The molar volume of any gas is equal. Finding the molar volume of gases

Names of acids are formed from the Russian name of the central atom of the acid with the addition of suffixes and endings. If the oxidation state of the central atom of the acid corresponds to the group number of the Periodic System, then the name is formed using the simplest adjective from the name of the element: H 2 SO 4 - sulfuric acid, HMnO 4 – permanganic acid. If acid-forming elements have two oxidation states, then the intermediate oxidation state is designated by the suffix –ist-: H 2 SO 3 – sulfurous acid, HNO 2 – nitrous acid. Various suffixes are used for the names of halogen acids that have many oxidation states: typical examples - HClO 4 - chlorine n acid, HClO 3 – chlorine novat acid, HClO 2 – chlorine ist acid, HClO – chlorine novatist ic acid (oxygen-free acid HCl is called hydrochloric acid - usually hydrochloric acid). Acids can differ in the number of water molecules that hydrate the oxide. Acids containing greatest number hydrogen atoms are called ortho acids: H 4 SiO 4 is orthosilicic acid, H 3 PO 4 is orthophosphoric acid. Acids containing 1 or 2 hydrogen atoms are called metaacids: H 2 SiO 3 - metasilicic acid, HPO 3 - metaphosphoric acid. Acids containing two central atoms are called di acids: H 2 S 2 O 7 – disulfuric acid, H 4 P 2 O 7 – diphosphoric acid.

The names of complex compounds are formed in the same way as names of salts, but the complex cation or anion is given a systematic name, that is, it is read from right to left: K 3 - potassium hexafluoroferrate(III), SO 4 - tetraammine copper(II) sulfate.

Names of oxides are formed using the word “oxide” and the genitive case of the Russian name of the central atom of the oxide, indicating, if necessary, the oxidation state of the element: Al 2 O 3 - aluminum oxide, Fe 2 O 3 - iron (III) oxide.

Names of bases are formed using the word "hydroxide" and genitive case Russian name of the central hydroxide atom indicating, if necessary, the oxidation state of the element: Al(OH) 3 – aluminum hydroxide, Fe(OH) 3 – iron(III) hydroxide.

Names of compounds with hydrogen are formed depending on the acid-base properties of these compounds. For gaseous acid-forming compounds with hydrogen, the following names are used: H 2 S – sulfane (hydrogen sulfide), H 2 Se – selan (hydrogen selenide), HI – hydrogen iodide; their solutions in water are called hydrogen sulfide, hydroselenic and hydroiodic acids, respectively. For some compounds with hydrogen, special names are used: NH 3 - ammonia, N 2 H 4 - hydrazine, PH 3 - phosphine. Compounds with hydrogen having an oxidation state of –1 are called hydrides: NaH is sodium hydride, CaH 2 is calcium hydride.

Names of salts are formed from the Latin name of the central atom of the acidic residue with the addition of prefixes and suffixes. The names of binary (two-element) salts are formed using the suffix - eid: NaCl – sodium chloride, Na 2 S – sodium sulfide. If the central atom of an oxygen-containing acidic residue has two positive oxidation states, then the highest oxidation state is denoted by the suffix – at: Na 2 SO 4 – sulf at sodium, KNO 3 – nitr at potassium, and the lowest oxidation state is the suffix - it: Na 2 SO 3 – sulf it sodium, KNO 2 – nitr it potassium To name oxygen-containing halogen salts, prefixes and suffixes are used: KClO 4 – lane chlorine at potassium, Mg(ClO 3) 2 – chlorine at magnesium, KClO 2 – chlorine it potassium, KClO – hypo chlorine it potassium

Covalent saturationsconnectionto her– manifests itself in the fact that in compounds of s- and p-elements there are no unpaired electrons, that is, all unpaired electrons of atoms form bonding electron pairs (exceptions are NO, NO 2, ClO 2 and ClO 3).

Lone electron pairs (LEP) are electrons that occupy atomic orbitals in pairs. The presence of NEP determines the ability of anions or molecules to form donor-acceptor bonds as donors of electron pairs.

Unpaired electrons are electrons of an atom, contained one in an orbital. For s- and p-elements, the number of unpaired electrons determines how many bonding electron pairs a given atom can form with other atoms through the exchange mechanism. The valence bond method assumes that the number of unpaired electrons can be increased by lone electron pairs if there are vacant orbitals within the valence electron level. In most compounds of s- and p-elements there are no unpaired electrons, since all unpaired electrons of atoms form bonds. However, molecules with unpaired electrons exist, for example, NO, NO 2, they have increased reactivity and tend to form dimers like N 2 O 4 due to unpaired electrons.

Normal concentration – this is the number of moles equivalents in 1 liter of solution.

Normal conditions - temperature 273K (0 o C), pressure 101.3 kPa (1 atm).

Exchange and donor-acceptor mechanisms of chemical bond formation. The formation of covalent bonds between atoms can occur in two ways. If the formation of a bonding electron pair occurs due to the unpaired electrons of both bonded atoms, then this method of formation of a bonding electron pair is called an exchange mechanism - the atoms exchange electrons, and the bonding electrons belong to both bonded atoms. If the bonding electron pair is formed due to the lone electron pair of one atom and the vacant orbital of another atom, then such formation of the bonding electron pair is a donor-acceptor mechanism (see. valence bond method).

Reversible ionic reactions – these are reactions in which products are formed that are capable of forming starting substances (if we keep in mind the written equation, then about reversible reactions we can say that they can proceed in one direction or another with the formation of weak electrolytes or poorly soluble compounds). Reversible ionic reactions are often characterized by incomplete conversion; since during a reversible ionic reaction, molecules or ions are formed that cause a shift towards the initial reaction products, that is, they seem to “slow down” the reaction. Reversible ionic reactions are described using the ⇄ sign, and irreversible ones - the → sign. An example of a reversible ionic reaction is the reaction H 2 S + Fe 2+ ⇄ FeS + 2H +, and an example of an irreversible one is S 2- + Fe 2+ → FeS.

Oxidizing agents – substances in which, during redox reactions, the oxidation states of some elements decrease.

Redox duality – the ability of substances to act in redox reactions as an oxidizing or reducing agent depending on the partner (for example, H 2 O 2, NaNO 2).

Redox reactions(OVR) – These are chemical reactions during which the oxidation states of the elements of the reacting substances change.

Redox potential – a value characterizing the redox ability (strength) of both the oxidizing agent and the reducing agent that make up the corresponding half-reaction. Thus, the redox potential of the Cl 2 /Cl - pair, equal to 1.36 V, characterizes molecular chlorine as an oxidizing agent and chloride ion as a reducing agent.

Oxides – compounds of elements with oxygen in which oxygen has an oxidation state of –2.

Orientation interactions– intermolecular interactions of polar molecules.

Osmosis – the phenomenon of transfer of solvent molecules on a semi-permeable (permeable only to solvent) membrane towards a lower solvent concentration.

Osmotic pressure – physicochemical property of solutions due to the ability of membranes to pass only solvent molecules. Osmotic pressure from a less concentrated solution equalizes the rate of penetration of solvent molecules into both sides of the membrane. The osmotic pressure of a solution is equal to the pressure of a gas in which the concentration of molecules is the same as the concentration of particles in the solution.

Arrhenius bases – substances that split off hydroxide ions during electrolytic dissociation.

Bronsted bases - compounds (molecules or ions of the S 2-, HS - type) that can attach hydrogen ions.

Reasons according to Lewis (Lewis bases) – compounds (molecules or ions) with lone pairs of electrons capable of forming donor-acceptor bonds. The most common Lewis base is water molecules, which have strong donor properties.

P1V1=P2V2, or, which is the same, PV=const (Boyle-Mariotte law). At constant pressure, the ratio of volume to temperature remains constant: V/T=const (Gay-Lussac's law). If we fix the volume, then P/T=const (Charles’ law). Combining these three laws gives a universal law which states that PV/T=const. This equation was established by the French physicist B. Clapeyron in 1834.

The value of the constant is determined only by the amount of substance gas. DI. Mendeleev derived an equation for one mole in 1874. So it is the value of the universal constant: R=8.314 J/(mol∙K). So PV=RT. In the case of an arbitrary quantity gasνPV=νRT. The amount of a substance itself can be found from mass to molar mass: ν=m/M.

Molar mass is numerically equal to relative molecular mass. The latter can be found from the periodic table; it is indicated in the cell of the element, as a rule, . The molecular weight is equal to the sum of the molecular weights of its constituent elements. In the case of atoms of different valences, an index is required. On at mer, M(N2O)=14∙2+16=28+16=44 g/mol.

Normal conditions for gases at It is commonly assumed that P0 = 1 atm = 101.325 kPa, temperature T0 = 273.15 K = 0°C. Now you can find the volume of one mole gas at normal conditions: Vm=RT/P0=8.314∙273.15/101.325=22.413 l/mol. This table value is the molar volume.

Under normal conditions conditions quantity relative to volume gas to molar volume: ν=V/Vm. For arbitrary conditions you need to use the Mendeleev-Clapeyron equation directly: ν=PV/RT.

Thus, to find the volume gas at normal conditions, you need the amount of substance (number of moles) of this gas multiply by the molar volume equal to 22.4 l/mol. Using the reverse operation, you can find the amount of a substance from a given volume.

To find the volume of one mole of a substance in solid or liquid state, find it molar mass and divide by density. One mole of any gas in normal conditions has a volume of 22.4 liters. If conditions change, calculate the volume of one mole using the Clapeyron-Mendeleev equation.

You will need

• Periodic table of Mendeleev, table of density of substances, pressure gauge and thermometer.

Instructions

Determining the volume of one mole or solid
Define chemical formula solid or liquid that is being studied. Then, using periodic table Find Mendeleev atomic masses elements that are included in the formula. If one is included in the formula more than once, multiply its atomic mass by that number. Add up the atomic masses and get the molecular mass of which it is composed solid or liquid. It will be numerically equal to the molar mass measured in grams per mole.

Using the table of substance densities, find this value for the material of the body or liquid being studied. After this, divide the molar mass by the density of the substance, measured in g/cm³ V=M/ρ. The result is the volume of one mole in cm³. If the substance remains unknown, it will be impossible to determine the volume of one mole of it.

The volume of 1 mole of a substance is called the Molar volume. Molar mass of 1 mole of water = 18 g/mol 18 g of water occupy a volume of 18 ml. This means the molar volume of water is 18 ml. 18 g of water occupy a volume equal to 18 ml, because the density of water is 1 g/ml CONCLUSION: Molar volume depends on the density of the substance (for liquids and solids).

1 mole of any gas under normal conditions occupies the same volume equal to 22.4 liters. Normal conditions and their designations no. (0 0 C and 760 mmHg; 1 atm.; 101.3 kPa). The volume of a gas with 1 mole of substance is called molar volume and is denoted by – V m

Solving problems Problem 1 Given: V(NH 3) n.s. = 33.6 m 3 Find: m - ? Solution: 1. Calculate the molar mass of ammonia: M(NH 3) = = 17 kg/kmol

CONCLUSIONS 1. The volume of 1 mole of a substance is called the molar volume V m 2. For liquid and solid substances, the molar volume depends on their density 3. V m = 22.4 l/mol 4. Normal conditions (n.s.): and pressure 760 mmHg, or 101.3 kPa 5. Molar volume gaseous substances expressed in l/mol, ml/mmol,

The volume of a gram-molecule of a gas, like the mass of a gram-molecule, is a derived unit of measurement and is expressed as the ratio of volume units - liters or milliliters to a mole. Therefore, the dimension of gram-molecular volume is equal to l/mol or ml/mol. Since the volume of a gas depends on temperature and pressure, the gram-molecular volume of a gas varies depending on the conditions, but since the gram-molecules of all substances contain the same number of molecules, the gram-molecules of all substances under the same conditions occupy the same volume. Under normal conditions. = 22.4 l/mol, or 22,400 ml/mol. Conversion of the gram-molecular volume of a gas under normal conditions to the volume under given conditions of production. is calculated according to the equation: J-t-tr from which it follows that where Vo is the gram-molecular volume of the gas under normal conditions, Umol is the desired gram-molecular volume of the gas. Example. Calculate the gram-molecular volume of the gas at 720 mm Hg. Art. and 87°C. Solution. The most important calculations related to the gram-molecular volume of a gas a) Converting the volume of gas to the number of moles and the number of moles to the volume of gas. Example 1. Calculate how many moles are contained in 500 liters of gas under normal conditions. Solution. Example 2. Calculate the volume of 3 mol of gas at 27*C 780 mm Hg. Art. Solution. We calculate the gram-molecular volume of the gas under the specified conditions: V - ™ ** RP st. - 22.A l/mol. 300 deg = 94 p. --273 vrad 780 mm Hg."ap.--24"° Calculate the volume of 3 moles GRAM MOLECULAR VOLUME OF GAS V = 24.0 l/mol 3 moles = 72 l b) Converting the mass of gas to its volume and volume of gas by its mass. In the first case, first calculate the number of moles of gas from its mass, and then the volume of gas from the found number of moles. In the second case, first calculate the number of moles of gas from its volume, and then, from the found number of moles, calculate the mass of the gas. Example 1, Calculate how much volume (at zero) 5.5 g of carbon dioxide CO* will occupy. Solution. |icoe ■= 44 g/mol V = 22.4 l/mol 0.125 mol 2.80 l Example 2. Calculate the mass of 800 ml (at zero) of carbon monoxide CO. Solution. |*co => 28 g/mol m « 28 g/lnm 0.036 did* =» 1.000 g If the mass of a gas is expressed not in grams, but in kilograms or tons, and its volume is expressed not in liters or milliliters, but in cubic meters , then a two-fold approach to these calculations is possible: either break up higher measures into lower ones, or calculate ae with moles, and with kilogram-molecules or tonne-molecules, using the following ratios: under normal conditions 1 kilogram-molecule-22,400 l/kmol , 1 ton molecule - 22,400 m*/tmol. Dimensions: kilogram-molecule - kg/kmol, ton-molecule - t/tmol. Example 1. Calculate the volume of 8.2 tons of oxygen. Solution. 1 ton-molecule Oa » 32 t/tmol. We find the number of tonne oxygen molecules contained in 8.2 tons of oxygen: 32 t/tmol ** 0.1 We calculate the volume of oxygen: Uo, = 22,400 m*/tmol 0.1 t/mol = 2240 l" Example 2. Calculate the mass of 1000 -k* ammonia (at standard conditions). Solution. We calculate the number of ton-molecules in the specified amount of ammonia: "-stag5JT-0.045 t/mol We calculate the mass of ammonia: 1 ton-molecule NH, 17 t/mol tyv, = 17 t/mol 0.045 t/mol * 0.765 t General principle of calculations, relating to gas mixtures is that calculations relating to individual components are carried out separately, and then the results are summed up. Example 1. Calculate how much volume it will occupy under normal conditions. gas mixture, consisting of 140 g of nitrogen and 30 g of hydrogen. Solution. We calculate the number of moles of nitrogen and hydrogen contained in the mixture (No. "= 28 e/mol; cn, = 2 g/mol): 140 £ 30 in 28 g/mol W Total 20 mol. GRAM MOLECULAR VOLUME OF GAS Calculate the volume of the mixture: Contained in 22"4 AlnoAb 20 mol « 448 l Example 2. Calculate the mass of 114 mixture (at zero) of carbon monoxide and carbon dioxide, the volumetric composition of which is expressed by the ratio: /lso: /iso, = 8:3. Solution. By the specified composition we find the volumes of each gas by the method of proportional division, after which we calculate the corresponding number of moles: t/ II l» 8 Q »» 11 J 8 Q Kcoe 8 + 3 8 * Va>"a & + & * VCQM grfc -0"36 ^- grfc "" 0.134 zhas * Calculating! the mass of each of the gases from the found number of moles of each of them. 1 "co 28 g/mol; jico. = 44 g/mol moo " 28 e! mol 0.36 mol "South tso . = 44 e/zham" - 0.134 "au> - 5.9 g By adding the found masses of each of the components, we find the mass of the mixture: t^m = 10 g -f 5.9 g = 15.9 e Calculation of the molecular mass of the gas per gram -molecular volume We discussed above the method of calculating the molecular mass of a gas by relative density. Now we will consider the method of calculating the molecular mass of a gas by gram-molecular volume. When calculating, we assume that the mass and volume of a gas are directly proportional to each other. It follows that the volume. gas and its mass are related to each other in the same way as the gram-molecular volume of a gas is to its gram-molecular mass, which is expressed in mathematical form as follows: molecular weight. Hence _ Uiol t r? Let's consider the calculation method using a specific example. "Example. The mass of 34$ ju gas at 740 mm Hg, pi and 21 ° C is equal to 0.604 g. Calculate the molecular mass of the gas. Solution. To solve, you need to know the gram-molecular volume of the gas. Therefore, before proceeding with the calculations, you need to stop at a certain gram-molecular volume of the gas. You can use the standard gram-molecular volume of the gas, which is equal to 22.4 l/mol. Then the volume of gas indicated in the problem statement should be reduced to normal conditions. But you can, on the contrary, calculate the gram-molecular volume of gas under the conditions specified in the problem. With the first method of calculation, the following design is obtained: 740 * mHg. 340 ml - 273 degrees ^ Q ^ 0 760 mm Hg 294 degrees ™ 1 l. 1 - 22.4 l/mol 0.604 v _ s i,pya. -tn-8 = 44 g, M0Аb With the second method we find: V - 22»4 A! mol No. mm Hg -29A deg 0A77 l1ylv. Uiol 273 vrad 740 mm Hg ~ R*0** In both cases, we calculate the mass of a gram molecule, but since the gram molecule is numerically equal to the molecular mass, we thereby find the molecular mass.

Along with mass and volume in chemical calculations often an amount of substance is used that is proportional to the number of structural units contained in the substance. In each case, it must be indicated which structural units (molecules, atoms, ions, etc.) are meant. The unit of quantity of a substance is the mole.

Mole is the amount of substance containing as many molecules, atoms, ions, electrons or other structural units as there are atoms in 12 g of the 12C carbon isotope.

The number of structural units contained in 1 mole of a substance (Avogadro's constant) is determined with great accuracy; in practical calculations it is taken equal to 6.02 1024 mol -1.

It is not difficult to show that the mass of 1 mole of a substance (molar mass), expressed in grams, is numerically equal to the relative molecular mass of this substance.

Thus, the relative molecular weight (or, for short, molecular weight) of free chlorine C1g is 70.90. Therefore, the molar mass of molecular chlorine is 70.90 g/mol. However, the molar mass of chlorine atoms is half as much (45.45 g/mol), since 1 mole of Cl chlorine molecules contains 2 moles of chlorine atoms.

According to Avogadro's law, equal volumes Any gas taken at the same temperature and the same pressure contains the same number of molecules. In other words, the same number of molecules of any gas occupies the same volume under the same conditions. At the same time, 1 mole of any gas contains the same number of molecules. Consequently, under the same conditions, 1 mole of any gas occupies the same volume. This volume is called the molar volume of the gas and under normal conditions (0°C, pressure 101, 425 kPa) is equal to 22.4 liters.

For example, the statement “the carbon dioxide content of the air is 0.04% (vol.)” means that at a partial pressure of CO 2 equal to the air pressure and at the same temperature, the carbon dioxide contained in the air will take up 0.04% of the total volume occupied by air.

Test task

1. Compare the number of molecules contained in 1 g of NH 4 and in 1 g of N 2. In what case and how many times is the number of molecules greater?

2. Express the mass of one sulfur dioxide molecule in grams.

4. How many molecules are there in 5.00 ml of chlorine under standard conditions?

4. What volume under normal conditions is occupied by 27 10 21 gas molecules?

5. Express the mass of one NO 2 molecule in grams -

6. What is the ratio of the volumes occupied by 1 mole of O2 and 1 mole of Oz (the conditions are the same)?

7. Equal masses of oxygen, hydrogen and methane are taken under the same conditions. Find the ratio of the volumes of gases taken.

8. To the question of how much volume 1 mole of water will occupy under normal conditions, the answer was: 22.4 liters. Is this the correct answer?

9. Express the mass of one HCl molecule in grams.

How many molecules of carbon dioxide are there in 1 liter of air if the volumetric content of CO 2 is 0.04% (normal conditions)?

10. How many moles are contained in 1 m 4 of any gas under normal conditions?

11. Express in grams the mass of one molecule of H 2 O-

12. How many moles of oxygen are in 1 liter of air, if the volume

14. How many moles of nitrogen are in 1 liter of air if its volumetric content is 78% (normal conditions)?

14. Equal masses of oxygen, hydrogen and nitrogen are taken under the same conditions. Find the ratio of the volumes of gases taken.

15. Compare the number of molecules contained in 1 g of NO 2 and in 1 g of N 2. In what case and how many times is the number of molecules greater?

16. How many molecules are contained in 2.00 ml of hydrogen under standard conditions?

17. Express in grams the mass of one molecule of H 2 O-

18. What volume under normal conditions is occupied by 17 10 21 gas molecules?

RATE OF CHEMICAL REACTIONS

When defining the concept speed chemical reaction it is necessary to distinguish between homogeneous and heterogeneous reactions. If a reaction occurs in a homogeneous system, for example, in a solution or in a mixture of gases, then it occurs throughout the entire volume of the system. Speed ​​of homogeneous reaction is the amount of a substance that reacts or is formed as a result of a reaction per unit time per unit volume of the system. Since the ratio of the number of moles of a substance to the volume in which it is distributed is the molar concentration of the substance, the rate of a homogeneous reaction can also be defined as change in concentration per unit time of any of the substances: the initial reagent or the reaction product. To ensure that the calculation result is always positive, regardless of whether it is based on a reagent or a product, the “±” sign is used in the formula:

Depending on the nature of the reaction, time can be expressed not only in seconds, as required by the SI system, but also in minutes or hours. During the reaction, the magnitude of its speed is not constant, but continuously changes: it decreases, as the concentrations of the starting substances decrease. The above calculation gives the average value of the reaction rate over a certain time interval Δτ = τ 2 – τ 1. True (instantaneous) speed is defined as the limit to which the ratio Δ tends WITH/ Δτ at Δτ → 0, i.e., the true speed is equal to the derivative of the concentration with respect to time.

For a reaction whose equation contains stoichiometric coefficients that differ from unity, the rate values ​​expressed in terms of different substances, are not the same. For example, for the reaction A + 4B = D + 2E, the consumption of substance A is one mole, the supply of substance B is three moles, and the input of substance E is two moles. That's why υ (A) = ⅓ υ (B) = υ (D) =½ υ (E) or υ (E) . = ⅔ υ (IN) .

If a reaction occurs between substances located in different phases of a heterogeneous system, then it can only occur at the interface between these phases. For example, the interaction between an acid solution and a piece of metal occurs only on the surface of the metal. Speed ​​of heterogeneous reaction is the amount of a substance that reacts or is formed as a result of a reaction per unit time per unit interface surface:

.

The dependence of the rate of a chemical reaction on the concentration of reactants is expressed by the law active masses: at a constant temperature, the rate of a chemical reaction is directly proportional to the product of the molar concentrations of the reacting substances raised to powers equal to the coefficients in the formulas of these substances in the reaction equation. Then for the reaction

2A + B → products

the ratio is valid υ ~ · WITH A 2 · WITH B, and to transition to equality a proportionality coefficient is introduced k, called reaction rate constant:

υ = k· WITH A 2 · WITH B = k·[A] 2 ·[B]

(molar concentrations in formulas can be denoted by the letter WITH with the corresponding index and the formula of the substance enclosed in square brackets). The physical meaning of the reaction rate constant is the reaction rate at concentrations of all reactants equal to 1 mol/l. The dimension of the reaction rate constant depends on the number of factors on the right side of the equation and can be c –1 ; s –1 ·(l/mol); s –1 · (l 2 /mol 2), etc., that is, such that in any case, in calculations, the reaction rate is expressed in mol · l –1 · s –1.

For heterogeneous reactions, the equation of the law of mass action includes the concentrations of only those substances that are in the gas phase or in solution. The concentration of a substance in the solid phase is a constant value and is included in the rate constant, for example, for the combustion process of coal C + O 2 = CO 2, the law of mass action is written:

υ = k I·const··= k·,

Where k= k I const.

In systems where one or more substances are gases, the rate of reaction also depends on pressure. For example, when hydrogen interacts with iodine vapor H 2 + I 2 = 2HI, the rate of the chemical reaction will be determined by the expression:

υ = k··.

If you increase the pressure, for example, by 4 times, then the volume occupied by the system will decrease by the same amount, and, consequently, the concentrations of each of the reacting substances will increase by the same amount. The reaction rate in this case will increase 9 times

Dependence of reaction rate on temperature described by van't Hoff's rule: with every 10 degree increase in temperature, the reaction rate increases by 2-4 times. This means that as the temperature rises in arithmetic progression the rate of a chemical reaction increases exponentially. The base in the progression formula is temperature coefficient of reaction rateγ, showing how many times the rate of a given reaction increases (or, which is the same thing, the rate constant) with an increase in temperature by 10 degrees. Mathematically, Van't Hoff's rule is expressed by the formulas:

or

where and are the reaction rates, respectively, at the initial t 1 and final t 2 temperatures. Van't Hoff's rule can also be expressed by the following relations:

; ; ; ,

where and are, respectively, the rate and rate constant of the reaction at temperature t; and – the same values ​​at temperature t +10n; n– number of “ten-degree” intervals ( n =(t 2 –t 1)/10), by which the temperature has changed (can be an integer or fractional number, positive or negative).

Test task

1. Find the value of the rate constant for the reaction A + B -> AB, if at concentrations of substances A and B equal to 0.05 and 0.01 mol/l, respectively, the reaction rate is 5 10 -5 mol/(l-min).

2. How many times will the rate of reaction 2A + B -> A2B change if the concentration of substance A is increased by 2 times, and the concentration of substance B is decreased by 2 times?

4. How many times should the concentration of the substance, B 2 in the system 2A 2 (g) + B 2 (g) = 2A 2 B (g), be increased so that when the concentration of substance A decreases by 4 times, the rate of the direct reaction does not change ?

4. Some time after the start of the reaction 3A+B->2C+D, the concentrations of substances were: [A] =0.04 mol/l; [B] = 0.01 mol/l; [C] =0.008 mol/l. What are the initial concentrations of substances A and B?

5. In the system CO + C1 2 = COC1 2, the concentration was increased from 0.04 to 0.12 mol/l, and the chlorine concentration was increased from 0.02 to 0.06 mol/l. How many times did the rate of the forward reaction increase?

6. The reaction between substances A and B is expressed by the equation: A + 2B → C. The initial concentrations are: [A] 0 = 0.04 mol/l, [B] o = 0.05 mol/l. The reaction rate constant is 0.4. Find initial speed reactions and the reaction rate after some time, when the concentration of substance A decreases by 0.01 mol/l.

7. How will the rate of the reaction 2CO + O2 = 2CO2, occurring in a closed vessel, change if the pressure is doubled?

8. Calculate how many times the reaction rate will increase if the temperature of the system is increased from 20 °C to 100 °C, taking the value of the temperature coefficient of the reaction rate equal to 4.

9. How will the reaction rate 2NO(r.) + 0 2 (g.) → 2N02(r.) change if the pressure in the system is increased by 4 times;

10. How will the reaction rate 2NO(r.) + 0 2 (g.) → 2N02(r.) change if the volume of the system is reduced by 4 times?

11. How will the rate of the reaction 2NO(r.) + 0 2 (g.) → 2N02(r.) change if the concentration of NO is increased by 4 times?

12. What is the temperature coefficient of the reaction rate if, with an increase in temperature by 40 degrees, the reaction rate

increases by 15.6 times?

14. . Find the value of the rate constant for the reaction A + B -> AB, if at concentrations of substances A and B equal to 0.07 and 0.09 mol/l, respectively, the reaction rate is 2.7 10 -5 mol/(l-min).

14. The reaction between substances A and B is expressed by the equation: A + 2B → C. The initial concentrations are: [A] 0 = 0.01 mol/l, [B] o = 0.04 mol/l. The reaction rate constant is 0.5. Find the initial reaction rate and the reaction rate after some time, when the concentration of substance A decreases by 0.01 mol/l.

15. How will the reaction rate 2NO(r.) + 0 2 (g.) → 2N02(r.) change if the pressure in the system is doubled;

16. In the system CO + C1 2 = COC1 2, the concentration was increased from 0.05 to 0.1 mol/l, and the chlorine concentration was increased from 0.04 to 0.06 mol/l. How many times did the rate of the forward reaction increase?

17. Calculate how many times the reaction rate will increase if the temperature of the system is increased from 20 °C to 80 °C, taking the value of the temperature coefficient of the reaction rate equal to 2.

18. Calculate how many times the reaction rate will increase if the temperature of the system is increased from 40 °C to 90 °C, taking the value of the temperature coefficient of the reaction rate equal to 4.

CHEMICAL BOND. FORMATION AND STRUCTURE OF MOLECULES

1.What types of chemical bonds do you know? Give an example of the formation of an ionic bond using the valence bond method.

2. Which one chemical bond called covalent? What is characteristic of the covalent type of bond?

4. What properties are characterized by a covalent bond? Show this with specific examples.

4. What type of chemical bond is in H2 molecules; Cl 2 HC1?

5.What is the nature of the bonds in molecules? NCI 4 CS 2, CO 2? Indicate for each of them the direction of displacement of the common electron pair.

6. What chemical bond is called ionic? What is characteristic of the ionic type of bond?

7. What type of bond is in the NaCl, N 2, Cl 2 molecules?

8. Picture everything possible ways overlap of the s-orbital with the p-orbital;. Indicate the direction of communication in this case.

9. Explain the donor-acceptor mechanism covalent bond using the example of the formation of phosphonium ion [PH 4 ]+.

10. In CO molecules, C0 2, is the bond polar or nonpolar? Explain. Describe hydrogen bonding.

11. Why are some molecules that have polar bonds generally nonpolar?

12.Covalent or ion type communication is typical for following connections: Nal, S0 2 , KF? Why ionic bond is the limiting case of covalent?

14. What is metal connection? How is it different from a covalent bond? What properties of metals does it determine?

14. What is the nature of the bonds between atoms in molecules; KHF 2, H 2 0, HNO ?

15. How can we explain the high bond strength between atoms in the nitrogen molecule N2 and the significantly lower strength in the phosphorus molecule P4?

16. What kind of bond is called a hydrogen bond? Why do molecules of H2S and HC1, in contrast to H2O and HF, form hydrogen bonds not typical?

17. What bond is called ionic? Does an ionic bond have the properties of saturation and directionality? Why is it an extreme case of covalent bonding?

18. What type of bond is in the molecules NaCl, N 2, Cl 2?