Electronic configuration atom is a formula showing the arrangement of electrons in an atom by levels and sublevels. After studying the article, you will learn where and how electrons are located, get acquainted with quantum numbers and be able to construct the electronic configuration of an atom by its number; at the end of the article there is a table of elements.
Why study the electronic configuration of elements?
Atoms are like a construction set: there is a certain number of parts, they differ from each other, but two parts of the same type are absolutely the same. But this construction set is much more interesting than the plastic one and here’s why. The configuration changes depending on who is nearby. For example, oxygen next to hydrogen Maybe turn into water, when near sodium it turns into gas, and when near iron it completely turns it into rust. To answer the question of why this happens and predict the behavior of an atom next to another, it is necessary to study the electronic configuration, which will be discussed below.
How many electrons are in an atom?
An atom consists of a nucleus and electrons rotating around it; the nucleus consists of protons and neutrons. In the neutral state, each atom has the number of electrons equal to the number of protons in its nucleus. The number of protons is designated by the atomic number of the element, for example, sulfur has 16 protons - the 16th element of the periodic table. Gold has 79 protons - the 79th element of the periodic table. Accordingly, sulfur has 16 electrons in the neutral state, and gold has 79 electrons.
Where to look for an electron?
By observing the behavior of the electron, certain patterns were derived; they are described by quantum numbers, there are four in total:
- Main quantum number
- Orbital quantum number
- Magnetic quantum number
- Spin quantum number
Orbital
Further, instead of the word orbit, we will use the term "orbital", orbital is wave function electron, roughly, is the region in which the electron spends 90% of its time.
N - level
L - shell
M l - orbital number
M s - first or second electron in the orbital
Orbital quantum number l
As a result of studying the electron cloud, they found that depending on the energy level, the cloud takes four main forms: a ball, dumbbells and two other, more complex ones. In order of increasing energy, these forms are called the s-, p-, d- and f-shell. Each of these shells can have 1 (on s), 3 (on p), 5 (on d) and 7 (on f) orbitals. The orbital quantum number is the shell in which the orbitals are located. The orbital quantum number for the s,p,d and f orbitals takes the values 0,1,2 or 3, respectively.
There is one orbital on the s-shell (L=0) - two electrons
There are three orbitals on the p-shell (L=1) - six electrons
There are five orbitals on the d-shell (L=2) - ten electrons
There are seven orbitals on the f-shell (L=3) - fourteen electrons
Magnetic quantum number m l
There are three orbitals on the p-shell, they are designated by numbers from -L to +L, that is, for the p-shell (L=1) there are orbitals “-1”, “0” and “1”. The magnetic quantum number is denoted by the letter m l.
Inside the shell, it is easier for electrons to be located in different orbitals, so the first electrons fill one in each orbital, and then a pair of electrons is added to each one.
Consider the d-shell:
The d-shell corresponds to the value L=2, that is, five orbitals (-2,-1,0,1 and 2), the first five electrons fill the shell taking the values M l =-2, M l =-1, M l =0 , M l =1,M l =2.
Spin quantum number m s
Spin is the direction of rotation of an electron around its axis, there are two directions, so the spin quantum number has two values: +1/2 and -1/2. One energy sublevel can only contain two electrons with opposite spins. The spin quantum number is denoted m s
Principal quantum number n
The main quantum number is the energy level at this moment seven energy levels are known, each indicated by an Arabic numeral: 1,2,3,...7. The number of shells at each level is equal to the level number: there is one shell on the first level, two on the second, etc.
Electron number
So, any electron can be described by four quantum numbers, the combination of these numbers is unique for each position of the electron, let's take the first electron, the lowest energy level this is N=1, at the first level there is one shell, the first shell at any level has the shape of a ball (s-shell), i.e. L=0, the magnetic quantum number can take only one value, M l =0 and the spin will be equal to +1/2. If we take the fifth electron (in whatever atom it is), then the main quantum numbers for it will be: N=2, L=1, M=-1, spin 1/2.
Algorithm for composing the electronic formula of an element:
1. Determine the number of electrons in an atom using the Periodic Table of Chemical Elements D.I. Mendeleev.
2. Using the number of the period in which the element is located, determine the number of energy levels; the number of electrons in the last electronic level corresponds to the group number.
3. Divide the levels into sublevels and orbitals and fill them with electrons in accordance with the rules for filling orbitals:
It must be remembered that the first level contains a maximum of 2 electrons 1s 2, on the second - a maximum of 8 (two s and six R: 2s 2 2p 6), on the third - a maximum of 18 (two s, six p, and ten d: 3s 2 3p 6 3d 10).
- Principal quantum number n should be minimal.
- First to fill s- sublevel, then р-, d- b f- sublevels.
- Electrons fill the orbitals in order of increasing energy of the orbitals (Klechkovsky's rule).
- Within a sublevel, electrons first occupy free orbitals one by one, and only after that they form pairs (Hund’s rule).
- There cannot be more than two electrons in one orbital (Pauli principle).
Examples.
1. Let's create the electronic formula of nitrogen. IN periodic table nitrogen is at number 7.
2. Let's create the electronic formula for argon. Argon is number 18 on the periodic table.
1s 2 2s 2 2p 6 3s 2 3p 6.
3. Let's create the electronic formula of chromium. Chromium is number 24 on the periodic table.
1s 2 2s 2 2p 6 3s 2 3p 6 4s 1 3d 5
Energy diagram of zinc.
4. Let's create the electronic formula of zinc. Zinc is number 30 on the periodic table.
1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10
Please note that part of the electronic formula, namely 1s 2 2s 2 2p 6 3s 2 3p 6, is electronic formula argon.
The electronic formula of zinc can be represented as:
When writing electronic formulas for atoms of elements, indicate energy levels (values of the main quantum number n in the form of numbers - 1, 2, 3, etc.), energy sublevels (orbital quantum number values l in the form of letters - s, p, d, f) and the number at the top indicate the number of electrons in a given sublevel.
The first element in the table is D.I. Mendeleev is hydrogen, therefore the charge of the nucleus of the atom N equals 1, an atom has only one electron per s-sublevel of the first level. Therefore, the electronic formula of the hydrogen atom has the form:
The second element is helium; its atom has two electrons, so the electronic formula of the helium atom is 2 Not 1s 2. The first period includes only two elements, since the first energy level is filled with electrons, which can only be occupied by 2 electrons.
The third element in order - lithium - is already in the second period, therefore, its second energy level begins to be filled with electrons (we talked about this above). The filling of the second level with electrons begins with s-sublevel, therefore the electronic formula of the lithium atom is 3 Li 1s 2 2s 1 . The beryllium atom is completed filling with electrons s-sublevel: 4 Ve 1s 2 2s 2 .
In subsequent elements of the 2nd period, the second energy level continues to be filled with electrons, only now it is filled with electrons R-sublevel: 5 IN 1s 2 2s 2 2R 1 ; 6 WITH 1s 2 2s 2 2R 2 … 10 Ne 1s 2 2s 2 2R 6 .
The neon atom completes filling with electrons R-sublevel, this element ends the second period, it has eight electrons, since s- And R-sublevels can only contain eight electrons.
The elements of the 3rd period have a similar sequence of filling the energy sublevels of the third level with electrons. The electronic formulas of the atoms of some elements of this period are as follows:
11 Na 1s 2 2s 2 2R 6 3s 1 ; 12 Mg 1s 2 2s 2 2R 6 3s 2 ; 13 Al 1s 2 2s 2 2R 6 3s 2 3p 1 ;
14 Si 1s 2 2s 2 2R 6 3s 2 3p 2 ;…; 18 Ar 1s 2 2s 2 2R 6 3s 2 3p 6 .
The third period, like the second, ends with an element (argon), which is completely filled with electrons R-sublevel, although the third level includes three sublevels ( s, R, d). According to the above order of filling energy sublevels in accordance with Klechkovsky's rules, the energy of sublevel 3 d more sublevel 4 energy s, therefore, the potassium atom next to argon and the calcium atom behind it are filled with electrons 3 s– sublevel of the fourth level:
19 TO 1s 2 2s 2 2R 6 3s 2 3p 6 4s 1 ; 20 Sa 1s 2 2s 2 2R 6 3s 2 3p 6 4s 2 .
Starting from the 21st element - scandium, sublevel 3 in the atoms of the elements begins to be filled with electrons d. The electronic formulas of the atoms of these elements are:
21 Sc 1s 2 2s 2 2R 6 3s 2 3p 6 4s 2 3d 1 ; 22 Ti 1s 2 2s 2 2R 6 3s 2 3p 6 4s 2 3d 2 .
In the atoms of the 24th element (chromium) and the 29th element (copper), a phenomenon called “leakage” or “failure” of an electron is observed: an electron from the outer 4 s– sublevel “falls” by 3 d– sublevel, completing filling it halfway (for chromium) or completely (for copper), which contributes to greater stability of the atom:
24 Cr 1s 2 2s 2 2R 6 3s 2 3p 6 4s 1 3d 5 (instead of...4 s 2 3d 4) and
29 Cu 1s 2 2s 2 2R 6 3s 2 3p 6 4s 1 3d 10 (instead of...4 s 2 3d 9).
Starting from the 31st element - gallium, the filling of the 4th level with electrons continues, now - R– sublevel:
31 Ga 1s 2 2s 2 2R 6 3s 2 3p 6 4s 2 3d 10 4p 1 …; 36 Kr 1s 2 2s 2 2R 6 3s 2 3p 6 4s 2 3d 10 4p 6 .
This element ends the fourth period, which already includes 18 elements.
A similar order of filling energy sublevels with electrons occurs in the atoms of elements of the 5th period. For the first two (rubidium and strontium) it is filled s– sublevel of the 5th level, for the next ten elements (from yttrium to cadmium) is filled d– sublevel of the 4th level; The period is completed by six elements (from indium to xenon), the atoms of which are filled with electrons R– sublevel of the external, fifth level. There are also 18 elements in a period.
For elements of the sixth period, this order of filling is violated. At the beginning of the period, as usual, there are two elements whose atoms are filled with electrons s– sublevel of the external, sixth, level. The next element behind them, lanthanum, begins to fill with electrons d–sublevel previous level, i.e. 5 d. This completes the filling with electrons 5 d-sublevel stops and the next 14 elements - from cerium to lutetium - begin to fill f-sublevel of the 4th level. These elements are all included in one cell of the table, and below is an expanded row of these elements, called lanthanides.
Starting from the 72nd element - hafnium - to the 80th element - mercury, filling with electrons continues 5 d-sublevel, and the period ends, as usual, with six elements (from thallium to radon), the atoms of which are filled with electrons R– sublevel of the external, sixth, level. This is the most long period, which includes 32 elements.
In the atoms of the elements of the seventh, incomplete, period, the same order of filling sublevels is visible as described above. We let students write the electronic formulas of atoms of elements of the 5th – 7th periods themselves, taking into account everything said above.
Note:In some textbooks A different order of writing the electronic formulas of atoms of elements is allowed: not in the order of their filling, but in accordance with the number of electrons given in the table at each energy level. For example, the electronic formula of the arsenic atom may look like: As 1s 2 2s 2 2R 6 3s 2 3p 6 3d 10 4s 2 4p 3 .
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3. Write an electronic formula and she thallium Tl 3+. For valence electrons atom Tl indicate the set of all four quantum numbers.
Solution:
According to Klechkovsky’s rule, filling of energy levels and sublevels occurs in the following sequence:
1s2s2p3s3p4s3d4p5s4d5p6s(5d 1)4f
5d6p7s (6d 3-2)5f6d7p.
The element thallium Tl has a nuclear charge of +81 (atomic number 81), respectively, 81 electrons. According to Klechkovsky’s rule, we distribute electrons among energy sublevels and obtain the electronic formula of the element Tl:
81 Tl thallium 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 4p 6 5s 2 4d 10 5p 6 6s 2 4f 14 5d 10 6p 1
The thallium ion Tl 3+ has a charge of +3, which means that the atom gave up 3 electrons, and since an atom can only give up valence electrons external level(for thallium these are two 6s and one 6p electrons), its electronic formula will look like this:
81 Tl 3+ thallium 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 4p 6 5s 2 4d 10 5p 6 6s 0 4f 14 5d 10 6p 0
Principal quantum number n defines total energy electron and the degree of its removal from the nucleus (energy level number); it accepts any integer values starting from 1 (n = 1, 2, 3, ...), i.e. corresponds to the period number.
Orbital (side or azimuthal) quantum number l defines the shape atomic orbital. It can take integer values from 0 to n-1 (l = 0, 1, 2, 3,..., n-1). Regardless of the energy level number, each value l The orbital quantum number corresponds to an orbital of a special shape.
Orbitals with l= 0 are called s-orbitals,
l= 1 – p-orbitals (3 types, differing in magnetic quantum number m),
l= 2 – d-orbitals (5 types),
l= 3 – f-orbitals (7 types).
Magnetic quantum number m l characterizes the position electron orbital in space and takes integer values from - l to + l, including 0. This means that for each orbital shape there is (2 l+ 1) energetically equivalent orientation in space.
The spin quantum number m S characterizes magnetic moment, which occurs when an electron rotates around its axis. Accepts only two values +1/2 and –1/2 corresponding opposite directions rotation.
Valence electrons are electrons in the outer energy level. Thallium has 3 valence electrons: 2 s electrons and 1 p electron.
Quantum numbers s – electrons:
Orbital quantum number l= 0 (s – orbital)
Magnetic quantum number m l = (2 l+ 1 = 1): m l = 0.
Spin quantum number m S = ±1/2
Quantum numbers p – electron:
Principal quantum number n = 6 (sixth period)
Orbital quantum number l= 1 (p – orbital)
Magnetic quantum number (2 l+ 1 = 3): m = -1, 0, +1
Spin quantum number m S = ±1/2
23. Specify those properties chemical elements, which change periodically. What is the reason for the periodic repetition of these properties? Using examples, explain the essence of the periodicity of changes in the properties of chemical compounds.
Solution:
The properties of elements, determined by the structure of the outer electronic layers of atoms, naturally change according to periods and groups periodic table. Moreover, the similarity electronic structures generates similarity of properties of analogous elements, but not the identity of these properties. Therefore, when moving from one element to another in groups and subgroups, what is observed is not a simple repetition of properties, but their more or less pronounced natural change. In particular, chemical behavior atoms of elements is manifested in their ability to lose and gain electrons, i.e. in their ability to oxidize and reduce. A quantitative measure of the ability of an atom lose electrons is ionization potential(E And ) , and a measure of their ability to reacquire– electron affinity (E With ). The nature of the change in these quantities during the transition from one period to another is repeated, and these changes are based on a change in the electronic configuration of the atom. Thus, completed electronic layers corresponding to atoms of inert gases exhibit increased stability and an increased value of ionization potentials within the period. At the same time, the s-elements of the first group (Li, Na, K, Rb, Cs) have the lowest ionization potential values.
Electronegativity is a measure of the ability of an atom of this element draw electrons toward itself relative to the atoms of other elements in the compound. According to one of the definitions (Mulliken), the electronegativity of an atom can be expressed as half the sum of its ionization energy and electron affinity: = (E and + E c).
In periods there is The general trend an increase in the electronegativity of the element, and in subgroups – its decrease. Lowest values The s-elements of group I have electronegativity, and the p-elements of group VII have the greatest electronegativity.
The electronegativity of the same element can vary depending on the valence state, hybridization, oxidation state, etc. Electronegativity significantly affects the nature of changes in the properties of compounds of elements. For example, sulfuric acid shows stronger acid properties than her chemical analogue– selenic acid, since in the latter the central selenium atom, due to its lower electronegativity compared to the sulfur atom, does not polarize the H–O bonds in the acid so much, which means a weakening of acidity.
H–O O
Another example: chromium(II) hydroxide and chromium(VI) hydroxide. Chromium (II) hydroxide, Cr(OH) 2, exhibits basic properties in contrast to chromium (VI) hydroxide, H 2 CrO 4, since the oxidation state of chromium +2 determines the weakness of the Coulomb interaction of Cr 2+ with the hydroxide ion and the ease of elimination of this ion, i.e. manifestation of basic properties. At the same time, the high oxidation state of chromium +6 in chromium (VI) hydroxide causes a strong Coulomb attraction between the hydroxide ion and the central chromium atom and the impossibility of dissociation along the bond 
– OH. On the other hand, the high oxidation state of chromium in chromium(VI) hydroxide enhances its ability to attract electrons, i.e. electronegativity, which determines high degree polarization of H–O bonds in this compound, being a prerequisite for an increase in acidity.
The next important characteristic of atoms is their radii. In periods, the radii of metal atoms with increasing serial number elements are reduced, because with an increase in the atomic number of an element within a period, the charge of the nucleus increases, and therefore the total charge of the electrons that balances it; as a consequence, the Coulomb attraction of electrons also increases, which ultimately leads to a decrease in the distance between them and the nucleus. The most pronounced decrease in radius is observed in elements of short periods, in which the outer energy level is filled with electrons.
In large periods, d- and f-elements exhibit a smoother decrease in radii with increasing charge of the atomic nucleus. Within each subgroup of elements, atomic radii tend to increase from top to bottom, since such a shift signifies a transition to a higher energy level.
The influence of the radii of element ions on the properties of the compounds they form can be illustrated by the example of an increase in the acidity of hydrohalic acids in the gas phase: HI > HBr > HCl > HF.
43. Name the elements for whose atoms only one thing is possible valence state, and indicate what it will be - basic or excited.
Solution:
Atoms of elements that have one unpaired electron at the outer valence energy level can have one valence state - these are elements of group I of the periodic system (H - hydrogen, Li - lithium, Na - sodium, K - potassium, Rb - rubidium, Ag - silver, Cs – cesium, Au – gold, Fr – francium), with the exception of copper since in the formation chemical bonds, the number of which is determined by the valence, d-electrons of the pre-external level also take part (the ground state of the copper atom 3d 10 4s 1 is due to the stability of the filled d-shell, however, the first excited state 3d 9 4s 2 exceeds the ground state in energy by only 1.4 eV ( about 125 kJ/mol).Therefore, in chemical compounds Both states manifest themselves to the same extent, giving rise to two series of copper compounds (I) and (II)).
Also, atoms of elements in which the outer energy level is completely filled and electrons do not have the opportunity to go into an excited state can have one valence state. These are the elements main subgroup VIII group – inert gases(He – helium, Ne – neon, Ar – argon, Kr – krypton, Xe – xenon, Rn – radon).
For all the listed elements, the only valence state is the ground state, because there is no possibility of transition to an excited state. In addition, the transition to an excited state determines the new valence state of the atom; accordingly, if such a transition is possible, the valence state of a given atom is not the only one.
63. Using the model of repulsion of valence electron pairs and the method valence bonds, consider the spatial structure of the proposed molecules and ions. Indicate: a) the number of bonding and lone electron pairs of the central atom; b) the number of orbitals involved in hybridization; c) type of hybridization; d) type of molecule or ion (AB m E n); e) spatial arrangement of electron pairs; f) spatial structure of a molecule or ion.
SO 3; 
Solution:
According to the valence bond method (using this method leads to the same result as using the OEPVO model), spatial configuration A molecule is determined by the spatial arrangement of hybrid orbitals of the central atom, which are formed as a result of interactions between orbitals.
To determine the type of hybridization of the central atom, it is necessary to know the number of hybridizing orbitals. It can be found by adding the number of bonding and lone electron pairs of the central atom and subtracting the number of π bonds.
In a SO 3 molecule
the total number of bonding pairs is 6. Subtracting the number of π-bonds, we obtain the number of hybridizing orbitals: 6 – 3 = 3. Thus, the type of hybridization is sp 2, the type of ion is AB 3, the spatial arrangement of electron pairs has the shape of a triangle, and the molecule itself is triangle:
In ion 
the total number of bonding pairs is 4. There are no π bonds. Number of hybridizing orbitals: 4. Thus, the type of hybridization is sp 3, the type of AB 4 ion, the spatial arrangement of electron pairs has the shape of a tetrahedron, and the ion itself is a tetrahedron:
83. Write the equations possible reactions interactions of KOH, H 2 SO 4, H 2 O, Be(OH) 2 with the compounds given below:
H 2 SO 3, BaO, CO 2, HNO 3, Ni(OH) 2, Ca(OH) 2;
Solution:
a) KOH reaction reactions
2KOH + H 2 SO 3 K 2 SO 3 + 2H 2 O
2K + + 2 OH - + 2H+ + SO 3 2- 2K + + SO 3 2- + H 2 O
OH - + H + H 2 O
KOH + BaO no reaction
2KOH + CO 2 K 2 CO 3 + H 2 O
2K + + 2 OH - + CO 2 2K + + CO 3 2- + H 2 O
2OH - + H 2 CO 3 CO 3 2- + H 2 O
KOH + HNO 3 no reaction, the solution contains ions at the same time:
K + + OH - + H + + NO 3 -
2KOH + Ni(OH) 2 K
2K + + 2 OH- + Ni(OH) 2 K + + -
KOH + Ca(OH) 2 no reaction
b) reaction reactions H 2 SO 4
H 2 SO 4 + H 2 SO 3 no reaction
H 2 SO 4 + BaO BaSO 4 + H 2 O
2H + + SO 4 2- + BaO BaSO 4 + H 2 O
H 2 SO 4 + CO 2 no reaction
H 2 SO 4 + HNO 3 no reaction
H 2 SO 4 + Ni(OH) 2 NiSO 4 + 2H 2 O
2H+ + SO 4 2- + Ni(OH) 2 Ni 2+ + SO 4 2- + 2 H 2 O
2H + + Ni(OH) 2 Ni 2+ + 2H 2 O
H 2 SO 4 + Ca(OH) 2 CaSO 4 + 2H 2 O
2H + + SO 4 2- + Ca(OH) 2 CaSO 4 + 2H 2 O
c) reaction reactions of H 2 O
H 2 O + H 2 SO 3 no reaction
H 2 O + BaO Ba(OH) 2
H 2 O + BaO Ba 2+ + 2OH -
H 2 O + CO 2 no reaction
H 2 O + HNO 3 no reaction
H 2 O + NO 2 no reaction
H 2 O + Ni(OH) 2 no reaction
H 2 O + Ca(OH) 2 no reaction
a) reaction reactions Be(OH) 2
Be(OH) 2 + H 2 SO 3 BeSO 3 + 2H 2 O
Be(OH) 2 + 2H+ + SO 3 2- Be 2+ + SO 3 2- + 2 H 2 O
Be(OH) 2 + 2H+ Be 2+ + 2 H 2 O
Be(OH) 2 + BaO no reaction
2Be(OH) 2 + CO 2 Be 2 CO 3 (OH) 2 ↓ + 2H 2 O
Be(OH) 2 + 2HNO 3 Be(NO 3) 2 + 2H 2 O
Be(OH) 2 + 2H+ + NO 3 - Be 2+ + 2NO 3 - + 2 H 2 O
Be(OH) 2 + 2H + Be 2+ + 2H 2 O
Be(OH) 2 + Ni(OH) 2 no reaction
Be(OH) 2 + Ca(OH) 2 no reaction
103. For the indicated reaction
b) explain which of the factors: entropy or enthalpy contributes to the spontaneous occurrence of the reaction in the forward direction;
c) in which direction (direct or reverse) the reaction will proceed at 298K and 1000K;
e) name all the ways to increase the concentration of products of an equilibrium mixture.
f) plot the dependence of ΔG p (kJ) on T (K)
Solution:
CO (g) + H 2 (g) = C (k) + H 2 O (g)
Standard enthalpy of formation, entropy and Gibbs energy of formation of substances
1. (ΔH 0 298) h.r. =
– 
= -241.84 + 110.5 = -131.34 kJ 2. (ΔS 0 298) c.r. = 
+ 
– 
– 
= 188.74+5.7-197.5-130.6 = -133.66 J/K = -133.66 10 -3 kJ/mol > 0. The direct reaction is accompanied by a decrease in entropy, the disorder in the system decreases - unfavorable factor for leakage chemical reaction in the forward direction.
3. Calculate the standard Gibbs energy of the reaction.
according to Hess's law:
(ΔG 0 298) h.r. = 
– 
= -228.8 +137.1 = -91.7 kJ
It turned out that (ΔН 0 298) ch.r. > (ΔS 0 298) c.r. ·T and then (ΔG 0 298) h.r.
4. 
≈ 
≈ 982.6 K.
≈ 982.6 K is the approximate temperature at which true chemical equilibrium is established; above this temperature a reverse reaction will occur. At a given temperature, both processes are equally probable.
5. Calculate the Gibbs energy at 1000K:
(ΔG 0 1000) h.r. ≈ ΔН 0 298 – 1000·ΔS 0 298 ≈ -131.4 – 1000·(-133.66)·10 -3 ≈ 2.32 kJ > 0.
Those. at 1000 K: ΔS 0 h.r. ·Т > ΔН 0 h.r.
The enthalpy factor became decisive; the spontaneous occurrence of a direct reaction became impossible. The reverse reaction occurs: from one mole of gas and 1 mole solid 2 moles of gas are formed.
log K 298 = 16.1; K 298 ≈ 10 16 >> 1.
The system is far from being true chemical equilibrium, it is dominated by reaction products.
Dependence of ΔG 0 on temperature for the reaction
CO (g) + H 2 (g) = C (k) + H 2 O (g)
K 1000 = 0.86 > 1 – the system is close to equilibrium, but at this temperature the starting substances predominate in it.
8. According to Le Chatelier’s principle, as the temperature increases, the equilibrium should shift towards the reverse reaction, and the equilibrium constant should decrease.
9. Let's consider how our calculated data agree with Le Chatelier's principle. Let us present some data showing the dependence of the Gibbs energy and the equilibrium constant of this reaction on temperature:
|
T, K |
ΔG 0 t, kJ |
K t |
|
298 |
-131,34 |
10 16 |
|
982,6 |
0 |
1 |
|
1000 |
2,32 |
0,86 |
Thus, the obtained calculated data correspond to our conclusions made on the basis of Le Chatelier’s principle.
123. Equilibrium in the system:
)
established at the following concentrations: [B] and [C], mol/l.
Determine the initial concentration of the substance [B] 0 and the equilibrium constant if the initial concentration of substance A is [A] 0 mol/l
From the equation it can be seen that the formation of 0.26 mol of substance C requires 0.13 mol of substance A and the same amount of substance B.
Then the equilibrium concentration of substance A is [A] = 0.4-0.13 = 0.27 mol/l.
The initial concentration of the substance B [B] 0 = [B] + 0.13 = 0.13+0.13 = 0.26 mol/l.
Answer: [B] 0 = 0.26 mol/l, Kp = 1.93.
143. a) 300 g of solution contains 36 g of KOH (solution density 1.1 g/ml). Calculate the percentage and molar concentration of this solution.
b) How many grams of crystalline soda Na 2 CO 3 ·10H 2 O must be taken to prepare 2 liters of 0.2 M Na 2 CO 3 solution?
Solution:
We find the percentage concentration using the equation:
The molar mass of KOH is 56.1 g/mol;
To calculate the molarity of the solution, we find the mass of KOH contained in 1000 ml (i.e., 1000 · 1.100 = 1100 g) of solution:
1100: 100 = at: 12; at= 12 1100 / 100 = 132 g
C m = 56.1 / 132 = 0.425 mol/l.
Answer: C = 12%, Cm = 0.425 mol/l
Solution:
1. Find the mass of anhydrous salt
m = cm·M·V, where M – molar mass, V – volume.
m = 0.2 106 2 = 42.4 g.
2. Find the mass of crystalline hydrate from the proportion
molar mass of crystalline hydrate 286 g/mol - mass X
molar mass of anhydrous salt 106g/mol - mass 42.4g
hence X = m Na 2 CO 3 10H 2 O = 42.4 286/106 = 114.4 g.
Answer: m Na 2 CO 3 10H 2 O = 114.4 g.
163. Calculate the boiling point of a 5% solution of naphthalene C 10 H 8 in benzene. The boiling point of benzene is 80.2 0 C.
|
Given: Average (C 10 H 8) = 5% tboil (C 6 H 6) = 80.2 0 C |
|
Find: tboil (solution) -? |
Solution:
From Raoult's second law
ΔT = E m = (E m B 1000) / (m A μ B)
Here E is the ebullioscopic constant of the solvent
E(C 6 H 6) = 2.57
m A is the weight of the solvent, m B is the weight of the solute, M B is its molecular weight.
Let the mass of the solution be 100 grams, therefore, the mass of the solute is 5 grams, and the mass of the solvent is 100 – 5 = 95 grams.
M (naphthalene C 10 H 8) = 12 10 + 1 8 = 128 g/mol.
We substitute all the data into the formula and find the increase in the boiling point of the solution compared to a pure solvent:
ΔT = (2.57 5 1000)/(128 95) = 1.056
The boiling point of a naphthalene solution can be found using the formula:
T k.r-ra = T k.r-la + ΔT = 80.2 + 1.056 = 81.256
Answer: 81.256 o C
183. Task 1. Write dissociation equations and dissociation constants for weak electrolytes.
Task 2. According to the given ionic equations write the appropriate molecular equations.
Task 3. Write the reaction equations for the following transformations in molecular and ionic forms.
|
No. |
Exercise 1 |
Task 2 |
Task 3 |
|
183 |
Zn(OH) 2 , H 3 AsO 4 |
Ni 2+ + OH – + Cl – = NiOHCl |
NaHSO 3 →Na 2 SO 3 →H 2 SO 3 →NaHSO 3 |
Solution:
Write dissociation equations and dissociation constants for weak electrolytes.
Ist.: Zn(OH) 2 ↔ ZnOH + + OH -
Kd 1 = 
= 1.5·10 -5
IIst.: ZnOH + ↔ Zn 2+ + OH -
Kd 2 = 
= 4.9·10 -7
Zn(OH) 2 – amphoteric hydroxide, acid-type dissociation is possible
Ist.: H 2 ZnO 2 ↔ H + + HZnO 2 -
Kd 1 = 
IIst.: HZnO 2 - ↔ H + + ZnO 2 2-
Kd 2 = 
H 3 AsO 4 – orthoarsenic acid – strong electrolyte, dissociates completely in solution:
H 3 AsO 4 ↔3Н + + AsO 4 3-
Given the ionic equations, write the corresponding molecular equations.
Ni 2+ + OH – + Cl – = NiOHCl
NiCl2 + NaOH(insufficient) = NiOHCl + NaCl
Ni 2+ + 2Cl - + Na + + OH - = NiOHCl + Na + + Cl -
Ni 2+ + Cl - + OH - = NiOHCl
Write the reaction equations for the following transformations in molecular and ionic forms.
NaHSO 3 →Na 2 SO 3 →H 2 SO 3 →NaHSO 3
1) NaHSO 3 + NaOH →Na 2 SO 3 + H 2 O
Na++ HSO 3 - +Na++ OH- → 2Na + + SO 3 2- + H 2 O
HSO 3 - + OH - → + SO 3 2- + H 2 O
2) Na 2 SO 3 + H 2 SO 4 → H 2 SO 3 + Na 2 SO 3
2Na + + SO 3 2- + 2N+ + SO 4 2- → H 2 SO 3 + 2Na + + SO 3 2-
SO 3 2- + 2N + → H 2 SO 3 + SO 3 2-
3) H 2 SO 3 (excess) + NaOH → NaHSO 3 + H 2 O
2 N + + SO 3 2- + Na + + OH- → Na + + HSO 3 - + H 2 O
2
N + + SO 3 2 + OH- → Na + + H 2 O
203. Task 1. Write equations for the hydrolysis of salts in molecular and ionic forms, indicate the pH of solutions (pH > 7, pH Task 2. Write equations for reactions occurring between substances in aqueous solutions
|
No. |
Exercise 1 |
Task 2 |
|
203 |
Na2S; CrBr 3 |
FeCl 3 + Na 2 CO 3; Na 2 CO 3 + Al 2 (SO 4) 3 |
Task 1. Write equations for the hydrolysis of salts in molecular and ionic forms, indicate the pH of solutions (pH > 7, pH
Na2S - salt formed strong foundation And weak acid, undergoes hydrolysis at the anion. The reaction of the medium is alkaline (pH > 7).
Ist. Na 2 S + HON ↔ NaHS + NaOH
2Na + + S 2- + HON ↔ Na + + HS - + Na + + OH -
IIst. NaHS + HOH ↔ H 2 S + NaOH
Na + + HS - + HOH ↔ Na + + H 2 S + OH -
CrBr 3 -
salt formed by a weak base and strong acid, undergoes hydrolysis at the cation. The reaction of the medium is acidic (pH
Ist. CrBr 3 + HOH ↔ CrOHBr 2 + HBr
Cr 3+ + 3Br - + HOH ↔ CrOH 2+ + 2Br - + H + + Br -
IIst. CrOHBr 2 + HON ↔ Cr(OH) 2 Br + HBr
CrOH 2+ + 2Br - + HOH ↔ Cr(OH) 2 + + Br - + H + + Br -
III Art. Cr(OH) 2 Br + HON↔ Cr(OH) 3 + HBr
Cr(OH) 2 + + Br - + HOH↔ Cr(OH) 3 + H + + Br -
Hydrolysis occurs predominantly in the first stage.
Task 2. Write equations for reactions occurring between substances in aqueous solutions
FeCl 3 + Na 2 CO 3
FeCl3 – salt formed by a strong acid and a weak base
Na 2 CO 3 – a salt formed by a weak acid and a strong base
2FeCl 3 + 3Na 2 CO 3 + 6H(OH) = 2Fe(OH) 3 + 3H 2 CO 3 + 6NaCl
2Fe 3+ + 6Cl - + 6Na + + 3 CO 3 2- + 6N(HE) = 2Fe( OH) 3 + 3H 2 CO 3 + 6Na + +6Cl -
2Fe 3+ + 3CO 3 2- + 6N(HE) = 2Fe( OH) 3 + 3H 2 O + 3CO 2
Na 2 CO 3 + Al 2 (SO 4) 3
Mutual enhancement of hydrolysis occurs
Al 2 (SO 4) 3 – a salt formed by a strong acid and a weak base
Na 2 CO 3 – salt formed by a weak acid and a strong base
When two salts are hydrolyzed together, a weak base and a weak acid are formed:
Ist: 2Na 2 CO 3 + Al 2 (SO 4) 3 + 2HOH => 4Na + + 2HCO 3 - + 2AlOH 2+ + 3 SO 4 2 -
IIst: 2HCO 3 - + 2AlOH 2+ + 2HOH => 2H 2 CO 3 + 2Al(OH) 2 +
IIIst: 2Al(OH) 2 + + 2HOH => 2Al(OH) 3 + 2H +
Summary hydrolysis equation
Al 2 (SO 4) 3 + 2 Na 2 CO 3 + 6H 2 O = 2Al(OH) 3 ↓ + 2H 2 CO 3 + 2 Na 2 SO 4 + H 2 SO 4
2Al 3+ + 3 SO 4 2 - + 2 Na + + 2 CABOUT 3 2- + 6H 2 O = 2Al(OH) 3 ↓ + 2H 2 C O 3 + 2 Na + + 2SO 4 2 - + 2H + + SO 4 2 -
2Al 3+ + 2CABOUT 3
2- + 6H 2 O = 2Al(OH) 3 ↓ + 2H 2 C O 3
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