In the study of chemicals, important concepts are such quantities as molar mass, substance density, molar volume. So, what is the molar volume, and how is it different 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 meters per mol (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 of 1 mole always occupies the same volume (if the same parameters are observed).

The volume of gas depends on temperature and pressure, so the calculation 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 mol 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 under normal conditions can be derived using the equation of state for an ideal gas, which is called the Claiperon-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 temperature (K) and pressure values ​​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 a gaseous form the distance between the 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 consequence of 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 the Avogadro constant.

To find out the number of moles of a solute contained in one liter of a solution, it is necessary to determine the molar concentration of a substance using the formula c \u003d n / V, where n is the amount of a solute, expressed in moles, V is the volume of the solution, expressed in liters C - molarity.

What have we learned?

In the school curriculum in chemistry of the 8th grade, 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.

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In order to know the composition of any gaseous substances, it is necessary to be able to operate with such concepts as molar volume, molar mass and density of a substance. In this article, we will consider what is molar volume, and how to calculate it?

Amount of substance

Quantitative calculations are carried out in order to actually carry out a particular process or find out the composition and structure of a certain substance. These calculations are inconvenient to make with the absolute values ​​of the masses of atoms or molecules due to the fact that they are very small. Relative atomic masses are also in most cases impossible to use, since they are not related to generally accepted measures of the mass or volume of a substance. Therefore, the concept of the amount of substance was introduced, which is denoted by the Greek letter v (nu) or n. The amount of a substance is proportional to the number of structural units (molecules, atomic particles) contained in the substance.

The unit of quantity of a substance is the mole.

A mole is the amount of a substance that contains as many structural units as there are atoms in 12 g of a carbon isotope.

The mass of 1 atom is 12 a. e. m., so the number of atoms in 12 g of the carbon isotope is:

Na \u003d 12g / 12 * 1.66057 * 10 to the power of -24g \u003d 6.0221 * 10 to the power of 23

The physical quantity Na is called the Avogadro constant. One mole of any substance contains 6.02 * 10 to the power of 23 particles.

Rice. 1. Avogadro's law.

Molar volume of gas

The molar volume of a gas is the ratio of the volume of a substance to the amount of that substance. This value is calculated by dividing the molar mass of a substance by its density according to the following formula:

where Vm is the molar volume, M is the molar mass, and p is the density of the substance.

Rice. 2. Molar volume formula.

In the international C system, the measurement of the molar volume of gaseous substances is carried out in cubic meters per mol (m 3 / mol)

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

The volume of gas depends on temperature and pressure, so the calculation 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 mol 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.

Rice. 3. Molar volume of gas under normal conditions.

Table "molar volume of gases"

The following table shows the volume of some gases:

Gas	Molar volume, l
H2	22,432
O2	22,391
Cl2	22,022
CO2	22,263
NH3	22,065
SO2	21,888
Ideal	22,41383

What have we learned?

The molar volume of a gas studied in chemistry (grade 8), along with the molar mass and density, are the necessary quantities to determine the composition of a particular chemical substance. A feature of a molar gas is that one mole of gas always contains the same volume. This volume is called the molar volume of the gas.

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Where m is mass, M is molar mass, V is volume.

4. Law of Avogadro. Established by the Italian physicist Avogadro in 1811. The same volumes of any gases, taken at the same temperature and the same pressure, contain the same number of molecules.

Thus, the concept of the amount of a substance can be formulated: 1 mole of a substance contains a number of particles equal to 6.02 * 10 23 (called the Avogadro constant)

The consequence of this law is that 1 mole of any gas occupies under normal conditions (P 0 \u003d 101.3 kPa and T 0 \u003d 298 K) a volume equal to 22.4 liters.

5. Boyle-Mariotte Law

At constant temperature, the volume of a given amount of gas is inversely proportional to the pressure under which it is:

6. Gay-Lussac's Law

At constant pressure, the change in the volume of a gas is directly proportional to the temperature:

V/T = const.

7. The relationship between gas volume, pressure and temperature can be expressed the combined law of Boyle-Mariotte and Gay-Lussac, which is used to bring gas volumes from one condition to another:

P 0 , V 0 ,T 0 - volume pressure and temperature under normal conditions: P 0 =760 mm Hg. Art. or 101.3 kPa; T 0 \u003d 273 K (0 0 C)

8. Independent assessment of the value of molecular masses M can be done using the so-called equations of state for an ideal gas or the Clapeyron-Mendeleev equations :

pV=(m/M)*RT=vRT.(1.1)

Where R - gas pressure in a closed system, V- volume of the system, T - mass of gas T - absolute temperature, R- universal gas constant.

Note that the value of the constant R can be obtained by substituting the values ​​characterizing one mole of gas at N.C. into equation (1.1):

r = (p V) / (T) \u003d (101.325kPa 22.4 l) / (1 mol 273K) \u003d 8.31J / mol.K)

Examples of problem solving

Example 1 Bringing the volume of gas to normal conditions.

What volume (n.s.) will occupy 0.4 × 10 -3 m 3 of gas at 50 0 C and a pressure of 0.954 × 10 5 Pa?

Solution. To bring the volume of gas to normal conditions, use the general formula that combines the laws of Boyle-Mariotte and Gay-Lussac:

pV/T = p 0 V 0 /T 0 .

The volume of gas (n.o.) is , where T 0 = 273 K; p 0 \u003d 1.013 × 10 5 Pa; T = 273 + 50 = 323 K;

m 3 \u003d 0.32 × 10 -3 m 3.

When (n.o.) gas occupies a volume equal to 0.32×10 -3 m 3 .

Example 2 Calculation of the relative density of a gas from its molecular weight.

Calculate the density of ethane C 2 H 6 from hydrogen and air.

Solution. It follows from Avogadro's law that the relative density of one gas over another is equal to the ratio of molecular masses ( M h) of these gases, i.e. D=M 1 /M 2. If M 1С2Н6 = 30, M 2 H2 = 2, the average molecular weight of air is 29, then the relative density of ethane with respect to hydrogen is D H2 = 30/2 =15.

Relative density of ethane in air: D air= 30/29 = 1.03, i.e. ethane is 15 times heavier than hydrogen and 1.03 times heavier than air.

Example 3 Determination of the average molecular weight of a mixture of gases by relative density.

Calculate the average molecular weight of a mixture of gases consisting of 80% methane and 20% oxygen (by volume) using the values ​​of the relative density of these gases with respect to hydrogen.

Solution. Often calculations are made according to the mixing rule, which is that the ratio of the volumes of gases in a two-component gas mixture is inversely proportional to the differences between the density of the mixture and the densities of the gases that make up this mixture. Let us denote the relative density of the gas mixture with respect to hydrogen through D H2. it will be greater than the density of methane, but less than the density of oxygen:

80D H2 - 640 = 320 - 20 D H2; D H2 = 9.6.

The hydrogen density of this mixture of gases is 9.6. average molecular weight of the gas mixture M H2 = 2 D H2 = 9.6×2 = 19.2.

Example 4 Calculation of the molar mass of a gas.

The mass of 0.327 × 10 -3 m 3 of gas at 13 0 C and a pressure of 1.040 × 10 5 Pa is 0.828 × 10 -3 kg. Calculate the molar mass of the gas.

Solution. You can calculate the molar mass of a gas using the Mendeleev-Clapeyron equation:

Where m is the mass of gas; M is the molar mass of the gas; R- molar (universal) gas constant, the value of which is determined by the accepted units of measurement.

If the pressure is measured in Pa, and the volume in m 3, then R\u003d 8.3144 × 10 3 J / (kmol × K).

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.

The first corollary of Avogadro's law:

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

V m \u003d V / ν (m 3 / mol)

What is called normal conditions (n.o.):

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

From the first consequence 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 n.o.

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 task 1: What is the volume at n.o.s. occupies 10 moles of gas?

V=V m ν=22.4 10=224 (l/mol)

Typical task 2: What is the volume at n.o.s. takes 16 g of oxygen?

V(O 2)=V m ·(m/M) M r (O 2)=32; M(O 2) \u003d 32 g / mol V (O 2) \u003d 22.4 (16/32) \u003d 11.2 l

The second corollary of Avogadro's law:

Knowing the density of the gas (ρ=m/V) at n.o., we can calculate the molar mass of this gas: M=22.4 ρ

The 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.

Sample problem 3: Determine the relative density of carbon dioxide from hydrogen and air.

D hydrogen (CO 2) \u003d M r (CO 2) / M r (H 2) \u003d 44/2 \u003d 22 D air \u003d 44/29 \u003d 1.5

• one volume of hydrogen and one volume of chlorine give two volumes of hydrogen chloride: H 2 + Cl 2 \u003d 2HCl
• two volumes of hydrogen and one volume of oxygen give two volumes of water vapor: 2H 2 + O 2 \u003d 2H 2 O

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

Solution:

M(CO 2) \u003d 12 + 16 2 \u003d 44 g / mol ν \u003d m / M \u003d 44/44 \u003d 1 mol N (CO 2) \u003d ν N A \u003d 1 6.02 10 23 \u003d 6.02 10 23

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

Solution:

M(O 3) \u003d 16 3 \u003d 48 g m (O 3) \u003d M (O 3) / N A \u003d 48 / (6.02 10 23) \u003d 7.97 10 -23 g M (Ar) \u003d 40 g m (Ar) \u003d M (Ar) / N A \u003d 40 / (6.02 10 23) \u003d 6.65 10 -23 g

Task 3 . What is the volume at n.o. occupies 2 moles of methane.

Solution:

ν \u003d V / 22.4 V (CH 4) \u003d ν 22.4 \u003d 2 22.4 \u003d 44.8 l

Task 4 . Determine the density and relative density of carbon monoxide (IV) for 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) \u003d 29 g / mol ρ \u003d m / V ρ (CO 2) \u003d 44 / 22.4 \u003d 1.96 g / mol D (CH 4) \u003d 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) \u003d 16 g / mol 22.4 cubic meters CH 4 \u003d 16 kg 2.8 cubic meters CH 4 \u003d x m (CH 4) \u003d x \u003d 2.8 16 / 22.4 \u003d 2 kg 22.4 cubic meters CO 2 \u003d 28 kg 1.12 cubic meters CO 2 \u003d x m (CO 2) \u003d x \u003d 1.12 28 / 22.4 \u003d 1.4 kg m (CH 4) + m (CO 2) \u003d 2 + 1, 4=3.4 kg

Task 6 . Determine the volumes of oxygen and air required for the combustion of 112 cubic meters of divalent carbon monoxide with the content of non-combustible impurities in it in volume fractions of 0.50.

Solution:

• determine the volume of pure CO in the mixture: V (CO) \u003d 112 0.5 \u003d 66 cubic meters
• determine the volume of oxygen required to burn 66 cubic meters of CO: 2CO + O 2 \u003d 2CO 2 2mol + 1mol 66m 3 + X m 3 V (CO) \u003d 2 22.4 \u003d 44.8 m 3 V (O 2) \u003d 22 .4 m 3 66 / 44.8 \u003d X / 22.4 X \u003d 66 22.4 / 44.8 \u003d 33 m 3 or 2V (CO) / V (O 2) \u003d V 0 (CO) / V 0 (O 2) V - molar volumes V 0 - calculated volumes V 0 (O 2) \u003d 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? Likewise for hydrogen and oxygen?

Solution:

• H 2 + Cl 2 \u003d 2HCl - as a result of the interaction of 1 mol of hydrogen and 1 mol of chlorine, 2 mol of hydrogen chloride is obtained: 1 (mol) + 1 (mol) \u003d 2 (mol), therefore, the pressure will not change, since the resulting volume of the gas mixture is the sum of the volumes of the components involved in the reaction.
• 2H 2 + O 2 \u003d 2H 2 O - 2 (mol) + 1 (mol) \u003d 2 (mol) - the pressure in the vessel will decrease by one and a half times, since 2 volumes of the gas mixture were obtained from 3 volumes of the components that entered into the reaction.

Task 8 . 12 liters of a gas mixture of ammonia and tetravalent carbon monoxide at n.o.s. have a mass of 18 g. How much is in the mixture of each of the gases?

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) \u003d x / (22.4 17) g m (CO 2) \u003d y / (22.4 44) g System of equations mixture volume: x + y \u003d 12 mixture mass: x / (22.4 ) 17)+y/(22.4 44)=18 After solving we get: x=4.62 l y=7.38 l

Task 9 . How much 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 \u003d 2H 2 O

It can be seen from the reaction equation that the number of reactants does not correspond to the ratio of stoichiometric coefficients in the equation. In such cases, calculations are carried out on the substance, which is less, i.e., this substance will finish first in the course of the reaction. To determine which of the components is in short supply, 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, for the reaction with 1.5 moles of oxygen, 3 moles of hydrogen (1.5 2) are needed, and we have only 2 moles of it, i.e., 1 mole of hydrogen is not enough for all one and a half moles of oxygen to react. Therefore, we will calculate the amount of water by hydrogen:

ν (H 2 O) \u003d ν (H 2) \u003d 2 mol m (H 2 O) \u003d 2 18 \u003d 36 g

Task 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 n.o.s.?

Solution:

From the Clapeyron equation:

P V/T = P n V n /T n V n = (PVT n)/(P n T) V n = (3 1 273)/(1 400) = 2.05 l

Along with mass and volume in chemical calculations, the amount of a substance is often used, which is proportional to the number of structural units contained in the substance. In this case, 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.

A mole is the amount of a 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 easy to show that the mass of 1 mole of a substance (molar mass), expressed in grams, is numerically equal to the relative molecular weight of this substance.

Thus, the relative molecular weight (or molecular weight for short) of free chlorine C1r is 70.90. Therefore, the molar mass of molecular chlorine is 70.90 g/mol. However, the molar mass of chlorine atoms is half that (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. However, 1 mole of any gas contains the same number of molecules. Therefore, under the same conditions, 1 mole of any gas occupies the same volume. This volume is called the molar volume of gas and under normal conditions (0 ° C, pressure 101, 425 kPa) is 22.4 liters.

For example, the statement "the content of carbon dioxide in 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 0.04% of the total volume occupied by air.

Control task

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

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

4. How many molecules are contained 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 in grams the mass of one NO 2 molecule -

6. What is the ratio of the volumes occupied by 1 mole of O 2 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. When asked how much volume 1 mole of water will take under normal conditions, the answer was received: 22.4 liters. Is this the correct answer?

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

How many molecules of carbon dioxide are in 1 liter of air if the volume 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 volume 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 numbers of molecules contained in 1 g of NO 2 and 1 g of N 2. In which case and how many times the number of molecules is greater?

16. How many molecules are contained in 2.00 ml of hydrogen under normal 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 chemical reaction rate it is necessary to distinguish between homogeneous and heterogeneous reactions. If the reaction proceeds in a homogeneous system, for example, in a solution or in a mixture of gases, then it takes place in the entire volume of the system. The rate of a homogeneous reaction called the amount of a substance that enters into a reaction or is formed as a result of a reaction per unit of time in a 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 the concentration per unit time of any of the substances: the initial reagent or reaction product. To ensure that the result of the calculation is always positive, regardless of whether it is produced by 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 value of its rate is not constant, but continuously changes: it decreases, since 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 . The true (instantaneous) speed is defined as the limit to which the ratio Δ WITH/ Δτ at Δτ → 0, i.e. the true velocity is equal to the time derivative of the concentration.

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 \u003d D + 2E, the consumption of substance A is one mole, substance B is three moles, the arrival 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 proceeds between substances that are in different phases of a heterogeneous system, then it can only take place at the interface between these phases. For example, the interaction of an acid solution and a piece of metal occurs only on the surface of the metal. The rate of a heterogeneous reaction called the amount of a substance that enters into a reaction or is formed as a result of a reaction per unit of time per unit of the interface between phases:

.

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 reactants 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 υ ~ · WITH A 2 WITH B, and for the transition to equality, the coefficient of proportionality is introduced k, called reaction rate constant:

υ = k· WITH A 2 WITH B = k[A] 2 [V]

(molar concentrations in formulas can be denoted as 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 from -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 = kI const = k·,

Where k= kI const.

In systems where one or more substances are gases, the reaction rate also depends on pressure. For example, when hydrogen interacts with iodine vapor H 2 + I 2 \u003d 2HI, the rate of a chemical reaction will be determined by the expression:

υ = k··.

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

Temperature dependence of the reaction rate is described by the van't Hoff rule: for every 10 degrees increase in temperature, the reaction rate increases by 2-4 times. This means that as the temperature increases exponentially, the rate of a chemical reaction increases exponentially. The base in the progression formula is reaction rate temperature coefficientγ, showing how many times the rate of a given reaction increases (or, what is the same, the rate constant) with an increase in temperature by 10 degrees. Mathematically, the van't Hoff 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 as follows:

; ; ; ,

where and are, respectively, the rate and rate constant of the reaction at a temperature t; and are the same values ​​at temperature t +10n; n is the 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).

Control task

1. Find the value of the reaction rate constant 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 reaction rate 2A + B -> A2B change if the concentration of substance A is increased by 2 times, and the concentration of substance B is reduced by 2 times?

4. How many times should the concentration of a substance be increased, B 2 in the system 2A 2 (g.) + B 2 (g.) \u003d 2A 2 B (g.), 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] \u003d 0.008 mol / l. What are the initial concentrations of substances A and B?

5. In the CO + C1 2 = COC1 2 system, the concentration was increased from 0.04 to 0.12 mol / l, and the concentration of chlorine - from 0.02 to 0.06 mol / l. By how much 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 \u003d 0.04 mol / l, [B] o \u003d 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 2СО + О2 = 2СО2, proceeding 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 raised from 20 °C to 100 °C, assuming the temperature coefficient of the reaction rate to be 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 reaction rate 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 reaction rate constant 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 \u003d 0.01 mol / l, [B] o \u003d 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 CO + C1 2 = COC1 2 system, the concentration was increased from 0.05 to 0.1 mol / l, and the concentration of chlorine - from 0.04 to 0.06 mol / l. By how much 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, assuming the value of the temperature coefficient of the reaction rate to be 2.

18. Calculate how many times the reaction rate will increase if the temperature of the system is raised from 40 ° C to 90 ° C, assuming the value of the temperature coefficient of the reaction rate to be 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 by the method of valence bonds.

2. What chemical bond is called covalent? What is characteristic of a covalent type of bond?

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

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

5. What is the nature of 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 an ionic bond?

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

8. Draw all possible ways of overlapping the s-orbital with the p-orbital;. Specify the direction of the connection in this case.

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

10. In CO, CO 2 molecules, is the bond polar or non-polar? Explain. Describe a hydrogen bond.

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

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

14. What is a metallic bond? How is it different from a covalent bond? What properties of metals does it cause?

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

15. How to explain the high strength of the bond between atoms in the nitrogen molecule N 2 and the much lower strength in the phosphorus molecule P 4?

16 . What is 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 the limiting case of a covalent bond?

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