The reactions of the compound are as follows. Introduction to General Chemistry
Chemical reaction- this is the “transformation” of one or more substances into another substance, with a different structure and chemical composition. The resulting substance or substances are called “reaction products.” During chemical reactions, nuclei and electrons form new compounds (redistributed), but their quantity does not change and the isotopic composition of chemical elements remains the same.
All chemical reactions are divided into simple and complex.
Based on the number and composition of the starting and resulting substances, simple chemical reactions can be divided into several main types.
Decomposition reactions are reactions in which several other substances are obtained from one complex substance. At the same time, the formed substances can be both simple and complex. As a rule, for a chemical decomposition reaction to occur, heating is necessary (this is an endothermic process, heat absorption).
For example, when malachite powder is heated, three new substances are formed: copper oxide, water and carbon dioxide:
Cu 2 CH 2 O 5 = 2CuO + H 2 O + CO 2
malachite → copper oxide + water + carbon dioxide
If only decomposition reactions occurred in nature, then all complex substances that can decompose would decompose and chemical phenomena could no longer occur. But there are other reactions.
In compound reactions, several simple or complex substances produce one complex substance. It turns out that the compound reactions are the reverse of the decomposition reactions.
For example, when copper is heated in air, it becomes covered with a black coating. Copper is converted to copper oxide:
2Cu + O 2 = 2CuO
copper + oxygen → copper oxide
Chemical reactions between a simple and a complex substance, in which the atoms that make up the simple substance replace the atoms of one of the elements of the complex substance, are called substitution reactions.
For example, if you dip an iron nail into a solution of copper chloride (CuCl 2), it (the nail) will begin to become covered with copper released on its surface. And by the end of the reaction, the solution turns from blue to greenish: instead of copper chloride, it now contains ferric chloride:
Fe + CuCl 2 = Cu + FeCl 2
Iron + copper chloride → copper + ferric chloride
The copper atoms in copper chloride were replaced by iron atoms.
An exchange reaction is a reaction in which two complex substances exchange their constituent parts. Most often, such reactions occur in aqueous solutions.
In the reactions of metal oxides with acids, two complex substances - an oxide and an acid - exchange their constituent parts: oxygen atoms for acid residues, and hydrogen atoms for metal atoms.
For example, if copper oxide (CuO) is combined with sulfuric acid H 2 SO 4 and heated, a solution is obtained from which copper sulfate can be isolated:
CuO + H 2 SO 4 = CuSO 4 + H 2 O
copper oxide + sulfuric acid → copper sulfate + water
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7.1. Basic types of chemical reactions
Transformations of substances, accompanied by changes 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 shape or physical state of substances changes or the composition of atomic nuclei changes are called physical. An example of physical phenomena is the heat treatment of metals, during which their shape changes (forging), the melting of the metal, the sublimation of iodine, the transformation of water into ice or steam, etc., as well as nuclear reactions, as a result of which atoms are formed from atoms of some elements other elements.
Chemical phenomena can be accompanied by physical transformations. For example, as a result of chemical reactions occurring in a galvanic cell, an electric current arises.
Chemical reactions are classified according to various criteria.
1. According to the sign of the thermal effect, all reactions are divided into endothermic(proceeding with heat absorption) and exothermic(flowing with the release of heat) (see § 6.1).
2. Based on the state of aggregation of the starting 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 (l),
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 (l).
3. According to the ability to flow only in the forward direction, as well as in the forward and reverse directions, they distinguish irreversible And reversible chemical reactions (see § 6.5).
4. Based on the presence or absence of catalysts, they distinguish catalytic And non-catalytic reactions (see § 6.5).
5. According to the mechanism of their occurrence, chemical reactions are divided into ionic, radical etc. (the mechanism of chemical reactions occurring with the participation of organic compounds is discussed in the course of organic chemistry).
6. According to the state of oxidation states of the atoms included in the composition of the reacting substances, reactions occurring without changing the oxidation state atoms, and with a change in the oxidation state of atoms ( redox reactions) (see § 7.2) .
7. Reactions are distinguished by changes in the composition of the starting substances and reaction products connection, decomposition, substitution and exchange. These reactions can occur both with and without changes in the oxidation states of elements, table . 7.1.
Table 7.1
Types of chemical reactions
General scheme |
Examples of reactions that occur 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 |
|
Decompositions (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, atoms of one substance replace atoms of another substance in a 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 mentioned above, all chemical reactions are divided into two groups:
Chemical reactions that occur with a change in the oxidation state of the atoms that make up the reactants are called redox reactions.
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 adding electrons to 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+ .
Atoms, molecules or ions that accept electrons are called oxidizing agents. Restorers are atoms, molecules or ions that donate electrons.
By 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. Thus, the number of electrons given up by the reducing agent is always equal to the number of electrons accepted by the oxidizing agent.
7.2.1. Oxidation state
The oxidation state is the conditional (formal) charge of an atom in a compound, calculated under the assumption that it consists only of ions. The oxidation state is usually denoted by an Arabic numeral above the element symbol with a “+” or “–” sign. For example, Al 3+, S 2–.
To find oxidation states, the following rules are used:
the oxidation state of atoms in simple substances is zero;
the algebraic sum of the oxidation states of atoms in a molecule is equal to zero, in a complex ion - the charge of the ion;
the oxidation state of alkali metal atoms is always +1;
the 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.);
The fluorine atom in compounds always exhibits an oxidation state of –1;
The oxidation state of the oxygen atom in 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 and oxygen, since their highest oxidation state is lower than the number of the group in which they are found. Elements of the copper subgroup form compounds in which their oxidation state exceeds the group number (CuO, AgF 5, AuCl 3).
The maximum negative oxidation state of elements located in the main subgroups of the periodic table can be determined by subtracting the group number from eight. For carbon it 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 secondary subgroups, on the contrary, from top to bottom the stability of higher oxidation states increases.
The conventionality of the concept of oxidation state can be demonstrated using the example of some inorganic and organic compounds. In particular, in phosphinic (phosphorous) 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 four.
Despite the fact that the oxidation state is a conventional concept, it is widely used in composing redox reactions.
7.2.2. The most important oxidizing and reducing agents
Typical oxidizing agents are:
1. Simple substances whose atoms have high electronegativity. These are, first of all, elements of the main subgroups VI and VII of groups of the periodic table: oxygen, halogens. Of the simple substances, the most powerful oxidizing agent is fluorine.
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 in which are in high positive oxidation states: 2–, –, etc.
Reducing agents include:
1. Simple substances whose atoms have low electronegativity are 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 simple ions such as I – , S 2– .
4. Compounds containing complex ions (S 4+ O 3) 2–, (НР 3+ O 3) 2–, in which elements can, by donating electrons, increase their positive oxidation state.
In laboratory practice, the following oxidizing agents are most often 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);
oxides of manganese (IV) and lead (IV) (MnO 2, PbO 2);
molten potassium nitrate (KNO 3) and melts of some other nitrates.
Reducing agents 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 reactions (self-oxidation-self-reduction).
Intermolecular reactions occur with a change in the oxidation state of atoms that are found in different molecules. For example:
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 are 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 .
IN disproportionation reactions(self-oxidation-self-reduction) 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 composing redox reactions
The composition of redox reactions is carried out according to the steps presented in table. 7.2.
Table 7.2
Stages of compiling equations for redox reactions
Action |
|
Determine the oxidizing agent and reducing agent. |
|
Identify the products of the redox reaction. |
|
Create an electron balance and use it to assign 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 coefficients by counting the amount of substance of the atoms (usually hydrogen and oxygen) located on the left and right sides of the reaction equation. |
Let's consider the rules for composing redox reactions using the example of the interaction of potassium sulfite with potassium permanganate in an acidic environment:
1. Determination of oxidizing agent and reducing agent
Manganese, which is in the highest oxidation state, cannot give up electrons. Mn 7+ will accept electrons, i.e. is an oxidizing agent.
The S 4+ ion can donate two electrons and go into 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. Establishment of reaction products
K2SO3 + KMnO4 + H2SO4?
By donating two electrons to an electron, S 4+ becomes S 6+. Potassium sulfite (K 2 SO 3) thus turns into sulfate (K 2 SO 4). In an acidic environment, Mn 7+ accepts 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 included in the permanganate), as well as water molecules. Thus, the reaction under consideration will be written as:
K 2 SO 3 + KMnO 4 + H 2 SO 4 = K 2 SO 4 + MnSO 4 + H 2 O.
3. Compiling electron balance
To compile an electron balance, 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 given up by the reducing agent must be equal to the number of electrons accepted by the oxidizing agent. Therefore, two Mn 7+ and five S 4+ must participate in the reaction:
Mn 7+ + 5 e = Mn 2+ 2,
S 4+ – 2 e = S 6+ 5.
Thus, the number of electrons given up by the reducing agent (10) will be equal to the number of electrons accepted by the oxidizing agent (10).
4. Arrangement of coefficients in the reaction equation
In accordance with the balance of electrons, it is necessary to put a coefficient of 5 in front of K 2 SO 3, and 2 in front of KMnO 4. On the right side, in front of potassium sulfate we set a coefficient of 6, since one molecule is added to the five molecules of K 2 SO 4 formed during the oxidation of potassium sulfite K 2 SO 4 as a result of the binding of potassium ions included in the permanganate. Since the reaction involves two permanganate molecules, on the right side are also formed two manganese sulfate molecules. To bind the reaction products (potassium and manganese ions included in the permanganate), it is necessary three molecules of sulfuric acid, 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 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
Below, as examples, we consider redox reactions involving potassium permanganate, potassium dichromate, hydrogen peroxide, potassium nitrite, potassium iodide and potassium sulfide. 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 environment (acidic, neutral, alkaline), potassium permanganate, acting as an oxidizing agent, gives various reduction products, Fig. 7.1.
Rice. 7.1. Formation of potassium permanganate reduction products in various media
Below are the reactions of KMnO 4 with potassium sulfide as a reducing agent in various environments, illustrating the scheme, Fig. 7.1. In these reactions, the product of sulfide ion oxidation is free sulfur. In an alkaline environment, KOH molecules do not take part in the reaction, but only determine the product of the reduction 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 (chromium) is widely used in laboratory practice as an oxidizing agent. Interacting with a reducing agent, one molecule of potassium dichromate 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 oxidizing agents (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 environment, is reduced according to the scheme, Fig. 7.2.
Rice. 7.2. Possible hydrogen peroxide reduction products
In this case, as a result of the 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 usually occur when substances whose molecules contain a reducing agent and an oxidizing agent are heated. 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 the same atom (ion) is both an oxidizing agent and a reducing agent. Let us consider the process of composing this type of reaction using the example of the interaction of sulfur with alkali.
Characteristic oxidation states of sulfur: – 2, 0, +4 and +6. Acting as a reducing agent, elemental sulfur donates 4 electrons:
S o – 4e = S 4+.
Sulfur – The oxidizing agent accepts two electrons:
S o + 2е = S 2– .
Thus, as a result of the reaction of sulfur disproportionation, compounds are formed whose oxidation states of the element are – 2 and right +4:
3 S + 6 KOH = 2 K 2 S + K 2 SO 3 + 3 H 2 O.
When nitrogen oxide (IV) is disproportioned 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,
Disproportionation of chlorine in a cold alkali solution leads to the formation of hypochlorite, and in a hot alkali 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 redox process that occurs in solutions or melts when a direct electric current is passed through them is called electrolysis. In this case, oxidation of anions occurs at the positive electrode (anode). Cations are reduced at the negative electrode (cathode).
2 Na 2 CO 3 4 Na + O 2 + 2CO 2 .
During the electrolysis of aqueous solutions of electrolytes, along with transformations of the dissolved substance, electrochemical processes can occur 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; hydrogen is reduced instead.
2. Metal cations located in the series of standard electrode potentials (in the voltage series) to the right of hydrogen are reduced to free metals at the cathode during electrolysis.
3. Metal cations located between Al 3+ and H + are reduced at the cathode simultaneously with the hydrogen cation.
The processes occurring in aqueous solutions at the anode depend on the substance from which the anode is made. There are insoluble anodes ( inert) and soluble ( active). Graphite or platinum is used as the material of 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. When halide ions are oxidized, free halogens are released.
2. During the electrolysis of solutions containing the anions SO 2 2–, NO 3 –, PO 4 3–, oxygen is released, i.e. It is not these ions that are oxidized at the anode, but water molecules.
Taking into account the above rules, let us consider, as an example, the electrolysis of aqueous solutions of NaCl, CuSO 4 and KOH with inert electrodes.
1). In solution, sodium chloride dissociates into ions.
1. Indicate the correct definition of a 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 constituents.
2. Indicate the correct definition of a substitution reaction: A. The reaction between a base and an acid; B. The reaction of interaction of two simple substances; B. A reaction between substances in which atoms of a simple substance replace atoms of one of the elements in a complex substance.
3. Indicate the correct definition of a 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 constituents; B. Reaction with the formation of oxygen and hydrogen molecules.
5. What type of reaction is the interaction of acidic oxides with basic oxides: 5. What type of reaction is the interaction of acidic oxides with basic oxides: A. Exchange reaction; B. Compound reaction; B. Decomposition reaction; D. Substitution reaction.
7. Substances whose formulas are KNO 3 FeCl 2, Na 2 SO 4 are called: 7. Substances whose formulas are KNO 3 FeCl 2, Na 2 SO 4 are called: A) salts; B) reasons; B) acids; D) oxides. A) salts; B) reasons; B) acids; D) oxides. 8. Substances whose formulas are HNO 3, HCl, H 2 SO 4 are called: 8. Substances whose formulas are HNO 3, HCl, H 2 SO 4 are called: A) salts; B) acids; B) reasons; D) oxides. A) salts; B) acids; B) reasons; D) oxides. 9. Substances whose formulas are KOH, Fe(OH) 2, NaOH are called: 9. Substances whose formulas are KOH, Fe(OH) 2, NaOH are called: A) salts; B) acids; B) reasons; D) oxides. 10. Substances whose formulas are NO 2, Fe 2 O 3, Na 2 O are called: A) salts; B) acids; B) reasons; D) oxides. 10. Substances whose formulas are NO 2, Fe 2 O 3, Na 2 O are called: A) salts; B) acids; B) reasons; D) oxides. A) salts; B) acids; B) reasons; D) oxides. 11. Indicate the metals that form alkalis: 11. Indicate the metals that form alkalis: Cu, Fe, Na, K, Zn, Li. Cu, Fe, Na, K, Zn, Li.
The concept of “compound reaction” is an antonym of the concept of “decomposition reaction”. Try, using the technique of contrast, to define the concept of “compound reaction.” Right! You have the following formulation.
Let's consider this type of reaction using another form of recording chemical processes that is new to you - the so-called chains of transitions, or transformations. For example, circuit
shows the transformation of phosphorus into phosphorus oxide (V) P 2 O 5, which, in turn, is then converted into phosphoric acid H 3 PO 4.
The number of arrows in the diagram of the transformation of substances corresponds to the minimum number of chemical transformations - chemical reactions. In the example under consideration, these are two chemical processes.
1st process. Obtaining phosphorus oxide (V) P 2 O 5 from phosphorus. Obviously, this is a reaction between phosphorus and oxygen.
Let's put some red phosphorus in a burning spoon and set it on fire. Phosphorus burns with a bright flame producing white smoke consisting of small particles of phosphorus (V) oxide:
4P + 5O 2 = 2P 2 O 5.
2nd process. Let's add a spoonful of burning phosphorus into the flask. It is filled with thick smoke from phosphorus (V) oxide. Take the spoon out of the flask, pour water into the flask and shake the contents, after closing the neck of the flask with a stopper. The smoke gradually thins out, dissolves in the water and finally disappears completely. If you add a little litmus to the solution obtained in the flask, it will turn red, which is evidence of the formation of phosphoric acid:
R 2 O 5 + ZN 2 O = 2H 3 PO 4.
The reactions that are carried out to carry out the transitions under consideration occur without the participation of a catalyst, which is why they are called non-catalytic. The reactions discussed above proceed only in one direction, i.e. they are irreversible.
Let us analyze how many and what substances entered into the reactions discussed above and how many and what substances were formed in them. In the first reaction, one complex substance was formed from two simple substances, and in the second, from two complex substances, each of which consists of two elements, one complex substance was formed, consisting of three elements.
One complex substance can also be formed as a result of the reaction of combining a complex and a simple substance. For example, in the production of sulfuric acid from sulfur oxide (IV) sulfur oxide (VI) is obtained:
This reaction proceeds both in the forward direction, i.e., with the formation of a reaction product, and in the reverse direction, i.e., the decomposition of the reaction product into the starting substances occurs, therefore, instead of the equal sign, they put the reversibility sign.
This reaction involves a catalyst - vanadium (V) oxide V 2 O 5, which is indicated above the reversibility sign:
A complex substance can also be obtained by combining three substances. For example, nitric acid is produced by a reaction whose scheme is:
NO 2 + H 2 O + O 2 → HNO 3.
Let's consider how to select coefficients to equalize the scheme of this chemical reaction.
There is no need to equalize the number of nitrogen atoms: there is one nitrogen atom in both the left and right parts of the diagram. Let's equalize the number of hydrogen atoms - before the acid formula we write the coefficient 2:
NO 2 + H 2 O + O 2 → 2HNO 3.
but in this case the equality of the number of nitrogen atoms will be violated - one nitrogen atom remains on the left side, and there are two on the right side. Let's write the coefficient 2 before the formula of nitric oxide (IV):
2NO 2 + H 2 O + O 2 → 2HNO 3.
Let's count the number of oxygen atoms: there are seven on the left side of the reaction diagram, and six on the right side. To equalize the number of oxygen atoms (six atoms in each part of the equation), remember that before the formulas of simple substances you can write the fractional coefficient 1/2:
2NO 2 + H 2 O + 1/2O 2 → 2HNO 3.
Let's make the coefficients integers. To do this, we rewrite the equation by doubling the coefficients:
4NO 2 + 2H 2 O + O 2 → 4HNO 3.
It should be noted that almost all reactions of the compound are exothermic reactions.
Laboratory experiment No. 15
Calcination of copper in the flame of an alcohol lamp
- Examine the copper wire (plate) given to you and describe its appearance. Heat the wire, holding it with crucible tongs, in the upper part of the flame of an alcohol lamp for 1 minute. Describe the reaction conditions. Describe a sign that indicates a chemical reaction has occurred. Write an equation for the reaction that took place. Name the starting materials and products of the reaction.
Explain whether the mass of the copper wire (plate) changed after the end of the experiment. Justify your answer using your knowledge of the law of conservation of mass of substances.
Key words and phrases
- Combination reactions are antonyms of decomposition reactions.
- Catalytic (including enzymatic) and non-catalytic reactions.
- Chains of transitions or transformations.
- Reversible and irreversible reactions.
Work with computer
- Refer to the electronic application. Study the lesson material and complete the assigned tasks.
- Find email addresses on the Internet that can serve as additional sources that reveal the content of keywords and phrases in the paragraph. Offer your help to the teacher in preparing a new lesson - make a report on the key words and phrases of the next paragraph.
Questions and tasks
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 give up 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 Q p– 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? |
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