How to equalize reactions. How to write an equation for a chemical reaction: sequence of actions. A chemical equation is called

A chemical equation is a recording of a reaction using the symbols of the elements and formulas of the compounds involved in it. The relative amounts of reactants and products, expressed in moles, are indicated by numerical coefficients in the complete (balanced) reaction equation. These coefficients are sometimes called stoichiometric coefficients. Currently, there is an increasing tendency to include indications of the physical states of reactants and products in chemical equations. This is done using the following symbols: (gas) or means gaseous state, (-liquid, ) - solid, (-water solution.

A chemical equation can be constructed based on experimentally established knowledge of the reactants and products of the reaction being studied, and by measuring the relative amounts of each reactant and product that participate in the reaction.

Writing a Chemical Equation

Writing a complete chemical equation involves the following four steps.

1st stage. Recording the reaction in words. For example,

2nd stage. Replacement of verbal names with formulas of reagents and products.

3rd stage. Balancing the equation (determining its coefficients)

This equation is called balanced or stoichiometric. The need to balance the equation is dictated by the fact that in any reaction the law of conservation of matter must be satisfied. In relation to the reaction we are considering as an example, this means that not a single atom of magnesium, carbon or oxygen can be formed or destroyed in it. In other words, the number of atoms of each element on the left and right sides of a chemical equation must be the same.

4th stage. Indication of the physical condition of each participant in the reaction.

Types of Chemical Equations

Consider the following complete equation:

This equation describes the entire reaction system as a whole. However, the reaction under consideration can also be represented in a simplified form using the ionic equation -.

This equation does not include information about sulfate ions, which are not listed because they do not participate in the reaction under consideration. Such ions are called observer ions.

The reaction between iron and copper(II) is an example of redox reactions (see Chapter 10). It can be divided into two reactions, one of which describes reduction, and the other - oxidation, occurring simultaneously in a general reaction:

These two equations are called half-reaction equations. They are especially often used in electrochemistry to describe processes occurring at electrodes (see Chapter 10).

Interpretation of chemical equations

Consider the following simple stoichiometric equation:

It can be interpreted in two ways. First, according to this equation, one mole of hydrogen molecules reacts with one mole of bromine molecules to form two moles of hydrogen bromide molecules. This interpretation of the chemical equation is sometimes called the molar interpretation.

However, this equation can also be interpreted in such a way that in the resulting reaction (see below) one molecule of hydrogen reacts with one molecule of bromine to form two molecules of hydrogen bromide. This interpretation of a chemical equation is sometimes called its molecular interpretation.

Both molar and molecular interpretations are equally valid. However, it would be completely wrong to conclude, based on the equation of the reaction in question, that one molecule of hydrogen collides with one molecule of bromine to form two molecules of hydrogen bromide. The fact is that this reaction, like most others, is carried out in several successive stages. The set of all these stages is usually called the reaction mechanism (see Chapter 9). In the example we are considering, the reaction includes the following stages:

Thus, the reaction in question is actually a chain reaction involving intermediates called radicals (see Chapter 9). The mechanism of the reaction under consideration also includes other stages and side reactions. Thus, the stoichiometric equation indicates only the resulting reaction. It does not provide information about the reaction mechanism.

Calculation using chemical equations

Chemical equations are the starting point for a wide variety of chemical calculations. Here and later in the book a number of examples of such calculations are given.

Calculation of the mass of reactants and products. We already know that a balanced chemical equation indicates the relative molar amounts of reactants and products involved in a reaction. These quantitative data allow the masses of reactants and products to be calculated.

Let us calculate the mass of silver chloride formed when an excess amount of sodium chloride solution is added to a solution containing 0.1 mol of silver in the form of ions

The first stage of all such calculations is to write the equation of the reaction in question: I

Since the reaction uses an excess amount of chloride ions, it can be assumed that all ions present in the solution are converted into The reaction equation shows that one mole of ions is obtained from one mole. This allows us to calculate the mass of the product as follows:

Hence,

Since g/mol, then

Determination of the concentration of solutions. Calculations based on stoichiometric equations underlie quantitative chemical analysis. As an example, consider determining the concentration of a solution based on the known mass of the product formed in the reaction. This type of quantitative chemical analysis is called gravimetric analysis.

A quantity of potassium iodide solution was added to the nitrate solution, which is sufficient to precipitate all the lead in the form of iodide. The mass of the iodide formed was 2.305 g. The volume of the initial nitrate solution was equal to. It is required to determine the concentration of the initial nitrate solution

We have already encountered the equation for the reaction in question:

This equation shows that one mole of lead(II) nitrate is required to produce one mole of iodide. Let us determine the molar amount of lead (II) iodide formed in the reaction. Because the

Reactions between various types of chemical substances and elements are one of the main subjects of study in chemistry. To understand how to create a reaction equation and use them for your own purposes, you need a fairly deep understanding of all the patterns in the interaction of substances, as well as processes with chemical reactions.

Writing equations

One way to express a chemical reaction is a chemical equation. It records the formula of the starting substance and the product, coefficients that show how many molecules each substance has. All known chemical reactions are divided into four types: substitution, combination, exchange and decomposition. Among them are: redox, exogenous, ionic, reversible, irreversible, etc.

Learn more about how to write equations for chemical reactions:

1. It is necessary to determine the name of the substances that interact with each other in the reaction. We write them on the left side of our equation. As an example, consider the chemical reaction that formed between sulfuric acid and aluminum. We place the reagents on the left: H2SO4 + Al. Next we write the equal sign. In chemistry, you may come across an “arrow” sign that points to the right, or two arrows directed in opposite directions, they mean “reversibility”. The result of the interaction of metal and acid is salt and hydrogen. Write the products obtained after the reaction after the equal sign, that is, on the right. H2SO4+Al= H2+ Al2(SO4)3. So, we can see the reaction scheme.
2. To compose a chemical equation, you must find the coefficients. Let's return to the previous diagram. Let's look at its left side. Sulfuric acid contains hydrogen, oxygen and sulfur atoms in an approximate ratio of 2:4:1. On the right side there are 3 sulfur atoms and 12 oxygen atoms in the salt. Two hydrogen atoms are contained in a gas molecule. On the left side the ratio of these elements is 2:3:12
3. To equalize the number of oxygen and sulfur atoms that are in the composition of aluminum (III) sulfate, it is necessary to put a factor of 3 in front of the acid on the left side of the equation. Now we have 6 hydrogen atoms on the left side. To equalize the number of elements of hydrogen, you need to put 3 in front of hydrogen on the right side of the equation.
4. Now all that remains is to equalize the amount of aluminum. Since the salt contains two metal atoms, we set a coefficient of 2 on the left side in front of aluminum. As a result, we obtain the reaction equation for this scheme: 2Al+3H2SO4=Al2(SO4)3+3H2

Having understood the basic principles of how to write a reaction equation chemical substances, in the future it will not be difficult to write down any reaction, even the most exotic one from the point of view of chemistry.

A reaction equation in chemistry is the recording of a chemical process using chemical formulas and mathematical symbols.

This notation is a diagram of a chemical reaction. When the "=" sign appears, it is called an "equation". Let's try to solve it.

In contact with

Example of analysis of simple reactions

There is one atom in calcium, since the coefficient is not worth it. The index is also not written here, which means one. On the right side of the equation, Ca is also one. We don't need to work on calcium.

Video: Coefficients in chemical reaction equations.

Let's look at the next element - oxygen. Index 2 indicates that there are 2 oxygen ions. There are no indices on the right side, that is, one particle of oxygen, and on the left there are 2 particles. What are we doing? No additional indices or corrections can be made to the chemical formula, since it is written correctly.

The coefficients are what is written before the smallest part. They have the right to change. For convenience, we do not rewrite the formula itself. On the right side, we multiply one by 2 to get 2 oxygen ions there.

After we set the coefficient, we got 2 calcium atoms. There is only one on the left side. This means that now we must put 2 in front of calcium.

Now let's check the result. If the number of atoms of an element is equal on both sides, then we can put the “equal” sign.

Another clear example: there are two hydrogens on the left, and after the arrow we also have two hydrogens.

• There are two oxygens before the arrow, but there are no indices after the arrow, which means there is one.
• There is more on the left and less on the right.
• We put coefficient 2 in front of water.

We multiplied the entire formula by 2, and now the amount of hydrogen has changed. We multiply the index by the coefficient, and we get 4. And on the left side there are two hydrogen atoms left. And to get 4, we have to multiply hydrogen by two.

Video: Arranging coefficients in a chemical equation

This is the case when the element in one and the other formula is on the same side, up to the arrow.

One sulfur ion on the left, and one ion on the right. Two oxygen particles, plus two more oxygen particles. This means that there are 4 oxygens on the left side. On the right there are 3 oxygens. That is, on one side there is an even number of atoms, and on the other, an odd number. If we multiply the odd number by two times, we get an even number. First we bring it to an even value. To do this, multiply the entire formula after the arrow by two. After multiplication, we get six oxygen ions, and also 2 sulfur atoms. On the left we have one microparticle of sulfur. Now let's equalize it. We put the equations on the left before gray 2.

Called.

Complex reactions

This example is more complex because there are more elements of matter.

This is called a neutralization reaction. What needs to be equalized here first:

• On the left side is one sodium atom.
• On the right side, the index says that there are 2 sodium.

The conclusion suggests itself is that you need to multiply the entire formula by two.

Video: Drawing up chemical reaction equations

Now let's see how much sulfur there is. One on the left and right sides. Let's pay attention to oxygen. On the left side we have 6 oxygen atoms. On the other hand - 5. Less on the right, more on the left. An odd number must be brought to an even number. To do this, we multiply the formula of water by 2, that is, from one oxygen atom we make 2.

Now there are already 6 oxygen atoms on the right side. There are also 6 atoms on the left side. Let's check the hydrogen. Two hydrogen atoms and 2 more hydrogen atoms. So there will be four hydrogen atoms on the left side. And on the other side there are also four hydrogen atoms. All elements are equal. We put the equal sign.

Video: Chemical equations. How to write chemical equations.

Next example.

Here the example is interesting because parentheses appear. They say that if a factor is behind the brackets, then each element in the brackets is multiplied by it. You need to start with nitrogen, since there is less of it than oxygen and hydrogen. On the left there is one nitrogen, and on the right, taking into account the brackets, there are two.

There are two hydrogen atoms on the right, but four are needed. We get out of this by simply multiplying water by two, resulting in four hydrogens. Great, hydrogen equalized. There is oxygen left. Before the reaction there are 8 atoms, after - also 8.

Great, all the elements are equal, we can set “equal”.

Last example.

Next up is barium. It is equalized, you don’t need to touch it. Before the reaction there are two chlorines, after it there is only one. What needs to be done? Place 2 in front of the chlorine after the reaction.

Video: Balancing Chemical Equations.

Now, due to the coefficient that was just set, after the reaction we got two sodiums, and before the reaction we also got two. Great, everything else is equalized.

You can also equalize reactions using the electronic balance method. This method has a number of rules by which it can be implemented. The next step is to arrange the oxidation states of all elements in each substance in order to understand where oxidation occurred and where reduction occurred.

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 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 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 as a special case of an exchange reaction - the 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. These 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? Numeric values valencies 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 the general principle of composing chemical equations based on given valences for certain elements. Most often, this method is 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 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

Already calling in a 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, 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 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 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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The calculator below is designed to equalize chemical reactions.

As is known, there are several methods for equalizing chemical reactions:

• Method for selecting coefficients
• Mathematical method
• Garcia method
• Electronic balance method
• Electron-ion balance method (half-reaction method)

The last two are used for redox reactions

This calculator uses mathematical method- as a rule, in the case of complex chemical equations, it is quite labor-intensive for manual calculations, but it works great if the computer calculates everything for you.

The mathematical method is based on the law of conservation of mass. 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. This makes it possible to balance the equations of any reactions (including redox ones). To do this, it is necessary to write down the reaction equation in general form, based on material balance (equality of masses of a certain chemical element in the original and resulting substances) create a system of mathematical equations and solve it.

Let's look at this method using an example:

Let the chemical reaction be given:

Let us denote the unknown coefficients:

Let's create equations for the number of atoms of each element participating in a chemical reaction:
For Fe:
For Cl:
For Na:
For P:
For O:

Let's write them in the form of a general system:

IN in this case we have five equations for four unknowns, and the fifth can be obtained by multiplying the fourth by four, so it can be safely discarded.

Let's rewrite this system of linear algebraic equations in the form of a matrix:

This system can be solved using the Gaussian method. Actually, it will not always be so lucky that the number of equations will coincide with the number of unknowns. However, the beauty of the Gauss method is that it allows you to solve systems with any number of equations and unknowns. A calculator was written specifically for this purpose. Solving a system of linear equations using the Gauss method with finding a general solution, which is used in equalizing chemical reactions.
That is, the calculator below parses the reaction formula, compiles the SLAE and passes it to the calculator at the link above, which solves the SLAE using the Gaussian method. The solution is then used to display the balanced equation.

Chemical elements should be written as they are written in the periodic table, i.e., take into account large and small letters (Na3PO4 - correct, na3po4 - incorrect).