Part I
1. Lomonosov-Lavoisier law - the law of conservation of the mass of substances:
2. The equations of a chemical reaction are conditional notation of a chemical reaction using chemical formulas and mathematical signs.
3. The chemical equation must comply with the law conservation of the mass of substances, which is achieved by arranging the coefficients in the reaction equation.
4. What does the chemical equation show?
1) What substances react.
2) What substances are formed as a result.
3) Quantitative ratios of substances in the reaction, i.e., the amount of reacting and formed substances in the reaction.
4) Type of chemical reaction.
5. Rules for arranging the coefficients in the scheme of a chemical reaction on 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 formed substances:
b) begin to equalize the reaction scheme with the salt formula (if available). At the same time, remember that several complex ions in the composition of a base or salt are indicated by brackets, and their number is indicated by indices outside the brackets:
c) equalize hydrogen in the penultimate turn:
d) equalize oxygen last - this is an indicator of the correct placement of the 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 the reaction equations, the schemes of which are:
2. Write the equations of chemical reactions:
3. Establish a correspondence between the scheme 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 have reacted;
2) Formed as a result of the reaction of salt and water;
3) Coefficients before starting substances 1 and 2.
6. Using the following diagram, write an equation for a chemical reaction using a doubling of 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 (V) oxide - 2 mol, 284 g;
3) oxygen - 5 mol or 160 l.
Let's talk about how to write a chemical equation, because they are the main elements of this discipline. Thanks to a deep awareness of all the patterns of interactions and substances, you can control them, apply them in various fields of activity.
Theoretical Features
Compilation of chemical equations is an important and crucial stage, considered in the eighth grade of secondary schools. What should precede this stage? Before the teacher tells his pupils how to make a chemical equation, it is important to introduce the schoolchildren to the term "valence", to teach them to determine this value for metals and non-metals using the periodic table of elements.
Compilation of binary formulas by valence
In order to understand how to write a chemical equation in terms of valence, you first need to learn how to formulate compounds consisting of two elements using valency. We propose an algorithm that will help to cope with the task. For example, you need to write a formula for sodium oxide.
First, it is important to consider that the chemical element that is mentioned last in the name should be in the first place in the formula. In our case, sodium will be written first in the formula, oxygen second. Recall that binary compounds are called oxides, in which the last (second) element must necessarily be oxygen with an oxidation state of -2 (valency 2). Further, according to the periodic table, it is necessary to determine the valencies of each of the two elements. To do this, we use certain rules.
Since sodium is a metal that is located in the main subgroup of group 1, its valence is a constant value, it is equal to I.
Oxygen is a non-metal, since it is the last in the oxide, to determine its valency, we subtract 6 from eight (the number of groups) (the group in which oxygen is located), we get that the valence of oxygen is II.
Between certain valences, we find the least common multiple, then divide it by the valency of each of the elements, we get their indices. We write down the finished formula Na 2 O.
Instructions for compiling an equation
Now let's talk more about how to write a chemical equation. First, let's look at the theoretical points, then move on to specific examples. So, the compilation of chemical equations involves a certain procedure.
- 1st stage. After reading the proposed task, it is necessary to determine which chemicals should be present on the left side of the equation. A "+" sign is placed between the original components.
- 2nd stage. After the equal sign, it is necessary to draw up a formula for the reaction product. When performing such actions, an algorithm for compiling formulas for binary compounds, which we discussed above, will be required.
- 3rd stage. We check the number of atoms of each element before and after the chemical interaction, if necessary, put additional coefficients in front of the formulas.
Combustion reaction example
Let's try to figure out how to make a chemical equation for the combustion of magnesium using an algorithm. On the left side of the equation, we write through the sum of magnesium and oxygen. Do not forget that oxygen is a diatomic molecule, so it must have an index of 2. After the equal sign, we draw up a formula for the product obtained after the reaction. They will be in which magnesium is written first, and we put oxygen second in the formula. Further, according to the table of chemical elements, we determine the valencies. Magnesium, which is in group 2 (the main subgroup), has a constant valency II, for oxygen, by subtracting 8 - 6, we also obtain valence II.
The process record will look like: Mg+O 2 =MgO.
In order for the equation to correspond to the law of conservation of mass of substances, it is necessary to arrange the coefficients. First, we check the amount of oxygen before the reaction, after the completion of the process. Since there were 2 oxygen atoms, and only one was formed, on the right side, before the magnesium oxide formula, you must add a factor of 2. Next, we count the number of magnesium atoms before and after the process. As a result of the interaction, 2 magnesium was obtained, therefore, on the left side, a coefficient of 2 is also required in front of a simple substance magnesium.
The final form of the reaction: 2Mg + O 2 \u003d 2MgO.
An example of a substitution reaction
Any abstract in chemistry contains a description of different types of interactions.
Unlike a compound, in a substitution there will be two substances on both the left and right sides of the equation. Suppose you need to write the interaction reaction between zinc and We use the standard writing algorithm. First, on the left side we write zinc and hydrochloric acid through the sum, on the right side we draw up the formulas of the resulting reaction products. Since in the electrochemical series of voltages of metals, zinc is located before hydrogen, in this process it displaces molecular hydrogen from the acid, forming zinc chloride. As a result, we get the following entry: Zn+HCL=ZnCl 2 +H 2 .
Now we turn to equalizing the number of atoms of each element. Since there was one atom on the left side of chlorine, and after the interaction there were two of them, a factor of 2 must be put in front of the hydrochloric acid formula.
As a result, we obtain a ready-made reaction equation corresponding to the law of conservation of mass of substances: Zn + 2HCL = ZnCl 2 +H 2.
Conclusion
A typical chemistry abstract necessarily contains several chemical transformations. Not a single section of this science is limited to a simple verbal description of transformations, processes of dissolution, evaporation, everything is necessarily confirmed by equations. The specificity of chemistry lies in the fact that all the processes that occur between different inorganic or organic substances can be described using coefficients, indices.
How is chemistry different from other sciences? Chemical equations help not only to describe the ongoing transformations, but also to carry out quantitative calculations on them, thanks to which it is possible to carry out laboratory and industrial production of various substances.
Reactions between various kinds of chemicals and elements are one of the main subjects of study in chemistry. To understand how to draw up 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 contains the formula of the starting substance and product, the 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:
- It is necessary to determine the name of the substances interacting with each other in the reaction. We write them on the left side of our equation. As an example, consider the chemical reaction that took place between sulfuric acid and aluminum. We have the reagents on the left: H2SO4 + Al. Next, write the equal sign. In chemistry, you can see an arrow sign that points to the right, or two opposite arrows that 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.
- To compile a chemical equation, it is imperative to find the coefficients. Let's go back to the previous diagram. Let's look at the left side of it. 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. There are two hydrogen atoms in a gas molecule. On the left side, the ratio of these elements is 2:3:12
- 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. In order to equalize the number of elements of hydrogen, you need to put 3 in front of hydrogen on the right side of the equation.
- Now it remains only to equalize the amount of aluminum. Since the composition of the salt includes two metal atoms, then on the left side in front of aluminum we set the coefficient 2. As a result, we will get the reaction equation of this scheme: 2Al + 3H2SO4 = Al2 (SO4) 3 + 3H2
Having understood the basic principles of how to write an equation for the reaction of chemicals, in the future it will not be difficult to write down any, even the most exotic, from the point of view of chemistry, reaction.
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 all got wet, as they say, “to the skin”, then is this not a transformation: after all, the clothes were dry, but became wet.
If, for example, you take an iron nail, process it with a file, and then assemble iron filings (Fe) , then this is also not a transformation: there was a nail - it became powder. But if after that to assemble the device and hold obtaining oxygen (O 2): heat up potassium permanganate(KMpo 4) and collect oxygen in a test tube, and then place these iron filings heated “to red” in 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 state of clothing (dry, wet) change, these are not transformations. The fact is that the nail itself, as it was a substance (iron), remained so, despite its different form, and our clothes soaked up the water from the rain, and then it evaporated into the atmosphere. The water itself has not changed. So what are transformations in terms of chemistry?
From the point of view of chemistry, transformations are such phenomena that are accompanied by a change in the composition of a substance. Let's take the same nail as an example. It does not matter what form it took after being filed, but after being collected from it iron filings placed in an atmosphere of oxygen - it turned into iron oxide(Fe 2 O 3 ) . So, has something really changed? Yes, it has. There was a nail substance, 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 a person uninitiated in chemistry, questions immediately arise. What is the "molecular equation", what is Fe? Why are there numbers "4", "3", "2"? What are the small numbers "2" and "3" in the formula Fe 2 O 3? This means that the time has come to sort things out in order.
Signs of chemical elements.
Despite the fact that they begin to study chemistry 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 "Mendeleev's Table".
In this table, in the appropriate order, the elements are located. 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 hesitation, identifying them with objects: an iron bolt, aluminum wire, oxygen in the atmosphere, a golden ring, etc. etc. But in fact, all these substances (bolt, wire, ring) consist of their respective elements. The whole paradox is that the element cannot be touched, 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, however, as in other sciences, for calculations, drawing up equations, and solving problems. Each element differs from the other in that it is characterized by its own 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 #1. Its atom consists of 1 proton and 1 electron. Helium is element number 2. Its atom consists of 2 protons and 2 electrons. Lithium is element number 3. Its atom consists of 3 protons and 3 electrons. Darmstadtium - element number 110. Its atom consists of 110 protons and 110 electrons.
Each element is denoted by a certain symbol, Latin letters, and has a certain reading in translation 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 designations can be found in any chemistry textbook for the 8th grade. For us now, the main thing is to understand that when compiling 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 iron and sulfur substances interact, then the equation will take the following 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). And you should pay attention
special attention to the fact that all metals are denoted by single symbols: K, Ba, Ca, Al, V, Mg, etc., and non-metals - either by 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 be of great importance in the formulation of equations. It is not at all difficult to guess that complex substances are substances formed from atoms of different types, for example,
one). Oxides:
aluminium oxide Al 2 O 3, 
sodium oxide Na 2 O
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 potassium alkali KOH,
sodium hydroxide NaOH.
3). Acids:
hydrochloric acid HCl
sulfurous acid H2SO3,
Nitric acid HNO3
four). Salts:
sodium thiosulfate Na 2 S 2 O 3,
sodium sulfate or Glauber's salt Na 2 SO 4,
calcium carbonate or limestone CaCO 3,
copper chloride CuCl 2
5). organic matter:
sodium acetate CH 3 COOHa,
methane CH 4,
acetylene C 2 H 2,
glucose C 6 H 12 O 6
Finally, after we have clarified the structure of various substances, we can begin to write chemical equations.
Chemical equation.
The word “equation” itself is derived from the word “equalize”, i.e. divide something into equal parts. In mathematics, equations are almost the very essence of this science. For example, you can give such 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 number of atoms, the 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 conditional record of a chemical reaction using chemical formulas and mathematical signs. A chemical equation inherently reflects a particular 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 that take part barium chloride BaCl 2 and sulphuric acid H 2 SO 4. As a result of this reaction, an insoluble precipitate is formed - barium sulfate BaSO 4 and hydrochloric acid Hcl:
ВаСl 2 + H 2 SO 4 = BaSO 4 + 2НCl (3)
First of all, it is necessary to understand that the large number “2” in front of the HCl substance is called the coefficient, and the small numbers “2”, “4” under the formulas ВаСl 2, H 2 SO 4, BaSO 4 are called indices. Both the coefficients and indices in chemical equations play the role of factors, not terms. In order to correctly write a chemical equation, it is necessary arrange the 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). Whence it follows that on the right side of the equation the number of hydrogen and chlorine atoms is half that 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 the number of atoms of the elements involved in this reaction, both on the left and on the right, we get the following balance:
In both parts of the equation, the number of atoms of the elements participating in the reaction are equal, therefore it is correct.
Chemical equation and chemical reactions
As we have already found out, chemical equations are a reflection of chemical reactions. Chemical reactions are such phenomena in the process of which the transformation of one substance into another occurs. Among their diversity, two main types can be distinguished:
one). Connection reactions
2). decomposition reactions.
The overwhelming majority of chemical reactions belong to addition reactions, since changes in its composition can rarely occur with a single substance if it is not subjected to external influences (dissolution, heating, light). Nothing characterizes a chemical phenomenon, or reaction, as much as the changes that occur when two or more substances interact. Such phenomena can occur spontaneously and be accompanied by an increase or decrease in temperature, light effects, color changes, sedimentation, release of gaseous products, noise.
For clarity, we present several equations that reflect the processes of compound reactions, during which we obtain sodium chloride(NaCl), zinc chloride(ZnCl 2), silver chloride precipitate(AgCl), aluminum chloride(AlCl 3)
Cl 2 + 2Nа = 2NaCl (4)
CuCl 2 + Zn \u003d ZnCl 2 + Cu (5)
AgNO 3 + KCl \u003d AgCl + 2KNO 3 (6)
3HCl + Al(OH) 3 \u003d AlCl 3 + 3H 2 O (7)
Among the reactions of the compound, the following should be especially noted : substitution (5), exchange (6), and as a special case of the exchange reaction, the reaction neutralization (7).
Substitution reactions include those in which atoms of a simple substance replace the 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 ZnCl 2 salt, and copper is released from the solution in the metallic state.
Exchange reactions are those reactions in which two complex substances exchange their constituents. In the case of reaction (6), the soluble salts of AgNO 3 and KCl, when both solutions are drained, form an insoluble precipitate of the AgCl salt. At the same time, they exchange their constituent parts - cations and anions. Potassium cations K + are attached to NO 3 anions, and silver cations Ag + - to Cl - anions.
A special, particular case of exchange reactions is the neutralization reaction. Neutralization reactions are reactions in which acids react with bases to form salt and water. In example (7), hydrochloric acid HCl reacts with base Al(OH) 3 to form AlCl 3 salt and water. In this case, aluminum cations Al 3+ from the base are exchanged with Cl anions - from the acid. As a result, it happens hydrochloric acid neutralization.
Decomposition reactions include those in which two or more new simple or complex substances, but of a simpler composition, are formed from one complex one. As reactions, one can cite those in the process of which 1) decompose. potassium nitrate(KNO 3) with the formation of potassium nitrite (KNO 2) and oxygen (O 2); 2). Potassium permanganate(KMnO 4): potassium manganate is formed (K 2 MnO 4), manganese oxide(MnO 2) and oxygen (O 2); 3). calcium carbonate or marble; in the process are formed carbonicgas(CO 2) and calcium oxide(Cao)
2KNO 3 \u003d 2KNO 2 + O 2 (8)
2KMnO 4 \u003d K 2 MnO 4 + MnO 2 + O 2 (9)
CaCO 3 \u003d CaO + CO 2 (10)
In reaction (8), one complex and one simple substance is 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 undergo decomposition:
one). 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: sulphuric acid H 2 SO 4 \u003d SO 3 + H 2 O (13)
four). Salts: calcium carbonate CaCO 3 \u003d CaO + CO 2 (14)
5). organic matter: alcoholic fermentation of glucose
C 6 H 12 O 6 \u003d 2C 2 H 5 OH + 2CO 2 (15)
According to another classification, all chemical reactions can be divided into two types: reactions that take place with the release of heat, they are called exothermic, and reactions that go 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. interactions with oxygen methane combustion:
CH 4 + 2O 2 \u003d 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 during the reaction (+Q) or absorbed (-Q):
CaCO 3 \u003d 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 \u003d 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 \u003d 2Mg +2 O -2
These types of reactions are redox . They will be considered separately. To formulate equations for reactions of this type, it is necessary to use half-reaction method and apply electronic balance equation.
After bringing various types of chemical reactions, you can proceed to the principle of compiling chemical equations, in other words, the selection of coefficients in their left and right parts.
Mechanisms for compiling chemical equations.
Whatever type this or that chemical reaction belongs to, its record (chemical equation) must correspond to the condition of equality of the number of atoms before the reaction and after the reaction.
There are equations (17) that do not require adjustment, 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 parts of the equation. What principles should be followed in such cases? Is there any system in the selection of coefficients? There is, and not one. These systems include:
one). Selection of coefficients according to given formulas.
2). Compilation according to the valencies of the reactants.
3). Compilation according to the oxidation states of the reactants.
In the first case, it is assumed that we know the formulas of the reactants 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 the equality between the atoms of the elements before and after the reaction is established, the equal sign (=) is not put in the equation, but is replaced by an arrow (→). Now let's get down to the actual balancing. 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). It is not necessary to equalize it by the number of nitrogen atoms, but by oxygen it is necessary to achieve equality, since two atoms participated before the reaction, and after the reaction there were three atoms. Let's make the following diagram:
before reaction after reaction
O 2 O 3
Let's define the smallest multiple between the given numbers of atoms, it will be "6".
O 2 O 3
\ 6 /
Divide this number on the left side of the oxygen equation by "2". We get the number "3", put it in 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", just put it in the equation to be solved:
N 2 + 3O 2 → 2N 2 O 3
The number of oxygen atoms in both the left and right parts of the equation became equal, respectively, 6 atoms:
But the number of nitrogen atoms in both sides of the equation will not match:
On the left side there are two atoms, on the right side there are four atoms. Therefore, in order to achieve equality, it is necessary to double the amount of nitrogen on the left side of the equation, putting the coefficient "2":
Thus, the equality for nitrogen is observed and, in general, the equation will take the form:
2N 2 + 3O 2 → 2N 2 O 3
Now in the equation, instead of an arrow, you can put an equal sign:
2N 2 + 3O 2 \u003d 2N 2 O 3 (20)
Let's take 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). It is not necessary to equalize it by the number of phosphorus atoms, but for chlorine it is necessary to achieve equality, since two atoms participated before the reaction, and after the reaction there were five atoms. Let's make the following diagram:
before reaction after reaction
Cl 2 Cl 5
Let's define 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 equation for chlorine by "2". We get the number "5", put it in the equation to be solved:
Р + 5Cl 2 → РCl 5
We also divide the number "10" for the right side of the equation by "5". We get the number "2", just put it in the equation to be solved:
Р + 5Cl 2 → 2РCl 5
The number of chlorine atoms in both the left and right parts of the equation became equal, respectively, 10 atoms:
But the number of phosphorus atoms in both sides of the equation will not match:
Therefore, in order to achieve equality, it is necessary to double the amount of phosphorus on the left side of the equation, putting the coefficient "2":
Thus, the equality for phosphorus is observed and, in general, the equation will take the form:
2Р + 5Cl 2 = 2РCl 5 (21)
When writing equations by valency must be given definition of valence and set values for the most famous elements. Valence is one of the previously used concepts, currently not used in a number of school programs. But with its help it is easier to explain the principles of compiling equations of chemical reactions. By valency is meant the number of chemical bonds that an atom can form with another, or other atoms . Valence has no 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 apply them in the preparation of chemical equations? The numerical values of the valencies of the elements coincide with their group number of the Periodic system of chemical elements of D. I. Mendeleev (Table 1).
For other elements valency values may have other values, but never greater than the number of the group in which they are located. Moreover, for even numbers of groups (IV and VI), the valences of elements take only even values, and for odd ones, they can have both even and odd values (Table.2).
Of course, there are exceptions to the valency values for some elements, but in each specific case, these points are usually specified. Now let's consider the general principle of compiling chemical equations for given valences for certain elements. Most often, this method is acceptable in the case of compiling equations for chemical reactions of combining simple substances, for example, when interacting with oxygen ( oxidation reactions). Suppose you want to display the oxidation reaction aluminum. But recall that metals are denoted by single atoms (Al), and non-metals that are in a gaseous state - with indices "2" - (O 2). First, we write the general scheme of the reaction:
Al + O 2 → AlO
At this stage, it is not yet known what the correct spelling should be for alumina. And it is precisely at this stage that knowledge of the valencies of the elements will come to our aid. For aluminum and oxygen, we put them above the proposed formula for this oxide:
III II
Al O
After that, "cross"-on-"cross" these symbols of the elements will put the corresponding indices below:
III II
Al 2 O 3
Composition of a chemical compound Al 2 O 3 determined. The further scheme of the reaction equation will take the form:
Al + O 2 →Al 2 O 3
It remains only to equalize the left and right parts of it. We proceed in the same way as in the case of formulating equation (19). We equalize the number of oxygen atoms, resorting to finding the smallest multiple:
before reaction after reaction
O 2 O 3
\ 6 /
Divide this number on the left side of the oxygen equation by "2". We get the number "3", put it in the equation to be solved. We also divide the number "6" for the right side of the equation by "3". We get the number "2", just put it in the equation to be solved:
Al + 3O 2 → 2Al 2 O 3
In order to achieve equality for aluminum, it is necessary to adjust its amount on the left side of the equation by setting the coefficient "4":
4Al + 3O 2 → 2Al 2 O 3
Thus, the equality for aluminum and oxygen is observed and, in general, the equation will take the final form:
4Al + 3O 2 \u003d 2Al 2 O 3 (22)
Using the valency method, it is possible to predict which substance is formed in the course of a chemical reaction, what its formula will look like. Suppose nitrogen and hydrogen with the corresponding valences III and I entered into the reaction of the compound. Let's write the general reaction scheme:
N 2 + H 2 → NH
For nitrogen and hydrogen, we put down the valencies over the proposed formula of this compound:
As before, "cross"-on-"cross" for these element symbols, we put the corresponding indices below:
III I
N H 3
The further scheme of the reaction equation will take the form:
N 2 + H 2 → NH 3
Equalizing in the already known way, through the smallest multiple for hydrogen, equal to "6", we obtain the desired coefficients, and the equation as a whole:
N 2 + 3H 2 \u003d 2NH 3 (23)
When compiling equations for oxidation states reacting substances, it must be recalled that the degree of oxidation of an element is the number of electrons received or given away in the process of a chemical reaction. The oxidation state in compounds basically, numerically coincides with the values of the element's valences. 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 valencies 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 commonly used in equations are shown in Table 3.
In the case of compound reactions, the principle of compiling equations in terms of oxidation states is the same as in compiling in terms of valencies. For example, let's give the reaction equation for the oxidation of chlorine with oxygen, in which chlorine forms a compound with an oxidation state of +7. Let's write the proposed equation:
Cl 2 + O 2 → ClO
We put the oxidation states of the corresponding atoms over the proposed ClO compound:
As in the previous cases, we establish that the desired 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
Equalizing for oxygen, finding the smallest multiple between two and seven, equal to "14", we finally establish the equality:
2Cl 2 + 7O 2 \u003d 2Cl 2 O 7 (24)
A slightly different method must be used with oxidation states when compiling 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 do you know what happens in a reaction?
Indeed, how do you know: what reaction products can arise in the course of a particular reaction? For example, what is formed when barium nitrate and potassium sulfate react?
Ba (NO 3) 2 + K 2 SO 4 →?
Maybe VAC 2 (NO 3) 2 + SO 4? Or Ba + NO 3 SO 4 + K 2? Or something else? Of course, during this reaction, compounds are formed: BaSO 4 and KNO 3. And how is this known? And how to write formulas of substances? Let's start with what is most often overlooked: the very concept of "exchange reaction". This means that in these reactions, the substances change with each other in constituent parts. Since the exchange reactions are mostly carried out between bases, acids or salts, the parts with which they will change 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 the metal cations (1) and (2), and An1 and An2 are the anions (1) and (2) corresponding to them. In this case, it must be taken into account that in compounds before and after the reaction, cations are always established in the first place, and anions in the second. Therefore, if it reacts potassium chloride and silver nitrate, both in solution
KCl + AgNO 3 →
then in the process of it substances KNO 3 and AgCl are formed and the corresponding equation will take the form:
KCl + AgNO 3 \u003d KNO 3 + AgCl (26)
In neutralization reactions, protons from acids (H +) will combine with hydroxyl anions (OH -) to form water (H 2 O):
HCl + KOH \u003d KCl + H 2 O (27)
The oxidation states of metal cations and the charges of anions of acid residues are indicated in the table of the solubility of substances (acids, salts and bases in water). Metal cations are shown horizontally, and anions of acid residues are shown vertically.
Based on this, when compiling the equation for the exchange reaction, it is first necessary to establish the oxidation states of the particles receiving in this chemical process in its left part. For example, you need to write an equation for the interaction between calcium chloride and sodium carbonate. Let's draw up the initial scheme for this reaction:
CaCl + NaCO 3 →
Ca 2+ Cl - + Na + CO 3 2- →
Having performed the already known “cross”-to-“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 establish the preliminary formulas of the substances formed during the reaction:
CaCl 2 + Na 2 CO 3 → CaCO 3 + NaCl
We put down the corresponding charges over their cations and anions:
Ca 2+ CO 3 2- + Na + Cl -
Substance formulas are written correctly, in accordance with the charges of cations and anions. Let's make a complete equation by equating the left and right parts of it in terms of sodium and chlorine:
CaCl 2 + Na 2 CO 3 \u003d CaCO 3 + 2NaCl (28)
As another example, here is the equation for the neutralization reaction between barium hydroxide and phosphoric acid:
VaON + NPO 4 →
We put the corresponding charges over cations and anions:
Ba 2+ OH - + H + RO 4 3- →
Let's define the real formulas of the starting materials:
Va (OH) 2 + H 3 RO 4 →
Based on the principle of exchange of cations and anions (25), we establish the preliminary formulas of the substances formed during the reaction, taking into account that in the exchange reaction, one of the substances must necessarily be water:
Ba (OH) 2 + H 3 RO 4 → Ba 2+ RO 4 3- + H 2 O
Let's determine the correct record of the formula of the salt formed during the reaction:
Ba (OH) 2 + H 3 RO 4 → Ba 3 (RO 4) 2 + H 2 O
Equate the left side of the equation for barium:
3VA (OH) 2 + H 3 RO 4 → Ba 3 (RO 4) 2 + H 2 O
Since on the right side of the equation the residue of phosphoric acid is taken twice, (PO 4) 2, then on the left it is also necessary to double its amount:
3VA (OH) 2 + 2H 3 RO 4 → Ba 3 (RO 4) 2 + H 2 O
It remains to match the number of hydrogen and oxygen atoms on the right side of the water. Since the total number of hydrogen atoms on the left is 12, on the right it must also correspond to twelve, therefore, before the water formula, it is necessary put a coefficient"6" (since there are already 2 hydrogen atoms in the water molecule). For oxygen, equality is also observed: on the left 14 and on the right 14. So, the equation has the correct form of writing:
3Ва (ОН) 2 + 2Н 3 РО 4 → Ва 3 (РО 4) 2 + 6Н 2 O (29)
Possibility of chemical reactions
The world is made up 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, assert that a chemical reaction will correspond to it? There is a misconception that if the right arrange odds in the equation, then it will be feasible in practice. For example, if we take sulfuric acid solution and drop into it zinc, then we can observe the process of hydrogen evolution:
Zn + H 2 SO 4 \u003d ZnSO 4 + H 2 (30)
But if copper is lowered 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 \u003d CuSO 4 + SO 2 + 2H 2 O (31)
In reaction (23) between nitrogen and hydrogen gases, thermodynamic balance, those. how many molecules ammonia NH 3 is formed per unit time, the same number of them will decompose back into nitrogen and hydrogen. Shift in chemical equilibrium can be achieved by increasing the pressure and decreasing the temperature
N 2 + 3H 2 \u003d 2NH 3
If you take potassium hydroxide solution and pour on it 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 attributed to a substitution reaction:
NaCl + Br 2 ≠
What are the reasons for such discrepancies? The fact is that it is not enough just to correctly define compound formulas, it is necessary to know the specifics of the interaction of metals with acids, to skillfully use the table of solubility of substances, to know the rules of substitution in the series of activity of metals and halogens. This article outlines only the most basic principles of how arrange the coefficients in the reaction equations, how write molecular equations, how determine the composition of a chemical compound.
Chemistry, as a science, is extremely diverse and multifaceted. This article reflects only a small part of the processes taking place in the real world. Types, thermochemical equations, electrolysis, organic synthesis processes and much, much more. But more on that in future articles.
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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 reagents (substances that enter into the reaction) are written, and to the right are 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 should be equal to what is on the right. Therefore, after drawing up a scheme of a chemical reaction (recording of reactants and products), the coefficients are substituted to equalize the number of atoms.
The coefficients are numbers in front of the formulas of substances, indicating the number of molecules that react.
For example, suppose that 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 are two hydrogen and oxygen atoms, and on the right are two hydrogen atoms and only one oxygen. Suppose that as a result of the reaction for one molecule of hydrogen and one oxygen, two molecules of water are formed:
H 2 + O 2 → 2H 2 O
Now the number of oxygen atoms before and after the reaction is equalized. However, hydrogen before the reaction is two times less than after. It should be concluded that for the formation of two water molecules, two molecules of hydrogen and one oxygen are needed. Then you get the following reaction scheme:
2H 2 + O 2 → 2H 2 O
Here, the number of atoms of different chemical elements is 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 equalized:
2H 2 + O 2 \u003d 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 includes three sodium atoms. Equalize the amount of sodium before the reaction:
3NaOH + H 3 PO 4 → Na 3 PO 4 + H 2 O
The amount of hydrogen before the reaction is six atoms (three in sodium hydroxide and three in phosphoric acid). After the reaction - only two hydrogen atoms. Dividing six by two gives three. So, before the water you need to put the number three:
3NaOH + H 3 PO 4 → Na 3 PO 4 + 3H 2 O
The number of oxygen atoms before and after the reaction is the same, which means that further calculation of the coefficients can be omitted.