EntranceHow to write equations in chemistry. Chemical reaction diagram
Topic: Chemical bond. Electrolytic dissociation
Lesson: Writing Equations for Ion Exchange Reactions
Let's create an equation for the reaction between iron (III) hydroxide and nitric acid.
Fe(OH) 3 + 3HNO 3 = Fe(NO 3) 3 + 3H 2 O
(Iron (III) hydroxide is an insoluble base, therefore it is not subjected to. Water is a poorly dissociated substance; it is practically not dissociated into ions in solution.)
Fe(OH) 3 + 3H + + 3NO 3 - = Fe 3+ + 3NO 3 - + 3H 2 O
Cross out the same number of nitrate anions on the left and right and write the abbreviated ionic equation:
Fe(OH) 3 + 3H + = Fe 3+ + 3H 2 O
This reaction proceeds to completion, because a slightly dissociable substance is formed - water.
Let's write an equation for the reaction between sodium carbonate and magnesium nitrate.
Na 2 CO 3 + Mg(NO 3) 2 = 2NaNO 3 + MgCO 3 ↓
Let's write this equation in ionic form:
(Magnesium carbonate is insoluble in water and therefore does not break down into ions.)
2Na + + CO 3 2- + Mg 2+ + 2NO 3 - = 2Na + + 2NO 3 - + MgCO 3 ↓
Let's cross out the same number of nitrate anions and sodium cations on the left and right, and write the abbreviated ionic equation:
CO 3 2- + Mg 2+ = MgCO 3 ↓
This reaction proceeds to completion, because a precipitate is formed - magnesium carbonate.
Let's write an equation for the reaction between sodium carbonate and nitric acid.
Na 2 CO 3 + 2HNO 3 = 2NaNO 3 + CO 2 + H 2 O
(Carbon dioxide and water are products of the decomposition of the resulting weak carbonic acid.)
2Na + + CO 3 2- + 2H + + 2NO 3 - = 2Na + + 2NO 3 - + CO 2 + H 2 O
CO 3 2- + 2H + = CO 2 + H 2 O
This reaction proceeds to completion, because As a result, gas is released and water is formed.
Let's create two molecular reaction equations, which correspond to the following abbreviated ionic equation: Ca 2+ + CO 3 2- = CaCO 3 .
The abbreviated ionic equation shows the essence of the ion exchange reaction. In this case, we can say that to obtain calcium carbonate, it is necessary that the composition of the first substance include calcium cations, and the composition of the second - carbonate anions. Let's create molecular equations for reactions that satisfy this condition:
CaCl 2 + K 2 CO 3 = CaCO 3 ↓ + 2KCl
Ca(NO 3) 2 + Na 2 CO 3 = CaCO 3 ↓ + 2NaNO 3
1. Orzhekovsky P.A. Chemistry: 9th grade: textbook. for general education establishment / P.A. Orzhekovsky, L.M. Meshcheryakova, L.S. Pontak. - M.: AST: Astrel, 2007. (§17)
2. Orzhekovsky P.A. Chemistry: 9th grade: general education. establishment / P.A. Orzhekovsky, L.M. Meshcheryakova, M.M. Shalashova. - M.: Astrel, 2013. (§9)
3. Rudzitis G.E. Chemistry: inorganic. chemistry. Organ. chemistry: textbook. for 9th grade. / G.E. Rudzitis, F.G. Feldman. - M.: Education, OJSC “Moscow Textbooks”, 2009.
4. Khomchenko I.D. Collection of problems and exercises in chemistry for high school. - M.: RIA “New Wave”: Publisher Umerenkov, 2008.
5. Encyclopedia for children. Volume 17. Chemistry / Chapter. ed. V.A. Volodin, Ved. scientific ed. I. Leenson. - M.: Avanta+, 2003.
Additional web resources
1. A unified collection of digital educational resources (video experiences on the topic): ().
2. Electronic version of the journal “Chemistry and Life”: ().
Homework
1. In the table, mark with a plus sign the pairs of substances between which ion exchange reactions are possible and proceed to completion. Write reaction equations in molecular, full and reduced ionic form.
|
Reacting substances |
K2 CO3 |
AgNO3 |
FeCl3 |
HNO3 |
|
|
CuCl2 |
|||||
2. p. 67 No. 10,13 from the textbook P.A. Orzhekovsky “Chemistry: 9th grade” / P.A. Orzhekovsky, L.M. Meshcheryakova, M.M. Shalashova. - M.: Astrel, 2013.
9.1. What are the chemical reactions?
Let us remember that we call any chemical phenomena in nature chemical reactions. During a chemical reaction, some chemical bonds are broken and others are formed. As a result of the reaction, other substances are obtained from some chemical substances (see Chapter 1).
While doing your homework for § 2.5, you became acquainted with the traditional selection of four main types of reactions from the entire set of chemical transformations, and then you also proposed their names: reactions of combination, decomposition, substitution and exchange.
Examples of compound reactions:
C + O 2 = CO 2; (1)
Na 2 O + CO 2 = Na 2 CO 3; (2)
NH 3 + CO 2 + H 2 O = NH 4 HCO 3. (3)
Examples of decomposition reactions:
2Ag 2 O 4Ag + O 2; (4)
CaCO 3 CaO + CO 2; (5)
(NH 4) 2 Cr 2 O 7 N 2 + Cr 2 O 3 + 4H 2 O. (6)
Examples of substitution reactions:
CuSO 4 + Fe = FeSO 4 + Cu; (7)
2NaI + Cl 2 = 2NaCl + I 2; (8)
CaCO 3 + SiO 2 = CaSiO 3 + CO 2. (9)
| Exchange reactions- chemical reactions in which starting substances seem to exchange their constituent parts. |
Examples of exchange reactions:
Ba(OH) 2 + H 2 SO 4 = BaSO 4 + 2H 2 O; (10)
HCl + KNO 2 = KCl + HNO 2; (eleven)
AgNO 3 + NaCl = AgCl + NaNO 3. (12)
The traditional classification of chemical reactions does not cover all their diversity - in addition to the four main types of reactions, there are also many more complex reactions.
The identification of two other types of chemical reactions is based on the participation in them of two important non-chemical particles: electron and proton.
During some reactions, complete or partial transfer of electrons from one atom to another occurs. In this case, the oxidation states of the atoms of the elements that make up the starting substances change; of the examples given, these are reactions 1, 4, 6, 7 and 8. These reactions are called redox.
In another group of reactions, a hydrogen ion (H +), that is, a proton, passes from one reacting particle to another. Such reactions are called acid-base reactions or proton transfer reactions.
Among the examples given, such reactions are reactions 3, 10 and 11. By analogy with these reactions, redox reactions are sometimes called electron transfer reactions. You will become acquainted with OVR in § 2, and with KOR in the following chapters.
COMPOUNDING REACTIONS, DECOMPOSITION REACTIONS, SUBSTITUTION REACTIONS, EXCHANGE REACTIONS, REDOX REACTIONS, ACID-BASE REACTIONS.
Write down reaction equations corresponding to the following schemes:
a) HgO Hg + O 2 ( t); b) Li 2 O + SO 2 Li 2 SO 3; c) Cu(OH) 2 CuO + H 2 O ( t);
d) Al + I 2 AlI 3; e) CuCl 2 + Fe FeCl 2 + Cu; e) Mg + H 3 PO 4 Mg 3 (PO 4) 2 + H 2 ;
g) Al + O 2 Al 2 O 3 ( t); i) KClO 3 + P P 2 O 5 + KCl ( t); j) CuSO 4 + Al Al 2 (SO 4) 3 + Cu;
l) Fe + Cl 2 FeCl 3 ( t); m) NH 3 + O 2 N 2 + H 2 O ( t); m) H 2 SO 4 + CuO CuSO 4 + H 2 O.
Indicate the traditional type of reaction. Label redox and acid-base reactions. In redox reactions, indicate which atoms of elements change their oxidation states.
9.2. Redox reactions
Let's consider the redox reaction that occurs in blast furnaces during the industrial production of iron (more precisely, cast iron) from iron ore:
Fe 2 O 3 + 3CO = 2Fe + 3CO 2.
Let us determine the oxidation states of the atoms that make up both the starting substances and the reaction products
| Fe2O3 | + | = | 2Fe | + |
As you can see, the oxidation state of carbon atoms increased as a result of the reaction, the oxidation state of iron atoms decreased, and the oxidation state of oxygen atoms remained unchanged. Consequently, the carbon atoms in this reaction underwent oxidation, that is, they lost electrons ( oxidized), and the iron atoms – reduction, that is, they added electrons ( recovered) (see § 7.16). To characterize OVR, the concepts are used oxidizer And reducing agent.
Thus, in our reaction the oxidizing atoms are iron atoms, and the reducing atoms are carbon atoms.
In our reaction, the oxidizing agent is iron(III) oxide, and the reducing agent is carbon(II) monoxide.
In cases where oxidizing atoms and reducing atoms are part of the same substance (example: reaction 6 from the previous paragraph), the concepts of “oxidizing substance” and “reducing substance” are not used.
Thus, typical oxidizing agents are substances that contain atoms that tend to gain electrons (in whole or in part), lowering their oxidation state. Of the simple substances, these are primarily halogens and oxygen, and to a lesser extent sulfur and nitrogen. From complex substances - substances that contain atoms in higher oxidation states that are not inclined to form simple ions in these oxidation states: HNO 3 (N +V), KMnO 4 (Mn +VII), CrO 3 (Cr +VI), KClO 3 (Cl +V), KClO 4 (Cl +VII), etc.
Typical reducing agents are substances that contain atoms that tend to completely or partially donate electrons, increasing their oxidation state. Simple substances include hydrogen, alkali and alkaline earth metals, and aluminum. Of the complex substances - H 2 S and sulfides (S –II), SO 2 and sulfites (S +IV), iodides (I –I), CO (C +II), NH 3 (N –III), etc.
In general, almost all complex and many simple substances can exhibit both oxidizing and reducing properties. For example:
SO 2 + Cl 2 = S + Cl 2 O 2 (SO 2 is a strong reducing agent);
SO 2 + C = S + CO 2 (t) (SO 2 is a weak oxidizing agent);
C + O 2 = CO 2 (t) (C is a reducing agent);
C + 2Ca = Ca 2 C (t) (C is an oxidizing agent).
Let's return to the reaction we discussed at the beginning of this section.
| Fe2O3 | + | = | 2Fe | + |
Please note that as a result of the reaction, oxidizing atoms (Fe + III) turned into reducing atoms (Fe 0), and reducing atoms (C + II) turned into oxidizing atoms (C + IV). But CO 2 is a very weak oxidizing agent under any conditions, and iron, although it is a reducing agent, is under these conditions much weaker than CO. Therefore, the reaction products do not react with each other, and the reverse reaction does not occur. The given example is an illustration of the general principle that determines the direction of the flow of the OVR:
Redox reactions proceed in the direction of the formation of a weaker oxidizing agent and a weaker reducing agent.
The redox properties of substances can only be compared under identical conditions. In some cases, this comparison can be made quantitatively.
While doing your homework for the first paragraph of this chapter, you became convinced that it is quite difficult to select coefficients in some reaction equations (especially ORR). To simplify this task in the case of redox reactions, the following two methods are used:
A) electronic balance method And
b) electron-ion balance method.
You will learn the electron balance method now, and the electron-ion balance method is usually studied in higher education institutions.
Both of these methods are based on the fact that electrons in chemical reactions neither disappear nor appear anywhere, that is, the number of electrons accepted by atoms is equal to the number of electrons given up by other atoms.
The number of given and accepted electrons in the electron balance method is determined by the change in the oxidation state of atoms. When using this method, it is necessary to know the composition of both the starting substances and the reaction products.
Let's look at the application of the electronic balance method using examples.
Example 1. Let's create an equation for the reaction of iron with chlorine. It is known that the product of this reaction is iron(III) chloride. Let's write down the reaction scheme:
Fe + Cl 2 FeCl 3 .
Let us determine the oxidation states of the atoms of all elements that make up the substances participating in the reaction:
Iron atoms donate electrons, and chlorine molecules accept them. Let us express these processes electronic equations:
Fe – 3 e– = Fe +III,
Cl2+2 e –= 2Cl –I.
In order for the number of electrons given to be equal to the number of electrons received, the first electronic equation must be multiplied by two, and the second by three:
| Fe – 3 e– = Fe +III, Cl2+2 e– = 2Cl –I |
2Fe – 6 e– = 2Fe +III, 3Cl 2 + 6 e– = 6Cl –I. |
By introducing coefficients 2 and 3 into the reaction scheme, we obtain the reaction equation:
2Fe + 3Cl 2 = 2FeCl 3.
Example 2. Let's create an equation for the combustion reaction of white phosphorus in excess chlorine. It is known that phosphorus(V) chloride is formed under these conditions:
| +V –I | ||||
| P 4 | + | Cl2 | PCl 5. |
White phosphorus molecules give up electrons (oxidize), and chlorine molecules accept them (reduce):
| P 4 – 20 e– = 4P +V Cl2+2 e– = 2Cl –I |
1 10 |
2 20 |
P 4 – 20 e– = 4P +V Cl2+2 e– = 2Cl –I |
P 4 – 20 e– = 4P +V 10Cl 2 + 20 e– = 20Cl –I |
The initially obtained factors (2 and 20) had a common divisor, by which (like future coefficients in the reaction equation) they were divided. Reaction equation:
P4 + 10Cl2 = 4PCl5.
Example 3. Let's create an equation for the reaction that occurs when iron(II) sulfide is roasted in oxygen.
Reaction scheme:
| +III –II | +IV –II | |||||
| + | O2 | + |
In this case, both iron(II) and sulfur(–II) atoms are oxidized. The composition of iron(II) sulfide contains atoms of these elements in a 1:1 ratio (see the indices in the simplest formula).
Electronic balance:
| 4 | Fe+II – e– = Fe +III S–II–6 e– = S +IV |
In total they give 7 e – |
| 7 | O 2 + 4e – = 2O –II |
Reaction equation: 4FeS + 7O 2 = 2Fe 2 O 3 + 4SO 2.
Example 4. Let's create an equation for the reaction that occurs when iron(II) disulfide (pyrite) is roasted in oxygen.
Reaction scheme:
| +III –II | +IV –II | |||||
| + | O2 | + |
As in the previous example, both iron(II) atoms and sulfur atoms are also oxidized here, but with an oxidation state of I. The atoms of these elements are included in the composition of pyrite in a ratio of 1:2 (see the indices in the simplest formula). It is in this regard that the iron and sulfur atoms react, which is taken into account when compiling the electronic balance:
| Fe+III – e– = Fe +III 2S–I – 10 e– = 2S +IV |
In total they give 11 e – | |
| O2+4 e– = 2O –II |
Reaction equation: 4FeS 2 + 11O 2 = 2Fe 2 O 3 + 8SO 2.
There are also more complex cases of ODD, some of which you will become familiar with while doing your homework.
OXIDIZING ATOM, REDUCING ATOM, OXIDIZING SUBSTANCE, REDUCING SUBSTANCE, ELECTRONIC BALANCE METHOD, ELECTRONIC EQUATIONS.
1. Compile an electronic balance for each OVR equation given in the text of § 1 of this chapter.
2. Make up equations for the ORRs that you discovered while completing the task for § 1 of this chapter. This time, use the electronic balance method to set the odds. 3.Using the electron balance method, create reaction equations corresponding to the following schemes: a) Na + I 2 NaI;
b) Na + O 2 Na 2 O 2;
c) Na 2 O 2 + Na Na 2 O;
d) Al + Br 2 AlBr 3;
e) Fe + O 2 Fe 3 O 4 ( t);
e) Fe 3 O 4 + H 2 FeO + H 2 O ( t);
g) FeO + O 2 Fe 2 O 3 ( t);
i) Fe 2 O 3 + CO Fe + CO 2 ( t);
j) Cr + O 2 Cr 2 O 3 ( t);
l) CrO 3 + NH 3 Cr 2 O 3 + H 2 O + N 2 ( t);
m) Mn 2 O 7 + NH 3 MnO 2 + N 2 + H 2 O;
m) MnO 2 + H 2 Mn + H 2 O ( t);
n) MnS + O 2 MnO 2 + SO 2 ( t)
p) PbO 2 + CO Pb + CO 2 ( t);
c) Cu 2 O + Cu 2 S Cu + SO 2 ( t);
t) CuS + O 2 Cu 2 O +SO 2 ( t);
y) Pb 3 O 4 + H 2 Pb + H 2 O ( t).
9.3. Exothermic reactions. Enthalpy
Why do chemical reactions occur?
To answer this question, let us remember why individual atoms combine into molecules, why an ionic crystal is formed from isolated ions, and why the principle of least energy applies when the electron shell of an atom is formed. The answer to all these questions is the same: because it is energetically beneficial. This means that during such processes energy is released. It would seem that chemical reactions should occur for the same reason. Indeed, many reactions can be carried out, during which energy is released. Energy is released, usually in the form of heat.
If during an exothermic reaction the heat does not have time to be removed, then the reaction system heats up.
For example, in the methane combustion reaction
CH 4 (g) + 2O 2 (g) = CO 2 (g) + 2H 2 O (g)
so much heat is released that methane is used as fuel.
The fact that this reaction releases heat can be reflected in the reaction equation:
CH 4 (g) + 2O 2 (g) = CO 2 (g) + 2H 2 O (g) + Q.
This is the so called thermochemical equation. Here the symbol "+ Q" means that when methane is burned, heat is released. This heat is called thermal effect of reaction.
Where does the released heat come from?
You know that during chemical reactions chemical bonds are broken and formed. In this case, the bonds between carbon and hydrogen atoms in CH 4 molecules, as well as between oxygen atoms in O 2 molecules, are broken. In this case, new bonds are formed: between carbon and oxygen atoms in CO 2 molecules and between oxygen and hydrogen atoms in H 2 O molecules. To break bonds, you need to expend energy (see “bond energy”, “atomization energy”), and when forming bonds, energy is released. Obviously, if the “new” bonds are stronger than the “old” ones, then more energy will be released than absorbed. The difference between the released and absorbed energy is the thermal effect of the reaction.
Thermal effect (amount of heat) is measured in kilojoules, for example:
2H 2 (g) + O 2 (g) = 2H 2 O (g) + 484 kJ.
This notation means that 484 kilojoules of heat will be released if two moles of hydrogen react with one mole of oxygen to produce two moles of gaseous water (water vapor).
Thus, in thermochemical equations, the coefficients are numerically equal to the amounts of substance of the reactants and reaction products.
What determines the thermal effect of each specific reaction?
The thermal effect of the reaction depends
a) on the aggregative states of the starting substances and reaction products,
b) on temperature and
c) on whether the chemical transformation occurs at constant volume or at constant pressure.
The dependence of the thermal effect of a reaction on the state of aggregation of substances is due to the fact that the processes of transition from one state of aggregation to another (like some other physical processes) are accompanied by the release or absorption of heat. This can also be expressed by a thermochemical equation. Example – thermochemical equation for condensation of water vapor:
H 2 O (g) = H 2 O (l) + Q.
In thermochemical equations, and, if necessary, in ordinary chemical equations, the aggregative states of substances are indicated using letter indices:
(d) – gas,
(g) – liquid,
(t) or (cr) – solid or crystalline substance.
The dependence of the thermal effect on temperature is associated with differences in heat capacities
starting materials and reaction products.
Since the volume of the system always increases as a result of an exothermic reaction at constant pressure, part of the energy is spent on doing work to increase the volume, and the heat released will be less than if the same reaction occurs at a constant volume.
Thermal effects of reactions are usually calculated for reactions occurring at constant volume at 25 °C and are indicated by the symbol Q o.
If energy is released only in the form of heat, and a chemical reaction proceeds at a constant volume, then the thermal effect of the reaction ( Q V) is equal to the change internal energy(D U) substances participating in the reaction, but with the opposite sign:
Q V = – U.
The internal energy of a body is understood as the total energy of intermolecular interactions, chemical bonds, the ionization energy of all electrons, the bond energy of nucleons in nuclei, and all other known and unknown types of energy “stored” by this body. The “–” sign is due to the fact that when heat is released, the internal energy decreases. That is
U= – Q V .
If the reaction occurs at constant pressure, then the volume of the system can change. Doing work to increase the volume also takes part of the internal energy. In this case
U = –(QP+A) = –(QP+PV),
Where Qp– the thermal effect of a reaction occurring at constant pressure. From here
Q P = – U–PV .
A value equal to U+PV got the name enthalpy change and denoted by D H.
H=U+PV.
Hence
Q P = – H.
Thus, as heat is released, the enthalpy of the system decreases. Hence the old name for this quantity: “heat content”.
Unlike the thermal effect, a change in enthalpy characterizes a reaction regardless of whether it occurs at constant volume or constant pressure. Thermochemical equations written using enthalpy change are called thermochemical equations in thermodynamic form. In this case, the value of the enthalpy change under standard conditions (25 °C, 101.3 kPa) is given, denoted H o. For example:
2H 2 (g) + O 2 (g) = 2H 2 O (g) H o= – 484 kJ;
CaO (cr) + H 2 O (l) = Ca(OH) 2 (cr) H o= – 65 kJ.
Dependence of the amount of heat released in the reaction ( Q) from the thermal effect of the reaction ( Q o) and the amount of substance ( n B) one of the participants in the reaction (substance B - the starting substance or reaction product) is expressed by the equation:
Here B is the amount of substance B, specified by the coefficient in front of the formula of substance B in the thermochemical equation.
Task
Determine the amount of hydrogen substance burned in oxygen if 1694 kJ of heat was released.
Solution
2H 2 (g) + O 2 (g) = 2H 2 O (g) + 484 kJ. |
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Q = 1694 kJ, 6. The thermal effect of the reaction between crystalline aluminum and gaseous chlorine is 1408 kJ. Write the thermochemical equation for this reaction and determine the mass of aluminum required to produce 2816 kJ of heat using this reaction. 9.4. Endothermic reactions. Entropy In addition to exothermic reactions, reactions are possible in which heat is absorbed, and if it is not supplied, the reaction system is cooled. Such reactions are called endothermic. The thermal effect of such reactions is negative. For example: Thus, the energy released during the formation of bonds in the products of these and similar reactions is less than the energy required to break bonds in the starting substances.
Let's take two flasks and fill one of them with nitrogen (colorless gas) and the other with nitrogen dioxide (brown gas) so that both the pressure and temperature in the flasks are the same. It is known that these substances do not react chemically with each other. Let's tightly connect the flasks with their necks and install them vertically, so that the flask with heavier nitrogen dioxide is at the bottom (Fig. 9.1). After some time, we will see that brown nitrogen dioxide gradually spreads into the upper flask, and colorless nitrogen penetrates into the lower one. As a result, the gases mix, and the color of the contents of the flasks becomes the same. Thus,
Equations of connection between entropy ( S) and other quantities are studied in physics and physical chemistry courses. Entropy unit [ S] = 1 J/K. G= H–T S Condition for spontaneous reaction: G< 0. At low temperatures, the factor determining the possibility of a reaction occurring is largely the energy factor, and at high temperatures it is the entropy factor. From the above equation, in particular, it is clear why decomposition reactions that do not occur at room temperature (entropy increases) begin to occur at elevated temperatures. ENDOTHERMIC REACTION, ENTROPY, ENERGY FACTOR, ENTROPY FACTOR, GIBBS ENERGY. 2CuO (cr) + C (graphite) = 2Cu (cr) + CO 2 (g) is –46 kJ. Write down the thermochemical equation and calculate how much energy is needed to produce 1 kg of copper from this reaction. CaCO 3 (cr) = CaO (cr) + CO 2 (g) – 179 kJ 24.6 liters of carbon dioxide were formed. Determine how much heat was wasted uselessly. How many grams of calcium oxide were formed? |
Redox reactions are the process of “flowing” electrons from one atom to another. As a result, oxidation or reduction of the chemical elements that make up the reagents occurs.
Basic Concepts
The key term when considering redox reactions is the oxidation state, which represents the nominal charge of the atom and the number of electrons redistributed. Oxidation is the process of losing electrons, which increases the charge of an atom. Reduction, on the other hand, is a process of electron gain in which the oxidation state decreases. Accordingly, the oxidizing agent accepts new electrons, and the reducing agent loses them, and such reactions always occur simultaneously.
Determination of oxidation state
Calculating this parameter is one of the most popular tasks in a school chemistry course. Finding the charges of atoms can be either an elementary question or a task requiring scrupulous calculations: it all depends on the complexity of the chemical reaction and the number of constituent compounds. I would like the oxidation states to be indicated in the periodic table and always at hand, but this parameter must either be remembered or calculated for a specific reaction. So, there are two unambiguous properties:
- The sum of the charges of a complex compound is always zero. This means that some atoms will have a positive degree, and some will have a negative degree.
- The oxidation state of elementary compounds is always zero. Simple compounds are those that consist of atoms of one element, that is, iron Fe2, oxygen O2 or octasulfur S8.
There are chemical elements whose electric charge is unambiguous in any connection. These include:
- -1 - F;
- -2 - O;
- +1 - H, Li, Ag, Na, K;
- +2 - Ba, Ca, Mg, Zn;
- +3 - Al.
Although clear, there are some exceptions. Fluorine F is a unique element whose oxidation state is always -1. Thanks to this property, many elements change their charge when paired with fluorine. For example, oxygen in combination with fluorine has a charge of +1 (O 2 F 2) or +2 (OF2). In addition, oxygen changes its degree in peroxide compounds (in hydrogen peroxide H202 the charge is -1). And, of course, oxygen has zero degree in its simple compound O2.
When considering redox reactions, it is important to consider substances that are made up of ions. Atoms of ionic chemical elements have an oxidation state equal to the charge of the ion. For example, in the sodium hydride compound NaH, the hydrogen is supposed to have a charge of +1, but the sodium ion also has a charge of +1. Since the compound must be electrically neutral, the hydrogen atom takes on a -1 charge. Metal ions stand out separately in this situation, since the atoms of such elements are ionized to different amounts. For example, iron F ionizes at both +2 and +3 depending on the composition of the chemical substance.
Example of determining oxidation states
For simple compounds that involve atoms with unambiguous charges, the distribution of oxidation states is not difficult. For example, for water H2O, the oxygen atom has a charge of -2 and the hydrogen atom has a charge of +1, which adds up to a neutral zero. In more complex compounds, there are atoms that may have different charges and the method of exclusion must be used to determine oxidation states. Let's look at an example.
Sodium sulfate Na 2 SO 4 contains a sulfur atom, the charge of which can take values of -2, +4 or +6. Which value should I choose? First of all, we determine that the sodium ion has a charge of +1. Oxygen in the vast majority of cases has a charge of –2. Let's make a simple equation:
1 × 2 + S + (–2) × 4 = 0
Thus, the charge of sulfur in sodium sulfate is +6.
Arrangement of coefficients according to the reaction scheme
Now that you know how to determine the charges of atoms, you can assign coefficients to redox reactions to balance them. Standard chemistry task: select reaction coefficients using the electron balance method. In these tasks, you do not need to determine what substances are formed at the end of the reaction, since the result is already known. For example, determine the proportions in a simple reaction:
Na + O2 → Na 2 O
So, let's determine the charge of the atoms. Since sodium and oxygen on the left side of the equation are simple substances, their charge is zero. In sodium oxide Na2O, oxygen has a charge of -2, and sodium has a charge of +1. We see that on the left side of the equation sodium has a zero charge, and on the right side it has a positive +1 charge. Same thing with oxygen, which changed its oxidation number from zero to -2. Let’s write this in “chemical” language, indicating the charges of the elements in parentheses:
Na(0) – 1e = Na(+1)
O(0) + 2e = O(–2)
To balance the reaction, you need to balance the oxygen and add a factor of 2 to the sodium oxide. We get the reaction:
Na + O2 → 2Na2O
Now we have an imbalance in sodium, let's balance it using a factor of 4:
4Na + O2 → 2Na2O
Now the number of atoms of the elements are the same on both sides of the equation, therefore the reaction is balanced. We did all this manually, and it was not difficult, since the reaction itself is elementary. But what if you need to balance the reaction of the form K 2 Cr 2 O 7 + KI + H 2 SO 4 → Cr 2 (SO 4)3 + I2 + H 2 O + K 2 SO 4? The answer is simple: use a calculator.
Redox Reaction Balancing Calculator
Our program allows you to automatically set coefficients for the most common chemical reactions. To do this, you need to enter a reaction in the program field or select it from the drop-down list. To solve the redox reaction presented above, you just need to select it from the list and click on the “Calculate” button. The calculator will instantly give the result:
K 2 Cr 2 O 7 + 6KI + 7H 2 SO 4 → Cr 2 (SO 4)3 + 3I2 + 7H 2 O + 4K 2 SO 4
Using a calculator will help you quickly balance the most complex chemical reactions.
Conclusion
The ability to balance reactions is necessary for all schoolchildren and students who dream of connecting their lives with chemistry. In general, calculations are performed according to strictly defined rules, for understanding which elementary knowledge of chemistry and algebra is sufficient: remember that the sum of the oxidation states of the atoms of a compound is always equal to zero and be able to solve linear equations.
To characterize a certain chemical reaction, you must be able to create a record that will display the conditions for the chemical reaction, show which substances reacted and which were formed. To do this, chemical reaction schemes are used.
Chemical reaction diagram– a conditional record showing which substances react, what reaction products are formed, as well as the conditions for the reaction. Let us consider, as an example, the reaction between coal and oxygen. Scheme this reaction is written as follows:
C + O2 → CO2
Coal reacts with oxygen to form carbon dioxide
Carbon and oxygen- in this reaction there are reactants, and the resulting carbon dioxide is the product of the reaction. Sign " → " indicates the progress of the reaction. Often the conditions under which the reaction occurs are written above the arrow.
- Sign « t° → » indicates that the reaction occurs when heated.
- Sign "R →" stands for pressure
- Sign "hv →"– that the reaction occurs under the influence of light. Additional substances involved in the reaction may also be indicated above the arrow.
- For example, "O2 →". If a gaseous substance is formed as a result of a chemical reaction, then in the reaction scheme, after the formula of this substance, write the sign “ → " If a precipitate is formed during the reaction, it is indicated by the sign “ → ».
- For example, when chalk powder is heated (it contains a substance with the chemical formula CaCO3), two substances are formed: quicklime CaO and carbon dioxide. The reaction scheme is written as follows:
СaCO3 t° → CaO + CO2
Thus, natural gas mainly consists of CH4 methane; when heated to 1500°C, it turns into two other gases: hydrogen H2 and acetylene C2H2. The reaction scheme is written as follows:
CH4 t° → C2H2 + H2.
It is important not only to be able to draw up diagrams of chemical reactions, but also to understand what they mean. Let's consider another reaction scheme:
H2O electric current → H2 + O2
This diagram means that under the influence of electric current, water decomposes into two simple gaseous substances: hydrogen and oxygen. The diagram of a chemical reaction is a confirmation of the law of conservation of mass and shows that chemical elements do not disappear during a chemical reaction, but are only rearranged into new chemical compounds.
Chemical Reaction Equations
According to the law of conservation of mass, the initial mass of products is always equal to the mass of the resulting reactants. The number of atoms of elements before and after the reaction is always the same; the atoms only rearrange and form new substances. Let's return to the reaction schemes recorded earlier:
СaCO3 t° → CaO + CO2
C + O2 CO2.
In these reaction schemes the sign “ → " can be replaced with the "=" sign, since it is clear that the number of atoms before and after the reactions is the same. The entries will look like this:
CaCO3 = CaO + CO2
C + O2 = CO2.
It is these records that are called equations of chemical reactions, that is, these are records of reaction schemes in which the number of atoms before and after the reaction is the same.
Chemical reaction equation– a conventional notation of a chemical reaction using chemical formulas, which corresponds to the law of conservation of mass of a substance
If we look at the other equation schemes given earlier, we can see that in At first glance, the law of conservation of mass does not hold true in them:
CH4 t° → C2H2 + H2.
It can be seen that on the left side of the diagram there is one carbon atom, and on the right there are two. There are equal numbers of hydrogen atoms and there are four of them on the left and right sides. Let's turn this diagram into an equation. For this it is necessary equalize number of carbon atoms. Chemical reactions are equalized using coefficients that are written before the formulas of substances. Obviously, in order for the number of carbon atoms to become the same on the left and right, on the left side of the diagram, before the methane formula, it is necessary to put coefficient 2:
2CH4 t° → C2H2 + H2
It can be seen that there are now equal numbers of carbon atoms on the left and right, two each. But now the number of hydrogen atoms is not the same. On the left side of the equation their 2∙4 = 8. On the right side of the equation there are 4 hydrogen atoms (two of them in the acetylene molecule, and two more in the hydrogen molecule). If you put a coefficient in front of acetylene, the equality of carbon atoms will be violated. Let's put a factor of 3 in front of the hydrogen molecule:
2CH4 = C2H2 + 3H2
Now the number of carbon and hydrogen atoms on both sides of the equation is the same. The law of conservation of mass is fulfilled! Let's look at another example. Reaction scheme Na + H2O → NaOH + H2 needs to be turned into an equation. In this scheme, the number of hydrogen atoms is different. On the left side there are two, and on the right side - three atoms. Let's put a factor of 2 in front of NaOH.
Na + H2O → 2NaOH + H2
Then there will be four hydrogen atoms on the right side, therefore, coefficient 2 must be added before the water formula:
Na + 2H2O → 2NaOH + H2
Let's equalize the number of sodium atoms:
2Na + 2H2O = 2NaOH + H2
Now the number of all atoms before and after the reaction is the same. Thus, we can conclude: To turn a chemical reaction diagram into a chemical reaction equation, it is necessary to equalize the number of all atoms that make up the reactants and reaction products using coefficients. Coefficients are placed before the formulas of substances. Let's summarize the equations of chemical reactions
- A chemical reaction diagram is a conventional notation showing which substances react, what reaction products are formed, as well as the conditions for the reaction to occur
- In reaction schemes, symbols are used that indicate the peculiarities of their occurrence.
- The equation of a chemical reaction is a conventional representation of a chemical reaction using chemical formulas, which corresponds to the law of conservation of mass of a substance
- A chemical reaction diagram is converted into an equation by placing coefficients in front of the formulas of substances
Class: 8
Presentation for the lesson
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The purpose of the lesson: help students develop knowledge of a chemical equation as a conditional recording of a chemical reaction using chemical formulas.
Tasks:
Educational:
- systematize previously studied material;
- teach the ability to compose equations of chemical reactions.
Educational:
- develop communication skills (work in pairs, ability to listen and hear).
Educational:
- develop educational and organizational skills aimed at accomplishing the task;
- develop analytical thinking skills.
Lesson type: combined.
Equipment: computer, multimedia projector, screen, assessment sheets, reflection card, “set of chemical symbols”, notebook with printed base, reagents: sodium hydroxide, iron(III) chloride, alcohol lamp, holder, matches, Whatman paper, multi-colored chemical symbols.
Lesson presentation (Appendix 3)
Lesson structure.
I. Organizing time.
II. Updating knowledge and skills.
III. Motivation and goal setting.
IV. Learning new material:
4.1 combustion reaction of aluminum in oxygen;
4.2 decomposition reaction of iron (III) hydroxide;
4.3 algorithm for arranging coefficients;
4.4 minutes of relaxation;
4.5 set the coefficients;
V. Consolidation of acquired knowledge.
VI. Summing up the lesson and grading.
VII. Homework.
VIII. Final words from the teacher.
During the classes
Chemical nature of a complex particle
determined by the nature of elementary
components,
their number and
chemical structure.
D.I.Mendeleev
Teacher. Hello guys. Sit down.
Please note: you have a printed notebook on your desk. (Appendix 2), in which you will work today, and a score sheet in which you will record your achievements, sign it.
Updating knowledge and skills.
Teacher. We got acquainted with physical and chemical phenomena, chemical reactions and signs of their occurrence. We studied the law of conservation of mass of substances.
Let's test your knowledge. I suggest you open your printed notebooks and complete task 1. You are given 5 minutes to complete the task.
Test on the topic “Physical and chemical phenomena. Law of conservation of mass of substances.”
1. How do chemical reactions differ from physical phenomena?
- Change in shape and state of aggregation of a substance.
- Formation of new substances.
- Change of location.
2. What are the signs of a chemical reaction?
- Precipitate formation, color change, gas evolution.
3. In accordance with what law are equations of chemical reactions drawn up?
- The law of constancy of the composition of matter.
- Law of conservation of mass of matter.
- Periodic law.
- Law of dynamics.
- The law of universal gravitation.
4. The law of conservation of mass of matter discovered:
- DI. Mendeleev.
- C. Darwin.
- M.V. Lomonosov.
- I. Newton.
- A.I. Butlerov.
5. A chemical equation is called:
- Conventional notation of a chemical reaction.
Teacher. You've done the job. I suggest you check it out. Exchange notebooks and check each other. Attention to the screen. For each correct answer - 1 point. Enter the total number of points on the evaluation sheets.
Motivation and goal setting.
Teacher. Using this knowledge, today we will draw up equations of chemical reactions, revealing the problem “Is the law of conservation of mass of substances the basis for drawing up equations of chemical reactions”
Learning new material.
Teacher. We are accustomed to thinking that an equation is a mathematical example where there is an unknown, and this unknown needs to be calculated. But in chemical equations there is usually nothing unknown: everything is simply written down in them using formulas: which substances react and which are obtained during this reaction. Let's see the experience.
(Reaction of sulfur and iron compound.) Appendix 3
Teacher. From the point of view of the mass of substances, the reaction equation for the compound of iron and sulfur is understood as follows
Iron + sulfur → iron (II) sulfide (task 2 tpo)
But in chemistry, words are reflected by chemical signs. Write this equation using chemical symbols.
Fe + S → FeS
(One student writes on the board, the rest in TVET.)
Teacher. Now read it.
Students. An iron molecule interacts with a sulfur molecule to produce one molecule of iron (II) sulfide.
Teacher. In this reaction, we see that the amount of starting substances is equal to the amount of substances in the reaction product.
We must always remember that when composing reaction equations, not a single atom should be lost or unexpectedly appear. Therefore, sometimes, having written all the formulas in the reaction equation, you have to equalize the number of atoms in each part of the equation - set the coefficients. Let's see another experiment
(Combustion of aluminum in oxygen.) Appendix 4
Teacher. Let's write the equation of a chemical reaction (task 3 in TPO)
Al + O 2 → Al +3 O -2
To write the oxide formula correctly, remember that
Students. Oxygen in oxides has an oxidation state of -2, aluminum is a chemical element with a constant oxidation state of +3. LCM = 6
Al + O 2 → Al 2 O 3
Teacher. We see that 1 aluminum atom enters into the reaction, two aluminum atoms are formed. Two oxygen atoms enter, three oxygen atoms are formed.
Simple and beautiful, but disrespectful to the law of conservation of mass of substances - it is different before and after the reaction.
Therefore, we need to arrange the coefficients in this chemical reaction equation. To do this, let's find the LCM for oxygen.
Students. LCM = 6
Teacher. We put coefficients in front of the formulas for oxygen and aluminum oxide so that the number of oxygen atoms on the left and right is equal to 6.
Al + 3 O 2 → 2 Al 2 O 3
Teacher. Now we find that as a result of the reaction, four aluminum atoms are formed. Therefore, in front of the aluminum atom on the left side we put a coefficient of 4
Al + 3O 2 → 2Al 2 O 3Let us once again count all the atoms before and after the reaction. We bet equal.
4Al + 3O 2 _ = 2 Al 2 O 3
Teacher. Let's look at another example
(The teacher demonstrates an experiment on the decomposition of iron (III) hydroxide.)
Fe(OH) 3 → Fe 2 O 3 + H 2 O
Teacher. Let's arrange the coefficients. One iron atom reacts and two iron atoms are formed. Therefore, before the formula of iron hydroxide (3) we put a coefficient of 2.
Fe(OH) 3 → Fe 2 O 3 + H 2 OTeacher. We find that 6 hydrogen atoms enter into the reaction (2x3), 2 hydrogen atoms are formed.
Students. NOC =6. 6/2 = 3. Therefore, we set the coefficient of 3 for the water formula
2Fe(OH) 3 → Fe 2 O 3 + 3 H 2 O
Teacher. We count oxygen.
Students. Left – 2x3 =6; on the right – 3+3 = 6
Students. The number of oxygen atoms that entered into the reaction is equal to the number of oxygen atoms formed during the reaction. You can bet equally.
2Fe(OH) 3 = Fe 2 O 3 +3 H 2 O
Teacher. Now let's summarize everything that was said earlier and get acquainted with the algorithm for arranging coefficients in the equations of chemical reactions.
- Count the number of atoms of each element on the right and left sides of the chemical reaction equation.
- Determine which element has a changing number of atoms and find the LCM.
- Divide the NOC into indices to obtain coefficients. Place them before the formulas.
- Recalculate the number of atoms and repeat the action if necessary.
- The last thing to check is the number of oxygen atoms.
Teacher. You've worked hard and you're probably tired. I suggest you relax, close your eyes and remember some pleasant moments in life. They are different for each of you. Now open your eyes and make circular movements with them, first clockwise, then counterclockwise. Now move your eyes intensively horizontally: right - left, and vertically: up - down.
Now let’s activate our mental activity and massage our earlobes.
Teacher. We continue to work.
In printed notebooks we will complete task 5. You will work in pairs. You need to place the coefficients in the equations of chemical reactions. You are given 10 minutes to complete the task.
- P + Cl 2 →PCl 5
- Na + S → Na 2 S
- HCl + Mg →MgCl 2 + H 2
- N 2 + H 2 →NH 3
- H 2 O → H 2 + O 2
Teacher. Let's check the completion of the task ( the teacher questions and displays the correct answers on the slide). For each correctly set coefficient - 1 point.
You completed the task. Well done!
Teacher. Now let's get back to our problem.
Guys, what do you think, is the law of conservation of mass of substances the basis for drawing up equations of chemical reactions?
Students. Yes, during the lesson we proved that the law of conservation of mass of substances is the basis for drawing up equations of chemical reactions.
Consolidation of knowledge.
Teacher. We have studied all the main issues. Now let's do a short test that will allow you to see how you have mastered the topic. You should only answer “yes” or “no”. You have 3 minutes to work.
Statements.
- In the reaction Ca + Cl 2 → CaCl 2, coefficients are not needed.(Yes)
- In the reaction Zn + HCl → ZnCl 2 + H 2, the coefficient for zinc is 2. (No)
- In the reaction Ca + O 2 → CaO, the coefficient for calcium oxide is 2.(Yes)
- In the reaction CH 4 → C + H 2 no coefficients are needed.(No)
- In the reaction CuO + H 2 → Cu + H 2 O, the coefficient for copper is 2. (No)
- In the reaction C + O 2 → CO, a coefficient of 2 must be assigned to both carbon monoxide (II) and carbon. (Yes)
- In the reaction CuCl 2 + Fe → Cu + FeCl 2 no coefficients are needed.(Yes)
Teacher. Let's check the progress of the work. For each correct answer - 1 point.
Lesson summary.
Teacher. You did a good job. Now calculate the total number of points scored for the lesson and give yourself a grade according to the rating that you see on the screen. Give me your evaluation sheets so that you can enter your grade into the journal.
Homework.
Teacher. Our lesson came to an end, during which we were able to prove that the law of conservation of mass of substances is the basis for composing reaction equations, and we learned how to compose chemical reaction equations. And as a final point, write down your homework
§ 27, ex. 1 – for those who received a rating of “3”
ex. 2 – for those who received a rating of “4”
ex. 3 – for those who received a rating“5”
Final words from the teacher.
Teacher. I thank you for the lesson. But before you leave the office, pay attention to the table (the teacher points to a piece of Whatman paper with an image of a table and multi-colored chemical symbols). You see chemical signs of different colors. Each color symbolizes your mood.. I suggest you create your own table of chemical elements (it will differ from D.I. Mendeleev’s PSHE) - a table of the mood of the lesson. To do this, you must go to the sheet of music, take one chemical element, according to the characteristic that you see on the screen, and attach it to a table cell. I will do this first by showing you how comfortable I am working with you.
F I felt comfortable in the lesson, I received answers to all my questions.
F I was bored in class, I didn’t learn anything new.