified State Examination and entering universities.

Chemistry is the science of substances and their transformations into each other.

Substances are chemically pure substances

A chemically pure substance is a collection of molecules that have the same qualitative and quantitative composition and the same structure.

CH 3 -O-CH 3 -

CH 3 -CH 2 -OH

Molecule - the smallest particles of a substance that have all its chemical properties; a molecule is made up of atoms.

An atom is the chemically indivisible particles that make up molecules. (for noble gases, the molecule and the atom are the same, He, Ar)

An atom is an electrically neutral particle consisting of a positively charged nucleus, around which negatively charged electrons are distributed according to their strictly defined laws. Moreover, the total charge of the electrons is equal to the charge of the nucleus.

The nucleus of atoms consists of positively charged protons (p) and neutrons (n) that do not carry any charge. The common name for neutrons and protons is nucleons. The mass of protons and neutrons is almost the same.

Electrons (e -) carry a negative charge equal to that of a proton. The mass e - is approximately 0.05% of the mass of the proton and neutron. Thus, the entire mass of an atom is concentrated in its nucleus.

The number p in the atom, equal to the charge of the nucleus, is called the serial number (Z), since the atom is electrically neutral, the number e is equal to the number p.

The mass number (A) of an atom is the sum of protons and neutrons in the nucleus. Accordingly, the number of neutrons in an atom is equal to the difference between A and Z. (the mass number of the atom and the serial number). (N=A-Z).

17 35 Cl p=17, N=18, Z=17. 17p + , 18n 0 , 17e - .

Nucleons

The chemical properties of atoms are determined by their electronic structure (number of electrons), which is equal to serial number atoms (the charge of the nucleus). Therefore, all atoms with the same nuclear charge behave chemically in the same way and are calculated as atoms of the same chemical element.

An element is a collection of atoms with the same nuclear charge. (110 chemical elements).

Atoms, having the same nuclear charge, can differ in mass number, which is associated with a different number of neutrons in their nuclei.

Atoms that have the same Z but different mass numbers are called isotopes.

17 35 Cl 17 37 Cl

Hydrogen isotopes H:

Designation: 1 1 N 1 2 D 1 3 T

Name: protium deuterium tritium

Core composition: 1p 1p+1n 1p+2n

Protium and deuterium are stable

Tritium-decays (radioactive) Used in hydrogen bombs.

Atomic mass unit. Avogadro's number. Moth.

The masses of atoms and molecules are very small (approximately 10 -28 to 10 -24 g), for the practical display of these masses it is advisable to introduce your own unit of measurement, which would lead to a convenient and familiar scale.

Since the mass of an atom is concentrated in its nucleus, consisting of protons and neutrons of almost identical mass, it is logical to take the mass of one nucleon as a unit mass of atoms.

We agreed to take one twelfth of the carbon isotope, which has a symmetrical structure of the nucleus (6p + 6n), as a unit of mass of atoms and molecules. This unit is called the atomic mass unit (amu), it is numerically equal to the mass of one nucleon. In this scale, the masses of atoms are close to integer values: He-4; Al-27; Ra-226 amu……

Calculate the mass of 1 amu in grams.

1/12 (12 C) \u003d \u003d 1.66 * 10 -24 g / a.u.m

Let us calculate how many amu is contained in 1g.

N A = 6.02 *-Avogadro's number

The resulting ratio is called the Avogadro number, it shows how many a.m.u. are contained in 1g.

Atomic masses given in the Periodic Table are expressed in amu

Molecular mass is the mass of a molecule, expressed in amu, is found as the sum of the masses of all the atoms that form this molecule.

m (1 molecule H 2 SO 4) \u003d 1 * 2 + 32 * 1 + 16 * 4 \u003d 98 amu

For the transition from a.m.u. to 1 g, which is practically used in chemistry, a portioned calculation of the amount of a substance was introduced, and each portion contains the number N A of structural units (atoms, molecules, ions, electrons). In this case, the mass of such a portion, called 1 mol, expressed in grams, is numerically equal to the atomic or molecular mass, expressed in amu.

Let's find the mass of 1 mol H 2 SO 4:

M (1 mol H 2 SO 4) \u003d

98a.u.m*1.66**6.02*=

As can be seen, the molecular molar mass are numerically equal.

1 mol- the amount of substance containing the Avogadro number of structural units (atoms, molecules, ions).

Molecular weight(M) is the mass of 1 mole of a substance, expressed in grams.

The amount of substance-V (mol); mass of substance m(g); molar mass M (g / mol) - related by the ratio: V =;

2H 2 O+ O 2 2H 2 O

2 mol 1 mol

2.Basic laws of chemistry

The law of constancy of the composition of a substance - a chemically pure substance, regardless of the method of preparation, always has a constant qualitative and quantitative composition.

CH3+2O2=CO2+2H2O

NaOH+HCl=NaCl+H2O

Substances with a constant composition are called daltonites. As an exception, substances of constant composition are known - bertolites (oxides, carbides, nitrides)

The law of conservation of mass (Lomonosov) - the mass of substances that have entered into a reaction is always equal to the mass of the reaction products. It follows from this that atoms do not disappear during the reaction and are not formed; they pass from one substance to another. This is the basis for the selection of coefficients in the chemical reaction equation, the number of atoms of each element in the left and right parts of the equation should be equal.

The law of equivalent-in chemical reactions substances react and are formed in quantities equal to the equivalent (How many equivalents of one substance are consumed, exactly the same number of equivalents are consumed or formed of another substance).

The equivalent is the amount of a substance that adds, replaces, releases one mole of H atoms (ions) during the reaction. The equivalent mass expressed in grams is called the equivalent mass (E).

Gas laws

Dalton's law - the total pressure of a mixture of gases is equal to the sum of the partial pressures of all components of the gas mixture.

Avogadro's law - equal volumes of different gases under the same conditions contain an equal number of molecules.

Consequence: one mole of any gas under normal conditions (t=0 degrees or 273K and P=1 atmosphere or 101255 Pascal or 760 mmHg. Pillar.) occupies V=22.4 liters.

V which occupies one mole of gas is called the molar volume Vm.

Knowing the volume of gas (gas mixture) and Vm under given conditions, it is easy to calculate the amount of gas (gas mixture) =V/Vm.

The Mendeleev-Clapeyron equation relates the amount of gas to the conditions under which it is located. pV=(m/M)*RT= *RT

When using this equation, all physical quantities must be expressed in SI: p-gas pressure (pascal), V-gas volume (liters), m- gas mass (kg.), M-molar mass (kg / mol), T- absolute temperature (K), Nu-amount of gas (mol), R- gas constant = 8.31 J / (mol * K).

D - the relative density of one gas in relation to another - the ratio of M gas to M gas, selected as a standard, shows how many times one gas is heavier than another D \u003d M1 / ​​M2.

Ways of expressing the composition of a mixture of substances.

Mass fraction W- the ratio of the mass of the substance to the mass of the entire mixture W \u003d ((m in-va) / (m solution)) * 100%

Mole fraction æ - the ratio of the number of in-va, to the total number of all centuries. in the mixture.

Most of the chemical elements in nature are present as a mixture of different isotopes; knowing the isotopic composition of a chemical element, expressed in mole fractions, calculate the weighted average value of the atomic mass of this element, which is translated into ISCE. А= Σ (æi*Аi)= æ1*А1+ æ2*А2+…+ æn*Аn , where æi is the mole fraction of the i-th isotope, Аi is the atomic mass of the i-th isotope.

Volume fraction (φ) - the ratio of Vi to the volume of the entire mixture. φi=Vi/VΣ

Knowing the volumetric composition of the gas mixture, the Mav of the gas mixture is calculated. Мav= Σ (φi*Mi)= φ1*М1+ φ2*М2+…+ φn*Мn