Storage of hydrogen in metals. Alloys hydrogen - metal
Inorganic chemistry
Joint hydrolysis of salts
For example:
Task 1.1.
Task 1.2
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Task 1.3.
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Reactions of oxides with water
For example:
Task 2.1
Mn 2 O 7 + H 2 O =
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Task 3.1
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Alkaline or acid hydrolysis of binary compounds
For a school course - an exotic thing, but in the Unified State Examination-2014 it met ... We are talking about such, for example, reactions:
Ca 3 N 2 + HCl \u003d
Here you can argue like this. An alkali (NaOH) or an acid (HCl) reacts with a binary compound in solution. And this means that in fact, the first reaction is with water (hydrolysis of a binary compound):
PCl 5 + H 2 O → H 3 PO 4 + HCl
Ca 3 N 2 + H 2 O → Сa (OH) 2 + NH 3
And then the hydrolysis products react with an alkali (in the first case) or with an acid (in the second case):
PCl 5 + H 2 O → H 3 PO 4 + HCl → (+ NaOH) → Na 3 PO 4 + NaCl + H 2 O
Ca 3 N 2 + H 2 O → Сa (OH) 2 + NH 3 → (+ HCl) → CaCl 2 + NH 4 Cl + (H 2 O)
As a result, the equations will look like this:
PCl 5 + 8NaOH \u003d Na 3 PO 4 + 5NaCl + 4 H 2 O
Ca 3 N 2 + 8HCl \u003d 3CaCl 2 + 2NH 4 Cl
Practice:
Task 3.2 Arguing similarly, determine what happens in the interaction:
Na 3 N + HCl →
PBr 3 + NaOH →
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Ammonia and its properties
Ammonia reacts with acids, adding a proton by the donor-acceptor mechanism and forming ammonium salts.
Task 4.1. Ammonia was passed through a solution of sulfuric acid. What two salts can form in this case? What does it depend on? Write reaction equations.
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An aqueous solution of ammonia has the properties of a weak alkali, so it can be used to precipitate insoluble metal hydroxides.
Task 4.2. An excess of ammonia was passed through an aqueous solution of chromium (III) sulfate. Write down the reaction equation.
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3) Ammonia is a reducing agent. In particular, it is able to restore metals from oxides.
Task 4.3. A stream of ammonia was passed through the oxide of copper (II) when heated. Write the reaction equation.
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4) Ammonia is capable of being a ligand and can form complexes - ammoniates. The mention of the ammonia complex of copper in the exam is especially likely, since it has a bright blue color and can be used to detect divalent copper compounds.
Task 4.4. To a solution of copper sulfate (II) was added an excess of aqueous ammonia. Write down the reaction equation.
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In general, those reactions that are accompanied by explosions proceed with the greatest speed. And under normal conditions - ion exchange reactions in aqueous solutions. Why? Because they involve electrolytes that are already dissociated, the bonds are broken. Therefore, nothing prevents the ions from instantly connecting with each other. We can assume that the activation barrier of such a reaction approaches zero.
For example:
Which substances react with each other at the highest rate at room temperature?
1) HCl(p-p) and NaOH(p-p)
2) S (solid) and H 2 (g)
3) CO 2 (g) and H 2 O (l)
4) FeS 2 (solid) and O 2 (g)
The correct answer is 1), since this is an ion exchange reaction.
Mixed oxides Fe 3 O 4 and Pb 3 O 4
Iron forms a mixed oxide - iron scale Fe 3 O 4 (FeO ∙ Fe 2 O 3) with oxidation states +2 and +3.
Lead forms a mixed oxide - minium Pb 3 O 4 (2PbO ∙ PbO 2) with oxidation states +2 and +4.
When these oxides react with acids, two salts can be obtained at once:
Fe 3 O 4 + 8HCl \u003d FeCl 2 + 2FeCl 3 + 4H 2 O
Pb 3 O 4 + 4HNO 3 \u003d 2Pb (NO 3) 2 + PbO 2 + H 2 O (PbO 2 is amphoteric, therefore it does not turn into a salt).
Transitions Fe +2 ↔ Fe +3 and Cu +1 ↔ Cu +2
Here are some difficult situations:
Fe 3 O 4 + HNO 3 = what happens?
It would seem that two salts and water should be obtained: Fe (NO 3) 2 + Fe (NO 3) 3 + H 2 O (see previous section), but HNO 3 is a strong oxidizing agent, therefore it will oxidize iron +2 as part of iron oxide to iron +3 and you get only one salt:
Fe 3 O 4 + 10HNO 3 (conc) = 3Fe (NO 3) 3 + NO 2 + 5H 2 O
Similarly, in the reaction of Cu 2 O + HNO 3, it may seem that the products will be CuNO 3 + H 2 O. But in fact, monovalent copper (Cu + 1 2 O) can be oxidized to divalent, so the redox reaction will go:
Cu 2 O + 6HNO 3 (conc) = 2Cu (NO 3) 2 + 2NO 2 + 3H 2 O
Task 7.1. Write down the reaction equations:
Fe 3 O 4 + H 2 SO 4 (diff) =
Fe 3 O 4 + H 2 SO 4 (conc) =
Fe 2 (SO 4) 3 + H 2 S =
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Decomposition of nitrates
In general, the decomposition of nitrates occurs according to a well-known scheme, and the composition of the products depends on the location of the metal in the activity series. But there are difficult situations:
Task 9.1 What products will be obtained from the decomposition of iron (II) nitrate? Write down the reaction equation.
Task 9.2 What products will be obtained from the decomposition of copper (II) nitrate? Write down the reaction equation.
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Organic chemistry
Trivial names
You need to know which organic substances correspond to the names:
isoprene, divinyl, vinylacetylene, toluene, xylene, styrene, cumene, ethylene glycol, glycerin, formaldehyde, acetaldehyde, propionaldehyde, acetone, the first six limiting monobasic acids (formic, acetic, propionic, butyric, valeric, caproic), acrylic acid, stearic acid, palmitic acid, oleic acid, linoleic acid, oxalic acid, benzoic acid, aniline, glycine, alanine. Do not confuse propionic acid with propenoic acid!! Salts of the most important acids: formic - formates, acetic - acetates, propionic - propionates, butyric - butyrates, oxalic - oxalates. The radical –CH=CH 2 is called vinyl!!
At the same time, some inorganic trivial names:
Table salt (NaCl), quicklime (CaO), slaked lime (Ca(OH) 2), lime water (Ca(OH) 2 solution), limestone (CaCO 3), quartz (aka silica or silicon dioxide - SiO 2 ), carbon dioxide (CO 2), carbon monoxide (CO), sulfur dioxide (SO 2), brown gas (NO 2), drinking or baking soda (NaHCO 3), soda ash (Na 2 CO 3), ammonia (NH 3), phosphine (PH 3), silane (SiH 4), pyrite (FeS 2), oleum (solution of SO 3 in concentrated H 2 SO 4), copper sulfate (CuSO 4 ∙ 5H 2 O).
Some rare reactions
1) Formation of vinylacetylene:
2) Direct oxidation reaction of ethylene to acetaldehyde:
This reaction is insidious in that we know well how acetylene turns into aldehyde (Kucherov's reaction), and if the ethylene → aldehyde transformation occurs in the chain, then this can confuse us. So, this is the reaction!
3) The reaction of direct oxidation of butane to acetic acid:
This reaction underlies the industrial production of acetic acid.
4) Lebedev's reaction:
Differences between phenols and alcohols
A huge number of errors in such tasks !!
1) It should be remembered that phenols are more acidic than alcohols (the O-H bond in them is more polar). Therefore, alcohols do not react with alkali, while phenols react with both alkali and some salts (carbonates, bicarbonates).
For example:
Task 10.1
Which of these substances react with lithium:
a) ethylene glycol, b) methanol, c) phenol, d) cumene, e) glycerin.
Task 10.2
Which of these substances react with potassium hydroxide:
a) ethylene glycol, b) styrene, c) phenol, d) ethanol, e) glycerin.
Task 10.3
Which of these substances react with cesium bicarbonate:
a) ethylene glycol, b) toluene, c) propanol-1, d) phenol, e) glycerin.
2) It should be remembered that alcohols react with hydrogen halides (this reaction proceeds via the C-O bond), but phenols do not (in them, the C-O bond is inactive due to the conjugation effect).
disaccharides
Main disaccharides: sucrose, lactose and maltose have the same formula C 12 H 22 O 11 .
They should be remembered:
1) that they are able to hydrolyze into those monosaccharides that make up: sucrose- for glucose and fructose, lactose- for glucose and galactose, maltose- two glucose.
2) that lactose and maltose have an aldehyde function, that is, they are reducing sugars (in particular, they give reactions of “silver” and “copper” mirrors), and sucrose, a non-reducing disaccharide, does not have an aldehyde function.
Reaction mechanisms
Let's hope that the following knowledge is enough:
1) for alkanes (including in the side chains of arenes, if these chains are limiting), reactions are characteristic free radical substitution (with halogens) that go along radical mechanism (chain initiation - the formation of free radicals, the development of the chain, chain termination on the walls of the vessel or during the collision of radicals);
2) reactions are characteristic for alkenes, alkynes, arenes electrophilic addition that go along ionic mechanism (through education pi-complex and carbocation ).
Features of benzene
1. Benzene, unlike other arenes, is not oxidized by potassium permanganate.
2. Benzene and its homologues are able to enter into addition reaction with hydrogen. But only benzene can also enter into addition reaction with chlorine (only benzene and only with chlorine!). At the same time, all arenas are able to enter into substitution reaction with halogens.
Zinin's reaction
Reduction of nitrobenzene (or similar compounds) to aniline (or other aromatic amines). This reaction in one of its types is almost certain to occur!
Option 1 - reduction with molecular hydrogen:
C 6 H 5 NO 2 + 3H 2 → C 6 H 5 NH 2 + 2H 2 O
Option 2 - reduction with hydrogen obtained by the reaction of iron (zinc) with hydrochloric acid:
C 6 H 5 NO 2 + 3Fe + 7HCl → C 6 H 5 NH 3 Cl + 3FeCl 2 + 2H 2 O
Option 3 - reduction with hydrogen obtained by the reaction of aluminum with alkali:
C 6 H 5 NO 2 + 2Al + 2NaOH + 4H 2 O → C 6 H 5 NH 2 + 2Na
Amine properties
For some reason, the properties of amines are the least remembered. Perhaps this is due to the fact that amines are studied last in the course of organic chemistry, and their properties cannot be repeated by studying other classes of substances. Therefore, the recipe is this: just learn all the properties of amines, amino acids and proteins.
Decomposition of acetates
For some reason, the compilers of the exam believe that you need to know how acetates decompose. Although this reaction is not in the textbooks. Different acetates decompose in different ways, but let's remember the reaction that comes across in the exam:
thermal decomposition of barium acetate (calcium) produces barium carbonate (calcium) and acetone!!!
Ba(CH 3 COO) 2 → BaCO 3 + (CH 3) 2 CO ( t0)
Ca(CH 3 COO) 2 → CaCO 3 + (CH 3) 2 CO ( t0)
In fact, when this occurs, decarboxylation occurs:
Answers:
1.1. During the joint hydrolysis of salts, one of which is hydrolyzed by the cation, and the other by the anion, the hydrolysis is mutually enhanced and proceeds to the formation of the final hydrolysis products of both salts: 2AlCl 3 + 3Na 2 S + 6H 2 O = 2Al(OH) 3 ↓ + 3H 2S + 6NaCl
1.2. Similarly: 2FeCl 3 + 3Na 2 CO 3 + 3H 2 O \u003d 2Fe (OH) 3 ↓ + 3CO 2 + 6NaCl
1.3. Reaction sequence:
2Al + 3I 2 = 2AlI 3
AlI 3 + 3NaOH \u003d Al (OH) 3 + 3NaI
Al(OH) 3 + 3HCl = AlCl 3 + 3H 2 O
2AlCl 3 + 3Na 2 CO 3 + 3H 2 O \u003d 2Al (OH) 3 + 3CO 2 + 6NaCl
NO + H 2 O = do not react (as non-salt-forming oxide)
BaO + H 2 O \u003d Ba (OH) 2 (react, as a soluble hydroxide is obtained)
CrO + H 2 O = (do not react, since chromium (II) hydroxide is insoluble)
SO 2 + H 2 O \u003d H 2 SO 3 (they react as a soluble hydroxide is obtained)
SiO 2 + H 2 O = (do not react, since silicon (IV) hydroxide, that is, silicic acid, is insoluble)
Mn 2 O 7 + H 2 O \u003d 2HMnO 4 (react, as a soluble hydroxide is obtained - manganese acid)
2NO 2 + H 2 O \u003d HNO 2 + HNO 3
3.1. The hydrolysis of binary compounds produces the hydroxide of the first element and the hydrogen compound of the second element. In the case of a hydride, the second product would simply be hydrogen:
NaH + H 2 O \u003d NaOH + H 2
MgH 2 + 2H 2 O \u003d Mg (OH) 2 + 2H 2
Na 3 N + 4HCl → 3NaCl + NH 4 Cl
PBr 3 + 6NaOH → Na3PO3 + 3NaBr + 3H 2 O
4.1 When passing ammonia through solutions of polybasic acids, medium or acidic salts can be obtained, depending on which of the reagents is in excess:
NH 3 + H 2 SO 4 \u003d NH 4 HSO 4 (acid in excess)
2NH 3 + H 2 SO 4 \u003d 2 (NH 4) 2 SO 4 (ammonia in excess)
Cr 2 (SO 4) 3 + 6NH 3 + 6H 2 O \u003d 2Cr (OH) 3 ↓ + 3 (NH 4) 2 SO 4
(In fact, this is the same reaction as:
Cr 2 (SO 4) 3 + 6NH 4 OH \u003d 2Cr (OH) 3 ↓ + 3 (NH 4) 2 SO 4,
but the formula NH 4 OH is not accepted now).
3CuO + 2NH 3 \u003d 3Cu + N 2 + 3H 2 O
CuSO 4 + 4NH 3 \u003d SO 4
(Although in fact, this reaction will go first:
CuSO 4 + 2NH 3 + 2H 2 O \u003d Cu (OH) 2 ↓ + (NH 4) 2 SO 4 (since ammonia acts as an alkali)
And then: Cu(OH) 2 ↓ + 4NH 3 = (OH) 2)
In general, in any case, with a sufficient amount of ammonia, you will get a complex and a bright blue color!
K 3 + 6HBr \u003d 3KBr + AlBr 3 + 6H 2 O
K 3 + 3HBr \u003d 3KBr + Al (OH) 3 ↓ + 3H 2 O
Na 2 + 2CO 2 \u003d 2NaHCO 3 + Zn (OH) 2 ↓
K \u003d KAlO 2 + 2H 2 O ( t0)
Cl + 2HNO 3 \u003d 2NH 4 NO 3 + AgCl ↓
2СuSO 4 + 4KI \u003d 2CuI + I 2 + 2K 2 SO 4 (divalent copper is reduced to monovalent)
Fe 2 O 3 + 6HI \u003d 2FeI 2 + I 2 + 3H 2 O
KNO 2 + NH 4 I \u003d KI + N 2 + 2H 2 O
H 2 O 2 + 2KI \u003d I 2 + 2KOH
Fe 3 O 4 + 4H 2 SO 4 (diff) = FeSO 4 + Fe 2 (SO 4) 3 + 4H2O
since dilute sulfuric acid is not a strong oxidizing agent, the usual exchange reaction takes place.
2Fe 3 O 4 + 10H 2 SO 4 (conc) = 3Fe 2 (SO 4) 3 + SO 2 + 10H 2 O
since concentrated sulfuric acid is a strong oxidizing agent, iron +2 is oxidized to iron +3.
Fe 2 (SO 4) 3 + H 2 S \u003d 2FeSO 4 + S + H 2 SO 4
since hydrogen sulfide is a reducing agent, iron +3 is reduced to iron +2.
NaHSO 4 + NaOH = Na 2 SO 4 + H 2 O
Na 2 SO 4 + NaOH - do not react
NaHSO 4 + Ba(OH) 2 = BaSO 4 + NaOH + H 2 O
Na 2 SO 4 + Ba(OH) 2 = BaSO 4 + 2NaOH
Cu + 2H 2 SO 4 (conc) = CuSO 4 + SO 2 + 2H 2 O
Сu + HCl - do not react
CuO + 2HCl = CuCl 2 + H2O
ZnS + 2HCl = ZnCl 2 + H 2 S
ZnO + 2HCl = ZnCl 2 + H 2 O
Cu 2 O + 3H 2 SO 4 \u003d 2CuSO 4 + SO 2 + 3H 2 O (the point is that, since the acid is concentrated, it oxidizes Cu +1 to Cu +2.
CuO + H 2 SO 4 \u003d CuSO 4 + H 2 O
It would seem that the decomposition of iron (II) nitrate should produce iron oxide (II), nitric oxide (IV) and oxygen. But the trick is that since iron (II) oxide does not have the highest oxidation state, but oxygen is released in the reaction, iron will be oxidized to +3 and iron (III) oxide will be obtained:
Fe(NO 3) 2 → Fe 2 O 3 + NO 2 + O 2
There are two reducing agents in this reaction: iron and oxygen. The coefficients will look like this:
4Fe(NO 3) 2 = 2Fe 2 O 3 + 8NO 2 + O 2
There is nothing special in this reaction, except that it is often forgotten that copper is also one of those metals, during the decomposition of which a metal oxide is obtained, and not the metal itself:
2Cu(NO 3) 2 \u003d 2CuO + 4NO 2 + O 2
But all the metals that are behind copper, when decomposing their nitrates, will give just a metal.
Correct answers: a, b, c, e (there is no hydroxyl group in cumene at all, it is an arene).
Correct answers: in (there is no hydroxyl group in styrene at all, it is an arene).
Correct answers: no correct answer (there is no hydroxyl group in toluene at all, it is an arene. Phenol is not acidic enough. Some carboxylic acid could react.).
Inorganic chemistry
Joint hydrolysis of salts
For example:
Task 1.1. What happens when aqueous solutions of aluminum chloride and sodium sulfide are combined (write the reaction equation)?
Task 1.2. What happens when aqueous solutions of iron (III) chloride and sodium carbonate are combined (write the reaction equation)?
Answers below
Co-hydrolysis is often found in C2 tasks, where it is not so easy to detect. Here is an example:
Task 1.3. Metallic aluminum powder was mixed with solid iodine and a few drops of water were added. Sodium hydroxide solution was added to the resulting salt until a precipitate formed. The resulting precipitate was dissolved in hydrochloric acid. Upon subsequent addition of sodium carbonate solution, precipitation was again observed. Write the equations for the four described reactions.
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Reactions of oxides with water
Question: When do oxides react with water?
Answer: only salt-forming oxides react with water and only if a soluble hydroxide is obtained.
For example:
Task 2.1. Write down the equations of feasible reactions:
Mn 2 O 7 + H 2 O =
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Metal hydrides and their properties
Hydrogen is able to react with active metals (mainly standing before aluminum in the series of metal activity, that is, these are alkali and alkaline earth metals). In this case, hydrides are formed, for example: LiH, CaH 2.
In hydrides, the oxidation state of hydrogen is -1!
Hydrides are binary compounds, and therefore are capable of hydrolyzing.
Task 3.1 Write down the hydrolysis equations for sodium hydride, magnesium hydride.
Let us name several distinctive characteristics of materials used in hydride systems.
1) All alloys bearing the HY-STOR trademark are manufactured by Energies, Inc. Much of the data presented in this section comes from the work of Houston and Sandrock. In chemical formulas, the symbol M stands for mischmetal - a mixture of rare earth metals, usually obtained from monazite dust. The influence of mischmetal on the pressure on the plateau depends strongly on the ratio of the amount of cerium and lanthanum in this mixture of metals.
plateau slope
In accordance with the simplified thermodynamic model of the hydride system described in the next paragraph, the plateau in the equilibrium dependence | pressure from concentration should be horizontal. However, in practice; the pressure on the plateau slightly increases with increasing hydrogen concentration in the solid phase.
The slope of the plateau can be quantified by the slope factor d n(pd)/d(H, M), where pd is the pressure on the plateau in the desorption isotherm. On fig. 9.7, the dotted line passing through the desorption isotherm corresponding to 25 °C intersects the vertical line H/M = 0 at the point pd = 9.1 atm, and the line H/M = 1.2 at the point pd = 14.8 atm. Then
dlnpd In 14.8-In 9.1
M) 1.2 ' ■ U '
This coefficient value is acceptable. The pressure plateau slope parameter for TiFe alloy, for example, is zero, while for some calcium alloys the value of this parameter exceeds three. When the alloy solidifies (at the manufacturing stage), there is a tendency to segregation, i.e., the separation of some elements that make up the alloy. Apparently, this phenomenon is the main reason for the appearance of the slope of the plateau, since from the standpoint of thermodynamics, the dependence of the equilibrium pressure on the hydrogen concentration for an ideally homogeneous alloy should have a horizontal plateau. Annealing the material prior to grinding it can reduce the slope of the plateau. The values of the slope coefficient and some other characteristics are given in Table. 9.4, 9.5 and 9.6.
Absorption-desorption hysteresis
As noted above, the pressure on the plateau during absorption is usually somewhat higher than during desorption. In other words, there is a hysteresis of the absorption and desorption processes during the cyclic charging and discharging of the alloy (see Fig. 9.7,
9.8, 9.10 and 9.11).
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Table 9.4. Thermodynamic properties of some metal hydrides
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Plateau slope8*, ^ |
Hysteresis factor Pa/Pd |
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Table 9.6. Maximum hydrogen content and heat capacity of some metal hydrides
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The phenomenon of hysteresis is associated with the irreversible process of heat release due to plastic deformation of the crystal lattice, namely, its expansion during absorption and compression during desorption of hydrogen.
The phenomenon of hysteresis is quantitatively characterized by the ratio of the values of the equilibrium pressure of hydrogen during absorption and desorption at a value of HM = 0.5 and usually at a temperature of 25 °C. It is generally accepted that this ratio does not depend on temperature.
Useful capacity is defined as the change in the number of absorbed hydrogen atoms per metal atom in the hydride, N/M, as the pressure changes from 10 times the plateau pressure to 0.1 times the plateau pressure. This method of determining the usable capacity gives somewhat overestimated values. A more realistic value is obtained if the range of pressure change is significantly narrowed.
On fig. 9.9 (alloy Fe0 8ІЧІ(| 2Ті), the pressure on the plateau at a temperature of 70 ° C is approximately 0.9 atm. At a pressure of 10 times the specified value, the H / M ratio is 0.65, and at a pressure 10 times less than pressure on the plateau, N/M = 0.02 Thus, the difference A(H/M) = 0.63 In other words, 0.63 kmol of atomic hydrogen (0.63 kg) can be extracted from 1 kmol of hydride.
FeTi alloy (cf. Fig. 9.4)
Heat capacity
Hydride systems are activated by changing the temperature. In order to design such systems, it is necessary to have information about the value of the heat capacity of various alloys. The values of the heat capacity of a number of alloys are given in table. 9.6.
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Stick and ball iron hydride molecule model |
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| names | |
|---|---|
| IUPAC systematic name Hydridoiron (3) |
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| Identifiers | |
| properties | |
| FeH3 | |
| Molar mass | 56.853 g mol -1 |
| thermochemistry | |
| 450.6 kJ mol -1 | |
| Related compounds | |
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Related compounds |
Iron hydrides, FeH 2 CrH, CaH, MgH |
| Infobox links | |
Iron(I) hydride, systematically named iron hydride and poly(hydridoiron) is a solid inorganic compound with the chemical formula (FeH)
n
(also written()
P
or FeH). It is both thermodynamically and kinetically unstable with respect to decomposition at ambient temperature, and as such, little is known about its bulk properties.
Iron(I) hydride is the simplest polymeric iron hydride. Due to its instability, it has no practical industrial application. However, in metallurgical chemistry, iron(I) hydride is the basis for some forms of iron-hydrogen alloys.
Nomenclature
Systematic name iron hydride, a valid IUPAC name, built according to compositional nomenclature. However, as the name is compositional in nature, it does not distinguish between compounds of the same stoichiometry, such as molecular particles that exhibit different chemical properties. Systematic names poly(hydridoiron) and poly, as well as valid IUPAC names, are constructed according to the additive and electron-deficient replacement nomenclatures, respectively. They distinguish the title compound from others.
Hydridoiron
Hydridoiron, also systematically named ferrane (1), is a compound related to the chemical formula FeH (also spelled ). It is also unstable at ambient temperature with an additional tendency to autopolymerize and therefore cannot be concentrated.
Hydridoiron is the simplest molecular iron hydride. Also, it can be considered as an iron(I) hydride monomer. It has only been found in isolation under extreme conditions, as trapped in frozen noble gases, in out cold stars, or as a gas at temperatures above the boiling point of iron. It is supposed to have three dangling valence bonds, and is therefore a free radical; its formula can be written FeH 3 to emphasize this fact.
At very low temperatures (below 10 ), FeH can complex with the molecular hydrogen FeH H 2 .
Hydridoiron was first discovered in the laboratory of B. Clément and L. Åkerlind in the 1950s.
properties
Radicality and acidity
One electron of other atomic or molecular species can be connected to the iron center in hydridoiron by substitution:
RR → R
Because of this single electron capture, hydridoiron has a radical nature. Hydridoiron is a strong radical.
The electron pair of the Lewis base can enter with the iron center by bringing:
+:L →
Due to this capture of attached electron pairs, hydridoiron has a Lewis-acid character. It should be expected that iron (I) hydride has significantly decreased radical properties, but has similar properties of acids, but the reaction rate and equilibrium constant are different.
Compound
In iron(I) hydride, the atoms form a network, the individual atoms are connected to each other by covalent bonds. Since it is a polymeric solid, a single crystal sample will not undergo transitions between states such as melting and dissolution, as this would require the rearrangement of molecular bonds and therefore change its chemical identity. Colloidal crystalline samples, in which intermolecular forces are relevant, are expected to undergo transitions between states.
(I) Iron hydride adopts a double hexagonal close-packed crystal structure with space group P6 3 / MMC, also referred to as epsilon-simple iron hydride in the context of the iron-hydrogen system. Predicted to exhibit polymorphism, transition at some temperature below -173 °C (-279 °F) to a face-centered crystal structure with a Pm 3 m space group.
Electromagnetic properties
FeH is predicted to have quartet and sext ground states.
The FeH molecule has at least four low electronic energy states, caused by non-binding electrons, taking positions in different orbitals: X 4 Δ, A 6 Δ b 6 Π, and c 6 Σ + . Higher energy states are called B 4 E - , C 4 Φ, D 4 Σ + , E 4 Π and F 4 Δ. Even higher levels are labeled G 4 P and N 4 D from the quartet system, and r - Σ - , e 6 Π, F 6 Δ, and r 6 Φ. In a quartet of states, the internal quantum number J takes on the values 1/2, 3/2, 5/2, and 7/2.
FeH plays an important absorption band (the so-called wing group-Ford) in the near infrared region from the edge of the band at 989.652 nm and the absorption maximum at 991 nm. It also has lines in blue at 470 to 502.5 nm, and in green from 520 to 540 nm.
The slight isotopic shift in deuterated FED compared to PE at this wavelength indicates that the group is due to a (0,0) transition from the state, namely F 4 D-X 4 D.
Various other groups exist in each part of the spectrum due to different vibrational transitions. The (1.0) band, also due to the F 4 Δ-X 4 Δ transitions, is around 869.0 nm and the (2.0) band is around 781.8 nm.
Each group has a large number of lines. This is due to the transition between different rotational states. The lines are grouped in subranges 4 Δ 7/2 - 4 Δ 7/2 (strong) and 4 Δ 5/2 - 4 Δ 5/2, 4 Δ 3/2 - 4 Δ 3/2 and 4 Δ 1/2 - 4 ∆1/2. Numbers like 7/2 are the ohm value of the component's spin. Each of them has two branches P and R, and some of them a branch Q. Inside each there is what is called Λ splitting, which leads to lower energy lines (denoted "a") and higher energy lines (called "b" ). For each of them there are a number of spectral lines depending on J, a rotational quantum number starting at 3.5 and going up in steps of 1. How high J gets is temperature dependent. In addition, there are 12 satellite branches 4 Δ 7/2 - 4 Δ 5/2 , 4 Δ 5/2 - 4 Δ 3/2 , 4 Δ 3/2 - 4 Δ 1/2 , 4 Δ 5/2 - 4 Δ 7/2, 4 Δ 3/2 - 4 Δ 5/2 and 4 Δ 1/2 - 4 Δ 3/2 with P and R branches.
Some lines are magnetically sensitive, such as 994.813 and 995.825 nm. They are expanded by the Zeeman effect yet others in the same frequency band are insensitive to magnetic fields such as 994.911 and 995.677 nm. There are 222 lines in the spectrum of the (0-0) group.
Entry into outer space
Iron hydride is one of the few molecules found in the Sun. Lines for PV in the blue-green part of the Sun's spectrum were recorded in 1972, including many absorption lines in 1972. In addition to sunspots, umbras show the Wing-Ford group prominently.
PV strips (and other hydrides
Let's start with the composition of interstitial compounds. Let us consider this issue using the example of transition element hydrides. If, during the formation of the interstitial phase, hydrogen atoms fall only into tetrahedral voids in the metal lattice, then the limiting hydrogen content in such a compound should correspond to the formula MeH 2 (where Me is a metal whose atoms form a close packing). After all, there are twice as many tetrahedral voids in the lattice as there are atoms forming a dense packing. If, on the other hand, hydrogen atoms fall only into octahedral voids, then it follows from the same considerations that the limiting hydrogen content must correspond to the formula MeH, - there are as many octahedral voids in a dense packing as there are atoms that make up this packing.
Usually, during the formation of compounds of transition metals with hydrogen, either octahedral or tetrahedral voids are filled. Depending on the nature of the initial substances and the conditions of the process, complete or only partial filling may occur. In the latter case, the composition of the compound will deviate from the integer formula, will be indefinite, for example, MeH 1-x; MeH 2-x. Embedding connections, therefore, by their very nature must be compounds of variable composition, i.e., those whose composition, depending on the conditions for their preparation and further processing, varies within fairly wide limits.
Let us consider some typical properties of interstitial phases using the example of compounds with hydrogen. To do this, we compare the hydrides of some transition elements with the hydride of an alkali metal (lithium).
When lithium is combined with hydrogen, a substance of a certain composition LiH is formed. In terms of physical properties, it has nothing to do with the original metal. Lithium conducts electric current, has a metallic luster, plasticity, in a word, the whole complex of metallic properties. Lithium hydride does not have any of these properties. It is a colorless salt-like substance, not at all like a metal. Like other alkali and alkaline earth metal hydrides, lithium hydride is a typical ionic compound, where the lithium atom has a significant positive charge, and the hydrogen atom has the same negative charge. The density of lithium is 0.53 g / cm 3, and the density of lithium hydride is 0.82 g / cm 3 - occurs noticeable increase in density. (The same is observed in the formation of hydrides of other alkali and alkaline earth metals).
Palladium (a typical transition element) undergoes completely different transformations when interacting with hydrogen. There is a well-known demonstration experiment in which a palladium plate coated on one side with a gas-tight varnish is bent when blown with hydrogen.
This is because the density of the resulting palladium hydride decreases. Such a phenomenon can take place only if the distance between the metal atoms increases. The introduced hydrogen atoms "push" the metal atoms, changing the characteristics of the crystal lattice.
The increase in the volume of metals during the absorption of hydrogen with the formation of interstitial phases is so noticeable that the density of the metal saturated with hydrogen turns out to be significantly lower than the density of the original metal (see Table 2)
Strictly speaking, the lattice formed by the atoms of a metal usually does not remain completely unchanged after absorption of hydrogen by this metal. No matter how small the hydrogen atom, it still introduces distortions into the lattice. In this case, usually there is not just a proportional increase in the distances between atoms in the lattice, but also some change in its symmetry. Therefore, it is often said, only for simplicity, that hydrogen atoms are introduced into voids in close packing - the dense packing of metal atoms itself is still violated when hydrogen atoms are introduced.
Table 2 Changes in the density of some transition metals during the formation of interstitial phases with hydrogen.
This is far from the only difference between typical and transition metal hydrides.
During the formation of interstitial hydrides, such typical properties of metals as metallic luster and electrical conductivity are preserved. True, they can be less pronounced than in the parent metals. Thus, interstitial hydrides are much more similar to parent metals than alkali and alkaline earth metal hydrides.
Such a property as ductility changes much more strongly - hydrogen-saturated metals become brittle, it is often difficult to turn the original metals into powder, and it is much easier to do this with hydrides of the same metals.
Finally, a very important property of interstitial hydrides should be noted. When transition metals interact with hydrogen, the metal sample is not destroyed. Moreover, it retains its original shape. The same happens during the reverse process - the decomposition of hydrides (loss of hydrogen).
A natural question may arise: can the process of formation of interstitial phases be considered chemical in the full sense of the word? Perhaps the formation of aqueous solutions - a process that has much more "chemistry"?
The answer is to use chemical thermodynamics.
It is known that the formation of chemical compounds from simple substances (as well as other chemical processes) is usually accompanied by noticeable energy effects. Most often, these effects are exothermic, and the more energy is released, the stronger the resulting connection.
Thermal effects are one of the most important signs that not just a mixture of substances is taking place, but a chemical reaction is taking place. Since the internal energy of the system changes, therefore, new bonds are formed.
Let us now see what energy effects are caused by the formation of interstitial hydrides. It turns out that the spread here is quite large. In metals of secondary subgroups III, IV and V of groups of the periodic system, the formation of interstitial hydrides is accompanied by a significant release of heat, of the order of 30–50 kcal / mol (when lithium hydride is formed from simple substances, about 21 kcal / mol is released). It can be recognized that interstitial hydrides, at least of the elements of the indicated subgroups, are quite "real" chemical compounds. However, it should be noted that for many metals located in the second half of each transition row (for example, for iron, nickel, copper), the energy effects of the formation of interstitial hydrides are small. For example, for a hydride with an approximate composition of FeH 2, the thermal effect is only 0.2 kcal / mol .
The small value of DN arr of such hydrides dictates the methods of their preparation - not the direct interaction of the metal with hydrogen, but an indirect way.
Let's look at a few examples.
Nickel hydride, whose composition is close to NiH 2 , can be obtained by acting on an ethereal solution of nickel chloride with phenylmagnesium bromide in a stream of H 2:
The nickel hydride obtained as a result of this reaction is a black powder, which readily gives off hydrogen (which is generally characteristic of interstitial hydrides), and ignites when slightly heated in an oxygen atmosphere.
In the same way, hydrides of nickel's neighbors in the periodic system, cobalt and iron, can be obtained.
Another method for obtaining transition hydrides is based on the use of lithium alanate LiAlH. When the chloride of the corresponding metal reacts with LiAlH 4 in an ether solution, an alanate of this metal is formed:
MeCl 2 + LiAlH 4 > Me(AlH 4 ) 2 + LiCl(5)
For many metals, alanates are fragile compounds that decompose with increasing temperature.
Me(AlH 4 ) 2 >MeH 2 + Al + H 2 (6)
But for some metals of the secondary subgroups, a different process occurs:
Me(AlH 4 ) 2 >MeH 2 +AlH 3 (7)
In this case, instead of a mixture of hydrogen and aluminum, aluminum hydride is formed, which is ether-soluble. By washing the reaction product with ether, pure transition metal hydride can be obtained as a residue. In this way, for example, low-stable hydrides of zinc, cadmium and mercury were obtained.
It can be concluded that the preparation of hydrides of elements of secondary subgroups is based on typical methods of inorganic synthesis: exchange reactions, thermal decomposition of fragile compounds under certain conditions, etc. These methods were used to obtain hydrides of almost all transition elements, even very fragile ones. The composition of the obtained hydrides is usually close to stoichiometric: FeH 2 , CoH 2 , NiH 2 ZnH 2 , CdH 2 , HgH 2 . Apparently, the achievement of stoichiometry is facilitated by the low temperature at which these reactions are carried out.
Let us now analyze the effect of reaction conditions on the composition of the resulting interstitial hydrides. It follows directly from Le Chatelier's principle. The higher the pressure of hydrogen and the lower the temperature, the closer to the limiting value of saturation of the metal with hydrogen. In other words, each specific temperature and each pressure corresponds to a certain degree of saturation of the metal with hydrogen. And vice versa, each temperature corresponds to a certain equilibrium pressure of hydrogen over the metal surface.
This leads to one of the possible applications of transition element hydrides. Suppose, in some system, it is necessary to create a strictly defined pressure of hydrogen. A metal saturated with hydrogen is placed in such a system (titanium was used in the experiments). By heating it to a certain temperature, it is possible to create the required pressure of hydrogen gas in the system.
Any class of compounds is interesting for its chemical nature, the composition and structure of the particles of which it consists, and the nature of the bond between these particles. Chemists devote their theoretical and experimental work to this. They are no exception to the implementation phase.
There is no final point of view on the nature of interstitial hydrides yet. Often different, sometimes opposing points of view successfully explain the same facts. In other words, so far there are no unified theoretical views on the structure and properties of interstitial compounds.
Let's consider some experimental facts.
The process of absorption of hydrogen by palladium has been studied in most detail. This transition metal is characterized by the fact that the concentration of hydrogen dissolved in it at a constant temperature is proportional to the square root of the external pressure of hydrogen.
At any temperature, hydrogen to some extent dissociates into free atoms, so there is an equilibrium:
The constant of this equilibrium is:
where R H -- pressure (concentration) of atomic hydrogen.
From here (11)
It can be seen that the concentration of atomic hydrogen in the gas phase is proportional to the square root of the pressure (concentration) of molecular hydrogen. But the concentration of hydrogen in palladium is also proportional to the same value.
From this we can conclude that palladium dissolves hydrogen in the form of individual atoms.
What, then, is the nature of the bond in palladium hydride? A number of experiments have been carried out to answer this question.
It was found that when an electric current is passed through hydrogen-saturated palladium, non-metal atoms move towards the cathode. It must be assumed that the hydrogen found in the metal lattice completely or partially dissociates into protons (i.e., H + ions) and electrons.
Data on the electronic structure of palladium hydride were obtained by studying the magnetic properties. The change in the magnetic properties of the hydride depending on the amount of hydrogen included in the structure was studied. Based on the study of the magnetic properties of a substance, it is possible to estimate how many unpaired electrons are contained in the particles that make up this substance. On average, there are approximately 0.55 unpaired electrons per atom of palladium. When palladium is saturated with hydrogen, the number of unpaired electrons decreases. And in a substance with the composition PdH 0.55, unpaired electrons are practically absent.
Based on these data, we can conclude that the unpaired electrons of palladium form pairs with the unpaired electrons of hydrogen atoms.
However, the properties of interstitial hydrides (in particular, electrical and magnetic) can also be explained on the basis of the opposite hypothesis. It can be assumed that interstitial hydrides contain H - ions, which are formed due to the capture by hydrogen atoms of part of the semi-free electrons present in the metal lattice. In this case, the electrons received from the metal would also form pairs with the electrons present on the hydrogen atoms. This approach also explains the results of magnetic measurements.
It is possible that both types of ions coexist in interstitial hydrides. Metal electrons and hydrogen electrons form pairs and, therefore, a covalent bond occurs. These electron pairs can be shifted to one degree or another to one of the atoms - metal or hydrogen.
The electron pair is more strongly biased towards the metal atom in the hydrides of those metals that are less likely to donate electrons, such as palladium or nickel hydrides. But in the hydrides of scandium and uranium, apparently, the electron pair is strongly shifted towards hydrogen. Therefore, hydrides of lanthanides and actinides are in many respects similar to hydrides of alkaline earth metals. By the way, lanthanum hydride reaches the composition LaH 3 . For typical interstitial hydrides, the hydrogen content, as we now know, is not higher than that corresponding to the formulas MeH or MeH 2 .
Another experimental fact shows the difficulty in determining the nature of the bond in interstitial hydrides.
If hydrogen is removed from palladium hydride at a low temperature, then it is possible to preserve the distorted ("expanded") lattice that was found in hydrogen-saturated palladium. The magnetic properties (note this), electrical conductivity and hardness of such palladium are generally the same as those of hydride.
Hence it follows that in the formation of interstitial hydrides, the change in properties is caused not only by the presence of hydrogen in them, but also simply by a change in the interatomic distances in the lattice.
We have to admit that the question of the nature of interstitial hydrides is very complicated and far from a final solution.
Mankind has always been famous for the fact that, even without fully knowing all the aspects of any phenomena, it was able to practically use these phenomena. This fully applies to interstitial hydrides.
The formation of interstitial hydrides in some cases is deliberately used in practice, in other cases, on the contrary, they try to avoid it.
Interstitial hydrides release hydrogen relatively easily when heated, and sometimes at low temperatures. Where can this property be used? Of course in redox processes. Moreover, the hydrogen given off by interstitial hydrides is in the atomic state at some stage of the process. This is probably related to the chemical activity of interstitial hydrides.
It is known that Group VIII metals (iron, nickel, platinum) are good catalysts for reactions in which hydrogen is added to some substance. Perhaps their catalytic role is associated with the intermediate formation of unstable interstitial hydrides. Further dissociating, hydrides provide the reaction system with a certain amount of atomic hydrogen.
For example, finely dispersed platinum (the so-called platinum black) catalyzes the oxidation of hydrogen with oxygen - in its presence, this reaction proceeds at a noticeable rate even at room temperature. This property of platinum black is used in fuel cells - devices where chemical reactions are used to directly produce electrical energy, bypassing the production of heat (combustion stage). The so-called hydrogen electrode, an important tool for studying the electrochemical properties of solutions, is based on the same property of finely dispersed platinum.
The formation of interstitial hydrides is used to obtain highly pure metal powders. Metallic uranium and other actinides, as well as very pure titanium and vanadium, are ductile, and therefore it is practically impossible to prepare powders from them by grinding the metal. To deprive the metal of plasticity, it is saturated with hydrogen (this operation is called "embrittlement" of the metal). The resulting hydride is easily triturated into powder. Some metals, when saturated with hydrogen, themselves pass into a powder state (uranium). Then, when heated in a vacuum, the hydrogen is removed and pure metal powder remains.
Thermal decomposition of some hydrides (UH 3 , TiH 2) can be used to produce pure hydrogen.
The most interesting areas of application of titanium hydride. It is used for the production of foam metals (for example, aluminum foam). To do this, the hydride is introduced into molten aluminum. At high temperatures, it decomposes, and the resulting hydrogen bubbles foam liquid aluminum.
Titanium hydride can be used as a reducing agent for some metal oxides. It can serve as a solder for joining metal parts, and as a substance that accelerates the process of sintering of metal particles in powder metallurgy. The last two cases also use the reducing properties of the hydride. A layer of oxides usually forms on the surface of metal particles and metal parts. It prevents adhesion of adjacent metal sections. Titanium hydride, when heated, reduces these oxides, thereby cleaning the metal surface.
Titanium hydride is used to produce some special alloys. If it is decomposed on the surface of a copper product, a thin layer of copper-titanium alloy is formed. This layer gives the surface of the product special mechanical properties. Thus, it is possible to combine several important properties in one product (electrical conductivity, strength, hardness, abrasion resistance, etc.).
Finally, titanium hydride is a very effective protection against neutrons, gamma rays and other hard radiation.
Sometimes, on the contrary, one has to struggle with the formation of interstitial hydrides. In metallurgy, chemical, oil and other industries, hydrogen or its compounds are under pressure and at high temperatures. Under such conditions, hydrogen can diffuse to a noticeable extent through the heated metal, simply "leave" the equipment. In addition (and this is perhaps the most important thing!), due to the formation of interstitial hydrides, the strength of metal equipment can be greatly reduced. And this is already fraught with a serious danger when working with high pressures.
The usual methods of storing (in cylinders) compressed or liquefied hydrogen is a rather dangerous occupation. In addition, hydrogen penetrates very actively through most metals and alloys, which makes shut-off and transport valves very expensive.
The property of hydrogen to dissolve in metals has been known since the 19th century, but only now the prospects for the use of metal hydrides and intermetallic compounds as compact hydrogen storage facilities have become visible.
Hydride types
Hydrides are classified into three types (some hydrides may have multiple bond properties, such as being metal-covalent): metallic, ionic, and covalent.
Ionic hydrides - as a rule, they are created at high pressures (~100 atm.) and at temperatures above 100°C. Typical representatives are alkali metal hydrides. An interesting feature of ionic hydrides is a greater degree of atomic density than in the original substance.
covalent hydrides- practically do not find application due to the low stability and high toxicity of the metals and intermetallic compounds used. A typical representative is beryllium hydride, obtained by the “wet chemistry” method by the reaction of dimethylberyllium with lithium aluminum hydride in a solution of diethyl ether.
Metal hydrides- can be considered as alloys of metallic hydrogen, these compounds are characterized by high electrical conductivity, like the parent metals. Metal hydrides form almost all transition metals. Depending on the types of bonds, metal hydrides can be covalent (for example, magnesium hydride) or ionic. Almost all metal hydrides require high temperatures for dehydrogenation (hydrogen release reaction).
Typical metal hydrides
- Lead hydride - PbH4 - a binary inorganic chemical compound of lead with hydrogen. Very active, in the presence of oxygen (in air) ignites spontaneously.
- Zinc hydroxide - Zn (OH) 2 - amphoteric hydroxide. It is widely used as a reagent in many chemical industries.
- Palladium hydride is a metal in which hydrogen is located between the palladium atoms.
- Nickel hydride - NiH - is often used with additives of lanthanum LaNi5 for battery electrodes.
Metal hydrides can form the following metals:
Ni, Fe, Ni, Co, Cu, Pd, Pt, Rh, Pd-Pt, Pd-Rh, Mo-Fe, Ag-Cu, Au-Cu, Cu-Ni, Cu-Pt, Cu-Sn.
Metals-record holders in terms of the volume of stored hydrogen
The best metal for storing hydrogen is palladium (Pd). One volume of palladium can contain almost 850 volumes of hydrogen. But the viability of such a repository raises strong doubts due to the high cost of this platinum group metal.
On the contrary, some metals (for example copper Cu) dissolve only 0.6 volume of hydrogen per volume of copper.
Magnesium hydride (MgH2) can store up to 7.6% mass fractions of hydrogen in the crystal lattice. Despite the tempting values and low specific gravity of such systems, an obvious obstacle is the high temperatures of the direct and reverse charge-discharge reactions and high endothermic losses during the dehydrogenation of the compound (about a third of the stored hydrogen energy).
Crystal structure of the β-phase of MgH2 hydride (figure)
Accumulation of hydrogen in metals
The reaction of hydrogen absorption by metals and intermetallics occurs at a higher pressure than its release. This is determined by the residual plastic deformations of the crystal lattice during the transition from a saturated α-solution (original substance) to a β-hydride (substances with stored hydrogen).
Metals that do not dissolve hydrogen
The following metals do not absorb hydrogen:
Ag, Au, Cd, Pb, Sn, Zn
Some of them are used as valves for storing compressed and liquefied hydrogen.
Low-temperature metal hydrides are among the most promising hydrides. They have low losses during dehydrogenation, high rates of charge-discharge cycles, are almost completely safe and have low toxicity. The limitation is the relatively low specific density of hydrogen storage. The theoretical maximum is the storage of 3%, but in reality 1-2% of the mass fraction of hydrogen.
The use of powdered metal hydrides imposes restrictions on the rate of "charge-discharge" cycles due to the low thermal conductivity of powders and requires a special approach to the design of containers for their storage. It is typical to introduce areas into the storage vessel to facilitate heat transfer and to produce thin and flat cylinders. Some increase in the rate of discharge-charge cycles can be achieved by introducing an inert binder into the metal hydride, which has a high thermal conductivity and a high threshold of inertness to hydrogen and the base substance.
Intermetallic hydrides
In addition to metals, storage of hydrogen in the so-called "intermetallic compounds" is promising. Such hydrogen storage facilities are widely used in household metal hydride batteries. The advantage of such systems lies in the relatively low cost of reagents and low environmental impact. At the moment, metal hydride batteries are almost universally replaced by lithium energy storage systems. The maximum stored energy of industrial designs in nickel-metal hydride batteries (Ni-MH) is 75 Wh / kg.
An important property of some intermetallic compounds is their high resistance to impurities contained in hydrogen. This property allows such compounds to be used in polluted environments and in the presence of moisture. Multiple "charge-discharge" cycles in the presence of impurities and water in hydrogen do not poison the working substance, but reduce the capacity of subsequent cycles. The decrease in useful capacity occurs due to contamination of the base substance with metal oxides.
Separation of intermetallic hydrides
Intermetallic hydrides are divided into high-temperature (dehydrogenating at room temperature) and high-temperature (more than 100°C). The pressure at which decomposition of the hydride phase occurs) is usually not more than 1 atm.
In real practice, complex intermetallic hydrides are used, consisting of three or more elements.
Typical intermetallic hydrides
Nickel lanthanum hydride, LaNi5, is a hydride in which one unit of LaNi5 contains more than 6 H atoms. Hydrogen desorption from nickel lanthanum is possible at room temperatures. However, the elements included in this intermetallic compound are also very expensive.
A unit volume of lanthanum-nickel contains one and a half times more hydrogen than liquid H2.
Features of intermetallic-hydrogen systems:
- high hydrogen content in the hydride (wt. %);
- exo (endo)-thermicity of the reaction of absorption (desorption) of hydrogen isotopes;
- change in the volume of the metal matrix in the process of absorption - desorption of hydrogen;
- reversible and selective absorption of hydrogen.
Areas of practical application of intermetallic hydrides:
- stationary storages of hydrogen;
- storage mobility and transportation of hydrogen;
- compressors;
- separation (purification) of hydrogen;
- heat pumps and air conditioners.
Application examples of metal-hydrogen systems:
- fine purification of hydrogen, all kinds of hydrogen filters;
- reagents for powder metallurgy;
- moderators and reflectors in nuclear fission systems (nuclear reactors);
- separation of isotopes;
- thermonuclear reactors;
- water dissociation installations (electrolyzers, vortex chambers for producing gaseous hydrogen);
- electrodes for batteries based on tungsten-hydrogen systems;
- metal hydride batteries;
- air conditioners (heat pumps);
- converters for power plants (nuclear reactors, thermal power plants);
- transportation of hydrogen.
The article mentions metals:
