2 volume ratios of gases in chemical reactions. Lesson topic. The ratio of the volumes of gases in chemical reactions. Calculation of volume ratios of gases by chemical equations. Inorganic chemistry

Volume ratios of gases in chemical reactions.

Target: consolidate knowledge of gases, be able to calculate the volumetric ratios of gases, according to chemical equations using the law of volumetric relations, apply Avogadro's law and the concept of molar volume when solving problems.

Equipment: Cards with tasks, Avogadro's law on the board.

During the classes:

I Org. moment

Repetition

1. What are the substances in the gaseous state?

(H 2, N 2, O 2, CH 4, C 2 H 6)

2. What concept is typical for these gases? ("Volume")

3. Which scientist suggested that the composition of gases includes 2 atoms and which ones?

(A. Avogadro, H 2, O 2, N 2 )

4. What law was discovered by Avogadro?

(In equal volumes of various gases under the same conditions (tand pressure) contains the same number of molecules)

5. According to Avogadro's law, 1 mole of any gas occupies a volume equal to (22.4 l / mol)

6. What law denotes the volume of gas? (Vm - molar volume)

7. By what formulas do we find:V, Vm, the amount of substance?

V m = V v = V V = V m ∙ v

v V m

II. Studying the material

When the reactant reacted and the resulting product are in a gaseous state, then their volume ratios can be determined from the reaction equation.

For example, consider the interaction of hydrogen with chlorine. For example, the reaction equation:

H 2 + CI 2 = 2NS I

1 mol 1 mol 2 mol

22.4 l/mol 22.4 l/mol 44.8 l/mol

As you can see, 1 mole of hydrogen and 1 mole of chlorine react to form 2 moles of hydrogen chloride. If we shorten these numerical values volumes by 22.4, we get a volumetric ratio of 1:1:2. In this way, it is also possible to determine the volumetric ratios of gaseous substances at normal conditions.

Avogadro's law playing important role in chemical calculations of gaseous substances, is formed as follows:

In equal volumes under the same external conditions ( t and pressure) contain the same number of molecules.

From this law follows the consequence that 1 mole of any gas under normal conditions always occupies the same volume (the molar volume of the gas). Equal to 22.4 liters.

The coefficients in the reaction equations show the number of moles and the number of volumes of gaseous substances.

Example: Calculate how much oxygen is consumed when 10m³ of hydrogen interacts with it.

Let's write the reaction equation

10 m³ x m³

2H 2 + O 2 \u003d 2H2O

2 mol 1 mol

2 m³ 1 m³

According to the reaction equation, it is known that hydrogen and oxygen react in volume ratios of 2:1.

Then 10:2 = X:1, X = 5 m³. Therefore, in order for 10 m³ of hydrogen to react, 5 m³ of oxygen is needed.

Calculations using Avogadro's law.

I task type.

Determining the amount of a substance from a known volume of gas and calculating the volume of gas (N.O.) from the production of the amount of substance.

Example 1Calculate the number of moles of oxygen, the volume of which at n.o. occupies 89.6 liters.

According to the formula V = V m ∙ vfind the amount of matterv = V

V m

v (O 2 ) = _____89.6l___= 4 mol

22.4 l/mol Answer: v(O 2) = 4 mol

Example 2 What is the volume of 1.5 mole of oxygen under normal conditions?

v (O 2 ) = V m ∙ v \u003d 22.4 l / mol ∙ 1.5 mol \u003d 33.6 l.

II task type.

Calculation of volume (n.s.) from the mass of a gaseous substance.

Example. Calculate the volume (at N.C.) occupied by 96 g of oxygen. First, find the molar mass of oxygen O 2. It is equal to M (O 2) \u003d 32 g / mol.

Now according to the formulam = M ∙ v find.

v (O 2 ) = m = 96 g____= 3 mol.

M 32 g/mol

Calculate the volume occupied by 3 moles of oxygen (n.c.) using the formulaV = V m ∙ v :

V(O 2 ) \u003d 22.4 l / mol ∙ 3 mol \u003d 67.2 l.

Answer: V(O 2) = 67.2 liters.

III. Consolidation of the lesson

1. work with ex. pp 80 (8.9)

2. d / z: paragraph 29 p. 80 ex. 10

Materials used in this section methodological manual"Teaching Problem Solving in Chemistry". Authors - compilers: teacher of chemistry of the highest category, methodologist of the Educational Establishment "Gymnasium No. 1 in Grodno" Tolkach L.Ya.; methodologist of the educational and methodological department of the Educational Institution "Grodno OIPK and PRR and SO" Korobova N.P.

Calculations using the molar volume of gases.

Calculation of the relative density of gases.

Volume ratios of gases

One mole of any gas under the same conditions occupies the same volume. So, under normal conditions (n.s.),those. at 0 °C and normal atmospheric pressure, equal to 101.3 kPa, one mole of any gas occupies a volume22.4 dm3.

Attitudevolume of a gas to the corresponding chemical quantity of a substance is a quantity calledmolar volume of gas (Vm):

Vm = V/ ndm 3 , whenceV = Vm · n

In order to determine whether a gas is lighter or heavier relative to another gas, it is enough to compare their densities:

gaseous state of matter. Avogadro's law. Molar volume of a gas.

Substances can be in three states of aggregation - solid, liquid and gaseous. The particles that make up solids are sufficiently strongly bonded to each other that solids have a definite shape. Particles of solids can be atoms, molecules, ions, forming crystalline structures. These particles oscillate with small amplitude around the nodes crystal lattice. In liquids, particles are less bound to each other and can move over fairly long distances. Therefore, liquids have fluidity and take the shape of the vessel in which they are located.

The transition of a substance from a solid to a liquid state occurs upon heating, as a result of which the amplitude of particle oscillations gradually increases. At a certain temperature, the particles of a substance acquire the ability to leave the lattice sites, and melting occurs. When cooled, on the contrary, the liquid particles lose their ability to move and are fixed in a certain position, forming a solid. At normal conditions liquids are usually molecular structure. At high temperature the structure of the liquid may be different (melts of salts and metals).

Interaction between molecules is much weaker than between ions in ionic crystal structures; atoms connected covalent bond V atomic structures; metal ions bound by electron gas in metallic structures.

The solid and liquid states of matter are common name condensed state. The densities of substances in the condensed state are in the range of approximately 0.5 - 22.5 g / cm 3. Substances in the gaseous state have much lower densities - about 10 -2 - 10 -3 g/cm 3 . The transition to the gaseous state is carried out as a result of heating of substances that are in a condensed state (boiling of liquids, sublimation of solids). Gaseous substances under normal conditions consist of molecules.

When passing into the gaseous state, the particles of a substance overcome the forces of intermolecular interaction. The volume that a gas occupies is essentially the volume of free space between randomly moving gas molecules. The size of this space is determined by temperature and pressure. In this case, the volume occupied by the molecules themselves can be neglected. this implies Avogadro's law :

Equal volumes of different gases under the same conditions contain the same number of molecules.

From Avogadro's law follow two main implications .

First consequence

One mole of any gas under the same conditions occupies the same volume. This volume is called molar volume of gas ( V m ) , which is measured in m 3 / mol (more often in dm 3 / mol). The molar volume of a gas is equal to the ratio of the gas volume to its quantity:

It is clear that the value of V m depends on the conditions (temperature, pressure). To solve problems, it is necessary to remember the value of V m at normal conditions (n.o.) - atmospheric pressure (101.3 kPa) and ice melting temperature (0 0 C or 273.15 K).

Under normal conditions, V m \u003d 22.4 dm 3 / mol, or

in SI units 0.0224 m 3 / mol.

Second consequence

The densities of gases (or the masses of their equal volumes) are related to each other as the molar masses of gases.

This is evident from the following considerations. Let there be two portions of different gases of the same volume (the volumes are measured under the same conditions). Let's find their masses:

The ratio of their masses:

If using densities:

According to Avogadro's law n 1 \u003d n 2, from here:

The ratio of gas densities, equal to the ratio of molar masses, is called relative density of one gas over another ( D ). D is a dimensionless quantity.

Knowing D and the molar mass of one gas, it is easy to find the molar mass of another gas:

; M1 = M2 × D.

Examples

M (x) \u003d M (H 2) × D=2 × 8.5 = 17 g/mol

A gas with this molar mass is ammonia NH 3 .

The density of some gaseous hydrocarbon in air is two. Determine the molar mass of the hydrocarbon.

The average molar mass of air is 29 g/mol.

M(x) = M(air) × D=29 × 2 = 58 g/mol

A hydrocarbon with such a molar mass is C 4 H 10, butane.

It should be noted that gases with a molar mass less than 29 are lighter than air, more than 29 are heavier.

In computational problems, relative densities for nitrogen, oxygen, and other gases can be given. In this case, to find the molar mass, it is necessary to multiply the relative density by the molar mass, respectively, of nitrogen (28 g / mol), oxygen (32 g / mol), etc.

Avogadro's law is widely used in chemical calculations. Since for gases the volumes are proportional to the quantities of substances, the coefficients in the reaction equation, reflecting the quantities of reacting substances, are proportional to the volumes of interacting gases. Obviously, the volumes must be measured under the same conditions.

Example

What volume of oxygen is required to burn 2 dm 3 propane? Volumes are measured at n. y.

C 3 H 8 + 5O 2  3CO 2 + 4H 2 O.

It follows from Avogadro's law that equal volumes of different gases contain the same number of molecules, and, accordingly, the same number of moles of substances. Let the volume of propane be 1 dm 3. Then, from the reaction equation, for burning 1 dm 3 of propane, 5 dm 3 of oxygen will be required, and for 2 dm 3 (two liters) - 10 dm 3 O 2.

Lesson Objectives:

1) educational: to acquaint students with the Gay-Lussac law of volumetric ratios of gases; to teach to solve calculation problems, applying the studied law;

2) educational: educate careful attitude students to human environment air environment; also to cultivate self-confidence, the ability to work independently with educational material;

3) developing: to deepen students' ideas about substances in a gaseous state of aggregation; improve the ability to solve calculation problems if gaseous substances are involved in the reaction or are formed.

Design problem 2. Calculation of volume ratios of gases.

Type of lesson: lesson of studying and primary consolidation of new knowledge.

Forms of work: conversation on questions, teacher's story, work of students in groups (pairs), individual work of students.

During the classes.

I. Organization of the class for the lesson.

II. Main part.

1. Peer review homework(from the right side of the board):

Answers to exercise 187 (p. 134):

Answers to exercise 199a* (p. 145):

Simultaneously with checking homework, two students work at the blackboard on assignments on cards (Appendix 2):

Student 2. Make formulas of substances by their names:

a) 4 - methyl - 2 - pentene;

b) 3 - methyl - 1 - butyne.

2. Update basic knowledge students.

Conversation on:

1. In what states of aggregation can substances be?

There are three states of aggregation in which substances can be: solid, liquid and gaseous.

2. Describe the features of the gaseous state of matter.

Large distances between molecules or atoms of a substance (about 1800-2000 particle sizes), the volume of gaseous substances does not depend on the size of their molecules.

3. Name simple substances gases, indicate the position of the atoms that form them in periodic system chemical elements DI. Mendeleev.

Simple substances-gases: (hydrogen), Hydrogen No. 1; (nitrogen), Nitrogen No. 7; (oxygen), Oxygen No. 8; (fluorine), Fluor #9; (chlorine), Chlorine No. 17.

Write down from the technological map of the lesson (Appendix 1) in the notebook the formulas of gaseous substances studied in the school chemistry course:

Formulas of gaseous substances:

1) simple substances -

3) oxides of non-metals -

4) organic substances -

3. Learning new material.

During chemical reactions, the volume of reactants can change.

This occurs most often when gaseous starting materials are involved in the reaction or gaseous reaction products are formed.

The volume of any gas under constant physical conditions (temperature and pressure) is proportional to the mass of this gas.

It is rather problematic to weigh gaseous substances, therefore, in practical calculations of the mass of gases, it is more convenient to replace them with volumes.

Analyzing the results of experiments with gaseous substances, the French scientist Joseph Louis Gay-Lussac in 1808 formulated the law of volumetric ratios.

P.146, write down in the notebook from the textbook (or the technological map of the lesson) the formulation of the law of volumetric relations:

4. Consolidation of the studied material. Solving calculation problems (tasks 1 and 2 are written on the front side of the right turn, the remaining 3 tasks are printed on cards and solved on the spot in groups).

1. One student works at the board with explanations, the rest work on the spot.

Task 1. What volume of hydrogen is released during the decomposition of 7 liters of methane?

Answer: V() = 10.5 l

Three groups (pairs) of students receive a task to complete on the spot (Appendix 3, distribute one per group or pair), with a report at the blackboard:

I remind students that we get oxygen for breathing and burning fuel from the air, and at the same time carbon dioxide is released, the increased content of which in the atmosphere leads to a “greenhouse” effect.

2. One student works at the board with explanations, the rest work on the spot.

III. Conclusion.

Summing up the lesson (reflection), homework.

Reflection questions: 1) At the lesson I worked (how?)

2) I am (satisfied, no) with my work in the lesson

3) The lesson for me was (what?)

4) At the lesson I learned (about what?)

5) The material of the lesson was (understandable, incomprehensible) to me

6) My mood after the lesson ...

In today's lesson, we studied the law of Gay-Lussac volumetric ratios and learned how to solve calculation problems when gaseous starting materials take part in the reaction or gaseous reaction products are formed.

Write down your homework (the assignment is written on the back of the board or in technological map lesson):

§23(p.145-149), ex. 206, 207* (p. 149)

Annex 1

Technological map of the lesson.

Subject. Volume ratios of gases in chemical reactions.

1a. Checking homework (from the right side of the board)

Exercise 187 (p. 134),

Exercise 199a* (p. 145).

1b. Work at the blackboard with formulas (2 students receive tasks on cards).

2. Write in a notebook:

Formulas of gaseous substances:

1) simple substances -

2) volatile hydrogen compounds -

3) oxides of non-metals -

4) organic substances -

3. Write in a notebook:

p.146 The law of volumetric relations:

The volumes of gases that react and are formed in

the result of the reaction are related as small integers.

For example: methane at high temperature decomposes into acetylene and hydrogen

4. Solving calculation problems (problems 1 and 2 on the board, the rest - in groups).

Task 1. What volume of hydrogen is released during the decomposition of 7 liters of methane C?

Problem 2. Calculate the volumes of oxygen and air needed to burn 8 liters of ethane.

Group 1. What volume of carbon (IY) oxide is released during the complete combustion of 2.75 liters of ethylene.

Group 2. Calculate the volumes of oxygen and air required to burn 4.5 liters of butene.

Group 3. What volume of oxygen is needed for the complete combustion of 12 liters of butane.

5. Homework: §23(p.145-149), exercise 206, 207*(p.149).

Appendix 2

student 1.

Name the substances according to their formulas:

student 2.

Write formulas of substances according to their names:

a) 4 - methyl - 2 - pentene;

b) 3 - methyl - 1 - butyne.

Appendix 3

What volume of carbon (IY) oxide is released during the complete combustion of 2.75 liters of ethylene.

Calculate the volumes of oxygen and air required to burn 4.5 liters of butene.

What volume of oxygen is required for the complete combustion of 12 liters of butane.

Lesson objectives: to form students' knowledge of the law of volumetric ratios for gaseous substances using the example of chemical reactions of organic substances; to form the ability to apply the law of volumetric ratios for calculations according to chemical equations.

Type of lesson: the formation of new skills and abilities.

Forms of work: execution training exercises(practice by examples, guided and independent practice).

Equipment: task cards.

II. Checking homework. Updating of basic knowledge. Motivation for learning activities

1. Frontal conversation

1) Compare physical properties alkanes, alkenes and alkyniv.

2) What are the general Chemical properties hydrocarbons.

3) What reactions (addition, substitution) are typical for alkanes? Why?

4) What reactions (addition, substitution) are typical for alkenes? Why?

5) From the substances given on the board, select those that decolorize bromine water. Give an example of reaction equations.

2. Checking homework

III. Learning new material

Frontal conversation on the material of grade 8

What is the molar volume of any gas under normal conditions?

All gases are equally compressed, have the same thermal expansion coefficient. The volumes of gases do not depend on the size of individual molecules, but on the distance between the molecules. The distances between molecules depend on the speed of their movement, energy and, accordingly, temperature.

Based on these laws and his research, the Italian scientist Amedeo Avogadro formulated the law:

Equal volumes of different gases contain the same number of molecules.

Under normal conditions, gaseous substances have a molecular structure. gas molecules are very small compared to the distance between them. Therefore, the volume of a gas is determined not by the size of particles (molecules), but by the distance between them, which is approximately the same for any gas.

A. Avogadro concluded that if we take 1 mol, that is, 6.02 1923 molecules of any gases, they will occupy the same volume. But at the same time, this volume is measured under the same conditions, that is, at the same temperature and pressure.

The conditions under which such calculations are carried out are called normal conditions.

Normal conditions (n. w.):

T = 273 K or t = 0 °C;

P = 101.3 kPa or P = 1 atm. = 760 mmHg Art.

The volume of 1 mol of a substance is called the molar volume (Vm). For gases under normal conditions, it is 22.4 l / mol.

According to Avogadro's law, 1 mole of any gas occupies the same volume under normal conditions equal to 22.4 l / mol.

Therefore, the volumes of gaseous reactants and reaction products are related as their coefficients in the reaction equation. This regularity is used for chemical calculations.

IV. Primary application of acquired knowledge

1. Practice with examples

Task 1. Calculate the amount of chlorine that can add 5 liters of ethylene.

Answer: 5 liters of chlorine.

Task 2. Calculate how much oxygen is needed to burn 1 m3 of methane.

Answer: 2 m3 of oxygen.

Task 3. Calculate the volume of acetylene, for the complete hydrogenation of which 20 liters of hydrogen were spent.

Answer: 10 liters of acetylene.

2. Guided practice

Task 4. Calculate the volume of oxygen required to burn 40 liters of a mixture containing 20% ​​methane, 40% ethane and 40% etene.

Answer: 104 liters of oxygen.

3. Independent practice

Task 5. Calculate the volume of hydrogen that will be required for the complete hydrogenation of substance X.

(Students fill in the table on their own, after finishing the work they check the answers.)

Task 6. Calculate the volume of air (oxygen content is assumed to be 20% by volume), which will be consumed for complete combustion of the mixture.

(Students independently solve one or two tasks for assessment on the instructions of the teacher.)