Law of conservation of mass of reacting substances. The law of conservation of mass of substances

In 1748, MV Lomonosov (Russia) and in 1789 A. Lavoisier (France) independently discovered the law of conservation of the mass of substances in chemical reactions. This law is formulated as follows:

The mass of all substances that enter into a chemical reaction is equal to the mass of all the products of the reaction.

CH 4 + O 2 \u003d CO 2 + H 2 O

According to the law of conservation of mass:

m(CH 4) + m(O 2) = m(CO 2) + m(H 2 O),

where m(CH 4) and m(O 2) - the masses of methane and oxygen that reacted; m(CO 2) and m(H 2 O) - masses carbon dioxide and water formed as a result of the reaction.

The conservation of the mass of substances in chemical reactions is explained by the fact that the number of atoms of each element does not change before and after the reaction. During a chemical reaction, only the rearrangement of atoms occurs. In a reaction, for example, in the starting materials - CH 4 and O 2 - the carbon atom combines with hydrogen atoms, and oxygen atoms with each other; in the molecules of the reaction products - CO 2 and H 2 O - both the carbon atom and the hydrogen atoms are combined with oxygen atoms. It is easy to calculate that in order to preserve the number of atoms of each element, 1 molecule of CH 4 and 2 molecules of O 2 should enter into this reaction, and as a result of the reaction, 1 molecule of CO 2 and 2 molecules of H 2 O should be formed:

CH 4 + 2O 2 \u003d CO 2 + 2H 2 O

This expression is the equation chemical reaction, or chemical equation.

The numbers in front of the formulas of substances in the reaction equation are called coefficients. In the equation, the coefficients in front of the formulas O 2 and H 2 O are equal to 2; the coefficients in front of the formulas CH 4 and CO 2 are equal to 1 (they are usually not written down).

chemical equation- this is an expression of a chemical reaction, in which the formulas of the starting substances (reagents) and reaction products are written, as well as coefficients showing the number of molecules of each substance.

If the reaction scheme is known, then to draw up a chemical equation, you need to find the coefficients.

Let us compose, for example, the reaction equation, which is expressed by the following scheme:

Al + HCl \u003d AlCl 3 + H 2

In the left part of the diagram, the atoms and are part of the HCl molecule in the ratio 1:1; the right side of the scheme contains 3 chlorine atoms in the composition of the AlC1 3 molecule and 2 hydrogen atoms in the composition of the H 2 molecule. The least common multiple of 3 and 2 is 6.

We write the coefficient "6" before the formula HCl, the coefficient "2" - before the formula AlC1 3 and the coefficient "3" - before the formula H;

Al + 6HCl \u003d 2AlCl 3 + 3H 2

Since now there are 2 atoms on the right side, we write the coefficient "2" in front of the formula Al on the left side of the diagram:

2Al + 6HC1 = 2AlC1 3 + 3H 2

As a result, we have obtained the equation for this reaction. The coefficients in the chemical equation show not only the number of molecules, but also the number of moles of the starting materials and reaction products. For example, this equation shows that 2 moles of aluminum Al and 6 moles enter into the reaction, and as a result of the reaction, 2 moles of aluminum chloride AlC1 3 and 3 moles of hydrogen H 2) are formed.

Lesson Objectives:

1. Empirically prove and formulate the law of conservation of mass of substances.
2. Give the concept of a chemical equation as a conditional record of a chemical reaction using chemical formulas.

Lesson type: combined

Equipment: scales, beakers, mortar and pestle, porcelain cup, spirit lamp, matches, magnet.

Reagents: paraffin, CuSO solutions 4 , NaOH, HCl, phenolphthalein, iron and sulfur powders.

During the classes.

I. organizational stage.

II. Goal setting.Message about the topic and purpose of the lesson.

III. Checking homework.

Review questions:

1. How are physical phenomena different from chemical ones?

2. What are the applications physical phenomena you know?

3. What are the signs that a chemical reaction has taken place?

4. What are exothermic and endothermic reactions? What conditions are necessary for them to occur?

5. Students report the results of their home experiment (No. 1,2 after §26)

Exercise. Find a match

1 option - chemical phenomena, option 2 - physical:

1. melting paraffin
2. Rotting plant debris
3. Metal forging
4. Burning alcohol
5. Souring fruit juice
6. Dissolving sugar in water
7. blackening copper wire when calcined
8. freezing water
9. Souring milk
10. frost formation

IV. Knowledge introduction.

1. The law of conservation of mass of substances.

Problem question:whether the mass of the reactants will change compared to the mass of the reaction products.

Demonstration experiments:

The teacher puts two cups on the scales:

a) one with freshly precipitated Cu(OH) 2 , another with HCl solution; weighs them, pours the solutions into one glass, puts the other side by side, and the guys note that the balance of the weights has not been disturbed, although the reaction has passed, as evidenced by the dissolution of the precipitate;

b) similarly, the neutralization reaction is also carried out - an excess of acid from another glass is added to the alkali colored with phenolphthalein.

Video experiment:heating copper.

Description of the experiment:Place 2 grams of crushed copper into a conical flask. Close the flask tightly with a stopper and weigh. Remember the mass of the flask. Gently heat the flask for 5 minutes and observe the changes that occur. Stop heating, and when the flask has cooled, weigh it. Compare the mass of the flask before heating with the mass of the flask after heating.

Conclusion: The mass of the flask after heating did not change.

Wording mass conservation law:the mass of substances that have entered into the reaction is equal to the mass of the formed substances(students write the wording in a notebook).

The law of conservation of mass was theoretically discovered in 1748 and experimentally confirmed in 1756 by the Russian scientist M.V. Lomonosov.

The French scientist Antoine Lavoisier in 1789 finally convinced the scientific world of the universality of this law. Both Lomonosov and Lavoisier used very precise scales in their experiments. They heated metals (lead, tin, and mercury) in sealed vessels and weighed the starting materials and reaction products.

2. Chemical Equations.

Demo experiment:Heating a mixture of iron and sulfur.

Description of the experiment:In a mortar, prepare a mixture of 3.5 grams of Fe and 2 grams of S. Transfer this mixture to a porcelain cup and heat it strongly on a burner flame, observing the changes that occur. Bring the magnet to the formed substance.

The resulting substance - iron (II) sulfide - is different from the original mixture. Neither iron nor sulfur can be visually detected in it. It is impossible to separate them with a magnet. A chemical transformation has taken place.

Substances that take part in chemical reactions are called reagents.

New substances formed as a result of a chemical reaction are called products.

Let's write the reaction in the form of a diagram:

iron + sulfur → iron(II) sulfide

chemical equation- This is a conditional record of a chemical reaction through chemical formulas.

We write the ongoing reaction in the form of a chemical equation:

Fe + S → FeS

Rules for compiling chemical equations

(screen presentation).

1. On the left side of the equation, write down the formulas of the substances that enter into the reaction (reagents). Then put an arrow.

a) N 2 + H 2 →

B) Al(OH) 3 →

C) Mg + HCl →

D) CaO + HNO 3 →

2. On the right side (after the arrow) write down the formulas of the substances formed as a result of the reaction (products). All formulas are compiled in accordance with the degree of oxidation.

a) N 2 + H 2 → NH 3

B) Al (OH) 3 → Al 2 O 3 + H 2 O

C) Mg + HCl → MgCl 2 + H 2

D) CaO + HNO 3 → Ca(NO 3) 2 + H 2 O

3. The reaction equation is compiled on the basis of the law of conservation of mass of substances, that is, the left and right must have the same number of atoms. This is achieved by placing the coefficients in front of the formulas of substances.

Algorithm for the placement of coefficients in the equation of a chemical reaction.

2. Determine which element has a changing number of atoms, find N.O.K.

3. Split N.O.K. on indices – get coefficients. Put coefficients before formulas.

5. It is better to start with O atoms or any other non-metal (unless O is in the composition of several substances).

A) N 2 + 3H 2 → 2NH 3 b) 2Al (OH) 3 → Al 2 O 3 + 3H 2 O

C) Mg + 2HCl → MgCl 2 + H 2 d) CaO + 2HNO 3 → Ca(NO 3) 2 + H 2 O

v. Homework.§ 27 (up to types of reactions); No. 1 after §27

VI. Summary of the lesson. Students formulate conclusions about the lesson.

In lesson 11 "" from the course " Chemistry for dummies» we will find out by whom and when the law of conservation of mass of substances was discovered; we will get acquainted with chemical equations and learn how to correctly place the coefficients in them.

So far, when considering chemical reactions we paid attention to them quality side, i.e., on how and under what conditions the starting materials are converted into reaction products. But in chemical phenomena there is another side - quantitative.

Does the mass of substances that enter into a chemical reaction change? In search of an answer to this question, the English scientist R. Boyle back in the 17th century. conducted many experiments on calcining lead in sealed vessels. After the end of the experiments, he opened the vessels and weighed the reaction products. As a result, Boyle came to the conclusion that the mass of the substance after the reaction is greater than the mass of the original metal. He explained this by attaching some "fiery matter" to the metal.

The experiments of R. Boyle on the calcination of metals were repeated by the Russian scientist M.V. Lomonosov in 1748. He carried out the calcination of iron in a special flask (retort) (Fig. 56), which was hermetically sealed. Unlike Boyle, after the reaction, he left the retort sealed. Weighing the retort after the reaction showed that its mass did not change. This indicated that, although a chemical reaction had taken place between the metal and the substance contained in the air, the sum of the masses of the starting substances was equal to the mass of the reaction product.

M. V. Lomonosov concluded: “ All the changes that occur in nature are the essence of such a state that how much of what is taken away from one body, so much will be added to another, so if some matter decreases somewhere, it will multiply in another place.».

In 1789, the French chemist A. Lavoisier proved that the calcination of metals is the process of their interaction with one of constituent parts air - oxygen. Based on the works of M. V. Lomonosov and A. Lavoisier, the law of conservation of mass of substances in chemical reactions.

The mass of substances that entered into a chemical reaction is equal to the mass of substances formed as a result of the reaction.

In chemical reactions, atoms do not disappear without a trace and do not arise from nothing. Their number remains unchanged. And since they have constant mass , then the mass of the substances formed by them also remains constant.

The law of conservation of mass of substances can be verified experimentally. To do this, use the device shown in Figure 57, a, b. Its main part is a two-leg test tube. In one knee we pour lime water, in the second - a solution of copper sulphate. We balance the device on the scales, and then mix both solutions in one knee. At the same time, we will see that a blue precipitate of a new substance precipitates. The formation of a precipitate confirms that a chemical reaction has taken place. The mass of the device remains the same. This means that as a result of a chemical reaction, the mass of substances does not change.

The law is important for a correct understanding of everything that happens in nature: nothing can disappear without a trace and arise from nothing.

Chemical reactions can be represented using chemical formula language. Chemical elements represent chemical symbols, the composition of substances is written using chemical formulas, chemical reactions are expressed using chemical equations, i.e., just as words are made up of letters, sentences are made up of words.

Chemical reaction equation (chemical equation)- this is a conditional record of the reaction using chemical formulasand signs "+" and "=".

The law of conservation of mass of substances in chemical reactions must also be observed when compiling equations of chemical reactions. As in mathematical equations, in the equations of chemical reactions there is a left side (where the formulas of the starting substances are written) and a right side (where the formulas of the reaction products are written). For example (Fig. 58):

When writing the equations of chemical reactions, the “+” (plus) sign connects the formulas of substances in the left and right parts equations. Since the mass of substances before the reaction is equal to the mass of the formed substances, the sign "=" (equal) is used, which connects the left and right sides of the equation. To equalize the number of atoms in the left and right parts of the equation, the numbers in front of the formulas of substances are used. These numbers are called coefficients of chemical equations and show the number of molecules or formula units. Since 1 mole of any substance consists of the same number structural units(6.02 * 10 23), then the coefficients also show the chemical quantities of each of the substances:

When writing chemical equations, special signs are also used, for example, the sign "↓", indicating that the substance forms a precipitate.

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Preview:

Lesson topic: " Chemical equations. The law of conservation of mass of substances "

Lesson type: Discovery of new knowledge

The main objectives of the lesson:

1) Introduce students to the signs and conditions of chemical reactions

2) Empirically prove and formulate the law of conservation of mass of matter

3) Give the concept of a chemical equation as a conditional record of a chemical reaction using chemical formulas

4) Start building skills in writing chemical equations

Demonstration material and equipment:scales, beakers, reagents (CuSO 4, NaOH, HCl, CaCO 3 , phenolphthalein, Ba Cl 2, H 2 SO 4 ), computer, projector, screen, presentation)

During the classes

1. Self-determination to learning activities:

Target:

Create motivation for learning activities by updating internal motives (I can and want to)

Determine the content of the lesson with students

Organization educational process at stage 1

1. As we already know, chemistry is the science of substances. What do we already know about substances? Is this knowledge enough for us to answer all the questions that interest us? Can we answer the question of how the transformations of substances occur? What are the laws of chemical reactions? What do you think about today's lesson?
2. Right! Today we will go with you to wonderful world chemical transformations! And the knowledge gained earlier in chemistry lessons will help us in this.

2. Updating knowledge and fixing an individual difficulty in a trial action:

Target:

Review the material covered in the previous lesson

Organize independent execution trial action and fix the difficulties that have arisen

Organization of the educational process at stage 2

1. Earlier we learned that all phenomena in nature can be divided into two groups. What are these groups? Let's remember with you how some phenomena differ from others and give examples (slide)

One student at the blackboard completes the task. The game "Tic-tac-toe". You should indicate the winning path, which consists of only chemical phenomena (slide).

What is another word for chemical phenomena? (Chemical reactions)

Do we all know about chemical reactions? (Not)

1. Today in the lesson we will continue to study chemical reactions. I propose to begin our journey into the world of chemical transformations.
2. As you absolutely rightly pointed out, hallmark the course of a chemical reaction is the formation of a new substance -reaction product- possessing other properties that they did not possessstarting materials.
3. What is always accompanied by the formation of a new substance? (signs of a chemical reaction)
4. Now we will again need the knowledge gained earlier. Let's remember what signs of chemical reactions we already know and try to demonstrate them.

Together with the students, the teacher shows experiments in test tubes. Students name the observed signs that appear simultaneously on the slide.

Precipitation (CuSO 4 and NaOH)

Dissolution of the precipitate (Cu(OH) 2 and HCl)

Color change (NaOH and phenolphthalein)

Gas evolution (CaCO 3 and H 2 SO 4 )

Emission of heat, light (combustion reaction)

1. What conclusion can we draw from what we see? (The course of a chemical reaction can be judged by the appearance of external signs).
2. I suggest you reflect on a piece of paper one of the following chemical reactions. Describe what is happening in the test tube using chemical formulas and mathematical symbols.
3. Let's look at your entries, consider the options received. Why are there different options?

3. Identification of the place and cause of the difficulty and setting the goal of the activity

Target:

1. correlate the trial action with the existing knowledge, skills and abilities of students
2. agree on the topic and individual objectives of the lesson

Organization of the educational process at stage 3

1. 1) Let's see why not everyone managed to record a chemical reaction? How was this assignment different from previous ones?
2. 2) So, what are the goals of the lesson today?
3. Do you know the name of a record that reflects the essence of a chemical reaction?
4. How do we formulate the topic of today's lesson?

4. Building a project for getting out of a difficulty

Target:

1. create conditions for students to make a conscious choice of a new way of obtaining knowledge through an experiment

Organization of the educational process at stage 4

1. So, we can describe a chemical reaction using chemical formulas and signs if we know the mechanism for the transformation of one substance into another. To solve this problem, I propose to scientific discovery! And for this we will go to the distant 18th century, to the laboratory of the great Russian scientist M.V. Lomonosov (slide), who, like you and me, was puzzled by the same question: “How do some substances turn into others and what happens to the mass of substances? Will the mass of the starting materials be equal to the mass of the products of the reaction?
2. Tell me, how did we get new knowledge earlier? (We used a textbook, tables, presentations, etc.)
3. Is it possible to conduct an experiment to gain new knowledge? (Yes)

5. Implementation of the constructed project

Target:

Conduct an experiment to discover new knowledge

Summarize observations and draw preliminary conclusions

Organization of the educational process at stage 5

1. I propose to conduct an experiment: (the teacher invites the student to the laboratory table)
2. We put two cups on the weighing platform - one with a solution of BaCl 2 , another with solution H 2 SO 4 . Mark the position of the scale arrow with a marker. We merge the solutions into one glass, and put the empty one next to it.
3. Did the reaction proceed when the two solutions were combined? (Yes)
4. What testifies to this? (Formation of a white precipitate)
5. Did the readings of the instrument pointer change at the same time? (Not)
6. What conclusion can we draw? Does the mass of the resulting reaction products differ from the mass of the starting materials? (Not)
7. Lomonosov also came to this conclusion, who from 1748 to 1756 did a great job and experimentally proved that the mass of substances before and after the reaction remains unchanged. His experiments were based on the reaction of the interaction of metals with oxygen from the air during calcination. Now we will watch a video illustrating such an experiment. (slide video)

Guys, what conclusion can we draw now? (The mass of substances before the reaction is equal to the mass of substances after the reaction)

This statement is the law of conservation of mass of substances. (Formulation on the slide). Now we can clarify how the theme of our today's lesson will sound completely? (Chemical equations. The law of conservation of mass of substances)

Let's turn to the textbook (p. 139) and read out the formulation of the law of conservation of the mass of substances.

What happens to substances during a chemical reaction? Are new atoms formed chemical elements? (No, they are not formed. Only their rearrangement occurs!)

And if the number of atoms before and after the reaction remains unchanged, then their total weight is also unchanged. We will verify the validity of this conclusion by watching the video (slide animation)

Now, knowing the law of conservation of mass of substances, we can reflect the essence of chemical reactions using the chemical formulas of compounds.

Guys, how is it customary to call a conditional record of a chemical reaction using chemical formulas and mathematical signs? (Chemical equation) (slide)

Let's try to describe the experiment with calcining copper that was viewed in the video. (the student at the blackboard writes the reaction equation).

On the left side of the equation, we write down the starting substances (formulas of the substances that reacted). What substances interact? (Copper and oxygen). As we remember, the “AND” union in mathematics is replaced by a “plus” sign (we connect the starting substances with a “plus” sign). On the right side, we write down the reaction products. (Copper II oxide). We put an arrow between the parts:

Cu + O 2 \u003d CuO

That's how simple and beautiful. but ... disrespectful to the law of conservation of mass of substances. Is it observed in this case? (No!) Are the masses of substances equal before and after the reaction? (Not).

How many oxygen atoms are on the left side? (2) , but on the right? (one). Therefore, before the copper oxide formula, we must put 2! - equalize oxygen.

But .. Now the equality for copper is violated. Obviously, you must also put 2 in front of the copper formula.

Have we equalized the number of atoms of each element on the left and right sides? (Yes!)

Did you get equality? (Yes)

What is such a record called? (by chemical equation)

6. Primary consolidation with pronunciation during external speech:

Target:

Create conditions for fixing the studied material in external speech

- Let's practice writing the equations of a chemical reaction and try to make an algorithm of actions. (a student at the blackboard makes up an equation for a chemical reaction)

1. Let us write the reaction for the formation of ammonia from a molecule of hydrogen and nitrogen.

1. On the left side of the equation, we write down the formulas of the substances that have reacted (reagents). Then we put an arrow:

H 2 + N 2 →

1. On the right side (after the arrow) we write down the formulas of the substances formed as a result of the reaction (products).

H 2 + N 2 → NH 3

1. The reaction equation is based on the law of conservation of mass.
2. Determine which element has a changing number of atoms? find the least common multiple (LCM), divide the LCM by indices - we get the coefficients.
3. We put down the coefficients in front of the formulas of the compounds.
4. We recalculate the number of atoms, if necessary, repeat the steps.

3H 2 + N 2 → 2NH 3

6. Independent work with self-test according to the standard:

Target:

Encourage students to complete assignments on their own new way self-checking activities.

Organize children's self-assessment of the correctness of the assignment (if necessary, correction of possible errors)

Organization of the educational process at stage 6

1. Ready to try your hand? Then make your own equation for the chemical reaction of water formation, placing the missing coefficients in the equation

(slide animation) - an example of the formation of water.

(the initial substances are displayed on the screen - a hydrogen molecule and an oxygen molecule, then the reaction product appears - a water molecule)

Check (missing coefficients appear on the screen in the reaction equation)

Who is having trouble? What remains unclear?

7. Reflection of educational activity in the lesson

Target:

Fix in speech new terms (chemical reaction, chemical equation) and the formulation of the law of conservation of mass

Fix unresolved difficulties in the lesson as a direction for future learning activities

Evaluate your own activity in the lesson

Coordinate homework

Organization of the educational process at stage 7

What was today's lesson about? What was the theme of the lesson? What goals have we set and have we achieved them?

Where can we apply what we have learned today?

What difficulties arose? Did you manage to overcome them? What remained unclear?

Whose work in the lesson would you highlight? How do you rate your work?

Homework:

P. 27, ex. 1, 2. Exercises on cards (at the next lesson, students do a self-test according to the standard slide on the screen).

The law of conservation of mass.

The mass of substances entering into a chemical reaction is equal to the mass of substances formed as a result of the reaction.

The law of conservation of mass is a special case of the general law of nature - the law of conservation of matter and energy. Based on this law, chemical reactions can be displayed using chemical equations, using the chemical formulas of substances and stoichiometric coefficients that reflect the relative quantities (number of moles) of the substances involved in the reaction.

For example, the combustion reaction of methane is written as follows:

The law of conservation of mass of substances

(M.V. Lomonosov, 1748; A. Lavoisier, 1789)

The mass of all substances involved in a chemical reaction is equal to the mass of all the products of the reaction.

The atomic-molecular theory explains this law as follows: as a result of chemical reactions, atoms do not disappear and do not arise, but they are rearranged (i.e., a chemical transformation is the process of breaking some bonds between atoms and the formation of others, as a result of which from the molecules of the original substances, molecules of reaction products are obtained). Since the number of atoms before and after the reaction remains unchanged, their total mass should also not change. Mass was understood as a quantity characterizing the amount of matter.

At the beginning of the 20th century, the formulation of the law of conservation of mass was revised in connection with the advent of the theory of relativity (A. Einstein, 1905), according to which the mass of a body depends on its speed and, therefore, characterizes not only the amount of matter, but also its movement. The energy E received by the body is related to the increase in its mass m by the relation E = m c 2 , where c is the speed of light. This ratio is not used in chemical reactions, because 1 kJ of energy corresponds to a mass change of ~10 -11 g and m can hardly be measured. AT nuclear reactions, where Е is ~10 6 times greater than in chemical reactions, m should be taken into account.

Based on the law of conservation of mass, it is possible to draw up equations for chemical reactions and use them to make calculations. It is the basis of quantitative chemical analysis.

Law of constancy of composition

Composition constancy law ( J.L. Proust, 1801 -1808.) - any specific chemically pure compound, regardless of the method of its preparation, consists of the same chemical elements, and the ratios of their masses are constant, and relative numbers them atoms expressed as whole numbers. This is one of the fundamental laws chemistry.

The law of constancy of composition does not hold for berthollids(compounds of variable composition). However, conventionally, for simplicity, the composition of many berthollides is recorded as constant. For example, the composition iron(II) oxide is written as FeO (instead of the more precise formula Fe 1-x O).

Law of multiple ratios

The law of multiple ratios is one of stoichiometric laws chemistry: if two substances (simple or difficult) form more than one compound with each other, then the masses of one substance per the same mass of another substance are related as whole numbers, usually small.

Examples

1) The composition of nitrogen oxides (in percent by mass) is expressed next numbers:

Dividing the bottom row numbers by 0.57, we see that they are related as 1:2:3:4:5.

2) Calcium chloride forms with water 4 crystalline hydrate, the composition of which is expressed by the formulas: CaCl 2 H 2 O, CaCl 2 2H 2 O, CaCl 2 4H 2 O, CaCl 2 6H 2 O, i.e. in all these compounds, the mass of water per CaCl molecule 2 are related as 1:2:4:6.

Law of Volumetric Relations

(Gay-Lussac, 1808)

"The volumes of gases entering into chemical reactions and the volumes of gases formed as a result of the reaction are related to each other as small integers."

Consequence. Stoichiometric coefficients in the equations of chemical reactions for molecules of gaseous substances show in what volume ratios gaseous substances react or are obtained.

2CO + O 2  2CO 2

When two volumes of carbon monoxide (II) are oxidized with one volume of oxygen, 2 volumes of carbon dioxide are formed, i.e. the volume of the initial reaction mixture is reduced by 1 volume.

b) In the synthesis of ammonia from the elements:

n 2 + 3h 2  2nh 3

One volume of nitrogen reacts with three volumes of hydrogen; in this case, 2 volumes of ammonia are formed - the volume of the initial gaseous reaction mass will decrease by 2 times.

Klaiperon-Mendeleev equation

If we write the combined gas law for any mass of any gas, then we get the Claiperon-Mendeleev equation:

where m is the mass of gas; M is the molecular weight; p - pressure; V - volume; T - absolute temperature (°K); R is the universal gas constant (8.314 J / (mol K) or 0.082 l atm / (mol K)).

For a given mass of a particular gas, the m/M ratio is constant, so the combined gas law is derived from the Claiperon-Mendeleev equation.

What volume will take up at a temperature of 17 ° C and a pressure of 250 kPa carbon monoxide (II) weighing 84 g?

The number of moles of CO is:

 (CO) \u003d m (CO) / M (CO) \u003d 84 / 28 \u003d 3 mol

CO volume at n.c. is

3 22.4 l = 67.2 l

From the combined gas law of Boyle-Mariotte and Gay-Lussac:

(P V) / T = (P 0 V 0) / T 2

V (CO) \u003d (P 0 T V 0) / (P T 0) \u003d (101.3 (273 + 17) 67.2) / (250 273) \u003d 28.93 l

The relative density of gases shows how many times 1 mole of one gas is heavier (or lighter) than 1 mole of another gas.

D A(B) = (B)  (A) = M (B) / M (A)

The average molecular weight of a mixture of gases is equal to the total mass of the mixture divided by the total number of moles:

M cf \u003d (m 1 + .... + m n) / ( 1 + .... +  n) \u003d (M 1 V 1 + .... M n V n) / ( 1 + .. .. +  n)

LAW OF ENERGY CONSERVATION : in isolation. the energy of the system remains constant, only transitions of one type of energy into another are possible. In the thermodynamics of energy conservation, the law corresponds to the first law of thermodynamics, which is expressed by the equation Q \u003d DU + W, where Q is the number of heat communicated to the system, DU is the change in ext. energy of the system, W is the work done by the system. A special case of the conservation of energy law is the Hessian law.

The concept of energy was revised in connection with the advent of the theory of relativity (A. Einstein, 1905): the total energy E is proportional to the mass m and is related to it by the relation E = mc2, where c is the speed of light. Therefore, the mass can be expressed in units of energy and formulate a more general law of conservation of mass and energy: in iso-lyre. In a system, the sum of masses and energy is constant, and only transformations in strictly equivalent ratios of some forms of energy into others and equivalently related changes in mass and energy are possible.

Law of Equivalents

substances interact with each other in amounts proportional to their equivalents. When solving some problems, it is more convenient to use a different formulation of this law: the masses (volumes) of substances reacting with each other are proportional to their equivalent masses (volumes).

equivalents: chemical elements combine with each other in strictly defined quantities corresponding to their equivalents. The mathematical expression of the law of equivalents has next view: where m1 and m2 are the masses of reacting or formed substances, m equiv (1) and m equiv (2) are the equivalent masses of these substances.

For example: a certain amount of metal, the equivalent mass of which is 28 g / mol, displaces 0.7 liters of hydrogen from the acid, measured at normal conditions. Determine the mass of the metal. Solution: knowing that the equivalent volume of hydrogen is 11.2 l / mol, it is a proportion: 28 g of metal are equivalent to 11.2 liters of hydrogen x g of metal are equivalent to 0.7 liters of hydrogen. Then x \u003d 0.7 * 28 / 11.2 \u003d 1.75 g.

To determine the equivalent or equivalent mass, it is not necessary to proceed from its combination with hydrogen. They can be determined by the composition of the compound of a given element with any other, the equivalent of which is known.

For example: when 5.6 g of iron was combined with sulfur, 8.8 g of iron sulfide was formed. It is necessary to find the equivalent mass of iron and its equivalent, if it is known that the equivalent mass of sulfur is 16 g/mol. Solution: from the conditions of the problem it follows that in iron sulfide, 5.6 g of iron accounts for 8.8-5.6 = 3.2 g of sulfur. According to the law of equivalents, the masses of interacting substances are proportional to their equivalent masses, that is, 5.6 g of iron is equivalent to 3.2 g of sulfur meq (Fe) is equivalent to 16 g/mol of sulfur. It follows from here that m3KB(Fe) = 5.6*16/3.2=28 g/mol. The iron equivalent is: 3=meq(Fe)/M(Fe)=28 g/mol:56 g/mol=1/2. Therefore, the equivalent of iron is 1/2 mole, that is, 1 mole of iron contains 2 equivalents.

Avogadro's Law

Consequences of the law

The first corollary of Avogadro's law: one mole of any gas under the same conditions occupies the same volume.

In particular, under normal conditions, i.e. at 0 ° C (273 K) and 101.3 kPa, the volume of 1 mole of gas is 22.4 liters. This volume is called the molar volume of gas V m . You can recalculate this value to other temperatures and pressures using the Mendeleev-Clapeyron equation:

.

The second corollary of Avogadro's law: the molar mass of the first gas is equal to the product of the molar mass of the second gas and the relative density of the first gas according to the second.

This position was of great importance for the development of chemistry, since it makes it possible to determine the partial weight of bodies capable of passing into a gaseous or vaporous state. If through m we denote the partial weight of the body, and through d is its specific gravity in the vapor state, then the ratio m / d should be constant for all bodies. Experience has shown that for all the bodies studied, passing into steam without decomposition, this constant is equal to 28.9, if, when determining the partial weight, we proceed from the specific gravity of air, taken as a unit, but this constant will be equal to 2, if we take the specific gravity of hydrogen as a unit. Denoting this constant, or, what is the same, the partial volume common to all vapors and gases through With, we have from the formula on the other hand m = dC. Since the specific gravity of steam is easily determined, then, substituting the value d in the formula, the unknown partial weight of the given body is also displayed.

Thermochemistry

Thermal effect of a chemical reaction

From Wikipedia, the free encyclopedia

Thermal effect of a chemical reaction or change enthalpy system due to the occurrence of a chemical reaction - the amount of heat related to the change in the chemical variable received by the system in which the chemical reaction took place and the reaction products took the temperature of the reactants.

In order for the thermal effect to be a quantity that depends only on the nature of the ongoing chemical reaction, the following conditions must be met:

The reaction must proceed either at a constant volume Q v (isochoric process), or at constant pressure Q p( isobaric process).

No work is done in the system, except for the expansion work that is possible with P = const.

If the reaction is carried out under standard conditions at T \u003d 298.15 K \u003d 25 ° C and P \u003d 1 atm \u003d 101325 Pa, the thermal effect is called the standard thermal effect of the reaction or the standard enthalpy of reaction Δ H r O . In thermochemistry, the standard thermal effect of a reaction is calculated using the standard enthalpies of formation.

Standard enthalpy of formation (standard heat of formation)

The standard heat of formation is understood as the heat effect of the reaction of formation of one mole of a substance from simple substances, its components, which are in stable standard states.

For example, the standard enthalpy of formation is 1 mol methane from carbon and hydrogen equal to the heat of the reaction:

C (tv) + 2H 2 (g) \u003d CH 4 (g) + 76 kJ / mol.

The standard enthalpy of formation is denoted Δ H fO . Here the index f means formation (education), and the crossed out circle, resembling the Plimsol disk - that the value refers to standard state substances. In the literature, another designation for the standard enthalpy is often found - ΔH 298,15 0 , where 0 indicates the pressure is equal to one atmosphere (or, somewhat more precisely, to the standard conditions ), and 298.15 is the temperature. Sometimes index 0 is used for quantities related to pure substance, stipulating that it is possible to designate standard thermodynamic quantities with it only when it is a pure substance that is chosen as the standard state . The standard can also be taken, for example, the state of matter in extremely dilute solution. "Plimsol disk" in this case means the actual standard state of matter, regardless of its choice.

The enthalpy of formation of simple substances is assumed to be zero, and the zero value of the enthalpy of formation refers to the state of aggregation, which is stable at T = 298 K. For example, for iodine in the crystalline state Δ H I2(tv) 0 = 0 kJ/mol, and for liquid iodine Δ H I2(l) 0 = 22 kJ/mol. The enthalpies of formation of simple substances under standard conditions are their main energy characteristics.

The thermal effect of any reaction is found as the difference between the sum of the heats of formation of all products and the sum of the heats of formation of all reactants in this reaction (corollary Hess' law):

Δ H reactions O = ΣΔ H f O (products) - ΣΔ H f O (reagents)

Thermochemical effects can be included in chemical reactions. Chemical equations in which the amount of released or absorbed heat is indicated are called thermochemical equations. Reactions accompanied by the release of heat into the environment have a negative thermal effect and are called exothermic. Reactions accompanied by the absorption of heat have a positive thermal effect and are called endothermic. The thermal effect usually refers to one mole of the reacted starting material, the stoichiometric coefficient of which is maximum.

Temperature dependence thermal effect(enthalpies) of reaction

To calculate the temperature dependence of the enthalpy of reaction, it is necessary to know the molar heat capacity substances involved in the reaction. The change in the enthalpy of the reaction with increasing temperature from T 1 to T 2 is calculated according to the Kirchhoff law (it is assumed that in this temperature range the molar heat capacities do not depend on temperature and there is no phase transformations):

If phase transformations occur in a given temperature range, then in the calculation it is necessary to take into account the heats of the corresponding transformations, as well as the change in the temperature dependence of the heat capacity of substances that have undergone such transformations:

where ΔC p (T 1 ,T f) is the change in heat capacity in the temperature range from T 1 to the phase transition temperature; ΔC p (T f ,T 2 ) is the change in heat capacity in the temperature range from the phase transition temperature to the final temperature, and T f is the phase transition temperature.

Standard enthalpy of combustion

Standard enthalpy of combustion - Δ H Gor o, the thermal effect of the reaction of combustion of one mole of a substance in oxygen to the formation of oxides in the highest degree oxidation. The heat of combustion of non-combustible substances is assumed to be zero.

Standard enthalpy of dissolution

Standard enthalpy of dissolution - Δ H solution, the thermal effect of the process of dissolving 1 mole of a substance in an infinitely large amount of solvent. Composed of the heat of destruction crystal lattice and warmth hydration(or heat solvation for non-aqueous solutions), released as a result of the interaction of solvent molecules with molecules or ions of the dissolved substance with the formation of compounds of variable composition - hydrates (solvates). The destruction of the crystal lattice, as a rule, is an endothermic process - Δ H resh > 0, and ion hydration is exothermic, Δ H hydra< 0. В зависимости от соотношения значений ΔH resh and Δ H hydr enthalpy of dissolution can have both positive and negative meaning. So the dissolution of the crystalline potassium hydroxide accompanied by the release of heat

Δ H solution KOH o \u003d Δ H resh o + Δ H hydrK + o + Δ H hydroOH −o = −59 kJ/mol

Under the enthalpy of hydration - Δ H hydr, refers to the heat that is released during the transition of 1 mole of ions from vacuum to solution.

Standard enthalpy of neutralization

Standard enthalpy of neutralization - Δ H neutral about the enthalpy of the reaction of interaction of strong acids and bases with the formation of 1 mole of water under standard conditions:

HCl + NaOH = NaCl + H2O

H + + OH - \u003d H 2 O, ΔH neutral ° \u003d -55.9 kJ / mol

Standard enthalpy of neutralization for concentrated solutions strong electrolytes depends on the concentration of ions, due to the change in the value of ΔH hydration ° ions when diluted.

Enthalpy

Enthalpy is a property of matter that indicates the amount of energy that can be converted into heat.

Enthalpy is a thermodynamic property of a substance that indicates the level of energy stored in its molecular structure. This means that while matter can have energy based on temperature and pressure, not all of it can be converted to heat. Part of the internal energy always remains in the substance and maintains its molecular structure. Part kinetic energy substance is not available when its temperature approaches the ambient temperature. Therefore, enthalpy is the amount of energy that is available to be converted into heat at a certain temperature and pressure. Enthalpy units- British thermal unit or joule for energy and Btu/lbm or J/kg for specific energy.

Enthalpy quantity

Quantity enthalpy substance based on its given temperature. Given temperature is the value chosen by scientists and engineers as the basis for calculations. This is the temperature at which the enthalpy of a substance is zero J. In other words, the substance has no available energy that can be converted into heat. This temperature at various substances different. For example, this temperature of water is the triple point (0°C), nitrogen is -150°C, and refrigerants based on methane and ethane are -40°C.

If the temperature of a substance is above its given temperature, or changes state to gaseous at a given temperature, the enthalpy is expressed as a positive number. Conversely, at a temperature below a given enthalpy of a substance is expressed as a negative number. Enthalpy is used in calculations to determine the difference in energy levels between two states. This is necessary to set up the equipment and determine coefficient usefulness of the process.

Enthalpy is often defined as the total energy of matter, since it is equal to the sum of its internal energy (u) in given state along with his ability to get the job done (pv). But in reality, enthalpy does not indicate full energy substances at a given temperature above absolute zero (-273°C). Therefore, rather than defining enthalpy as the total heat of a substance, it is more accurate to define it as the total amount of available energy of a substance that can be converted to heat. H=U+pV

Internal energy

The internal energy of a body (denoted as E or U) is the sum of the energies of molecular interactions and thermal motions of a molecule. The internal energy is a single-valued function of the state of the system. This means that whenever the system is in a given state, its internal energy takes on the value inherent in this state, regardless of the history of the system. Consequently, the change in internal energy during the transition from one state to another will always be equal to the difference between its values ​​in the final and initial states, regardless of the path along which the transition was made.

The internal energy of a body cannot be measured directly. Only the change in internal energy can be determined:

Attached to the body heat, measured in joules

- Job, performed by the body against external forces, measured in joules

This formula is a mathematical expression first law of thermodynamics

For quasi-static processes the following relation holds:

-temperature, measured in kelvins

-entropy, measured in joules/kelvin

-pressure, measured in Pascals

-chemical potential

Number of particles in the system

Ideal gases

According to Joule's law, derived empirically, the internal energy ideal gas independent of pressure or volume. Based on this fact, one can obtain an expression for the change in the internal energy of an ideal gas. A-priory molar heat capacity at a constant volume . Since the internal energy of an ideal gas is a function of temperature only, then

.

The same formula is also true for calculating the change in the internal energy of any body, but only in processes with a constant volume ( isochoric processes); in general case C V (T,V) is a function of both temperature and volume.

If we neglect the change in molar heat capacity with a change in temperature, we get:

Δ U = ν C V Δ T,

where ν is the amount of substance, Δ T- temperature change.

INTERNAL ENERGY OF A SUBSTANCE, BODY, SYSTEM

(Greek: ένέργια - activity, energy). Internal energy is part total body energy (systems tel): E = E k + E p + U, where E k - kinetic energy macroscopic movements systems, E p - potential energy, due to the presence of external force fields(gravitational, electric, etc.), U- internal energy. Internal energy substances, bodies, systems of bodies - function states, defined as the total energy reserve of the internal state of a substance, body, system, changing (released) in process chemical reactions, heat transfer and performance work. Components of internal energy: (a) kinetic energy of thermal probabilistic movement of particles (atoms, molecules, ions etc.), constituting a substance (body, system); (b) potential energy of particles due to their intermolecular interaction; (c) energy of electrons in electron shells, atoms and ions; (d) intranuclear energy. Internal energy is not related to the process of changing the state of the system. With any changes in the system, the internal energy of the system, together with its environment, remains constant. That is, internal energy is neither lost nor gained. At the same time, energy can move from one part of the system to another or be transformed from one forms to another. This is one of the expressions law conservation of energy - the first law of thermodynamics. Part of the internal energy can be converted into work. This part of the internal energy is called free energy - G. (AT chemical compounds they call it chemical potential). The rest of the internal energy, which cannot be converted into work, is called bound energy - W b .

Entropy

Entropy (from Greekἐντροπία - turn, transformation) into natural sciences- measure of disorder systems, consisting of many elements. In particular, in statistical physics - measure probabilities realization of any macroscopic state; in information theory- a measure of the uncertainty of any experience (test), which can have different outcomes, and hence the number information; in historical science, for explications phenomenon alternative history (invariance and variability historical process).