How to write chemical reaction equations. How to balance a chemical equation: rules and algorithm. chemical solution of ionic equations
Story
Title page of Tyrocinium Chymicum.
At first there was no idea about chemical equations, the basic chemical laws were not yet known, but already in ancient times, during the alchemical period of the development of chemistry, they began to designate chemical elements with symbols.
With the further development of chemistry, ideas about the symbolism of chemical elements changed, and knowledge about their compounds expanded. With the discovery of many chemical phenomena, the need arose to move from their verbal description to a more convenient mathematical notation using chemical formulas. The first to propose the use of chemical equations was Jean Beguin in 1615 in the first chemistry textbook, Tyrocinium Chymicum (Principles of Chemistry).
Late XVIII - early XIX centuries - the formation of the laws of stoichiometry. The origins of these studies were the German scientist I.V. Richter. During his student years, he was greatly impressed by the words of his teacher, the philosopher I. Kant, that in certain areas of the natural sciences there is as much true science as there is mathematics in it. Richter devoted his dissertation to the use of mathematics in chemistry. Not being essentially a chemist, Richter introduced the first quantitative equations chemical reactions, began to use the term stoichiometry.
Compilation rules
On the left side of the equation, write down the formulas of the substances that reacted, connecting them with a plus sign. On the right side of the equation, write down the formulas of the resulting substances, also connected by a plus sign. An arrow is placed between the parts of the equation. Then they find odds- numbers placed before formulas of substances so that the number of atoms of identical elements on the left and right sides of the equation is equal.
To draw up equations of chemical reactions, in addition to knowing the formulas of the reagents and reaction products, it is necessary to select the correct coefficients. This can be done using simple rules:
1. Before the formula of a simple substance, you can write a fractional coefficient, which shows the amount of substance of the reacting and resulting substances.
2. If the reaction scheme contains a salt formula, then first the number of ions forming the salt is equalized.
3. If the substances involved in the reaction contain hydrogen and oxygen, then the hydrogen atoms are equalized in the penultimate order, and the oxygen atoms in the last place.
4. If there are several salt formulas in the reaction scheme, then it is necessary to begin the equation with the ions that are part of the salt containing a larger number of them.
Symbols in chemical equations
To indicate various types reactions the following symbols are used:
Arrangement of coefficients in equations
The law of conservation of mass states that the amount of matter of each element before a reaction is equal to the amount of matter of each element after the reaction. Thus, the left and right sides of a chemical equation must have the same number of atoms of a particular element. The chemical equation must be electrically neutral, that is, the sum of the charges on the left and right sides of the equation must add up to zero. One way to equalize the number of atoms in a chemical equation is to select coefficients by trial and error. For more complex cases a system of linear algebraic equations should be used. As a rule, chemical equations are written with the smallest integer coefficients. If there is no coefficient in front of a chemical formula, it is assumed that it is equal to one. Checking the material balance, that is, the number of atoms on the left and right sides, can be as follows: a coefficient of 1 is placed in front of the most complex chemical formula. Next, the coefficients are placed in front of the formulas in such a way that the number of atoms of each element in the left and right sides of the equation is equal . If one of the coefficients is fractional, then all coefficients should be multiplied by the number in the denominator of the fractional coefficient. If there is a coefficient 1 in front of the formula, then it is omitted. Example, arrangement of coefficients in the chemical reaction combustion of methane:
1CH 4 + O 2 CO 2 + H 2 O
The number of carbon atoms on the left and right sides is the same. The next element to equalize is hydrogen. There are 4 hydrogen atoms on the left, 2 on the right, to equalize the number of hydrogen atoms, you should put a factor of 2 in front of water, as a result:
1CH 4 + O 2 CO 2 + 2H 2 O
Checking the correct placement of coefficients in any chemical equation is done by counting the number of oxygen atoms; if the number of oxygen atoms on the left and right sides is the same, then the coefficients are placed correctly.
1CH 4 + 2O 2 CO 2 + 2H 2 O
In front of the CH 4 and CO 2 molecules, the coefficient 1 is omitted.
Redox reactions
Oxidation-reduction reactions (ORR) are back-to-back chemical reactions that occur with a change in the oxidation states of the atoms that make up the reacting substances, realized by the redistribution of electrons between the oxidizing atom and the reducing atom.
During the redox reaction, the reducing agent gives up electrons, that is, it is oxidized; The oxidizing agent gains electrons, that is, it is reduced. Moreover, any redox reaction represents the unity of two opposite transformations - oxidation and reduction, occurring simultaneously and without separating one from the other.
Oxidation is the process of losing electrons, with an increase in the degree of oxidation. When a substance is oxidized, its oxidation state increases as a result of the loss of electrons. The atoms of the substance being oxidized are called electron donors, and the atoms of the oxidizing agent are called electron acceptors. The oxidizing agent, accepting electrons, acquires reducing properties, turning into a conjugate reducing agent.
Reduction is the process of adding electrons to an atom of a substance, while its oxidation state decreases. During reduction, atoms or ions gain electrons. In this case, the oxidation state of the element decreases. The reducing agent, giving up electrons, acquires oxidizing properties, turning into a conjugate oxidizing agent.
When composing the equation for a redox reaction, it is necessary to determine the reducing agent, the oxidizing agent, and the number of electrons given and received. Typically, coefficients are selected using either the electron balance method or the electron-ion balance method (sometimes the latter is called the half-reaction method).
Selection of coefficients using the electronic balance method.
IN simple equations coefficients are selected element by element in accordance with the formula of the final product. In more complex equations of redox reactions, the selection of coefficients is carried out using the electronic balance method:
1. Write down the reaction scheme (formula of reactants and products), and then find the elements that increase and decrease their oxidation states and write them down separately;
2. Compose equations for half-reactions of reduction and oxidation, observing the laws of conservation of the number of atoms and charge in each half-reaction;
3. Additional factors are selected to equalize the half-reactions so that the law of conservation of charge is satisfied for the reaction as a whole, for which the number of elements accepted in the reduction half-reactions is made equal to the number given up elements in the oxidation half-reaction;
4. Enter (based on the found factors) the stoichiometric coefficients in the reaction scheme (coefficient 1 is omitted);
5. Equalize the numbers of atoms of those elements that do not change their oxidation state during the reaction (if there are two such elements, then it is enough to equalize the number of atoms of one of them and check for the second). Obtain the equations of a chemical reaction;
6. They check for an element that has not changed its oxidation state (most often it is oxygen).
Arrangement of coefficients in ionic equations
Ionic equations are chemical equations in which electrolytes are written as dissociated ions. Ionic equations are used to write substitution and exchange reactions in aqueous solutions. Example, exchange reaction, interaction of calcium chloride and silver nitrate with the formation of a silver chloride precipitate:
CaCl 2 (l) + 2AgNO 3 (l) Ca(NO 3) 2 (l) + 2AgCl (s)
complete ionic equation:
Ca 2+ + 2Cl − + 2Ag + + 2NO 3 − Ca 2+ + 2NO 3 − + 2AgCl(s)
abbreviated ionic equation:
2Cl − (l) + 2Ag + (l) 2AgCl(s)
ionic equation:
Ag + + Cl − AgCl(s)
The Ca 2+ and NO 3 − ions remain in solution, therefore they are not participants in the chemical reaction. In neutralization reactions, the ionic reaction equation is as follows:
H + + OH − H 2 O
There are several neutralization reactions that produce another slightly dissociating substance in addition to water. An example is the reaction of barium hydroxide with phosphoric acid, since barium phosphate, which is insoluble in water, is formed.
Literature
- Levitsky M. The language of chemists // Chemistry and life. – 2000. –№1. – P.50-52.
- Kudryavtsev A.A. Writing chemical equations - 4th edition, revised. and additional, 1968 - 359 p.
- Berg L.G. Gromakov S.D. Zoroatskaya I.V. Averko-Antonovich I.N. Methods for selecting coefficients in chemical equations - Kazan: Kazan University Publishing House, 1959. - 148 p.
- Leenson I.A. Even or odd - M.: Chemistry, 1987. - 176 p.
- Chemistry, 8th grade textbook. ARC Publishing. 2003.
- Chemistry, 8th grade textbook. Publishing house Bustard. 2009.
- Chemistry, 8th grade textbook. Publishing house "Mektep" Almaty. 2012.
- Chemistry, 9th grade textbook. Publishing house "Enlightenment" 2008.
see also
Links
- // Encyclopedic Dictionary of Brockhaus and Efron: In 86 volumes (82 volumes and 4 additional ones). - St. Petersburg. , 1890-1907.
9.1. What are the chemical reactions?
Let us remember that we call any chemical phenomena in nature chemical reactions. During a chemical reaction, some break down and others form. chemical bonds. As a result of the reaction, other substances are obtained from some chemical substances (see Chapter 1).
Carrying out homework by § 2.5, you became familiar with the traditional selection from the entire set chemical transformations reactions of four main types, then you also proposed their names: reactions of combination, decomposition, substitution and exchange.
Examples of compound reactions:
C + O 2 = CO 2; (1)
Na 2 O + CO 2 = Na 2 CO 3; (2)
NH 3 + CO 2 + H 2 O = NH 4 HCO 3. (3)
Examples of decomposition reactions:
2Ag 2 O 4Ag + O 2; (4)
CaCO 3 CaO + CO 2; (5)
(NH 4) 2 Cr 2 O 7 N 2 + Cr 2 O 3 + 4H 2 O. (6)
Examples of substitution reactions:
CuSO 4 + Fe = FeSO 4 + Cu; (7)
2NaI + Cl 2 = 2NaCl + I 2; (8)
CaCO 3 + SiO 2 = CaSiO 3 + CO 2. (9)
| Exchange reactions- chemical reactions in which starting substances seem to exchange their constituent parts. |
Examples of exchange reactions:
Ba(OH) 2 + H 2 SO 4 = BaSO 4 + 2H 2 O; (10)
HCl + KNO 2 = KCl + HNO 2; (eleven)
AgNO 3 + NaCl = AgCl + NaNO 3. (12)
The traditional classification of chemical reactions does not cover all their diversity - in addition to the four main types of reactions, there are also many more complex reactions.
The identification of two other types of chemical reactions is based on the participation in them of two important non-chemical particles: electron and proton.
During some reactions, complete or partial transfer of electrons from one atom to another occurs. In this case, the oxidation states of the atoms of the elements that make up the starting substances change; of the examples given, these are reactions 1, 4, 6, 7 and 8. These reactions are called redox.
In another group of reactions, a hydrogen ion (H +), that is, a proton, passes from one reacting particle to another. Such reactions are called acid-base reactions or proton transfer reactions.
Among the examples given, such reactions are reactions 3, 10 and 11. By analogy with these reactions, redox reactions are sometimes called electron transfer reactions. You will become acquainted with OVR in § 2, and with KOR in the following chapters.
COMPOUNDING REACTIONS, DECOMPOSITION REACTIONS, SUBSTITUTION REACTIONS, EXCHANGE REACTIONS, REDOX REACTIONS, ACID-BASE REACTIONS.
Write down reaction equations corresponding to the following schemes:
a) HgO Hg + O 2 ( t); b) Li 2 O + SO 2 Li 2 SO 3; c) Cu(OH) 2 CuO + H 2 O ( t);
d) Al + I 2 AlI 3; e) CuCl 2 + Fe FeCl 2 + Cu; e) Mg + H 3 PO 4 Mg 3 (PO 4) 2 + H 2 ;
g) Al + O 2 Al 2 O 3 ( t); i) KClO 3 + P P 2 O 5 + KCl ( t); j) CuSO 4 + Al Al 2 (SO 4) 3 + Cu;
l) Fe + Cl 2 FeCl 3 ( t); m) NH 3 + O 2 N 2 + H 2 O ( t); m) H 2 SO 4 + CuO CuSO 4 + H 2 O.
Indicate the traditional type of reaction. Label redox and acid-base reactions. In redox reactions, indicate which atoms of elements change their oxidation states.
9.2. Redox reactions
Let's consider the redox reaction that occurs in blast furnaces during the industrial production of iron (more precisely, cast iron) from iron ore:
Fe 2 O 3 + 3CO = 2Fe + 3CO 2.
Let us determine the oxidation states of the atoms that make up both the starting substances and the reaction products
| Fe2O3 | + | = | 2Fe | + |
As you can see, the oxidation state of carbon atoms increased as a result of the reaction, the oxidation state of iron atoms decreased, and the oxidation state of oxygen atoms remained unchanged. Consequently, the carbon atoms in this reaction underwent oxidation, that is, they lost electrons ( oxidized), and the iron atoms – reduction, that is, they added electrons ( recovered) (see § 7.16). To characterize OVR, the concepts are used oxidizer And reducing agent.
Thus, in our reaction the oxidizing atoms are iron atoms, and the reducing atoms are carbon atoms.
In our reaction, the oxidizing agent is iron(III) oxide, and the reducing agent is carbon(II) monoxide.
In cases where oxidizing atoms and reducing atoms are part of the same substance (example: reaction 6 from the previous paragraph), the concepts of “oxidizing substance” and “reducing substance” are not used.
Thus, typical oxidizing agents are substances that contain atoms that tend to gain electrons (in whole or in part), lowering their oxidation state. Of the simple substances, these are primarily halogens and oxygen, and to a lesser extent sulfur and nitrogen. From complex substances - substances that contain atoms in higher oxidation states that are not inclined to form simple ions in these oxidation states: HNO 3 (N +V), KMnO 4 (Mn +VII), CrO 3 (Cr +VI), KClO 3 (Cl +V), KClO 4 (Cl +VII), etc.
Typical reducing agents are substances that contain atoms that tend to completely or partially donate electrons, increasing their oxidation state. From simple substances these are hydrogen, alkali and alkaline earth metals, as well as aluminum. Of the complex substances - H 2 S and sulfides (S –II), SO 2 and sulfites (S +IV), iodides (I –I), CO (C +II), NH 3 (N –III), etc.
In general, almost all complex and many simple substances can exhibit both oxidizing and reducing properties. For example:
SO 2 + Cl 2 = S + Cl 2 O 2 (SO 2 is a strong reducing agent);
SO 2 + C = S + CO 2 (t) (SO 2 is a weak oxidizing agent);
C + O 2 = CO 2 (t) (C is a reducing agent);
C + 2Ca = Ca 2 C (t) (C is an oxidizing agent).
Let's return to the reaction we discussed at the beginning of this section.
| Fe2O3 | + | = | 2Fe | + |
Please note that as a result of the reaction, oxidizing atoms (Fe + III) turned into reducing atoms (Fe 0), and reducing atoms (C + II) turned into oxidizing atoms (C + IV). But CO 2 is a very weak oxidizing agent under any conditions, and iron, although it is a reducing agent, is under these conditions much weaker than CO. Therefore, the reaction products do not react with each other, and the reverse reaction does not occur. The given example is an illustration of the general principle that determines the direction of the flow of OVR:
Redox reactions proceed in the direction of the formation of a weaker oxidizing agent and a weaker reducing agent.
The redox properties of substances can only be compared under identical conditions. In some cases, this comparison can be made quantitatively.
While doing your homework for the first paragraph of this chapter, you became convinced that it is quite difficult to select coefficients in some reaction equations (especially ORR). To simplify this task in the case of redox reactions, the following two methods are used:
A) electronic balance method And
b) electron-ion balance method.
You will learn the electron balance method now, and the electron-ion balance method is usually studied in higher education institutions.
Both of these methods are based on the fact that electrons in chemical reactions neither disappear nor appear anywhere, that is, the number of electrons accepted by atoms is equal to the number of electrons given up by other atoms.
The number of given and accepted electrons in the electron balance method is determined by the change in the oxidation state of atoms. When using this method, it is necessary to know the composition of both the starting substances and the reaction products.
Let's look at the application of the electronic balance method using examples.
Example 1. Let's create an equation for the reaction of iron with chlorine. It is known that the product of this reaction is iron(III) chloride. Let's write down the reaction scheme:
Fe + Cl 2 FeCl 3 .
Let us determine the oxidation states of the atoms of all elements that make up the substances participating in the reaction:
Iron atoms give up electrons, and chlorine molecules accept them. Let us express these processes electronic equations:
Fe – 3 e– = Fe +III,
Cl2+2 e –= 2Cl –I.
In order for the number of electrons given to be equal to the number of electrons received, the first electronic equation must be multiplied by two, and the second by three:
| Fe – 3 e– = Fe +III, Cl2+2 e– = 2Cl –I |
2Fe – 6 e– = 2Fe +III, 3Cl 2 + 6 e– = 6Cl –I. |
By introducing coefficients 2 and 3 into the reaction scheme, we obtain the reaction equation:
2Fe + 3Cl 2 = 2FeCl 3.
Example 2. Let's create an equation for the combustion reaction of white phosphorus in excess chlorine. It is known that phosphorus(V) chloride is formed under these conditions:
| +V –I | ||||
| P 4 | + | Cl2 | PCl 5. |
White phosphorus molecules give up electrons (oxidize), and chlorine molecules accept them (reduce):
| P 4 – 20 e– = 4P +V Cl2+2 e– = 2Cl –I |
1 10 |
2 20 |
P 4 – 20 e– = 4P +V Cl2+2 e– = 2Cl –I |
P 4 – 20 e– = 4P +V 10Cl 2 + 20 e– = 20Cl –I |
The initially obtained multipliers (2 and 20) had common divisor, into which (as future coefficients in the reaction equation) they were divided. Reaction equation:
P4 + 10Cl2 = 4PCl5.
Example 3. Let's create an equation for the reaction that occurs when iron(II) sulfide is roasted in oxygen.
Reaction scheme:
| +III –II | +IV –II | |||||
| + | O2 | + |
In this case, both iron(II) and sulfur(–II) atoms are oxidized. The composition of iron(II) sulfide contains atoms of these elements in a 1:1 ratio (see the indices in the simplest formula).
Electronic balance:
| 4 | Fe+II – e– = Fe +III S–II–6 e– = S +IV |
In total they give 7 e – |
| 7 | O 2 + 4e – = 2O –II |
Reaction equation: 4FeS + 7O 2 = 2Fe 2 O 3 + 4SO 2.
Example 4. Let's create an equation for the reaction that occurs when iron(II) disulfide (pyrite) is roasted in oxygen.
Reaction scheme:
| +III –II | +IV –II | |||||
| + | O2 | + |
As in the previous example, both iron(II) atoms and sulfur atoms are also oxidized here, but with an oxidation state of I. The atoms of these elements are included in the composition of pyrite in a ratio of 1:2 (see the indices in the simplest formula). It is in this regard that the iron and sulfur atoms react, which is taken into account when compiling the electronic balance:
| Fe+III – e– = Fe +III 2S–I – 10 e– = 2S +IV |
In total they give 11 e – | |
| O2+4 e– = 2O –II |
Reaction equation: 4FeS 2 + 11O 2 = 2Fe 2 O 3 + 8SO 2.
There are also more complex cases of ODD, some of which you will become familiar with while doing your homework.
OXIDIZING ATOM, REDUCING ATOM, OXIDIZING SUBSTANCE, REDUCING SUBSTANCE, ELECTRONIC BALANCE METHOD, ELECTRONIC EQUATIONS.
1. Compile an electronic balance for each OVR equation given in the text of § 1 of this chapter.
2. Make up equations for the ORRs that you discovered while completing the task for § 1 of this chapter. This time, use the electronic balance method to set the odds. 3.Using the electron balance method, create reaction equations corresponding to the following schemes: a) Na + I 2 NaI;
b) Na + O 2 Na 2 O 2;
c) Na 2 O 2 + Na Na 2 O;
d) Al + Br 2 AlBr 3;
e) Fe + O 2 Fe 3 O 4 ( t);
e) Fe 3 O 4 + H 2 FeO + H 2 O ( t);
g) FeO + O 2 Fe 2 O 3 ( t);
i) Fe 2 O 3 + CO Fe + CO 2 ( t);
j) Cr + O 2 Cr 2 O 3 ( t);
l) CrO 3 + NH 3 Cr 2 O 3 + H 2 O + N 2 ( t);
l) Mn 2 O 7 + NH 3 MnO 2 + N 2 + H 2 O;
m) MnO 2 + H 2 Mn + H 2 O ( t);
n) MnS + O 2 MnO 2 + SO 2 ( t)
p) PbO 2 + CO Pb + CO 2 ( t);
c) Cu 2 O + Cu 2 S Cu + SO 2 ( t);
t) CuS + O 2 Cu 2 O + SO 2 ( t);
y) Pb 3 O 4 + H 2 Pb + H 2 O ( t).
9.3. Exothermic reactions. Enthalpy
Why do chemical reactions occur?
To answer this question, let us remember why individual atoms combine into molecules, why an ionic crystal is formed from isolated ions, and why the principle of least energy applies when the electron shell of an atom is formed. The answer to all these questions is the same: because it is energetically beneficial. This means that during such processes energy is released. It would seem that chemical reactions should occur for the same reason. Indeed, many reactions can be carried out, during which energy is released. Energy is released, usually in the form of heat.
If during an exothermic reaction the heat does not have time to be removed, then the reaction system heats up.
For example, in the methane combustion reaction
CH 4 (g) + 2O 2 (g) = CO 2 (g) + 2H 2 O (g)
so much heat is released that methane is used as fuel.
The fact that this reaction releases heat can be reflected in the reaction equation:
CH 4 (g) + 2O 2 (g) = CO 2 (g) + 2H 2 O (g) + Q.
This is the so called thermochemical equation. Here the symbol "+ Q" means that when methane is burned, heat is released. This heat is called thermal effect of reaction.
Where does the released heat come from?
You know that during chemical reactions chemical bonds are broken and formed. IN in this case bonds between carbon and hydrogen atoms in CH 4 molecules are broken, as well as between oxygen atoms in O 2 molecules. In this case, new bonds are formed: between carbon and oxygen atoms in CO 2 molecules and between oxygen and hydrogen atoms in H 2 O molecules. To break bonds, you need to expend energy (see “bond energy”, “atomization energy”), and when forming bonds, energy is released. Obviously, if the “new” bonds are stronger than the “old” ones, then more energy will be released than absorbed. The difference between the released and absorbed energy is the thermal effect of the reaction.
Thermal effect (amount of heat) is measured in kilojoules, for example:
2H 2 (g) + O 2 (g) = 2H 2 O (g) + 484 kJ.
This notation means that 484 kilojoules of heat will be released if two moles of hydrogen react with one mole of oxygen to produce two moles of gaseous water (water vapor).
Thus, in thermochemical equations, the coefficients are numerically equal to the amounts of substance of the reactants and reaction products.
What determines the thermal effect of each specific reaction?
The thermal effect of the reaction depends
a) from states of aggregation starting materials and reaction products,
b) on temperature and
c) on whether the chemical transformation occurs at constant volume or at constant pressure.
The dependence of the thermal effect of a reaction on the state of aggregation of substances is due to the fact that the processes of transition from one state of aggregation to another (like some other physical processes) are accompanied by the release or absorption of heat. This can also be expressed by a thermochemical equation. Example - thermo chemical equation water vapor condensation:
H 2 O (g) = H 2 O (l) + Q.
In thermochemical equations, and, if necessary, in ordinary chemical equations, the aggregative states of substances are indicated using letter indices:
(d) – gas,
(g) – liquid,
(t) or (cr) – solid or crystalline substance.
The dependence of the thermal effect on temperature is associated with differences in heat capacities
starting materials and reaction products.
Since the volume of the system always increases as a result of an exothermic reaction at constant pressure, part of the energy is spent on doing work to increase the volume, and the heat released will be less than if the same reaction occurs at a constant volume.
Thermal effects of reactions are usually calculated for reactions occurring at constant volume at 25 °C and are indicated by the symbol Q o.
If energy is released only in the form of heat, and a chemical reaction proceeds at a constant volume, then the thermal effect of the reaction ( Q V) is equal to the change internal energy(D U) substances participating in the reaction, but with the opposite sign:
Q V = – U.
The internal energy of a body is understood as the total energy of intermolecular interactions, chemical bonds, the ionization energy of all electrons, the bond energy of nucleons in nuclei, and all other known and unknown types of energy “stored” by this body. The “–” sign is due to the fact that when heat is released, the internal energy decreases. That is
U= – Q V .
If the reaction occurs at constant pressure, then the volume of the system can change. Part of the internal energy is also spent on doing the work to increase the volume. In this case
U = –(QP+A) = –(QP+PV),
Where Qp– the thermal effect of a reaction occurring at constant pressure. From here
Q P = – U–PV .
A value equal to U+PV got the name enthalpy change and denoted by D H.
H=U+PV.
Hence
Q P = – H.
Thus, as heat is released, the enthalpy of the system decreases. Hence the old name for this quantity: “heat content”.
Unlike the thermal effect, a change in enthalpy characterizes a reaction regardless of whether it occurs at constant volume or constant pressure. Thermochemical equations written using enthalpy change are called thermochemical equations in thermodynamic form. In this case, the value of the enthalpy change under standard conditions (25 °C, 101.3 kPa) is given, denoted H o. For example:
2H 2 (g) + O 2 (g) = 2H 2 O (g) H o= – 484 kJ;
CaO (cr) + H 2 O (l) = Ca(OH) 2 (cr) H o= – 65 kJ.
Dependence of the amount of heat released in the reaction ( Q) from the thermal effect of the reaction ( Q o) and the amount of substance ( n B) one of the participants in the reaction (substance B - the starting substance or reaction product) is expressed by the equation:
Here B is the amount of substance B, specified by the coefficient in front of the formula of substance B in the thermochemical equation.
Task
Determine the amount of hydrogen substance burned in oxygen if 1694 kJ of heat was released.
Solution
2H 2 (g) + O 2 (g) = 2H 2 O (g) + 484 kJ. |
|
|
Q = 1694 kJ, 6. The thermal effect of the reaction between crystalline aluminum and gaseous chlorine is 1408 kJ. Write the thermochemical equation for this reaction and determine the mass of aluminum required to produce 2816 kJ of heat using this reaction. 9.4. Endothermic reactions. Entropy In addition to exothermic reactions, reactions are possible in which heat is absorbed, and if it is not supplied, the reaction system is cooled. Such reactions are called endothermic. The thermal effect of such reactions is negative. For example: Thus, the energy released during the formation of bonds in the products of these and similar reactions is less than the energy required to break bonds in the starting substances.
Let's take two flasks and fill one of them with nitrogen (colorless gas) and the other with nitrogen dioxide (brown gas) so that both the pressure and temperature in the flasks are the same. It is known that these substances do not react chemically with each other. Let's tightly connect the flasks with their necks and install them vertically, so that the flask with heavier nitrogen dioxide is at the bottom (Fig. 9.1). After some time, we will see that brown nitrogen dioxide gradually spreads into the upper flask, and colorless nitrogen penetrates into the lower one. As a result, the gases mix, and the color of the contents of the flasks becomes the same. Thus,
Equations of connection between entropy ( S) and other quantities are studied in physics and physical chemistry courses. Entropy unit [ S] = 1 J/K. G= H–T S Condition for spontaneous reaction: G< 0. At low temperatures, the factor determining the possibility of a reaction occurring is largely the energy factor, and at high temperatures it is the entropy factor. From the above equation, in particular, it is clear why the room temperature decomposition reactions (entropy increases) begin to occur at elevated temperatures. ENDOTHERMIC REACTION, ENTROPY, ENERGY FACTOR, ENTROPY FACTOR, GIBBS ENERGY. 2CuO (cr) + C (graphite) = 2Cu (cr) + CO 2 (g) is –46 kJ. Write down the thermochemical equation and calculate how much energy is needed to produce 1 kg of copper from this reaction. CaCO 3 (cr) = CaO (cr) + CO 2 (g) – 179 kJ 24.6 liters of carbon dioxide were formed. Determine how much heat was wasted uselessly. How many grams of calcium oxide were formed? |
Let's talk about how to create a chemical equation, because they are the main elements of this discipline. Thanks to a deep understanding of all the patterns of interactions and substances, you can control them and apply them in various fields of activity.
Theoretical features
Drawing up chemical equations is an important and responsible stage, considered in the eighth grade. secondary schools. What should precede this stage? Before the teacher tells his students how to create a chemical equation, it is important to introduce schoolchildren to the term “valence” and teach them to determine this value for metals and non-metals using the periodic table of elements.
Compilation of binary formulas by valence
In order to understand how to create a chemical equation by valence, you first need to learn how to create formulas for compounds consisting of two elements using valency. We propose an algorithm that will help cope with the task. For example, you need to create a formula for sodium oxide.
First, it is important to take into account that the chemical element that is mentioned last in the name should be in first place in the formula. In our case, sodium will be written first in the formula, oxygen second. Let us recall that oxides are binary compounds in which the last (second) element must be oxygen with an oxidation state of -2 (valency 2). Next, using the periodic table, it is necessary to determine the valence of each of the two elements. To do this we use certain rules.
Since sodium is a metal that is located in the main subgroup of group 1, its valence is a constant value, it is equal to I.
Oxygen is a non-metal, since it is the last one in the oxide; to determine its valency, we subtract 6 from eight (the number of groups) (the group in which oxygen is located), we obtain that the valency of oxygen is II.
Between certain valences we find the least common multiple, then divide it by the valency of each of the elements to obtain their indices. We write down the finished formula Na 2 O.
Instructions for composing an equation
Now let's talk in more detail about how to write a chemical equation. First, let's look at the theoretical aspects, then move on to specific examples. So, composing chemical equations presupposes a certain procedure.
- 1st stage. After reading the proposed task, you need to determine which chemical substances must be present on the left side of the equation. A “+” sign is placed between the original components.
- 2nd stage. After the equal sign, you need to create a formula for the reaction product. When performing such actions, you will need the algorithm for composing formulas for binary compounds, which we discussed above.
- 3rd stage. Checking the number of atoms of each element before and after chemical interaction, if necessary, we put additional coefficients in front of the formulas.
Example of a combustion reaction
Let's try to figure out how to create a chemical equation for the combustion of magnesium using an algorithm. On the left side of the equation we write the sum of magnesium and oxygen. Do not forget that oxygen is a diatomic molecule, so it must be given an index of 2. After the equal sign, we compose the formula for the product obtained after the reaction. It will be in which magnesium is written first, and oxygen is written second in the formula. Further according to the table chemical elements determine valency. Magnesium, which is in group 2 (the main subgroup), has a constant valency II; for oxygen, by subtracting 8 - 6 we also get valence II.
The process record will look like: Mg+O 2 =MgO.
In order for the equation to comply with the law of conservation of mass of substances, it is necessary to arrange the coefficients. First, we check the amount of oxygen before the reaction, after the process is completed. Since there were 2 oxygen atoms, but only one was formed, a coefficient of 2 must be added on the right side before the magnesium oxide formula. Next, we count the number of magnesium atoms before and after the process. As a result of the interaction, 2 magnesium was obtained, therefore, on the left side in front of the simple substance magnesium, a coefficient of 2 is also required.
The final type of reaction: 2Mg+O 2 =2MgO.
Example of a substitution reaction
Any chemistry summary contains a description of different types of interactions.
Unlike a compound, in a substitution there will be two substances on both the left and right sides of the equation. Let's say we need to write the reaction of interaction between zinc and We use the standard writing algorithm. First, on the left side we write zinc and hydrochloric acid through the sum, and on the right side we write the formulas for the resulting reaction products. Since zinc is located before hydrogen in the electrochemical voltage series of metals, in this process it displaces molecular hydrogen from the acid and forms zinc chloride. As a result, we get the following entry: Zn+HCL=ZnCl 2 +H 2.
Now we move on to equalizing the number of atoms of each element. Since there was one atom on the left side of chlorine, and after the interaction there were two, it is necessary to put a factor of 2 in front of the formula of hydrochloric acid.
As a result, we obtain a ready-made reaction equation corresponding to the law of conservation of mass of substances: Zn+2HCL=ZnCl 2 +H 2 .
Conclusion
A typical chemistry note necessarily contains several chemical transformations. Not a single section of this science is limited to simple verbal description transformations, dissolution processes, evaporation, everything is necessarily confirmed by equations. The specificity of chemistry lies in the fact that all processes that occur between different inorganic or organic substances can be described using coefficients and indices.
How else does chemistry differ from other sciences? Chemical equations help not only to describe the transformations that occur, but also to carry out quantitative calculations on them, thanks to which it is possible to carry out laboratory and industrial production different substances.
Part I
1. Lomonosov-Lavoisier law – the law of conservation of mass of substances:
2. Chemical reaction equations are conventional notation of a chemical reaction using chemical formulas and mathematical symbols.
3. The chemical equation must correspond to the law preservation of the mass of substances, which is achieved by arranging the coefficients in the reaction equation.
4. What does a chemical equation show?
1) What substances react.
2) What substances are formed as a result.
3) Quantitative relationships substances in the reaction, i.e. the amount of reacting and resulting substances in the reaction.
4) Type of chemical reaction.
5. Rules for arranging coefficients in a chemical reaction scheme using the example of the interaction of barium hydroxide and phosphoric acid with the formation of barium phosphate and water.
a) Write down the reaction scheme, i.e. the formulas of the reacting and resulting substances:
b) start balancing the reaction scheme with the formula of the salt (if available). Remember that several complex ions in a base or salt are indicated by brackets, and their number is indicated by indices outside the brackets:
c) equalize hydrogen next to last:
d) equalize oxygen last - this is an indicator of the correct placement of coefficients.
Before the formula of a simple substance, it is possible to write a fractional coefficient, after which the equation must be rewritten with doubled coefficients.
Part II
1. Make up reaction equations, the schemes of which are:
2. Write the equations of chemical reactions:
3. Establish a correspondence between the diagram and the sum of the coefficients in the chemical reaction.
4. Establish a correspondence between the starting materials and the reaction products.
5. What does the equation of the following chemical reaction show:
1) Copper hydroxide and hydrochloric acid reacted;
2) Salt and water were formed as a result of the reaction;
3) Coefficients before starting substances 1 and 2.
6. Using the following diagram, create an equation for a chemical reaction using doubling the fractional coefficient:
7. Chemical reaction equation:
4P+5O2=2P2O5
shows the amount of substance of the starting substances and products, their mass or volume:
1) phosphorus – 4 mol or 124 g;
2) phosphorus oxide (V) – 2 mol, 284 g;
3) oxygen – 5 mol or 160 l.
Chemistry is the science of substances, their properties and transformations
.
That is, if nothing happens to the substances around us, then this does not apply to chemistry. But what does “nothing happens” mean? If a thunderstorm suddenly caught us in the field, and we were all wet, as they say, “to the skin,” then isn’t this a transformation: after all, the clothes were dry, but they became wet.
If, for example, you take an iron nail, file it, and then assemble iron filings (Fe) , then isn’t this also a transformation: there was a nail - it became powder. But if you then assemble the device and carry out obtaining oxygen (O 2): heat up potassium permanganate(KMpO 4) and collect oxygen in a test tube, and then place these red-hot iron filings into it, then they will flare up with a bright flame and after combustion will turn into a brown powder. And this is also a transformation. So where is the chemistry? Despite the fact that in these examples the shape (iron nail) and the condition of the clothing (dry, wet) change, these are not transformations. The fact is that the nail itself was a substance (iron), and remained so, despite its different shape, and our clothes absorbed the water from the rain and then evaporated it into the atmosphere. The water itself has not changed. So what are transformations from a chemical point of view?
From a chemical point of view, transformations are those phenomena that are accompanied by a change in the composition of a substance. Let's take the same nail as an example. It doesn’t matter what shape it took after filing, but after collecting from it iron filings placed in an oxygen atmosphere - it turned into iron oxide(Fe 2 O 3 ) . So, something has changed after all? Yes, it has changed. There was a substance called a nail, but under the influence of oxygen a new substance was formed - element oxide gland. Molecular equation This transformation can be represented by the following chemical symbols:
4Fe + 3O 2 = 2Fe 2 O 3 (1)
For someone uninitiated in chemistry, questions immediately arise. What is "molecular equation", what is Fe? Why are the numbers “4”, “3”, “2”? What are the little numbers “2” and “3” in the formula Fe 2 O 3? This means it’s time to sort everything out in order.
Signs of chemical elements.
Despite the fact that chemistry begins to be studied in the 8th grade, and some even earlier, many people know the great Russian chemist D.I. Mendeleev. And of course, his famous “Periodic Table of Chemical Elements”. Otherwise, more simply, it is called the “Periodical Table”.
In this table, the elements are arranged in the appropriate order. To date, about 120 of them are known. The names of many elements have been known to us for a long time. These are: iron, aluminum, oxygen, carbon, gold, silicon. Previously, we used these words without thinking, identifying them with objects: an iron bolt, an aluminum wire, oxygen in the atmosphere, a gold ring, etc. etc. But in fact, all these substances (bolt, wire, ring) consist of their corresponding elements. The whole paradox is that the element cannot be touched or picked up. How so? They are in the periodic table, but you can’t take them! Yes exactly. A chemical element is an abstract (that is, abstract) concept, and is used in chemistry, as well as in other sciences, for calculations, drawing up equations, and solving problems. Each element differs from the other in that it has its own characteristic electronic configuration of an atom. The number of protons in the nucleus of an atom is equal to the number of electrons in its orbitals. For example, hydrogen is element No. 1. Its atom consists of 1 proton and 1 electron. Helium is element #2. Its atom consists of 2 protons and 2 electrons. Lithium is element #3. Its atom consists of 3 protons and 3 electrons. Darmstadtium – element No. 110. Its atom consists of 110 protons and 110 electrons.
Each element is designated by a certain symbol, Latin letters, and has a certain reading translated from Latin. For example, hydrogen has the symbol "N", read as "hydrogenium" or "ash". Silicon has the symbol "Si" read as "silicium". Mercury has a symbol "Hg" and is read as "hydrargyrum". And so on. All these notations can be found in any 8th grade chemistry textbook. The main thing for us now is to understand that when composing chemical equations, it is necessary to operate with the indicated symbols of the elements.
Simple and complex substances.
Denoting various substances with single symbols of chemical elements (Hg mercury, Fe iron, Cu copper, Zn zinc, Al aluminum) we essentially denote simple substances, that is, substances consisting of atoms of the same type (containing the same number of protons and neutrons in an atom). For example, if the substances iron and sulfur interact, then the equation will take the following writing form:
Fe + S = FeS (2)
Simple substances include metals (Ba, K, Na, Mg, Ag), as well as non-metals (S, P, Si, Cl 2, N 2, O 2, H 2). Moreover, one should pay attention
special attention to the fact that all metals are designated by single symbols: K, Ba, Ca, Al, V, Mg, etc., and non-metals are either simple symbols: C, S, P or may have different indices that indicate their molecular structure: H 2, Cl 2, O 2, J 2, P 4, S 8. In the future this will have a very great importance when writing equations. It is not at all difficult to guess that complex substances are substances formed from atoms different types, For example,
1). Oxides:
aluminium oxide Al 2 O 3, 
sodium oxide Na2O,
copper oxide CuO,
zinc oxide ZnO,
titanium oxide Ti2O3,
carbon monoxide or carbon monoxide (+2) CO,
sulfur oxide (+6) SO 3
2). Reasons:
iron hydroxide(+3) Fe(OH) 3,
copper hydroxide Cu(OH)2,
potassium hydroxide or alkali potassium KOH,
sodium hydroxide NaOH.
3). Acids:
hydrochloric acid HCl,
sulfurous acid H2SO3,
Nitric acid HNO3
4). Salts:
sodium thiosulfate Na 2 S 2 O 3 ,
sodium sulfate or Glauber's salt Na2SO4,
calcium carbonate or limestone CaCO 3,
copper chloride CuCl2
5). Organic matter:
sodium acetate CH 3 COONa,
methane CH 4,
acetylene C 2 H 2,
glucose C 6 H 12 O 6
Finally, after we have figured out the structure of various substances, we can begin to write chemical equations.
Chemical equation.
The word “equation” itself is derived from the word “equalize”, i.e. divide something into equal parts. In mathematics, equations constitute almost the very essence of this science. For example, you can give a simple equation in which the left and right sides will be equal to “2”:
40: (9 + 11) = (50 x 2) : (80 – 30);
And in chemical equations the same principle: the left and right sides of the equation must correspond to the same numbers of atoms and elements participating in them. Or, if an ionic equation is given, then in it number of particles must also meet this requirement. A chemical equation is a conventional representation of a chemical reaction using chemical formulas and mathematical symbols. A chemical equation inherently reflects one or another chemical reaction, that is, the process of interaction of substances, during which new substances arise. For example, it is necessary write a molecular equation reactions in which they take part barium chloride BaCl 2 and sulfuric acid H 2 SO 4. As a result of this reaction, an insoluble precipitate is formed - barium sulfate BaSO 4 and hydrochloric acid HCl:
BaCl 2 + H 2 SO 4 = BaSO 4 + 2HCl (3)
First of all, it is necessary to understand that big number The “2” in front of the substance HCl is called a coefficient, and the small numbers “2”, “4” under the formulas BaCl 2, H 2 SO 4, BaSO 4 are called indices. Both coefficients and indices in chemical equations act as multipliers, not summands. To write a chemical equation correctly, you need assign coefficients in the reaction equation. Now let's start counting the atoms of the elements on the left and right sides of the equation. On the left side of the equation: the substance BaCl 2 contains 1 barium atom (Ba), 2 chlorine atoms (Cl). In the substance H 2 SO 4: 2 hydrogen atoms (H), 1 sulfur atom (S) and 4 oxygen atoms (O). On the right side of the equation: in the BaSO 4 substance there is 1 barium atom (Ba), 1 sulfur atom (S) and 4 oxygen atoms (O), in the HCl substance: 1 hydrogen atom (H) and 1 chlorine atom (Cl). It follows that on the right side of the equation the number of hydrogen and chlorine atoms is half as much as on the left side. Therefore, before the HCl formula on the right side of the equation, it is necessary to put the coefficient “2”. If we now add up the numbers of atoms of the elements participating in this reaction, both on the left and on the right, we obtain the following balance:
In both sides of the equation, the numbers of atoms of the elements participating in the reaction are equal, therefore it is composed correctly.
Chemical equation and chemical reactions
As we have already found out, chemical equations are a reflection of chemical reactions. Chemical reactions are those phenomena during which the transformation of one substance into another occurs. Among their diversity, two main types can be distinguished:
1). Compound reactions
2). Decomposition reactions.
The overwhelming majority of chemical reactions belong to addition reactions, since changes in its composition can rarely occur with an individual substance if it is not exposed to external influences (dissolution, heating, exposure to light). Nothing characterizes a chemical phenomenon or reaction better than the changes that occur during the interaction of two or more substances. Such phenomena can occur spontaneously and be accompanied by an increase or decrease in temperature, light effects, color changes, sediment formation, release of gaseous products, and noise.
For clarity, we present several equations reflecting the processes of compound reactions, during which we obtain sodium chloride(NaCl), zinc chloride(ZnCl2), silver chloride precipitate(AgCl), aluminum chloride(AlCl 3)
Cl 2 + 2Nа = 2NaCl (4)
CuCl 2 + Zn = ZnCl 2 + Cu (5)
AgNO 3 + KCl = AgCl + 2KNO 3 (6)
3HCl + Al(OH) 3 = AlCl 3 + 3H 2 O (7)
Among the reactions of the compound, special mention should be made of the following: : substitution (5), exchange (6), and how special case exchange reactions - reaction neutralization (7).
Substitution reactions include those in which atoms of a simple substance replace atoms of one of the elements in a complex substance. In example (5), zinc atoms replace copper atoms from the CuCl 2 solution, while zinc passes into the soluble salt ZnCl 2, and copper is released from the solution in the metallic state.
Exchange reactions include those reactions in which two complex substances exchange their components. In the case of reaction (6), the soluble salts AgNO 3 and KCl, when both solutions are merged, form an insoluble precipitate of the AgCl salt. At the same time, they exchange their constituent parts - cations and anions. Potassium cations K + are added to the NO 3 anions, and silver cations Ag + are added to the Cl - anions.
A special, special case of exchange reactions is the neutralization reaction. Neutralization reactions include those reactions in which acids react with bases, resulting in the formation of salt and water. In example (7), hydrochloric acid HCl reacts with the base Al(OH) 3 to form the salt AlCl 3 and water. In this case, aluminum cations Al 3+ from the base are exchanged with Cl - anions from the acid. What happens in the end neutralization of hydrochloric acid.
Decomposition reactions include those in which two or more new simple or complex substances, but of a simpler composition, are formed from one complex substance. Examples of reactions include those in the process of which 1) decomposes. Potassium nitrate(KNO 3) with the formation of potassium nitrite (KNO 2) and oxygen (O 2); 2). Potassium permanganate(KMnO 4): potassium manganate (K 2 MnO 4) is formed, manganese oxide(MnO 2) and oxygen (O 2); 3). Calcium carbonate or marble; in the process are formed carbonicgas(CO2) and calcium oxide(CaO)
2KNO 3 = 2KNO 2 + O 2 (8)
2KMnO 4 = K 2 MnO 4 + MnO 2 + O 2 (9)
CaCO 3 = CaO + CO 2 (10)
In reaction (8), one complex and one simple substance are formed from a complex substance. In reaction (9) there are two complex and one simple. In reaction (10) there are two complex substances, but simpler in composition
All classes of complex substances are subject to decomposition:
1). Oxides: silver oxide 2Ag 2 O = 4Ag + O 2 (11)
2). Hydroxides: iron hydroxide 2Fe(OH) 3 = Fe 2 O 3 + 3H 2 O (12)
3). Acids: sulfuric acid H 2 SO 4 = SO 3 + H 2 O (13)
4). Salts: calcium carbonate CaCO 3 = CaO + CO 2 (14)
5). Organic matter: alcoholic fermentation of glucose
C 6 H 12 O 6 = 2C 2 H 5 OH + 2CO 2 (15)
According to another classification, all chemical reactions can be divided into two types: reactions that release heat are called exothermic, and reactions that occur with the absorption of heat - endothermic. The criterion for such processes is thermal effect of the reaction. As a rule, exothermic reactions include oxidation reactions, i.e. interaction with oxygen, for example methane combustion:
CH 4 + 2O 2 = CO 2 + 2H 2 O + Q (16)
and to endothermic reactions - decomposition reactions already given above (11) - (15). The Q sign at the end of the equation indicates whether heat is released (+Q) or absorbed (-Q) during the reaction:
CaCO 3 = CaO+CO 2 - Q (17)
You can also consider all chemical reactions according to the type of change in the degree of oxidation of the elements involved in their transformations. For example, in reaction (17), the elements participating in it do not change their oxidation states:
Ca +2 C +4 O 3 -2 = Ca +2 O -2 +C +4 O 2 -2 (18)
And in reaction (16), the elements change their oxidation states:
2Mg 0 + O 2 0 = 2Mg +2 O -2
Reactions of this type are redox . They will be considered separately. To compose equations for reactions of this type, you must use half-reaction method and apply electronic balance equation.
After presenting the various types of chemical reactions, you can proceed to the principle of composing chemical equations, or, in other words, selecting coefficients on the left and right sides.
Mechanisms for composing chemical equations.
Whatever type a chemical reaction belongs to, its recording (chemical equation) must correspond to the condition that the number of atoms before and after the reaction is equal.
There are equations (17) that do not require equalization, i.e. placement of coefficients. But in most cases, as in examples (3), (7), (15), it is necessary to take actions aimed at equalizing the left and right sides of the equation. What principles should be followed in such cases? Is there any system for selecting odds? There is, and not only one. Such systems include:
1). Selection of coefficients according to given formulas.
2). Compilation by valences of reacting substances.
3). Arrangement of reacting substances according to oxidation states.
In the first case, it is assumed that we know the formulas of the reacting substances both before and after the reaction. For example, given the following equation:
N 2 + O 2 →N 2 O 3 (19)
It is generally accepted that until equality is established between the atoms of the elements before and after the reaction, the equal sign (=) is not placed in the equation, but is replaced by an arrow (→). Now let's get down to the actual adjustment. On the left side of the equation there are 2 nitrogen atoms (N 2) and two oxygen atoms (O 2), and on the right side there are two nitrogen atoms (N 2) and three oxygen atoms (O 3). There is no need to equalize it in terms of the number of nitrogen atoms, but in terms of oxygen it is necessary to achieve equality, since before the reaction there were two atoms involved, and after the reaction there were three atoms. Let's make the following diagram:
before reaction after reaction
O 2 O 3
Let's determine the smallest multiple between the given numbers of atoms, it will be “6”.
O 2 O 3
\ 6 /
Let's divide this number on the left side of the oxygen equation by “2”. We get the number “3” and put it into the equation to be solved:
N 2 + 3O 2 →N 2 O 3
We also divide the number “6” for the right side of the equation by “3”. We get the number “2”, and also put it in the equation to be solved:
N 2 + 3O 2 → 2N 2 O 3
The numbers of oxygen atoms on both the left and right sides of the equation became equal, respectively, 6 atoms each:
But the number of nitrogen atoms on both sides of the equation will not correspond to each other:
The left one has two atoms, the right one has four atoms. Therefore, in order to achieve equality, it is necessary to double the amount of nitrogen on the left side of the equation, setting the coefficient to “2”:
Thus, equality in nitrogen is observed and, in general, the equation takes the form:
2N 2 + 3О 2 → 2N 2 О 3
Now in the equation you can put an equal sign instead of an arrow:
2N 2 + 3О 2 = 2N 2 О 3 (20)
Let's give another example. The following reaction equation is given:
P + Cl 2 → PCl 5
On the left side of the equation there is 1 phosphorus atom (P) and two chlorine atoms (Cl 2), and on the right side there is one phosphorus atom (P) and five oxygen atoms (Cl 5). There is no need to equalize it in terms of the number of phosphorus atoms, but in terms of chlorine it is necessary to achieve equality, since before the reaction there were two atoms involved, and after the reaction there were five atoms. Let's make the following diagram:
before reaction after reaction
Cl 2 Cl 5
Let's determine the smallest multiple between the given numbers of atoms, it will be “10”.
Cl 2 Cl 5
\ 10 /
Divide this number on the left side of the chlorine equation by “2”. Let’s get the number “5” and put it into the equation to be solved:
P + 5Cl 2 → PCl 5
We also divide the number “10” for the right side of the equation by “5”. We get the number “2”, and also put it in the equation to be solved:
P + 5Cl 2 → 2РCl 5
The numbers of chlorine atoms on both the left and right sides of the equation became equal, respectively, 10 atoms each:
But the number of phosphorus atoms on both sides of the equation will not correspond to each other:
Therefore, in order to achieve equality, it is necessary to double the amount of phosphorus on the left side of the equation by setting the coefficient “2”:
Thus, equality for phosphorus is observed and, in general, the equation takes the form:
2Р + 5Cl 2 = 2РCl 5 (21)
When composing equations by valencies must be given valency determination and set values for the most famous elements. Valence is one of the previously used concepts, currently in a number of school programs not used. But with its help it is easier to explain the principles of drawing up equations of chemical reactions. Valence is understood as the number of chemical bonds that an atom can form with another or other atoms . Valency does not have a sign (+ or -) and is indicated by Roman numerals, usually above the symbols of chemical elements, for example:
Where do these values come from? How to use them when writing chemical equations? The numerical values of the valences of elements coincide with their group number Periodic table chemical elements by D.I. Mendeleev (Table 1).
For other elements valence values may have other values, but never greater than the number of the group in which they are located. Moreover, for even group numbers (IV and VI), the valences of elements take only even values, and for odd ones they can have both even and odd values (Table 2).
Of course, there are exceptions to the valency values for some elements, but in each specific case these points are usually specified. Now let's consider general principle compiling chemical equations based on given valences for certain elements. More often this method acceptable in the case of drawing up equations of chemical reactions of compounds of simple substances, for example, when interacting with oxygen ( oxidation reactions). Let's say you need to display an oxidation reaction aluminum. But let us recall that metals are designated by single atoms (Al), and non-metals in the gaseous state are designated by the indices “2” - (O 2). First we'll write general scheme reactions:
Al + О 2 →AlО
At this stage, it is not yet known what the correct spelling should be for aluminum oxide. And it is precisely at this stage that knowledge of the valences of elements will come to our aid. For aluminum and oxygen, let’s put them above the expected formula of this oxide:
III II
Al O
After that, “cross”-on-“cross” for these element symbols we will put the corresponding indices at the bottom:
III II
Al 2 O 3
Composition of a chemical compound Al 2 O 3 determined. The further diagram of the reaction equation will take the form:
Al+ O 2 →Al 2 O 3
All that remains is to equalize its left and right parts. Let us proceed in the same way as in the case of composing equation (19). Let's equalize the numbers of oxygen atoms by finding the smallest multiple:
before reaction after reaction
O 2 O 3
\ 6 /
Let's divide this number on the left side of the oxygen equation by “2”. Let’s get the number “3” and put it into the equation being solved. We also divide the number “6” for the right side of the equation by “3”. We get the number “2”, and also put it in the equation to be solved:
Al + 3O 2 → 2Al 2 O 3
To achieve equality in aluminum, it is necessary to adjust its quantity on the left side of the equation by setting the coefficient to “4”:
4Al + 3O 2 → 2Al 2 O 3
Thus, equality for aluminum and oxygen is observed and, in general, the equation will take its final form:
4Al + 3O 2 = 2Al 2 O 3 (22)
Using the valence method, you can predict what substance is formed during a chemical reaction and what its formula will look like. Let’s assume that the compound reacted with nitrogen and hydrogen with the corresponding valences III and I. Let’s write the general reaction scheme:
N 2 + N 2 → NH
For nitrogen and hydrogen, let’s put the valencies above the expected formula of this compound:
As before, “cross”-on-“cross” for these element symbols, let’s put the corresponding indices below:
III I
NH 3
The further diagram of the reaction equation will take the form:
N 2 + N 2 → NH 3
Equating in the already known way, through the smallest multiple for hydrogen equal to “6”, we obtain the required coefficients and the equation as a whole:
N 2 + 3H 2 = 2NH 3 (23)
When composing equations according to oxidation states reactants, it is necessary to recall that the oxidation state of a particular element is the number of electrons accepted or given up during a chemical reaction. Oxidation state in compounds Basically, it numerically coincides with the valence values of the element. But they differ in sign. For example, for hydrogen, the valence is I, and the oxidation state is (+1) or (-1). For oxygen, the valence is II, and the oxidation state is -2. For nitrogen, the valences are I, II, III, IV, V, and the oxidation states are (-3), (+1), (+2), (+3), (+4), (+5), etc. . The oxidation states of the elements most often used in equations are given in Table 3.
In the case of compound reactions, the principle of compiling equations by oxidation states is the same as when compiling by valences. For example, let us give the equation for the oxidation of chlorine with oxygen, in which chlorine forms a compound with an oxidation state of +7. Let's write down the proposed equation:
Cl 2 + O 2 → ClO
Let us place the oxidation states of the corresponding atoms over the proposed compound ClO:
As in previous cases, we establish that the required compound formula will take the form:
7 -2
Cl 2 O 7
The reaction equation will take the following form:
Cl 2 + O 2 → Cl 2 O 7
Equating for oxygen, finding the smallest multiple between two and seven, equal to “14,” we ultimately establish the equality:
2Cl 2 + 7O 2 = 2Cl 2 O 7 (24)
A slightly different method must be used with oxidation states when composing exchange, neutralization, and substitution reactions. In some cases, it is difficult to find out: what compounds are formed during the interaction of complex substances?
How to find out: what will happen in the reaction process?
Indeed, how do you know what reaction products may arise during a particular reaction? For example, what is formed when barium nitrate and potassium sulfate react?
Ba(NO 3) 2 + K 2 SO 4 → ?
Maybe BaK 2 (NO 3) 2 + SO 4? Or Ba + NO 3 SO 4 + K 2? Or something else? Of course, during this reaction the following compounds are formed: BaSO 4 and KNO 3. How is this known? And how to write the formulas of substances correctly? Let's start with what is most often overlooked: the very concept of “exchange reaction.” This means that in these reactions substances change their constituent parts with each other. Since exchange reactions are mostly carried out between bases, acids or salts, the parts with which they will be exchanged are metal cations (Na +, Mg 2+, Al 3+, Ca 2+, Cr 3+), H + ions or OH -, anions - acid residues, (Cl -, NO 3 2-, SO 3 2-, SO 4 2-, CO 3 2-, PO 4 3-). IN general view The exchange reaction can be given in the following notation:
Kt1An1 + Kt2An1 = Kt1An2 + Kt2An1 (25)
Where Kt1 and Kt2 are metal cations (1) and (2), and An1 and An2 are their corresponding anions (1) and (2). In this case, it is necessary to take into account that in compounds before and after the reaction, cations are always installed in first place, and anions are in second place. Therefore, if the reaction occurs potassium chloride And silver nitrate, both in dissolved state
KCl + AgNO 3 →
then in its process the substances KNO 3 and AgCl are formed and the corresponding equation will take the form:
KCl + AgNO 3 =KNO 3 + AgCl (26)
During neutralization reactions, protons from acids (H +) will combine with hydroxyl anions (OH -) to form water (H 2 O):
HCl + KOH = KCl + H 2 O (27)
The oxidation states of metal cations and the charges of anions of acidic residues are indicated in the table of solubility of substances (acids, salts and bases in water). The horizontal line shows metal cations, and the vertical line shows the anions of acid residues.
Based on this, when drawing up an equation for an exchange reaction, it is necessary first to establish on the left side the oxidation states of the hosts in this chemical process particles. For example, you need to write an equation for the interaction between calcium chloride and sodium carbonate. Let’s create the initial diagram of this reaction:
CaCl + NaCO 3 →
Ca 2+ Cl - + Na + CO 3 2- →
Having performed the already known “cross”-on-“cross” action, we determine the real formulas of the starting substances:
CaCl 2 + Na 2 CO 3 →
Based on the principle of exchange of cations and anions (25), we will establish preliminary formulas for the substances formed during the reaction:
CaCl 2 + Na 2 CO 3 → CaCO 3 + NaCl
Let us place the corresponding charges above their cations and anions:
Ca 2+ CO 3 2- + Na + Cl -
Substance formulas written correctly, in accordance with the charges of cations and anions. Let's compose complete equation, equalizing its left and right parts for sodium and chlorine:
CaCl 2 + Na 2 CO 3 = CaCO 3 + 2NaCl (28)
As another example, here is the equation for the neutralization reaction between barium hydroxide and phosphoric acid:
VaON + NPO 4 →
Let us place the corresponding charges over the cations and anions:
Ba 2+ OH - + H + PO 4 3- →
Let's determine the real formulas of the starting substances:
Ba(OH) 2 + H 3 PO 4 →
Based on the principle of exchange of cations and anions (25), we will establish preliminary formulas for the substances formed during the reaction, taking into account that during an exchange reaction one of the substances must necessarily be water:
Ba(OH) 2 + H 3 PO 4 → Ba 2+ PO 4 3- + H 2 O
Let us determine the correct notation for the formula of the salt formed during the reaction:
Ba(OH) 2 + H 3 PO 4 → Ba 3 (PO 4) 2 + H 2 O
Let's equalize the left side of the equation for barium:
3Ba (OH) 2 + H 3 PO 4 → Ba 3 (PO 4) 2 + H 2 O
Since on the right side of the equation the orthophosphoric acid residue is taken twice, (PO 4) 2, then on the left it is also necessary to double its amount:
3Ba (OH) 2 + 2H 3 PO 4 → Ba 3 (PO 4) 2 + H 2 O
It remains to match the number of hydrogen and oxygen atoms on the right side of water. Since on the left the total number of hydrogen atoms is 12, on the right it must also correspond to twelve, therefore before the formula of water it is necessary set the coefficient“6” (since the water molecule already has 2 hydrogen atoms). For oxygen, equality is also observed: on the left is 14 and on the right is 14. So, the equation has correct form entries:
3Ba (OH) 2 + 2H 3 PO 4 → Ba 3 (PO 4) 2 + 6H 2 O (29)
Possibility of chemical reactions
The world consists of a great variety of substances. The number of variants of chemical reactions between them is also incalculable. But can we, having written this or that equation on paper, say that a chemical reaction will correspond to it? There is a misconception that if it is correct set the odds in the equation, then it will be feasible in practice. For example, if we take sulfuric acid solution and put it in zinc, then you can observe the process of hydrogen evolution:
Zn+ H 2 SO 4 = ZnSO 4 + H 2 (30)
But if copper is dropped into the same solution, then the process of gas evolution will not be observed. The reaction is not feasible.
Cu+ H 2 SO 4 ≠
If concentrated sulfuric acid is taken, it will react with copper:
Cu + 2H 2 SO 4 = CuSO 4 + SO 2 + 2H 2 O (31)
In reaction (23) between the gases nitrogen and hydrogen, we observe thermodynamic equilibrium, those. how many molecules ammonia NH 3 is formed per unit time, the same amount of them will decompose back into nitrogen and hydrogen. Chemical equilibrium shift can be achieved by increasing pressure and decreasing temperature
N 2 + 3H 2 = 2NH 3
If you take potassium hydroxide solution and pour it on him sodium sulfate solution, then no changes will be observed, the reaction will not be feasible:
KOH + Na 2 SO 4 ≠
Sodium chloride solution when interacting with bromine, it will not form bromine, despite the fact that this reaction can be classified as a substitution reaction:
NaCl + Br 2 ≠
What are the reasons for such discrepancies? The point is that it is not enough just to correctly determine compound formulas, it is necessary to know the specifics of the interaction of metals with acids, skillfully use the table of solubility of substances, and know the rules of substitution in the activity series of metals and halogens. This article outlines only the most basic principles of how assign coefficients in reaction equations, How write molecular equations, How determine the composition of a chemical compound.
Chemistry, as a science, is extremely diverse and multifaceted. The above article reflects only a small part of the processes occurring in real world. Types, thermochemical equations, electrolysis, processes of organic synthesis and much, much more. But more on that in future articles.
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