Chemical properties of simple substances of metals. Chemical properties of the main classes of chemicals
For the last 200 years of mankind studied the properties of substances better than in the entire history of the development of chemistry. Naturally, the number of substances is also growing rapidly, this is due primarily to the development of various methods for obtaining substances.
V Everyday life we are exposed to many things. Among them are water, iron, aluminum, plastic, soda, salt and many others. Substances that exist in nature, such as oxygen and nitrogen contained in the air, substances dissolved in water, and having a natural origin, are called natural substances. Aluminum, zinc, acetone, lime, soap, aspirin, polyethylene and many other substances do not exist in nature.
They are obtained in the laboratory and produced by the industry. Artificial substances do not occur in nature, they are created from natural substances. Some substances that exist in nature can also be obtained in a chemical laboratory.
So, when potassium permanganate is heated, oxygen is released, and when chalk is heated - carbon dioxide. Scientists have learned how to turn graphite into diamond, grow crystals of ruby, sapphire and malachite. So, along with substances of natural origin, there is a huge variety of artificially created substances that are not found in nature.
Substances that are not found in nature are produced at various enterprises: factories, plants, combines, etc.
In exhaustion natural resources of our planet, chemists now face an important task: to develop and implement methods by which it is possible to obtain artificially, in a laboratory or industrial production, substances that are analogues of natural substances. For example, reserves of fossil fuels in nature are running out.
There may come a time when oil and natural gas run out. Already, new types of fuels are being developed that would be just as efficient, but would not pollute environment. To date, mankind has learned to artificially obtain various precious stones, such as diamonds, emeralds, beryls.
Aggregate state of matter
Substances can exist in several states of aggregation, three of which you know: solid, liquid, gaseous. For example, water in nature exists in all three states of aggregation: solid (in the form of ice and snow), liquid (liquid water) and gaseous (water vapor). Known substances that cannot exist in normal conditions in all three aggregate states. An example of this is carbon dioxide. At room temperature, it is an odorless and colorless gas. At -79°С this substance "freezes" and passes into a solid state of aggregation. The household (trivial) name for such a substance is "dry ice". This name is given to this substance due to the fact that “dry ice” turns into carbon dioxide without melting, that is, without transitioning to a liquid state of aggregation, which is present, for example, in water.
Thus, an important conclusion can be drawn. When a substance passes from one state of aggregation to another, it does not change into other substances. The very process of some change, transformation, is called a phenomenon.
physical phenomena. Physical properties of substances.
Phenomena in which substances change their state of aggregation, but do not turn into other substances, are called physical. Each individual substance has certain properties. The properties of substances can be different or similar to each other. Each substance is described using a set of physical and chemical properties. Let's take water as an example. Water freezes and turns into ice at a temperature of 0°C, and boils and turns into steam at a temperature of +100°C. These phenomena are physical, since water has not turned into other substances, only a change in the state of aggregation occurs. These freezing and boiling points are physical properties specific to water.
The properties of substances that are determined by measurements or visually in the absence of the transformation of some substances into others are called physical
The evaporation of alcohol, like the evaporation of water- physical phenomena, substances at the same time change the state of aggregation. After the experiment, you can make sure that alcohol evaporates faster than water - these are the physical properties of these substances.
The main physical properties of substances include the following: state of aggregation, color, smell, solubility in water, density, boiling point, melting point, thermal conductivity, electrical conductivity. Such physical properties as color, smell, taste, shape of crystals can be determined visually, using the senses, and density, electrical conductivity, melting and boiling points are determined by measurement. Information about physical properties ah many substances are collected in specialized literature, for example, in reference books. The physical properties of a substance depend on its state of aggregation. For example, the density of ice, water and water vapor is different.
Gaseous oxygen is colorless, and liquid oxygen is blue. Knowledge of physical properties helps to "recognize" a lot of substances. For instance, copper- the only red metal. Only table salt has a salty taste. iodine- an almost black solid that turns into a purple vapor when heated. In most cases, to define a substance, several of its properties must be considered. As an example, we characterize the physical properties of water:
- color - colorless (in a small volume)
- odor - odorless
- state of aggregation - under normal conditions, liquid
- density - 1 g / ml,
- boiling point – +100°С
- melting point - 0°С
- thermal conductivity - low
- electrical conductivity - pure water does not conduct electricity
Crystalline and amorphous substances
When describing the physical properties of solids, it is customary to describe the structure of the substance. If you look at a sample of table salt under a magnifying glass, you will notice that the salt consists of many tiny crystals. Very large crystals can also be found in salt deposits. Crystals - solid bodies, having the form of regular polyhedra Crystals can be of various shapes and sizes. Crystals of certain substances, such as table salt – fragile, easy to break. There are crystals quite hard. For example, one of the hardest minerals is diamond. If you look at salt crystals under a microscope, you will notice that they all have a similar structure. If we consider, for example, glass particles, then they will all have a different structure - such substances are called amorphous. Amorphous substances include glass, starch, amber, beeswax. Amorphous substances - substances that do not have a crystalline structure
chemical phenomena. Chemical reaction.
If at physical phenomena substances, as a rule, only change the state of aggregation, then with chemical phenomena, some substances are transformed into other substances. Here are a few simple examples: the burning of a match is accompanied by the charring of wood and the release of gaseous substances, that is, the irreversible transformation of wood into other substances occurs. Another example: over time, bronze sculptures become covered with a green coating. This is because bronze contains copper. This metal reacts slowly with oxygen, carbon dioxide and air moisture, as a result, new green substances are formed on the surface of the sculpture Chemical phenomena - the phenomena of the transformation of one substance into another The process of interaction of substances with the formation of new substances is called a chemical reaction. Chemical reactions take place all around us. Chemical reactions take place in ourselves. In our body, transformations of many substances are constantly taking place, substances react with each other, forming reaction products. Thus, in chemical reaction there are always reacting substances, and substances formed as a result of the reaction.
- Chemical reaction- the process of interaction of substances, as a result of which new substances with new properties are formed
- Reagents- substances that enter into a chemical reaction
- Products- substances formed as a result of a chemical reaction
The chemical reaction is depicted in general view reaction scheme REAGENTS -> PRODUCTS
- reagents– initial substances taken for the reaction;
- products- new substances formed as a result of the reaction.
Any chemical phenomena (reactions) are accompanied by certain signs, with the help of which chemical phenomena can be distinguished from physical ones. Such signs include a change in the color of substances, the release of gas, the formation of a precipitate, the release of heat, and the emission of light.
Many chemical reactions are accompanied by the release of energy in the form of heat and light. As a rule, such phenomena are accompanied by combustion reactions. In combustion reactions in air, substances react with oxygen contained in the air. So, for example, magnesium metal flares up and burns in air with a bright blinding flame. That is why magnesium flash was used to create photographs in the first half of the twentieth century. In some cases, it is possible to release energy in the form of light, but without the release of heat. One of the species of Pacific plankton is able to emit a bright blue light, clearly visible in the dark. The release of energy in the form of light is the result of a chemical reaction that occurs in the organisms of this type of plankton.
Summary of the article:
- There are two large groups of substances: substances of natural and artificial origin.
- Under normal conditions, substances can be in three states of aggregation
- The properties of substances that are determined by measurements or visually in the absence of the transformation of some substances into others are called physical
- Crystals are solid bodies that have the shape of regular polyhedra.
- Amorphous substances - substances that do not have a crystalline structure
- Chemical phenomena - the phenomena of the transformation of one substance into another
- Reagents are substances that enter into a chemical reaction.
- Products - substances formed as a result of a chemical reaction
- Chemical reactions may be accompanied by the release of gas, sediment, heat, light; color change of substances
- Combustion is a complex physical and chemical process of transformation of starting materials into combustion products during a chemical reaction, accompanied by intense release of heat and light (flame)
2NaOH + Zn + 2H 2 O \u003d Na 2 + H 2
2KOH + 2Al + 6H 2 O = 2K + 3H 2
salt
1. Salt of a weak acid + strong acid \u003d salt of a strong acid + weak acid
Na 2 SiO 3 + 2HNO 3 \u003d 2NaNO 3 + H 2 SiO 3
BaCO 3 + 2HCl \u003d BaCl 2 + H 2 O + CO 2 (H 2 CO 3)
2. Soluble salt + soluble salt = insoluble salt + salt
Pb(NO 3) 2 + K 2 S = PbS + 2KNO 3
CaCl 2 + Na 2 CO 3 \u003d CaCO 3 + 2NaCl
3. Soluble salt + alkali \u003d salt + insoluble base
Cu(NO 3) 2 + 2NaOH = 2NaNO 3 + Cu(OH) 2
2FeCl 3 + 3Ba(OH) 2 = 3BaCl 2 + 2Fe(OH) 3
4. Soluble metal salt (*) + metal (**) = metal salt (**) + metal (*)
Zn + CuSO 4 \u003d ZnSO 4 + Cu
Cu + 2AgNO 3 \u003d Cu (NO 3) 2 + 2Ag
Important: 1) metal (**) must be in the voltage series to the left of metal (*), 2) metal (**) must NOT react with water.
Example 1 Zinc hydroxide can react with each substance in a pair:
1) calcium sulfate, sulfur oxide (VI);
2) sodium hydroxide (solution), hydrochloric acid;
3) water, sodium chloride;
4) barium sulfate, iron (III) hydroxide.
Solution- 2) Zinc hydroxide - amphoteric. It reacts with both acids and alkalis.
Example 2 A solution of copper(II) sulfate reacts with each of the two substances:
1) HCl and H 2 SiO 3;
2) H 2 O and Cu (OH) 2;
3) O 2 and HNO 3;
4) NaOH and BaCl 2 .
Solution- 4) In solutions, the reaction proceeds if the conditions are met: a precipitate forms, a gas is released, a low-dissociating substance is formed, for example, water.
Example 3 The scheme of transformations E -> E 2 O 3 -> E (OH) 3 corresponds to the genetic series:
1) sodium -> sodium oxide -> sodium hydroxide;
2) aluminum -> aluminum oxide -> aluminum hydroxide;
3) calcium -> calcium oxide -> calcium hydroxide;
4) nitrogen -> nitric oxide (V) -> nitric acid.
Solution- 2) According to the scheme, you can find out that the element is a trivalent metal, which forms the corresponding oxide and hydroxide.
Example 4 How to make the following transformations:
Ca → Ca(OH) 2 → CaCO 3 → CaO → CaSO 4 → CaCl 2 → Ca?
Solution:
Ca + 2H 2 O \u003d Ca (OH) 2 + H 2
Ca(OH) 2 + H 2 CO 3 = CaCO 3 + 2H 2 O
CaCO 3 == t CaO + CO 2
CaO + SO 3 \u003d CaSO 4
CaSO 4 + BaCl 2 \u003d CaCl 2 + BaSO 4
CaCl 2 + Ba = BaCl 2 + Ca
Assignments on topic 5
161-170. Confirm the acid properties of oxides with the reaction equations in molecular and ionic form. Name the resulting substances.
181-190. Write the reaction equations that can be used to carry out the following transformations of substances:
| № | Transformation scheme |
| Potassium→potassium hydroxide→potassium carbonate→potassium nitrate→potassium sulfate | |
| Zinc→zinc chloride→zinc hydroxide→zinc oxide→zinc nitrate | |
| Copper(II)→copper oxide→copper sulfate→copper hydroxide→copper oxide→copper chloride | |
| Carbon→carbon dioxide→sodium carbonate→calcium carbonate→carbon dioxide | |
| Hydrogen→water→sodium hydroxide→sodium carbonate→sodium nitrate | |
| Sulfur→hydrogen sulfide→sodium sulfide→iron(II) sulfide→hydrogen sulfide | |
| Sodium→sodium hydroxide→sodium sulfide→sodium chloride→sodium sulfate | |
| Magnesium→magnesium sulfate→magnesium hydroxide→magnesium oxide→magnesium chloride | |
| Lead→lead(II) oxide→lead nitrate→lead hydroxide→lead oxide→lead sulfate | |
| Sulfur→hydrogen sulfide→potassium sulfide→potassium chloride→hydrochloric acid | |
| Calcium→calcium hydroxide→calcium carbonate→calcium nitrate→nitric acid | |
| Aluminum→aluminum sulfate→aluminum hydroxide→aluminum oxide→aluminum nitrate | |
| Sulfur → sulfur(IV) oxide → sulfurous acid → sodium sulfite → sulfurous acid | |
| Oxygen→aluminum oxide→aluminum sulfate→aluminum hydroxide→sodium metaaluminate | |
| Aluminum → aluminum chloride → aluminum nitrate → aluminum hydroxide → aluminum sulfate | |
| Copper→copper(II) chloride→copper→copper(II) oxide→copper nitrate | |
| Iron→iron(II) chloride→iron(II) hydroxide→iron(II) sulfate→iron | |
| Iron→iron(III) chloride→iron(III) nitrate→iron(III) sulfate→iron | |
| Aluminum→aluminum nitrate→aluminum hydroxide→aluminum oxide→sodium aluminate→aluminum sulfate | |
| Zinc→sodium tetrahydroxozincate→zinc nitrate→zinc hydroxide→zinc oxide→potassium zincate |
Chemical reactions.
One of the types of interaction of atoms, molecules and ions are reactions in whichreagentsgive and others receiveelectrons. During these reactions, called redox, atoms of one or more elements change their oxidation state.
Under oxidation state is understood as the conditional charge that would arise on a given atom, if we assume that all bonds in a particle (molecule, complex ion) are ionic. In this case, it is believed that the electrons are completely displaced to a more electronegative atom, which attracts them more strongly. The concept of the oxidation state is formal and often does not coincide with either the effective charges of atoms in compounds, or the actual number of bonds that an atom forms. However, it is convenient in the formulation of equations for redox processes and is useful in describing the redox properties of chemical compounds.
The oxidation states of atoms are calculated based on the following basic rules: The oxidation state is indicated by a superscript above the atom, with its sign first and then the value. It can be either an integer or a fractional number. For example, if in H 2 O and H 2 O2 for oxygen the oxidation state is (-2) and (-1), then in KO2 and KO3- respectively (-1/2) and (-1/3).
1) the oxidation state of an atom in simple substances ax is zero, for example:
Na 0 ; H20; Cl 0 2; O 2 0, etc.;
2) the oxidation state of a simple ion, for example: Na +; Ca+2; Fe+3; Cl-; S-2 is equal to its charge, i.e., respectively, (+1); (+2); (+3); (-one); (-2);
3) in most compounds, the oxidation state of the hydrogen atom is (+1) (except for the hydrides Me - LiH; CaH, etc., in which it is (-1));
4) the oxidation state of the oxygen atom in most compounds is
(-2), except for peroxides (-1), oxygen fluoride OF2 (+2), etc.;
5) the algebraic sum of the values of the oxidation states of all atoms in the molecule is zero, and in a complex ion - the charge of this ion. For example, the oxidation state of nitrogen in a nitric acid molecule - HNO3 is determined as follows: the oxidation state of hydrogen is (+1), oxygen (-2), nitrogen (x). Compiling algebraic equation: (+1) + x + (-2) 3 = 0, get x = +5.
Returning to the definition of redox reactions, we note that Oxidation is the process of giving up electrons, and reduction is the process of their addition. An oxidizing agent is a substance containing an element whose oxidation state decreases during the reaction. Reducing agent - a substance containing an element whose oxidation state increases during the reaction. It should be emphasized that the oxidation and reduction reactions are impossible one without the other ( coupled reactions). Thus, as a result of a redox reaction, the oxidizing agent is reduced, and the reducing agent is oxidized.
Typical reducing agents:
1) metals, for example: K, Mg, Al, Zn and some non-metals in the free state - C, H (in most cases), etc.;
2) simple ions corresponding to the lowest oxidation state of the element: S2-; I; Cl-and others;
3) complex ions and molecules containing atoms in the lowest degree of oxidation
ions: N in the NH4 ion, S in the H 2 S molecule, I in the KI molecule, etc.
Typical oxidizers:
1) atoms and molecules of some non-metals: F2; Cl and O2 (in most cases), etc.;
2) simple ions corresponding to the highest oxidation states of the element: Hg+2; Au+3; Pb and others;
3) complex ions and molecules containing atoms in the highest degree oxidation: Pb +4 in PbO2; N+5 in HNO3; S +6 in HSO4; Cr +6 in Cr2O7 2- or CrO4 2- ; Mn +7 in MnO - etc.
Some substances have dual redox function, exhibiting (depending on the conditions) either oxidative or restorative properties. These include molecules of certain substances, simple and complex ions, in which the atoms are in an intermediate oxidation state: C +2 in the CO molecule, O - in the H 2 O 2 molecule, in the S +4 ion SO 3 2-, in the N ion +3 in the NO 2 ion - and others.
In a redox reaction, electrons are transferred from the reducing agent to the oxidizing agent.
Example 1 Write the equation for the oxidation of iron (II) disulfide with concentrated nitric acid. Make up: schemes of electronic and electron-ionic balance.
Solution. HNO 3 is a strong oxidizing agent, therefore sulfur will be oxidized to the maximum oxidation state S +6, and iron to Fe +3, while HNO 3 can be reduced to NO or NO 2. Consider the case of recovery to NO 2 .
FeS 2 + HNO 3 (conc) → Fe (NO 3) 3 + H 2 SO 4 + NO 2.
Where H 2 O will be located (on the left or right side) is still unknown.
Equalize this reaction method electronic balance. The recovery process is described by the diagram:
N +5 + e → N +4
Two elements enter into the oxidation half-reaction at once - Fe and S. Iron in disulfide has an oxidation state of +2, and sulfur -1. It is necessary to take into account that there are two S atoms for one Fe atom:
Fe +2 - e → Fe +3
2S - - 14e → 2S +6 .
Together, iron and sulfur donate 15 electrons.
The full balance looks like:
15 HNO 3 molecules go to the oxidation of FeS 2, and 3 more HNO 3 molecules are necessary for the formation of Fe (NO 3) 3:
FeS 2 + 18HNO 3 → Fe (NO 3) 3 + 2H 2 SO 4 + 15NO 2.
To equalize hydrogen and oxygen, 7 H 2 O molecules must be added to the right side:
FeS 2 + 18HNO 3 (conc) \u003d Fe (NO 3) 3 + 2H 2 SO 4 + 15NO 2 + 7H 2 O.
Let us now use the method of electron-ion balance. Consider the oxidation half-reaction. The FeS 2 molecule turns into a Fe 3+ ion (Fe (NO 3) 3 completely dissociates into ions) and two SO 4 2- ions (dissociation of H 2 SO 4):
FeS 2 → Fe 3+ + 2SO 2 4-.
In order to equalize oxygen, add 8 H 2 O molecules to the left side, and 16 H + ions to the right side (acidic environment!):
FeS 2 + 8H 2 O → Fe 3+ + 2SO 4 2- + 16H +.
The charge of the left side is 0, the charge of the right side is +15, so FeS 2 must give 15 electrons:
FeS 2 + 8H 2 O - 15e → Fe 3+ + 2SO 4 2- + 16H +.
Consider now the reduction half-reaction of the nitrate ion:
NO -3 → NO 2.
It is necessary to take one O atom from NO 3. To do this, add 2 H + ions (acidic medium) to the left side, and one H 2 O molecule to the right side:
NO 3 - + 2H + → NO 2 + H 2 O.
To equalize the charge to the left side (charge +1), add one electron:
NO 3 - + 2H + + e → NO 2 + H 2 O.
The complete electron-ion balance has the form:
Reducing both parts by 16H + and 8H 2 O, we get the reduced ionic equation of the redox reaction:
FeS 2 + 15NO 3 - + 14H + = Fe 3+ + 2SO 4 2- + 15NO 2 + 7H 2 O.
Adding to both sides of the equation the appropriate number of ions, three ions NO 3 - and H +, we find the molecular reaction equation:
FeS 2 + 18HNO 3 (conc) \u003d Fe (NO 3) 3 + 2H 2 SO 4 + 15NO 2 + 7H 2 O.
Chemical kinetics studies the rates and mechanisms of chemical processes, as well as their dependence on various factors. The rate of chemical reactions depends on: 1) the nature of the reactants; 2) reaction conditions: concentration of reactants; pressure, if gaseous substances are involved in the reaction; temperature; the presence of a catalyst.
EXAMPLE 2 . Calculate how many times the reaction rate will increase with an increase in temperature by 40 °, if the temperature coefficient of the reaction rate is 3.
SOLUTION. The dependence of the reaction rate on temperature is expressed by the empirical van't Hoff's rule, according to which for every 10 ° increase in temperature, the rate of most homogeneous reactions increases by 2-4 times, or
where is the temperature coefficient of the reaction rate, often takes values of 2-4, shows how many times the reaction rate will increase with an increase in temperature by 10 degrees;
v T 1 , v T2 - chemical reaction rates at temperatures T1 and T2. In this example:
The reaction rate will increase by 81 times
EXAMPLE 3. Oxidation of carbon monoxide (II) and graphite proceeds according to the equations: a) 2CO(g) + O= 2CO2(g);
b) 2C(t) + O2(g)= 2CO(g).
Calculate how the rates of these reactions will change if you increase three times: 1) the concentration of oxygen; 2) the volume of the reaction space; 3) pressure in the system.
Solution: Reaction a) proceeds in homogeneous system - all substances are in the same phase (all substances are gases), reaction b) proceeds in heterogeneous system - the reactants are in different phases (O2 and CO - gases, C - solid). Therefore, the reaction rates for these systems, according to the ZDM, are equal to:
a) 2CO(g) + O2(g) = 2CO; b) 2C(t) + O2(g) = 2CO(g);
a) 
b) 
After increasing the oxygen concentration, the rates of reactions a) and b) will be equal to:
a) 
b) 
The increase in the reaction rate in relation to the initial one is determined by the ratio:
A) 
b) 
Therefore, after an increase in the oxygen concentration by a factor of 3, the rates of reactions a) and b) will increase by a factor of 3.
2) An increase in the volume of the system by 3 times will cause a decrease in the concentration of each gaseous substance by 3 times. Therefore, the reaction rates will decrease by 27 times (a) and 3 times (b), respectively:
A) 
b) 
3) An increase in pressure in the system by 3 times will cause a decrease in volume by 3 times and an increase in the concentration of gaseous substances by 3 times. So:
A) 
b) 
EXAMPLE 4. The decomposition reaction of phosphorus pentachloride proceeds according to the equation:
PCl5(g)= PCl3(g)+ Cl2(g); H = +92.59 kJ.
In what direction will the equilibrium of this reaction shift with: a) an increase in the concentration of PCl5; b) increasing the concentration of Cl2; c) pressure increase; d) lowering the temperature; e) the introduction of a catalyst.
SOLUTION. A shift or shift in chemical equilibrium is a change in the equilibrium concentrations of reactants as a result of a change in one of the conditions for the reaction to occur. The direction of equilibrium shift is determined by Le Chatelier's principle: if any external influence is exerted on a system in equilibrium (change concentration, pressure, temperature), then the equilibrium will shift in the direction of that reaction (direct or reverse) that counteracts the effect.
a) An increase in the concentration of reactants (PCl5) increases the rate of the forward reaction compared to the rate of the reverse reaction, and the equilibrium shifts towards the forward reaction, i.e. right;
b) an increase in the concentration of products (Cl2) of the reaction increases the rate of the reverse reaction compared to the rate of the forward reaction, and the equilibrium shifts to the left;
c) an increase in pressure shifts the equilibrium towards the reaction proceeding with the formation of a smaller amount of gaseous substances. In this example, the forward reaction is accompanied by the formation of 2 mol of gases (1 mol of PCl3 and 1 mol of Cl2), and the reverse reaction is accompanied by the formation of 1 mol of PCl5. Therefore, an increase in pressure will lead to a shift of equilibrium to the left, i.e. in the direction of the reverse reaction;
d) since the direct reaction proceeds with the absorption of heat), then lowering the temperature shifts the equilibrium in the opposite direction (exothermic reaction);
e) the introduction of a catalyst into the system does not affect the equilibrium shift, because equally increases the rate of forward and reverse reactions.
Topic 6 assignments
201-220. According to these schemes, make up the equations of redox reactions, indicate the oxidizing agent and reducing agent:
| № | Reaction scheme |
| KBr + KBrO 3 + H 2 SO 4 →Br 2 + K 2 SO 4 + H 2 O | |
| KClO 3 + Na 2 SO 3 → Na 2 SO 4 + MnO 2 + KOH | |
| PbS+HNO 3 →S+Pb(NO 3) 2 +NO+H 2 O | |
| KMnO 4 + Na 2 SO 3 +KOH → K 2 MnO 4 + Na 2 SO 4 + H 2 O | |
| P+ HNO 3 + H 2 O → H 3 PO 4 + NO | |
| Cu 2 O+ HNO 3 →Cu(NO 3) 2 +NO+ H 2 O | |
| KClO 3 + Na 2 SO 3 → S+ K 2 SO 4 + MnSO 4 + H 2 O | |
| HNO 3 + Ca → NH 4 NO 3 + Ca (NO 3) 2 + H 2 O | |
| NaCrO 2 + PbO 2 + NaOH → Na 2 CrO 4 + Na 2 PbO 2 + H 2 O | |
| K 2 Cr 2 O 7 + H 2 S+ H 2 SO 4 → S + Cr 2 (SO 4) 3 + K 2 SO 4 + H 2 O | |
| KClO 3 + Na 2 SO 3 → KCl + Na 2 SO 4 | |
| KMnO 4 + HBr → Br 2 + KBr + MnBr 2 + H 2 O | |
| H 3 AsO 3 + KMnO 4 + H 2 SO 4 → H 3 AsO 4 + MnSO 4 + K 2 SO 4 + H 2 O | |
| P + HClO 3 + H 2 O → H 3 PO 4 + HCl | |
| NaCrO 2 + Br 2 + NaOH → Na 2 CrO 4 + NaBr + H 2 O | |
| FeS+ HNO 3 →Fe(NO 3) 2 +S+ NO+ H 2 O | |
| HNO 3 + Zn → N 2 O + Zn (NO 3) 2 + H 2 O | |
| FeSO 4 + KClO 3 + H 2 SO 4 → Fe 2 (SO 4) 3 + KCl + H 2 O | |
| K 2 Cr 2 O 7 + HCl → Cl 2 + CrCl 3 + KCl + H 2 O | |
| Au+ HNO 3 + HCl→AuCl 3 +NO+ H 2 O |
221-230. How many times will the rate of the direct reaction change if the temperature regime is changed from T 1 to T 2? The temperature coefficient is given in the table.
| № | ||||||||||
| T 1, K | ||||||||||
| T 2 , K | ||||||||||
| γ |
231-240. Calculate how many times the reaction rate will change if the process conditions are changed.
236-240. How should a) temperature be changed, b) pressure, c) concentration, in order to shift the chemical equilibrium in the direction of a direct reaction?
Metals and non-metals.
The set of OVR that occur on the electrodes in solutions or melts of electrolytes when passed through them electric current is called electrolysis.
At the cathode of the current source, the process of transferring electrons to cations from a solution or melt occurs, therefore the cathode is a "reducing agent". At the anode, electrons are given off by anions, so the anode is an “oxidizing agent”. During electrolysis, competing processes can occur both at the anode and at the cathode.
When electrolysis is carried out using an inert (non-consumable) anode (for example, graphite or platinum), as a rule, two oxidation and reduction processes are competing:
- at the anode— oxidation of anions and hydroxide ions,
- on the cathode— reduction of hydrogen cations and ions.
When electrolysis is carried out using an active (consumable) anode, the process becomes more complicated and the competing reactions on the electrodes are as follows:
- at the anode- oxidation of anions and hydroxide ions, anodic dissolution of the metal - the material of the anode;
- on the cathode- recovery of the salt cation and hydrogen ions, recovery of metal cations obtained by dissolving the anode. When choosing the most probable process at the anode and cathode, it is assumed that the reaction proceeds that requires the least energy. In the electrolysis of salt solutions with an inert electrode, the following rules are used.
1. The following products can form at the anode:
a) during the electrolysis of solutions containing anions F -, SO 4 2-, NO 3 -, PO 4 3-, OH - oxygen is released;
b) during the oxidation of halide ions, free halogens are released;
c) during the oxidation of anions organic acids process is going on:
2RCOO - - 2е → R-R + 2СО 2.
2. During the electrolysis of salt solutions containing ions located in a series of voltages to the left of Al 3+, hydrogen is released at the cathode; if the ion is located to the right of hydrogen, then a metal is released.
3. During the electrolysis of salt solutions containing ions located between Al 3+ and H + on the cathode, competing processes of both cation reduction and hydrogen evolution can occur.
The dependence of the amount of substance formed during electrolysis on time and current strength is described by the generalized Faraday law:
m = (E / F) . I. t = (M / (n . F)) . I. t,
where m is the mass of the substance formed during electrolysis (g); E is the equivalent mass of the substance (g/mol); M is the molar mass of the substance (g/mol); n is the number of given or received electrons; I - current strength (A); t is the duration of the process (s); F is the Faraday constant, which characterizes the amount of electricity required to release 1 equivalent mass of a substance (F = 96500 C / mol = 26.8 A. h / mol).
Example 1 Electrolysis of sodium chloride melt:
NaCl = Na + + Cl - ;
cathode (-) (Na +): Na + + e= Na 0 ,
anode (-) (Cl -): Cl - - e\u003d Cl 0, 2Cl 0 \u003d Cl 2;
2NaCl \u003d 2Na + Cl 2.
Example 2 Electrolysis of sodium chloride solution:
NaCl \u003d Na + + Cl -,
H 2 O \u003d H + + OH -;
cathode (-) (Na + ; H +): H + + e= H 0 , 2H 0 = H 2
(2H 2 O + 2 e\u003d H 2 + 2OH -),
anode (+) (Cl - ; OH -): Cl - - e\u003d Cl 0, 2Cl 0 \u003d Cl 2;
2NaCl + 2H 2 O \u003d 2NaOH + Cl 2 + H 2.
Example 3 Electrolysis of copper(II) nitrate solution:
Cu(NO 3) 2 \u003d Cu 2+ + NO 3 -
H 2 O \u003d H + + OH -;
cathode (-) (Cu 2+; H +): Cu 2+ + 2 e= Cu 0 ,
anode (+) (OH -): OH - - e=OH0,
4H 0 \u003d O 2 + 2H 2 O;
2Cu(NO 3) 2 + 2H 2 O \u003d 2Cu + O 2 + 4HNO 3.
Assignments on topic 7
241-250. Make up the electronic equations of the processes occurring on inert electrodes during the electrolysis of a) a melt, b) a solution of a substance:
| № | ||||||||||
| Substance | NaOH | Kcl | AgNO3 | Cu(NO 3) 2 | FeSO4 | K 2 S | KOH | Fe(NO 3) 2 | ZnSO4 | Zn(NO 3) 2 |
251-260. What substances and in what quantity will be released on the carbon electrodes during the electrolysis of the solution during the time t (h) at the current strength I (A).
271-280. Write an equation for the reaction between substances, given that the transfer of electrons is maximum.
| № | Substances | № | Substances |
| P+HNO 3 (conc) | H 2 S+ H 2 SO 4 (conc) | ||
| P + H 2 SO 4 (conc) | PH 3 +HNO 3 (conc) | ||
| S+HNO 3 (conc) | PH 3 + H 2 SO 4 (conc) | ||
| S+ H 2 SO 4 (conc) | HClO+HNO 3 (conc) | ||
| H 2 S+HNO 3 (conc) | HClO+ H 2 SO 4 (conc) |
Main:
1. Erokhin Yu.M. "Chemistry": A textbook for secondary vocational schools. - M .: Publishing Center "Academy", 2004.
2. Rudzitis G.E., Feldman F.G. "Chemistry" 10 cells-M.: Enlightenment. 1995.
3. Rudzitis G.E., Feldman F.G. "Chemistry" 11 cells. -M.: Enlightenment. 1995.
4. Akhmetov M.S. "Laboratory and seminar classes on general and not organic chemistry» M.: graduate School. 2002.
Additional:
1. Petrov M.M., Mikhilev L.A., Kukushkin Yu.N. "Inorganic chemistry". M.: Chemistry. 1989.
2. Potapov V.M. "Organic Chemistry". - M .: Education. 1983.
3. Mikhilev L.A., Passet N.F., Fedotova M.I. "Problems and Exercises in Inorganic Chemistry". M.: Chemistry. 1989.
4. Potapov V.M., Tatarinchik S.N., Averina A.V. "Problems and exercises in organic chemistry" -M.: Chemistry. 1989.
5. Khomchenko I.G. " general chemistry". -M.: New wave. -ONYX 1999.
6. Khomchenko G.P. "Collection of problems in chemistry for applicants to the university." -M.: New wave. 1999.
God gave man iron, and the devil slipped rust.
Proverb
Change of properties in decades. Since the d-elements are characterized positive st.ok., then in the form of simple substances they exhibit reducing properties, which in aqueous solutions are characterized by the value of the redox potential E . 0 In decades, from left to right, its value, correlating with the value of I 1 , growing, but upon transition to manganese and to the zinc subgroup, despite a sharp increase in I 1, it decreases due to a decrease in the value of I 2 and a decrease in the energy of the crystal lattice upon transition to these metals (from those located to the left of them in the periodic table).
In the compact state at rev. even M of the first decade, having negative values of E (0 from Sc to Mn E 0< −0,90 B), с водой не реагируют вследствие образованияpassivating oxide films on their surface. However, at red heat temperatures, even less active metals (iron, nickel, vanadium and titanium analogues) displace hydrogen from water. The reactivity of M also sharply increases when they are transferred to finely dispersed state, for example, manganese and chromium powders interact with water at about. (with the formation of MnO 2 and Cr 2 O 3).
All metals of the first decade, for which E 0< 0, кроме ванадия. Наиболее активные М: цинк и марганец – растворяются даже в уксусной кислоте, а медь (в ряду напряжений стоит правее водорода) лишь в т.н. кислотах-окислителях. При указанных взаимодействиях только Sc и Тi образуют соединения в ст.ок. (+3), остальные – в (+2), хотя хром(II) и (гораздо медленнее) железо(II) на воздухе затем окисляются до (+3).
The anomalous passivity of vanadium (E 0 \u003d -1.20 V) in dilute acids is explained special density its oxide film. It dissolves only in HF or concentrated HNO, 3 with which this metal reacts:
V + HNO 3 \u003d HVO 3 + NO.
Other active M depending on solubility their oxide films in concentrated nitric acid either react with it, reducing nitrogen to (-3) (these are zinc, manganese and the scandium subgroup), or are passivated by it due to the thickening of the oxide film, such as Cr 124.
Passivation can also be carried out artificially. Thus, the treatment of chromium (located in a series of voltages between zinc and iron) with concentrated nitric acid increases its potential from -0.56 V to +1.2 V, i.e. makes Cr almost as noble as Pt. (Chromium is especially easily deactivated in the composition of stainless steel and other alloys 125 .) Concentrated H 2 SO 4 and HNO 3 also passivate iron.
Cobalt and nickel are similar to Fe in chemical activity due to the proximity of the atomic radii (therefore, they are combined into familygland). However, if iron reacts with dilute HCl and H 2 SO 4 at vol., then Co and Ni when heated. In addition, they are deactivated by nitric acid to a lesser extent than iron, due to the greater solubility of their oxides in this acid.
Note that for the elements of the second and third decades, the nature of the change in the value of E 0 remains approximately the same as in the first.
Changes to properties in subgroups. The value of I 1 in d-subgroups is mainly grows andincreased strength bonds in the M lattice (compare mp). As a consequence (in contrast to the main subgroups and the Sc subgroup), the value of E 0 becomes more positive, and the reactivity of metals decreases.
So, in subgroup IB, if copper dissolves in concentrated sulfuric acid at o.c., then silver only at t> 160 0 C. However, silver, like copper, at roomtemperature interacts with nitric acid, and gold - only with "aqua regia" (as well as with selenic acid (see above) and with chlorine water in the presence of HCl).
In the IIB subgroup, Zn is soluble even in acetic acid, Cd in HCl, and Hg (E 0 > 0) only in HNO (3 with a lack of acid, oxidation goes to Hg, 2 2 + and with an excess - to Hg). 2+
Similarly, in the VIIB subgroup, Mn also reacts with CHCOOH 3, and Tc and Re (their values
E 0: 0.47 V and 0.37 V, respectively), at about. dissolve only in oxidizing acids, for example, in nitric acid (products of NO and HEO 4).
In the VIIIB subgroup, the iron family metals all interact with dilute acids. And their analogues, i.e. platinum metals (E 0 > 0) are oxidized only in tough conditions, and the proximity of their radii causes a large resemblance in chemical behavior, but there are also differences.
So, the most active of them - palladium - belongs to acids, like silver; and rhodium and iridium, unlike the others, do not dissolve even in "royal vodka" 126 . They react with a solution of sodium chloride saturated with chlorine at a red heat temperature by forming sustainable complexes Na 3 [ECl 6 ]. However, in the form of black, these metals easily react with hot sulfuric acid and even with hydrochloric acid in the presence of oxygen. Note that under these conditions, osmium, due to its high affinity for oxygen (?), also dissolves in a compact form.
In IV, V and VI side subgroups in M of the second and third decades E 0< 0 , но за счет влиянияdense oxide film on their surface, they react with acids only under harsh conditions. So, Zr and Hf are soluble only in complexing acids: in hot sulfuric acid (the product is H 2 [E (SO 4) ] 3) and in hydrofluoric acid (H 4 [EF 8 ]); molybdenum interacts only with oxidizing acids when heated, while tungsten, niobium, and tantalum interact only with a mixture of HF and HNO (3 products of NO and H 2 WF 8 or H 2 EF 7, respectively).
So, regardless of whether there is an overlay of the kinetic factor (passivating film) or not, the activity of d-metals in relation to acids in the subgroups decreases. An exception, as already noted, is scandium subgroup, in which there is no effect of f-compression and the nature of the change in the values of the atomic radius, I 1 and E 0 is the same as in the main subgroups. As a result, lanthanum (unlike scandium and yttrium, which are soluble at vol. only in acids) interacts even with water:
La + H 2 O → La (OH) 3 + H 2.
The ratio of d-metals to alkalis. Silver is most resistant to alkali 127, and zinc is the least resistant: it oxidizes even solution alkali, reducing the hydrogen of water and forming a complex 128 - . The remaining d-metals, if they tend to exist in an anionic form, react with alkalis (or soda) when fused, For example:
Ti⎫ ⎧Na 2 TiO 3 ⎬ + NaOH→ H 2 + ⎨ .
⎭ ⎩Na 3 VO 4
In the case of others, it is necessary to have oxidizing agent:
Cr + NaNO 3 + NaOH → Na 2 CrO 4 + NaNO 2,
O 2 + Na 2 CO 3 → Na 2 WO 4 + CO 2.
Moreover, W and Mo interact with alkali more actively than Cr, because their surface is coated during the reaction with a more acidic oxide (EO), 3 than in the case of chromium (Cr 2 O 3).
Interaction of d-metals with simple substances. Corrosion. Under room conditions, only fluorine oxidizes most d-metals, except for noble ones (but reactions with Cu, Ni, Fe (as well as with Pb, Al) are limited to the formation of protective fluoride films). In addition, at about. gold interacts with bromine, and mercury with iodine and sulfur due to the formation of thermodynamically very sustainable products: AuBr, 3 HgI 2 and HgS (see section "Halogens").
In air, in a finely dispersed state, rather active metals (Ti, Cr, Mn, Fe, Co, Ni) pyrophoric 2 (i.e., they light up when pouring out in air), but in a compact form, most Ms are stable due to passivation. Especially dense surface films form metals of the vanadium and titanium subgroups, therefore they have high corrosion resistance (even in sea water).
Other metals are not as stable. Under the influence of air components (what?), Zinc and copper are slowly corroded (with the formation of E 2 (OH) 2 CO 3); even silver darkens, becoming covered with sulfide (under the combined influence of O 2, H 2 O and H 2 S; what is the role of each of them?).
Iron corrodes especially quickly. True, in a dry atmosphere, its oxidation proceeds only until the formation dense passivating FeO film. But in the presence of moisture, the product obtained by the reaction:
Fe + H 2 O → FeO + H 2,
oxidized by oxygen activated by H 2 O molecules to Fe 2 O 3 . At the same time, water sorbed by the metal surface, partially dissolving the oxidation products in itself, hinders formation dense oxide structure, as a result of which the corrosion of iron proceeds deep into.
The addition of alkali reduces the oxidizing potential of oxygen, and therefore the process goes to a lesser extent. Note that verypure iron, which adsorbs hydrogen well and thus passivates its surface, does not oxidize.
Technical iron for corrosion protection is painted or subjected to tinning, zinc plating, chromium plating, nickel plating, nitriding (Fe 4 N coating), cementing (Fe C 3) and other processing methods. In particular, vitrification metal surface with the help of a laser increases corrosion resistance by 12 times, but when M is heated above 200 0 C, this effect is lost. A more reliable, but expensive way to combat the oxidation of iron in air is to obtain stainless steel (18% Cr and 9% Ni).
However, corrosion is a slow process, and enough quickly d-metals react with non-metals only when heated, even the most active M subgroups of scandium (oxidized to (+3)). (However, from Sc to La, the interaction activity increases (?), and lanthanum, for example, ignites in chlorine at rev.)
In the case of less reactive (?) metals of the titanium subgroup, it is required more heating (above 150 0 С). In this case, Hf transforms into Hf + 4 , and Ti and Zr can form products in inferior st.ok.: Ti 2 O 3, ZrCl 2, etc. However, they are strong reducing agents, especially in the case of Zr (?) - they oxidize in air or dismutate:
ZrCl 2 → Zr+ ZrCl 4 .
With even less active metals of the vanadium subgroup, reactions proceed at t > 400 0 C, and with the formation of products only in the highest st. (+5).
When moving to the chromium subgroup, the reactivity of M growing(due to the greater volatility of oxides), but decreases from Cr to W (?). So, chromium interacts with all G 2, molybdenum does not react with I 2, and tungsten does not react with Br 2 either. Moreover, the oxidation of chromium goes up to (+3), and its analogues - up to (+6). (Note that WF is 6 the heaviest gas at N.C.)
Similar patterns are observed in other subgroups of d-metals. So, technetium and rhenium do not interact with iodine, but with other halogens - only at t > 400 0 C, forming EG 7 . At the same time, manganese oxidizes with slight heating.
even gray and up to st. (+2).
Copper reacts with wet chlorine at vol., silver - with slight heating, and gold - only at t> 200 0 C. When heated, oxygen acts only on copper (CuO product, at higher temperatures - Cu 2 O (?)), and silver oxidized (unlike gold) by ozone (to AgO).
Zinc also burns in CO 2, and mercury at about. even an oxide film is not covered. When heated to 300 0 C, it forms a mixture of HgO and Hg 2 O oxides, which at t> 400 0 C split off O, turning into Hg, while the decomposition temperature of cadmium oxide is 1813 0 C, and ZnO - 1950 0 C.
Platinum metals and gold are the most chemically stable, but with sufficient heating they react with almost all non-metals (G 2, O 2, S, P, As), although with different activity and selectivity; namely: in periods from left to right, resistance to O 2 and F 2 increases, and to Cl 2 and S decreases (in accordance with the electronic structure of the atoms of the elements (?)).
So, if fluorine reacts with platinum only at t > 400 0 C, then chlorine reacts at 250 0 C (PtCl 2 product). Or if we consider the interaction with oxygen: osmium in the form of black is oxidized in air at vol. (up to OsO 4), ruthenium - with a slight heating, and the rest - at a red heat temperature. Products: IrO 2 , PdO, PtO 2 , Rh 2 O 3 .
(When heated more strongly, these oxides decompose, moreover, if the reaction:
PtO 2 → Pt+ O 2
goes at 500 0 C, then decomposition:
RuO 2 → Ru+ O 2
occurs only when t > 1300 0 C).
A similar increase in the resistance of the metal to oxygen is also observed in the transition from iron to nickel (see Table 14).
Table 14 Characteristics of the interaction of metals of the iron family with oxygen
Formation of solid solutions. A feature of d-metals is their tendency due to the wide variety of art. and valence states to form compounds non-stoichiometric composition: intermetallides (AlNi, etc.) or metallides (Fe S 3, VN, LaB, ZrC 6, etc.). As well as solid solutions, in particular, solutions implementation gases. So, metals of the scandium and titanium subgroups absorb hydrogen at vol. to the composition: EH 2 and EH (3 when heated, the solubility of H 2 decreases).
Nickel and palladium have a special affinity for hydrogen (1 V Pd dissolves 1000 V H 2), which are therefore catalysts for reactions hydrogenation. And, for example, platinum mainly sorbs O 2 (up to 700 V) and therefore is used as a catalyst for processes that occur with the participation of oxygen: oxidation NH 3 to NO, SO 2 to SO, 3 for afterburning car exhaust gases (in this case, in particular, NO turns into N 2, and CO into CO 2), etc.
The mechanism of the catalytic action of these metals is that, as it is assumed, gases dissolved in M atomized. Thus, hydrogen released when its solution in a metal is heated is a stronger reducing agent than molecular hydrogen.
In addition, for example, when absorbing H 2, palladium retains metallic properties up to a certain limit, but loses paramagnetism. This means that at least some of the hydrogen atoms donate their valence electrons to the conduction band of the metal.
There are also data on the partial formation of hydride ions, for example, when hydrogen is dissolved in iron. Received and so-called. unconventional hydrides in which H 2 molecules are coordinated as a whole on a d-metal atom. (They serve as models in the study of intermediates arising during catalysis.)
Chemistry preparation for ZNO and DPA
Comprehensive Edition
PART AND
GENERAL CHEMISTRY
CHEMISTRY OF THE ELEMENTS
HALOGENS
Simple substances
Chemical properties of Fluorine
Fluorine is the strongest oxidizing agent in nature. Directly it does not react only with helium, neon and argon.
During the reaction with metals, fluorides are formed, ionic type compounds:
Fluorine reacts vigorously with many non-metals, even with some inert gases:
Chemical properties of chlorine. Interaction with complex substances
Chlorine is a stronger oxidizing agent than bromine or iodine, so chlorine displaces heavy halogens from their salts:
Dissolving in water, chlorine partially reacts with it, resulting in the formation of two acids: chloride and hypochlorite. In this case, one chlorine atom increases the degree of oxidation, and the other atom reduces it. Such reactions are called disproportionation reactions. Disproportionation reactions are self-healing-self-oxidation reactions, i.e. reactions in which one element exhibits the properties of both an oxide and a reducing agent. With disproportionation, compounds are simultaneously formed in which the element is in a more oxidized and reduced state compared to the primitive one. The oxidation state of the Chlorine atom in the hypochlorite acid molecule is +1:
The interaction of chlorine with alkali solutions proceeds similarly. In this case, two salts are formed: chloride and hypochlorite.
Chlorine interacts with various oxides:
Chlorine oxidizes some salts in which the metal is not in the maximum oxidation state:
Molecular chlorine reacts with many organic compounds. In the presence of ferrum(III) chloride as a catalyst, chlorine reacts with benzene to form chlorobenzene, and when irradiated with light, hexachlorocyclohexane is formed as a result of the same reaction:
Chemical properties bromine and iodine
Both substances react with hydrogen, fluorine and alkalis:
Iodine is oxidized by various strong oxidizing agents:
Methods for the extraction of simple substances
Extraction of fluorine
Since fluorine is the strongest chemical oxide, it is impossible to isolate it by chemical reactions from compounds in a free form, and therefore fluorine is mined by the physicochemical method - electrolysis.
To extract fluorine, potassium fluoride melt and nickel electrodes are used. Nickel is used due to the fact that the surface of the metal is passivated by fluorine due to the formation of insoluble
NiF2, therefore, the electrodes themselves are not destroyed by the action of the substance that is released on them:Extraction of chlorine
Chlorine is commercially produced by electrolysis of sodium chloride solution. As a result of this process, sodium hydroxide is also extracted:
In small quantities, chlorine is obtained by oxidizing a solution of hydrogen chloride by various methods:
Chlorine is a very important product of the chemical industry.
Its world production is millions of tons.
Extraction of bromine and iodine
For industrial use, bromine and iodine are obtained from the oxidation of bromides and iodides, respectively. For oxidation, molecular chlorine, concentrated sulfate acid or manganese dioxide are most often used:
Fluorine and some of its compounds are used as an oxidizing agent for rocket fuel. Large amounts of fluorine are used to produce various refrigerants (freons) and some polymers that are characterized by chemical and thermal resistance (Teflon and some others). Fluorine is used in nuclear technology to separate uranium isotopes.
Most of the chlorine is used to produce hydrochloric acid, and also as an oxidizing agent for the extraction of other halogens. In industry, it is used to bleach fabrics and paper. In larger quantities than fluorine, it is used for the production of polymers (PVC and others) and refrigerants. Disinfect with chlorine drinking water. It is also needed to extract some solvents such as chloroform, methylene chloride, carbon tetrachloride. And it is also used to produce many substances, such as potassium chlorate (bertolet salt), bleach and many other compounds containing chlorine atoms.
Bromine and iodine are not used in industry on the same scale as chlorine or fluorine, but the use of these substances is increasing every year. Bromine is used in the manufacture of various sedative medicines. Iodine is used in the manufacture of antiseptic preparations. Bromine and Iodine compounds are widely used in the quantitative analysis of substances. With the help of iodine, some metals are purified (this process is called iodine refining), such as titanium, vanadium and others.
