Upon reaction, a compound is formed. Chemical reactions
Many processes, without which it is impossible to imagine our life (such as respiration, digestion, photosynthesis and the like), are associated with various chemical reactions organic compounds(and inorganic). Let's look at their main types and dwell in more detail on the process called connection (connection).
What is called a chemical reaction
First of all it is worth giving general definition this phenomenon. The phrase under consideration means different reactions substances of varying complexity, as a result of which different from the initial products are formed. The substances involved in this process are called "reagents".
In writing, the chemical reaction of organic compounds (and inorganic) is written using specialized equations. Outwardly, they look a little like mathematical examples by addition. However, instead of the equal sign ("="), arrows ("→" or "⇆") are used. In addition, there can sometimes be more substances on the right side of the equation than on the left. Everything before the arrow is the substance before the start of the reaction (left side of the formula). Everything after it (right side) are compounds formed as a result of the chemical process that has occurred.
As an example chemical equation you can consider water to hydrogen and oxygen under the action electric current: 2H 2 O → 2H 2 + O 2. Water is the starting reagent, and oxygen and hydrogen are products.
As another, but already more complex example of the chemical reaction of compounds, we can consider a phenomenon familiar to every housewife who has baked sweets at least once. It's about extinguishing baking soda using table vinegar. The occurring action is illustrated using the following equation: NaHCO 3 + 2 CH 3 COOH → 2CH 3 COONa + CO 2 + H 2 O. From it it is clear that in the process of interaction of sodium bicarbonate and vinegar sodium salt acetic acid, water and carbon dioxide.
By its nature, it occupies an intermediate place between physical and nuclear.
Unlike the former, the compounds involved in chemical reactions are capable of changing their composition. That is, from the atoms of one substance, several others can be formed, as in the above equation for the decomposition of water.
Unlike nuclear reactions, chemical reactions do not affect the atomic nuclei of the interacting substances.
What are the types of chemical processes
The distribution of the reactions of compounds by type occurs according to different criteria:
- Reversibility / irreversibility.
- The presence / absence of catalytic substances and processes.
- By absorption / release of heat (endothermic / exothermic reactions).
- By the number of phases: homogeneous / heterogeneous and their two hybrid varieties.
- By changing the oxidation states of the interacting substances.
Types of chemical processes in inorganic chemistry by the method of interaction
This criterion is special. With its help, four types of reactions are distinguished: compound, substitution, decomposition (cleavage) and exchange.
The name of each of them corresponds to the process that it describes. That is, they unite, in substitution, they change to other groups, in decomposition, several are formed from one reagent, and in exchange, the participants in the reaction exchange atoms with each other.
Types of processes by the method of interaction in organic chemistry
Despite the great complexity, the reactions of organic compounds proceed according to the same principle as inorganic ones. However, they have slightly different names.
So, the reactions of compound and decomposition are called "addition", as well as "elimination" (elimination) and directly organic decomposition (in this section of chemistry there are two types of decomposition processes).
Other reactions of organic compounds are substitution (the name does not change), rearrangement (exchange) and redox processes. Despite the similarity of the mechanisms of their course, in organics they are more multifaceted.
Chemical reaction of a compound
Having considered different kinds processes in which substances enter into organic and not organic chemistry, it is worth dwelling in more detail on the connection.
This reaction differs from all the others in that, regardless of the number of reagents at its beginning, in the end they all combine into one.
As an example, we can recall the process of slaking lime: CaO + H 2 O → Ca (OH) 2. In this case, the reaction of the compound of calcium oxide (quicklime) with hydrogen oxide (water) occurs. The result is calcium hydroxide (slaked lime) and warm steam. By the way, this means that this process really exothermic.
Compound reaction equation
The process under consideration can be schematically depicted as follows: A + BV → ABC. In this formula, ABC is a newly formed A - a simple reagent, and BV is a variant of a complex compound.
It should be noted that this formula is also typical for the process of joining and joining.
Examples of the reaction under consideration are the interaction of sodium oxide and carbon dioxide(NaO 2 + CO 2 (t 450-550 ° C) → Na 2 CO 3), as well as sulfur oxide with oxygen (2SO 2 + O 2 → 2SO 3).
Also, several complex compounds are capable of reacting with each other: AB + VG → ABVG. For example, the same sodium oxide and hydrogen oxide: NaO 2 + H 2 O → 2NaOH.
Reaction conditions in inorganic compounds
As shown in the previous equation, substances of varying degrees of complexity are capable of entering the interaction under consideration.
In this case, for simple reagents of inorganic origin, redox reactions of the compound (A + B → AB) are possible.
As an example, we can consider the process of obtaining trivalent. For this, a compound reaction is carried out between chlorine and ferum (iron): 3Cl 2 + 2Fe → 2FeCl 3.
If it comes on the interaction of complex inorganic substances(AB + VG → ABVG), the processes in them are capable of occurring, both influencing and not affecting their valence.
As an illustration of this, it is worth considering the example of the formation of calcium bicarbonate from carbon dioxide, hydrogen oxide (water) and white food coloring E170 (calcium carbonate): CO 2 + H 2 O + CaCO 3 → Ca (CO 3) 2. In this case, it has place is a classic connection reaction. During its implementation, the valence of the reagents does not change.
A slightly more perfect (than the first) chemical equation 2FeCl 2 + Cl 2 → 2FeCl 3 is an example of a redox process in the interaction of simple and complex inorganic reagents: gas (chlorine) and salt (ferric chloride).
Types of addition reactions in organic chemistry
As already indicated in the fourth paragraph, in substances of organic origin, the reaction under consideration is called "addition". As a rule, complex substances with a double (or triple) bond take part in it.
For example, the reaction between dibromium and ethylene, leading to the formation of 1,2-dibromoethane: (C 2 H 4) CH 2 = CH 2 + Br 2 → (C₂H₄Br₂) BrCH 2 - CH 2 Br. By the way, signs similar to equals and minus ("=" and "-"), in this equation show the bonds between the atoms of a complex substance. This is a feature of recording the formulas of organic substances.
Depending on which of the compounds act as reagents, there are several varieties of the addition process under consideration:
- Hydrogenation (hydrogen H molecules are added at multiple bonds).
- Hydrohalogenation (hydrogen halide is added).
- Halogenation (addition of halogens Br 2, Cl 2 and the like).
- Polymerization (the formation of high molecular weight substances from several low molecular weight compounds).
Examples of the addition reaction (connection)
After listing the varieties of the process under consideration, it is worth learning in practice some examples of the compound reaction.
As an illustration of hydrogenation, you can pay attention to the equation of interaction of propene with hydrogen, which will result in propane: (C 3 H 6) CH 3 —CH = CH 2 + H 2 → (C 3 H 8) CH 3 —CH 2 —CH 3.
In organic chemistry, a compound (addition) reaction can occur between hydrochloric acid and ethylene to form chloroethane: (C 2 H 4) CH 2 = CH 2 + HCl → CH 3 - CH 2 —Cl (C 2 H 5 Cl). The equation presented is an example of hydrohalogenation.
As for halogenation, it can be illustrated by the reaction between dichlorine and ethylene, leading to the formation of 1,2-dichloroethane: (C 2 H 4) CH 2 = CH 2 + Cl 2 → (C₂H₄Cl₂) ClCH 2 -CH 2 Cl.
A bunch of nutrients formed by organic chemistry. The reaction of connection (addition) of ethylene molecules with a radical initiator of polymerization under the influence of ultraviolet radiation is a confirmation of this: n CH 2 = CH 2 (R and UV light) → (-CH 2 -CH 2 -) n. The substance formed in this way is well known to every person under the name of polyethylene.
Various types of packaging, bags, dishes, pipes, insulation materials and much more are made from this material. A feature of this substance is the possibility of its recycling. Polyethylene owes its popularity to the fact that it does not decompose, which is why environmentalists have a negative attitude towards it. However, in last years a method was found for the safe disposal of polyethylene products. For this, the material is processed nitric acid(HNO 3). Then certain types bacteria are able to decompose this substance into safe components.
The connection (addition) reaction plays important role in nature and human life. In addition, it is often used by scientists in laboratories to synthesize new substances for various important research.
Compound reactions (the formation of one complex substance from several simple or complex substances) A + B = AB
Decomposition reactions (decomposition of one complex substance into several simple or complex substances) AB = A + B
Substitution reactions (between simple and complex substances, in which atoms of a simple substance replace the atoms of one of the elements in a complex substance): AB + C = AC + B
Exchange reactions (between two complex substances in which substances exchange their constituent parts) AB + SD = AD + SV
1. Indicate the correct definition of the compound reaction:
A. The reaction of the formation of several substances from one simple substance;
B. A reaction in which one complex substance is formed from several simple or complex substances.
B. A reaction in which substances exchange their constituent parts.
2. Indicate the correct definition of the substitution reaction:
A. Reaction between base and acid;
B. Reaction of interaction of two simple substances;
C. A reaction between substances in which atoms of a simple substance replace the atoms of one of the elements in a complex substance.
3. What is the correct definition of the decomposition reaction:
A. A reaction in which several simple or complex substances are formed from one complex substance;
B. A reaction in which substances exchange their constituent parts;
B. Reaction with the formation of oxygen and hydrogen molecules.
4. Indicate the signs of the exchange reaction:
A. Water formation;
B. Gas formation only;
B. Only precipitation;
D. Sedimentation, gas formation or weak electrolyte formation.
5. What type of reactions does the interaction of acidic oxides with basic oxides refer to:
A. Reaction of exchange;
B. Reaction of the compound;
B. Decomposition reaction;
D. Substitution reaction.
6. What type of reactions does the interaction of salts with acids or bases refer to:
A. Substitution reactions;
B. Decomposition reactions;
B. Reactions of exchange;
D. Compound reactions.
7. Substances whose formulas are KNO3 FeCl2, Na2SO4 are called:
A) salts; B) grounds; B) acids; D) oxides.
8 ... Substances whose formulas are HNO3, HCl, H2SO4 are called:
9 ... Substances whose formulas are KOH, Fe (OH) 2, NaOH are called:
A) salts; B) acids; B) grounds; D) oxides. 10 ... Substances whose formulas are NO2, Fe2O3, Na2O are called:
A) salts; B) acids; B) grounds; D) oxides.
11 ... Indicate the metals that form alkalis:
Cu, Fe, Na, K, Zn, Li.
Answers:
Na, K, Li.
7.1. The main types of chemical reactions
Transformations of substances, accompanied by a change in their composition and properties, are called chemical reactions or chemical interactions. During chemical reactions, there is no change in the composition of the atomic nuclei.
Phenomena in which the form or physical state of substances changes or the composition of the nuclei of atoms changes are called physical. An example physical phenomena is an heat treatment metals, at which there is a change in their shape (forging), melting of metal, sublimation of iodine, transformation of water into ice or steam, etc., as well as nuclear reactions, as a result of which atoms of other elements are formed from the atoms of some elements.
Chemical phenomena can be accompanied by physical transformations. For example, as a result of chemical reactions in a galvanic cell, an electric current is generated.
Chemical reactions classified according to various criteria.
1. According to the sign of the heat effect, all reactions are divided into endothermic(flowing with heat absorption) and exothermic(flowing with the release of heat) (see § 6.1).
2. According to the state of aggregation of the initial substances and reaction products, they are distinguished:
homogeneous reactions , in which all substances are in the same phase:
2 KOH (p-p) + H 2 SO 4 (p-p) = K 2 SO (p-p) + 2 H 2 O (g),
CO (g) + Cl 2 (g) = COCl 2 (g),
SiO 2 (k) + 2 Mg (k) = Si (k) + 2 MgO (k).
heterogeneous reactions, substances in which are in different phases:
CaO (k) + CO 2 (g) = CaCO 3 (k),
CuSO 4 (solution) + 2 NaOH (solution) = Cu (OH) 2 (k) + Na 2 SO 4 (solution),
Na 2 SO 3 (solution) + 2HCl (solution) = 2 NaCl (solution) + SO 2 (g) + H 2 O (g).
3. By the ability to flow only in the forward direction, as well as in the forward and reverse directions, they are distinguished irreversible and reversible chemical reactions (see § 6.5).
4. By the presence or absence of catalysts are distinguished catalytic and non-catalytic reactions (see § 6.5).
5. According to the mechanism of occurrence, chemical reactions are divided into ionic, radical and others (the mechanism of chemical reactions involving organic compounds is considered in the course of organic chemistry).
6. According to the state of the oxidation states of the atoms that make up the reactants, there are reactions that take place without changing the oxidation state atoms, and with a change in the oxidation state of atoms ( redox reactions) (see § 7.2).
7. According to the change in the composition of the starting substances and reaction products, reactions are distinguished compound, decomposition, substitution and exchange... These reactions can proceed with or without a change in the oxidation states of the elements, table . 7.1.
Table 7.1
Types of chemical reactions
Examples of reactions proceeding without changing the oxidation state of elements |
Examples of redox reactions |
||
Connections (one new substance is formed from two or more substances) |
HCl + NH 3 = NH 4 Cl; SO 3 + H 2 O = H 2 SO 4 |
H 2 + Cl 2 = 2HCl; 2Fe + 3Cl 2 = 2FeCl 3 |
|
Decomposition (several new substances are formed from one substance) |
A = B + C + D |
MgCO 3 MgO + CO 2; H 2 SiO 3 SiO 2 + H 2 O |
2AgNO 3 2Ag + 2NO 2 + O 2 |
Substitutions (when substances interact, the atoms of one substance replace the atoms of another substance in the molecule) |
A + BC = AB + C |
CaCO 3 + SiO 2 CaSiO 3 + CO 2 |
Pb (NO 3) 2 + Zn = Mg + 2HCl = MgCl 2 + H 2 |
|
(two substances exchange their constituent parts, forming two new substances) |
AB + CD = AD + CB |
AlCl 3 + 3NaOH = Ca (OH) 2 + 2HCl = CaCl 2 + 2H 2 O |
7.2. Redox reactions
As stated above, all chemical reactions fall into two groups:
Chemical reactions proceeding with a change in the oxidation state of the atoms that make up the reactants are called redox.
Oxidation Is the process of giving up electrons by an atom, molecule or ion:
Na o - 1e = Na +;
Fe 2+ - e = Fe 3+;
H 2 o - 2e = 2H +;
2 Br - - 2e = Br 2 o.
Recovery Is the process of attaching electrons by an atom, molecule or ion:
S o + 2e = S 2–;
Cr 3+ + e = Cr 2+;
Cl 2 o + 2e = 2Cl -;
Mn 7+ + 5e = Mn 2+.
The atoms, molecules or ions that accept electrons are called oxidants. Restorers are atoms, molecules or ions donating electrons.
Accepting electrons, the oxidizing agent is reduced during the reaction, and the reducing agent is oxidized. Oxidation is always accompanied by reduction and vice versa. In this way, the number of electrons donated by the reducing agent is always equal to the number of electrons donated by the oxidizing agent.
7.2.1. Oxidation state
The oxidation state is the conditional (formal) charge of an atom in a compound, calculated on the assumption that it consists only of ions. It is customary to denote the oxidation state with an Arabic numeral at the top of the element symbol with a “+” or “-” sign. For example, Al 3+, S 2–.
To find the oxidation states, the following rules are followed:
the oxidation state of atoms in simple substances is zero;
the algebraic sum of the oxidation states of atoms in a molecule is zero, in a complex ion - the charge of the ion;
the oxidation state of alkali metal atoms is always +1;
a hydrogen atom in compounds with non-metals (CH 4, NH 3, etc.) exhibits an oxidation state of +1, and with active metals, its oxidation state is –1 (NaH, CaH 2, etc.);
a fluorine atom in compounds always exhibits an oxidation state of –1;
the oxidation state of the oxygen atom in the compounds is usually –2, except for peroxides (H 2 O 2, Na 2 O 2), in which the oxidation state of oxygen is –1, and some other substances (superoxides, ozonides, oxygen fluorides).
The maximum positive oxidation state of elements in a group is usually equal to the group number. The exceptions are fluorine, oxygen, since their highest oxidation state is lower than the number of the group in which they are located. Elements of the copper subgroup form compounds in which their oxidation state exceeds the group number (CuO, AgF 5, AuCl 3).
Maximum negative oxidation state of elements in the main subgroups periodic system can be determined by subtracting the group number from eight. For carbon, this is 8 - 4 = 4, for phosphorus - 8 - 5 = 3.
In the main subgroups, when moving from elements from top to bottom, the stability of the highest positive oxidation state decreases, in side subgroups, on the contrary, the stability of higher oxidation states increases from top to bottom.
The conventionality of the concept of the oxidation state can be demonstrated by the example of some inorganic and organic compounds. In particular, in phosphinic (hypophosphorous) H 3 PO 2, phosphonic (phosphorous) H 3 PO 3 and phosphoric H 3 PO 4 acids, the oxidation states of phosphorus are, respectively, +1, +3, and +5, while in all these compounds phosphorus is pentavalent. For carbon in methane CH 4, methanol CH 3 OH, formaldehyde CH 2 O, formic acid HCOOH and carbon monoxide (IV) CO 2, the oxidation states of carbon are –4, –2, 0, +2 and +4, respectively, while as the valency of the carbon atom in all these compounds is equal to four.
Despite the fact that the oxidation state is a conventional concept, it is widely used in the preparation of redox reactions.
7.2.2. The most important oxidizing and reducing agents
Typical oxidizing agents are:
1. Simple substances, the atoms of which are highly electronegative. These are, first of all, elements of the main subgroups VI and VII of groups of the periodic system: oxygen, halogens. Of the simple substances, fluorine is the strongest oxidizing agent.
2. Compounds containing some metal cations in high oxidation states: Pb 4+, Fe 3+, Au 3+, etc.
3. Compounds containing some complex anions, the elements of which are in high positive oxidation states: 2–, - -, etc.
Reducing agents include:
1. Simple substances, atoms of which have low electronegativity - active metals. Non-metals, such as hydrogen and carbon, can also exhibit reducing properties.
2. Some metal compounds containing cations (Sn 2+, Fe 2+, Cr 2+), which, by donating electrons, can increase their oxidation state.
3. Some compounds containing such simple ions as, for example, I -, S 2–.
4. Compounds containing complex ions (S 4+ O 3) 2–, (НР 3+ O 3) 2–, in which the elements can, by donating electrons, increase their positive degree oxidation.
In laboratory practice, the following oxidizing agents are most commonly used:
potassium permanganate (KMnO 4);
potassium dichromate (K 2 Cr 2 O 7);
nitric acid (HNO 3);
concentrated sulfuric acid (H 2 SO 4);
hydrogen peroxide (H 2 O 2);
manganese (IV) and lead (IV) oxides (MnO 2, PbO 2);
molten potassium nitrate (KNO 3) and melts of some other nitrates.
Reducing agents that are used in laboratory practice include:
- magnesium (Mg), aluminum (Al) and other active metals;
- hydrogen (H 2) and carbon (C);
- potassium iodide (KI);
- sodium sulfide (Na 2 S) and hydrogen sulfide (H 2 S);
- sodium sulfite (Na 2 SO 3);
- tin chloride (SnCl 2).
7.2.3. Classification of redox reactions
Redox reactions are usually divided into three types: intermolecular, intramolecular, and disproportionation (self-oxidation-self-healing) reactions.
Intermolecular reactions proceed with a change in the oxidation state of atoms that are in different molecules. For instance:
2 Al + Fe 2 O 3 Al 2 O 3 + 2 Fe,
C + 4 HNO 3 (conc) = CO 2 + 4 NO 2 + 2 H 2 O.
TO intramolecular reactions These include reactions in which the oxidizing agent and the reducing agent are part of the same molecule, for example:
(NH 4) 2 Cr 2 O 7 N 2 + Cr 2 O 3 + 4 H 2 O,
2 KNO 3 2 KNO 2 + O 2.
V disproportionation reactions(self-oxidation-self-healing) an atom (ion) of the same element is both an oxidizing agent and a reducing agent:
Cl 2 + 2 KOH KCl + KClO + H 2 O,
2 NO 2 + 2 NaOH = NaNO 2 + NaNO 3 + H 2 O.
7.2.4. Basic rules for compiling redox reactions
The composition of redox reactions is carried out according to the steps presented in table. 7.2.
Table 7.2
Stages of drawing up the equations of redox reactions
Action |
|
Determine the oxidizing and reducing agent. |
|
Establish the products of the redox reaction. |
|
Draw up the balance of electrons and, with its help, arrange the coefficients for substances that change their oxidation states. |
|
Arrange the coefficients for other substances that take part and are formed in the redox reaction. |
|
Check the correctness of the arrangement of the coefficients by counting the amount of substance atoms (usually hydrogen and oxygen) located in the left and right sides reaction equations. |
Let us consider the rules for drawing up redox reactions using the example of the interaction of potassium sulfite with potassium permanganate in an acidic medium:
1. Determination of oxidizing agent and reducing agent
Manganese, which is in the highest oxidation state, cannot donate electrons. Mn 7+ will accept electrons, i.e. is an oxidizing agent.
The S 4+ ion can donate two electrons and go to S 6+, i.e. is a reducing agent. Thus, in the reaction under consideration, K 2 SO 3 is a reducing agent, and KMnO 4 is an oxidizing agent.
2. Determination of reaction products
K 2 SO 3 + KMnO 4 + H 2 SO 4?
Donating two electrons an electron, S 4+ goes into S 6+. Potassium sulfite (K 2 SO 3) is thus converted to sulfate (K 2 SO 4). In an acidic medium, Mn 7+ takes 5 electrons and in a solution of sulfuric acid (medium) forms manganese sulfate (MnSO 4). As a result of this reaction, additional molecules of potassium sulfate are also formed (due to the potassium ions that make up the permanganate), as well as water molecules. Thus, the reaction under consideration will be written in the form:
K 2 SO 3 + KMnO 4 + H 2 SO 4 = K 2 SO 4 + MnSO 4 + H 2 O.
3. Compilation of the balance of electrons
To compile the balance of electrons, it is necessary to indicate those oxidation states that change in the reaction under consideration:
K 2 S 4+ O 3 + KMn 7+ O 4 + H 2 SO 4 = K 2 S 6+ O 4 + Mn 2+ SO 4 + H 2 O.
Mn 7+ + 5 e = Mn 2+;
S 4+ - 2 e = S 6+.
The number of electrons donated by the reducing agent should be equal to the number of electrons donated by the oxidizing agent. Therefore, the reaction should involve two Mn 7+ and five S 4+:
Mn 7+ + 5 e = Mn 2+ 2,
S 4+ - 2 e = S 6+ 5.
Thus, the number of electrons donated by the reducing agent (10) will be equal to the number of electrons donated by the oxidizing agent (10).
4. Arrangement of coefficients in the reaction equation
In accordance with the balance of electrons in front of K 2 SO 3, it is necessary to put a factor of 5, and in front of KMnO 4 - 2. On the right side, in front of potassium sulfate, we set a factor of 6, since one molecule is added to the five K 2 SO 4 molecules formed during the oxidation of potassium sulfite K 2 SO 4 as a result of the binding of potassium ions that make up the permanganate. Since the reaction is involved as an oxidizing agent two permanganate molecules, on the right side are also formed two manganese sulfate molecules. To bind the reaction products (potassium and manganese ions that make up the permanganate), it is necessary three sulfuric acid molecules, therefore, as a result of the reaction, three water molecules. Finally, we get:
5 K 2 SO 3 + 2 KMnO 4 + 3 H 2 SO 4 = 6 K 2 SO 4 + 2 MnSO 4 + 3 H 2 O.
5. Checking the correctness of the arrangement of the coefficients in the reaction equation
The number of oxygen atoms on the left side of the reaction equation is:
5 3 + 2 4 + 3 4 = 35.
On the right side, this number will be:
6 4 + 2 4 + 3 1 = 35.
The number of hydrogen atoms on the left side of the reaction equation is six and corresponds to the number of these atoms on the right side of the reaction equation.
7.2.5. Examples of redox reactions involving typical oxidizing and reducing agents
7.2.5.1. Intermolecular oxidation-reduction reactions
Redox reactions involving potassium permanganate, potassium dichromate, hydrogen peroxide, potassium nitrite, potassium iodide and potassium sulfide are considered below as examples. Redox reactions involving other typical oxidizing and reducing agents are discussed in the second part of the manual (Inorganic Chemistry).
Redox reactions involving potassium permanganate
Depending on the medium (acidic, neutral, alkaline), potassium permanganate, acting as an oxidizing agent, gives various reduction products, Fig. 7.1.
Rice. 7.1. Formation of products of reduction of potassium permanganate in various media
Below are the reactions of KMnO 4 with potassium sulfide as a reducing agent in various media, illustrating the scheme, Fig. 7.1. In these reactions, the oxidation product of the sulfide ion is free sulfur. V alkaline environment KOH molecules do not take part in the reaction, but only determine the reduction product of potassium permanganate.
5 K 2 S + 2 KMnO 4 + 8 H 2 SO 4 = 5 S + 2 MnSO 4 + 6 K 2 SO 4 + 8 H 2 O,
3 K 2 S + 2 KMnO 4 + 4 H 2 O 2 MnO 2 + 3 S + 8 KOH,
K 2 S + 2 KMnO 4 – (KOH) 2 K 2 MnO 4 + S.
Redox reactions involving potassium dichromate
In an acidic environment, potassium dichromate is a strong oxidizing agent. A mixture of K 2 Cr 2 O 7 and concentrated H 2 SO 4 (chromic peak) is widely used in laboratory practice as an oxidizing agent. Interacting with a reducing agent, one potassium dichromate molecule accepts six electrons, forming trivalent chromium compounds:
6 FeSO 4 + K 2 Cr 2 O 7 +7 H 2 SO 4 = 3 Fe 2 (SO 4) 3 + Cr 2 (SO 4) 3 + K 2 SO 4 +7 H 2 O;
6 KI + K 2 Cr 2 O 7 + 7 H 2 SO 4 = 3 I 2 + Cr 2 (SO 4) 3 + 4 K 2 SO 4 + 7 H 2 O.
Redox reactions involving hydrogen peroxide and potassium nitrite
Hydrogen peroxide and potassium nitrite exhibit predominantly oxidizing properties:
H 2 S + H 2 O 2 = S + 2 H 2 O,
2 KI + 2 KNO 2 + 2 H 2 SO 4 = I 2 + 2 K 2 SO 4 + H 2 O,
However, when interacting with strong oxidants (such as, for example, KMnO 4), hydrogen peroxide and potassium nitrite act as reducing agents:
5 H 2 O 2 + 2 KMnO 4 + 3 H 2 SO 4 = 5 O 2 + 2 MnSO 4 + K 2 SO 4 + 8 H 2 O,
5 KNO 2 + 2 KMnO 4 + 3 H 2 SO 4 = 5 KNO 3 + 2 MnSO 4 + K 2 SO 4 + 3 H 2 O.
It should be noted that hydrogen peroxide, depending on the medium, is reduced according to the scheme, Fig. 7.2.
Rice. 7.2. Possible reduction products of hydrogen peroxide
In this case, as a result of reactions, water or hydroxide ions are formed:
2 FeSO 4 + H 2 O 2 + H 2 SO 4 = Fe 2 (SO 4) 3 + 2 H 2 O,
2 KI + H 2 O 2 = I 2 + 2 KOH.
7.2.5.2. Intramolecular oxidation-reduction reactions
Intramolecular redox reactions occur, as a rule, when substances are heated, the molecules of which contain a reducing agent and an oxidizing agent. Examples of intramolecular reduction-oxidation reactions are the processes of thermal decomposition of nitrates and potassium permanganate:
2 NaNO 3 2 NaNO 2 + O 2,
2 Cu (NO 3) 2 2 CuO + 4 NO 2 + O 2,
Hg (NO 3) 2 Hg + NO 2 + O 2,
2 KMnO 4 K 2 MnO 4 + MnO 2 + O 2.
7.2.5.3. Disproportionation reactions
As noted above, in disproportionation reactions one and the same atom (ion) is both an oxidizing agent and a reducing agent. Let us consider the process of compiling this type of reactions using the example of the interaction of sulfur with alkali.
Typical oxidation states of sulfur: – 2, 0, +4 and +6. Acting as a reducing agent, elemental sulfur gives up 4 electrons:
S o – 4e = S 4+.
Sulfur – the oxidizer accepts two electrons:
S o + 2е = S 2–.
Thus, as a result of the sulfur disproportionation reaction, compounds are formed, the oxidation states of the element in which – 2 and on the right +4:
3 S + 6 KOH = 2 K 2 S + K 2 SO 3 + 3 H 2 O.
With the disproportionation of nitrogen oxide (IV) in alkali, nitrite and nitrate are obtained - compounds in which the oxidation states of nitrogen are +3 and +5, respectively:
2 N 4+ O 2 + 2 KOH = KN 3+ O 2 + KN 5+ O 3 + H 2 O,
The disproportionation of chlorine in a cold alkali solution leads to the formation of hypochlorite, and in a hot solution, chlorate:
Cl 0 2 + 2 KOH = KCl - + KCl + O + H 2 O,
Cl 0 2 + 6 KOH 5 KCl - + KCl 5+ O 3 + 3H 2 O.
7.3. Electrolysis
The oxidation-reduction process that occurs in solutions or melts by passing a constant electric current through them is called electrolysis. In this case, anions are oxidized on the positive electrode (anode). Cations are reduced on the negative electrode (cathode).
2 Na 2 CO 3 4 Na + О 2 + 2CO 2.
In the electrolysis of aqueous solutions of electrolytes, along with the transformations of the solute, electro chemical processes with the participation of hydrogen ions and hydroxide ions of water:
cathode (-): 2 Н + + 2е = Н 2,
anode (+): 4 OH - - 4e = O 2 + 2 H 2 O.
In this case, the reduction process at the cathode occurs as follows:
1. Cations of active metals (up to Al 3+ inclusive) are not reduced at the cathode; instead, hydrogen is reduced.
2. Metal cations located in the series of standard electrode potentials (in the series of voltages) to the right of hydrogen are reduced at the cathode to free metals during electrolysis.
3. The metal cations located between Al 3+ and H + are reduced at the cathode simultaneously with the hydrogen cation.
The processes taking place in aqueous solutions at the anode depend on the substance from which the anode is made. Distinguish between insoluble anodes ( inert) and soluble ( active). Graphite or platinum is used as a material for inert anodes. Soluble anodes are made from copper, zinc, and other metals.
During the electrolysis of solutions with an inert anode, the following products can be formed:
1. Oxidation of halide ions liberates free halogens.
2. During electrolysis of solutions containing SO 2 2–, NO 3 -, PO 4 3– anions, oxygen is released; it is not these ions that are oxidized at the anode, but water molecules.
Considering the above rules, consider as an example the electrolysis of aqueous solutions of NaCl, CuSO 4 and KOH with inert electrodes.
one). In solution, sodium chloride dissociates into ions.
9.1. What are the chemical reactions
Let's remember that we call any chemical phenomena of nature by chemical reactions. During a chemical reaction, some are broken and other chemical bonds are formed. As a result of the reaction, other substances are obtained from some chemical substances (see Chapter 1).
By doing homework to § 2.5, you got acquainted with the traditional separation of four main types of reactions from the whole set of chemical transformations, at the same time you proposed their names: reactions of combination, decomposition, substitution and exchange.
Examples of compound reactions:
C + O 2 = CO 2; (one)
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; (eight)
CaCO 3 + SiO 2 = CaSiO 3 + CO 2. (9)
| Exchange reactions- chemical reactions in which the initial 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 selection of two other types of chemical reactions is based on the participation in them of two most important non-chemical particles: an electron and a proton.
In the course of some reactions, there is a complete or partial transfer of electrons from one atom to another. In this case, the oxidation states of the atoms of the elements that make up the starting materials 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 familiar with RR in § 2, and RR in the following chapters.
COMPOUNDS REACTIONS, DECOMPOSITION REACTIONS, SUBSTITUTION REACTIONS, EXCHANGE REACTIONS, REDUCING-REDUCING REACTIONS, ACID-BASIC REACTIONS.
Make up the 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; f) 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. Note the redox and acid-base reactions. In redox reactions, indicate the atoms of which elements change their oxidation states.
9.2. Redox reactions
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 initial substances and the reaction products
| Fe 2 O 3 | + | = | 2Fe | + |
As you can see, the oxidation state of carbon atoms as a result of the reaction increased, 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 iron atoms - reduction, that is, they added electrons ( recovered) (see § 7.16). To characterize RVR, the concepts are used oxidizing agent 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) oxide.
In cases where oxidizing and reducing atoms are part of the same substance (example: reaction 6 from the previous paragraph), the terms "oxidizing agent" and "reducing agent" are not used.
Thus, typical oxidizing agents are substances that contain atoms that tend to attach electrons (in whole or in part), lowering their oxidation state. Of the simple substances, these are primarily halogens and oxygen, to a lesser extent sulfur and nitrogen. Of complex substances - substances that include 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 give up electrons in whole or in part, increasing their oxidation state. Of the simple substances, these are hydrogen, alkali and alkaline earth metals, and aluminum. Of 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 instance:
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 go back to the reaction we analyzed at the beginning of this section.
| Fe 2 O 3 | + | = | 2Fe | + |
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 under any conditions is a very weak oxidizing agent, 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 ORR flow:
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 be compared only under the same conditions. In some cases, this comparison can be made quantitatively.
Doing your homework for the first paragraph of this chapter, you became convinced that it is rather difficult to find the coefficients in some reaction equations (especially the OVR). 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 study the electronic balance method now, and the electronic ion balance method is usually studied in higher education institutions.
Both of these methods are based on the fact that electrons in chemical reactions do not disappear anywhere and do not appear from anywhere, that is, the number of electrons received by atoms is equal to the number of electrons donated by other atoms.
The number of donated and received electrons in the electronic 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 materials and the reaction products.
Let's consider the application of the electronic balance method using examples.
Example 1. Let's compose the 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 receive them. Let us express these processes electronic equations:
Fe - 3 e- = Fe + III,
Cl 2 + 2 e -= 2Cl –I.
In order for the number of electrons donated 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, Cl 2 + 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 compose the equation of the combustion reaction white phosphorus in excess of chlorine. It is known that phosphorus (V) chloride is formed under these conditions:
| + V –I | ||||
| P 4 | + | Cl 2 | PCl 5. |
White phosphorus molecules donate electrons (are oxidized), and chlorine molecules receive them (are reduced):
| P 4 - 20 e- = 4P + V Cl 2 + 2 e- = 2Cl –I |
1 10 |
2 20 |
P 4 - 20 e- = 4P + V Cl 2 + 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 the future coefficients in the reaction equation) they were divided. Reaction equation:
P 4 + 10Cl 2 = 4PCl 5.
Example 3. Let us compose the equation of the reaction occurring during the roasting of iron (II) sulfide in oxygen.
Reaction scheme:
| + III –II | + IV –II | |||||
| + | O 2 | + |
In this case, both iron (II) and sulfur (- II) atoms are oxidized. The atoms of these elements are in the composition of iron (II) sulfide in a ratio of 1: 1 (see the indices in the simplest formula).
Electronic balance:
| 4 | Fe + II - e- = Fe + III S –II - 6 e- = S + IV |
Total give 7 e – |
| 7 | O 2 + 4e - = 2O –II |
Reaction equation: 4FeS + 7O 2 = 2Fe 2 O 3 + 4SO 2.
Example 4. Let us compose the equation of the reaction occurring during the roasting of iron (II) disulfide (pyrite) in oxygen.
Reaction scheme:
| + III –II | + IV –II | |||||
| + | O 2 | + |
As in the previous example, here, too, iron (II) atoms and sulfur atoms are also oxidized, but with an oxidation state of I. The atoms of these elements are in the composition of pyrite in a ratio of 1: 2 (see the indices in the simplest formula). It is in this respect that the atoms of iron and sulfur enter into a reaction, which is taken into account when compiling the electronic balance:
| Fe + III - e- = Fe + III 2S –I - 10 e- = 2S + IV |
Total give 11 e – | |
| O 2 + 4 e- = 2O –II |
Reaction equation: 4FeS 2 + 11O 2 = 2Fe 2 O 3 + 8SO 2.
There are also more complex cases of OVR, some of which you will get to know while doing your homework.
ATOM-OXIDIZER, ATOM-REDUCER, SUBSTANCE-OXIDIZER, SUBSTANCE-REDUCER, ELECTRONIC BALANCE METHOD, ELECTRONIC EQUATIONS.
1. Make an electronic balance for each ORP equation given in the text of § 1 of this chapter.
2. Make the equations of the OVR, found by you during the assignment to § 1 of this chapter. This time, use the electronic balance method to set the odds. 3. Using the electronic balance method, make up the 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);
f) 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);
m) 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 take place?
To answer this question, let us recall why individual atoms combine into molecules, why an ionic crystal is formed from isolated ions, why the principle of the least energy operates during the formation of the electron shell of an atom. The answer to all these questions is the same: because it is energetically beneficial. This means that energy is released during these processes. It would seem that chemical reactions should proceed for the same reason. Indeed, many reactions can be carried out, during the course of 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 combustion reaction of methane
CH 4 (g) + 2O 2 (g) = CO 2 (g) + 2H 2 O (g)
so much heat is released that methane is used as a fuel.
The fact that heat is released in this reaction 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 chemical reactions break and form chemical bonds. In this case, bonds are broken between the carbon and hydrogen atoms in the CH 4 molecules, as well as between the oxygen atoms in the O 2 molecules. In this case, new bonds are formed: between the carbon and oxygen atoms in the CO2 molecules and between the oxygen and hydrogen atoms in the H2O molecules. To break the bonds, you need to spend energy (see "bond energy", "atomization energy"), and during the formation bonds the 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.
Such a record means that 484 kilojoules of heat will be released if two moles of hydrogen react with one mole of oxygen to form two moles of gaseous water (water vapor).
In this way, in thermochemical equations, the coefficients are numerically equal to the amounts of the substance of the reactants and reaction products.
What determines the thermal effect of each specific reaction?
The heat effect of the reaction depends
a) from the states of aggregation of the initial 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 associated with 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. It can also be expressed by a thermochemical equation. Example - thermochemical equation condensation of water vapor:
H 2 O (g) = H 2 O (g) + Q.
In thermochemical equations, and, if necessary, in ordinary chemical equations, the state of aggregation of substances is indicated using letter indices:
(g) - 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 performing work to increase the volume, and the released heat will be less than in the case of the same reaction at constant volume.
Heat effects of reactions are usually calculated for reactions proceeding at a constant volume at 25 ° C and are denoted by the symbol Q o.
If energy is released only in the form of heat, and the chemical reaction proceeds at a constant volume, then the heat 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 proceeds at constant pressure, then the volume of the system can change. Part of the internal energy also goes to work to increase the volume. In this case
U = -(Q P + A) = –(Q P + PV),
where Q p- the thermal effect of the reaction proceeding 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, with the release of heat, the enthalpy of the system decreases. Hence the old name for this quantity: "heat content".
In contrast to the thermal effect, the change in enthalpy characterizes the 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 change in enthalpy under standard conditions (25 ° C, 101.3 kPa) is given, denoted H about... For instance:
2H 2 (g) + O 2 (g) = 2H 2 O (g) H about= - 484 kJ;
CaO (cr) + H 2 O (l) = Ca (OH) 2 (cr) H about= - 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 initial 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 for substance B in the thermochemical equation.
Task
Determine the amount of hydrogen substance burned in oxygen if 1694 kJ of heat was released.
Solution
2H 2 (g) + O 2 (g) = 2H 2 O (g) + 484 kJ. |
|
|
Q = 1694 kJ, 6. The thermal effect of the reaction of interaction of crystalline aluminum with gaseous chlorine is 1408 kJ. Write down the thermochemical equation for this reaction and determine the mass of aluminum required to obtain 2816 kJ of heat using this reaction. 9.4. Endothermic reactions. Entropy In addition to exothermic reactions, reactions are possible, during the course of which heat is absorbed, and if it is not supplied, then the reaction system is cooled. Such reactions are called endothermic. The thermal effect of such reactions is negative. For instance: 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 initial substances.
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 enter into a chemical reaction with each other. Let us tightly connect the flasks with their necks and set them vertically, so that the flask with the heavier nitrogen dioxide is at the bottom (Fig. 9.1). After a while, 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. In this way,
The coupling equations between entropy ( S) and other quantities are studied in physics and physical chemistry courses. Unit of measurement of entropy [ S] = 1 J / K. G = H - T S Spontaneous reaction condition: G< 0. At low temperatures the factor determining the possibility of the reaction proceeding is to a greater extent the energy factor, and at high - the entropy one. From the above equation, in particular, it is clear why the room temperature decomposition reactions (entropy increases) begin to proceed at elevated temperatures. ENDOTHERMAL REACTION, ENTROPY, ENERGY FACTOR, ENTROPIC 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 needs to be expended to obtain 1 kg of copper by such a reaction. CaCO 3 (cr) = CaO (cr) + CO 2 (g) - 179 kJ formed 24.6 liters of carbon dioxide. Determine how much heat was wasted. How many grams of calcium oxide were formed? |
DEFINITION
Chemical reaction called the transformation of substances in which there is a change in their composition and (or) structure.
Most often, chemical reactions are understood as the process of converting initial substances (reagents) into final substances (products).
Chemical reactions are written using chemical equations containing the formulas of the starting materials and reaction products. According to the law of conservation of mass, the number of atoms of each element on the left and right sides of the chemical equation is the same. Usually, the formulas of the starting materials are written on the left side of the equation, and the formulas for the products are on the right. The equality of the number of atoms of each element in the left and right sides of the equation is achieved by placing integer stoichiometric coefficients in front of the formulas of substances.
Chemical equations may contain additional information about the features of the reaction: temperature, pressure, radiation, etc., which is indicated by the corresponding symbol above (or “below”) the equal sign.
All chemical reactions can be grouped into several classes, which have certain characteristics.
Classification of chemical reactions by the number and composition of the starting and resulting substances
According to this classification, chemical reactions are subdivided into reactions of combination, decomposition, substitution, exchange.
As a result compound reactions one new substance is formed from two or more (complex or simple) substances. V general view the equation for such a chemical reaction will look like this:
For instance:
CaCO 3 + CO 2 + H 2 O = Ca (HCO 3) 2
SO 3 + H 2 O = H 2 SO 4
2Mg + O 2 = 2MgO.
2FеСl 2 + Сl 2 = 2FеСl 3
The reactions of the compound are in most cases exothermic, i.e. proceed with the release of heat. If the reaction involves simple substances, then such reactions are most often redox (ORR), i.e. proceed with a change in the oxidation states of the elements. It is impossible to say unequivocally whether the reaction of a compound between complex substances belongs to the OVR.
Reactions as a result of which several other new substances (complex or simple) are formed from one complex substance are referred to as decomposition reactions... In general, the chemical decomposition equation will look like this:
For instance:
CaCO 3 CaO + CO 2 (1)
2H 2 O = 2H 2 + O 2 (2)
CuSO 4 × 5H 2 O = CuSO 4 + 5H 2 O (3)
Cu (OH) 2 = CuO + H 2 O (4)
H 2 SiO 3 = SiO 2 + H 2 O (5)
2SO 3 = 2SO 2 + O 2 (6)
(NH 4) 2 Cr 2 O 7 = Cr 2 O 3 + N 2 + 4H 2 O (7)
Most decomposition reactions occur on heating (1,4,5). Decomposition by electric current possible (2). The decomposition of crystalline hydrates, acids, bases and salts of oxygen-containing acids (1, 3, 4, 5, 7) proceeds without changing the oxidation states of the elements, i.e. these reactions do not belong to OVR. OVP decomposition reactions include the decomposition of oxides, acids and salts, formed by elements v higher degrees oxidation (6).
Decomposition reactions are also found in organic chemistry, but under other names - cracking (8), dehydrogenation (9):
C 18 H 38 = C 9 H 18 + C 9 H 20 (8)
C 4 H 10 = C 4 H 6 + 2H 2 (9)
At substitution reactions a simple substance interacts with a complex substance, forming a new simple and new complex substance. In general terms, the equation for the chemical reaction of substitution will look like this:
For instance:
2Аl + Fe 2 O 3 = 2Fе + Аl 2 О 3 (1)
Zn + 2HCl = ZnCl 2 + H 2 (2)
2KBr + Cl 2 = 2KCl + Br 2 (3)
2KSlO 3 + l 2 = 2KlO 3 + Сl 2 (4)
CaCO 3 + SiO 2 = CaSiO 3 + CO 2 (5)
Ca 3 (PO 4) 2 + 3SiO 2 = 3CaSiO 3 + P 2 O 5 (6)
CH 4 + Cl 2 = CH 3 Cl + HCl (7)
Substitution reactions are mostly redox reactions (1 - 4, 7). Examples of decomposition reactions in which no change in oxidation states occurs are few (5, 6).
Exchange reactions call the reactions that occur between complex substances, in which they exchange their constituent parts. Usually this term is used for reactions involving ions located in aqueous solution... In general, the equation of the chemical exchange reaction will look like this:
AB + CD = AD + CB
For instance:
CuO + 2HCl = CuCl 2 + H 2 O (1)
NaOH + HCl = NaCl + H 2 O (2)
NaHCO 3 + HCl = NaCl + H 2 O + CO 2 (3)
AgNO 3 + KBr = AgBr ↓ + KNO 3 (4)
СrСl 3 + ЗNаОН = Сr (ОН) 3 ↓ + ЗNаСl (5)
Metabolic reactions are not redox. A special case of these exchange reactions are neutralization reactions (reactions of interaction of acids with alkalis) (2). Exchange reactions proceed in the direction where at least one of the substances is removed from the reaction sphere in the form of a gaseous substance (3), a precipitate (4, 5), or a low-dissociating compound, most often water (1, 2).
Classification of chemical reactions by changes in oxidation states
Depending on the change in the oxidation states of the elements that make up the reagents and reaction products, all chemical reactions are subdivided into redox (1, 2) and proceeding without a change in the oxidation state (3, 4).
2Mg + CO 2 = 2MgO + C (1)
Mg 0 - 2e = Mg 2+ (reducing agent)
C 4+ + 4e = C 0 (oxidizing agent)
FeS 2 + 8HNO 3 (conc) = Fe (NO 3) 3 + 5NO + 2H 2 SO 4 + 2H 2 O (2)
Fe 2+ -e = Fe 3+ (reducing agent)
N 5+ + 3e = N 2+ (oxidizing agent)
AgNO 3 + HCl = AgCl ↓ + HNO 3 (3)
Ca (OH) 2 + H 2 SO 4 = CaSO 4 ↓ + H 2 O (4)
Thermal classification of chemical reactions
Depending on whether heat (energy) is released or absorbed during the reaction, all chemical reactions are conventionally divided into exo - (1, 2) and endothermic (3), respectively. The amount of heat (energy) released or absorbed during the reaction is called the heat effect of the reaction. If the amount of released or absorbed heat is indicated in the equation, then such equations are called thermochemical.
N 2 + 3H 2 = 2NH 3 +46.2 kJ (1)
2Mg + O 2 = 2MgO + 602.5 kJ (2)
N 2 + O 2 = 2NO - 90.4 kJ (3)
Classification of chemical reactions according to the direction of the reaction
According to the direction of the reaction, reversible (chemical processes, the products of which are able to react with each other under the same conditions in which they were obtained, with the formation of initial substances) and irreversible (chemical processes, the products of which are not able to react with each other to form ).
For reversible reactions the equation in general form is usually written as follows:
A + B ↔ AB
For instance:
CH 3 COOH + C 2 H 5 OH↔ H 3 COOC 2 H 5 + H 2 O
Examples of irreversible reactions include the following reactions:
2KSlO 3 → 2KSl + 3O 2
С 6 Н 12 О 6 + 6О 2 → 6СО 2 + 6Н 2 О
Evidence of the irreversibility of the reaction can be the release of a gaseous substance, a precipitate or a low-dissociating compound, most often water, as the reaction products.
Classification of chemical reactions by the presence of a catalyst
From this point of view, catalytic and non-catalytic reactions are distinguished.
A catalyst is a substance that accelerates the course of a chemical reaction. Reactions involving catalysts are called catalytic. Some reactions are generally impossible without the presence of a catalyst:
2H 2 O 2 = 2H 2 O + O 2 (catalyst MnO 2)
Often, one of the reaction products serves as a catalyst that accelerates this reaction (autocatalytic reactions):
MeO + 2HF = MeF 2 + H 2 O, where Me is a metal.
Examples of problem solving
EXAMPLE 1
