Introduction 2
1.Avogadro's Law 3
2. Gas laws 6
3. Consequences from Avogadro’s law 7
4.Problems on Avogadro's law 8
Conclusion 11
References 12
Introduction
Anticipate the results of the experiment, feel general beginning, predict a pattern - this marks the creativity of many scientists. Most often, forecasting extends only to the area in which the researcher is engaged, and not everyone has the determination to bravely step far forward in their predictions. Sometimes courage can give the ability to reason logically.
1.Avogadro's Law
In 1808, Gay-Lussac (together with the German naturalist Alexander Humboldt) formulated the so-called law of volumetric relations, according to which the relationship between the volumes of reacting gases is expressed in simple integers. For example, 2 volumes of hydrogen combine with 1 volume of hydrogen to produce 2 volumes of water vapor; 1 volume of chlorine combines with 1 volume of hydrogen, giving 2 volumes of hydrogen chloride, etc. This law gave scientists little benefit at that time, since there was no consensus about what particles of different gases are made of. There was no clear distinction between such concepts as atom, molecule, corpuscle.
In 1811, Avogadro, having carefully analyzed the results of experiments by Gay-Lussac and other scientists, came to the conclusion that the law of volumetric relations allows us to understand how gas molecules are “structured.” “The first hypothesis,” he wrote, “which arises in connection with this and which seems to be the only acceptable one, is the assumption that the number of constituent molecules of any gas is always the same in the same volume...” And “composite molecules "(now we simply call them molecules), according to Avogadro, consist of smaller particles - atoms.
Three years later, Avogadro stated his hypothesis even more clearly and formulated it in the form of a law that bears his name: “Equal volumes of gaseous substances at the same pressure and temperature contain the same number of molecules, so that the density of different gases serves as a measure of the mass of their molecules ..." This addition was very important: it meant that by measuring the density of different gases, it was possible to determine the relative masses of the molecules of which these gases consist. Indeed, if 1 liter of hydrogen contains the same number of molecules as 1 liter of oxygen, then the ratio of the densities of these gases is equal to the ratio of the masses of the molecules. Avogadro emphasized that molecules in gases do not necessarily have to consist of single atoms, but can contain several atoms - identical or different. (In fairness, it should be said that in 1814 the famous French physicist A.M. Ampere, independently of Avogadro, came to the same conclusions.)
In Avogadro's time, his hypothesis could not be proven theoretically. But this hypothesis provided a simple opportunity to experimentally determine the composition of the molecules of gaseous compounds and determine their relative mass. Let's try to trace the logic of such reasoning. The experiment shows that the volumes of hydrogen, oxygen and water vapor formed from these gases are in the ratio 2:1:2. Different conclusions can be drawn from this fact. First: hydrogen and oxygen molecules consist of two atoms (H 2 and O 2), and a water molecule consists of three, and then the equation 2H 2 + O 2 → 2H 2 O is true. But the following conclusion is also possible: hydrogen molecules are monatomic, and oxygen and water molecules are diatomic, and then the equation 2H + O 2 → 2HO with the same volume ratio 2:1:2 is true. In the first case, from the ratio of the masses of hydrogen and oxygen in water (1:8) it followed that the relative atomic mass of oxygen was equal to 16, and in the second - that it was equal to 8. By the way, even 50 years after Gay-Lussac’s work, some scientists continued to insist on the fact that the formula of water is HO, and not H 2 O. Others believed that the correct formula was H 2 O 2. Accordingly, in a number of tables the atomic mass of oxygen was taken equal to 8.
However, there was a simple way to choose the correct one from two assumptions. To do this, it was only necessary to analyze the results of other similar experiments. Thus, it followed from them that equal volumes of hydrogen and chlorine give twice the volume of hydrogen chloride. This fact immediately rejected the possibility of hydrogen being monoatomic: reactions such as H + Cl → HCl, H + Cl 2 → HCl 2 and the like do not produce a double volume of HCl. Therefore, hydrogen molecules (and also chlorine) consist of two atoms. But if hydrogen molecules are diatomic, then oxygen molecules are also diatomic, and water molecules have three atoms, and its formula is H 2 O. It is surprising that such simple arguments for decades could not convince some chemists of the validity of Avogadro’s theory, which for several remained virtually unnoticed for decades.
This is partly due to the lack in those days of a simple and clear recording of formulas and equations of chemical reactions. But the main thing is that the opponent of Avogadro’s theory was the famous Swedish chemist Jens Jakob Berzelius, who had unquestioned authority among chemists all over the world. According to his theory, all atoms have electrical charges, and molecules are formed by atoms with opposite charges that attract each other. It was believed that oxygen atoms have a strong negative charge, and hydrogen atoms have a positive charge. From the point of view of this theory, it was impossible to imagine an oxygen molecule consisting of two equally charged atoms! But if oxygen molecules are monatomic, then in the reaction of oxygen with nitrogen: N + O → NO the volume ratio should be 1:1:1. And this contradicted the experiment: 1 liter of nitrogen and 1 liter of oxygen gave 2 liters of NO. On this basis, Berzelius and most other chemists rejected Avogadro's hypothesis as inconsistent with experimental data!
The young Italian chemist Stanislao Cannizzaro (1826–1910) revived Avogadro's hypothesis and convinced chemists of its validity in the late 1850s. He accepted the correct (double) formulas for the molecules of gaseous elements: H 2, O 2, Cl 2, Br 2, etc. and reconciled Avogadro's hypothesis with all experimental data. “The cornerstone of modern atomic theory,” wrote Cannizzaro, “is the theory of Avogadro... This theory represents the most logical starting point for the explanation of the basic ideas about molecules and atoms and for the proof of the latter... At first it seemed that physical facts were in disagreement with the theory of Avogadro and Ampere, so that it was left aside and soon forgotten; but then the chemists, by the very logic of their research and as a result of the spontaneous evolution of science, imperceptibly for them, were led to the same theory... Who does not see in this long and unconscious whirling of science around and in the direction of the set goal a decisive proof in favor of the theory of Avogadro and Ampere? A theory that was arrived at starting from different and even opposite points, a theory that made it possible to foresee many facts confirmed by experience, must be something more than a simple scientific invention. It must be... the truth itself."
D.I. Mendeleev wrote about the heated discussions of that time: “In the 50s, some took O = 8, others O = 16, if H = 1. Water for the first was HO, hydrogen peroxide HO 2, for the second, as now , water H 2 O, hydrogen peroxide H 2 O 2 or H O. Confusion and confusion reigned. In 1860, chemists from all over the world gathered in Karlsruhe in order to reach agreement and uniformity at a congress. Having been present at this congress, I remember well how great the disagreement was, how the conditional agreement was guarded with the greatest dignity by the luminaries of science, and how then the followers of Gerard, led by the Italian professor Cannizzaro, ardently pursued the consequences of Avogadro’s law.”
After Avogadro's hypothesis became generally accepted, scientists were able not only to correctly determine the composition of the molecules of gaseous compounds, but also to calculate atomic and molecular masses. This knowledge helped to easily calculate the mass ratios of reagents in chemical reactions. Such relationships were very convenient: by measuring the mass of substances in grams, scientists seemed to be operating with molecules. An amount of a substance numerically equal to the relative molecular mass, but expressed in grams, was called a gram molecule or mole (the word “mole” was coined at the beginning of the 20th century by the German physical chemist Nobel Prize laureate Wilhelm Ostwald (1853–1932); it contains the same the root is the same as the word “molecule” and comes from the Latin moles - bulk, mass with a diminutive suffix). The volume of one mole of a substance in a gaseous state was also measured: under normal conditions (i.e. at a pressure of 1 atm = 1.013 10 5 Pa and a temperature of 0°C) it is equal to 22.4 liters (provided that the gas close to ideal). The number of molecules in one mole began to be called Avogadro's constant (it is usually denoted N A). This definition of mole persisted for almost a century.
Currently, a mole is defined differently: it is the amount of a substance containing the same amount structural elements(this could be atoms, molecules, ions or other particles) how many are in 0.012 kg of carbon-12. In 1971, by decision of the 14th General Conference on Weights and Measures, the mole was introduced into the International System of Units (SI) as the 7th base unit.
Even in Cannizzaro's time it was obvious that since atoms and molecules are very small and no one had ever seen them, Avogadro's constant must be very large. Over time, they learned to determine the size of molecules and the value N A - at first very roughly, then more and more precisely. First of all, they understood that both quantities are related to each other: the smaller the atoms and molecules, the larger Avogadro’s number. The size of atoms was first assessed by the German physicist Joseph Loschmidt (1821–1895). Based on the molecular kinetic theory of gases and experimental data on the increase in the volume of liquids during their evaporation, in 1865 he calculated the diameter of the nitrogen molecule. He came up with 0.969 nm (1 nanometer is a billionth of a meter), or, as Loschmidt wrote, “the diameter of an air molecule is rounded equal to one millionth of a millimeter.” This is approximately three times the modern value, which was a good result for that time. Loschmidt's second article, published in the same year, also gives the number of molecules in 1 cm 3 of gas, which has since been called the Loschmidt constant ( N L). It is easy to get the value from it N A, multiplied by the molar volume of an ideal gas (22.4 l/mol).
Avogadro's constant has been determined by many methods. For example, from the blue color of the sky it follows that sunlight dissipates in the air. As Rayleigh showed, the intensity of light scattering depends on the number of air molecules per unit volume. By measuring the ratio of the intensities of direct sunlight and scattered light from the blue sky, Avogadro's constant can be determined. For the first time, such measurements were carried out by the Italian mathematician and prominent political figure Quintino Sella (1827–1884) on the top of Monte Rosa (4634 m), in southern Switzerland. Calculations made on the basis of these and similar measurements showed that 1 mole contains approximately 6·10 23 particles.
Another method was used by the French scientist Jean Perrin (1870–1942). Under a microscope, he counted the number of tiny (about 1 micron in diameter) balls of gum, a substance related to rubber and obtained from the sap of some tropical trees, suspended in water. Perrin believed that the same laws that govern gas molecules apply to these balls. In this case, it is possible to determine the “molar mass” of these balls; and knowing the mass of an individual ball (unlike the mass of real molecules, it can be measured), it was easy to calculate Avogadro’s constant. Perrin got about 6.8 10 23.
The modern meaning of this constant N A = 6.0221367·10 23.
Avogadro's constant is so large that it is difficult to imagine. For example, if a soccer ball is enlarged by N And since it’s in volume, the globe will fit in it. If in N And if you increase the diameter of the ball, then the largest galaxy containing hundreds of billions of stars will fit in it! If you pour a glass of water into the sea and wait until this water is evenly distributed over all seas and oceans, to their very bottom, then, scooping up a glass of water anywhere on the globe, several dozen molecules of water that were once there will certainly fall into it. in a glass. If you take a mole of dollar bills, they will cover all continents with a 2-kilometer dense layer...
2. Gas laws
The relationship between pressure and volume of an ideal gas at constant temperature is shown in Fig. 1.
The pressure and volume of a gas sample are inversely proportional, i.e. their products are a constant value: pV = const. This relationship can be written in a form more convenient for solving problems:
p1V1 = p2V2 (Boyle-Mariotte law).
Let's imagine that 50 liters of gas (V1), under a pressure of 2 atm (p1), are compressed to a volume of 25 liters (V2), then its new pressure will be equal to:
Z
The dependence of the properties of ideal gases on temperature is determined by the Gay-Lussac law: the volume of a gas is directly proportional to its absolute temperature (at constant mass: V = kT, where k is the proportionality coefficient). This ratio is usually written in more convenient form to solve problems:
For example, if 100 liters of gas at a temperature of 300K are heated to 400K without changing the pressure, then at a higher temperature the new volume of gas will be equal to
Z
the writing of the combined gas law pV/T= = const can be transformed into the Mendeleev-Clapeyron equation:
where R is the universal gas constant, a is the number of moles of gas.
U
The Mendeleev-Clapeyron equation allows for a wide variety of calculations. For example, you can determine the number of moles of gas at a pressure of 3 atm and a temperature of 400 K, occupying a volume of 70 l:
One of the consequences of the unified gas law: Equal volumes of different gases at the same temperature and pressure contain the same number of molecules. This is Avogadro's law.
An important corollary also follows from Avogadro’s law: the masses of two identical volumes of different gases (naturally, at the same pressure and temperature) are related as their molecular masses:
m1/m2 = M1/M2 (m1 and m2 are the masses of the two gases);
M1IM2 represents relative density.
Avogadro's law applies only to ideal gases. Under normal conditions, gases that are difficult to compress (hydrogen, helium, nitrogen, neon, argon) can be considered ideal. For carbon monoxide (IV), ammonia, sulfur oxide (IV), deviations from ideality are observed already under normal conditions and increase with increasing pressure and decreasing temperature.
3. Consequences from Avogadro's law
4.Problems on Avogadro's law
Problem 1
At 25 °C and a pressure of 99.3 kPa (745 mm Hg), a certain gas occupies a volume of 152 cm3. Find what volume the same gas will occupy at 0 °C and a pressure of 101.33 kPa?
Solution
Substituting the problem data into equation (*) we get:
Vo = PVTo / TPo = 99.3*152*273 / 101.33*298 = 136.5 cm3.
Problem 2
Express the mass of one CO2 molecule in grams.
Solution
The molecular weight of CO2 is 44.0 amu. Therefore, the molar mass of CO2 is 44.0 g/mol. 1 mole of CO2 contains 6.02*1023 molecules. From here we find the mass of one molecule: m = 44.0 / 6.02-1023 = 7.31 * 10-23 g.
Task 3
Determine the volume that nitrogen weighing 5.25 g will occupy at 26 °C and a pressure of 98.9 kPa (742 mm Hg).
Solution
Determine the amount of N2 contained in 5.25 g: 5.25 / 28 = 0.1875 mol,
V, = 0.1875*22.4 = 4.20 dm3. Then we bring the resulting volume to the conditions specified in the problem: V = PoVoT / PTo = 101.3 * 4.20 * 299 / 98.9 * 273 = 4.71 dm3.
Problem 4
Carbon monoxide (“carbon monoxide”) is a dangerous air pollutant. It reduces the ability of blood hemoglobin to carry oxygen, causes diseases of the cardiovascular system, and reduces brain activity. Due to incomplete combustion of natural fuels, 500 million tons of CO are formed annually on Earth. Determine what volume (at normal conditions) will be occupied by carbon monoxide formed on Earth for this reason.
Solution
Let us write the problem condition in formula form:
m(CO) = 500 million tons = 5. 1014 g
M(CO) = 28 g/mol
VM = 22.4 l/mol (n.s.)
V(CO) = ? (Well.)
To solve the problem, equations are used that relate the amount of a substance, mass and molar mass:
m(CO) / M(CO) = n(CO),
as well as the amount of gaseous substance, its volume and molar volume:
V (CO) / VM = n(CO)
Therefore: m(CO) / M(CO) = V (CO) / VM, hence:
V(CO) = (VM . m(CO)) / M(CO) = (22.4 . 5 . 1014) / 28
[(l/mol) . g / (g/mol)] = 4 . 1014 l = 4. 1011 m3 = 400 km3
Problem 5
Calculate the volume occupied (at zero) by a portion of the gas required for breathing if this portion contains 2.69 . 1022 molecules of this gas. What gas is this?
Solution.
The gas needed for breathing is, of course, oxygen. To solve the problem, we first write its condition in formula form:
N(O2) = 2.69. 1022 (molecules)
VM = 22.4 l/mol (n.s.)
NA = 6.02. 1023 mol--1
V(O2) = ? (Well.)
To solve the problem, equations are used that relate the number of particles N(O2) in a given portion of a substance n(O2) and Avogadro’s number NA:
n(O2) = N(O2) / NA,
as well as the amount, volume and molar volume of the gaseous substance (n.s.):
n(O2) = V(O2) / VM
Hence: V(O2) = VM. n(O2) = (VM . N(O2)) / NA = (22.4 . 2.69 . 1022) : (6.02 . 1023) [(l/mol) : mol--1] = 1, 0 l
Answer. A portion of oxygen, which contains the number of molecules specified in the condition, occupies at no. volume 1 l.
Problem 6
Carbon dioxide with a volume of 1 liter under normal conditions has a mass of 1.977 g. What is the real volume of a mole of this gas (at normal conditions)? Explain your answer.
Solution
Molar mass M (CO2) = 44 g/mol, then volume of mole 44/1.977 = 22.12 (l). This value is less than that accepted for ideal gases (22.4 l). The decrease in volume is associated with an increase in the interaction between CO2 molecules, i.e., a deviation from ideality.
Problem 7
Gaseous chlorine weighing 0.01 g, located in a sealed ampoule with a volume of 10 cm3, is heated from 0 to 273oC. What is the initial pressure of chlorine at 0oC and at 273oC?
Solution
Mr(Cl2) =70.9; hence 0.01 g of chlorine corresponds to 1.4 10-4 mol. The volume of the ampoule is 0.01 l. Using the Mendeleev-Clapeyron equation pV=vRT, we find the initial pressure of chlorine (p1) at 0oC:
similarly we find the pressure of chlorine (p2) at 273oC: p2 = 0.62 atm.
Task 8
What is the volume occupied by 10 g of carbon monoxide (II) at a temperature of 15oC and a pressure of 790 mm Hg? Art.?
Solution
Problem 8
Firemine gas or CH 4 methane is a real disaster for miners. Its explosions in mines lead to great destruction and loss of life. G. Davy invented a safe miner's lamp. In it, the flame was surrounded by a copper mesh and did not escape beyond its limits, so the methane did not heat up to the ignition temperature. The victory over firedamp is considered a civil feat by G. Davy.
If the amount of methane substance at no. equals 23.88 moles, then what is the volume of this gas, calculated in liters?
Solution
V = 23.88 mol * 22.4 l/mol = 534.91 l
Problem 9
Anyone who has ever lit a match knows the smell of sulfur dioxide SO2. This gas is highly soluble in water: 42 liters of sulfur dioxide can be dissolved in 1 liter of water. Determine the mass of sulfur dioxide that can be dissolved in 10 liters of water.
Solution
ν = V/V m V=ν * V m m = ν * M
42 l SO 2 dissolves in 1 l water
x l SO 2 - in 10 l of water
x = 42* 10/1 = 420 l
ν = 420 l/ 22.4 l/mol = 18.75 mol
m = 18.75 mol * 64 g/mol = 1200 g
Problem 10
In an hour, an adult exhales approximately 40 g of carbon dioxide. Determine the volume (no.s.) of a given mass of this gas.
Solution
m = ν * M ν = m/M V=ν * V m
ν(CO 2) = 40 g / 44 g/mol = 0.91 mol
V(CO 2) = 0.91 mol * 22.4 l/mol = 20.38 l
Conclusion
Avogadro's merits as one of the founders of molecular theory have since received universal recognition. Avogadro's logic turned out to be impeccable, which was later confirmed by J. Maxwell with calculations based on the kinetic theory of gases; then experimental confirmation was obtained (for example, based on the study of Brownian motion), and it was also found how many particles are contained in a mole of each gas. This constant - 6.022 1023 - was called Avogadro's number, immortalizing the name of the insightful researcher.
References
Butskus P.F. Reading book on organic chemistry. Manual for students of 10th grade / comp. Butskus P.F. – 2nd. ed., revised. – M.: Education, 1985.
Bykov G.V. Amedeo Avogadro: Sketch of life and work. M.: Nauka, 1983
Glinka N.L. General chemistry. Uch. manual for universities. – L.: Chemistry, 1983.
Kritsman V.A. Robert Boyle, John Dalton, Amedeo Avogadro. The creators of molecular science in chemistry. M., 1976
Kuznetsov V.I. General chemistry. Development trends. – M.: Higher school.
Makarov K. A. Chemistry and health. Enlightenment, 1985.
Mario Liuzzi. History of physics. M., 1970
Poller Z. Chemistry on the way to the third millennium. Translation from German / translation and preface by Vasina N.A. – M.: Mir, 1982.
Avogadro's law, discovered in 1811, played a major role in the development of chemistry. First of all, he contributed to the recognition of the atomic-molecular doctrine, formulated for the first time in the middle of the 18th century. M.V. Lomonosov. So, for example, using Avogadro’s number:
It turned out to be possible to calculate not only the absolute masses of atoms and molecules, but also the actual linear dimensions of these particles. According to Avogadro's law:
“Equal volumes of different gases at constant pressure and temperature contain the same number of molecules, equal to”
A number of important consequences regarding the molar volume and density of gases follow from Avogadro’s law. Thus, it directly follows from Avogadro’s law that the same number of molecules of different gases will occupy the same volume, equal to 22.4 liters. This volume of gases is called molar volume. The opposite is also true - the molar volume of various gases is the same and equal to 22.4 liters:
Indeed, since 1 mole of any substance contains the same number of molecules, equal to , then obviously their volumes in the gaseous state under the same conditions will be the same. Thus, under normal conditions (n.s.), i.e. at pressure and temperature the molar volume of various gases will be 
. The amount of substance, volume and molar volume of gases can be related to each other in the general case by a relationship of the form:
from where, respectively:
In general, normal conditions (n.s.) are distinguished:
Standard conditions include:
In order to convert temperature on the Celsius scale to temperature on the Kelvin scale, use the following relationship:
The mass of the gas itself can be calculated from the value of its density, i.e.
Because as shown above:
then it's obvious:
from where, respectively:
From the above relations of the form:
after substitution into the expression:
it also follows that:
from where, respectively:
and thus we have:
Since under normal conditions 1 mole of anything occupies a volume equal to:
then accordingly:
The relationship obtained in this way is quite important for understanding the 2nd corollary of Avogadro’s law, which in turn is directly related to such a concept as the relative density of gases. In general, the relative density of gases is a value that shows how many times one gas is heavier or lighter than another, i.e. How many times is the density of one gas greater or less than the density of another, i.e. we have a relation of the form:
So, for the first gas we have:
respectively for the second gas:
then it's obvious:
and thus:
In other words, the relative density of a gas is the ratio of the molecular mass of the gas under study to the molecular mass of the gas with which the comparison is made. The relative density of a gas is a dimensionless quantity. Thus, in order to calculate the relative density of one gas from another, it is enough to know the relative molecular masses of these gases. In order to make it clear with which gas the comparison is being made, an index is given. For example, it means that a comparison is made with hydrogen and then they talk about the density of the gas in terms of hydrogen, without using the word “relative”, taking this as if by default. Measurements are carried out similarly, using air as a reference gas. In this case, indicate that the comparison of the gas under study is carried out with air. In this case, the average molecular mass of air is taken to be 29, and since the relative molecular mass and molar mass are numerically the same, then:
Chemical formula The gas being tested is placed side by side in parentheses, for example:
and reads as - the density of chlorine by hydrogen. Knowing the relative density of one gas in relation to another, it is possible to calculate the molecular as well as molar mass of the gas, even if the formula of the substance is unknown. All the above ratios refer to the so-called normal conditions.
The lesson is devoted to the study of Avogadro's law, which applies only to gaseous substances and allows you to compare the number of molecules in portions of gaseous substances. You will learn how, based on this law, you can draw a conclusion about the composition of gas molecules, and get acquainted with the models of molecules of some substances.
Topic: Initial chemical ideas
Lesson: Avogadro's Law. Composition of molecules
IN solids, in comparison with liquids and especially gases, particles of matter are in close interconnection, at short distances. In gaseous substances, the distances between molecules are so large that interaction between them is practically eliminated.
Rice. 1. Models of the structure of matter in different states of aggregation
In the absence of interaction between molecules, their individuality does not appear. This means that we can assume that the distances between molecules in any gases are the same. But provided that these gases are under the same conditions - at the same pressure and temperature.
Since the distances between gas molecules are equal, it means that equal volumes of gases contain equal number molecules. This assumption was made in 1811 by the Italian scientist Amedeo Avogadro. Subsequently, his assumption was proven and called Avogadro's law.
Avogadro used his hypothesis to explain the results of experiments with gaseous substances. In the process of reasoning, he was able to draw important conclusions about the composition of the molecules of certain substances.
Let's consider the results of experiments on the basis of which Avogadro was able to model the molecules of some substances.
You already know that when passed through water electric current, water decomposes into two gaseous substances - hydrogen and oxygen.
We will conduct an experiment on water decomposition in an electrolyzer. When an electric current is passed through water, gases will begin to be released at the electrodes, which will displace the water from the test tubes. The gases will turn out clean, because there is no air in the test tubes filled with water. Moreover, the volume of hydrogen released will be 2 times greater than the volume of oxygen released.
What conclusion did Avogadro draw from this? If the volume of hydrogen is twice the volume of oxygen, then there are also 2 times more hydrogen molecules formed. Therefore, in a water molecule there is one oxygen atom for every two hydrogen atoms.
Let us consider the results of other experiments that allow us to make assumptions about the structure of the molecules of substances. It is known that the decomposition of 2 liters of ammonia produces 1 liter of nitrogen and 3 liters of hydrogen (Fig. 2).
Rice. 2. Ratio of volumes of gases participating in the reaction
From this we can conclude that in an ammonia molecule there are three hydrogen atoms per nitrogen atom. But why then did the reaction require not 1 liter of ammonia, but 2 liters?
If we use the models of hydrogen and ammonia molecules proposed by D. Dalton, we get a result that contradicts experiment, because From 1 nitrogen atom and three hydrogen atoms you will get only 1 molecule of ammonia. Thus, according to Avogadro’s law, the volume of decomposed ammonia in this case will be equal to 1 liter.
Rice. 3. Explanation of the experimental results from the perspective of D. Dalton’s theory
If we assume that each molecule of hydrogen and nitrogen consists of two atoms, then the model will not contradict the experimental result. In this case, one nitrogen molecule and three hydrogen molecules are formed from two ammonia molecules.
Rice. 4. Model of the ammonia decomposition reaction
Let's consider the results of another experiment. It is known that when 1 liter of oxygen interacts with 2 liters of hydrogen, 2 liters of water vapor are formed (since the reaction is carried out at a temperature of more than 100 C). What conclusion can be drawn about the composition of the molecules of oxygen, hydrogen and water? This relationship can be explained if we assume that the molecules of hydrogen and oxygen consist of two atoms:
Rice. 5. Model of the reaction between hydrogen and oxygen
From two hydrogen molecules and 1 oxygen molecule, 2 water molecules are formed.
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.
2. Ushakova O.V. Workbook in chemistry: 8th grade: to the textbook 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. 26-27)
3. Chemistry: 8th grade: textbook. for general education institutions / P.A. Orzhekovsky, L.M. Meshcheryakova, L.S. Pontak. M.: AST: Astrel, 2005.(§11)
4. Encyclopedia for children. Volume 17. Chemistry / Chapter. ed.V.A. Volodin, Ved. scientific ed. I. Leenson. – M.: Avanta+, 2003.
Additional web resources
1. Unified collection of digital educational resources ().
2. Electronic version of the journal “Chemistry and Life” ().
Homework
1. p.67 No. 2 from the textbook “Chemistry: 8th grade” (P.A. Orzhekovsky, L.M. Meshcheryakova, L.S. Pontak. M.: AST: Astrel, 2005).
2. №45 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.
Amedeo Avogadro was one of the Italian physicists and chemists in the nineteenth century. It must be said that he received a legal education, but his desire for mathematics and physics prompted him to independently study these sciences. And in this matter he succeeded.
At the age of thirty, Avogadro became a physics teacher at one of the university lyceums of that time. He would later become a professor of mathematics at the university. However, Avogadro is not known at all for his successful career as a teacher of the exact sciences, which he mastered independently, he is known primarily as a scientist, and as a person who expressed one of the fundamental hypotheses physical chemistry. He suggested that if we take equal volumes of two different ideal gases at the same pressure and temperature, then these volumes will contain the same number of molecules. Subsequently, the hypothesis was confirmed, and today it can be proven using theoretical calculations. Today this rule is called Avogadro's law. In addition, a certain constant number was named after him, the so-called Avogadro number, which will be discussed below.
Avogadro's number
All substances consist of some kind of structural elements, as a rule, these are either molecules or atoms, but this is not important. What should happen when we mix two substances and they react? It is logical that one structural element, a brick, of one substance should react with one structural element, a brick, of another substance. Therefore, during a complete reaction, the number of elements for both substances should be the same, although the weight and volume of the preparations may differ. Thus, any chemical reaction must contain the same number of structural elements of each substance, or these numbers must be proportional to some number. The value of this number is completely unimportant, but later they decided to take twelve grams of carbon-12 as a basis and calculate the number of atoms in it. It's about six times ten to the twenty-third power. If a substance contains such a number of structural elements, then we speak of one mole of the substance. Accordingly, everything chemical reactions in theoretical calculations they are written in moles, that is, moles of substances are mixed.
As mentioned above, the value of Avogadro’s number is, in principle, unimportant, but it is determined physically. Since experiments on at the moment have insufficient accuracy, then given number is being clarified all the time. One can, of course, hope that someday it will be calculated absolutely accurately, but for now this is far from happening. To date, the last clarification was made in 2011. In addition, in the same year a resolution was adopted on how to correctly write this number. Since it is constantly being refined, today it is written as 6.02214X multiplied by ten to the twenty-third power. This number of structural elements is contained in one mole of a substance. The letter “X” in this entry indicates that the number is being specified, that is, the value of X will be specified in the future.
Avogadro's law
At the very beginning of this article we mentioned Avogadro's Law. This rule says that the number of molecules is the same. In this case, it makes sense to connect this law with Avogadro's number or mole. Then Avogadro's law will state that a mole of each ideal gas at the same temperature and pressure occupies the same volume. It is estimated that under normal conditions this volume is about twenty-four and a half liters. Eat exact value this figure is 22.41383 liters. And since the processes occurring under normal conditions are important and occur very often, there is a name for this volume, the molar volume of the gas.
In theoretical calculations, very often, molar volumes of gas are considered. If there is a need to move to other temperatures or pressure, then the volume, of course, will change, but there are corresponding formulas from physics that allow you to calculate it. You just have to always remember that a mole of gas always refers to normal conditions, that is, it is some specific temperature and some specific pressure, and according to the 1982 decree, under normal conditions, the gas pressure is ten to the fifth Pascal, and the temperature is 273.15 Kelvin .
In addition to the obvious practical significance of the two concepts that were discussed above, there are more interesting consequences that follow from them. So, knowing the density of water and taking one mole of it, we can estimate the size of the molecule. Here we assume that we know the atomic mass of water and carbon molecules. Thus, if we take twelve grams for carbon, then the mass of water is determined according to the proportional relationship, it is equal to eighteen grams. Since the density of water is easy to determine, the necessary data to estimate the size of a water molecule is now sufficient. Calculations show that the size of a water molecule is on the order of tenths of a nanometer.
Interesting and further development Avogadro's law. Thus, Van't Hoff extended the laws of ideal gases to solutions. The essence comes down to the analogy of laws, but in the end this made it possible to find out the molecular masses of substances that would be very difficult to obtain otherwise.
The study of the properties of gases allowed the Italian physicist A. Avogadro in 1811. put forward a hypothesis, which was subsequently confirmed by experimental data, and became known as Avogadro’s law: equal volumes of different gases under the same conditions (temperature and pressure) contain the same number of molecules.
An important corollary follows from Avogadro’s law: a mole of any gas under normal conditions (0C (273 K) and a pressure of 101.3 kPa ) occupies a volume of 22.4 liters. This volume contains 6.02 10 23 gas molecules (Avogadro's number).
It also follows from Avogadro’s law that the masses of equal volumes of different gases at the same temperature and pressure are related to each other as the molar masses of these gases:
where m 1 and m 2 are masses,
M 1 and M 2 are the molecular masses of the first and second gases.
Since the mass of a substance is determined by the formula
where ρ is the gas density,
V – volume of gas,
then the densities of various gases under the same conditions are proportional to their molar masses. The simplest method for determining the molar mass of substances in a gaseous state is based on this corollary of Avogadro’s law.
.
From this equation we can determine the molar mass of the gas:
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2.4 Law of volumetric relations
The first quantitative studies of reactions between gases belonged to the French scientist Gay-Lussac, the author of the famous law on the thermal expansion of gases. By measuring the volumes of gases that reacted and those formed as a result of reactions, Gay-Lussac came to a generalization known as the law of simple volume ratios: the volumes of gases that reacted relate to each other and the volumes of the resulting gaseous reaction products as small integers equal to their stoichiometric coefficients .
For example, 2H 2 + O 2 = 2H 2 O, when two volumes of hydrogen and one volume of oxygen interact, two volumes of water vapor are formed. The law is valid in the case when volume measurements were carried out at the same pressure and the same temperature.
2.5 Law of equivalents
The introduction into chemistry of the concepts of “equivalent” and “molar mass of equivalents” made it possible to formulate a law called the law of equivalents: The masses (volumes) of substances reacting with each other are proportional to the molar masses (volumes) of their equivalents .
It is worth dwelling on the concept of the volume of a mole of gas equivalents. As follows from Avogadro's law, a mole of any gas under normal conditions occupies a volume equal to 22,4 l. Accordingly, to calculate the volume of a mole of gas equivalents, it is necessary to know the number of moles of equivalents in one mole. Since one mole of hydrogen contains 2 moles of hydrogen equivalents, 1 mole of hydrogen equivalents occupies the volume under normal conditions:
3 Solving typical problems
3.1 Mol. Molar mass. Molar volume
Task 1. How many moles of iron (II) sulfide are contained in 8.8 g of FeS?
Solution Determine the molar mass (M) of iron (II) sulfide.
M(FeS)= 56 +32 = 8 8 g/mol
Let's calculate how many moles are contained in 8.8 g of FeS:
n = 8.8 ∕ 88 = 0.1 mol.
Task 2. How many molecules are there in 54 g of water? What is the mass of one water molecule?
Solution Determine the molar mass of water.
M(H 2 O) = 18 g/mol.
Therefore, 54 g of water contains 54/18 = 3 mol H 2 O. One mole of any substance contains 6.02 10 23 molecules. Then 3 moles (54g H 2 O) contain 6.02 10 23 3 = 18.06 10 23 molecules.
Let's determine the mass of one water molecule:
m H2O = 18 ∕ (6.02 10 23) = 2.99 10 23 g.
Task 3. How many moles and molecules are contained in 1 m 3 of any gas under normal conditions?
Solution 1 mole of any gas under normal conditions occupies a volume of 22.4 liters. Therefore, 1 m3 (1000 l) will contain 44.6 moles of gas:
n = 1000/ 22.4 = 44.6 mol.
1 mole of any gas contains 6.02 10 23 molecules. It follows from this that 1 m 3 of any gas under normal conditions contains
6.02 10 23 44.6 = 2.68 10 25 molecules.
Task 4. Express in moles:
a) 6.02 10 22 molecules C 2 H 2;
b) 1.80 10 24 nitrogen atoms;
c) 3.01 10 23 NH 3 molecules.
What is the molar mass of these substances?
Solution A mole is the amount of a substance that contains the number of particles of any certain type, equal to Avogadro's constant. From here
a)n C2H2 = 6.02 · 10 22 /6.02 · 10 23 = 0.1 mol;
b) n N = 1.8 · 10 24 / 6.02 · 10 23 = 3 moles;
c) n NH3 = 3.01 · 10 23 / 6.02 · 10 23 = 0.5 mol.
The molar mass of a substance in grams is numerically equal to its relative molecular (atomic) mass.
Therefore, the molar masses of these substances are equal:
a) M(C 2 H 2) = 26 g/mol;
b) M(N) = 14 g/mol;
c) M(NH 3) = 17 g/mol.
Task 5. Determine the molar mass of the gas if, under normal conditions, 0.824 g of it occupy a volume of 0.260 liters.
Solution Under normal conditions, 1 mole of any gas occupies a volume of 22.4 liters. By calculating the mass of 22.4 liters of this gas, we find out its molar mass.
0.824 g of gas occupies a volume of 0.260 l
X g of gas occupy a volume of 22.4 liters
X = 22.4 · 0.824 ∕ 0.260 = 71 g.
Therefore, the molar mass of the gas is 71 g/mol.
3.2 Equivalent. Equivalence factor. Molar mass equivalents
Task 1. Calculate the equivalent, equivalence factor and molar mass of H 3 PO 4 equivalents during exchange reactions that result in the formation of acidic and normal salts.
Solution Let us write down the reaction equations for the interaction of phosphoric acid with alkali:
H 3 PO 4 + NaOH = NaH 2 PO 4 + H 2 O; (1)
H 3 PO 4 + 2NaOH = Na 2 HPO 4 + 2H 2 O; (2)
H 3 PO 4 + 3NaOH = Na 3 PO 4 + 3H 2 O. (3)
Since phosphoric acid is a tribasic acid, it forms two acid salts (NaH 2 PO 4 - sodium dihydrogen phosphate and Na 2 HPO 4 - sodium hydrogen phosphate) and one middle salt (Na 3 PO 4 - sodium phosphate).
In reaction (1), phosphoric acid exchanges one hydrogen atom for the metal, i.e. behaves like a monobasic acid, therefore f e (H 3 PO 4) in reaction (1) is equal to 1; E(N 3 PO 4) = H 3 PO 4; M e (H 3 PO 4) = 1· M (H 3 PO 4) = 98 g/mol.
In reaction (2), phosphoric acid exchanges two hydrogen atoms for the metal, i.e. behaves like a dibasic acid, therefore f e (H 3 PO 4) in reaction (2) is equal to 1/2; E(N 3 PO 4) = 1/2H 3 PO 4; M e (H 3 PO 4) = 1/2 · M (H 3 PO 4) = 49 g/mol.
In reaction (3), phosphoric acid behaves like a tribasic acid, therefore f e (H 3 PO 4) in this reaction is equal to 1/3; E(N 3 PO 4) = 1/3H 3 PO 4; M e (H 3 PO 4) = 1/3 · M (H 3 PO 4) = 32.67 g/mol.
Problem 2. Excess potassium hydroxide was applied to solutions of: a) potassium dihydrogen phosphate; b) dihydroxobismuth (III) nitrate. Write equations for the reactions of these substances with KOH and determine their equivalents, equivalence factors and molar masses of equivalents.
Solution Let us write down the equations of the reactions occurring:
KN 2 RO 4 + 2KON = K 3 RO 4 + 2 H 2 O;
Bi(OH) 2 NO 3 + KOH = Bi(OH) 3 + KNO 3.
Various approaches can be used to determine the equivalent, equivalence factor, and molar mass equivalent.
The first is based on the fact that substances react in equivalent quantities.
Potassium dihydrogen phosphate reacts with two equivalents of potassium hydroxide, since E(KOH) = KOH. 1/2 KH 2 PO 4 interacts with one equivalent of KOH, therefore, E(KH 2 PO 4) = 1/2KH 2 PO 4 ; f e (KH 2 PO 4) = 1/2; Me (KH 2 PO 4) = 1/2 · M (KH 2 PO 4) = 68 g/mol.
Dihydroxobismuth (III) nitrate reacts with one equivalent of potassium hydroxide, therefore, E(Bi(OH) 2 NO 3) = Bi(OH) 2 NO 3 ; f e (Bi(OH) 2 NO 3) = 1; M e (Bi(OH) 2 NO 3) = 1 · M (Bi(OH) 2 NO 3) = 305 g/mol.
The second approach is based on the fact that the equivalence factor of a complex substance is equal to one divided by the equivalence number, i.e. the number of formed or restructured connections.
Potassium dihydrogen phosphate, when interacting with KOH, exchanges two hydrogen atoms for the metal, therefore, f e (KH 2 PO 4) = 1/2; E(KN 2 RO 4) = 1/2 KN 2 RO 4; M e (1/2 KN 2 PO 4) = 1/2 · M (KH 2 PO 4) = 68 g/mol.
Dihydroxobismuth (III) nitrate, when reacting with potassium hydroxide, exchanges one NO 3 – group, therefore, (Bi(OH) 2 NO 3) = 1; E(Bi(OH) 2 NO 3) = Bi(OH) 2 NO 3; Me (Bi(OH) 2 NO 3) = 1 · Me (Bi(OH) 2 NO 3) = 305 g/mol.
Task 3. The oxidation of 16.74 g of divalent metal produced 21.54 g of oxide. Calculate the molar masses of the equivalents of the metal and its oxide. What are the molar and atomic mass of the metal?
Rdecision According to the law of conservation of mass of substances, the mass of metal oxide formed during the oxidation of a metal with oxygen is equal to the sum of the masses of the metal and oxygen.
Consequently, the mass of oxygen required to form 21.5 g of oxide during the oxidation of 16.74 g of metal will be:
21.54 – 16.74 = 4.8 g.
According to the law of equivalents
m Me ∕ M e (Me) = mO 2 ∕ M e (O 2); 16.74 ∕ M e (Me) = 4.8 ∕ 8.
Therefore, M e (Me) = (16.74 8) ∕ 4.8 = 28 g/mol.
The molar mass of the oxide equivalent can be calculated as the sum of the molar masses of the metal and oxygen equivalents:
Me(MeO) = M e (Me) + M e (O 2) = 28 + 8 + 36 g/mol.
The molar mass of a divalent metal is:
M (Me) = Me (Me) ∕ fe(Me) = 28 ∕ 1 ∕ 2 = 56 g/mol.
The atomic mass of the metal (A r (Me)), expressed in amu, is numerically equal to the molar mass A r (Me) = 56 amu.
