For theoretical material, see the page "Molar volume of gas".

Basic formulas and concepts:

From Avogadro's law, for example, it follows that under the same conditions, 1 liter of hydrogen and 1 liter of oxygen contain the same number of molecules, although their sizes vary greatly.

First corollary of Avogadro's law:

The volume occupied by 1 mole of any gas under normal conditions (n.s.) is 22.4 liters and is called molar volume of gas(Vm).

V m =V/ν (m 3 /mol)

What are called normal conditions (n.s.):

• normal temperature = 0°C or 273 K;
• normal pressure = 1 atm or 760 mm Hg. or 101.3 kPa

From the first corollary of Avogadro’s law it follows that, for example, 1 mole of hydrogen (2 g) and 1 mole of oxygen (32 g) occupy the same volume, equal to 22.4 liters at ground level.

Knowing V m, you can find the volume of any quantity (ν) and any mass (m) of gas:

V=V m ·ν V=V m ·(m/M)

Typical problem 1: What is the volume at no. occupies 10 moles of gas?

V=V m ·ν=22.4·10=224 (l/mol)

Typical problem 2: What is the volume at no. takes up 16 g of oxygen?

V(O 2)=V m ·(m/M) M r (O 2)=32; M(O 2)=32 g/mol V(O 2)=22.4·(16/32)=11.2 l

Second corollary of Avogadro's law:

Knowing the gas density (ρ=m/V) at normal conditions, we can calculate the molar mass of this gas: M=22.4·ρ

Density (D) of one gas is otherwise called the ratio of the mass of a certain volume of the first gas to the mass of a similar volume of the second gas, taken under the same conditions.

Typical task 3: Determine the relative density of carbon dioxide compared to hydrogen and air.

D hydrogen (CO 2) = M r (CO 2)/M r (H 2) = 44/2 = 22 D air = 44/29 = 1.5

• one volume of hydrogen and one volume of chlorine give two volumes of hydrogen chloride: H 2 +Cl 2 =2HCl
• two volumes of hydrogen and one volume of oxygen give two volumes of water vapor: 2H 2 + O 2 = 2H 2 O

Task 1. How many moles and molecules are contained in 44 g of carbon dioxide?

Solution:

M(CO 2) = 12+16 2 = 44 g/mol ν = m/M = 44/44 = 1 mol N(CO 2) = ν N A = 1 6.02 10 23 = 6.02 ·10 23

Task 2. Calculate the mass of one molecule of ozone and an argon atom.

Solution:

M(O 3) = 16 3 = 48 g m(O 3) = M(O 3)/N A = 48/(6.02 10 23) = 7.97 10 -23 g M(Ar) = 40 g m(Ar) = M(Ar)/N A = 40/(6.02 10 23) = 6.65 10 -23 g

Task 3. What is the volume at standard conditions? occupies 2 moles of methane.

Solution:

ν = V/22.4 V(CH 4) = ν 22.4 = 2 22.4 = 44.8 l

Task 4. Determine the density and relative density of carbon monoxide (IV) from hydrogen, methane and air.

Solution:

M r (CO 2)=12+16·2=44; M(CO 2)=44 g/mol M r (CH 4)=12+1·4=16; M(CH 4)=16 g/mol M r (H 2)=1·2=2; M(H 2)=2 g/mol M r (air)=29; M(air)=29 g/mol ρ=m/V ρ(CO 2)=44/22.4=1.96 g/mol D(CH 4)=M(CO 2)/M(CH 4)= 44/16=2.75 D(H 2)=M(CO 2)/M(H 2)=44/2=22 D(air)=M(CO 2)/M(air)=44/24= 1.52

Task 5. Determine the mass of the gas mixture, which includes 2.8 cubic meters of methane and 1.12 cubic meters of carbon monoxide.

Solution:

M r (CO 2)=12+16·2=44; M(CO 2)=44 g/mol M r (CH 4)=12+1·4=16; M(CH 4) = 16 g/mol 22.4 cubic meters CH 4 = 16 kg 2.8 cubic meters CH 4 = x m(CH 4) = x = 2.8 16/22.4 = 2 kg 22.4 cubic meters CO 2 = 28 kg 1.12 cubic meters CO 2 = x m(CO 2)=x=1.12·28/22.4=1.4 kg m(CH 4)+m(CO 2)=2+1, 4=3.4 kg

Task 6. Determine the volumes of oxygen and air required to burn 112 cubic meters of divalent carbon monoxide when it contains non-combustible impurities in a volume fraction of 0.50.

Solution:

• determine the volume of pure CO in the mixture: V(CO)=112·0.5=66 cubic meters
• determine the volume of oxygen required to burn 66 cubic meters of CO: 2CO+O 2 =2CO 2 2mol+1mol 66m 3 +X m 3 V(CO)=2·22.4 = 44.8 m 3 V(O 2)=22 .4 m 3 66/44.8 = X/22.4 X = 66 22.4/44.8 = 33 m 3 or 2V(CO)/V(O 2) = V 0 (CO)/V 0 (O 2) V - molar volumes V 0 - calculated volumes V 0 (O 2) = V(O 2)·(V 0 (CO)/2V(CO))

Task 7. How will the pressure change in a vessel filled with hydrogen and chlorine gases after they react? Is it the same for hydrogen and oxygen?

Solution:

• H 2 +Cl 2 =2HCl - as a result of the interaction of 1 mole of hydrogen and 1 mole of chlorine, 2 moles of hydrogen chloride are obtained: 1 (mol) + 1 (mol) = 2 (mol), therefore, the pressure will not change, since the resulting volume of the gas mixture is equal to the sum of the volumes of the components that reacted.
• 2H 2 + O 2 = 2H 2 O - 2 (mol) + 1 (mol) = 2 (mol) - the pressure in the vessel will decrease by one and a half times, since from 3 volumes of the components that reacted, 2 volumes of the gas mixture are obtained.

Task 8. 12 liters of a gas mixture of ammonia and tetravalent carbon monoxide at no. have a mass of 18 g. How much of each gas is in the mixture?

Solution:

V(NH 3)=x l V(CO 2)=y l M(NH 3)=14+1 3=17 g/mol M(CO 2)=12+16 2=44 g/mol m( NH 3)=x/(22.4·17) g m(CO 2)=y/(22.4·44) g System of equations volume of mixture: x+y=12 mass of mixture: x/(22.4· 17)+y/(22.4·44)=18 After solving we get: x=4.62 l y=7.38 l

Task 9. What amount of water will be obtained as a result of the reaction of 2 g of hydrogen and 24 g of oxygen?

Solution:

2H 2 +O 2 =2H 2 O

From the reaction equation it is clear that the number of reactants does not correspond to the ratio of the stoichiometric coefficients in the equation. In such cases, calculations are carried out using a substance that is less abundant, i.e., this substance will end up first during the reaction. To determine which of the components is deficient, you need to pay attention to the coefficient in the reaction equation.

Amounts of starting components ν(H 2)=4/2=2 (mol) ν(O 2)=48/32=1.5 (mol)

However, there is no need to rush. In our case, to react with 1.5 moles of oxygen, 3 moles of hydrogen (1.5 2) are needed, but we only have 2 moles, i.e., 1 mole of hydrogen is missing for all one and a half moles of oxygen to react. Therefore, we will calculate the amount of water using hydrogen:

ν(H 2 O)=ν(H 2)=2 mol m(H 2 O) = 2 18=36 g

Problem 10. At a temperature of 400 K and a pressure of 3 atmospheres, the gas occupies a volume of 1 liter. What volume will this gas occupy at zero level?

Solution:

From the Clapeyron equation:

P·V/T = Pn ·Vn/Tn Vn = (PVT n)/(Pn T) Vn = (3·1·273)/(1·400) = 2.05 l

When studying chemical substances, important concepts are such quantities as molar mass, density of a substance, and molar volume. So, what is molar volume, and how does it differ for substances in different states of aggregation?

Molar volume: general information

To calculate the molar volume of a chemical substance, it is necessary to divide the molar mass of this substance by its density. Thus, the molar volume is calculated by the formula:

where Vm is the molar volume of the substance, M is the molar mass, p is the density. In the International SI System, this value is measured in cubic meter per mole (m 3 /mol).

Rice. 1. Molar volume formula.

The molar volume of gaseous substances differs from substances in liquid and solid states in that a gaseous element with an amount of 1 mole always occupies the same volume (if the same parameters are met).

The volume of gas depends on temperature and pressure, so when calculating, you should take the volume of gas under normal conditions. Normal conditions are considered to be a temperature of 0 degrees and a pressure of 101.325 kPa.

The molar volume of 1 mole of gas under normal conditions is always the same and equal to 22.41 dm 3 /mol. This volume is called the molar volume of an ideal gas. That is, in 1 mole of any gas (oxygen, hydrogen, air) the volume is 22.41 dm 3 /m.

The molar volume at normal conditions can be derived using the equation of state for an ideal gas, called the Clayperon-Mendeleev equation:

where R is the universal gas constant, R=8.314 J/mol*K=0.0821 l*atm/mol K

Volume of one mole of gas V=RT/P=8.314*273.15/101.325=22.413 l/mol, where T and P are the value of temperature (K) and pressure under normal conditions.

Rice. 2. Table of molar volumes.

Avogadro's law

In 1811, A. Avogadro put forward the hypothesis that equal volumes of different gases under the same conditions (temperature and pressure) contain the same number of molecules. Later the hypothesis was confirmed and became a law bearing the name of the great Italian scientist.

Rice. 3. Amedeo Avogadro.

The law becomes clear if we remember that in gaseous form the distance between particles is incomparably greater than the size of the particles themselves.

Thus, the following conclusions can be drawn from Avogadro’s law:

• Equal volumes of any gases taken at the same temperature and at the same pressure contain the same number of molecules.
• 1 mole of completely different gases under the same conditions occupies the same volume.
• One mole of any gas under normal conditions occupies a volume of 22.41 liters.

The corollary to Avogadro's law and the concept of molar volume are based on the fact that a mole of any substance contains the same number of particles (for gases - molecules), equal to Avogadro's constant.

To find out the number of moles of solute contained in one liter of solution, it is necessary to determine the molar concentration of the substance using the formula c = n/V, where n is the amount of solute, expressed in moles, V is the volume of the solution, expressed in liters C is molarity.

What have we learned?

In the 8th grade chemistry school curriculum, the topic “Molar volume” is studied. One mole of gas always contains the same volume, equal to 22.41 cubic meters/mol. This volume is called the molar volume of the gas.

Test on the topic

Evaluation of the report

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The molar volume of a gas is equal to the ratio of the volume of the gas to the amount of substance of this gas, i.e.

V m = V(X) / n(X),

where V m is the molar volume of gas - a constant value for any gas under given conditions;

V(X) – volume of gas X;

n(X) – amount of gas substance X.

The molar volume of gases under normal conditions (normal pressure p n = 101,325 Pa ≈ 101.3 kPa and temperature T n = 273.15 K ≈ 273 K) is V m = 22.4 l/mol.

Ideal gas laws

In calculations involving gases, it is often necessary to switch from these conditions to normal ones or vice versa. In this case, it is convenient to use the formula following from the combined gas law of Boyle-Mariotte and Gay-Lussac:

pV / T = p n V n / T n

Where p is pressure; V - volume; T - temperature on the Kelvin scale; the index “n” indicates normal conditions.

Volume fraction

The composition of gas mixtures is often expressed using the volume fraction - the ratio of the volume of a given component to the total volume of the system, i.e.

φ(X) = V(X) / V

where φ(X) is the volume fraction of component X;

V(X) - volume of component X;

V is the volume of the system.

Volume fraction is a dimensionless quantity; it is expressed in fractions of a unit or as a percentage.

Example 1. What volume will ammonia weighing 51 g occupy at a temperature of 20°C and a pressure of 250 kPa?

	1. Determine the amount of ammonia substance: n(NH 3) = m(NH 3) / M(NH 3) = 51 / 17 = 3 mol. 2. The volume of ammonia under normal conditions is: V(NH 3) = V m n(NH 3) = 22.4 3 = 67.2 l. 3. Using formula (3), we reduce the volume of ammonia to these conditions (temperature T = (273 + 20) K = 293 K): V(NH 3) = pn Vn (NH 3) / pT n = 101.3 293 67.2 / 250 273 = 29.2 l. Answer: V(NH 3) = 29.2 l.

Example 2. Determine the volume that a gas mixture containing hydrogen, weighing 1.4 g, and nitrogen, weighing 5.6 g, will occupy under normal conditions.

1. Find the amounts of hydrogen and nitrogen substances: n(N 2) = m(N 2) / M(N 2) = 5.6 / 28 = 0.2 mol n(H 2) = m(H 2) / M(H 2) = 1.4 / 2 = 0.7 mol 2. Since under normal conditions these gases do not interact with each other, the volume of the gas mixture will be equal to the sum of the volumes of the gases, i.e. V(mixtures) = V(N 2) + V(H 2) = V m n(N 2) + V m n(H2) = 22.4 0.2 + 22.4 0.7 = 20.16 l. Answer: V(mixture) = 20.16 l.

Law of volumetric relations

How to solve a problem using the “Law of Volumetric Relations”?

Law of Volume Ratios: The volumes of gases involved in a reaction are related to each other as small integers equal to the coefficients in the reaction equation.

The coefficients in the reaction equations show the numbers of volumes of reacting and formed gaseous substances.

Example. Calculate the volume of air required to burn 112 liters of acetylene.

1. We compose the reaction equation:

2. Based on the law of volumetric relations, we calculate the volume of oxygen:

112 / 2 = X / 5, from where X = 112 5 / 2 = 280l

3. Determine the volume of air:

V(air) = V(O 2) / φ(O 2)

V(air) = 280 / 0.2 = 1400 l.

Along with mass and volume, chemical calculations often use the amount of a substance 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 of any gases taken at the same temperature and the same pressure contain 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 normal 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 is occupied by 17 10 21 gas molecules under normal conditions?

RATE OF CHEMICAL REACTIONS

When defining the concept chemical reaction rate 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 for different substances are not the same. For example, for the reaction A + 4B = D + 2E, the consumption of substance A is one mole, that of substance B is three moles, and the supply 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 of mass action: 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 increases in an 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 the initial reaction rate 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. What chemical bond is 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. Draw all possible ways of overlapping the s-orbital with the p-orbital;. Indicate the direction of communication in this case.

9. Explain the donor-acceptor mechanism of covalent bonds 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 ionic type of bond is typical for the following compounds: Nal, S0 2, KF? Why is an ionic bond an extreme case of a covalent bond?

14. What is a metal bond? 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 is the formation of hydrogen bonds not typical for H2S and HC1 molecules, unlike H2O and HF?

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?

In chemistry, they do not use the absolute masses of molecules, but use the relative molecular mass. It shows how many times the mass of a molecule is greater than 1/12 the mass of a carbon atom. This quantity is denoted by Mr.

Relative molecular mass is equal to the sum of the relative atomic masses of its constituent atoms. Let's calculate the relative molecular mass of water.

You know that a water molecule contains two hydrogen atoms and one oxygen atom. Then its relative molecular mass will be equal to the sum of the products of the relative atomic mass of each chemical element and the number of its atoms in a water molecule:

Knowing the relative molecular masses of gaseous substances, one can compare their densities, that is, calculate the relative density of one gas from another - D(A/B). The relative density of gas A to gas B is equal to the ratio of their relative molecular masses:

Let's calculate the relative density of carbon dioxide to hydrogen:

Now we calculate the relative density of carbon dioxide to hydrogen:

D(arc/hydr) = Mr(arc) : Mr(hydr) = 44:2 = 22.

Thus, carbon dioxide is 22 times heavier than hydrogen.

As you know, Avogadro's law applies only to gaseous substances. But chemists need to have an idea of ​​the number of molecules and in portions of liquid or solid substances. Therefore, to compare the number of molecules in substances, chemists introduced the value - molar mass .

Molar mass is denoted M, it is numerically equal to the relative molecular weight.

The ratio of the mass of a substance to its molar mass is called amount of substance .

The amount of substance is indicated n. This is a quantitative characteristic of a portion of a substance, along with mass and volume. The amount of a substance is measured in moles.

The word "mole" comes from the word "molecule". The number of molecules in equal amounts of a substance is the same.

It has been experimentally established that 1 mole of a substance contains particles (for example, molecules). This number is called Avogadro's number. And if we add a unit of measurement to it - 1/mol, then it will be a physical quantity - Avogadro's constant, which is denoted N A.

Molar mass is measured in g/mol. The physical meaning of molar mass is that this mass is 1 mole of a substance.

According to Avogadro's law, 1 mole of any gas will occupy the same volume. The volume of one mole of gas is called molar volume and is denoted Vn.

Under normal conditions (which is 0 °C and normal pressure - 1 atm. or 760 mm Hg or 101.3 kPa), the molar volume is 22.4 l/mol.

Then the amount of gas substance at ground level is can be calculated as the ratio of gas volume to molar volume.

TASK 1. What amount of substance corresponds to 180 g of water?

TASK 2. Let us calculate the volume at zero level that will be occupied by carbon dioxide in an amount of 6 mol.

Bibliography

1. Collection of problems and exercises in chemistry: 8th grade: to the textbook by P.A. Orzhekovsky and others. “Chemistry, 8th grade” / P.A. Orzhekovsky, N.A. Titov, F.F. Hegel. - M.: AST: Astrel, 2006. (p. 29-34)
2. Ushakova O.V. Chemistry workbook: 8th grade: to the textbook by P.A. Orzhekovsky and others. “Chemistry. 8th grade” / O.V. Ushakova, P.I. Bespalov, P.A. Orzhekovsky; under. ed. prof. P.A. Orzhekovsky - M.: AST: Astrel: Profizdat, 2006. (p. 27-32)
3. Chemistry: 8th grade: textbook. for general education institutions / P.A. Orzhekovsky, L.M. Meshcheryakova, L.S. Pontak. M.: AST: Astrel, 2005. (§§ 12, 13)
4. Chemistry: inorg. chemistry: textbook. for 8th grade. general education institution / G.E. Rudzitis, F.G. Feldman. - M.: Education, OJSC “Moscow Textbooks”, 2009. (§§ 10, 17)
5. Encyclopedia for children. Volume 17. Chemistry / Chapter. ed.V.A. Volodin, Ved. scientific ed. I. Leenson. - M.: Avanta+, 2003.

1. Unified collection of digital educational resources ().
2. Electronic version of the journal “Chemistry and Life” ().
3. Chemistry tests (online) ().

Homework

1.p.69 No. 3; p.73 No. 1, 2, 4 from the textbook “Chemistry: 8th grade” (P.A. Orzhekovsky, L.M. Meshcheryakova, L.S. Pontak. M.: AST: Astrel, 2005).

2. №№ 65, 66, 71, 72 from the Collection of problems and exercises in chemistry: 8th grade: to the textbook by P.A. Orzhekovsky and others. “Chemistry, 8th grade” / P.A. Orzhekovsky, N.A. Titov, F.F. Hegel. - M.: AST: Astrel, 2006.