Bases (hydroxides)– complex substances whose molecules contain one or more hydroxy OH groups. Most often, bases consist of a metal atom and an OH group. For example, NaOH is sodium hydroxide, Ca(OH) 2 is calcium hydroxide, etc.
There is a base - ammonium hydroxide, in which the hydroxy group is attached not to the metal, but to the NH 4 + ion (ammonium cation). Ammonium hydroxide is formed when ammonia is dissolved in water (the reaction of adding water to ammonia):
NH 3 + H 2 O = NH 4 OH (ammonium hydroxide).
The valency of the hydroxy group is 1. The number of hydroxyl groups in the base molecule depends on the valency of the metal and is equal to it. For example, NaOH, LiOH, Al (OH) 3, Ca(OH) 2, Fe(OH) 3, etc.
All reasons - solids that have different colors. Some bases are highly soluble in water (NaOH, KOH, etc.). However, most of them are not soluble in water.
Bases soluble in water are called alkalis. Alkali solutions are “soapy”, slippery to the touch and quite caustic. Alkalies include hydroxides of alkali and alkaline earth metals (KOH, LiOH, RbOH, NaOH, CsOH, Ca(OH) 2, Sr(OH) 2, Ba(OH) 2, etc.). The rest are insoluble.
Insoluble bases- these are amphoteric hydroxides, which act as bases when interacting with acids, and behave like acids with alkali.
Different bases have different abilities to remove hydroxy groups, so they are divided into strong and weak bases.
Strong bases in aqueous solutions easily give up their hydroxy groups, but weak bases do not.
Chemical properties of bases
The chemical properties of bases are characterized by their relationship to acids, acid anhydrides and salts.
1. Act on indicators. Indicators change color depending on interaction with different chemicals. In neutral solutions they have one color, in acid solutions they have another color. When interacting with bases, they change their color: the methyl orange indicator turns yellow, the litmus indicator turns blue, and phenolphthalein becomes fuchsia.
2. Interact with acid oxides with formation of salt and water:
2NaOH + SiO 2 → Na 2 SiO 3 + H 2 O.
3. React with acids, forming salt and water. The reaction of a base with an acid is called a neutralization reaction, since after its completion the medium becomes neutral:
2KOH + H 2 SO 4 → K 2 SO 4 + 2H 2 O.
4. Reacts with salts forming a new salt and base:
2NaOH + CuSO 4 → Cu(OH) 2 + Na 2 SO 4.
5. When heated, they can decompose into water and the main oxide:
Cu(OH) 2 = CuO + H 2 O.
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Chemical properties of the main classes of inorganic compounds
Acidic oxides
- Acidic oxide + water = acid (exception - SiO 2)
SO 3 + H 2 O = H 2 SO 4
Cl 2 O 7 + H 2 O = 2HClO 4 - Acidic oxide + alkali = salt + water
SO 2 + 2NaOH = Na 2 SO 3 + H 2 O
P 2 O 5 + 6KOH = 2K 3 PO 4 + 3H 2 O - Acidic oxide + basic oxide = salt
CO 2 + BaO = BaCO 3
SiO 2 + K 2 O = K 2 SiO 3Basic oxides
- Basic oxide + water = alkali (alkali and alkaline earth metal oxides react)
CaO + H 2 O = Ca(OH) 2
Na 2 O + H 2 O = 2NaOH - Basic oxide + acid = salt + water
CuO + 2HCl = CuCl 2 + H 2 O
3K 2 O + 2H 3 PO 4 = 2K 3 PO 4 + 3H 2 O - Basic oxide + acidic oxide = salt
MgO + CO 2 = MgCO 3
Na 2 O + N 2 O 5 = 2NaNO 3Amphoteric oxides
- Amphoteric oxide + acid = salt + water
Al 2 O 3 + 6HCl = 2AlCl 3 + 3H 2 O
ZnO + H 2 SO 4 = ZnSO 4 + H 2 O - Amphoteric oxide + alkali = salt (+ water)
ZnO + 2KOH = K 2 ZnO 2 + H 2 O (More correct: ZnO + 2KOH + H 2 O = K 2)
Al 2 O 3 + 2NaOH = 2NaAlO 2 + H 2 O (More correct: Al 2 O 3 + 2NaOH + 3H 2 O = 2Na) - Amphoteric oxide + acidic oxide = salt
ZnO + CO 2 = ZnCO 3 - Amphoteric oxide + basic oxide = salt (if fused)
ZnO + Na 2 O = Na 2 ZnO 2
Al 2 O 3 + K 2 O = 2KAlO 2
Cr 2 O 3 + CaO = Ca(CrO 2) 2Acids
- Acid + basic oxide = salt + water
2HNO 3 + CuO = Cu(NO 3) 2 + H 2 O
3H 2 SO 4 + Fe 2 O 3 = Fe 2 (SO 4) 3 + 3H 2 O - Acid + amphoteric oxide = salt + water
3H 2 SO 4 + Cr 2 O 3 = Cr 2 (SO 4) 3 + 3H 2 O
2HBr + ZnO = ZnBr 2 + H 2 O - Acid + base = salt + water
H 2 SiO 3 + 2KOH = K 2 SiO 3 + 2H 2 O
2HBr + Ni(OH) 2 = NiBr 2 + 2H 2 O - Acid + amphoteric hydroxide = salt + water
3HCl + Cr(OH) 3 = CrCl 3 + 3H 2 O
2HNO 3 + Zn(OH) 2 = Zn(NO 3) 2 + 2H 2 O - Strong acid + salt of weak acid = weak acid + salt of strong acid
2HBr + CaCO 3 = CaBr 2 + H 2 O + CO 2
H 2 S + K 2 SiO 3 = K 2 S + H 2 SiO 3 - Acid + metal (located in the voltage series to the left of hydrogen) = salt + hydrogen
2HCl + Zn = ZnCl 2 + H 2
H 2 SO 4 (diluted) + Fe = FeSO 4 + H 2
Important: oxidizing acids (HNO 3, conc. H 2 SO 4) react with metals differently.
Amphoteric hydroxides
- Amphoteric hydroxide + acid = salt + water
2Al(OH) 3 + 3H 2 SO 4 = Al 2 (SO 4) 3 + 6H 2 O
Be(OH) 2 + 2HCl = BeCl 2 + 2H 2 O - Amphoteric hydroxide + alkali = salt + water (when fused)
Zn(OH) 2 + 2NaOH = Na 2 ZnO 2 + 2H 2 O
Al(OH) 3 + NaOH = NaAlO 2 + 2H 2 O - Amphoteric hydroxide + alkali = salt (in aqueous solution)
Zn(OH) 2 + 2NaOH = Na 2
Sn(OH) 2 + 2NaOH = Na 2
Be(OH) 2 + 2NaOH = Na 2
Al(OH) 3 + NaOH = Na
Cr(OH) 3 + 3NaOH = Na 3Alkalis
- Alkali + acid oxide = salt + water
Ba(OH) 2 + N 2 O 5 = Ba(NO 3) 2 + H 2 O
2NaOH + CO 2 = Na 2 CO 3 + H 2 O - Alkali + acid = salt + water
3KOH + H3PO4 = K3PO4 + 3H2O
Ba(OH) 2 + 2HNO 3 = Ba(NO 3) 2 + 2H 2 O - Alkali + amphoteric oxide = salt + water
2NaOH + ZnO = Na 2 ZnO 2 + H 2 O (More correct: 2NaOH + ZnO + H 2 O = Na 2) - Alkali + amphoteric hydroxide = salt (in aqueous solution)
2NaOH + Zn(OH) 2 = Na 2
NaOH + Al(OH) 3 = Na - Alkali + soluble salt = insoluble base + salt
Ca(OH) 2 + Cu(NO 3) 2 = Cu(OH) 2 + Ca(NO 3) 2
3KOH + FeCl 3 = Fe(OH) 3 + 3KCl - Alkali + metal (Al, Zn) + water = salt + hydrogen
2NaOH + Zn + 2H 2 O = Na 2 + H 2
2KOH + 2Al + 6H 2 O = 2K + 3H 2Salts
- Salt of a weak acid + strong acid = salt of a strong acid + weak acid
Na 2 SiO 3 + 2HNO 3 = 2NaNO 3 + H 2 SiO 3
BaCO 3 + 2HCl = BaCl 2 + H 2 O + CO 2 (H 2 CO 3) - Soluble salt + soluble salt = insoluble salt + salt
Pb(NO 3) 2 + K 2 S = PbS + 2KNO 3
СaCl 2 + Na 2 CO 3 = CaCO 3 + 2NaCl - Soluble salt + alkali = salt + insoluble base
Cu(NO 3) 2 + 2NaOH = 2NaNO 3 + Cu(OH) 2
2FeCl 3 + 3Ba(OH) 2 = 3BaCl 2 + 2Fe(OH) 3 - Soluble metal salt (*) + metal (**) = metal salt (**) + metal (*)
Zn + CuSO 4 = ZnSO 4 + Cu
Cu + 2AgNO 3 = Cu(NO 3) 2 + 2Ag
Important: 1) the metal (**) must be in the voltage series to the left of the metal (*), 2) the metal (**) must NOT react with water.You may also be interested in other sections of the chemistry reference book:
- Salt of a weak acid + strong acid = salt of a strong acid + weak acid
- Alkali + acid oxide = salt + water
- Acid + basic oxide = salt + water
- Amphoteric oxide + acid = salt + water
- Basic oxide + water = alkali (alkali and alkaline earth metal oxides react)
2NaOH + Zn + 2H 2 O = Na 2 + H 2
2KOH + 2Al + 6H 2 O = 2K + 3H 2
Salts
1. Salt of a weak acid + strong acid = salt of a strong acid + weak acid
Na 2 SiO 3 + 2HNO 3 = 2NaNO 3 + H 2 SiO 3
BaCO 3 + 2HCl = 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 = CaCO 3 + 2NaCl
3. Soluble salt + alkali = 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 = ZnSO 4 + Cu
Cu + 2AgNO 3 = Cu(NO 3) 2 + 2Ag
Important: 1) the metal (**) must be in the voltage series to the left of the metal (*), 2) the metal (**) must NOT react with water.
Example 1 Zinc hydroxide can react with each substance in pairs:
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 is amphoteric. It reacts with both acids and alkalis.
Example 2 A solution of copper(II) sulfate reacts with each of 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, a reaction occurs if the following conditions are met: a precipitate forms, a gas is released, and a slightly dissociating substance, for example, water, is formed.
Example 3 The transformation scheme 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 diagram, 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 = 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 = CaSO 4
CaSO 4 + BaCl 2 = CaCl 2 + BaSO 4
CaCl 2 + Ba = BaCl 2 + Ca
Assignments on topic 5
161-170. Confirm the acidic properties of oxides with reaction equations in molecular and ionic form. Name the substances obtained.
181-190. Write 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 tetrahydroxyzincate→zinc nitrate→zinc hydroxide→zinc oxide→potassium zincate |
Chemical reactions.
One of the types of interactions of atoms, molecules and ions is reactions in which onereagentsgive, and others acquireelectrons. During such reactions, called redox, atoms of one or more elements change their oxidation state.
Under oxidation state refers to the conventional 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 shifted to a more electronegative atom, which attracts them more strongly. The concept of 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 when drawing up equations of 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, and its sign is indicated first, and then its value. It can be either an integer or a fraction. For example, if in H 2 O and H 2 O2 the oxidation state for oxygen is (-2) and (-1), then in KO2 and KO3 - (-1/2) and (-1/3), respectively.
1) the oxidation state of an atom in simple substances is zero, for example:
Na 0 ; H 2 0 ; 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., accordingly, (+1); (+2); (+3); (-1); (-2);
3) in most compounds the oxidation state of the hydrogen atom is equal to (+1) (except for the hydrides Me - LiH; CaH, etc., in which it is equal to (-1));
4) the oxidation state of the oxygen atom in most compounds is equal to
(-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 a molecule is equal to 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). Having composed the algebraic equation: (+1) + x + (-2) 3 = 0, we get x = +5.
Returning to the definition of redox reactions, we note that Oxidation is the process of losing electrons, and reduction is the process of adding electrons. An oxidizing agent is a substance containing an element whose oxidation state decreases during a reaction. A reducing agent is a substance containing an element whose oxidation state increases during a reaction. It should be emphasized that oxidation and reduction reactions are impossible without each 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 a free state - C, H (in most cases), etc.;
2) simple ions corresponding to the lowest oxidation state of the element: S2-; I; Cl-etc.;
3) complex ions and molecules containing atoms in the lowest oxidation degree
leniya: N in the NH4 ion, S in the H2S molecule, I in the KI molecule, etc.
Typical oxidizing agents:
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, etc.;
3) complex ions and molecules containing atoms in the highest oxidation state: 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 conditions) either oxidizing or reducing 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 SO 3 2- ion, in the N ion +3 in the NO 2 - ion, etc.
In a redox reaction, electrons are transferred from a reducing agent to an oxidizing agent.
Example 1 Write the equation for the oxidation of iron (II) disulfide with concentrated nitric acid. Make up: electronic and electron-ion balance diagrams.
Solution. HNO 3 is a strong oxidizing agent, so sulfur will be oxidized to a maximum oxidation state of S +6, and iron to Fe +3, while HNO 3 can be reduced to NO or NO 2. Let's consider the case of reduction to NO 2.
FeS 2 + HNO 3 (conc) → Fe (NO 3) 3 + H 2 SO 4 + NO 2.
It is not yet known where H 2 O will be located (on the left or right side).
Let's equalize this reaction using the electronic balance method. The recovery process is described by the following diagram:
N +5 + e → N +4
Two elements enter the oxidation half-reaction at once - Fe and S. Iron in the disulfide has an oxidation state of +2, and sulfur -1. It is necessary to take into account that for one Fe atom there are two S atoms:
Fe +2 - e → Fe +3
2S - - 14e → 2S +6 .
Together, iron and sulfur give up 15 electrons.
The full balance looks like:
15 HNO 3 molecules are used for the oxidation of FeS 2, and another 3 HNO 3 molecules are needed 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, you need to add 7 molecules of H2O to the right side:
FeS 2 + 18HNO 3(conc) = Fe(NO 3) 3 + 2H 2 SO 4 + 15NO 2 + 7H 2 O.
We now use the electron-ion balance method. Let's 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 H2O molecules to the left side, and 16 H+ ions to the right side (the environment is acidic!):
FeS 2 + 8H 2 O → Fe 3+ + 2SO 4 2- + 16H +.
The charge on the left side is 0, the charge on the right is +15, so FeS 2 must give up 15 electrons:
FeS 2 + 8H 2 O - 15e → Fe 3+ + 2SO 4 2- + 16H +.
Let us now consider the half-reaction of the reduction of nitrate ion:
NO -3 → NO 2 .
It is necessary to subtract 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, we add one electron to the left side (charge +1):
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 obtain the abbreviated ionic equation of the redox reaction:
FeS 2 + 15NO 3 - + 14H + = Fe 3+ + 2SO 4 2- + 15NO 2 + 7H 2 O.
By adding to both sides of the equation the corresponding number of ions, three ions NO 3 - and H +, we find the molecular equation of the reaction:
FeS 2 + 18HNO 3(conc) = 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; presence of a catalyst.
EXAMPLE 2 . Calculate how many times the reaction rate will increase when the temperature increases by 40°, if the temperature coefficient of the rate of this reaction is 3.
SOLUTION. The dependence of the reaction rate on temperature is expressed empirically van't Hoff's rule, according to which, with 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 2-4, shows how many times the reaction rate will increase when the temperature increases by 10 degrees;
v T 1, v T2 - rates of chemical reaction at temperatures T1 and T2. In this example:
The reaction speed will increase 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) oxygen concentration; 2) volume of the reaction space; 3) pressure in the system.
Solution: Reaction a) occurs in homogeneous system - all substances are in one phase (all substances are gases), reaction b) proceeds in heterogeneous system - the reacting substances are in different phases (O2 and CO are gases, C is solid). Therefore, the reaction rates for these systems according to the ZDM are equal:
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:
a) 
b) 
The increase in the reaction rate relative to the initial one is determined by the ratio:
A) 
b) 
Consequently, after increasing the oxygen concentration by 3 times, the rates of reactions a) and b) will increase by 3 times.
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 respectively by 27 times (a) and 3 times (b):
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. That's why:
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) increased pressure; d) decrease in temperature; e) introducing a catalyst.
SOLUTION. A displacement or shift in chemical equilibrium is a change in the equilibrium concentrations of reacting substances as a result of a change in one of the conditions for the reaction. The direction of the equilibrium shift is determined by Le Chatelier's principle: if any external influence is exerted on a system that is in equilibrium (change concentration, pressure, temperature), then the equilibrium will shift towards the reaction (direct or reverse) that counteracts the effect.
A) An increase in the concentration of reagents (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 that occurs with the formation of a smaller amount of gaseous substances. In this example, the direct 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. towards the opposite reaction;
d) since the direct reaction occurs with the absorption of heat), then a decrease in temperature shifts the equilibrium towards the opposite (exothermic reaction);
d) introduction of a catalyst into the system does not affect the equilibrium shift, because equally increases the rate of forward and reverse reactions.
Assignments on topic 6
201-220. Using these diagrams, create equations for redox reactions, indicate the oxidizing agent and the 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 change.
236-240. How should a) temperature, b) pressure, c) concentration be changed in order to shift the chemical equilibrium towards a direct reaction?
Metals and non-metals.
The set of redox reactions that occur on electrodes in solutions or melts of electrolytes when an electric current is passed through them is called electrolysis.
At the cathode of the current source, the process of transferring electrons to cations from a solution or melt occurs, so the cathode is a “reducing agent”. At the anode, electrons are given away by anions, so the anode is an “oxidizing agent.” During electrolysis, competing processes can occur at both the anode and 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:
- on the anode- oxidation of anions and hydroxide ions,
- on the cathode— reduction of cations and hydrogen ions.
When electrolysis is carried out using an active (consumable) anode, the process becomes more complicated and the competing reactions on the electrodes are the following:
- on the anode- oxidation of anions and hydroxide ions, anodic dissolution of the metal - anode material;
- on the cathode- reduction of salt cation and hydrogen ions, reduction 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 that requires the least amount of energy occurs. When electrolyzing 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 of organic acids, the process occurs:
2RCOO - - 2е → R-R + 2СО 2.
2. During the electrolysis of salt solutions containing ions located in the voltage series to the left of Al 3+, hydrogen is released at the cathode; if the ion is located to the right of hydrogen, then the metal is released.
3. During the electrolysis of salt solutions containing ions located between Al 3+ and H + at 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 electrons given or received; I—current strength (A); t—process duration (s); F is Faraday's constant, characterizing 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=Na0,
anode (-) (Cl -): Cl - - e= Cl 0, 2Cl 0 = Cl 2;
2NaCl = 2Na + Cl2.
Example 2 Electrolysis of sodium chloride solution:
NaCl = Na + + Cl - ,
H 2 O = H + + OH -;
cathode (-) (Na + ; H +): H + + e= H 0 , 2H 0 = H 2
(2H 2 O + 2 e= H 2 + 2OH -),
anode (+) (Cl - ; OН -): Cl - - e= Cl 0, 2Cl 0 = Cl 2;
2NaCl + 2H 2 O = 2NaOH + Cl 2 + H 2.
Example 3 Electrolysis of copper(II) nitrate solution:
Cu(NO 3) 2 = Cu 2+ + NO 3 --
H 2 O = H + + OH -;
cathode (-) (Cu 2+ ; H +): Cu 2+ + 2 e= Cu 0 ,
anode (+) (OH -): OH - - e=OH 0,
4H 0 = O 2 + 2H 2 O;
2Cu(NO 3) 2 + 2H 2 O = 2Cu + O 2 + 4HNO 3.
Assignments on topic 7
241-250. Compose electronic equations for the processes occurring on inert electrodes during the electrolysis of a) a melt, b) a solution of a substance:
| № | ||||||||||
| Substance | NaOH | KCl | AgNO3 | Cu(NO3)2 | FeSO4 | K2S | 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 for a time t(h) at a current strength I(A).
271-280. Write an equation for the reaction between substances, taking into account 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”: Textbook for secondary vocational educational institutions. - M.: Publishing Center “Academy”, 2004.
2. Rudzitis G.E., Feldman F.G. “Chemistry” 10th grade-M.: Enlightenment. 1995.
3. Rudzitis G.E., Feldman F.G. "Chemistry" 11th grade. -M.: Enlightenment. 1995.
4. Akhmetov M.S. “Laboratory and seminar classes in general and inorganic chemistry” M.: Higher 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. “Tasks 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 those entering the university.” -M.: New wave. 1999.
All chemical elements are divided into metals And nonmetals depending on the structure and properties of their atoms. Also, simple substances formed by elements are classified into metals and non-metals, based on their physical and chemical properties.
In the Periodic Table of Chemical Elements D.I. Mendeleev's non-metals are located diagonally: boron - astatine and above it in the main subgroups.
Metal atoms are characterized by relatively large radii and a small number of electrons on the outer level from 1 to 3 (exceptions: germanium, tin, lead - 4; antimony and bismuth - 5; polonium - 6 electrons).
Non-metal atoms, on the contrary, are characterized by small atomic radii and the number of electrons on the outer level from 4 to 8 (with the exception of boron, it has three such electrons).
Hence the tendency of metal atoms to give up external electrons, i.e. reducing properties, and for non-metal atoms - the desire to accept electrons missing to a stable eight-electron level, i.e. oxidizing properties.
Metals
In metals there is a metal bond and a metal crystal lattice. At the lattice sites there are positively charged metal ions, connected through shared external electrons belonging to the entire crystal.
This determines all the most important physical properties of metals: metallic luster, electrical and thermal conductivity, plasticity (the ability to change shape under external influence) and some others characteristic of this class of simple substances.
Metals of group I of the main subgroup are called alkali metals.
Metals of group II: calcium, strontium, barium - alkaline earth.
Chemical properties of metals
In chemical reactions, metals exhibit only reducing properties, i.e. their atoms give up electrons, resulting in positive ions.
1. Interact with non-metals:
a) oxygen (with the formation of oxides)
Alkali and alkaline earth metals oxidize easily under normal conditions, so they are stored under a layer of petroleum jelly or kerosene.
4Li + O 2 = 2Li 2 O
2Ca + O2 = 2CaO
Please note: when sodium reacts, peroxide is formed, potassium - superoxide
2Na + O 2 = Na 2 O 2, K + O2 = KO2
and oxides are obtained by calcination of peroxide with the corresponding metal:
2Na + Na 2 O 2 = 2Na 2 O
Iron, zinc, copper and other less active metals oxidize slowly in air and actively when heated.
3Fe + 2O 2 = Fe 3 O 4 (a mixture of two oxides: FeO and Fe 2 O 3)
2Zn + O 2 = 2ZnO
2Cu + O 2 = 2CuO
Gold and platinum metals are not oxidized by atmospheric oxygen under any conditions.
b) hydrogen (with the formation of hydrides)
2Na + H 2 = 2NaH
Ca + H 2 = CaH 2
c) chlorine (with the formation of chlorides)
2K + Cl 2 = 2KCl
Mg + Cl 2 = MgCl 2
2Al + 3Cl 2 = 2AlCl 3
Please note: when iron reacts, iron (III) chloride is formed:
2Fe + 3Cl 2 = 2FeCl 3
d) sulfur (with the formation of sulfides)
2Na + S = Na 2 S
Hg + S = HgS
2Al + 3S = Al 2 S 3
Please note: when iron reacts, iron (II) sulfide is formed:
Fe + S = FeS
e) nitrogen (with the formation of nitrides)
6K + N 2 = 2K 3 N
3Mg + N 2 = Mg 3 N 2
2Al + N 2 = 2AlN
2. Interact with complex substances:
It must be remembered that, according to their reducing ability, metals are arranged in a series, which is called the electrochemical series of voltages or activities of metals (displacement series of Beketov N.N.):
Li, K, Ba, Ca, Na, Mg, Al, Mn, Zn, Cr, Fe, Co, Ni, Sn, Pb, (H 2), Cu, Hg, Ag, Au, Pt
a) water
Metals located in the series up to magnesium, under normal conditions, displace hydrogen from water, forming soluble bases - alkalis.
2Na + 2H 2 O = 2NaOH + H 2
Ba + H 2 O = Ba(OH) 2 + H 2
Magnesium reacts with water when boiled.
Mg + 2H 2 O = Mg(OH) 2 + H 2
When removing the oxide film, aluminum reacts violently with water.
2Al + 6H 2 O = 2Al(OH) 3 + 3H 2
The remaining metals in the series up to hydrogen, under certain conditions, can also react with water to release hydrogen and form oxides.
3Fe + 4H 2 O = Fe 3 O 4 + 4H 2
b) acid solutions
(Except concentrated sulfuric acid and nitric acid of any concentration. See the section "Oxidation-reduction reactions.")
Please note: insoluble silicic acid is not used for reactions.
Metals in the series from magnesium to hydrogen displace hydrogen from acids.
Mg + 2HCl = MgCl 2 + H 2
Please note: ferrous iron salts are formed.
Fe + H 2 SO 4 (diluted) = FeSO 4 + H 2
The formation of an insoluble salt prevents the reaction from proceeding. For example, lead practically does not react with a solution of sulfuric acid due to the formation of insoluble lead sulfate on the surface.
Metals in the row after hydrogen do NOT displace hydrogen.
c) salt solutions
Metals in the series up to magnesium and which actively react with water are not used to carry out such reactions.
For other metals the following rule applies:
Each metal displaces from salt solutions other metals located in the series to the right of it, and can itself be displaced by metals located to the left of it.
Cu + HgCl 2 = Hg + CuCl 2
Fe + CuSO 4 = FeSO 4 + Cu
As with acid solutions, the formation of an insoluble salt prevents the reaction from proceeding.
d) alkali solutions
Metals whose hydroxides are amphoteric react.
Zn + 2NaOH + 2H 2 O = Na 2 + H 2
2Al + 2KOH + 6H 2 O = 2K + 3H 2
e) with organic substances
Alkali metals with alcohols and phenol.
2C 2 H 5 OH + 2Na = 2C 2 H 5 ONa + H 2
2C 6 H 5 OH + 2Na = 2C 6 H 5 ONa + H 2
Metals participate in reactions with haloalkanes, which are used to obtain lower cycloalkanes and for syntheses during which the carbon skeleton of the molecule becomes more complex (A. Wurtz reaction):
CH 2 Cl-CH 2 -CH 2 Cl + Zn = C 3 H 6 (cyclopropane) + ZnCl 2
2CH 2 Cl + 2Na = C 2 H 6 (ethane) + 2NaCl
Nonmetals
In simple substances, nonmetal atoms are connected by a covalent nonpolar bond. In this case, single (in molecules H 2, F 2, Cl 2, Br 2, I 2), double (in O 2 molecules), triple (in N 2 molecules) covalent bonds are formed.
The structure of simple substances - non-metals:
1. molecular
Under normal conditions, most of these substances are gases (H 2, N 2, O 2, O 3, F 2, Cl 2) or solids (I 2, P 4, S 8) and only bromine (Br 2) is liquid. All these substances have a molecular structure and are therefore volatile. In the solid state, they are fusible due to the weak intermolecular interaction that holds their molecules in the crystal, and are capable of sublimation.
2. atomic
These substances are formed by crystals, at the nodes of which there are atoms: (Bn, Cn, Sin, Gen, Sen, Ten). Due to the great strength of covalent bonds, they usually have high hardness, and any changes associated with the destruction of covalent bonds in their crystals (melting, evaporation) occur with a large expenditure of energy. Many such substances have high melting and boiling points, and their volatility is very low.
Many elements - nonmetals - form several simple substances - allotropic modifications. Allotropy can be associated with different compositions of molecules: oxygen O 2 and ozone O 3 and with different crystal structures: allotropic modifications of carbon are graphite, diamond, carbyne, fullerene. Elements are non-metals that have allotropic modifications: carbon, silicon, phosphorus, arsenic, oxygen, sulfur, selenium, tellurium.
Chemical properties of non-metals
Non-metal atoms have predominant oxidizing properties, that is, the ability to gain electrons. This ability is characterized by the value of electronegativity. In the series of non-metals
At, B, Te, H, As, I, Si, P, Se, C, S, Br, Cl, N, O, F
electronegativity increases and oxidizing properties increase.
It follows that simple substances - non-metals - will be characterized by both oxidizing and reducing properties, with the exception of fluorine - the strongest oxidizing agent.
1. Oxidizing properties
a) in reactions with metals (metals are always reducing agents)
2Na + S = Na 2 S (sodium sulfide)
3Mg + N 2 = Mg 3 N 2 (magnesium nitride)
b) in reactions with non-metals located to the left of this, that is, with a lower electronegativity value. For example, when phosphorus and sulfur interact, the oxidizing agent will be sulfur, since phosphorus has a lower electronegativity value:
2P + 5S = P 2 S 5 (phosphorus sulfide V)
Most nonmetals will be oxidizing agents in reactions with hydrogen:
H 2 + S = H 2 S
H 2 + Cl 2 = 2HCl
3H 2 + N 2 = 2NH 3
c) in reactions with some complex substances
Oxidizing agent – oxygen, combustion reactions
CH 4 + 2O 2 = CO 2 + 2H 2 O
2SO2 + O2 = 2SO3
Oxidizing agent – chlorine
2FeCl 2 + Cl 2 = 2FeCl 3
2KI + Cl 2 = 2KCl + I 2
CH 4 + Cl 2 = CH 3 Cl + HCl
Ch 2 =CH 2 + Br 2 = CH 2 Br-CH 2 Br
2. Restorative properties
a) in reactions with fluorine
S + 3F 2 = SF 6
H 2 + F 2 = 2HF
Si + 2F 2 = SiF 4
b) in reactions with oxygen (except fluorine)
S + O 2 = SO 2
N2 + O2 = 2NO
4P + 5O 2 = 2P 2 O 5
C + O 2 = CO 2
c) in reactions with complex substances - oxidizing agents
H 2 + CuO = Cu + H 2 O
6P + 5KClO 3 = 5KCl + 3P 2 O 5
C + 4HNO 3 = CO 2 + 4NO 2 + 2H 2 O
H 2 C=O + H 2 = CH 3 OH
3. Disproportionation reactions: the same non-metal is both an oxidizing agent and a reducing agent
Cl 2 + H 2 O = HCl + HClO
3Cl 2 + 6KOH = 5KCl + KClO 3 + 3H 2 O
For the last 200 years of humanity 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, first of all, to the development of various methods for obtaining substances.
In everyday life we come across many substances. 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 of 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 industry. Artificial substances are not found in nature; they are created from natural substances. Some substances that exist in nature can also be obtained in a chemical laboratory. Thus, when potassium permanganate is heated, oxygen is released, and when chalk is heated, oxygen is released.
carbon dioxide. Scientists have learned to turn graphite into diamond; they are growing crystals of ruby, sapphire and malachite. So, along with substances of natural origin, there are a huge number of artificially created substances that are not found in nature.
Substances not found in nature are produced in various enterprises: factories, factories, combines, etc.
In the context of depletion of the natural resources of our planet, chemists now face an important task: to develop and implement methods by which it is possible to artificially, in a laboratory or industrial production, obtain 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 fuel are being developed that would be just as efficient, but would not pollute the environment. Today, humanity has learned to artificially obtain various precious stones, for example, diamonds, emeralds, and beryls. solid (in the form of ice and snow), liquid (liquid water) and gaseous (water vapor). There are known substances that cannot exist under normal conditions in all three states of aggregation. For example, such a substance is carbon dioxide. At room temperature it is an odorless and colorless gas. At a temperature of –79°C this substance “freezes” and turns into a solid state of aggregation. The everyday (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. A substance, when transitioning from one state of aggregation to another, does not transform into other substances. The process of a certain change, transformation, is called a phenomenon.
Physical phenomena. Physical properties of substances.
Phenomena in which substances change their state of aggregation, but do not transform into other substances, are called physical. Each individual substance has certain properties. The properties of substances may 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 considered 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.
Properties of substances that are determined by measurements or visually in the absence of transformation of some substances into others are called physical Evaporation of alcohol, like evaporation of water
The main physical properties of substances include the following: state of aggregation, color, odor, solubility in water, density, boiling point, melting point, thermal conductivity, electrical conductivity.
Physical properties such as color, smell, taste, crystal shape can be determined visually using the senses, and density, electrical conductivity, melting and boiling points are determined by measurement. Information about the physical properties of many substances is collected in specialized literature, for example, in reference books. The physical properties of a substance depend on its state of aggregation. For example, the densities of ice, water and water vapor are different. Gaseous oxygen is colorless, but liquid oxygen is blue. Knowledge of physical properties helps to “recognize” many substances. For example, copper - The only metal that is red in color. Only table salt has a salty taste. Iodine - An almost black solid that turns into a purple vapor when heated. In most cases, to identify a substance, you need to consider several of its properties.
- As an example, let us characterize the physical properties of water:
- color – colorless (in small volumes)
- smell - no smell
- state of aggregation - liquid under normal conditions
- density – 1 g/ml,
- boiling point – +100°С
- melting point – 0°C
- 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 examine a sample of table salt under a magnifying glass, you will notice that the salt consists of many tiny crystals. In salt deposits you can also find very large crystals. Crystals are solids in the shape of regular polyhedra. Crystals can have different shapes and sizes. Crystals of certain substances, such as table salt – salt fragile and easy to break . There are crystals that are quite hard. For example, diamond is considered one of the hardest minerals. If you examine table salt crystals under a microscope, you will notice that they all have a similar structure. If we consider, for example, glass particles, they will all have a different structure - such substances are called amorphous.
Amorphous substances include glass, starch, amber, and beeswax.
If during physical phenomena substances, as a rule, only change their state of aggregation, then during chemical phenomena the transformation of some substances into other substances occurs. Here are some simple examples: burning of a match is accompanied by charring of wood and the release of gaseous substances, that is, an irreversible transformation of wood into other substances occurs. Another example: Over time, bronze sculptures become covered with a green coating. The fact is that bronze contains copper. This metal slowly interacts with oxygen, carbon dioxide and air moisture, as a result of which new green substances are formed on the surface of the sculpture Chemical phenomena - phenomena of 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 occur all around us. Chemical reactions also occur within ourselves. In our body, transformations of many substances continuously occur; substances react with each other, forming reaction products. Thus, in a 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
A chemical reaction is represented in general form by a reaction diagram REAGENTS -> PRODUCTS
- reagents– starting materials taken for the reaction;
- products– new substances formed as a result of a 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 changes in the color of substances, the release of gas, the formation of sediment, 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. For example, the metal magnesium flares up and burns in air with a bright, blinding flame. This is why magnesium flash was used to create photographs in the first half of the 20th century. In some cases, it is possible to release energy in the form of light, but without releasing heat. One type of Pacific plankton is capable of emitting 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 exist in three states of aggregation
- Properties of substances that are determined by measurements or visually in the absence of transformation of some substances into others are called physical
- Crystals are solids in the shape of regular polyhedra.
- Amorphous substances are substances that do not have a crystalline structure
- Chemical phenomena - phenomena of transformation of one substance into another
- Reagents are substances that enter into a chemical reaction.
- Products are substances formed as a result of a chemical reaction
- Chemical reactions can be accompanied by the release of gas, sediment, heat, light; change in color of substances
- Combustion is a complex physicochemical process of converting starting substances into combustion products during a chemical reaction, accompanied by intense release of heat and light (flame)