Hydrogen chloride is produced in industry 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° boil = –84°C, t° crisp = +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 phases it is found in the form of separate molecules (no hydrogen bonds). Strong compound E St = 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 the electrons. All hydrogen halides are formed similarly and are strong compounds.
Hydrogen chloride is highly soluble in water in any ratio (up to 450 volumes of HCl are dissolved in one volume of H 2 O), with water it forms several hydrates and gives an azeotropic mixture - 20.2% HCl and boiling point = 108.6°C.
Formation of hydrogen chloride from elements:
Cl 2 + H 2 = 2HCl
A mixture of hydrogen and chlorine explodes when illuminated, indicating a chain reaction.
At the beginning of 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
Open circuit: Cl + + ē → Cl
Cl + Cl → Cl 2
But ē was not found in the vessel.
In 1918, Nernst proposed a different mechanism:
Initiation: Cl 2 + hν → Cl + Cl
Chain: Cl + H 2 → HCl + H
H + Cl 2 → HCl + Cl
Open circuit: H + Cl → HCl
Subsequently, this mechanism was further developed and supplemented.
Stage 1 – initiation
reaction Cl 2 + hν → Cl + Cl
Initiated photochemically, i.e. by absorbing a light quantum hν. According to principle of equivalence According to Einstein, each quantum of light can cause the transformation of only one molecule. A 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 ordinary 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, during the decomposition of H 2 O 2 γ = 4.
If a Cl 2 molecule absorbs a light quantum, then it is in an excited state
10 -8 -10 -3 sec and, if the energy received with the light quantum was sufficient for conversion, then a reaction occurs, if not, then the molecule will again go into the ground state, either with the emission of a light quantum (fluorescence or phosphorescence), or electronic excitation is converted into vibration or rotation energy.
Let's see what happens in our case:
E dis H 2 = 426.4 kJ/mol
E dis Cl 2 = 239.67 kJ/mol
E arr HCl = 432.82 kJ/mol - the reaction does not proceed without irradiation.
A quantum of light has an energy E q = 41.1 * 10 -20 J. The energy required to start the reaction (activation energy) is equal to the energy expended in the dissociation of the Cl 2 molecule:
those. E Cl2<Е кв и энергии кванта достаточно для преодоления потенциального барьера реакции и реакция начинается.
Unlike catalysis, in which the potential barrier is reduced, in the case of photochemical reactions it is simply overcome by the energy of the light quantum.
Another possibility for initiating the reaction is the addition of Na vapor to the H 2 +Cl 2 mixture. The reaction occurs at 100°C in the dark:
Na + Cl 2 → NaCl + Cl
Cl + H 2 → HCl + H………
and up to 1000 HCl is formed per 1 Na atom.
Stage 2 – continuation of the chain
Chain continuation reactions in the production of HCl are of the following types:
1. Cl + H 2 → HCl + H E a =2.0 kJ/mol
2. H + Cl 2 → HCl + Cl E a =0.8 kJ/mol
These are chain links.
The rate of these reactions can be represented as follows:
W 1 = K 1 [H 2 ]
W 2 = K 2 [Cl 2 ]
Because The activation energies of these reactions are small, then their rates are high. The chains in this case are unbranched, and according to the theory of unbranched chains:
W chain development = W is initiated photochemically, i.e. by absorbing the light cutoff quantum,
Cl + Cl +M → Cl 2 + M,
then W arr = K 2
The rate of HCl production depends on reactions 1 and 2
in this case W 1 =W 2, since 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, the kinetic equation will be different. In particular
W arr = K, i.e. non-quadratic break and the course of the process completely changes.
Because There are substances that are chain reaction inhibitors. The inhibitor of the reaction of HCl formation 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 = O 2 + H 2 O 2
Calculations show that in the presence of 1% O 2 the reaction slows down 1000 times. The rate of the process is further slowed down by the presence of NCl 3 , which slows down the reaction 10 5 times more than oxygen. Because Nitrogen chloride may be present in chlorine during its production process in industry; careful purification of the initial chlorine is necessary before the synthesis of HCl.
When composing equations of redox reactions using this method, it is recommended to adhere to the following order:
1. Write down the reaction scheme indicating the starting and resulting substances, identify the elements that change the oxidation state as a result of the reaction, find the oxidizing agent and the reducing agent.
2. Make electronic equations based on what the oxidizing agent accepts electrons, and the restorer gives them away.
3. Select factors (main coefficients) for electronic equations so that the number of electrons given up during oxidation was equal to the number of electrons gained during reduction.
4. Arrange the coefficients in the reaction equation.
EXAMPLE 3: Write down the equation for the reduction reaction of iron (III) oxide carbon. The reaction proceeds according to the scheme:
Fe 2 O 3 + C → Fe + CO
Solution: Iron is reduced, lowering the oxidation state from +3 to 0; carbon is oxidized, its oxidation state increases from 0 to +2.
Let's draw up diagrams of these processes.
reducing agent 1| 2Fe +3 + 6e = 2Fe 0, oxidation process
oxidizing agent 3| C 0 -2e = C +2, recovery process
The total number of electrons given up 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 molecules and two oxidizing molecules, i.e. we find the corresponding coefficients in the reaction equation before the reducing agent, the oxidizing agent and the products of oxidation and reduction.
The equation will look like:
Fe 2 O 3 + 3C = 2Fe + 3CO
Method of electron-ion equations (half-reactions).
When composing electron-ion equations, the form of existence of substances in solution is taken into account (simple or complex ion, atom or molecule of an insoluble or difficult-to-dissociate substance in water).
To create equations for redox reactions using this method, it is recommended to adhere to the following order:
1.Make a reaction diagram 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 the reducing agent.
2. Draw up diagrams of oxidation and reduction half-reactions, indicating the initial ions or molecules formed under the reaction conditions.
3. Equal the number of atoms of each element in the left and right sides 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 oxygen reducing agents occur differently, depending on the pH of the environment. In acidic solutions, excess oxygen is bound by 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 (acidic environment)
NO 3 - + 6H 2 O + 8e = NH 3 + 9OH - (neutral or alkaline medium).
The addition of oxygen by a 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 medium)
4. Equal 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 sides of the half-reaction.
5. Select multipliers (main coefficients) for half-reactions so that the number of electrons given up during oxidation is equal to the number of electrons accepted during reduction.
6. Add up the half-reaction equations taking into account the found main coefficients.
7. Arrange the coefficients in the reaction equation.
EXAMPLE 4: Write an oxidation equation hydrogen sulfide chlorine water.
The reaction proceeds according to the scheme:
H 2 S + Cl 2 + H 2 O → H 2 SO 4 + HCl
Solution. The reduction of chlorine corresponds to the following half-reaction equation: Cl 2 + 2e = 2Cl - .
When composing 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 bonds with 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.
A total of 10 hydrogen ions are formed:
The left side of the diagram contains only uncharged particles, and the total charge of the ions on the right side of the diagram is +8. Therefore, eight electrons are released as a result of oxidation:
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 donated during the oxidation of sulfur is 8?2 or 4?1, then, when adding the equations for the half-reactions of reduction and oxidation, the first of them must be multiplied by 4, and the second by 1.
We get:
Cl 2 + 2e = 2Cl - | 4
H 2 S + 4H 2 O = SO 4 2- + 10H + +8e - | 1
4Cl 2 + H 2 S + 4H 2 O = 8Cl - + SO 4 2- +10H +
In molecular form, the resulting equation is as follows:
4Cl 2 + H 2 S + 4H 2 O = 8HCl + H 2 SO 4
The same substance under different conditions can be oxidized or reduced to different oxidation states of the corresponding element, therefore the equivalent value 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 adds in a given reaction.
For example, in the reduction reaction Cl 2 + 2e = 2Cl - . n = 2 Therefore, the equivalent mass of Cl 2 is equal to M/2, i.e. 71/2 = 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 a given reaction.
For example, in the oxidation reaction H 2 S + 4H 2 O - 8е = SO 4 2- + 10 H +
n = 8. Therefore, the equivalent mass of H 2 S is equal to M/8, i.e. 34.08/8 = 4.26 g/mol.
Chain reactions include in their mechanism many sequentially repeating elementary acts of the same type (chain).
Consider the reaction:
H2 + Cl2 = 2HCl
It consists of the following stages, common to all chain reactions:
1) Initiation, or chain initiation
Cl 2 = 2Cl
The decomposition of the chlorine molecule into atoms (radicals) occurs during UV irradiation or heating.
2) The essence of the initiation stage is the formation of active, reactive particles.
Chain development
Cl + H 2 = HCl + H
H + Cl 2 = HCl + Cl
3) 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. Recombination
, or open circuit
2Cl = Cl 2
2H = H 2
H + Cl = HCl
Radicals that happen to be nearby can recombine, forming a stable particle (molecule). They give excess energy to a “third particle” - for example, the walls of a vessel or impurity molecules. Considered chain reaction is unbranched , since in the elementary act of chain development the number of radicals does not increase . Chain reaction of hydrogen with oxygen, because the number of radicals in the elementary act of chain development increases:
H + O 2 = OH + O
O + H 2 = OH + H
OH + H 2 = 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 unbranched 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 regions: from 1 to 100 mm Hg. and above 1000 mm Hg. This follows from the reaction mechanism.
At low pressure, most of the resulting radicals recombine on the walls of the vessel, and the reaction proceeds slowly. When the pressure rises to 1 mm Hg. radicals reach the walls less often, because react more often with molecules. In these reactions, radicals multiply and an explosion occurs. However, at pressure 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 explosion (stationary flow). Above 1000 mm Hg. concentrations become very high, and even triple collisions are not enough to prevent the proliferation of radicals.
You know the branched chain reaction of fission of uranium-235, 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), it is also possible for it to have a steady flow or an explosion. This is another example of the correlation between the kinetics of chemical and nuclear processes.
Applications
The following substances are given: aqueous solutions of potassium tetrahydroxoaluminate K[Al(OH)4], aluminum chloride, potassium carbonate, chlorine. Write 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
The following 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
2Na2O2 + 2CO = 2Na2CO3 + 2СO2
The following 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
The substances given are: sodium carbonate, concentrated sodium hydroxide solution, aluminum oxide, phosphorus(V) fluoride, water. Let's write the equations for 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
The substances given are: concentrated nitric acid, phosphorus, sulfur dioxide, concentrated solution of ammonium sulfite. 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
The substances given are: 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
The following substances are given: concentrated perchloric 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
The following 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
The following 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