Vanadium hydride. Vanadium(I) Hydride Composition and Molar Mass. Thermochemical values ​​for VH(g)

The molecular constants used to calculate the VH thermodynamic functions are presented.

The symmetry of the ground state VH, vibrational and rotational constants are not experimentally determined. Quantum mechanical calculations of the molecule [ 74SCO/RIC, 75HEN/DAS, 81DAS, 83WAL/BAU, 86CHO/LAN, 96FUJ/IWA, 97BAR/ADA, 2004KOS/ISH, 2006FUR/PER, 2008GOE/MAS ] give a ground state symmetry of 5 Δ , equilibrium internuclear distance in the range of 1.677 – 1.79 Å, values ​​of the vibrational constant in the range of 1550 – 1659 cm -1 .

To calculate the thermodynamic functions, the averaged values ​​w e and r e according to the results of quantum mechanical calculations. Constants B e , w e x e , D e and a 1 are calculated further according to formulas 1.38, 1.67, 1.68 and 1.69, respectively. In table. V.D1 ground state constants are given against the lower Ω-component X 5 ∆ 0 . The energies of the spin-orbit components X 5 Δ calculated in [2004KOS/ISH], in Table. V.D1 the average values ​​are given for two calculation options [2004KOS/ISH].

Excited states VH are calculated in [74SCO/RIC, 75HEN/DAS, 81DAS, 83WAL/BAU, 96FUJ/IWA, 2004KOS/ISH, 2008GOE/MAS]. The resulting energies of quintet states have a noticeable spread: 5 Π (753 - 2260 cm -1), 5 Σ - (1694 - 4762 cm -1), 5 Φ (2629 - 5816 cm -1). In table. V.D1 shows the rounded average values ​​of the energies of these three states. The energies of low-lying triplet states are calculated in [75HEN/DAS, 2004KOS/ISH, 2008GOE/MAS]. The results of [75HEN/DAS, 2004KOS/ISH] are close to each other, while the calculation of [2008GOE/MAS] gives a much lower energy for the lower triplet state. In table. V.D1, the energies of triplet states are taken based on the graph of potential curves [2004KOS/ISH].

The following were included in the calculation of thermodynamic functions: a) ground state X 5 ∆ 0 ; b) other components of the spin-orbit splitting X 5 Δ as separate Ω-states; c) low-lying quintet and triplet states obtained in quantum mechanical calculations; d) synthetic (estimated) states that combine other excited states of the molecule with an estimated energy of up to 40,000 cm -1 .

The statistical weights of the synthetic states were estimated using the V + H - ionic model. The lower quintet states of the molecule correspond to the splitting components of the ground term of the ion V + 5 D(3d 4) (5 Δ, 5 Π, 5 Σ +) and the first excited term 5 F(3d 3 4s) (5 Φ, 5 Δ, 5 Π, 5 Σ –), however, the relative positions of terms of different configurations can change in the ligand field. Quantum-mechanical calculations of the molecule yielded quintet low-lying states 5 Φ, 5 Δ, 5 Π, 5 Σ – , of which 5 Φ and 5 Σ – can definitely be assigned to the 5 F(3d 3 4s) term. The energy difference 5 Φ and 5 Σ characterizes the splitting of the term 5 F(3d 3 4s) in the ligand field. The states 5 Δ and 5 Π do not fall into the interval between 5 Φ and 5 Σ due to repulsion with the second pair of states 5 Δ and 5 Π, which belongs to the term 5 D(3d 4). The unperturbed component of the splitting of the term 5 D(3d 4) is the state 5 Σ + , whose energy is estimated at 5000 cm -1 (the first synthetic state in Table V.D1). The second pair of states 5 Δ and 5 Π is included in (forms) the synthetic state 10000 cm -1 . The low-lying triplet states 3 Φ, 3 Δ, 3 Π, 3 Σ obtained in quantum mechanical calculations can be interpreted as splitting components of the 3 F(3d 3 4s) term. Other terms of the 3d 4 and 3d 3 4s configurations give higher lying states, their statistical weights are distributed over the synthetic states in accordance with the energy of the terms in the [ 71MOO ] ion plus a correction for the energy of the lower term of the configuration in the molecule. Correction for 5 D(3d 4) is estimated at 5500 cm -1 (~ energy 5 Σ + plus half of the assumed value of term splitting) and for 5 F(3d 3 4s) at 4000 cm -1 (average energy of states 5 Σ, 5 Σ -). Synthetic states of 20000 cm -1 and above also include the statistical weights of the terms of the 3d 3 4p configuration. The lower states of this configuration are placed in the region of 21000 cm -1 in accordance with the hypothetical interpretation of the VH absorption spectrum observed in [73SMI].

The thermodynamic functions VH(r) were calculated using equations (1.3) - (1.6) , (1.9) , (1.10) , (1.93) - (1.95) . Values Q ext and its derivatives were calculated by equations (1.90) - (1.92) taking into account nineteen excited states under the assumption that Q no.vr ( i) = (p i /p X)Q no.vr ( X) . Vibrational-rotational partition function of the state X 5 Δ 0 and its derivatives were calculated by equations (1.70) - (1.75) by direct summation over energy levels. The calculations took into account all energy levels with values J< J max,v , where J max,v was found from conditions (1.81) . Vibrational-rotational levels of state X 5 Δ 0 were calculated by equations (1.65) , the values ​​of the coefficients Y kl in these equations were calculated using relations (1.66) for the isotopic modification corresponding to the natural mixture of vanadium and hydrogen isotopes from the 51 V 1 H molecular constants given in Table V.D1. Coefficient values Y kl , as well as the quantities v max and J lim are given in Table V.D2.

At room temperature, the following values ​​are obtained:

C p o (298.15 K) = 32.256 ± 3.02 J × K -1 × mol -1

S o (298.15 K) = 215.030 ± 1.67 J × K -1 × mol -1

H o (298.15 K) - H o (0) = 9.832 ± 0.346 kJ× mol -1

The main contribution to the error of the calculated thermodynamic functions VH(r) over the entire temperature range comes from the uncertainty in the energies of low-lying electronic states. Within the error Φº( T) a comparable contribution is also made by the inaccuracy of the rotational and vibrational constants. At 3000 and 6000 K, a significant contribution to the error of the functions (in C p o already at 1000 K) introduces the calculation method. Errors in the values ​​of Φº( T) at T= 298.15, 1000, 3000 and 6000 K are estimated at 0.7, 1.6, 1.2 and 1.2 J×K -1 × mol -1 , respectively.

Other calculations of the thermodynamic functions VH(r) have not been found in the literature.

Thermochemical values ​​for VH(g).

The equilibrium constant of the reaction VH(g)=V(g)+H(g) is calculated from the accepted value of the dissociation energy

D° 0 (VH) \u003d 182 ± 23 kJ × mol -1 \u003d 15200 ± 1900 cm -1.

(54,. (57) METHODS OF VANADIUM, include allic vanadium irradiated prn demetal o.t., that, for the purpose of 1 p. use of temperature and sap of the process, pressure 5-30 attenuated at razlotermetalicheskoy AHF, 8 or h chemistry (proto-judiciary department of the USSR inventions and related (71) Institute of New Chemical Problems of the Academy of Sciences of the USSR (56) 1, Mikheeva V.I. Transition metal hydrides. Academy of Sciences of the USSR, I. 1946, p97-99.2, "1. Aveg, Spev. 1961, 83R 17, p. priming with hydrogen, semi"zhenin of hydrides of incompounds composition 1 General Order 10312/24 Circulation 471 Subscription VNIIPI of the State Committee of the USSR for Inventions and Discoveries 113035, Moscow, Zh, Raushskaya nab., d, 4/5 The invention relates to methods for producing vanadium dihydride, which can be used in powder metallurgy, as well as a source of hydrogen and a catalyst for the hydrogenation of organic substances. as well as a low (1.2-2 wt.b) hydrogen content in it. There is also a method for obtaining vanadium dihydride by treating vanadium hydride of the composition CN O with hydrogen under a pressure of 70 atm at room temperature for 6 hours. , about 5 G 2 E The disadvantage of the method is that the composition of the obtained hydride does not reach the maximum. metal vanadium hydrogen formed during the thermal decomposition of titanium hydride. Hydrogen treatment is carried out at first 30 at room temperature to the composition; corresponding to vanadium monohydride, after which hydrogen treatment is carried out at a temperature of -70 to -20 C. Hydrogen pressure is 1 atm. Process duration is 8-10 days. ) temperature and reduction of its duration. The goal is achieved by the fact that the treatment of metallic vanadium is carried out at a pressure of 5-30 atm with hydrogen obtained by the decomposition of intermetallic compounds of the composition LaI 1 Hbz or T 1 Ren 2, When the hydrides of intermetallic compounds of the composition 50 LaB 1 Hb B or T 1 GeH hydrogen is released with a purity of 99.9999. Hydrogen of this purity is able to easily penetrate through the oxide film located on the surface of the metal into the depth of the sample and interact with the unoxidized metal. It has a large diffusion coefficient and high mobility. This allows the hydrogenation process to be carried out at a high rate and at a sufficient depth without the use of low temperatures necessary to reduce the dissociation pressure of the resulting vanadium dihydride. When the hydrogen pressure drops below 5 atm, the hydrogenation time increases. An increase in pressure above 30 atm does not affect the speed of the process, but leads to its complication. the sample is pumped out for 0.5 h at 250 C. After cooling to 20 C, the autoclave is filled with hydrogen from a canister with a hydride of the composition La 1 R 1 H n to a pressure of 10 atm. The reaction starts immediately and lasts 1 hour. The end of the reaction is set when the pressure drop in the autoclave ceases. As a result of hydrogenation, vanadium dihydride of the composition ChNo is obtained, which is established on the basis of data from x-ray phase, gas volumetric and chemical analyzes, EXAMPLE 2. Similarly to example 1, from 4 g of vanadium powder at 20 ° C and under a hydrogen pressure of 5 atm for 1.5 h get vanadium hydride composition ChNdr. PRI me R 3. Similarly to example 1 from 8 g of vanadium in the form of a piece at 20 C and under a hydrogen pressure of 30 atm for 2 h receive vanadium hydride composition ChN. Thus, the invention makes it possible to simplify the process by eliminating the need to use low (minus) temperatures and reduce its duration from 8-10 days to 1-2 hours.

Application

3421538, 13.04.1982

INSTITUTE OF NEW CHEMICAL PROBLEMS AS USSR

SEMENENKO KIRILL NIKOLAEVICH, FOKINA EVELINA ERNESTOVNA, FOKIN VALENTIN NAZAROVICH, TROITSKAYA STELLA LEONIDOVNA, BURNASHEVA VENIANNA VENEDIKTOVNA, VERBETSKY VIKTOR NIKOLAEVICH, MITROKHIN SERGEY VLADILENOVICH

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Method for obtaining vanadium dihydride

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Chemical formula

Molar mass of VH, Vanadium(I) Hydride 51.94944 g/mol

Mass fractions of elements in the compound

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Molar mass calculator

mole

All substances are made up of atoms and molecules. In chemistry, it is important to accurately measure the mass of substances entering into a reaction and resulting from it. By definition, the mole is the SI unit for the amount of a substance. One mole contains exactly 6.02214076×10²³ elementary particles. This value is numerically equal to the Avogadro constant N A when expressed in units of moles⁻¹ and is called Avogadro's number. Amount of substance (symbol n) of a system is a measure of the number of structural elements. A structural element can be an atom, a molecule, an ion, an electron, or any particle or group of particles.

Avogadro's constant N A = 6.02214076×10²³ mol⁻¹. Avogadro's number is 6.02214076×10²³.

In other words, a mole is the amount of a substance equal in mass to the sum of the atomic masses of the atoms and molecules of the substance, multiplied by the Avogadro number. The mole is one of the seven basic units of the SI system and is denoted by the mole. Since the name of the unit and its symbol are the same, it should be noted that the symbol is not declined, unlike the name of the unit, which can be declined according to the usual rules of the Russian language. One mole of pure carbon-12 equals exactly 12 grams.

Molar mass

Molar mass is a physical property of a substance, defined as the ratio of the mass of that substance to the amount of the substance in moles. In other words, it is the mass of one mole of a substance. In the SI system, the unit of molar mass is kilogram/mol (kg/mol). However, chemists are accustomed to using the more convenient unit g/mol.

molar mass = g/mol

Molar mass of elements and compounds

Compounds are substances made up of different atoms that are chemically bonded to each other. For example, the following substances, which can be found in the kitchen of any housewife, are chemical compounds:

  • salt (sodium chloride) NaCl
  • sugar (sucrose) C₁₂H₂₂O₁₁
  • vinegar (acetic acid solution) CH₃COOH

The molar mass of chemical elements in grams per mole is numerically the same as the mass of the element's atoms expressed in atomic mass units (or daltons). The molar mass of compounds is equal to the sum of the molar masses of the elements that make up the compound, taking into account the number of atoms in the compound. For example, the molar mass of water (H₂O) is approximately 1 × 2 + 16 = 18 g/mol.

Molecular mass

Molecular weight (the old name is molecular weight) is the mass of a molecule, calculated as the sum of the masses of each atom that makes up the molecule, multiplied by the number of atoms in this molecule. The molecular weight is dimensionless a physical quantity numerically equal to the molar mass. That is, the molecular weight differs from the molar mass in dimension. Although the molecular mass is a dimensionless quantity, it still has a value called the atomic mass unit (amu) or dalton (Da), and is approximately equal to the mass of one proton or neutron. The atomic mass unit is also numerically equal to 1 g/mol.

Molar mass calculation

The molar mass is calculated as follows:

  • determine the atomic masses of the elements according to the periodic table;
  • determine the number of atoms of each element in the compound formula;
  • determine the molar mass by adding the atomic masses of the elements included in the compound, multiplied by their number.

For example, let's calculate the molar mass of acetic acid

It consists of:

  • two carbon atoms
  • four hydrogen atoms
  • two oxygen atoms
  • carbon C = 2 × 12.0107 g/mol = 24.0214 g/mol
  • hydrogen H = 4 × 1.00794 g/mol = 4.03176 g/mol
  • oxygen O = 2 × 15.9994 g/mol = 31.9988 g/mol
  • molar mass = 24.0214 + 4.03176 + 31.9988 = 60.05196 g/mol

Our calculator does just that. You can enter the formula of acetic acid into it and check what happens.

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