Study questions

  1. Balance state

    2. Equilibrium constant

    Calculation of equilibrium concentrations

    Bias chemical equilibrium. Le Chatelier's principle

  1. Balance state

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

Consider a reversible reaction that takes place in a closed system

The rate of the direct reaction is described by the equation:

pr = k pr [A] [B],

Where pr is the rate of the direct reaction;

k pr is the rate constant of the direct reaction.

Over time, the concentrations of the reagents A And IN decrease, the reaction rate drops (Fig. 1, curve etc).

Reaction between A And IN leads to the formation of substances C And D, whose molecules in collisions can again give substances A And IN.

The rate of the reverse reaction is described by the equation:

arr = k arr [C] [D],

Where arr is the rate of the reverse reaction;

k arr is the rate constant of the reverse reaction.

As the concentrations 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 reverse reactions in time

Over time the rates of the forward and reverse 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 cease to change in time . Such concentrations are called balanced .

Chemical equilibrium – This dynamic balance. The constancy of the concentrations of substances present in a closed system is a consequence of continuously ongoing chemical processes. The rates of the forward and reverse reactions are not equal to zero, but 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 rates 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] are equilibrium concentrations of substances.

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

The ratio of the rate constants of the forward and reverse reactions ( k etc / k arr ) is 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, but in a state of equilibrium they must ensure the stoichiometric transition of the starting materials into the reaction products and back. This means that the algebraic sum of all stages of the process is equal to the stoichiometric reaction equation, i.e. stoichiometric coefficients are the sum of the molecularities of all stages of the mechanism.

For a complex reaction

aA + bB  cC + dD

K c =

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

This is the second formulation of the law of mass action.

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 remain, as a rule, constant.

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

CO 2 (g) + C (tv)  2CO (g)

is written like this:

TO c =
.

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

A change in the concentration of any of these substances entails a change in the concentrations 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 temperature.

Equilibrium constant expressed in terms of molar concentrations of reactants ( TOWith) and the equilibrium constant expressed in terms of 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 the Gibbs energy is

G T = - RT ln Kp,

G T =  H – TS.

After equating the right parts of the equations:

- RT ln Kp =  H – TS

ln K R = -  H / ( RT) +  S/ R .

The equation not only establishes the type of 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 concentrations (as well as the rate constant of the reaction), the reaction mechanism, activation energy, and the presence of catalysts. A change in the mechanism, for example, upon 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.

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

The essence of chemical equilibrium

Equilibrium is a state that means things are always moving. Products are decomposed into reagents, and reagents are combined into products. Things move, but concentrations remain the same. The reaction is written with a double arrow instead of an equals sign to show that it is reversible.

Classic patterns

Back in the last century, chemists discovered certain patterns that provide for the possibility of changing the direction of the reaction in the same container. Knowledge of how chemical reactions proceed is incredibly important, both for laboratory research and industrial production. Wherein great importance has the ability to control all of these phenomena. It is human nature to intervene in many natural processes, especially reversible ones, in order to later use them for their own benefit. From knowledge of chemical reactions will be more useful if you are fluent in the levers of controlling 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 completed if it takes place in a closed system. The molecules of the resulting substances are in constant and random motion, and a reverse reaction may soon occur in which the molecules of the starting material will be restored.

Most often used in industry open systems. 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 fires, cement is produced in kilns. open type, blast furnaces operate with a constant supply of air, and ammonia is synthesized by the continuous removal of ammonia itself.

Reversible and irreversible chemical reactions

Based on the name, it is possible to give the appropriate definitions: irreversible reactions are those that are brought to completion, 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 may leave the reaction sphere. Thus, for example, it is possible to obtain gas (CaCO 3 \u003d CaO + CO 2), a precipitate (Cu (NO 3) 2 + H 2 S \u003d CuS + 2HNO 3) or others will also be considered irreversible if during the process it is released a large number of thermal energy, for example: 4P + 5O 2 \u003d 2P 2 O 5 + Q.

Almost all reactions that occur in nature are reversible. Regardless of such external conditions, like pressure and temperature, almost all processes can proceed simultaneously in different directions. As the law of mass action in chemistry says, the amount of heat absorbed 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 - the activities that chemists study. Many reactions go to their completion and then stop, which means that the reactants are completely converted into products, with no way 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 reactants in the second.

The dynamic state in which the concentrations of reactants and products remain constant is called equilibrium. It is possible to predict the behavior of substances with the help of certain laws that are applied in industries seeking to reduce the cost of producing specific chemicals. The concept of chemical equilibrium is also useful in understanding processes that maintain or potentially threaten human health. The chemical equilibrium constant is the value of a reaction factor that depends on ionic strength and temperature and is independent of the concentrations of reactants and products in solution.

Calculation of 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 a 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, not one phase is present, but at least two. For example, liquids and gases or and liquids.

The value of the equilibrium constant

For any given temperature, there is only one value for the equilibrium constant, which only changes 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 reactants. The reaction in this case can be called "total" 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 number of products formed. If this value tends to zero, we can assume that the reaction does not occur. If the values ​​of the equilibrium constant for the direct and reverse reactions are almost the same, then the amount of reactants and products will also be almost the same. This type of reaction is considered to be reversible.

Consider a specific reversible reaction

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

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

v 1 \u003d k 1 * c (H 2) * c (I 2),

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

When mixing iodine (I 2) and hydrogen (H 2) molecules, their interaction begins. At the initial stage, the concentration of these elements is maximum, but by the end of the reaction, the concentration of a 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, there comes a moment when these values ​​are equal, and this is the state called chemical equilibrium.

The expression of the chemical equilibrium constant, as a rule, is denoted using square brackets: , , . Since at equilibrium the speeds are equal, then:

k 1 \u003d k 2 2,

so we get the equation of the chemical equilibrium constant:

k 1 /k 2 = 2 / = K.

Le Chatelier-Brown principle

There is the following regularity: if a certain effect is made on a system that is in equilibrium (change the conditions of chemical equilibrium by changing temperature or pressure, for example), then the balance will shift in order 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 its expression

The equilibrium expression can be expressed in terms of the concentration of products and reactants. Only chemical substances 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 \u003d 1, and accordingly ceases to be taken into account, with the exception of highly concentrated solutions. For example, pure water has activity 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 the balance include the addition of a reactant or product (changes in concentration affect the balance). The addition of a reagent can bring equilibrium to the right in chemical equation where more product shapes appear. The addition of product can bring equilibrium to the left as more reactant forms become available.

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

Dynamic equilibrium is an example of a steady state function. For a system in steady state the behavior currently observed will continue into the future. Therefore, once the reaction reaches equilibrium, the ratio of product to reactant concentrations will remain the same even though the reaction continues.

How easy is it to talk about complex things?

Concepts such as chemical equilibrium and chemical equilibrium constant are quite difficult to understand. Let's take an example from life. Have you ever been 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 were measured and at the same speed moving 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.

Majority chemical reactions reversible, i.e. flow simultaneously in opposite directions. In cases where the forward and reverse reactions proceed at the same rate, chemical equilibrium occurs. For example, in a reversible homogeneous reaction: H 2 (g) + I 2 (g) ↔ 2HI (g), the ratio of the rates of direct and reverse reactions according to the law of mass action depends on the ratio of the concentrations of the reactants, namely: the rate of the direct reaction: υ 1 = k 1 [Н 2 ]. The rate of the reverse reaction: υ 2 \u003d k 2 2.

If H 2 and I 2 - starting materials, then at the first moment the rate of the direct 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 velocities are equalized, 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 at chemical equilibrium the rates of direct and reverse reactions are equal to V 1 \u003d V 2, then k 1 \u003d 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 general case for a reaction of the form аА+bB+…↔dD+eE+… the equilibrium constant is equal to . For the interaction between gaseous substances, the expression is often used, in which the reactants are represented by equilibrium partial pressures p. For the mentioned reaction .

The state of equilibrium characterizes the limit to which, under given 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 the equilibrium concentrations does not depend on which 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 reactants and on the temperature; the equilibrium constant does not depend on pressure (if it is too high) and on the concentration of reagents.

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

It shows 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 predominate in the equilibrium mixture. This equation allows us to calculate K from the value of ∆G o and then the equilibrium concentrations (partial pressures) of the reagents. If we take into account that ∆G o =∆Н o -Т∆S o , then after some transformation we get . It can be seen from this equation that the equilibrium constant is very sensitive to changes in temperature. The influence of the nature of the reagents 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 the conditions change, the state of equilibrium is disturbed, 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 corresponding to the new changed conditions.

The shift of equilibrium depending on changes in conditions is generally determined by the Le Chatelier principle (or the principle of moving equilibrium): if a system in equilibrium is influenced from outside by changing any of the conditions that determine the equilibrium position, then it is shifted in the direction of the process, the flow of which weakens the effect of the effect 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 the absorption of heat, and 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 the system in equilibrium (or vice versa, 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 the Le Chatelier principle, the introduction of additional amounts of a reagent into the equilibrium system causes a shift in the equilibrium in the direction in which the concentration of this substance decreases and, accordingly, the concentration of the products of its interaction increases.

The study of chemical equilibrium is of great importance both for theoretical research and for solving practical problems. By determining the equilibrium position for various temperatures and pressures, one can choose the most favorable conditions for conducting a chemical process. In the final choice of process conditions, their influence on the process rate is also taken into account.

Example 1 Calculation of the equilibrium constant of the reaction from the equilibrium concentrations of the reactants.

Calculate the equilibrium constant of the reaction A + B 2C, if the equilibrium concentrations [A] = 0.3 mol ∙ l -1; [B]=1.1 mol∙l -1; [C] \u003d 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. The reaction proceeds according to the equation A + 2B C.

Determine the equilibrium concentrations of the reactants if the initial concentrations of substances A and B are respectively 0.5 and 0.7 mol∙l -1, and the equilibrium constant of the reaction 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. The equilibrium concentrations of the reactants will be:

C A \u003d (o.5-x) mol ∙ l -1; C B \u003d (0.7-x) mol ∙ l -1; C C \u003d 2x mol ∙ l -1

x 1 \u003d 0.86; x 2 \u003d 0.44

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

C A \u003d 0.5-0.44 \u003d 0.06 mol ∙ l -1; C B \u003d 0.7-0.44 \u003d 0.26 mol ∙ l -1; C C \u003d 0.44 ∙ 2 \u003d 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 Kp=1.0685∙10 -4. The partial pressure of all reacting substances is the same and equal to 101325 Pa.

Solution.∆G 700 =2.303∙RT .

For this process:

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

Example 4. Shift in chemical equilibrium. In which direction will the equilibrium shift in the N 2 + 3H 2 2NH 3 -22 kcal 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) when the pressure decreases?

Solution. An increase in the concentration of substances on the left side of the reaction equation, according to the Le Chatelier 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 equilibrium to the left - towards an endothermic reaction that weakens the impact (case c).

A decrease in pressure (case d) will favor the 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 forward and reverse reactions in the system 2SO 2 (g) + O 2 (g) 2SO 3 (r) change if the volume of the gas mixture decreases three times? In which direction will the equilibrium of the system shift?

Solution. Let us denote the concentrations of reacting substances: = A, =b,=With. According to the law of mass action, the rates of the forward and reverse reactions before a change in volume are

v pr \u003d Ka 2 b, v arr \u003d K 1 s 2

After reducing the volume of a homogeneous system by a factor of three, the concentration of each of the reactants will increase by a factor of three: 3a,[O 2] = 3b; = 3s. At new concentrations of the rate v "np of the direct 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 rate of the forward reaction increased 27 times, and the reverse - only nine times. The equilibrium of the system has 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 with an increase in temperature from 30 to 70 0 C, 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 empirical rule according to the formula

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

Example 7 The equilibrium constant of a homogeneous system

CO (g) + H 2 O (g) CO 2 (g) + H 2 (g) at 850 ° C is 1. Calculate the concentrations of all substances at equilibrium if the initial concentrations are: [CO] ISC = 3 mol / l, [H 2 O] ISH \u003d 2 mol / l.

Solution. At equilibrium, the rates of the forward 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 r includes only the equilibrium concentrations of all substances in the system. Let us assume that by the moment of equilibrium the concentration [СО 2 ] Р = X mol/l. According to the equation of the system, the number of moles of hydrogen formed in this case will also be X mol/l. The same number of prayers (X mol / l) CO and H 2 O are consumed for the formation of X moles of CO 2 and H 2. Therefore, the equilibrium concentrations of all four substances (mol / l):

[CO 2] P \u003d [H 2] p \u003d 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 \u003d 6-2x-3x + x 2; 5x \u003d 6, l \u003d 1.2 mol / l.

EXAMPLE

Calculate the chemical equilibrium constant for a reversible homogeneous reaction, CO + H 2 O \u003d CO 2 + H 2, based on the fact that the equilibrium of the concentration of substances:

[CO] p = 0.045 mol/l,

[H 2 O] p \u003d 0.064 mol / l,

[CO 2 ] p \u003d 0.18 mol / l.

Given:

[CO] p = 0.045 mol/l

[H 2 O] p \u003d 0.064 mol / l

[CO 2] p \u003d 0.18 mol / l

Solution:

The molar ratio of the reaction products is 1:1, therefore

[CO 2] p \u003d [H 2] p \u003d 0.18 mol / l.

Based on expression (2.1), we calculate the value of the chemical equilibrium constant:

K x.r. = [CO 2] p [H 2] p / [CO 2] p [H 2 O] p \u003d 0.18 0.18 / 0.045 0.064

Answer: 11,25.

2. Calculation of equilibrium concentrations from the initial concentrations of reactants and vice versa

EXAMPLE 1.

The reversible gas reaction proceeds according to the equation:

CO + CI 2 \u003d COCI 2.

Initial concentrations of reactants:

[CO] 0 \u003d 0.03 mol / l;

0 \u003d 0.02 mol / l.

After the onset of equilibrium, the concentration of carbon monoxide became:

[CO] p = 0.021 mol/L.

Calculate the equilibrium concentrations of the remaining substances and the value of the chemical equilibrium constant.

Given:

[CO] 0 \u003d 0.03 mol / l

[С1 2 ] 0 = 0.02 mol/l

[CO] p = 0.021 mol/l

P , p , K x . p-?

Solution:

By the time of equilibrium, the change in CO concentration was:

∆[CO] \u003d [CO] 0 - [CO] p \u003d 0.03 - 0.021 \u003d 0.009 mol / l.

Since the molar ratio of the substances involved in the reaction is 1:1:1, the change in the concentration of all substances is the same:

[C1 2] p \u003d [C1 2] 0 - ∆ [C1 2] \u003d 0.02 - 0.009 \u003d 0.011 mol / l,

[COS1 2 ] p = 0.009 mol/l,

K x p \u003d [COS1 2] P / [CO] P [C1 2] p \u003d 0.009 / 0.021 0.011 \u003d 39.

The results of the calculations will be entered in the table, where the signs "+" and "-" mean, respectively, an increase or decrease in the concentration of a substance.

Answer:[C1 2 ] p = 0.011 mol/l; [COS1 2 ] p = 0.009 mol/l; K xp = 39.

EXAMPLE 2.

The equilibrium concentrations of substances involved in the reversible reaction 2NO + O 2 \u003d 2NO 2 are as follows (mol / l):

P = 0.056;

[O 2 ] = 0.028;

Given:

P = 0.056 mol/l

P = 0.028 mol/l

P = 0.044 mol/l

0 , [О 2 ] 0 – ?

Solution:

The initial concentration of nitric oxide (IV) was 0 = 0, and its change by the moment of equilibrium is ∆ = 0.044 mol/l.

The molar ratio of NO and NO 2 in the reaction is 2:2 (1:1), therefore, the initial concentration of NO will be:

0 \u003d p + 0.044 \u003d 0.056 + 0.044 \u003d 0.1 mol / l.

The molar ratio of O 2 and NO 2 is 1:2, hence the initial concentration of O 2 will be:

[O 2] 0 \u003d [O 2] p + 0.044 / 2 \u003d 0.028 + 0.022 \u003d 0.05 mol / l.

We write the results of calculations in a table

Answer: 0 = 0.1 mol/l; [O 2] 0 \u003d 0.05 mol / l.

EXAMPLE 3.

The ammonia synthesis reaction proceeds according to the equation ZH 2 + N 2 = 2NH 3. Initial concentrations of initial substances are equal (mol/l): hydrogen – 0.05; nitrogen - 0.04: the reaction rate constant is 0.3. Calculate: a) the initial rate of the reaction; b) the reaction rate when the ammonia concentration became equal to 0.02 mol/l.

Given:

a) [H 2] 0 \u003d 0.05 mol / l

0 = 0.04 mol/l

b) = 0.02 mol/l

Solution:

a) In accordance with the law of mass action, we find the initial reaction rate:

υ 0 \u003d k 0 3 0 \u003d 3 10 -1 3 \u003d 1.5 10 -6 mol / l s.

b) Based on the reaction equation, the molar ratio of hydrogen and ammonia is 3:2. An increase in ammonia concentration by 0.02 mol/l causes a decrease in hydrogen concentration by 0.03 mol/l (0.02 - 3/2 = 0.03).

Thus, by the time when the ammonia concentration increased by 0.02 mol/l, the hydrogen concentration decreased to 0.02 mol/l (0.05 - 0.03 = 0.02). The molar ratio of nitrogen and ammonia is 1:2. The nitrogen concentration will decrease by 0.01 mol (0.02 - 1/2 = = 0.01) and become equal to 0.03 mol / l (0.04 - 0.01 = 0.03). The reaction rate will also decrease as the concentration of the reactants decreases:

υ \u003d k 3 \u003d 3 10 -1 3 \u003d 7.2 10 -8 mol / l s.

Answer: a) 1.5 10–6 mol/l s; b) 7.2 10–8 mol/l s.

EXAMPLE 4

The reaction proceeds according to the equation 2NO + O 2 \u003d 2NO 2, some time after the start of the reaction, the concentrations of all substances participating in the reaction became: \u003d 0.04 mol / l; [O 2 ] = 0.01 mol/l; = 0.02 mol/l. Calculate the initial concentrations of starting materials and the initial reaction rate if the reaction rate constant k = 1.

Given:

0.04 mol/l

[O 2] \u003d 0.01 mol / l

0.02 mol/l

0 , 0 , x 0 – ?

Solution:

According to the reaction equation, the molar ratio of NO and NO 2 is 2:2 (1:1).

An increase in the concentration of the reaction product NO 2 to 0.02 mol/l caused a decrease in the concentration of NO by 0.02 mol. Therefore, the initial concentration of nitric oxide (II) was:

0 \u003d +0.02 \u003d 0.04 + 0.02 \u003d 0.06 mol / l.

The molar ratio of O 2 and NO 2 is 1:2, therefore, an increase in the concentration of NO 2 to 0.02 mol caused a decrease in the oxygen concentration by 0.01 mol (0.02 1/2 \u003d 0.01). As a result, the initial oxygen concentration was:

[O 2] 0 \u003d [O 2] + 0.01 \u003d 0.01 + 0.01 \u003d 0.02 mol / l.

Initial reaction rate

υ 0 \u003d k 0 2 0 \u003d 1 2 \u003d 7.2 10 -5 mol / l s.

Answer: 0 = 0.06 mol/l; [O 2] 0 \u003d 0.02 mol / l;

x 0 \u003d 7.2 10 -5 mol / l s.

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§ 3.2. Equilibrium constant
and isobaric potential of the reaction

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 starting materials 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 derive them using the simplest methods of mathematical logic, so the derivation of this formula is given below. After reading this material, you will get acquainted with the simplest representations of probability theory, with the entropy of activation, etc.

Not only the activation energy determines the rate of a chemical reaction. A huge role is played by the size and shape of the reacting molecules and the arrangement of reactive atoms or their groups in them. In this regard, when two particles collide, their specific orientation is important, i.e., the contact of precisely those centers that are reactive.

Let us denote the probability of the orientation of molecules necessary for the interaction in a collision as W:

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

From this expression follows:

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

The greater the probability of the necessary orientation for the reaction to proceed, the higher its rate 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:

We introduce the proportionality coefficient Z and put the 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 entropy of activation, i.e., increases the probability of the orientation of the reacting particles appropriate for the interaction.

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

Another feature of the catalytic action is that the catalyst takes excess energy from the newly formed particle, and it does not decompose into the original particles due to its high energy 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 of the basic equation of chemical kinetics:

The ratio of the two coefficients of proportionality Z pr / Z arr is a constant value that we will 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 (verify this by drawing the enthalpy diagram of a reaction that proceeds with the release of heat, and not forgetting that in this case D N< 0 ):

Similarly, the difference denote D S:

Explain why there is a minus sign before the brackets.

We get the equation:

Let's take the logarithm of both sides of this equation:

Where do we get:

This equation is so important to chemistry and other sciences that many foreign chemistry students wear shirts with this formula on them.

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

This formula has one feature: if the equilibrium constant is determined through the pressures of gaseous substances, then the pressures of these substances in atmospheres are substituted into the expression of the equilibrium constant (1 atm \u003d 101325 Pa \u003d 760 mm Hg).

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 the equilibrium reaction mixture contains reaction products (substances on the right side of the reaction equation), 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 equilibrium mixture contains reaction products and the more starting substances, the smaller 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.

When 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 DG>0, say that the reaction does not pass.

Thus,

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

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

DG>0, That TO<1 - the equilibrium is shifted towards the starting substances, to the left.

If you need to find out if the reaction you are interested in is possible (for example, to find out if the synthesis of the desired dye is possible, whether the given mineral composition will be sintered, the effect of atmospheric oxygen on 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 needs to be done to calculate the change in isobaric potential and the equilibrium constant at different temperatures (calculation algorithm).

1. Write out from the reference tables the values ​​\u200b\u200b(for a temperature of 298 K) of the enthalpies of formation from simple substances D H arr and entropy S all substances written in the equation of a chemical reaction. If D H arr expressed in kJ/mol, they should be converted to J/mol (why?).

2. Calculate the enthalpy change 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 entropy change 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 for the dependence of the change in the isobaric potential on changes in the enthalpy of reaction, entropy and temperature, substituting the numerical values ​​​​just obtained into the equation known to you D Н r-tion And D S:

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

6. By sign D G, 298 make a conclusion about the possibility of passing the reaction at 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 at the temperature T you are interested in:

and conclude how the change in temperature affects the possibility of passing the reaction. If it turns out that at this temperature the change in the isobaric potential has become less positive or more negative compared to D G 298, then, therefore, at this temperature the reaction becomes more probable.

8. Calculate the equilibrium constant K from the equation known to you at the temperature T of interest to you:

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

To conclude that the reaction can proceed at a negative value of the change in the isobaric potential ( D G<0 ) thermodynamic data alone is often insufficient. A thermodynamically possible reaction may turn out to be kinetically retarded and feasible under changing conditions (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 luster 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:

Calculate the enthalpy change in this reaction, remembering that the enthalpies of simple substances are zero:

We 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 occurs spontaneously. However, it is possible to confidently say that the reaction is spontaneous only by the sign of the change in the 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, so 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 the isobaric potential on the changes in enthalpy, entropy and temperature:

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

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

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

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

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 r-tion \u003d 0(the equilibrium constant is 1):

Where do we get:

T=150300/168.2=894K, or 621°C.

At this temperature, the reaction is equally likely to proceed both in the forward and reverse directions. At temperatures above 621°C, the reverse reaction of reduction of Fe 3 O 4 with hydrogen begins to predominate. 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 placed with his fat (why is this important?) hands on the surface of the water in a bowl. The pebble did not drown and began to float on 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 this way, and the mineral got its name from the name of 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 magnetite formula is depicted as follows: FeO Fe 2 O 3, implying that magnetite consists of two iron oxides. This is wrong: magnetite is an individual substance.

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

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