Chemical reaction equation scheme h2 cl2. Getting HCl
Substances are given: aqueous solutions of potassium tetrahydroxoaluminate K[Al(OH)4], aluminum chloride, potassium carbonate, chlorine. Write the equations for four possible reactions between these substances
(*answer*) 3K + AlCl3 = 4Al(OH)3 + 3KCl
(*answer*) 3K2CO3 + 2AlCl3 + 3H2O = 2Al(OH)3 + 3CO2 + 6KCl
(*answer*) K + CO2 = KHCO3 + Al(OH)3
(*answer*) 3K2CO3 + 3Cl2 = 5KCl + KClO3 + 3CO2
2AlCl3 + 2CO2 + 3H2O = Al(OH)3 + 2H2CO3 + 2HCl
Substances are given: aqueous solutions of potassium tetrahydroxozincate K2, sodium peroxide, coal, carbon dioxide. Let's write the equations of four possible reactions between these substances
(*answer*) K2 + CO2 = K2CO3 + Zn(OH)2 + H2O
(*answer*) 2Na2O2 + 2CO2 = 2Na2CO3 + O2
(*answer*) CO2 + C 2CO
(*answer*) 2Na2O2 + C Na2CO3 + Na2O
2Na2O2 + 2CO = 2Na2CO3 + 2CO2
Substances are given: an aqueous solution of potassium hexahydroxochromate K3[Cr(OH)6], solid potassium hypochlorite, manganese(IV) oxide, concentrated hydrochloric acid. Let's write the equations of four possible reactions between these substances: _
(*answer*) 2K3 + 3KClO = 2K2CrO4 + 3KCl + 2KOH + 5H2O
(*answer*) K3 + 6HCl = 3KCl + CrCl3 + 6H2O
(*answer*) 4HCl + MnO2 = Cl2 + MnCl2 + 2H2O
(*answer*) 2HCl + KClO = Cl2 + KCl + H2O
MnO2 + KClO = MnCl4 + KO
Substances given: sodium carbonate, concentrated sodium hydroxide solution, aluminum oxide, phosphorus(V) fluoride, water. Let's write the equations of four possible reactions between these substances:
(*answer*) PF5 + 4H2O = H3PO4 + 5HF
(*answer*) PF5 + 8NaOH = Na3PO4 + 5NaF + 4H2O
(*answer*) Na2CO3 + Al2O3 2NaAlO2 + CO2
(*answer*) Al2O3 + 2NaOH + 3H2O = 2Na
PF5 + 2Na2CO3 = Na3PO4 + 2CO2 + NaF
Substances are given: concentrated nitric acid, phosphorus, sulfur dioxide, concentrated ammonium sulfite solution. Let's write the equations of four possible reactions between these substances. As a result, we get: _
(*answer*) P + 5HNO3 = H3PO4 + 5NO2 + H2O
(*answer*) 2HNO3 + SO2 = H2SO4 + 2NO2
(*answer*) (NH4)2SO3 + SO2 + H2O = 2NH4HSO3
(*answer*) 2HNO3 + (NH4)2SO3 = (NH4)2SO4 + 2NO2 + H2O
P + SO2 = PS + O2
Given substances: concentrated sulfuric acid, sulfur, silver, sodium chloride. Let's write the equations of four possible reactions between these substances. As a result, we get: _
(*answer*) 2H2SO4 + S = 3SO2 + 2H2O
(*answer*) H2SO4 + 2NaCl = Na2SO4 + 2HCl (or NaHSO4 + HCl)
(*answer*) 2Ag + 2H2SO4 = Ag2SO4 + SO2 + 2H2O
(*answer*) 2Ag+S = Ag2S
3H2SO4 + 2NaCl = 2Na + 2HCl + 3SO2 + 2H2O+ O2
Substances are given: concentrated chloric acid, solutions of chromium(III) chloride, sodium hydroxide. Let's write the equations of four possible reactions between these substances. As a result, we get: _
(*answer*) HClO3 + 2CrCl3 + 4H2O = H2Cr2O7 + 7HCl
(*answer*) HClO3 + NaOH = NaClO3 + H2O
(*answer*) CrCl3 + 3NaOH = Cr(OH)3 + 3NaCl
(*answer*) CrCl3 + 6NaOH = Na3 + 3NaCl
CrCl3 + 8NaOH = Na4 + 4NaCl
Substances are given: chlorine, concentrated nitric acid, solutions of iron(II) chloride, sodium sulfide. Let's write the equations of four possible reactions between these substances. As a result, we get: _
(*answer*) 2FeCl2 + Cl2 = 2FeCl3
(*answer*) Na2S + FeCl2 = FeS + 2NaCl
(*answer*) Na2S + 4HNO3 = S + 2NO2 + 2NaNO3 + 2H2O
(*answer*) FeCl2 + 4HNO3 = Fe(NO3)3 + NO2 + 2HCl + H2O
2HNO3 + Cl2 = 2HCl + 2NO2 + H2O
Substances are given: phosphorus(III) chloride, concentrated sodium hydroxide solution, chlorine. Let's write the equations of four possible reactions between these substances. As a result, we get: _
(*answer*) PCl3 + 5NaOH = Na2PHO3 + 3NaCl + 2H2O
(*answer*) PCl3 + Cl2 = PCl5
(*answer*) 2NaOH + Cl2 = NaCl + NaClO + H2O
(*answer*) 6NaOH (hot) + 3Cl2 = 5NaCl + NaClO3 + 3H2O
4NaOH + 2Cl2 = 4NaCl + H2O + O3
Using the electron balance method, we will compose the reaction equation: Cl2 + NaI + H2O ® NaIO3 + … and determine the oxidizing agent and reducing agent. As a result, we get: _
(*answer*) reaction equation 3Cl2 + NaI + 3H2O = NaIO3 + 6HCl
(* answer *) oxidizing agent - chlorine
(* answer *) reducing agent - iodine
reaction equation 2Cl2 + NaI + 2H2O = NaIO3 + 4HCl
reducing agent - chlorine
oxidizing agent - iodine
When compiling the equations of redox reactions by this method, it is recommended to adhere to the following order:
1. Write down the reaction scheme indicating the initial and resulting substances, determine the elements that change the oxidation state as a result of the reaction, find the oxidizing agent and reducing agent.
2. Make electronic equations based on the fact that the oxidizing agent accepts electrons, and the reducing agent gives them away.
3. Select multipliers (basic coefficients) for electronic equations so that the number of electrons donated during oxidation was equal to the number of electrons obtained during reduction.
4. Arrange the coefficients in the reaction equation.
EXAMPLE 3: Write an equation for the reduction of iron oxide (III) with carbon. The reaction proceeds according to the scheme:
Fe 2 O 3 + C → Fe + CO
Solution: Iron is reduced by lowering the oxidation state from +3 to 0; carbon is oxidized, its oxidation state rises from 0 to +2.
Let's make schemes of these processes.
reducing agent 1| 2Fe +3 + 6e = 2Fe 0, oxidation process
oxidizing agent 3| C 0 -2e \u003d C +2, the recovery process
The total number of electrons donated by the reducing agent must be equal to the total number of electrons accepted by the oxidizing agent. Having found the least common multiple between the numbers 2 and 6, we determine that there should be three reducing agent molecules, and two oxidizing molecules, i.e. we find the corresponding coefficients in the reaction equation in front of the reducing agent, oxidizing agent and oxidation and reduction products.
The equation will look like:
Fe 2 O 3 + 3C \u003d 2Fe + 3CO
Method of electron-ionic equations (half-reactions).
When compiling electron-ionic equations, the form of existence of substances in solution is taken into account (a simple or complex ion, an atom or a molecule of a substance insoluble or difficult to dissociate in water).
To compose the equations of redox reactions by this method, it is recommended to adhere to the following order:
1. Draw up a reaction scheme indicating the starting materials and reaction products, mark the ions that change the oxidation state as a result of the reaction, determine the oxidizing agent and reducing agent.
2. Make schemes of oxidation and reduction half-reactions indicating the initial and formed ions or molecules under the reaction conditions.
3. Equalize the number of atoms of each element in the left and right parts of the half-reactions; it should be remembered that in aqueous solutions, water molecules, H + or OH - ions can participate in reactions.
It should be remembered that in aqueous solutions, the binding of excess oxygen and the addition of oxygen by the reducing agent occur differently, depending on the pH of the medium. In acidic solutions, excess oxygen binds with hydrogen ions to form water molecules, and in neutral and alkaline solutions, by water molecules to form hydroxide ions. For example,
MnO 4 - + 8H + + 5e = Mn 2+ + 4H 2 O (acid medium)
NO 3 - + 6H 2 O + 8e = NH 3 + 9OH - (neutral or alkaline medium).
The addition of oxygen by the reducing agent is carried out in acidic and neutral environments due to water molecules with the formation of hydrogen ions, and in an alkaline environment - due to hydroxide ions with the formation of water molecules. For example,
I 2 + 6H 2 O - 10e = 2IO 3 - + 12H + (acidic or neutral medium)
CrO 2 - + 4OH - - 3e = CrO 4 2- + 2H 2 O (alkaline)
4. Equalize the total number of charges in both parts of each half-reaction; to do this, add the required number of electrons to the left and right parts of the half-reaction.
5. Select multipliers (basic coefficients) for half-reactions so that the number of electrons donated during oxidation is equal to the number of electrons received during reduction.
6. Add up the equations of half-reactions, taking into account the found main coefficients.
7. Arrange the coefficients in the reaction equation.
EXAMPLE 4: Write an equation for the oxidation of hydrogen sulfide with chlorine water.
The reaction proceeds according to the scheme:
H 2 S + Cl 2 + H 2 O → H 2 SO 4 + HCl
Solution. The following half-reaction equation corresponds to the reduction of chlorine: Cl 2 + 2e = 2Cl - .
When compiling the equation for the half-reaction of sulfur oxidation, we proceed from the scheme: H 2 S → SO 4 2-. During this process, a sulfur atom is bound to four oxygen atoms, the source of which is water molecules. In this case, eight H + ions are formed; in addition, two H + ions are released from the H 2 S molecule.
In total, 10 hydrogen ions are formed:
The left side of the diagram contains only uncharged particles, while the total charge of the ions in the right side of the diagram is +8. Therefore, as a result of oxidation, eight electrons are released:
H 2 S + 4H 2 O → SO 4 2- + 10 H +
Since the ratio of the numbers of electrons accepted during the reduction of chlorine and given away during the oxidation of sulfur is 8 × 2 or 4 × 1, then, by adding the equations of the reduction and oxidation half-reactions, the first of them must be multiplied by 4, and the second by 1.
We get:
Cl 2 + 2e = 2Cl - | four
H 2 S + 4H 2 O \u003d SO 4 2- + 10H + + 8e - | one
4Cl 2 + H 2 S + 4H 2 O \u003d 8Cl - + SO 4 2- + 10H +
In molecular form, the resulting equation has the following form:
4Cl 2 + H 2 S + 4H 2 O \u003d 8HCl + H 2 SO 4
The same substance under different conditions can be oxidized or reduced to different oxidation states of the corresponding element, so the value of the equivalent of the oxidizing agent and reducing agent can also have different values.
The equivalent mass of an oxidizing agent is equal to its molar mass divided by the number of electrons n that one molecule of the oxidizing agent attaches in this reaction.
For example, in the reduction reaction Cl 2 + 2e = 2Cl - . n = 2 Therefore, the equivalent mass of Cl 2 is M/2, i.e. 71/2 \u003d 35.5 g / mol.
The equivalent mass of a reducing agent is equal to its molar mass divided by the number of electrons n that one molecule of the reducing agent gives up in this reaction.
For example, in the oxidation reaction H 2 S + 4H 2 O - 8e \u003d SO 4 2- + 10 H +
n = 8. Therefore, the equivalent mass of H 2 S is M/8, i.e. 34.08/8 = 4.26g/mol.
In industry, hydrogen chloride is obtained either by direct synthesis from chlorine and hydrogen, or from by-products during the chlorination of alkanes (methane). We will consider direct synthesis from elements.
HCl is a colorless gas with a pungent, characteristic odor.
t° pl = -114.8°C, t° bp = -84°C, t° crist = +57°C, i.e. Hydrogen chloride can be obtained at room temperature in liquid form by increasing the pressure to 50 - 60 atm. In the gas and liquid phase is in the form of separate molecules (absence of hydrogen bonds). Strong connection E sv \u003d 420 kJ / mol. Begins to decompose into elements at t>1500°C.
2HCl Cl 2 + H 2
Effective radius of HCl = 1.28 , dipole - 1.22 .
R Cl - = 1.81, i.e. the proton is introduced into the electron cloud of the chlorine ion by a third of the effective radius, and at the same time, the compound itself is strengthened due to an increase in the positive charge near the nucleus of the chlorine ion and balancing the repulsive effect of electrons. All hydrogen halides are formed in a similar way and are strong compounds.
Hydrogen chloride is highly soluble in water in any ratio (in one volume of H 2 O dissolves up to 450 volumes of HCl), forms several hydrates with water and gives an azeotropic mixture - 20.2% HCl and t ° kip = 108.6 ° C.
The formation of hydrogen chloride from the elements:
Cl 2 + H 2 \u003d 2HCl
A mixture of hydrogen and chlorine explodes when illuminated, indicating a chain reaction.
Early in the century, Badenstein proposed the following reaction mechanism:
Initiation: Cl 2 + hν → ē + Cl 2 +
Chain: Cl 2 + + H 2 → HCl + H + Cl +
H + Cl 2 → HCl + Cl
Chain termination: Cl + + ē → Cl
Cl + Cl → Cl2
But ē was not found in the vessel.
In 1918, Nernst proposed another mechanism:
Initiation: Cl 2 + hν → Cl + Cl
Chain: Cl + H 2 → HCl + H
H + Cl 2 → HCl + Cl
Chain termination: H + Cl → HCl
In the future, this mechanism was further developed and supplemented.
Stage 1 - initiation
reaction Cl 2 + hν → Cl + Cl
Initiated by photochemical means, i.e. by absorbing a light quantum hν. According to equivalence principle Einstein, each quantum of light can cause the transformation of only one molecule. The quantitative characteristic of the equivalence principle is the quantum yield of the reaction:
- the number of reacted molecules per 1 quantum of light.
γ in conventional photochemical reactions ≤1. However, in the case of chain reactions γ>>1. For example, in the case of the synthesis of HCl γ=10 5 , in the decay of H 2 O 2 γ=4.
If a Cl 2 molecule has absorbed a quantum of light, then it is in an excited state
10 -8 -10 -3 sec and, if the energy received with a quantum of light was enough for the transformation, then a reaction occurs, if not, then the molecule will again go into the ground state, either with the emission of a quantum of light (fluorescence or phosphorescence), or the electronic excitation is converted into vibrational or rotational energy.
Let's see what happens in our case:
E dis H 2 \u003d 426.4 kJ / mol
E dis Cl 2 = 239.67 kJ / mol
E arr HCl = 432.82 kJ / mol - without irradiation, the reaction does not proceed.
A quantum of light has an energy E kv \u003d 41.1 * 10 -20 J. The energy required to start the reaction (activation energy) is equal to the energy spent on the dissociation of the Cl 2 molecule:
those. E Cl2<Е кв и энергии кванта достаточно для преодоления потенциального барьера реакции и реакция начинается.
In contrast to catalysis, in which the potential barrier is lowered, in the case of photochemical reactions it is simply overcome by the energy of a light quantum.
Another possibility of initiating the reaction is the addition of Na vapor to the H 2 +Cl 2 mixture. The reaction proceeds at 100°C in the dark:
Na + Cl 2 → NaCl + Cl
Cl + H 2 → HCl + H ………
and up to 1000 HCl per 1 Na atom is formed.
Stage 2 - chain continuation
Chain propagation reactions in the production of HCl are of the following types:
1. Cl + H 2 → HCl + H E a \u003d 2.0 kJ / mol
2. H + Cl 2 → HCl + Cl E a \u003d 0.8 kJ / mol
These are links in the chain.
The rate of these reactions can be represented as follows:
W 1 = K 1 [ H 2 ]
W 2 \u003d K 2 [Cl 2]
Because the activation energies of these reactions are small, their rates are high. The chains in this case are unbranched, and according to the theory of unbranched chains:
W chain development = W is photochemically initiated, i.e. by absorbing a quantum of light from a break,
Cl + Cl + M → Cl 2 + M,
then W arr \u003d K 2
The rate of HCl production depends on reactions 1 and 2
in this case, W 1 \u003d W 2, because the chains are quite long (from the theory of chain reactions)
This kinetic equation is valid in the absence of impurities in the H 2 + Cl 2 mixture. If air gets into the system, then the kinetic equation will be different. In particular
W arr \u003d K, i.e. non-quadratic termination and the course of the process is reversed.
Because there are substances that are inhibitors of chain reactions. The inhibitor of the HCl formation reaction is oxygen:
O 2 + H → O 2 H
This radical is inactive and can only react with the same radical, regenerating oxygen.
O 2 H + O 2 H \u003d O 2 + H 2 O 2
Calculations show that in the presence of 1% O 2 the reaction slows down by a factor of 1000. The presence of NCl 3 slows down the rate of the process even more strongly, which slows down the reaction 10 5 times more than oxygen. Because Nitrogen chloride may be present in chlorine during its production in industry, careful purification of the initial chlorine is necessary before the synthesis of HCl.
chain reactions include in their mechanism a set of consecutively repeating elementary acts of the same type (a chain).
Consider the reaction:
H 2 + Cl 2 \u003d 2HCl
It consists of the following steps common to all chain reactions:
1) Initiation, or the origin of the chain
Cl 2 \u003d 2Cl
The decomposition of a chlorine molecule into atoms (radicals) occurs during UV irradiation or heating. The essence of the initiation stage is the formation of active, reactive particles.
2) Chain development
Cl + H 2 \u003d HCl + H
H + Cl 2 \u003d HCl + Cl
As a result of each elementary act of chain development, a new chlorine radical is formed, and this stage is repeated again and again, theoretically, until the reagents are completely consumed.
3) Recombination, or open circuit
2Cl = Cl 2
2H = H2
H + Cl = HCl
Radicals that are nearby can recombine, forming a stable particle (molecule). They give excess energy to the "third particle" - for example, the walls of a vessel or impurity molecules.
Considered chain reaction is unbranched, since the number of radicals does not increase in the elementary act of chain development. Chain reaction of interaction of hydrogen with oxygen is branched, because the number of radicals in the elementary act of chain development increases:
H + O 2 \u003d OH + O
O + H 2 \u003d OH + H
OH + H 2 \u003d H 2 O + H
Branched chain reactions include many combustion reactions. An uncontrolled increase in the number of free radicals (both as a result of chain branching and for straight chain reactions in the case of too rapid initiation) can lead to a strong acceleration of the reaction and an explosion.
It would seem that the greater the pressure, the higher the concentration of radicals and the more likely an explosion. But in fact, for the reaction of hydrogen with oxygen, an explosion is possible only in certain pressure areas: from 1 to 100 mm Hg. and above 1000 mm Hg. This follows from the mechanism of the reaction. At low pressure, most of the resulting radicals recombine on the walls of the vessel, and the reaction proceeds slowly. With an increase in pressure up to 1 mm Hg. radicals rarely reach the walls, because more likely to react with molecules. In these reactions, radicals multiply and an explosion occurs. However, at pressures above 100 mm Hg. the concentrations of substances increase so much that the recombination of radicals begins as a result of triple collisions (for example, with a water molecule), and the reaction proceeds calmly, without an explosion (stationary flow). Above 1000 mmHg the concentrations become very high, and even triple collisions are not enough to prevent the multiplication of radicals.
You know the branched chain reaction of uranium-235 fission, in each elementary act of which 1 neutron is captured (playing the role of a radical) and up to 3 neutrons are emitted. Depending on the conditions (for example, on the concentration of neutron absorbers), a stationary flow or explosion is also possible for it. This is another example of the correlation between the kinetics of chemical and nuclear processes.
