Study questions

1. State of equilibrium

Equilibrium constant

Calculation of equilibrium concentrations

Chemical equilibrium shift. Le Chatelier's principle

1. State of equilibrium

Reactions proceeding under the same conditions simultaneously in opposite directions are called reversible.

Consider a reversible reaction that occurs in a closed system

The forward reaction rate is described by the equation:

pr = k pr [A] [B],

where pr is the speed of the direct reaction;

k pr is the rate constant of the direct reaction.

Concentrations of reagents over time A and V decrease, the reaction rate decreases (Fig. 1, curve etc).

The reaction between A and V leads to the formation of substances C and D, the molecules of which, upon collisions, can again give substances A and V.

The back reaction rate is described by the equation:

arr = k arr [C] [D],

where obr - the speed of the reverse reaction;

k obr - rate constant of the reverse reaction.

As the concentration of substances C and D increase, the rate of the reverse reaction increases (Fig. 1, curve arr).

Fig. 1. Change in the rates of forward and backward reactions in time

Over time the rates of forward and backward reactions become equal:

pr = arr

This state of the system is called state of equilibrium .

In a state of equilibrium, the concentrations of all its participants stop changing over time. . Such concentrations are called equilibrium .

Chemical equilibrium – it dynamic balance. The invariability of the concentrations of substances present in a closed system is a consequence of continuously running chemical processes. The rates of the forward and reverse reactions are not equal to zero, and the observed rate of the process is equal to zero.

The equality of the rates of forward and reverse reactions is the kinetic condition of chemical equilibrium.

2. Equilibrium constant

When the speeds of the forward and reverse reactions are equal

pr = arr

fair equality

k pr [A] [B] = k arr [C] [D],

where [ A], [B], [WITH], [D] - equilibrium concentrations of substances.

Since the rate constants do not depend on the concentration, the equality can be written differently:

The ratio of the rate constants of the forward and reverse reactions ( k NS / k arr ) called the constant of chemical equilibrium:

True chemical equilibrium can be established only if all the elementary stages of the reaction mechanism are in equilibrium. No matter how complex the mechanisms of direct and reverse reactions are, in a state of equilibrium they should provide a stoichiometric transition of the initial substances into the reaction products and vice versa. This means that the algebraic sum of all stages of the process is equal to the stoichiometric equation of the reaction, i.e. stoichiometric coefficients are the sum of the molecularities of all stages of the mechanism.

For a complex reaction

aA + bB  cC + dD

For the same temperature, the ratio of the product of the equilibrium concentrations of the reaction products in powers equal to the stoichiometric coefficients to the product of the equilibrium concentrations of the starting substances in powers equal to the stoichiometric coefficients is a constant value.

This is the second formulation of the law of the masses at work.

The expression for the equilibrium constant of a heterogeneous reaction includes only the concentrations of substances in the liquid or gaseous phase, since the concentrations of solids, as a rule, remain constant.

For example, the expression for the equilibrium constant of the following reaction

СО 2 (g) + С (tv)  2СО (g)

written like this:

TO c =
.

The equilibrium constant equation shows that under equilibrium conditions the concentrations of all substances participating in the reaction are related. The numerical value of the equilibrium constant determines what the ratio of the concentrations of all reacting substances should be at equilibrium.

A change in the concentration of any of these substances entails changes in the concentration of all other substances. As a result, new concentrations are established, but the ratio between them again corresponds to the equilibrium constant.

The value of the equilibrium constant depends on the nature of the reactants and the temperature.

Equilibrium constant expressed in terms of molar concentrations of reactants ( TOWith) and the equilibrium constant, expressed in terms of the equilibrium partial pressures ( TOR) (see "Fundamentals of Chemical Thermodynamics") are interconnected by the relations:

TOR= KWithRT  , Kc = KR / (RT)  ,

where  is the change in the number of gaseous moles in the reaction.

The standard change in Gibbs energy is

G Т = - RT ln Kp,

G T =  H – T  S.

After equating the right-hand sides of the equations:

- RT ln Kp =  H – T  S

ln K R = -  H / ( RT) +  S/ R .

The equation not only establishes the form of the dependence of the constant on temperature, but also shows that the constant is determined by the nature of the reacting substances.

The equilibrium constant does not depend on the concentration (as well as the reaction rate constant), the reaction mechanism, activation energy, and the presence of catalysts. Changing the mechanism, for example, with the introduction of a catalyst, does not affect the numerical value of the equilibrium constant, but, of course, changes the rate at which the equilibrium state is reached.

GOU VPO "Ural State Technical University - UPI"

Determination of the equilibrium constants of chemical

reactions and calculation of chemical equilibrium

on the course of physical chemistry

for full-time students

Yekaterinburg 2007

UDC 544 (076) S79

Compiled by

Scientific editor, Ph.D., associate professor

Determination of equilibrium constants of chemical reactions and calculation of chemical equilibrium: methodological instructions for laboratory work No. 4 in the course of physical chemistry / comp. - Yekaterinburg: GOU VPO USTU-UPI, 20s.

Methodological instructions are intended for additional in-depth study of the material on chemical equilibrium within the framework of the analytical and analytical laboratory work. Contains 15 options for individual tasks, which contributes to the achievement of the set goal.

Bibliography: 5 titles. Rice. Tab.

© GOU VPO "Ural State

Technical University - UPI ", 2007

Introduction

This work, although it is carried out within the framework of a laboratory workshop, refers to computational and analytical work and consists in mastering theoretical material and solving a number of problems on the topic of the course of physical chemistry "Chemical equilibrium".

The need for its implementation is caused by the complexity of this topic on the one hand and the insufficient amount of study time allocated for its study on the other.

The main part of the topic "Chemical equilibrium": the derivation of the law of chemical equilibrium, consideration of the isobar equation and isotherm of a chemical reaction, etc. is presented in lectures and studied in practical classes (therefore, this material is not presented in this work). In this manual, a section of the topic concerning the experimental determination of equilibrium constants and the determination of the equilibrium composition of a system with a chemical reaction occurring in it is considered in detail.

So, the implementation of this work by students will allow solving the following tasks:

1) get acquainted with the methods for determining and calculating the equilibrium constants of chemical reactions;

2) learn how to calculate the equilibrium composition of a mixture based on a variety of experimental data.

1. THEORETICAL INFORMATION ABOUT THE METHODS

DEFINITIONS OF EQUILIBRIUM CONSTANTS OF CHEMICAL REACTIONS

Let us dwell briefly on the basic concepts used below. The equilibrium constant of a chemical reaction is the quantity

https://pandia.ru/text/78/005/images/image002_169.gif "width =" 51 "height =" 29 "> is the standard molar Gibbs energy of the reaction r.

Equation (1) is a definitive equation for the equilibrium constant of a chemical reaction. It should be noted that the equilibrium constant of a chemical reaction is a dimensionless quantity.

The chemical equilibrium law is written as follows

, (2)

where https://pandia.ru/text/78/005/images/image005_99.gif "width =" 23 "height =" 25 "> - activity k- a participant in the reaction; - dimension of activity; stoichiometric coefficient k- participant in the reaction r.

Experimental determination of equilibrium constants is a rather difficult task. First of all, it is necessary to be sure that equilibrium is achieved at a given temperature, i.e., the composition of the reaction mixture corresponds to an equilibrium state - a state with a minimum Gibbs energy, zero reaction affinity and equality of the rates of forward and reverse reactions. At equilibrium, the pressure, temperature and composition of the reaction mixture will be constant.

At first glance, it seems that the composition of an equilibrium mixture can be determined using methods of quantitative analysis with characteristic chemical reactions. However, the introduction of an extraneous reagent that binds one of the components of the chemical process shifts (i.e. changes) the equilibrium state of the system. This method can only be used if the reaction rate is slow enough. That is why, very often, when studying equilibrium, various physical methods are also used to determine the composition of a system.

1.1 Chemical methods

Distinguish between static chemical methods and dynamic chemical methods. Let's look at specific examples given in.

1.1.1 Static methods.

Static methods consist in the fact that the reaction mixture is placed in a reactor at a constant temperature and then, upon reaching equilibrium, the composition of the system is determined. The studied reaction should be slow enough so that the introduction of an extraneous reagent practically does not disturb the state of equilibrium. To slow down the process, you can cool the reaction flask quickly enough. A classic example of such research is the reaction between iodine and hydrogen.

H2 (g) + I2 (g) = 2HI (g) (3)

Lemoyne placed in glass containers either a mixture of iodine with hydrogen or hydrogen iodide. At 200 ° C, the reaction practically does not take place; at 265 ° C, the duration of equilibration is several months; at 350 ° C, equilibrium is established within a few days; at 440 ° C - for several hours. In this regard, to study this process, a temperature range of 300 - 400 ° C was chosen. The analysis of the system was carried out as follows. The reaction balloon was quickly cooled by immersion in water, then the tap was opened and hydrogen iodide was dissolved in water. The amount of hydroiodic acid was determined by titration. At each temperature, the experiment was carried out until the concentration reached a constant value, which indicates the establishment of chemical equilibrium in the system.

1.1.2 Dynamic methods.

Dynamic methods consist in the fact that the gas mixture is continuously circulated, then it is rapidly cooled for subsequent analysis. These methods are most useful for reasonably fast reactions. Reactions are accelerated, as a rule, either by performing them at elevated temperatures, or by introducing a catalyst into the system. The dynamic method was used, in particular, in the analysis of the following gas reactions:

2H2 + O2 ⇄ 2H2O. (4)

2CO + O2 ⇄ 2CO2. (5)

2SO2 + O2 ⇄ 2SO

3H2 + N2 ⇄ 2NH

1.2 Physical methods

These methods are mainly based on measuring the pressure or mass density of the reaction mixture, although other properties of the system can be used.

1.2.1 Pressure measurement

Each reaction, which is accompanied by a change in the number of moles of gaseous reactants, is accompanied by a change in pressure at a constant volume. If the gases are close to ideal, then the pressure is directly proportional to the total number of moles of gaseous reactants.

As an illustration, consider the following gas reaction, written per molecule of the starting substance

Number of moles

at the initial moment 0 0

in equilibrium

where https://pandia.ru/text/78/005/images/image016_35.gif "width =" 245 "height =" 25 src = ">, (9)

where https://pandia.ru/text/78/005/images/image018_30.gif "width =" 20 "height =" 21 src = ">. gif" width = "91" height = "31">.

There are ratios between these pressures:

https://pandia.ru/text/78/005/images/image022_24.gif "width =" 132 "height =" 52 src = ">. (11)

https://pandia.ru/text/78/005/images/image024_21.gif "width =" 108 "height =" 52 src = ">. (13)

The equilibrium constant expressed in p-scale will have the form

. (14)

Therefore, having measured the equilibrium pressure, using formula (13), one can determine the degree of dissociation, and then, using formula (14), calculate the equilibrium constant.

1.2.2 Measurement of density of mass

Each reaction, which is accompanied by a change in the number of moles of gaseous participants in the process, is characterized by a change in the density of the mass at constant pressure.

For example, for reaction (8), it is true

, (15)

where https://pandia.ru/text/78/005/images/image028_20.gif "width =" 16 "height =" 19 "> is the volume of the system in equilibrium. As a rule, in real experiments, not the volume is measured, but the density the mass of the system, which is inversely proportional to the volume..gif "width =" 37 height = 21 "height =" 21 "> is the density of the mass of the system at the initial moment and at the moment of equilibrium, respectively. By measuring the mass density of the system, one can use formula (16) to calculate the degree of dissociation, and then the equilibrium constant.

1.2.3 Direct measurement of partial pressure

The most direct way to determine the equilibrium constant of a chemical reaction is to measure the partial pressures of each participant in the process. In general, this method is very difficult to apply in practice; most often it is used only in the analysis of gas mixtures containing hydrogen. In this case, the property of platinum group metals is used to be permeable to hydrogen at high temperatures. The preheated gas mixture is passed at a constant temperature through a cylinder 1, which contains an empty iridium reservoir 2, connected to a pressure gauge 3 (Fig. 1). Hydrogen is the only gas that can pass through the walls of the iridium reservoir.

Thus, it remains to measure the total pressure of the gas mixture and the partial pressure of hydrogen in order to calculate the equilibrium constant of the reaction. This method allowed Lowenstein and Wartenberg (1906) to study the dissociation of water, HCl, HBr, HI and H2S, as well as a reaction such as:

https://pandia.ru/text/78/005/images/image033_14.gif "width =" 89 height = 23 "height =" 23 ">. (17)

1.2.4 Optical methods

There are methods for studying equilibrium based on the measurement of adsorption, which are especially effective in the case of colored gases. You can also determine the composition of a gas binary mixture by measuring the refractive index (refractometric). For example, Chadron (1921) studied the reduction of metal oxides by carbon monoxide by measuring the refractometric composition of a gas mixture of carbon monoxide and carbon dioxide.

1.2.5 Thermal conductivity measurement

This method was used to study dissociation reactions in the gas phase, for example

Suppose that a mixture of N2O4 and NO2 is placed in a vessel, the right side of which has a temperature of T2, and the left one is T1, and T2> T1 (Fig. 2). Dissociation of N2O4 will be greater in that part of the vessel that has a higher temperature. Consequently, the concentration of NO2 in the right side of the vessel will be higher than in the left, and diffusion of NO2 molecules from right to left and N2O4 from left to right will be observed. However, upon reaching the right side of the reaction vessel, N2O4 molecules again dissociate with the absorption of energy in the form of heat, and NO2 molecules, reaching the left side of the vessel, dimerize with the release of energy in the form of heat. That is, there is a superposition of ordinary thermal conductivity and thermal conductivity associated with the course of the dissociation reaction. This problem is solved quantitatively and allows you to determine the composition of the equilibrium mixture.

1.2.6 Measurement of electromotive force (EMF) of a galvanic cell

Measuring the EMF of galvanic cells is a simple and accurate method for calculating the thermodynamic functions of chemical reactions. It is only necessary 1) to compose such a galvanic cell so that the final reaction in it would coincide with the investigated one, the equilibrium constant of which must be determined; 2) measure the EMF of a galvanic cell in a thermodynamic equilibrium process. To do this, it is necessary that the corresponding current-forming process is carried out infinitely slowly, that is, for the element to work at an infinitely small current strength, which is why a compensation method is used to measure the EMF of a galvanic cell, which is based on the fact that the investigated galvanic cell is connected in series against the external potential difference , and the latter was selected in such a way that there was no current in the circuit. The value of the EMF, measured by the compensation method, corresponds to the thermodynamically equilibrium process in the element and the useful work of the process is maximum and is equal to the decrease in the Gibbs energy

https://pandia.ru/text/78/005/images/image035_12.gif "width =" 181 "height =" 29 src = "> (20)

at p, T = const, where F–Faraday number = 96500 C / mol, n- the smallest total multiple of the number of electrons involved in electrode reactions, Eo- standard EMF, V.

The value of the equilibrium constant can be found from the relation (21)

(21)

2. EXAMPLE OF LABORATORY WORK ON DETERMINING THE VALUE OF THE EQUILIBRIUM CONSTANT

In workshops on physical chemistry, laboratory work is often found concerning the study of the dissociation reaction of metal carbonates. Here is a brief summary of such work.

purpose of work – determination of the equilibrium constant and calculation of the basic thermodynamic values ​​of the carbonate decomposition reaction.

Calcium carbonate https://pandia.ru/text/78/005/images/image038_12.gif "width =" 192 "height =" 29 ">, (22)

in this case, gaseous carbon monoxide (IV), solid calcium oxide is formed, and some part of the non-dissociated calcium carbonate remains.

The equilibrium constant of reaction (22) will be written as:

, (23)

where https://pandia.ru/text/78/005/images/image041_11.gif "width =" 68 "height =" 51 "> in general form or; the activities of pure solid or liquid phases are equal to https: // pandia. ru / text / 78/005 / images / image044_10.gif "width =" 76 "height =" 28 src = ">.

If pressure is measured in atmospheres, then = https://pandia.ru/text/78/005/images/image046_9.gif "width =" 87 "height =" 53 ">. (24)

The equilibrium pressure of carbon dioxide over calcium carbonate is called the dissociation elasticity of CaCO3.

That is, the equilibrium constant of the dissociation reaction of calcium carbonate will be numerically equal to the dissociation elasticity of carbonate, if the latter is expressed in atmospheres. Thus, having determined experimentally the dissociation elasticity of calcium carbonate, it is possible to determine the value of the equilibrium constant of this reaction.

experimental part

The static method is used to determine the dissociation elasticity of calcium carbonate. Its essence lies in the direct measurement of the pressure of carbon dioxide in the installation at a given temperature.

Equipment. The main units of the installation are: a reaction vessel (1) made of heat-resistant material and placed in an electric furnace (2); a mercury manometer (3), hermetically connected to the reaction vessel and through a valve (4) with a manual vacuum pump (5). The temperature in the furnace is maintained using a regulator (6), the temperature is controlled using a thermocouple (7) and a voltmeter (8). A certain amount of the investigated powdery substance (9) (metal carbonates) is placed in the reaction vessel.

Work order... After checking the tightness of the system, turn on the furnace and use the regulator to set the required initial temperature of the reaction vessel. The first readings of the thermocouple and gauge are recorded. After that, using the regulator (6), the temperature in the furnace is increased by 10-20 degrees, they wait for a new constant temperature value to be established and the pressure value corresponding to this temperature is recorded. Thus, gradually increasing the temperature, at least 4-5 measurements are taken. After the end of the experiment, the furnace is cooled and the system is connected to the atmosphere through a tap (4). Then turn off the oven and voltmeter. Having processed the obtained experimental data, it is possible to calculate the equilibrium constant of the dissociation reaction.

Fig. 3. Installation for determining the elasticity of dissociation

metal carbonates.

3. DETERMINATION OF EQUILIBRIUM CONSTANTS

WITHOUT EXPERIMENT

3.1 Calculation of the equilibrium constant of a chemical reaction from

the value of the standard molar Gibbs function of the reaction

This method does not involve experiment at all. If the standard molar enthalpy and entropy of the reaction at a given temperature are known, then, using the corresponding equations, one can calculate the standard molar Gibbs function of the reaction under study at the desired temperature, and through it the value of the equilibrium constant.

If the values ​​of the standard molar entropy and enthalpy at a given temperature are unknown, then the Temkin and Shvartsman method can be used, that is, by the value of the standard molar enthalpy and entropy at a temperature of 298 K and the values ​​of the coefficients of the temperature dependence of the molar heat capacity of the reaction, calculate the standard molar Gibbs energy of the reaction for any temperature.

https://pandia.ru/text/78/005/images/image051_7.gif "width =" 137 "height =" 25 src = "> - reference coefficients that do not depend on the nature of the reaction and are determined only by temperature values.

3.2 Method of Combining Equilibria

This method is used in practical chemical thermodynamics. For example, experimentally at the same temperature, the equilibrium constants of two reactions were found

1.CH3OH (g) + CO ⇄ HCOOCH3 (g) . (26)

2.H2 + 0.5 HCOOCH3 (g) ⇄ CH3OH (g) . (27)

The equilibrium constant of the methanol synthesis reaction

3..gif "width =" 31 "height =" 32 "> and:

. (29)

3.3 Calculation of the equilibrium constant of a chemical reaction at a certain temperature from the known values ​​of the equilibrium constants of the same reaction at two other temperatures

This method of calculation is based on solving the equation of the isobar of a chemical reaction (Van't Hoff isobar)

, (30)

where https://pandia.ru/text/78/005/images/image060_3.gif "width =" 64 "height =" 32 "> and looks like:

. (31)

Using this equation, knowing the equilibrium constants at two different temperatures, you can calculate the standard molar enthalpy of the reaction, and knowing it and the equilibrium constant at one temperature, you can calculate the equilibrium constant at any other temperature.

4. EXAMPLES OF SOLVING PROBLEMS

Find the equilibrium constant of ammonia synthesis y N2 + "H2 ⇄ NH3 if the equilibrium mole fraction of ammonia is 0.4 at 1 atm and 600K. The initial mixture is stoichiometric; there is no product in the initial mixture.

Given: Reaction y N2 + "H2 ⇄ NH3, 1 atm, 600 K. = 1.5 mol; = 0.5 mol; = 0 mol = 0.4 Find: -?

Solution

From the condition of the problem, we know the stoichiometric equation, as well as the fact that at the initial moment of time the number of nitrogen moles is equal to the stoichiometric one, that is, 0.5 mol (https://pandia.ru/text/78/005/images/image069_3.gif " width = "247" height = "57 src =">

We write down the reaction, under the symbols of the elements we indicate the initial and equilibrium quantities of moles of substances

y N2 + ”H2 ⇄ NH3

0.5 - 0.5ξ 1.5 - 1.5 ξ ξ

The total number of moles of all participants in the reaction in the system at the moment of equilibrium

https://pandia.ru/text/78/005/images/image073_4.gif "width =" 197 "height =" 56 src = ">. gif" width = "76" height = "48 src =">

https://pandia.ru/text/78/005/images/image077_0.gif "width =" 120 "height =" 47 ">

= 3,42

The solution to the direct problem of chemical equilibrium is to calculate the equilibrium composition of the system in which a given reaction (several reactions) takes place. Obviously, the solution is based on the law of chemical equilibrium. It is only necessary to express all the variables included in this law through any one: for example, through the depth of a chemical reaction, through the degree of dissociation, or through some equilibrium mole fraction. It is better to choose which variable is convenient to use based on the specific conditions of the problem.

Task 2

Equilibrium constant of the gas reaction for the synthesis of hydrogen iodide

H2 + I2 ⇄ 2HI at a temperature of 600 K and the expression for pressure in atmospheres, is Kr= 45.7. Find the equilibrium depth of this reaction and the equilibrium yield of the product at a given temperature and pressure of 1 atm, if at the initial moment of time the amounts of starting materials correspond to stoichiometric ones, and there are no reaction products at the initial moment.

Given Kr= 45.7. = 1 mol; https://pandia.ru/text/78/005/images/image081_1.gif "width =" 68 "height =" 27 src = "> mol. Find: -? -?

Solution

Let us write down the reaction itself, and under the symbols of the elements, the number of moles of each participant at the initial moment and at the moment of established equilibrium according to the formula (4)

1 - ξ 1 - ξ 2ξ

1 - ξ + 1 - ξ + 2ξ = 2

Equilibrium mole fractions and partial pressures of all participants in the reaction, we express through a single variable - the depth of the chemical reaction

https://pandia.ru/text/78/005/images/image085_1.gif "width =" 144 "height =" 47 src = ">.

The law of mass action or the law of chemical equilibrium

https://pandia.ru/text/78/005/images/image082_1.gif "width =" 13 "height =" 23 src = "> = 0.772.

Problem 3

Its condition differs from problem 2 only in that the initial amounts of moles of hydrogen and iodine are equal to 3 and 2 moles, respectively. Calculate the molar composition of the equilibrium mixture.

Given: Possible reaction: H2 + I2 = 2HI. 600 K, 1 atm. Kr = 45,7 .

3 mol; mole; mole. Find: -? .Gif "width =" 32 "height =" 27 "> 1 1 0

3 - ξ 2 - ξ 2ξ

The total number of moles of all participants in the reaction at the moment of equilibrium is

3 - ξ + 2 - ξ + 2ξ = 5

Equilibrium mole fractions and partial pressures of all participants in the reaction, expressed through a single variable - the depth of the chemical reaction

Substitution of partial pressures into the law of chemical equilibrium gives:

https://pandia.ru/text/78/005/images/image090_1.gif "width =" 13 "height =" 21 "> and calculate the equilibrium constant, then build a graph and use it to determine the reaction depth that corresponds to the one found the value of the equilibrium constant.

= 1,5 = 12

https://pandia.ru/text/78/005/images/image067_4.gif "width =" 29 "height =" 29 src = "> =29,7

https://pandia.ru/text/78/005/images/image067_4.gif "width =" 29 "height =" 29 src = "> = 54

https://pandia.ru/text/78/005/images/image083_1.gif "width =" 35 height = 25 "height =" 25 "> = 0.712

To carry out the work, you need to complete the following tasks

Exercise 1

1. Describe a method for the experimental determination of the elasticity of carbon dioxide in the study of the dissociation reaction of CaCO3⇄CaO + CO2

(options 1 - 15, table 3);

2. Write down the law of chemical equilibrium for the studied reaction; determine the values ​​of the equilibrium constants of the dissociation reaction of calcium carbonate according to the experimental data (Table 3) at different temperatures;

3. Write down the determinant expression for the equilibrium constant and calculate theoretically the equilibrium constant of the studied reaction at the last temperature indicated in the table.

Assignment 2

1. Prepare an answer to question 1 (options 1-15, table 4)

2. Solve problems 2 and 3.

Reference data required to get the job done

Quantity for calculating the standard molar change of the Gibbs energy by the method of Temkin and Shvartsman

Table 1

Thermodynamic data for calculating the standard molar Gibbs energy

table 2

Experimental data for task 1

Table 3

Conditions for tasks for completing task 2

Table 4

It seems appropriate to include the following sections in the laboratory report: introduction, part 1, part 2, conclusions.

1. In the introduction you can briefly summarize theoretical information on one of the following issues: either about the law of the masses at work, the history of its discovery and about its authors; or about the basic concepts and definitional relationships of the section "Chemical equilibrium"; or deduce the law of chemical equilibrium in its modern formulation; or talk about the factors that affect the value of the equilibrium constant, etc.

The end of the "Introduction" section is followed by a statement of the objectives of the work.

In part 1 necessary

2.1. Give a diagram of the installation for determining the elasticity of dissociation of metal carbonates and describe the course of the experiment.

2.2 ... Provide the results of calculating the equilibrium constant according to the presented experimental data

2.3. Provide the calculation of the equilibrium constant from thermodynamic data

In part 2 necessary

3.1 ... Provide a complete substantiated answer to question 1 of task 2.

3.2 ... Provide the solution of tasks 2 and 3 of task 2. The condition of the tasks must be written in a symbolic notation.

In conclusions it is advisable to reflect the fulfillment of the goals set in the work, as well as compare the values ​​of the equilibrium constants calculated in 2.2 and 2.3.

Bibliographic list

1. Karjakin of Chemical Thermodynamics: Textbook. manual for universities. M .: Academy., 20s.

2. Modern thermodynamics. From heat engines to dissipative structures. M .: Mir, 20s.

3., Cherepanov in physical chemistry. Toolkit. Yekaterinburg: USU publishing house, 2003.

4. A short reference book of physical and chemical quantities / Ed. and. L .: Chemistry, 20s.

5. Tasks in physical chemistry: textbook. manual for universities /, etc. M .: Examination, 20s.

Computer layout

In 1885, the French physicist and chemist Le Chatelier was derived, and in 1887 the German physicist Brown substantiated the law of chemical equilibrium and the constant of chemical equilibrium, and also studied their dependence on the influence of various external factors.

The essence of chemical equilibrium

Equilibrium is a state that means things are always in motion. Products are decomposed into reagents and reagents are combined into products. Things move, but the concentration remains the same. The reaction is written with a double arrow instead of an equal sign to indicate that it is reversible.

Classic patterns

Even in the last century, chemists discovered certain patterns that provide for the likelihood of a change in the direction of the reaction in the same container. Knowledge of how chemical reactions take place is incredibly important, both for laboratory research and industrial production. At the same time, the ability to control all these phenomena is of great importance. It is natural for a person to interfere in many natural processes, especially reversible ones, in order to then use them for his own good. Knowledge of chemical reactions will be more useful if you have perfect control over them.

The law of mass action in chemistry is used by chemists to correctly calculate the rates of reactions. It gives a clear idea that none will be followed through in case it takes place in a closed system. The molecules of the resulting substances are in constant and disorderly movement, and a reverse reaction may soon occur, in which the molecules of the starting material will be reduced.

In industry, open systems are most often used. Vessels, apparatus and other containers where chemical reactions take place remain unlocked. This is necessary so that during these processes it is possible to extract the desired product and get rid of useless reaction products. For example, coal is burned in open furnaces, cement is produced in open-type furnaces, blast furnaces operate with a constant supply of air, and ammonia is synthesized with the continuous removal of ammonia itself.

Reversible and irreversible chemical reactions

Based on the name, one can give appropriate definitions: reactions are considered irreversible if they are carried through to the end, do not change their direction and proceed along a given trajectory, regardless of pressure drops and temperature fluctuations. Their distinctive feature is that some products can leave the reaction sphere. Thus, for example, it is possible to obtain gas (CaCO 3 = CaO + CO 2), precipitate (Cu (NO 3) 2 + H 2 S = CuS + 2HNO 3) or others will also be considered irreversible if a large amount is released during the process thermal energy, for example: 4P + 5O 2 = 2P 2 O 5 + Q.

Almost all reactions that occur in nature are reversible. Regardless of external conditions such as pressure and temperature, almost all processes can proceed simultaneously in different directions. According to the law of mass action in chemistry, the amount of absorbed heat will be equal to the amount released, which means that if one reaction was exothermic, then the second (reverse) will be endothermic.

Chemical equilibrium: chemical equilibrium constant

Reactions are the "verbs" of chemistry - activities that chemists study. Many reactions go to their completion and then stop, which means that the reagents are completely converted into products, without being able to return to their original state. In some cases, the reaction is indeed irreversible, for example, when combustion changes both physical and chemical. However, there are many other circumstances in which it is not only possible, but also continuous, since the products of the first reaction become reagents in the second.

A dynamic state in which the concentrations of reactants and products remain constant is called equilibrium. It is possible to predict the behavior of substances using certain laws that are applied in industries seeking to reduce the production costs of specific chemicals. The concept of chemical equilibrium is also useful for understanding the processes that preserve or potentially threaten human health. Chemical equilibrium constant is the value of the reaction factor, which depends on ionic strength and temperature, and does not depend on the concentrations of reagents and products in solution.

Calculating the equilibrium constant

This value is dimensionless, that is, it does not have a certain number of units. Although the calculation is usually written for two reactants and two products, it works for any number of reaction participants. The calculation and interpretation of the equilibrium constant depends on whether the chemical reaction is associated with homogeneous or heterogeneous equilibrium. This means that all reacting components can be pure liquids or gases. For reactions that reach heterogeneous equilibrium, as a rule, there is not one phase, but at least two. For example, liquids and gases or liquids.

Equilibrium constant value

For any given temperature, there is only one value for the equilibrium constant, which changes only if the temperature at which the reaction occurs changes in one direction or another. Some predictions about a chemical reaction can be made based on whether the equilibrium constant is large or small. If the value is very large, then the equilibrium favors the reaction to the right and more products are obtained than there were reagents. The reaction in this case can be called "complete" or "quantitative".

If the value of the equilibrium constant is small, then it favors the reaction to the left, where the amount of reactants was greater than the resulting products. If this value tends to zero, we can assume that the reaction does not occur. If the values ​​of the equilibrium constant for the forward and reverse reactions are almost the same, then the amount of reagents and products will also be almost the same. This type of reaction is considered reversible.

Consider a specific reversible reaction

Let's take two chemical elements such as iodine and hydrogen, which, when mixed, give a new substance - hydrogen iodide.

For v 1 we take the speed of the forward reaction, for v 2 - the speed of the reverse reaction, k - the equilibrium constant. Using the law of mass action, we get the following expression:

v 1 = k 1 * c (H 2) * c (I 2),

v 2 = k 2 * c 2 (HI).

When iodine (I 2) and hydrogen (H 2) molecules are mixed, their interaction begins. At the initial stage, the concentration of these elements is maximum, but by the end of the reaction, the concentration of the new compound, hydrogen iodide (HI), will be maximum. Accordingly, the reaction rates will also be different. At the very beginning, they will be maximum. Over time, a moment comes when these values ​​will be equal, it is a state called chemical equilibrium.

The expression for the chemical equilibrium constant is usually denoted using square brackets:,,. Since in a state of equilibrium the speeds are equal, then:

k 1 = k 2 2,

so we get the equation for the constant of chemical equilibrium:

k 1 / k 2 = 2 / = K.

Le Chatelier-Brown principle

There is the following pattern: if a certain effect is made on a system that is in equilibrium (to change the conditions of chemical equilibrium by changing temperature or pressure, for example), then the balance will shift to partially counteract the effect of the change. In addition to chemistry, this principle also applies in several different forms to the fields of pharmacology and economics.

Chemical equilibrium constant and ways of expressing it

Equilibrium expression can be expressed in terms of the concentration of products and reagents. Only chemicals in the aqueous and gaseous phases are included in the equilibrium formula, since the concentrations of liquids and solids do not change. What factors affect chemical equilibrium? If a pure liquid or solid is involved in it, it is considered that it has K = 1, and, accordingly, ceases to be taken into account, with the exception of highly concentrated solutions. For example, pure water has an activity of 1.

Another example is solid carbon, which can be formed by the reaction of two molecules of carbon monoxide to form carbon dioxide and carbon. Factors that can affect equilibrium include the addition of reagent or product (changes in concentration affect balance). The addition of a reagent can lead to equilibrium on the right in the chemical equation where more product forms appear. The addition of product can lead to equilibrium on the left as more forms of reactants become.

Equilibrium occurs when a reaction in both directions has a constant ratio of products and reactants. In general, chemical equilibrium is static, since the quantitative ratio of products and reagents is constant. However, a closer look reveals that equilibrium is actually a very dynamic process, as the reaction moves in both directions at an equal pace.

Dynamic equilibrium is an example of a steady state function. For a system in a steady state, the currently observed behavior continues into the future. Therefore, as soon as the reaction reaches equilibrium, the concentration ratio of the product and reagent will remain the same, although the reaction continues.

How easy is it to tell about the difficult?

Concepts such as chemical equilibrium and chemical equilibrium constant are difficult to understand. Let's take a real life example. Have you ever got stuck on a bridge between two cities and noticed that the traffic in the other direction is smooth and measured, while you are hopelessly stuck in traffic? This is not good.

What if the cars moved steadily and at the same speed on both sides? Would the number of cars in both cities remain constant? When the speed of entry and exit to both cities is the same, and the number of cars in each city is stable over time, this means that the whole process is in dynamic equilibrium.

Most chemical reactions are reversible, i.e. flow simultaneously in opposite directions. In cases where direct and reverse reactions proceed at the same rate, chemical equilibrium occurs. For example, in a reversible homogeneous reaction: H2 (g) + I 2 (g) ↔ 2HI (g), the ratio of the rates of direct and reverse reactions according to the law of effective masses depends on the ratio of the concentrations of the reacting substances, namely: the rate of the direct reaction: υ 1 = k 1 [H 2]. Feedback rate: υ 2 = k 2 2.

If H 2 and I 2 are initial substances, then at the first moment the rate of the forward reaction is determined by their initial concentrations, and the rate of the reverse reaction is zero. As H 2 and I 2 are consumed and HI is formed, the rate of the forward reaction decreases and the rate of the reverse reaction increases. After some time, both speeds become equal, and chemical equilibrium is established in the system, i.e. the number of formed and consumed HI molecules per unit time becomes the same.

Since in chemical equilibrium the rates of forward and reverse reactions are equal to V 1 = V 2, then k 1 = k 2 2.

Since k 1 and k 2 are constant at a given temperature, their ratio will be constant. Denoting it by K, we get:

K - is called the constant of chemical equilibrium, and the above equation is called the law of mass action (Guldberg - Vaale).

In the general case, for a reaction like aA + bB +… ↔dD + eE +… the equilibrium constant is ... For the interaction between gaseous substances, an expression is often used in which the reactants are represented by the equilibrium partial pressures p. For the mentioned reaction .

The equilibrium state characterizes the limit to which, under these conditions, the reaction proceeds spontaneously (∆G<0). Если в системе наступило химическое равновесие, то дальнейшее изменение изобарного потенциала происходить не будет, т.е. ∆G=0.

The ratio between equilibrium concentrations does not depend on what substances are taken as starting materials (for example, H 2 and I 2 or HI), i.e. equilibrium can be approached from both sides.

The chemical equilibrium constant depends on the nature of the reagents and on temperature; the equilibrium constant does not depend on the pressure (if it is too high) and on the concentration of the reagents.

Influence on the equilibrium constant of temperature, enthalpy and entropy factors... The equilibrium constant is associated with a change in the standard isobaric-isothermal potential of a chemical reaction ∆G o by a simple equation ∆G o = -RT ln K.

It can be seen from it that large negative values ​​of ∆G o (∆G o<<0) отвечают большие значения К, т.е. в равновесной смеси преобладают продукты взаимодействия. Если же ∆G o характеризуется большими положительными значениями (∆G o >> 0), then the initial substances prevail in the equilibrium mixture. The above equation allows the value of ∆G o to calculate K, and then the equilibrium concentrations (partial pressures) of the reagents. If we take into account that ∆G o = ∆H o -T∆S o, then after some transformation we obtain ... It can be seen from this equation that the equilibrium constant is very sensitive to temperature changes. The influence of the nature of the reactants on the equilibrium constant determines its dependence on the enthalpy and entropy factors.

Le Chatelier's principle

The state of chemical equilibrium is maintained under these constant conditions at any time. When conditions change, the state of equilibrium is violated, since in this case the rates of opposite processes change to different degrees. However, after some time, the system again comes to a state of equilibrium, but already meeting the new changed conditions.

An equilibrium shift depending on changes in conditions is generally determined by the Le Chatelier principle (or the principle of mobile equilibrium): if a system in equilibrium is influenced from the outside by changing any of the conditions that determine the equilibrium position, then it shifts in the direction of the process, the course of which weakens the effect of the impact produced.

Thus, an increase in temperature causes a shift in equilibrium in the direction of that of the processes, the course of which is accompanied by heat absorption, while a decrease in temperature acts in the opposite direction. Similarly, an increase in pressure shifts the equilibrium in the direction of a process accompanied by a decrease in volume, and a decrease in pressure acts in the opposite direction. For example, in the equilibrium system 3H 2 + N 2 2H 3 N, ∆H o = -46.2 kJ, an increase in temperature enhances the decomposition of H 3 N into hydrogen and nitrogen, since this process is endothermic. An increase in pressure shifts the equilibrium towards the formation of H 3 N, because the volume decreases.

If a certain amount of any of the substances participating in the reaction is added to a system in a state of equilibrium (or, on the contrary, removed from the system), then the rates of the forward and reverse reactions change, but gradually become equal again. In other words, the system again comes to a state of chemical equilibrium. In this new state, the equilibrium concentrations of all substances present in the system will differ from the initial equilibrium concentrations, but the ratio between them will remain the same. Thus, in a system in equilibrium, it is impossible to change the concentration of one of the substances without causing a change in the concentrations of all the others.

In accordance with Le Chatelier's principle, the introduction of additional amounts of a reagent into an equilibrium system causes a shift in the equilibrium in the direction at which the concentration of this substance decreases and, accordingly, the concentration of its interaction products increases.

The study of chemical equilibrium is of great importance both for theoretical research and for solving practical problems. Determining the position of equilibrium for various temperatures and pressures, you can choose the most favorable conditions for the chemical process. In the final choice of the conditions for carrying out the process, their influence on the speed of the process is also taken into account.

Example 1. Calculation of the reaction equilibrium constant based on the equilibrium concentrations of reactants.

Calculate the equilibrium constant of the reaction A + B 2C, if the equilibrium concentration [A] = 0.3 mol ∙ l -1; [B] = 1.1 mol ∙ l -1; [C] = 2.1 mol ∙ l -1.

Solution. The expression for the equilibrium constant for this reaction is:. Let us substitute here the equilibrium concentrations indicated in the condition of the problem: = 5.79.

Example 2... Calculation of equilibrium concentrations of reactants. This reaction takes place according to the equation A + 2B C.

Determine the equilibrium concentrations of the reacting substances if the initial concentrations of substances A and B are 0.5 and 0.7 mol ∙ l -1, respectively, and the equilibrium constant of the reaction is K p = 50.

Solution. For each mole of substances A and B, 2 moles of substance C are formed. If the decrease in the concentration of substances A and B is denoted by X mol, then the increase in the concentration of the substance will be equal to 2X mol. Equilibrium concentrations of reactants will be:

WITH A = (o, 5-x) mol ∙ l -1; C B = (0.7-x) mol ∙ l -1; С С = 2х mol ∙ l -1

x 1 = 0.86; x 2 = 0.44

By the condition of the problem, the value x 2 is valid. Hence, the equilibrium concentrations of reactants are equal:

C A = 0.5-0.44 = 0.06 mol ∙ l -1; C B = 0.7-0.44 = 0.26 mol ∙ l -1; С С = 0.44 ∙ 2 = 0.88 mol ∙ l -1.

Example 3. Determination of the change in the Gibbs energy ∆G o of the reaction by the value of the equilibrium constant K p. Calculate the Gibbs energy and determine the possibility of the reaction CO + Cl 2 = COCl 2 at 700K, if the equilibrium constant is equal to Kp = 1.0685 ∙ 10 -4. The partial pressure of all reactants is the same and equal to 101325Pa.

Solution.∆G 700 = 2.303 ∙ RT .

For this process:

Since ∆Gо<0, то реакция СО+Cl 2 COCl 2 при 700К возможна.

Example 4... Chemical equilibrium shift. In which direction will the equilibrium shift in the N 2 + 3H 2 2NH 3 -22kcal system:

a) with an increase in the concentration of N 2;

b) with an increase in the concentration of H 2;

c) when the temperature rises;

d) with decreasing pressure?

Solution. An increase in the concentration of substances on the left side of the reaction equation, according to Le Chatelier's rule, should cause a process that tends to weaken the effect, lead to a decrease in concentrations, i.e. the equilibrium will shift to the right (cases a and b).

The ammonia synthesis reaction is exothermic. An increase in temperature causes a shift in the equilibrium to the left - towards an endothermic reaction, which weakens the effect exerted (case c).

A decrease in pressure (case d) will favor a reaction leading to an increase in the volume of the system, i.e. towards the formation of N 2 and H 2.

Example 5. How many times will the rate of the forward and reverse reaction in the 2SO 2 (g) + O 2 (g) 2SO 3 (r) system change if the volume of the gas mixture decreases three times? In which direction will the equilibrium of the system shift?

Solution. Let's designate the concentration of reactants: = a, =b,=With. According to the law of mass action, the rates of forward and reverse reactions before the volume change are equal

v pr = Ka 2 b, v arr = K 1 s 2

After reducing the volume of the homogeneous system by three times, the concentration of each of the reacting substances will increase three times: = 3a,[About 2] = 3b; = 3c. At new concentrations, the rates v "np of the forward and reverse reactions:

v "np = K (3a) 2 (3b) = 27 Ka 2 b; v o 6 p = K 1 (3c) 2 = 9K 1 c 2.

;

Consequently, the speed of the forward reaction increased by 27 times, and the reverse - only nine times. The equilibrium of the system shifted towards the formation of SO 3.

Example 6. Calculate how many times the rate of the reaction proceeding in the gas phase will increase when the temperature rises from 30 to 70 0 С, if the temperature coefficient of the reaction is 2.

Solution. The dependence of the rate of a chemical reaction on temperature is determined by the van't Hoff rule according to the formula

Consequently, the reaction rate at 70 ° C is 16 times greater than the reaction rate at 30 ° C.

Example 7. Equilibrium constant of a homogeneous system

CO (g) + H 2 O (g) CO 2 (g) + H 2 (g) at 850 ° C is equal to 1. Calculate the concentrations of all substances at equilibrium if the initial concentrations: [CO] RR = 3 mol / L, [H 2 O] ISC = 2 mol / l.

Solution. In equilibrium, the rates of direct and reverse reactions are equal, and the ratio of the constants of these rates is constant and is called the equilibrium constant of the given system:

V np = K 1[CO] [H 2 O]; V o b p = TO 2 [CO 2] [H 2];

In the condition of the problem, the initial concentrations are given, while in the expression K p only equilibrium concentrations of all substances in the system are included. Let us assume that by the moment of equilibrium the concentration of [CO 2] P = NS mol / l. According to the equation of the system, the number of moles of the formed hydrogen will also be NS mol / l. The same number of moles (X mol / l) CO and H 2 O is consumed for the formation of NS moles of CO 2 and H 2. Therefore, the equilibrium concentrations of all four substances (mol / l):

[CO 2] P = [H 2] p = X;[CO] P = (3 –x); P = (2-x).

Knowing the equilibrium constant, we find the value X, and then the initial concentrations of all substances:

; x 2 = 6-2x-3x + x 2; 5x = 6, l = 1.2 mol / l.

FOR SECONDARY SCHOOL TEACHERS, STUDENTS OF PEDAGOGICAL UNIVERSITIES AND PUPILS OF 9-10 CLASSES WHO DECIDED TO DEDICATE CHEMISTRY AND NATURAL SCIENCES

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§ 3.2. Equilibrium constant
and isobaric reaction potential

The equilibrium constant can be easily found from the value of the isobaric potential, which is calculated from tabular data on the enthalpy of formation and entropy of the initial substances and reaction products

You will need this formula when you need to calculate the equilibrium constant of the reaction under study.

In this tutorial, we try not to give ready-made formulas, but to deduce them using the simplest methods of mathematical logic, therefore, the derivation of this formula is given below. After reading this material, you will get acquainted with the simplest concepts of the theory of probability, with the entropy of activation, etc.

Activation energy is not the only determinant of the rate of a chemical reaction. The size and shape of the reacting molecules and the arrangement of reactive atoms or their groups in them play a huge role. In this regard, when two particles collide, their definite orientation is important, that is, the contact of precisely those centers that are reactive.

Let us denote the probability of the orientation of molecules necessary for the interaction during collision W:

The natural logarithm of W multiplied by the gas constant R is called the activation entropy S a:

This expression implies:

Whence, by the definition of the logarithm, we obtain the probability of the required orientation:

The greater the probability of the required orientation for the reaction to proceed, the higher its speed and, accordingly, the rate constant, which can be written:

Earlier we learned that the rate constant depends on the activation energy and temperature:

Thus, the rate constant depends on the activation energy, temperature, and activation entropy:

Let's enter the proportionality coefficient Z and put an equal sign:

The resulting expression is called the basic equation of chemical kinetics.

This equation explains some aspects of catalysis: the catalyst lowers the activation energy of the reaction and increases the activation entropy, that is, it increases the probability of the appropriate orientation of the reacting particles for the interaction.

It is interesting to note that the activation entropy takes into account not only a certain orientation of the particles, but also the duration of the contact at the moment of collision. If the duration of the contact of the particles is very short, then their electron densities do not have time to redistribute to form new chemical bonds, and the particles, repelling, diverge in different directions. The catalyst also greatly increases the contact time of the reacting particles.

Another feature of the catalytic action: the catalyst takes away excess energy from the newly formed particle, and it does not decompose into original particles due to its high energetic activity.

You know that the equilibrium constant is the ratio of the rate constants of the forward and reverse reactions:

Let us replace the rate constants of the forward and reverse reactions with the expressions for the basic equation of chemical kinetics:

The ratio of the two proportionality coefficients Z pr / Z arr is a constant value, which we introduce into the value of the equilibrium constant, which is why it will remain, as before, a constant.

If you remember the rules of action with exponential functions, you will understand the transformation of the formula:

In accordance with Hess's law, the difference between the activation energies of the reverse and direct reactions is a change in enthalpy (make sure of this by drawing the enthalpy diagram of the reaction proceeding with the release of heat, and not forgetting that in this case D H< 0 ):

Likewise, the difference denote D S:

Explain why there is a minus sign in front of the parentheses.

We get the equation:

Let us logarithm both sides of this equation:

Where do we get:

This equation is so important to chemistry and other natural sciences that many overseas chemistry students wear shirts featuring the formula.

If D G is expressed in J / mol, then the formula takes the form:

This formula has one feature: if the equilibrium constant is determined through the pressure of gaseous substances, then the pressure of these substances in atmospheres (1 atm = 101325 Pa = 760 mm Hg) is substituted into the expression for the equilibrium constant.

This formula allows for a known value D G reaction, calculate the equilibrium constant and thus find out the composition of the equilibrium system at a given temperature. The formula shows that the higher the equilibrium constant and the more reaction products (substances on the right side of the reaction equation) are contained in the equilibrium reaction mixture, the more negative is the change in the isobaric potential of the reaction. And vice versa, the lower the value of the equilibrium constant and the less the reaction products are contained in the equilibrium mixture and the more initial substances, the less the negative value D G.

When the equilibrium constant is greater than 1 and the isobaric potential is negative, it is customary to say that the equilibrium is shifted towards the reaction products, or to the right. When the equilibrium constant is less than 1 and the isobaric potential is positive, it is customary to say that the equilibrium is shifted towards the starting substances, or to the left.

If the equilibrium constant is equal to 1, the isobaric potential is equal to 0. This state of the system is considered to be the boundary between the shift of equilibrium to the right or to the left. When for a given reaction the change in isobaric potential is negative ( D G<0 ), it is customary to say that the reaction can proceed in the forward direction; if D G> 0, they say that the reaction does not go away.

Thus,

D G<0 - the reaction can take place (thermodynamically possible);

D G<0 , then K> 1- the balance is shifted towards the products, to the right;

D G> 0, then TO<1 - the equilibrium is shifted towards the starting substances, to the left.

If you need to find out whether the reaction you are interested in is possible (for example, to find out whether the synthesis of the desired dye is possible, whether the given mineral composition will sinter, the effect of air oxygen on the color, etc.), it is enough to calculate for this reaction D G... If it turns out that the change in isobaric potential is negative, then the reaction is possible, and you can mix different starting materials to obtain the desired product.

Read what you need to do to calculate the change in isobaric potential and the equilibrium constant at different temperatures (calculation algorithm).

1. Write down the values ​​(for a temperature of 298 K) of the enthalpies of formation from simple substances from the look-up tables D H arr and entropy S of all substances written in the equation of a chemical reaction. If D H arr are expressed in kJ / mol, then they should be converted to J / mol (why?).

2. Calculate the change in enthalpy in the reaction (298 K) as the difference between the sum of the enthalpies of formation of the products and the sum of the enthalpies of formation of the starting materials, keeping in mind the stoichiometric coefficients:

3. Calculate the change in entropy in the reaction (298 K) as the difference between the sum of the entropies of the products and the sum of the entropies of the starting materials, keeping in mind the stoichiometric coefficients:

4. Make an equation of the dependence of the change in isobaric potential on changes in the enthalpy of reaction, entropy and temperature, substituting the just obtained numerical values ​​into the equation known to you D N p-tion and D S p-tions:

5. Calculate the change in isobaric potential at a standard temperature of 298 K:

6. By sign D G p-tions, 298 draw a conclusion about the possibility of the reaction at a standard temperature: if the sign is "minus", then the reaction is thermodynamically possible; if the sign is "plus", then the reaction is impossible.

7. Count D G p-tions at the temperature T you are interested in:

and draw a conclusion how the temperature change affects the possibility of the reaction. If it turns out that at this temperature the change in isobaric potential has become less positive or more negative in comparison with D G 298, then, therefore, at this temperature, the reaction becomes more likely.

8. Calculate from the equation known to you the equilibrium constant K at the temperature T you are interested in:

9. Make a conclusion about the displacement of the equilibrium towards the starting substances (K<1) или в сторону продуктов (К>1).

To conclude about the possibility of a reaction with a negative value of the change in isobaric potential ( D G p-tions<0 ) thermodynamic data alone are often insufficient. A thermodynamically possible reaction can turn out to be kinetically inhibited and feasible when conditions change (concentration of substances, pressure, temperature), through other reaction paths, or in the presence of a properly selected catalyst.

Consider the example of the reaction of crystalline iron with gaseous water (water vapor):

how to find out about the thermodynamic possibility of a reaction.

This reaction is interesting in that it shows the reasons for the decrease in the gloss of a metal product and its destruction from corrosion.

First of all, we select the stoichiometric coefficients of the reaction equation:

Let us write out from the reference tables the thermodynamic data (temperature 298 K) for all participants in the reaction:

Let's calculate the change in enthalpy in this reaction, remembering that the enthalpies of simple substances are equal to zero:

Let us express the change in enthalpy in J:

The reaction is accompanied by the release of heat, Q> 0, Q = + 50 300 J / mol, and this makes it possible to assume that it takes place spontaneously. However, it is safe to say that the reaction is spontaneous, it is possible only by the sign of the change in isobaric potential.

Let's calculate the change in entropy in this reaction, not forgetting about the stoichiometric coefficients:

The entropy of the system decreases as a result of the reaction; therefore, it can be noted that an increase in order occurs in the system.

Now we will compose the equation of the dependence of the change in isobaric potential on the changes in enthalpy, entropy and temperature:

Let us calculate the change in isobaric potential in the reaction at a standard temperature of 298 K:

A high negative value of the change in isobaric potential indicates that iron can be oxidized with oxygen at room temperature. If you could get the finest iron powder, you would see how iron burns in air. Why don't iron products, figurines, nails, etc., burn in the air? The calculation results show that iron corrodes in air, that is, it breaks down, turning into iron oxides.

Now let's see how an increase in temperature affects the possibility of this reaction. Let us calculate the change in isobaric potential at a temperature of 500 K:

A result was obtained showing that as the temperature rises, the change in the isobaric potential of the reaction becomes less negative. This means that as the temperature rises, the reaction becomes less thermodynamically probable, i.e., the equilibrium of the reaction shifts more and more towards the starting substances.

It is interesting to know at what temperature the equilibrium is equally shifted towards the reaction products and towards the starting materials. This happens when D G p-tion = 0(the equilibrium constant is 1):

From where we get:

T = 150300 / 168.2 = 894K, or 621 ° C.

At this temperature, the reaction is equally probable both in the forward and in the opposite direction. At temperatures above 621 ° C, the reverse reaction of Fe 3 O 4 reduction with hydrogen begins to prevail. This reaction is one of the ways to obtain pure iron (in metallurgy, iron oxides are reduced with carbon).

At a temperature of 298 K:

Thus, as the temperature rises, the equilibrium constant decreases.

Iron oxide Fe 3 O 4 is called magnetite (magnetic iron ore). This iron oxide, unlike the oxides FeO (wustite) and Fe 2 O 3 (hematite), is attracted by a magnet. There is a legend that in ancient times a shepherd named Magnus found a very small oblong pebble, which he put with his greasy (why is it important?) Hands on the surface of the water in a bowl. The pebble did not drown and began to float in the water, and no matter how the shepherd turned the bowl, the pebble always pointed only in one direction. As if the compass was invented, and the mineral was named after this shepherd. Although, perhaps, magnetite was so named after the ancient city of Asia Minor - Magnesia. Magnetite is the main ore from which iron is mined.

Sometimes the formula of magnetite is depicted as: FeO Fe 2 O 3, implying that magnetite consists of two iron oxides. This is wrong: magnetite is an individual substance.

Another oxide Fe 2 O 3 (hematite) - red iron ore - is so named because of its red color (in the translation from Greek - blood). Iron is obtained from hematite.

FeO oxide is almost never found in nature and has no industrial value.