How to make a chemical equation. Classification of chemical reactions. Solving chemistry problems using proportions

To describe the ongoing chemical reactions, equations of chemical reactions are compiled. In them, to the left of the equal sign (or arrow →) the formulas of the reactants (substances that react) are written, and to the right - the reaction products (substances that are obtained after a chemical reaction). Since we are talking about an equation, the number of atoms on the left side of the equation must be equal to that, which is in the right. Therefore, after drawing up a chemical reaction diagram (recording reactants and products), coefficients are substituted to equalize the number of atoms.

Coefficients are numbers before the formulas of substances that indicate the number of molecules that react.

For example, suppose in a chemical reaction hydrogen gas (H 2) reacts with oxygen gas (O 2). As a result, water (H 2 O) is formed. Reaction scheme will look like this:

H 2 + O 2 → H 2 O

On the left there are two hydrogen and oxygen atoms, and on the right there are two hydrogen atoms and only one oxygen. Suppose that the reaction of one hydrogen molecule and one oxygen produces two water molecules:

H 2 + O 2 → 2H 2 O

Now the number of oxygen atoms before and after the reaction is equal. However, there is two times less hydrogen before the reaction than after. It should be concluded that to form two molecules of water, two molecules of hydrogen and one of oxygen are needed. Then we get the following reaction scheme:

2H 2 + O 2 → 2H 2 O

Here is the number of different atoms chemical elements the same before and after the reaction. This means that this is no longer just a reaction scheme, but reaction equation. In reaction equations, the arrow is often replaced with an equal sign to emphasize that the number of atoms of different chemical elements is equal:

2H 2 + O 2 = 2H 2 O

Consider this reaction:

NaOH + H 3 PO 4 → Na 3 PO 4 + H 2 O

After the reaction, a phosphate was formed, which contains three sodium atoms. Let's equalize the amount of sodium before the reaction:

3NaOH + H3PO4 → Na3PO4 + H2O

The amount of hydrogen before the reaction is six atoms (three in sodium hydroxide and three in phosphoric acid). After the reaction there are only two hydrogen atoms. Dividing six by two gives three. This means that you need to put the number three in front of water:

3NaOH + H3PO4 → Na3PO4 + 3H2O

The number of oxygen atoms before and after the reaction is the same, which means that further calculation of the coefficients need not be done.

Class: 8

Presentation for the lesson

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The purpose of the lesson: help students develop knowledge about a chemical equation as a conventional notation of a chemical reaction using chemical formulas.

Tasks:

Educational:

systematize previously studied material;
teach the ability to compose equations of chemical reactions.

Educational:

develop communication skills (work in pairs, ability to listen and hear).

Educational:

develop educational and organizational skills aimed at accomplishing the task;
develop analytical thinking skills.

Lesson type: combined.

Equipment: computer, multimedia projector, screen, assessment sheets, reflection card, “set of chemical symbols”, notebook with printed base, reagents: sodium hydroxide, iron(III) chloride, alcohol lamp, holder, matches, Whatman paper, multi-colored chemical symbols.

Lesson presentation (Appendix 3)

Lesson structure.

I. Organizing time.
II. Updating knowledge and skills.
III. Motivation and goal setting.
IV. Learning new material:
4.1 combustion reaction of aluminum in oxygen;
4.2 decomposition reaction of iron (III) hydroxide;
4.3 algorithm for arranging coefficients;
4.4 minutes of relaxation;
4.5 set the coefficients;
V. Consolidation of acquired knowledge.
VI. Summing up the lesson and grading.
VII. Homework.
VIII. Final word teachers.

During the classes

Chemical nature of a complex particle
determined by the nature of elementary
components,
their number and
chemical structure.
D.I.Mendeleev

Teacher. Hello guys. Sit down.
Please note: you have a printed notebook on your desk. (Appendix 2), in which you will work today, and a score sheet in which you will record your achievements, sign it.

Updating knowledge and skills.

Teacher. We got acquainted with physical and chemical phenomena, chemical reactions and signs of their occurrence. We studied the law of conservation of mass of substances.
Let's test your knowledge. I suggest you open your printed notebooks and complete task 1. You are given 5 minutes to complete the task.

Test on the topic “Physical and chemical phenomena. Law of conservation of mass of substances.”

1. How do chemical reactions differ from physical phenomena?

Changing shape state of aggregation substances.
Formation of new substances.
Change of location.

2. What are the signs of a chemical reaction?

Precipitate formation, color change, gas evolution.

Magnetization, evaporation, vibration.

Growth and development, movement, reproduction.

3. In accordance with what law are equations of chemical reactions drawn up?

The law of constancy of the composition of matter.
Law of conservation of mass of matter.
Periodic law.
Law of dynamics.
The law of universal gravitation.

4. The law of conservation of mass of matter discovered:

DI. Mendeleev.
C. Darwin.
M.V. Lomonosov.
I. Newton.
A.I. Butlerov.

5. A chemical equation is called:

Conventional notation of a chemical reaction.

Conventional notation of the composition of a substance.

Recording the conditions of a chemical problem.

Teacher. You've done the job. I suggest you check it out. Exchange notebooks and check each other. Attention to the screen. For each correct answer - 1 point. Enter the total number of points on the evaluation sheets.

Motivation and goal setting.

Teacher. Using this knowledge, today we will draw up equations of chemical reactions, revealing the problem “Is the law of conservation of mass of substances the basis for drawing up equations of chemical reactions”

Learning new material.

Teacher. We are accustomed to thinking that an equation is a mathematical example where there is an unknown, and this unknown needs to be calculated. But in chemical equations there is usually nothing unknown: everything is simply written down in them using formulas: which substances react and which are obtained during this reaction. Let's see the experience.

(Reaction of sulfur and iron compound.) Appendix 3

Teacher. From the point of view of the mass of substances, the reaction equation for the compound of iron and sulfur is understood as follows

Iron + sulfur → iron (II) sulfide (task 2 tpo)

But in chemistry, words are reflected by chemical signs. Write this equation using chemical symbols.

Fe + S → FeS

(One student writes on the board, the rest in TVET.)

Teacher. Now read it.
Students. An iron molecule interacts with a sulfur molecule to produce one molecule of iron (II) sulfide.
Teacher. In this reaction we see that the amount starting materials equal to the amount of substances in the reaction product.
We must always remember that when composing reaction equations, not a single atom should be lost or unexpectedly appear. Therefore, sometimes, having written all the formulas in the reaction equation, you have to equalize the number of atoms in each part of the equation - set the coefficients. Let's see another experiment

(Combustion of aluminum in oxygen.) Appendix 4

Teacher. Let's write the equation of a chemical reaction (task 3 in TPO)

Al + O 2 → Al +3 O -2

To write the oxide formula correctly, remember that

Students. Oxygen in oxides has an oxidation state of -2, aluminum is a chemical element with a constant oxidation state of +3. LCM = 6

Al + O 2 → Al 2 O 3

Teacher. We see that 1 aluminum atom enters into the reaction, two aluminum atoms are formed. Two oxygen atoms enter, three oxygen atoms are formed.
Simple and beautiful, but disrespectful to the law of conservation of mass of substances - it is different before and after the reaction.
Therefore, we need to arrange the coefficients in given equation chemical reaction. To do this, let's find the LCM for oxygen.

Students. LCM = 6

Teacher. We put coefficients in front of the formulas for oxygen and aluminum oxide so that the number of oxygen atoms on the left and right is equal to 6.

Al + 3 O 2 → 2 Al 2 O 3

Teacher. Now we find that as a result of the reaction, four aluminum atoms are formed. Therefore, in front of the aluminum atom on the left side we put a coefficient of 4

Al + 3O 2 → 2Al 2 O 3

Let us once again count all the atoms before and after the reaction. We bet equal.

4Al + 3O 2 _ = 2 Al 2 O 3

Teacher. Let's look at another example

(The teacher demonstrates an experiment on the decomposition of iron (III) hydroxide.)

Fe(OH) 3 → Fe 2 O 3 + H 2 O

Teacher. Let's arrange the coefficients. One iron atom reacts and two iron atoms are formed. Therefore, before the formula of iron hydroxide (3) we put a coefficient of 2.

Fe(OH) 3 → Fe 2 O 3 + H 2 O

Teacher. We find that 6 hydrogen atoms enter into the reaction (2x3), 2 hydrogen atoms are formed.

Students. NOC =6. 6/2 = 3. Therefore, we set the coefficient of 3 for the water formula

2Fe(OH) 3 → Fe 2 O 3 + 3 H 2 O

Teacher. We count oxygen.

Students. Left – 2x3 =6; on the right – 3+3 = 6

Students. The number of oxygen atoms that entered into the reaction is equal to the number of oxygen atoms formed during the reaction. You can bet equally.

2Fe(OH) 3 = Fe 2 O 3 +3 H 2 O

Teacher. Now let's summarize everything that was said earlier and get acquainted with the algorithm for arranging coefficients in the equations of chemical reactions.

Count the number of atoms of each element on the right and left sides of the chemical reaction equation.

Determine which element has a changing number of atoms and find the LCM.

Divide the NOC into indices to obtain coefficients. Place them before the formulas.

Recalculate the number of atoms and repeat the action if necessary.

The last thing to check is the number of oxygen atoms.

Teacher. You've worked hard and you're probably tired. I suggest you relax, close your eyes and remember some pleasant moments in life. They are different for each of you. Now open your eyes and make circular movements with them, first clockwise, then counterclockwise. Now move your eyes intensively horizontally: right - left, and vertically: up - down.
Now let's activate mental activity and massage your earlobes.

Teacher. We continue to work.
In printed notebooks we will complete task 5. You will work in pairs. You need to place the coefficients in the equations of chemical reactions. You are given 10 minutes to complete the task.

P + Cl 2 →PCl 5
Na + S → Na 2 S
HCl + Mg →MgCl 2 + H 2
N 2 + H 2 →NH 3
H 2 O → H 2 + O 2

Teacher. Let's check the completion of the task ( the teacher questions and displays the correct answers on the slide). For each correctly set coefficient - 1 point.
You completed the task. Well done!

Teacher. Now let's get back to our problem.
Guys, what do you think, is the law of conservation of mass of substances the basis for drawing up equations of chemical reactions?

Students. Yes, during the lesson we proved that the law of conservation of mass of substances is the basis for drawing up equations of chemical reactions.

Consolidation of knowledge.

Teacher. We have studied all the main issues. Now let's do a short test that will allow you to see how you have mastered the topic. You should only answer “yes” or “no”. You have 3 minutes to work.

Statements.

In the reaction Ca + Cl 2 → CaCl 2, coefficients are not needed.(Yes)
In the reaction Zn + HCl → ZnCl 2 + H 2, the coefficient for zinc is 2. (No)
In the reaction Ca + O 2 → CaO, the coefficient for calcium oxide is 2.(Yes)
In the reaction CH 4 → C + H 2 no coefficients are needed.(No)
In the reaction CuO + H 2 → Cu + H 2 O, the coefficient for copper is 2. (No)
In the reaction C + O 2 → CO, a coefficient of 2 must be assigned to both carbon monoxide (II) and carbon. (Yes)
In the reaction CuCl 2 + Fe → Cu + FeCl 2 no coefficients are needed.(Yes)

Teacher. Let's check the progress of the work. For each correct answer - 1 point.

Lesson summary.

Teacher. You did a good job. Now calculate the total number of points scored for the lesson and give yourself a grade according to the rating that you see on the screen. Give me your evaluation sheets so that you can enter your grade into the journal.

Homework.

Teacher. Our lesson came to an end, during which we were able to prove that the law of conservation of mass of substances is the basis for composing reaction equations, and we learned how to compose chemical reaction equations. And as a final point, write down homework

§ 27, ex. 1 – for those who received a rating of “3”
ex. 2 – for those who received a rating of “4”
ex. 3 – for those who received a rating“5”

Final words from the teacher.

Teacher. I thank you for the lesson. But before you leave the office, pay attention to the table (the teacher points to a piece of Whatman paper with an image of a table and multi-colored chemical symbols). You see chemical signs of different colors. Each color symbolizes your mood.. I suggest you create your own table of chemical elements (it will differ from D.I. Mendeleev’s PSHE) - a table of the mood of the lesson. To do this, you must go to the sheet of music, take one chemical element, according to the characteristic that you see on the screen, and attach it to a table cell. I will do this first by showing you how comfortable I am working with you.

F I felt comfortable in the lesson, I received answers to all my questions.

F I achieved half the goal in the lesson.
F I was bored in class, I didn’t learn anything new.

To characterize a certain chemical reaction, you must be able to create a record that will display the conditions for the chemical reaction, show which substances reacted and which were formed. To do this, chemical reaction schemes are used.

Chemical reaction diagram– a conditional record showing which substances react, what reaction products are formed, as well as the conditions for the reaction. Let us consider, as an example, the reaction between coal and oxygen. Scheme this reaction is written as follows:

C + O2 → CO2

Coal reacts with oxygen to form carbon dioxide

Carbon and oxygen- in this reaction there are reactants, and the resulting carbon dioxide is the product of the reaction. Sign " → " indicates the progress of the reaction. Often the conditions under which the reaction occurs are written above the arrow.

Sign « t° → » indicates that the reaction occurs when heated.
Sign "R →" stands for pressure
Sign "hv →"– that the reaction occurs under the influence of light. Additional substances involved in the reaction may also be indicated above the arrow.
For example, "O2 →". If a gaseous substance is formed as a result of a chemical reaction, then in the reaction scheme, after the formula of this substance, write the sign “ → " If a precipitate is formed during the reaction, it is indicated by the sign “ → ».
For example, when chalk powder is heated (it contains a substance with the chemical formula CaCO3), two substances are formed: quicklime CaO and carbon dioxide. The reaction scheme is written as follows:

СaCO3 t° → CaO + CO2

Thus, natural gas mainly consists of CH4 methane; when heated to 1500°C, it turns into two other gases: hydrogen H2 and acetylene C2H2. The reaction scheme is written as follows:

CH4 t° → C2H2 + H2.

It is important not only to be able to draw up diagrams of chemical reactions, but also to understand what they mean. Let's consider another reaction scheme:

H2O electric current → H2 + O2

This diagram means that under the influence of electric current, water decomposes into two simple gaseous substances: hydrogen and oxygen. The diagram of a chemical reaction is a confirmation of the law of conservation of mass and shows that chemical elements do not disappear during a chemical reaction, but are only rearranged into new chemical compounds.

Chemical Reaction Equations

According to the law of conservation of mass, the initial mass of products is always equal to the mass of the resulting reactants. The number of atoms of elements before and after the reaction is always the same; the atoms only rearrange and form new substances. Let's return to the reaction schemes recorded earlier:

СaCO3 t° → CaO + CO2

C + O2 CO2.

In these reaction schemes the sign “ → " can be replaced with the "=" sign, since it is clear that the number of atoms before and after the reactions is the same. The entries will look like this:

CaCO3 = CaO + CO2

C + O2 = CO2.

It is these records that are called equations of chemical reactions, that is, these are records of reaction schemes in which the number of atoms before and after the reaction is the same.

Chemical reaction equation– a conventional notation of a chemical reaction using chemical formulas, which corresponds to the law of conservation of mass of a substance

If we look at the other equation schemes given earlier, we can see that in At first glance, the law of conservation of mass does not hold true in them:

CH4 t° → C2H2 + H2.

It can be seen that on the left side of the diagram there is one carbon atom, and on the right there are two. There are equal numbers of hydrogen atoms and there are four of them on the left and right sides. Let's turn this diagram into an equation. For this it is necessary equalize number of carbon atoms. Chemical reactions are equalized using coefficients that are written before the formulas of substances. Obviously, in order for the number of carbon atoms to become the same on the left and right, on the left side of the diagram, before the methane formula, it is necessary to put coefficient 2:

2CH4 t° → C2H2 + H2

It can be seen that there are now equal numbers of carbon atoms on the left and right, two each. But now the number of hydrogen atoms is not the same. On the left side of the equation their 2∙4 = 8. On the right side of the equation there are 4 hydrogen atoms (two of them in the acetylene molecule, and two more in the hydrogen molecule). If you put a coefficient in front of acetylene, the equality of carbon atoms will be violated. Let's put a factor of 3 in front of the hydrogen molecule:

2CH4 = C2H2 + 3H2

Now the number of carbon and hydrogen atoms on both sides of the equation is the same. The law of conservation of mass is fulfilled! Let's look at another example. Reaction scheme Na + H2O → NaOH + H2 needs to be turned into an equation. In this scheme, the number of hydrogen atoms is different. On the left side there are two, and on the right side - three atoms. Let's put a factor of 2 in front of NaOH.

Na + H2O → 2NaOH + H2

Then there will be four hydrogen atoms on the right side, therefore, coefficient 2 must be added before the water formula:

Na + 2H2O → 2NaOH + H2

Let's equalize the number of sodium atoms:

2Na + 2H2O = 2NaOH + H2

Now the number of all atoms before and after the reaction is the same. Thus, we can conclude: To turn a chemical reaction diagram into a chemical reaction equation, it is necessary to equalize the number of all atoms that make up the reactants and reaction products using coefficients. Coefficients are placed before the formulas of substances. Let's summarize the equations of chemical reactions

A chemical reaction diagram is a conventional notation showing which substances react, what reaction products are formed, as well as the conditions for the reaction to occur
In reaction schemes, symbols are used that indicate the peculiarities of their occurrence.
The equation of a chemical reaction is a conventional representation of a chemical reaction using chemical formulas, which corresponds to the law of conservation of mass of a substance
A chemical reaction diagram is converted into an equation by placing coefficients in front of the formulas of 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 ratios of substances in a reaction, i.e., the amounts of reacting and resulting substances in a 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 being filed, but after the pieces collected 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 compiling chemical equations, it is necessary to operate with the specified element symbols.

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 figured out the structure various substances, you can begin to compile 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 the large number “2” standing 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 constituent parts. 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 bringing various types chemical reactions, you can proceed to the principle of compiling chemical equations, otherwise, selecting the 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 in 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 number chemical bonds, which one or another 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 valence 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, let's write the general reaction scheme:

Al + О 2 →AlО

At this stage it is not yet known which correct writing should be 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 a well-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 first necessary to establish on the left side the oxidation states of the particles receiving in this chemical process. 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 create a complete equation, equalizing its left and right sides 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 the correct written form:

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 it 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 the 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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