economics The MINISTRY OF EDUCATION And SCIENCES УКРАИНЫ ВОСТОЧНОУКРАИНСКИЙ NATIONAL UNIVERSITY

Document Sample
economics The MINISTRY OF EDUCATION And SCIENCES УКРАИНЫ ВОСТОЧНОУКРАИНСКИЙ NATIONAL UNIVERSITY
The MINISTRY OF EDUCATION And SCIENCES УКРАИНЫ

ВОСТОЧНОУКРАИНСКИЙ NATIONAL UNIVERSITY

Name ВЛАДИМИРА ДАЛЯ









THE BRIEF ABSTRACT OF LECTURES IN CHEMISTRY

For students of the correspondence form of training.









Lugansk ВНУ 2003









4

The MINISTRY OF EDUCATION And SCIENCES УКРАИНЫ

ВОСТОЧНОУКРАИНСКИЙ NATIONAL UNIVERSITY

Name ВЛАДИМИРА ДАЛЯ









THE BRIEF ABSTRACT OF LECTURES IN CHEMISTRY

For students of the correspondence form of training of all engineering specialities









УТВЕЖДЕНО

On faculty meeting of chemistry

The report ____ from..









Lugansk ВНУ 2003









УДК54

THE BRIEF ABSTRACT OF LECTURES IN CHEMISTRY

For students of the correspondence form of training of all engineering specialities.

/ Сост: S.P.Bugrim, I.A.Horuzhaja.-Lugansk: Publishing house Vostochnoukr. нац. ф Ў-that, 2003.–104с.









The brief abstract of lectures in chemistry includes 12 lectures with examples of the decision of typical problems and

the instruction of references on the basic themes: « Стехиометрические laws of chemistry », « Квантовомеханическое

representation about a structure of atom », « Rules and the order of filling nuclear орбиталей », « D.I.Mendeleeva's

Periodic law. Laws of periodic system », « Chemical thermodynamics », « the Second beginning of chemical

thermodynamics », « Chemical кинетика », « Solutions of electrolits, weak electrolits », « Complex connections », «

Oxidation-reduction reactions », « Bases of electrochemistry », « Chemical sources of a current », « Corrosion of metals

and protection of metals against corrosion », «Электролиз». The brief abstract of lectures meets the requirements of the

program of a rate of chemistry for technical specialities. It is recommended for students of the correspondence form of

training of all engineering specialities.





.







Composers: S.P.Bugrim, item преп., I.A.Horuzhaja, доц.кафедры

Chemistry ВНУ it. V.Dalja



Отв. For release: V.L.Abramenko, доц.

Reviewers: A.A.Grigoriev, доц.





5

STOICHIOMETRY LAWS.



Obgectives After studying this chapter, students should

be able to:

1. The law of preservation of a matter.

2. The law of Proust.

3. The Law of Аvоgadro. Consequences from law Аvоgadro.

4. The law of equivalent attitudes.



STEHIOMETRI is the part of a chemistry in which we studie quantitative attitudes

between compounds in chemical processes. Stehiometri is the base of the chemical

analysis and is the base in development of theoretical chemistry.

The basic stehiometric laws are

1. The law of safety of a matter.

The weight of the substances entering chemical reaction is equal to weight of the

substances formed as a result of it.

The law was found in 1748 and proved in 1756 year by Russian scientist

M.Lomonosov.

From the point of view of the atomic-molecular doctrine the sense of the law of

preservation of weight of substances consists that nuclears have constant weight, and

their number does not change during reaction, therefore the weight of substances before

the reaction and after it remains a constant. For example:

t 0C

2Mg + O2   2MgO t 0C

 ( NH4) 2 Cr2O7   N2 + Cr2O3 +4H2O







2∙24 + 16.2 t 0C

  2(24+16)

 28 + 152 + 72



80 

t 0C

  80 252  252







On the bases of the law of preservation of weight it is possible to make the chemical

equations and on them to make necessary calculations.

Problem.

How many will be allocated lead (Pb) if the weight of the immersing zinc plates in a

solution of the lead nitrate Pb (NO3) 2 is increased on 7, 1g?

For the decision of a problem it is necessary to write down the equation of reaction:

Pb (NO3) 2 + Zn Zn (NO3) 2 + Pb,

65 207



Where 65 – molar weight of zinc; 207 – molar weight of Pb. According to this equation

of reaction we can write down this proportion

Whan weight of a zinc plate is increased on 142 grams (207-65) it means thet will be

allocated 207g of lead Pb. So, Whan weight of a zinc plate is increased on 7, 1 g it

means thet will be allocated X g of lead Pb. Will weight of Pb be allocated?





6

7,1  207

X = (mPb) =  10,35 g .

142

2. The law of a constancy of structure.

Any pure chemical compound always has constant quantitative and qualitative

structure are not dependent from method of its obtaining. The law has been

formulated by French scientist Joseph Proust in 1808.

For example, the molecular of water quantitative (on two atoms of hydrogen it

is necessary one atom of oxygen) and has constant qualitative structure (Н; O) are not

dependent on way of reaction:

CuO + H2  H2O + Cu

CH4 + 2O2  2H2O + CO2

C2H5OH + 3O2  3H2O + 2CO2

H2SO4 + 2NaOH  2H2O + Na2SO4



Cleanliness of substances is defined on t0C fusion, t0Сboiling, density, to specters.

3. Law of the Аvоgadro.

In equal volumes of various gases under identical conditions

(t0С, P) the same number of molecules contains. The law is formulated by Italian

physicist Avogadro in 1811 and has consequences.

Consequence 1. One mol of any substance contains 6, 02 10 23 molecules.

number 6, 02 10 23 mol–1 can be called as number of the Аvоgadro (NA).

Consequence 2. The meaning of the volume’ one mol of any gas under normal

conditions (n.c). (00С (273 K), 1,03.105 Па) equals to 22,4 liter/mol.

1mol О2 (M (О2) =32 g/mol) – contains 6, 02.10 23 molecules of oxygen, it volume is

22,4 liter/mol.

1 mol Н2 (M (Н2) =2 g/mol) – contains 6, 02 .10 23 molecules of hydrogen, it is volume

22,4 liter/mol.



Problem. How many hydrogen will be allocated after the reaction 3,25 Zn g with a

hydrochloric acid?

For the decision of a problem we write down the equation of the reaction:

Zn + 2 HCl  ZnCl2 + H2

65 g/mol 22,4 liter/mol



When molar weight of zinc equals to 65 g/mol; molar volume of hydrogen is

22,4 liter/mol. During etching of 65 g zinc allocates 22,4 л/mol hydrogen, and during

etching of the 32,5 г zinc allocates X л hydrogen.Let’s find the volume of the allocated

hydrogen at n.c.

3,25  22,4

V (H 2 ) = X =  1,12 л

65

As a result of works of German chemist Richter, English scientists Dalton and

Vollaston (1792 – 1800) “connecting weights” or equivalents of reacting substances

was established.









7

Chemical equivalent of substance is such substance quantity which can to

connect with one mol of hydrogen atoms or replaces the same quantity of hydrogen

in its compounds.

For example, in compounds such as НС1, H2О, NH3 equivalents of chlorine, oxygen,

nitrogen are accordingly equal to 1 mol, 1/2 mol, 1/3 mol. For unit of an equivalent

takes equivalent of hydrogen Э (H) =1 mol of atoms). An equivalent express in mol.

Notion ‘molar weight of an equivalent’ is used too except ‘equivalent’ notion.

The mоlar weight of an equivalent is a weight of one chemical equivalent of a

substance. МE (H) = 1 g/mol; МE (N) = 14∙1/3 = 4,67 g/mol; МE (Cl) = 35,45 g/mol; МЭ

(S) = 32∙1/2 = 16 g/mol.



Element’ mоlar weight of an equivalent - it is the ratio of mоlar weight of an

element (M) per its valency (V.) For example:

S+4O2, valency of sulfur equals to IV; ME (S+4) = M (S)/V = 32/4 = 8 g/mol;



S+6O3, valency of sulfur equals to VI; ME (S+6) = (S)/V = 32/6 = 5,3 g/mol.

The notion “Equivalent volume” is used for gaseous compounds.

The equivalent volume (Vэ) is the volume of one chemical equivalent of any gas

under normal conditions. The equivalent weight of hydrogen МЭ (H) = 1 g/mol equals

to 1/2 of its molar weight( M(H2) =2 g/mol). The molar volume of hydrogen equals to

22,4 liter (n.c.), then the equivalent volume of hydrogen equals to 11,2 л/mol. The

equivalent weight of oxygen equals to 5,6 liter/mol.



The molar volume of oxygen equals to 22,4 л/mol (n.c.), then the equivalent volume

of oxygen is equal 5,6 liter/mol.



Molar weights of equivalents of compounds can be fined



For оxide:

Ì ( Al2 O3 )

The first way: МE (Al2O3) = , where n – number of element’s atoms in a

n V

molecule; V – its valency. МE (Al2O3) = 102/2∙3 = 17 g/mol



The second way: МE (Al2O3) = МE (O-2) + МE (Al+3) =16/2 + 27/3 =17 g/mol





For the basis:

М (Са (ОН ) 2 ) 74

The first way: МE (Са (OH) 2) =   37 g/mol , where acidity (acidness) of

acidity 2

the basis-it is number (ОН-) in a molecule of the basis.

The second way: МE (Са (OH) 2) = МE (Са2 +) + МE (ОН-) = 40/2 + 17 = 37 g/mol





For the acid:





8

М ( Н 3 РО4 )

The first way: МE (H3PO4) =  98 / 3  32 ,6 g/mol , where basicity of the acid-it

basicity

is number cathions of the hydrogen Н + in a molecule of the acid.

The second way:

МE (Н3РО4) = МЭ (Н+) + МE (РО4)3- = 1 + 95/3 = 1 + 31,6 = 32,6 g/mol

.

For the salt:



М (Са3 ( РО4 ) 2 ) 310

The first way: МE (Са3 (РО4) 2) =   51,7 g/mol

n V 3 2

М ( NaHCO3 ) 84

МE (Nа HCO3) =   84 g/mol

n V 11



М ( А1(ОН )С12 ) 115

МE (A1 (OH) C12) =   57,5 g/mol ,

n V 1 2



Where V – valency of metal or a cathion charge of the basic salt.

n – number of atoms of metal in a molecule of salt.

The second way:

МE (Са3 (РО4) 2) = МE (Са2+) + МE (РО4)3- = 40/2 + 95/3 = 51,7 g/mol;

МE (NaHCO3) = МE (Na+) + МE (HCO)3- = 23 + 61 = 84 g/mol;

МE (А1ОНС12) = МE (А1ОН)2 + + МE (С1-) = 44/2 + 35,5 = 57,5 g/mol.



Finding molar weight of equivalent of compounds in the chemical reaction.



For acids.

Two ions of hydrogen was replaced in this reaction, so the basicity of phosphoric

acid in this case equals to two.

М ( Н 3 РО4 ) 98

H3PO4 + 2NaOH = Na2HPO4 + 2H2O; МE (Н3РО4) =   49 g/mol ;

2 2

Three ions of hydrogen was replaced in this reaction, so the basicity of an acid in this

case equals to three

М ( Н 3 РО4 ) 98

H3PO4 + 3NaOH = Na3PO4 + 3H2O; МE (Н3РО4) =   32,2 g/mol .

3 3







For the bases:

А1 (OH) 3 + НС1  А1 (OH) 2С1 + Н2О, acidity of this basis equals to 1;

МE (А1 (OH) 3) = М ( А1(ОН ) 3 )  78 g/mol

1



А1 (OH) 3 + 2НС1  А1ОНС12 + 2Н2О, acidity of the basis equals to 2



МE (А1 (OH) 3) = 78/2 = 39 g/mol

.





9

Finding molar weights of equivalents of substances during of the oxidation-

reduction reactions:

2KMnO4 + 5SnCl2 + 16HCl = 5SnCl4 + 2MnCl2 + 2KCl + 8H2O

Reducer Sn+2 - 2 Sn+4 5 is oxidized



Oxidizer Mn+7 + 5 Mn+2 2 is restored





Oxidizer’ molar weight of an equivalent equals to the ratio of the oxidizer’ molar

weight (M (KMnO4)) per quantity of the accepted electrons.

М ( KMnO4 ) 158

МE (KMnO4) =   31,6 g/mol .

5 5

Reducer’s molar weight of an equivalent equals to the ratio of the reducer’ molar

weight (M (SnCl2)) to quantity of the efficienced electrons.

(M E (SnCl2)) = s.

4. The law of the ratio of equivalents.

The ratio of the weight of the compounds the m (A) and m (B) is such the ratio as

their molar weight of equivalents: МE (A) and МE (B).

Mathematical expressions of this law:



m( A) M E ( A) m( A) M E ( A) V ( A) V E ( A)

 ;  ;  .

m( B ) M E ( B ) V ( B) VE ( B) V ( B) VE ( B)

Let's consider examples of the problems.

The problem 1. The оxid contains 52 % of the metal. Calculate metal’s molar weight of

an equivalent in this оxid. Is this metal called, if its valensy equals to six? Write down

oxid‘s formula.

The decision: We can calculate quantity of oxygen in this оxid as

100 % – 52 % = 48 %. Using the law of the ratio of equivalent attitudes we can write

down the expression:



m(Me) M E (Me) 52  8

 ; M E (Me)   8,66 g/mol

m(O) M E (O) 48

m (Me) = ME (Ме) ∙Valensy = 8,66∙6 = 52 g/mol



Six-valent metal with nuclear weight 52 а.u.м. Chrome is. Its оxid is CrO3.

The problem 2. For restoration 7,09 g оxid of the metal it is required 2,24 liter

hydrogen (n.c.). Calculate моlar weight of an equivalent of metal and its оxid.

The decision.

Using the law of the ratio of equivalent attitudes we can write down the expression:



m( MexOy ) M E ( MexOy ) 7,09  11,02

 ; M E ( МеxОy )   35,45 g/mol

V (H 2 ) VE ( H 2 ) 2,24





10

ME (Ме xОy) = ME (Ме) +ME (O); (МЭ (=8 g/mol), then ME (Ме) = 35,45 – 8 = 27,45

g/mol.



The problem 3. From 3,85 g nitrate of metal Me (NO3) x were received 1,60 g of its

hydroxide Me (OH) x. Calculate equivalent weight of metal.

The decision.

Using the law of the ratio of equivalent attitudes we can write down expression:



m( M E ( NO3 ) x ) M E (Me( NO3 ) х ) M E ( Mex )  M E ( NO3 )

  ;

m(M E (OH ) x ) M E (Ме(ОН ) х ) M E ( Mex )  M E (OH  )



3,85 M E ( Меx )  62

 x

; M E ( Mex )  15 g/mol

1,60 M E ( Me )  17

Problem 4. At interaction 3,24 g trivalent metal with an acid it is allocated 4,03 л Н2

(n.c.) Calculate equivalent’s molar weight of metal and molar weight of metal. Is this

metal called?

The decision.

Using the law of the ratio of equivalent attitudes we can write down expression:





m( M E ) M E ( Ме) m(Me)  VE ( H 2) 3,24  11,2

  M E (Me)    9 g/mol

V (H 2 ) VE ( H 2 ) V (H 2 ) 4,03

M ( Me)

M E ( Me)   Ì ( Ìå )  M E ( Me)  Val  9  3  27 g/mol

Val

The metal is called aluminium.



Review Questions

1. What do you know about the law of preservation of a matter?

2. What do you know about the law of Proust?

3. What do you know about the law of Аvоgadro?

4. How many molecules contain into one mol of any substance?

5. What is the volume of one mol any gas under normal conditions?

6. What is the definition of this notion “substance’s chemical equivalent”?

7. What is the definition of this notion “element’s molar weight of an equivalent”?

8. What is the definition of this notion “base’s molar weight of an equivalent”?

9. What is the definition of this notion “acid’s molar weight of an equivalent”?

10. What is the definition of this notion “salt’s molar weight of an equivalent”?

11. What is the definition of this notion “oxide’s molar weight of an equivalent”?

12. What is the definition of this notion “the equivalent volume” (Vэ)?

13. What do you know about the law of the ratio of equivalents?

Further Reading

1. Frolov V.V.Chemistry. Part V, §51-56.

2. Luchinsky G.P,rate of chemistry. Part. V, §8-12,

Part. VI, §13-18





11

QUANTUM-МECHANICAL INTRODUCION ABOUT THE STRUCTURE OF

AN ATOM. RULES AND THE ORDER OF FILLING NUCLEAR ОRBITALS.



Obgectives After studying this chapter, students should

be able to know:



1. Various theories about a structure of an atom.

2. Quantum numbers, their physical sense.

3. Principles of filling nuclear орбиталей.





The works by Russian scientist M.V.Lomonosov, the French chemists Lavoisier

and Prust, English chemist Dalton and others are scientific bases of the аtomic-

molecular doctrine. However till the twentieth century, the atom was considered as

not divisible. Only a series of discoveries 1896 – 1898 of the natural radio-activity by

Anry Becquerel (1896 - the radio-activity of uranium), Maria Skladovska-Kurie and

Pierre Kurie, have allowed to change representations on indivisibility of an atom.

The atom is the electroneutral particle consisting of a nuclear which is charged

positively and the electrons ( e ) rotating around it. Electrons are charged negatively.

1

Nuclears of atoms have a complex structure and consist of nucleons (protons ( 1 р ) and

1

neutrons ( 0n )).



Nucleons

1 1

Atom [the nuclear [∑ 1 р + ∑ 0n] + ∑ē] ← Subatomic particles

mass number (A)

The weight of a nuclear is less than the sum of weights of protons and neutrons.

The name of this difference is colled the defect of the weight. It characterizes stability

1

of the atoms nuclears and energy of connection of nucleons (protons ( 1 р ) and neutrons

1

( 0 n )).

The connection’s energy of nuclons in the nuclear is 7.106 eV. It is more then the

connection’s energy of atoms in a molecule (5eV) in millions times.Therefore in

chemical reactions a nuclear of atoms does not change.

Table 1

International System (SI) System of atomic units

The name Weight, Charge, Weight, Charge,

Subatomic particle Kg Кl а.u.w. а.u.ch.

-31

electron e 9,109•10 1,602•10-19 0,0005486 -1

Nucleons Proton 1,673•10-27 1,602•10-19 1,007277 +1

1











12

Neutron 1,673•10-27 0 1,008695 0

1

0n



(Here you can see the weight of an electron in International System (SI) 9,109•10-31………………

and in System of atomic units……0,0005486……………………………………….

The charge of an electron in International System

-19

is…………………………………………….1,602•10 .it is constant.

In system of atomic units the electron’s charge is minus one(-1).

So about nucleons:

- the weight of a proton in International System is……………………………………………………..

- the weight of a neutron in International System is……………………………………………too

- the charge of a electron in International System is………………………………………it is constant.

- a neutron hasn’t got the charge

In system of atomic units:

- the proton’s charge is plus one(+1)

- the neutron’s charge is zero(0).





Chemical element - is the kind of atoms with an identical charge of a nuclear and with

An identical nuclear number.

Each chemical element has some isotopes.

Isotopes - is the kind of atoms with an identical charge of a nuclear , but

different mass number (A).





Table №2

1

The name of an isotope 1р 1

0n

mass number (A)



1

Isotopes 1H - protium 1 0 1

of

2

1Н  Д -deuterium

hydrogen 1 Н  Т - tritium

3 1 1 2



1 2 3



12

6 С Carbon-12 6 6 12

13

Isotopes 6 С Carbon-13

14

С Carbon-14 6 7 13

of carbon 6





6 8 14

(Write down, please the isotopes of hydrogen

1

1H - protium - has only one proton and it mass number (A) equals to one.

2

1 Н  Д - deuterium - has one proton too and one neutron more, it mass number (A) equals to two.

3

1 Н  Т - tritium - has one proton too and two neutron, it mass number (A) equals to three(3).





13

Except for isotopes there are isotones and isobars.

1

isotones - are atoms with identical number of neutrons 0 n , but various

1

quantity of protons 1 р ( with a different charge of a nuclear) ,

for example,

1 230 231 0

88 Ra ( radium, 228( À)  88( 1 ð )  140( n )), 90Th (thorium), 91 Pa ( protoactinium) ( n=140)

228 1

0

Isobars - are atoms with identical mass number (A), but different nuclear

40 40 40

numbers, for example, 18

Ar; K ; Ca

19 20



1 1

After that as the three fundamental particles (е; 1 р ; 0 n ) have been opened, many

models of atom’s structure has been extended.

1. Tomson's Model, 1903(model « a pudding with raisin ») represents

the positively charged atom’s sphere with an electrons inclusions.









Fig. 1

2. In 1911 as a result of the well-known experiments on dispersion by a gold foile - α

particles have been established by E.Rutherford, the atom:

- Has massive positively charged nuclear having very small size and electrons which is

surround it;

- The atom is electroneutral, a positive charge of a nuclear is equal to the sum of a

negative charge of the electrons;

- The meaning of a nuclear’s charge equals to a serial number of the elements in the

D.I.Mendeleev's periodic system.









Fig.2

However, according to laws of classical mechanics and electrodynamics, rotation of

electrons around a nuclear should be accompanied by electromagnetic radiation with a

continuous spectrum. This theoretical fact contradicted descrepted spectra was received

during the expirements in 1880 by the student of Rutherford, Nils Bor.









14

3. N.Bor has developed theplanetary model of an atom in 1913 which is used now.

Such scientists as Rutherford, N.Bor considered thet the atom has massive positively

charged nuclear having very small sizes and the electrons wich surround it. However

N.Bor said, that the electrons were moving around a nuclear on their stability orbits.

These orbits had various energy. Passing from one orbit on another, electron could get

or loose energy. N.Bor сould explain and calculate theoretically descrepted spectrum

of hydrogen’s atom, and also a series of lines in x-ray spectrs of elements.







рис.3

4.In 1924 Луи де Бройль showed, that an elementary particle, moving with the speed,

could be not only as a particle having the weight of calmness, but also as a wave with

the frequency of fluctuations ():

In the case when the electrons is a particle, having the weight of calmness you can write Einstein’s

equation of Energy : E  mc2 ,roughly that energy equals mass times the square of the speed of light.

In the case when the electrons are the wave you can write an equation of

Energy as creation Planck's constant (h) and frequency of fluctuations (), write down E  h ,

If a particle on the one hand has the weight of calmness, and on the other hand it is a wave then

Energy’s weight of calmness equals the Energy of a wave: mc2  h

E  mc 2 

 h

E  h , mc 2  h ;  

mc

  c/  

Planck's constant =6,626. 10–34 Dj.;

с=3. 10 8 km/s (speed of light);

We can see at this equation λ=h/mc

the length of a wave (λ) is result attitudes of numerator Planck's constant(h) to a denominator, which

is a product of a weight of calmness and speed of light. If the mianing of the weight of calmness or

denominator increases, the length of a wave (λ) will decrease. In this situation, the properties of

macrobodies have been studied by classical mechanics.

And If the meaning of the weight of calmness or denominator are very small as an electron then the

length of a wave (λ) will increase.

The result of де -Brol, Dirak, Gejzenberg, Shrjedinger’s works and others have made

the scientific bases of the new physical theory - QUANTUM MECHANICS. This new

physical theory has declared Corpuscle-wave dualism of microparticles, for example, an

electron has a weight of calmness of 9,109•10-31 kg., showing properties of a particle,

and in experiences on diffraction it shows properties of a wave). In the quantum

mechanics the classical concept "trajectory" is replaced with concept « wave function »

or « atomic orbital (AO) ».

The atomic orbital (AO) is an area about nuclear space where electron can be with

enough high degree of probability. The term orbital is conformable to the term an

orbit, however their sense is the different. The orbit is a trajectory of a movement,

nuclear orbital – wave function. If a wave function (ψ) of the particles are known, it is

possible to calculate probability (ψ2) findings of a particle in various areas of space.







15

In 1925 Эрвин Shrjedinger has offered the Mathematics equation allowing to

describe wave function of particles.

h 2  d 2 d 2 d 2 

H   E  - Where Н – Hamilton H      U

8 2 m  dx2 dy 2 dz 2 

 



h 2  d 2 d 2 d 2 

 2   2  2  2  - The operator of kinetic energy,

8 m  dx

 dy dz  

U - the operator of the potential energy.

This differential linear equation of the second order in partial derivatives has many

decisions. From them we are interested in only such meanings, which do not contradict

to physical representations. These are quantum numbers n, l, ml, s.

n - THE MAIN QUANTUM NUMBER

n – characterizes a power condition of an electron in an atom. Accepts positive integer

meanings from 1 up to (infinite) (n=1... 7 …).

n 12345

Power KLNO

Level, cover



The number of the period in periodic system of D.I. Mendeleyev is such as meaning of

the main quantum number.

With the increase of the meanings of the main quantum number (n ) the energy of AO

increases too. If n=1 then energy is minimal, electron is in a stationary condition

,steadiest of all.  - orbital QUANTUM NUMBER

(COLLATERAL; AZIMUTHAL)

 – characterizes:

-defines energy of an electron to a power sublevel (subshell);

- The form of the nuclears orbital

(l=0, there corresponds a s-sublevel, the spherical form nuclear orbital;

l =1, corresponds р - the sublevel, the form of nuclear orbital reminds a dumbbell;

l = 2, there corresponds a d-sublevel, the form orbital of nuclear represents « four

petal » a figure).

- quantity orbital of nuclear (sublevels) on the level.

-admissible values 0,1,2,3.., n-1,where n - THE MAIN QUANTUM NUMBER .

Let's make the table. This table will contain four lines and five vertical columns.

In the first lines, please, write down the meanings of the n - THE MAIN QUANTUM

NUMBER

In the second lines, please, write down the meanings of the  - orbital QUANTUM

NUMBER

In The third lines, please, write down the Subshell





16

In The fourth lines, please write down the maximal number of an electrons at the level.

The maximum quantity of an electrons at a power level you can calculate using this

equation: 2n2



If electrons situate at the first power level it means that this power level is characterized

by the first MAIN QUANTUM NUMBER n=1.

If n=1 then  will be equal to zero.

If  =0 it means that first power level has only one s - sublevel.

Electron-clouds at this nuclear orbital have a spherical form

The maximum quantity of an electrons at s - sublevel is 2 , then at first power level it is

2 too.

You can use this equation: 2n2 for your calculations: n=1  2.12=2







If electrons situate at the second power level it means that this second power level is

characterized by the n=2.

If n=2 then  will be equal to zero and one. It means thet this second power level has

two sublevels: s – sublevel and p- sublevel. Electron-clouds at p- sublevel have the

form of remind a dumbbell.

The maximum quantity of an electrons at s - sublevel is 2, at p- sublevel -6 , then at the

second power level will be 8 electrons .

You can use this equation: 2n2 for your calculations: n=2  2.22=8









If electrons situate at the third power level it means that this third power level is

characterized of the n=3

If n=3 then  will be equal zero, one and two. It means thet this third power level has

three sublevels: s – sublevel, p- sublevel and d- sublevel. Electron-clouds at d- sublevel

have the form represents « four petal » a figure.

The maximum quantity of electrons at s - sublevel is 2, at p- sublevel -6, at d- sublevel -

10, then at the third power level will be 18 electrons.

You can use this equation: 2n2 for your calculations: n=2  2.32=18









.









17

If electrons situate at the fourth power level it means that this fourth power level is

characterized by the n=4

If n=4 then  will be equal to zero, one, two and three. It means that this fourth power

level has four sublevels: s – sublevel, p- sublevel, d- sublevel and f- sublevel.

The maximum quantity of electrons at s - sublevel is 2, at p- sublevel -6, at d- sublevel -

10, at f- sublevel -14 then at fourth power level will be 32 electrons.

You can use this equation: 2n2 for your calculations: n=2  2.42=32



Table №3

Environment K L M. N

Power level

n 1 2 3 4

l 0 0 1 0 1 2 0 1 2 3

Subshell s s p s p d s p d f

(Sublevel)

Maximal 2 2 6 2 6 1 2 6 1 1

Number the electrons

0 0 4

2

at the level, 2n

2 18 18 32





Alphabetic symbols s -sublevel; p-sublevel; d-sublevel; f-sublevel -were entered in

1890г. At the description of a spectrum of alkaline metals: (sharp-острый) l=0 (s-

sublevel); l=1 (p-sublevel) (main- главный); l=2 (d-sublevel) (diffuse-диффузный);

(fundamental) l=3 (f-sublevel). These letters are not reductions of the words describing

"form" орбитали.





ml – MAGNETIC QUANTUM NUMBER

Defines orientation nuclear orbital in space.

Give by a Klechkovky’s rule we shall calculate the energy of a sublevel



make the table. This table will contain 6 vertical columns.

In the first column, please write down the meanings of the  - orbital QUANTUM NUMBER

In the second lines, please write down the alphabetic meanings of the  - orbital

QUANTUM NUMBER

-In The third column, please write down the meanings ml – MAGNETIC QUANTUM NUMBER

-In The fourth column, please write down the number of the orbital at Subshell.





18

You can use this equation: ml =2l+1 for your calculations number of the orbital at

Subshell.

-In The fifth column, please write down the number of free orbitals (the number a

section). This is graphic representation of the number of free orbitals.

-In The sixth column, please write down the maximal number an electron on the atomic

orbital.

What does it mean if n=1?

How many meanings has  - orbital QUANTUM NUMBER got?  =0

What does it mean if  =0? (It means that an electron situats at the first power level, s –

sublevel, electron-clouds at this nuclear orbital have spherical form)

If  =0 it means that then m  will be equal to zero. You can use this equation: ml =2l+1

for your calculations number of the orbital at s – subshell: ml =2.0+1=1 - it means m 

has only one meaning. This is zero, and then at s – subshell there is only one free

orbital.

Graphic representation of this frees orbital at s – subshell is the one section.

m  =0 - it means that spherical electron-clouds are oriented in the space equal all axis of

coordinates, accordingly: x, y,z.





What does it mean if  =1? (It means that an electron situats at the, p – sublevel,

electron-clouds at this nuclear orbital have the form reminds a dumbbell.

spherical form)

If  =1 it means that then m  will be equal to -1, 0, +1(Minus one, zero and plus one). You

can use this equation: ml =2l+1 for your calculations the number of the orbital at p –

subshell: ml =2.1+1=3 - it means m  has three meanings. These are -1, 0, +1, then at p

– subshell there are three free orbitals.

The graphic representation of this free orbital at p – subshell is the three sections.

m  =-1,0,+1 - it means that electron-clouds are oriented variously in the space along

axis of coordinates x, y, z.





What does it mean if  =2? (It means that electrons situate at the, d – sublevel, electron-

clouds at this nuclear orbital have the form represents «four petals» a figure.)

If  =2 it mean that then m  will be equal to -2,-1, 0,+1,+2(Minus two, minus one, zero and

plus one, plus two). You can use this equation: ml =2l+1 for your calculations number of





19

the orbital at d – subshell: ml =2.2+1=5 - it means m  has five meanings. Thse are -2,-1,

0, +1, +2, then at d – subshell there are five free orbitals.The graphic representation of

these free orbitals at d– subshell are the five sections.

m  =-2,-1,0,+1,+2 - it means that electron-clouds are oriented variously in the space

along axis of coordinates x, y, z,xy,yz ,z2.



Table №4

ℓ Subshell ml Number of the

Graphic representation of The maximal number an

number free orbital electrons on atomic orbital

of the

orbital





(AO)

ml =2l+1



s 0 1  2

0





p -1 

1 0 3 6

+1

2 d -2 

-1

0 5 10

+1

+2

3 f -3 

-2

-1

0 7 14

+1

+2

+3









20

In absence of an external magnetic field electrons on orbitals with identical value of

orbital quantum number  are energetically equivalent. However in a constant magnetic

field some spectral lines are split. It means, that electrons become energetically not

equivalent. For example, p-sublevel has three values (рх, рy, pz) of conditions in a

magnetic field, d -sublevel has 5 values conditions.

S - SPIN QUANTUM NUMBER

It is not connected with the movement of an electron around a nucleus;

-characterizes only the moment of the rotation of an electron about its axis;

- meanings: +1/2 (the movement of an electron сlocwise direction); -1/2 (the movement

of an electron counter сlocwise direction).

Dutch physicists Ulenbek and Goudsmith opened the spin of an electron in 1925

SPIN of an electron is not a simple rotation electron about its axis; it is a complex of

physical phenomena. Дирак in 1928 showed presence of parallel () and antiparallel

()spins.





The condition electron in an atom is defined by these laws:

1. a principle of a minimum of the energy:

The orbital with smaller value of energy is filled at the first !

(1s2s2p3s4s3-rd4p5s4d5p6s4f5d6p7s);

2. PRINCIPLE Pauli :

- In an atom there cannot be two electrons whith all four identical quantum

numbers (1925).

Let's consider an example, using Pauli's principle.

Hydridgen is situated in Mendeleev’s periodic system in the first period. The number of

the period is the same as a value of the main quantum number. Therefore for elements

of the first period of hydrogen and helium n=1. It means thet their electrons situated at

the first power level, s – sublevel.

It mean that Hydridgen’s only one electron is characterized by such quantum numbers:

n=1,  =0, ml=0 ,S=+1/2 or n=1,  =0, ml=0 ,S=-1/2

its electronic formula 1s1 , its electrono- graphic formula

1s ↑

Helium has two electrons situated at the first power level, s – sublevel.

One helium’s electron is characterized by such quantum numbers as n=1,  =0, ml=0,

S=+1/2

if second helium’s electron will be characterized by such quantum numbers: n=1,  =0,

ml=0 ,S=+1/2 then helium’s electrons will have all four identical quantum numbers .

Such situation contradicts Pauli's principle. The second helium’s electron must be

characterized only by such quantum numbers as: n=1,  =0, ml=0, S=-1/2

Two electrons with any value opposite spins can be on this s-orbital





21

Helium’s electronic formula 1s2, its electrono- graphic formula 1s ↓↑





Thus, at the first power level there are no more than two electrons!

3.RULE ХУНДА - The steady condition of an atom is a condition with the

maximum number not coupled electron.

An electron must fill the power sublevel that their absolute value of the sum of a their

spins |S | is maximum

For example: the electronic structure of the last power sublevel of carbon’s atom is 6С

2s22p2 It can be described various by variants of electron-graphic formulas and-in.



It is forbidden

а) 2р ↓↑ |S|=0 б)2р ↑ ↓ | S|=0





It is authorized:

в) 2р ↑ ↑ |S|=1





The condition в) is more steady, because their absolute value of the sum of a their spins

|S | is maximal.

5. The first Klechkovsky’s rule:

Electrons are filling the energy levels from orbital with smaller value of the sum

(n+l) to оrbital with great value of this sum.



For example, the atom of potassium(K) is in Mendeleev’s periodic system in the four

the period. The number of the period is such as the value of the main quantum number.

Therefore for elements of the fourth period n=4. It means that their electrons are at the

power level 1, s – sublevel: 1s

power level 2, s,p – sublevels: 2s 2p electronic formula is 1s2 2s2 p6 3s2 p6 d0 4s1

power level 3, s,p,d – sublevels: 3s 3p 3d

power level 4, s,p,d,f – sublevels: 4s 4p 4d 4f



Serial number калия is 19. It means nuclear’s charge equal to19 too. The atom is

electroneutral therefore an electrons in the atom of potassium(калий) must be 19 too.

How does an electrons fill in the energy levels? To answer this question,

Let’s calculate energy of a sublevel by Klechkovsky’s rule:

for ls – sublevel: the sum (n+  )=1+0=1

for 2s– sublevel: the sum (n+  )=2+0=2

for 2p– sublevel: the sum (n+  )=2+1=3

for 3s– sublevel: the sum (n+  )=3+0=3

for 3p– sublevel: the sum (n+  )=3+1=4

for 3d– sublevel: the sum (n+  )=3+2=5

for 4s– sublevel: the sum (n+  )=4+0=4

for 4p– sublevel: the sum (n+  )=4+1=5

for 4d– sublevel: the sum (n+  )=4+2=6





22

for 4f– sublevel: the sum (n+  )=4+3=7

REMEMBER!

For s – sublevel according to  =0

For p – sublevel according to  =1

For d – sublevel according to  =2

For f – sublevel according to  =3

1s – sublevel will be filled at first because it has the smallest value of energy, then

will be filled 2s, then 2p, then 3s, then 3p, then 4s

electron fills 4s- sublevel, instead of 3d – sublevel, because the sum (n+l) for4s -

sublevel(4+0=4) smallest of energy then the sum (n+l) for Зd (3+2=5), according to first

rule Клечковского.

6. The second Klechkovsky’s rule: If the sum of the main quantum number

and orbital quantum number (n+l) of the оrbitals is identical, the orbital with

smaller value of the main quantum number (n) is filled before.

At filling electron orbital at atom of scandium which electronic formula

1s22s2p63s2p6d14s2, электрон fills 3-d (3+2=5), instead of 4р-sublevel (4+1=5), in

conformity with second rule Клечковского.

At some atoms the phenomenon « electronic jump », as, for example, at atom

of chrome is observed: Cr 3d44s2→3d54s1 or at atom of copper: Cu 3d94s2→3d104s1.

It is said that the raised power stability of electronic configurations with completely

(atom Сu) or half filled sublevel (Cr).

Review Questions

1. What do you know about a structure of an atom.?

2. What do you know about various theories about a structure of an atom?

3. What do you know about the main quantum number, it physical sense?

4. What do you know about the orbital quantum number, it physical sense?

5. What do you know about the magnetic quantum number, it physical sense?

6. What do you know about the spin quantum number, it physical sense?

7. What do you know about the principle of a minimum of the energy?

8. What do you know about the principle of Pauli?

9. What do you know about the rule Hunda?

10. What do you know about the first rule of Klechkovsky?

11. What do you know about the second rule of Klechkovsky?

Further Reading:



1. Frolov V.V.chemistry. Ch. V, §51-56.

2. Luchinsky G.P.rate of chemistry. Ch. V, §8-12, ch. VI, §13-18

3. Ahmetov N.S.general and inorganic chemistry. Section V, ch.3,4.

4. The general chemistry under ред. Sokolovskoj E.M., etc. Ch.6, §1-11.









23

PERIODIC D.I.MENDELEEV’S LAW.

LAW OF the PERIODIC SYSTEM.



Obgectives

After studying this chapter, students should

be able to know:







1. D.I.Mendeleyev's periodic law.

2. A structure of periodic system of D.I.Mendeleyev.

3. Physical sense of periodic system value.

4. Laws of periodic system. Radius of the atom. Valency of an element.

5. Energy of ionization, energy of electron affinity, electronegativity.

6. The plan of the characteristic of elements’s properties by their position in periodic

system.



The periodic law has been formulated on March, 1 first, 1869 by great Russian

scientist Dmitry Ivanovich Mendeleyev. D.I.Mendeleyev, the outstanding Russian

scientist, was born in Tobolsk in 1834. In 1850 at the age of 16 he entered the

Pedagogical Institute in Petersburg to study chemistry. Five years later he graduated

from it with a gold medal and was invited to lecture on theoretical and organic

chemistry at Petersburg University.To continue his studies and research Mendeleyev

was sent to Germany in 1859. While living abroad he made a number of important

investigations.

The year 1868 was the beginning of his highly important work “Fundamentals of

Chemistry”. When working at the subject D.I.Mendeleyev analysed an enormous

amount of literature, made thousands of experiments and calculations. This tremendous

work resulted in the Table of Elements consisting of vertical groups and horizontal

periods. Mendeleyev was the first to suggest a system of classification in which the

elements are arranged in the order of increasing atomic weights. The main idea of the

periodic System is the idea of periodic repetition of properties with the increase of the

atomic weights.

The periodic law in D.I.Mendeleyev's interpretation:

properties of simple bodies, and forms and properties of their chemical compounds

also are in periodic dependence of the value of its atomic weights.

Arranging all the existing elements in the Table Mendeleyev had to overcome great

difficulties, as a considerable number of elements were unknown at that time and the

atomic weights of 9 elements (out of 63) were wrongly determined.

Thanksn to his investigations Mendeleyev was able to predict not only the

extence of a few unknown elements but their properties as well. Later these elements

were discovered.h

Mendeleyev was engaged not only in the study of chemistry. His more than

350 works deal with many subjects. Combining theory with practical activity he carried







24

out enormous reserch in coal, petroleum, iron and steel industries. He died in 1907 at

the age of 73.

The modern formulation of the periodic law:

properties of simple bodies, and also forms and properties of their connections are

in periodic dependence on value of charges of their atomic nuclears.



The periodic system of elements is a graphic representation of the periodic law.

Some hundreds forms of periodic system are known, however, the short variant of

periodic system is used in Ukraine. The periodic system consists of the periods and

groups.

The period is a horizontal row of the elements, where the elements are located in

order of the increasing serial number from the first s-element (ns1) to a p-element

(ns2np6). Each period (except the first one) begins active alkaline metal and ends the

inert gas before which there is a nonmetal (halogen). The last (n) level, and also

before last (n-1) and (n-2) levels of atoms in the periods are filled by electrons. There is

: small (1,2,3) and greater (4,5,6,7) periods. In the period from left to right metal

properties of elements decrease, and nonmetallic – increase.

Number of the period and number of power levels in atom and value of the main

quantum number (n) an external power level are equel.

Groups is a vertical row of the elements, where the the elements have same

electronic structure. Elements of periodic system are subdivided into eight groups.

There are main (A) and collateral (B) subgroups in the group. The main (A)

subgroup there are s-, р - elements of the small and greater periods.Collateral (B)

subgroup have the elements only greater periods.

The number of group corresponds to the maximum valency of an element.

Valency of a chemical element is its ability to formate the chemical communications. In

representation of a method of valent communications value of valency is the quantity

of unpaired electrons. In periodic system each element has strictly certain serial

number and takes strictly a certain place.









The serial number of an element corresponds to a positive charge of a nuclear of

atom (to quantity of protons- 1 р ) and to sum of the electrons (  e ) on it orbitals.

1



Valency electrons have the greatest reserve of energy; they define the chemical

properties of atoms. Atoms, entering into chemical reactions, can to give off valent

electrons or accept them or form the general electronic pairs.

The radius of atom (Rаt) defines the ability of atoms to give off or accept electrons. The

more Rаt, the easier the atom loses the valency electrons and it is more difficult accepts

them.

In period Rат decreases from left to right. The serial number of an element

increases, then the quantity electrons of an element increases too, but the quantity of





25

power layers does not vary.As resalt: a positive charge of a nuclear in atom (to quantity

of protons- 1 р ) increases and a sum of negative charge of the electrons (  e ) on all

1



orbitals increases too. Electrostatic interaction increases. It leads to compression of

electronic density and as consequence, to reduction Rаt.

In the group Rat increases from top to down (as with increase in a serial number

of an element, the quantity of power layers increases too and the electrostatic interaction

decreases.) The atoms easier loses valent electrons and more difficultly them accepts.

The atoms lose valent electrons, show regenerative properties.

The quantitative characteristic of regenerative properties of elements is energy of

ionization (I, кDj/mol, eV/atom).

Energy of ionization (I) is the quantity of the energy which is necessary for

valent electrons detachment from еlectroneutral atom with transformation of last in

positively charged ion (кation). The first energy of ionization characterizes the

ability of atom to give of one electron:

Э0(еlectroneutral atom) + I(energy of ionization)  Э+( positively charged ion

(кation))+ one electron.

For multielectronic atoms of energy of ionization I1; I2; I3... characterizes the first

valent electron detachment, the second electron detachment … Thus I1 I2 I3...,

becouse the increase the quantityof the loseing valent electrons leads to increase of a

positive charge of an ion.

Potential of ionization -is the attitude energy of ionization to a electron’s charge

(is measured in electron-volt, (eV)). The potential of ionization is numerically equal to

energy of ionization.

In the period value of energy of ionization from alkaline metals to inert gases

increases, because the Rаt in the period decreases, and the charge of a nuclear of atom

increases, therefore there is a compression of electronic density, and it is necessary to

spend a lot of energy for tearing off electron from neutral atom.

The value of energy of ionization decreases from top to down in groups, because the

radius of atom (Rаt) increases in groups from top to down, therefore it isn’t necessary

to spend a lot of energy to tear off electron from neutral atom.

Francium 223Fr has the smallest energy of ionization.

87

The atoms accepting valent electrons, show oxidizing properties. The quantitative

characteristic of oxidizing properties of an element is energy of affinity to electron (Е;

eV/atom).

Energy of affinity to electron is the energy allocated at connection electron to

neutral atom; it characterizes the ability of atom to formation of negatively

charged ions:

Э0 (not raised atom) + one electron  Э- (negatively charged ions) +E(Energy

of affinity to electron).

Fluorine atom (F) has the greatest Energy of electron affinity. Energy of electron

affinity is much less than ionization energy of the same atoms.

Both these valuas (I and E) are depended on a charge of a nuclear and the atom’s

size: they should grow with increase in a nuclear’s charge, and with increase in radius of

atom – to decrease.





26

American scientist Poling entered the the special characteristic – electronegativity

(ЭN) 1932. It is characterizes the ability of atoms in molecule to attract the electronic

density.

ЭN = 1/2 (I +E).

The more valua of the ЭN, the nonmetallic properties of an element are expressed

more strongly. There are more strongly expressed metal properties of an elements if the

valua of the ЭN to less. Electronegativity (ЭN) has no strictly physical sense, but it is

used for the estimation of a real condition of chemical communication in molecules.

Lithium’s electronegativity is accepted for unit of the ЭN . The most electronegative

element is fluorine.

The elements of D.I.Mendeleev’s periodic system can be divided into four types

according to the atom structure.

s - elements (the s-sublevel of the last level is filled at the end):... ns1, ns2

p - elements (the p-sublevel of the last level is filled at the end)... np1-np6.

d - elements (the d-sublevel of the before last level is filled at the end)... (N-1) d1ns2 –

(n-1) d10ns1.

f - elements (the f-sublevel – the second of the before last level is filled at the end)...

(N-2) f1 ns2 - (n-2) f14 ns2.

s - elements settle down in IА and IIА groups, the main subgroups. The maximum

quantity of an electron at s - sublevel is 2.

These are typical metals, have similar chemical properties as are electronic analogues.

Electronic analogues are elements having the electronic formula of the last level

general for all, described by s – elements of the IА group:

Li (Lithium) Na(sodium) K(potassium) Pb, Cs, Fr

2s1 3s1 4s1 5s1 6s1 7s1

ns1 – electronic analogues (these elements have only one electron at s – sublevel)

Table №5

Reducer оxide The basis asid Salt

At atom 11Na

11 эlectrons, from them one

is valent, therefore Na is Na0  Na+

e

Na2O NaOH - --- NaCl

metal ( (is oxidized)

I valent basic= 1

s – elements of the IIА group:

Be(beryllium) Mg(magnesium) Ca(calcium) Sr Ba

2 2 2 2

2s 3s 4s 5s 6s2

ns2 – electronic analogues(these elements have two electrons at s – sublevel)

Table №6

Окси The basis Acid Salt

Reducer

д

56Ba – metal 

Ba0  Ba2+

е

2

valentbasic =0 - - ---

V* (возб.сост). = 2 BaO Barium BaSO4

(OH) 2





27

p - elements arrages from IIIА to VIIIА groups, the main subgroups. At p –

subshell there is three free orbitals, wich of them can consist of two electrons

The maximum quantity of an electrons at p - sublevel is 6.

Thus, the main subgroups has only s-, p – elements. Among p-elements there are

metals, but they can be amphoteric often:

Table №7

Reducer оxide The basis Acid Salt

13Al

AlCl3

valent basic=0 HAlO2

Al0 3e



 Al 3+

Al2O3 Al(OH)3 NaAlO2

V* (возб.сост). H3AlO3

Na3[Al(OH)6]

=3



p-elements are nonmetals if there are 4-8 electrons at last power level.

Table №8

Oxidizer оxide The basis Acid Salt

0 e

-

F  F

9F F2O – HF NaF

()

The element 17Cl has vacant 3d – a sublevel, to which can pass valent electrons from 3s

and 3p- sublevels. As resalt - the element 17Cl shows variable valency:



Table №9

оxide Acid Salt

0 e -

Cl  Cl HClchlorohydric CaCl2-chloride of calcium



Cl0  Cl+1

e NaClO–

Cl2O HClO hypochlorite of sodium

hypochlorous

17Cl

valent basic=1 Cl2O3 HClO2 Chlorous NaClO2–sodium chlorite

V*(возб.сост).= Cl0  Cl3+



3e





3,5,7 Cl0  Cl+5



5e

HClO3chloric NaClO3-

Cl2O5 хлорноватая chlorate of sodium





Cl0  Cl+7

7e



Cl2O7 HСlO4 perchloric NaClO4–perchlorate of

sodium



The inert gases (He, a neon) have the p-sublevel is filled completely by its

electrons. The valency of these inert gases (He, a neon) is equal to zero, there is no

place for any electrons.

The inert gas 18Ar has vacant 3d – a sublevel. valencies are possible:





28

valent норм=0; V * = 2,4,6,8.

The inert gas 54Xe burns in fluorine with formation: XeF2, XeF4, XeF6, XeF8.

d – elements are metals. Its have two (2) eletctrons at the last electronic level ns2. A

vacant оrbitals of the preexternal layer is filled by valency of electrons. d – elements

are named transitive, because it is in periodic system between s- and thet the p-

elements, they form collateral groups.

Table №10

оxide The basis Salt

21Sс

(3d14s2) Sc0  2 е Sc+2



ScO Sc (OH) 2 ScCl2



 

valent Sc0  Sc+3



3e



basic=1 Sc2O3 Sc (OH) 3 ScCl3

V*(возб.

сост).=

2,3

The plan of the characteristic of properties of elements by position in periodic

system (PS).

1. Position of an element in PS (a serial number, the period, group, and a subgroup).

2. A structure of atom (a charge of a nuclear and its structure) – number of protons,

neutrons, number of electrons in atom, structure of an electronic level of atom, the

electronic formula of valent levels, valency basically and the raised conditions.

3. Type of an element (s-, p-, d-, a f-element), (metal, nonmetal), possible degrees of

oxidation, the oxide’s formula and gidroxide for each valent condition.

Example: to give the characteristic of sodium. Its serial number is 11, a charge of a

nuclear of atom is equal to +11, around of a nuclear to rotate 11 electrons. Sodium is

located in the third period, therefore it electrons settle down at three power levels.

Sodium is located in the first group, the main subgroup; atom Na has one valent

electron at an external level. Na – metal ( the atom is metal if it has at an external power

level 1,2,3 electrons), oxside of sodium Na2O has the basic character. To it there

corresponds basis NaOH.

Opening of the periodic law by D.I.Mendeleyev and creation of periodic system of

chemical elements by it was triumph in the development of chemistry of theXIX

century.

The knowledge of properties 63 chemical elements which has been collected has

been resulted by D.I.Mendeleyev in the strict order. With opening the periodic law there

was an opportunity to predict property of new elements and their connections.

D.I.Mendeleev on the basis of the periodic law has predicted properties then still

unknown elements: Sc, Ga, Germany, Tc, Rе,Po, At, Fr, Rа, Аc, Pa. D.I.Mendeleev

could correct nuclear weights of elements already known at that time: Be, Ti, Y, In, La.

The achievements in chemistry and physics at the end of the 19th and the

beginning of the 20th century made it necessary to reconstruct the Periodic Table taking

into account new descoveries. This progress resulted in the discovery of the inert gases

of the so- called zero group, and the study of 14 rare earth elements. In the last few





29

decades 11 new radio- active elements were obtained. Two of them were named in

honour of Russian scientists: the 101st was called mendelevium and the 104th –

kurchatovium (in memory of Igor Kurchatov)

Time is the severest judge in science. After the 100 years of its existence, the

Periodic Law has preserved its full value and is being constantly developed with each

new discovery.

Review Questions

1. Definition of notion “group” of periodic system of D.I.Mendeleev.

2. Definition of notion “period” of periodic system of D.I.Mendeleev.

3. Definition of notion “radius” of atom. How is it chang in the group, period?



4. Definition of notion “energy of ionization” of atom. How is it chang in the group,

period?

5.Definition of notion “energy of affinity” of atom. How is it chang in the group,

period?

6. Definition of notion “electronegativity” of atom. How is it chang in the group,

period?

7. Definition of notion “Electronic analogues”

8. What do you know about s – elements?

9. What do you know about p – elements?

10. What do you know about d – elements

11. What do you know about f – elements

12 Fluorine and chlorine are electronic analogues. Why chlorine has more valence than

fluorine?

13 Why Ar or Xe has more compounds than He or Ne?





Further Reading:



1. Frolov V.V.Chemistry. v. V, §51-56.

2. Luchinsky G.P.rate of chemistry. v. V, §8-12, Ch. VI, §13-18

3. Ahmetov N.S.general and inorganic chemistry. Section V, ch.3,4.

4. The general chemistry under ред. Sokolovskoj E.M., etc. Ch.6, §1-11.









30

CHEMICAL THERMODYNAMICS.

Obgectives After studying this chapter, students should

be able to know:





1. Definition of this notion « chemical thermodynamic ».

2. The basic notions of section « chemical thermodynamic »:

Thermodynamic system, kinds of systems;

Thermodynamic parameters;

Process, classification of processes;

Cycle;

Thermodynamic functions.

3. The first law of thermodynamics:

izohorny process, internal energy of system

Izobarny process, definition of notion «entalpiy».

4. Thermochemistry:

Definition of notion "thermochemistry";

Thermal effect of chemical reaction;

Gess's law is important laws of thermochemistry.







The thermodynamics – as a science has arisen in the beginning of XIX century (in

connection with problems of perfection of thermal machines). In translation with Greek

"thermos" means heat, and «dynamic» – force and power. The classical thermodynamic

is researching of energy and work in macroscopical systems. It has value for such

sciences, as physics, chemistry, biology, geology, for numerous branches of technics as

any nature’s processes, wich are accompanied by changes of energy.

The chemical thermodynamics is a section of chemistry, which

studies 1) transitions of energy from one form in another at chemical processes and

2) establishes a direction and limits of their spontaneous course under the given

conditions.

Thermodynamic chemical system – the complex of substances cooperating

among themselves mentally isolated from an environment. For example, system can be

the chemical glass containing certain quantity of water, or heat- exchanger, used at the

chemical enterprise, etc.

There are three types of thermodynamic systems (Table.11).

The isolated systems – cannot change with environment energy, weight. For

example, isolated termostat, the Universe as a whole.

The closed systems – can change with an environment with energy only, but not

with weight. For example, molecules of the dissolved substance can be considered as

the closed system, and as an environment there can be all rest (probably solvent if it







31

does not participate in reaction). Therefore in chemical thermodynamics the closed

systems are more often used.

The open systems are systems which can change with an environment and

energy, and weight. For example, alive objects of an animal or flora.



Table №11



The isolated The closed The open

systems systems systems



E E E



m m m









All systems can be in various conditions. The thermodynamic characteristics are

used to describe their conditions. (рис.4). A condition of systems can be considered by

thermodynamic parameters of a condition. It is temperature (T), pressure (P) and

quantity of substance (n).

Other thermodynamic characteristics depend on these three parameters (Т, Р, n), and

from a condition of system at whole. Therefore them name are functions of a

condition. They are: U- internal energy, S-entropy, H-entalpy, Energy of

Гельмгольца,Energy of Гиббса.





Термодинамические характеристики









параметры состояния состояние я

функции состоя иия









Рис.4





A condition of any systems can be considered by U- internal energy.

Internal energy of system (U) is the sum of energy of thermal movement of molecules,

intramolecular energy and energy of intermolecular interaction. Whether is it possible to

calculate absolute value of internal energy? Whether is it possible to calculate kinetic or

potential energy of atoms and molecules from which any system consists of? No.







32

Absolute value of the Internal energy of system (U) is not known. So, what can we

calculate?



Let’s consider ourselves as open system. Imagine that you got up in the morning, had

breakfast, were full of forces and energy. It means thet you possess a stock of internal

energy U1. Then you went to university, attended lectures, someone can be also a sports

hall, and someone had visit to the girl. It depends on morning condition, from a

reference value of internal energy. In the evening you already feel weariness. The stock

of your internal energy has been spent for cares of day. Therefore your evening

condition will be characterized by a smaller quantity

of internal energy U2. Thus, internal energy characterizes a condition of system during

the certain moment of time, for example in the morning or in the evening as has been

shown in our example. And so we cannot calculate morning absolute value of internal

energy or evening absolute value of internal energy. But what can we calculate? We can

calculate changes of internal energy in processes–U. U does not depend on the a way

of transition of system from one condition in to another. Internal energy is a function of

a condition of system. Internal energy is defined by kinetic energy it mean, that it

depends on temperature. Internal energy is defined by potential energy, it mean, that it

depends on volume. Internal energy is a function of a condition of system, and depends

on the temperature and the volume. Therefore there is a sense to write down equation of

full differential for internal energy.

U U

dU= ( )V dT  ( ) T dV

T V



Tell, please: What was between a morning condition, full of forces and energy and an

evening condition of weariness? You have worked! Thus during the transition of system

from one condition in another there is a sense to speak about work(A) or heat (Q)

which is allocated or absorbed as a result of performance of this or that process.

Work and heat are not functions of a condition of system. These notions have sense only

when processes are. Therefore is possible to speak only about infinitesimal quantity of

heat and work and isn’t possible to speak of it full differential.

As to heat (Q) and works (A) it is two unique forms of transfer of energy from

system to an environment and back. Signs on heat and work are defined by the scheme

5.

Q0

Q0

System

A0 A0





scheme.5



1. Q0; A0 if the system receives heat or above it work is made.

2. Q 0; A0 if the system gives heat or itself makes work.









33

The corresponding sign (plus or a minus) isn’t wrote before letters Q and A, and it has

been written before digital value of the corresponding size, for example, Q =-300 кДж

or A= + 50 кДж.



All three concepts heat, work and internal energy are interconnected among themselves.

It is at the first law of thermodynamics.

Heat is transferred to system (Q) goes on change of its internal energy (U) and to

work



Q =U+A

The first law of thermodynamics has been formulated by Joule in the middle of the

XIX century. Inherently it is the law of conservation of energy









Thermodynamic processes are changing of any condition’s parameters ( transition

of a system from one condition in to another). Processes are:

- Isohorniy (V=const);

- Izobarniy (P=const);

- Isothermal (T=const);

- Adiabatic (there is no exchange of energy of system with an environment);

- Izobarniy - isothermal (P; Т=const);

- Isohorny - isothermal (V; T=const).



Let’s write down the differential equation of the first law of thermodynamics for

isohorny processes (V=const).  QV=dU+  A, where  A is the work of expansion:

 A=P dV

.



 QV=dU+ P dV, but there isn’t the changes of volume for isohorny

.

Then

processes (V=const). P.dV=0

the differential equation of the first law of thermodynamics for isohorny processes

(V=const) is  QV=dU

the integrated kind equation of the first law of thermodynamics for isohorny processes

(V=const) is QV = U.

For example, heating of gas under the fixed piston, the work is not made (А=0),

therefore QV = U.



Conclusions:

1. the heat of isohorny processes (QV) is function of a condition of system such

asU.

2. QV = U.This first mathematical equation of tht Gess’s law (see further).

Izobarniy processes (P=const).

For example, heating gas makes work of expansion against constant external

pressure. The work of expansion (А) is defined as sum of pressure(P) and changes of





34

volume (V). Therefore the differential equation of the first law of thermodynamics

 QP =dU+ P dV

.

for izobarniy processes (P=const) is

The pressure(P) is constant, then you can write down:  QP =dU+ d P.V. The sum

of the differential’s U and P.V is the differential of their sum:  QP =d(U+ P.V)

If the sum U+ P.V to designates as one letter H (H= U+ P.V) you can write down

 QP =dH it is

the differential equation of the first law of thermodynamics for Izobarniy processes

(P=const).

the integrated kind equation of the first law of thermodynamics for isohorny processes

(V=const) is QP = H.



Conclusions:



1. The sum H= U+ P.V is system’s heat or system’s energy at whole.

2. This letter H means ENTALPIY

3. Entalpiy is function of a system’s condition such as the heat of isobarny

processes QP.( as work against constant external pressure does not depend on a

way of fulfilment of the process, it is defined only by an initial and

final condition of the system V. The work is function of a

condition during izobarniy process only the done.



4. QP = H this second mathematical Equation of the Gess’s law

(see further).

5. For exothermal reactions QP  0, the system gives of heat to the

Environment and its internal energy decreases  Н  0. (entalpiy is belong

ziro)

For endоthermal reactions QP  0, the heat is given to a system and its

internal energy increases  H  0(entalpiy is above ziro).





The thermochemistry- is the section of thermodynamics studying thermal effects of

chemical reactions.

Gess's law is the law of thermochemistry (1840):

The thermal effect of chemical reaction does not depend on the mechanism of a

reaction, and depends only on nature and a physical condition of initial substances

and products of reaction( or of finel(конечные) substances) (under condition of

V=const; P=const).

QV = U - this first mathematical equation of the Gess’s law

QP = H - this second mathematical Equation of the Gess’s law



Let's illustrate the meaning of this law in an example.

Димер оксида nitrogen’s(IV) di oxid can be received in several ways.

The first, 1) N2 + O2 = (is) 2NO, H1 = 180900 Djoul/mol ;( Djoul per mol)







35

2) 2NO + O2 = 2NO2, H2 =-77100 Djoul/mol ;( if we add one molecula O2

to two NO molecula then we receve NO2)

3) 2NO2 = N2O4, H3 =-10800 Djoul/mol.

Or, the second,

4) N2 + 2O2 = N2O4, H4=93000 Djoul/mol.

H1 + H2 + H3 = H4.

If illustrate it to Graphic:









Рис.6

The entalpy’s definition includes internal energy, so absolute value of entalpy of a

system (or any substance) is not known. Substances’s enthalpy is characterized by

entalpy of substance’s formations.

Н0298 (form) (standard enthalpy of substance’s formations) is a heat of a reaction of

formation complex substance (X) from simple substances (or elements) under standard

conditions: simple substances  X complex substance.

For simple substances Н0form equals zero (in steady modular conditions).

Values Н0form there are in tables.



Reaction’s enthalpiy of any substances can’t be defined at the laboratory. For

example, it is impossible to define ethanol’s enthalpy in the laboratory, becouse ethanol

cannot be synthesized from atoms C, Н, O. Entalpy can be calculated in such case by

law of Gess.

Consequences from Gess's law:

The thermal effect (or hte enthalpiy) of chemical reaction can be found, as a

difference between the sum of heats of formation of ( finel(конечные) substances)

products and the sum of heats of formation of initial substances, accounting chemical

reaction’s coefficients.

Нc.r. = Н0 (products) – Н0(initial substances),

accounting chemical reaction’s stehiometric coefficients!

 - The Greek letter “sigma”, shows operation of summation.

Problem: To define standard enthalpy of formations Н0 (РН3), from the equation.

2РН3 (g) + 4О2 (g) = Р2О5 (с) + 3Н2О (l), Нc.r =-2360 кlJoul.

(с) - a сrystal condition, (l) a-liquid condition, (g) - a gaseous condition.



The decision:

0 0 0

Нc. r. = [Н (Р2О5 (c) +3Н (Н2О (l)] - 2Н (РН3);

2Н0 (РН3) = [Н0 (Р2О5 (c) +3Н0(Н2О (l)]- Н0c. r.,

Н (Р2О5 (c) =-1492 кDj/mol, Н (Н2О (l) =-285,8 кDj/mol.

0 0









36

2Н0 (РН3) = [(-1492 + 3 (-285,8))] – (-2360) = 10.6 кDj/mol

Н0 (РН3) = 10,6/2=5,3 кDj/mol.





Standard enthalpy of the reaction of neutralization.

The heat is identical for reaction of neutralization.

Let's calculate it:

the reaction of neutralization is the reaction between strong acids and the strong alkali.

for example the reaction between a hydrochloric acid and sodium alkali is

HCl + NaOH NaCl + H2O

The full ionic reaction of neutralization: Н + +Cl- +Na+ + ОН-  Н2О + Cl- +Na+

The reduced ionic reaction of neutralization: Н + + ОН-  Н2О



for example the reaction between a hydrochloric acid and spotasum alkali is

HCl + KOH KCl + H2O

The full ionic reaction of neutralization: Н + +Cl- +K+ + ОН-  Н2О + Cl- +K+

The reduced ionic reaction of neutralization: Н + + ОН-  Н2О









for example the reaction between a Nitric acid and calcium alkali is

2HNO3 + Ca (OH) 2 Ca (NO3)2+ 2H2O

The full ionic reaction of neutralization: 2Н + + 2NO3- +Ca2+ +2 ОН-  2Н2О + 2NO3-

+Ca2+

The reduced ionic reaction of neutralization: Н + + ОН-  Н2О

What is identical in these chemical reactions of neutralization?

The chemical meaning of these chemical reactions of neutralization is identical. It is the

reaction to receive a molecule of water.

Let's calculate it: Н + + ОН-  Н2О Н0neutral. = - 55.8 кDj/mol. It is the constant.



standard enthalpy of phase transformations.

Heat of phase transition is equal to a difference standard enthalpy of substances (Н0)

in one and in the other condition. For evaporation of water (250С) it has:

Н2O(l)  Н2O(g)

Н (Н2O(l))= -241.8 кDj/mol, Н (Н2O(g))= -285.8 кDj/mol

0 0



Н evaporation. =-241.8 + 285.8 = +44 кDj/mol (endothermal process).









Review Questions

1. What is the definition of this notion « chemical thermodynamic »

2. What do you know about thermodynamic system, kinds of systems?

3. What is the definition of this notion “thermodynamic process” What do you know

about classification of processes?



37

4. What do you know about thermodynamic parameters?

5. What is the definition of this notion “thermodynamic cycle”?

6. What are thermodynamic functions? Named them,please.

7. What do you know about the first law of thermodynamics?

8. What do you know about izohorny process?

9. What is internal energy of a system?

10. What do you know about izobarny process, the definition of notion «entalpiy».?

11. Define the notion of "thermochemistry".

12. What do you know about Gess's law?

13. Can you name the consequence from Gess's law?

14. Can you name two mathematical equations of the Gess’s law

15. Define the notion Н0298 (form) “standard enthalpy of formations of substance”

16. Define the notion “standard enthalpy of reaction of neutralization”.

17. Define the notion “standard enthalpy of phase transformations”.



Further Reading:

1. Frolov V.V.chemistry. ChaterV, §51-56.

2. Luchinsky G.P.rate of chemistry. Ch. V, §8-12, гл. VI, §13-18

3. Ahmetov N.S.general and inorganic chemistry. Section V, ch.3,4.

4. The general chemistry under ред. Sokolovskoj E.M., etc. Ch.6, §1-11.









38

THE SECOND LAW OF CHEMICAL THERMODYNAMICS.



Obgectives After studying this chapter, students should

be able to know:



1. Conception of thermodynamic probability.

2. Entropy.

3. Formulations of the second law of thermodynamics.

4. Energy of Helmholts.

5. Energy of Gibbs.



Essence the first law of thermodynamic, which you have studied at the last lecture,

is such as essence of the law of preservation.



Dear friends, let me tell you some examples which could happen in our life.



Have you ever met such phenomenon in your life: All the gases composing the air we

breathe (such as oxygen, nitrogen) without any external influence have been divided. If

there were the molecules of oxygen in one part of our audience, and the molecules of

nitrogen in another, which one would we breathe in such a situation? What do you

think?

For example you stroked your clothes, and switched off an iron. Instead of cooling

down, its temperature was increased. The iron took energy from an environment.



Tell me please do these phenomena contradict to the first law of thermodynamics?



The first law only allows counting the power of the process.

However it does not solve the problem of direction of the process and its possibility. For

example, the direction of such processes as spontaneous division of gases and the

transition of heat from a cold body to the hot one is impossible in nature. However,

such phenomena don’t contradict the first law of thermodynamics.

So, these problems are solved by second law of thermodynamics.

There are f lot of processes have to spontaneous go only one direction. For example,

time. Ray Bradbury in “A sound of thunder” wos riten:

- A touch of the hand, and this time on the instant, beautifully reverse itself. The old

years, the green years,might leap;roses sweeten the air, white hair turn black, wrinkles

vanish; all, everything fly back to seed, flee deas, rush down to their beginninggs, suns

rise in western skies end set in glorious easts. Can be such things in our life? It is

impossible as the certain processes have the certain direction. And time is going only

one direction from birth to death, unfortunately.





There are formulations, of the second law of thermodynamics in the textbooks for

example:





39

1. Klauzius’s formulation Клаузиус

« Heat cannot spontaneously be transferred from colder body to the hotter one».

2. Thomson’s formulation

« Heat of the coldest body cannot be a source of work ».



3. « The efficiency of the steam machine always is less than the unit », etc.

But these formulations are important for physical processes. And for chemical

processes it is important that the the second law of thermodynamics gives criteria

which helps us to solve the problem of possibility of the process and its direction.

The first criteria: The thermodynamic probability, W

And now there are some examples.



We use notion of probability in our life, to characterize a lot of processes. For example,

the low bird’s flying means the possible rain.

If you do well during the semester, you will possible pass the examination successfully.



So imagine, there are two rooms in a house, they were divided by a door. In the first

room there are the molecules of ‘He’, in the other one there are the molecules of ‘H2’. If

you opened the door, what would it be?



a) Probably these gases will mix. Let's put the letter W1 as a probability of this process

b) Probably these gases will not mix. Let's put the letter W2 as a probability of this

process.This is possible, if there are only a few molecules in a room. But our

experience, and practice have proved that the mixture is more possible. The

thermodynamic probability, W1> W2





He He W1

H2 H2 HHe

2

H2 HeHe He

H 2

He H2 H2

He H2 He H2

He He H2

He H2 W2

He H2









The thermodynamic probability, W -is a quantity of microconditions of the

system.

Let’s defermine the notion “microconditions of the system”



The microconditions of the system are a concrete position in space of separate

particles (molecules) at present time which corresponds to the steadiest condition

of system.





40

It is the function of a condition and it is maximal at the chemical equilibrium.

However W is connected with mechanical characteristics of the system: kinetic rate of

molecules, their position in space and so on. And for chemical thermodynamics it is

important to find out the criterion of a direction of the process, defined in

thermodynamic parameters: Т, Р, V (temperature, pressure, and volume).





The second criteria: Entropy S



Entropy S is a measure of the power disorder in the system.

Entropy S is connected with thermodynamic probability a parity: S = klnW, where



k -is Boltsman constant =R/NA  1,381.10-23 Dj/ degree

Dj * K

R- Gas constant = 8,31

mol

NA- number of the Аvоgadro = 6, 02 .10 23 моль–1.



k- Больцмана constant helps to transite the system from the simple "disorder" to « the

power disorder ». Counting upon 1 mol particles: S = R lnW.

Change Entropy is defined by the change of the number system’s microconditions

S = S2 – S1 = R ln W2/W1.

The mathematical equation of the second law of thermodynamics: S  Q/T

For reversible S = Q reversible/Т.

For irreversible processes: S Q irreversible /Т (the algebraic sum of the resulted heats

for irreversible processes: Q irreversible proc.  Q irreversible pros.



Experience shows, that many spontaneously continuing processes go with

allocation of heat (Н0). Among spontaneous processes there are also absorption’s

processes (Н0). For example, there is dissolution nitrate of ammonium in water:

NH4NO3 (c)  NH4NO3 (l); Н0c.r. = + 27 кDj/mol.

So, allocation of heat (Н0) is not the solving criterion of an opportunity of

spontaneous course of processes.

There are a lot of spontaneously continuing processes whiсh are characterized by

increase Entropy. But there are spontaneously continuing processes whiсh are

characterized to reduce Entropy(S0): C + О2  СО2.

Chemical communication has formated between molecules of carbon and oxygen О2.

As result the degree of the power disorder decreases.

What is criterion of spontaneity – Entalpy (heat) or Entropy?

Balance between Entalpy (heat) and Entropy (S) is the criterion to define an

opportunity of course of process. Spontaneous processes with absorption of

Entalpy (heat) can be, if growth of power "disorder" is incrising Entropy(S>0), and

processes with S0 are possible too, if it is accompanied by stronger allocation of heat.



41

For the isolated systems where the exchange of energy with an environment is

excluded Q=0, the inequality of the second law of thermodynamics is: S 0.

Thus, for the isolated systems there is only one criterion of spontaneous course

of processes – increase Entropy. The processes lead to increase Entropy is only

spontaneous.

So:

1. Entropy is a function of a system condition; therefore you can calculate

Entropy’s change using consequence from Gess's law. Entropy’s change of chemical

reaction is equal to a difference between the sums Entropy’s products of reaction

and Entropy’s initial substances, accounting chemical reaction’s coefficients.

Sc.r. = S0 (products) – S0 (initial substances),

accounting chemical reaction’s stehiometric coefficients!

 - The Greek letter “sigma”, shows operation of summation.



Problem: To calculate Entropy’s change this chemical reactions (Sх.р.):

3С2Н2 (g)  С6Н6 (l).

The decision: Sх.р. = S0 (C6H6 (l)) – 3 S0 (С2Н2 (g)).

Sх.р. = 269,2 – 3.200,8 = - 333,2 Dg/mol.

2. Entropy is measured by the attitude Q (heat) /T or H/T.

Problem: What is Entropy’s change of transition from crystal in a liquid condition one

mol iron (Fe) if Тfusion =15360С (1809K) at, if Entalpy Нfusion = 13765 Dj/mol.

At the temperature of fusion for equilibrium condition this system Fe(с)  Fe(l) 

T=const = Тfusion =15360С (1809K)





Sfusion = S (Fe (l))-S (Fe (c)) =Нfusion/Тfusion = 13765/1809 = 7,61 Dg/K mol.

S0  Process of transition from crystal in to a liquid condition at Тfusion=15360С

(1809K) is spontaneously.

3. Entropy is criterion of spontaneous course of processes only for the isolated

systems. Processes go spontaneously in the isolated systems with only increase

Entropy.

Состояние равновесия

S









S0 S0



изолированные системы





42

4. S is connected with the number of possible microconditions in the system (with

thermodynamic probability, W), characterizes a measure of the power disorder in the

system. S = klnW.

5. S is a thermodynamic function which unlike from thermodynamic probability

W is connected with thermodynamic values Q; T, it is more comprehensible to chemical

thermodynamics.

However, more often there are no systems isolated in natural, in technics. Entropy can

not be used for the closed systems.

For izothorno -isothermal conditions (T, V=const.) the first law of thermodynamics

has mathematical form possible to write down so:

QV = U (= РV; V = a constant  A=0)

For the second law of thermodynamics, write down so: S  QV/T  QV  ТS.

Uniting two expressions, we shall receive: TS  U; U - TS  0.

Let’s name a difference U - TS to name a new value F. It is the thermodynamic

function of a condition – free Helmholts’s energy, U - TS =F.

For izothorno -isothermal conditions (T, V=const.) the second law of

thermodynamics has mathematical form possible to write down so: F 0.

We can use only one criterion of a direction of the process for izothorno -isothermal

conditions (T, V=const.) (Such as Entropy (S) for the isolated systems). However this

criterion F of a direction of the process is only balance same criterias: U and S.

So:

1. For izothorno -isothermal conditions F is criterion of spontaneous course of

processes. If the Helmholts’s energy decreases (F 0), such processes are

spontaneous.

F





F 0 F 0





F = 0

Закрытые системы при постоя нном V и T

Состоя ние равновесия







рис10

F0, processes can go spontaneously;

F = 0, equilibrium condition of thisprocess;

F  0, processes cannot pass.





43

2. F =U - TS this is incorporated first and second laws of thermodynamics (T,

V=const.)

3. Helmholts’s energy is a function of a system condition; therefore you can

calculate the change of the Helmholts’s energy using consequence from Gess's law.

change of the Helmholts’s energy of the chemical reaction is equal to a difference

between the sums of Helmholts’s energy of the products of reaction and

Helmholts’s energy of the initial substances, accounting chemical reaction’s

coefficients.

Fc.r. = F0 (products) – F0(initial substances),

accounting chemical reaction’s stehiometric coefficients!

 - The Greek letter “sigma”, shows operation of summation.



For izobarno -isothermal conditions (P, T =const.) the first law of thermodynamics

has mathematical form possible to write down so:

QP = U + РV =H

For second law of thermodynamics to write down so: S  QP/T  QP  ТS. 

 H  ТS

Uniting two expressions, we shall receive: TS  H; H - TS  0.

Let’s name a difference H - TS to name a new value G. It is the thermodynamic

function of a condition – free Gibbs’s energy, H - TS =G.

For izobarno-isothermal conditions (P, T=const.) the second law of

thermodynamics has mathematical form possible to write down so: G 0.

We can to use only one criterion G of a direction of the process for izobarno -

isothermal conditions (P, T=const.) (Such as Entropy(S) for the isolated systems, such

as criterion F for the closed systems (V, T=const.)). However this criterion G, such

as criterion F for the closed systems (V, T=const.) is only are determined by the

balance of the same criteria: Н and S.

So,

1. For izobarno -isothermal conditions G is criterion of spontaneous course of

processes. If the Gibbs’s energy decreases (G 0), such processes are

spontaneous.

G



G 0 G 0





G = 0

Состояние равновесия



н

Закрытые системы при постоя ных Т и Р

44

рис.11

G0, processes can go spontaneously;

G = 0, equilibrium condition of thisprocess;

G  0, processes cannot pass.

2. G =Н - TS this is incorporated first and second laws of thermodynamics (T,

P=const.)





3. Gibbs’s energy is a function of a system condition; therefore you can to calculate

change of the Gibbs’s energy using consequence from Gess's law. Change of the

Gibbs’s energy of chemical reaction is equal to a difference between the sums

Gibbs’s energy of the products of reaction and Gibbs’s energy of the initial

substances, accounting chemical reaction’s coefficients.

Gc.r. = G0 (products) – G0 (initial substances),

accounting chemical reaction’s stehiometric coefficients!

 - The Greek letter “sigma”, shows operation of summation.



G0 (X) – standard of Gibbs’s energy of substance X under standard conditions

and in the certain condition.

G0 (X) standard of Gibbs’s energy of simple substances and elements (such as

Н0 (X)) is equal to zero.

Problem: Can these processes go spontaneously?

С6Н12О6 (l) ( Glucose)  2С2Н5 (OH) (l)( spirit) + 2СО2 (g)( сarbonic gas)

G0(С6Н12О6) =-915кDg/mol; G0(С2Н5 (OH)) =-174кDg/mol ; G0(СО2) =

-394кDg/mol

The decision: using consequence from Gess's law let’s write down:

G0c.r.=[2G0(С2Н5 (OH)) +2G0(СО2)] - G0(С6Н12О6)

G0c.r.=[ 2 (-174) + 2 (-394)] – (-915) =-221 кDg/mol

Н0c.r. this process is -79 кDg/mol. Hence, due to энтропийного the

Ability of fermentation process of spirit to make useful work increases in three times,

due to

Entropy’s contribution.

Fermentation process of spirit is source of the energy. It is very important for anaerobic

organisms, and on animals and the people in during intensive physical work.

4. G equal to useful maximal work which could be during isobarno-isothermal

process.

Review Questions

1. What is the definition of this notion « chemical thermodynamic »?



45

2. What do you know about the first law of thermodynamics?

3. What do you know about the second law of thermodynamics??

4. What are thermodynamics functions? Named them, please?

5. What do you know about izobarny process?

6. What do you know about «entropy»?

7. Whan the entropy is criterion of spontaneous course of processes only?

8. Whan Gibbs’s energy is criterion of spontaneous course of processes only?

9. Whan Gelmgolce’s energy is criterion of spontaneous course of processes only?

10. What the thermodynamic function is used for definition of an opportunity of

spontaneous course of process for the isolated systems?

11. What the thermodynamic function is used for definition of an opportunity of

Spontaneous course of process under condition Р, Т=const?

12. Shall the process are, if G0?

13. Can you name incorporated first and second laws of thermodynamics (T,

P=const.)?

14. Can you to name incorporated first and second laws of thermodynamics (T,

V=const.)?

15. The definition of this notion G0298 (form) “standard Gibbs’s energy” of substance.

16. What is the definition of this notion “Gibbs’s energy” of reaction.

17. Define the notion “entropy” of reaction.



Further Reading:

1. Frolov V.V.chemistry. Ch. VI, §6.1 – 6.11.

2. Luchinsky G.P.rate of chemistry. Ch. IV, §1-7.

3. Ahmetov N.S.general and inorganic chemistry. Section V, ch. 1,2.

4. The general chemistry under ред. Sokolovskoj E.M., etc. Ch. 5, 1-6.









46

CHEMICAL KINETICS.



Obgectives After studying this chapter, students should

be able to know:





1. Homogeneous and heterogeneous chemical processes.

2. The factors influencing on the rate of chemical reactions.

2.1 natures of reacting substances;

2.2 concentrations (pressure). The law of action of weights.

2.3 temperature. Rule of the Vant-Goph. Equation of the Аrrenius.

Energy of activation.

2.4 catalyst.

3. Chemical balance. A constant of balance.

4. Displacement of balance. Principle of the Le-Shatelye.



The chemical thermodynamics studies an opportunity, a direction and limits of

spontaneous course of chemical processes. However, the mechanism and rate of

processes in chemical thermodynamics are not considered. At the same time

representation of a rate of chemical reactions and the factors influencing on it, is

exclusively important for management of chemical processes.

Chemical kinetic is the area of chemistry studying the mechanism and rate of

chemical reactions.



Rate of the chemical reactions are various. One can be instantly: 2H2 + O2 

2H2O, or several seconds, minutes: CuSO4 + 2NaOH  Na2SO4+Cu (OH)2, FeCl3 +

3NaOH  3NaCl+Fe (OH)3.

Rate of the biological reactions can procced for years, decades, for example, the

transformation of wood into coal occurs during centuries, millenia. When you are

studying rate of chemical reactions it is necessary to know the definition of notions:

“homogeneous” and “heterogeneous” reactions.

Homogeneous reactions are such reactions in which initial substances and

products have identical phase’s conditions: solids(s) or liquids (l) or gases (g), and also

there isn’t not an interface among them.

Heterogeneous reactions are such reactions in which initial substances and

products of the reaction have an interface among them and they have various phases’

conditions.

Chemical kinetic has two main notions an instant rate and an average rate.

What is the rate of the chemical reactions?









47

The Instant rate of the chemical reaction is defined as the first derivative of

concentration to time.


By registering with docstoc.com you agree to our
privacy policy and terms of service

Successfully added document to cart!

Successfully added document to cart!