(4.45), replacing the designation of concentrations: A: [A]""[B]" = ?[D] /; p = 0.004; [H 2 O] p = 0.064; [CO 2 ]p = 0.016; [H 2] p \u003d 0.016,
What are the initial concentrations of water and CO? Answer: Kp = 1; ref = 0.08 mol/l; [CO] ref = 0.02 mol/l.
Solution:
The reaction equation is:
CO (g) + H 2 O (g) ⇔ CO 2 (g) + H 2 (g)
Equation constant this reaction has the expression:
Kp = .
/ .
Substituting the task data into the expression, we get:
K p = (0.016 .
0,016)/(0,004 .
0,064) = 1.
To find the initial concentrations of substances H 2 O and CO, we take into account that, according to the reaction equation, from 1 mol of CO and 1 mol of H 2 O, 1 mol of CO 2 and 1 mol of H 2 are formed. Since, according to the condition of the problem, 0.016 mol CO 2 and 0.016 mol H 2 were formed in each liter of the system, then 0.016 mol CO and H 2 O were consumed. Thus, the desired initial concentrations are:
Ref \u003d [H 2 O] P + 0.016 \u003d 0.004 + 0.016 \u003d 0.02 mol / l;
[CO] ref \u003d [CO] P + 0.016 \u003d 0.064 + 0.016 \u003d 0.08 mol / l.
Answer: Kp = 1; ref = 0.08 mol/l; [CO] ref = 0.02 mol/l.
Task 136.
Equilibrium constant of a homogeneous system:
CO (g) + H 2 O (g) ⇔ CO 2 (g) + H 2 (g)
at some temperature is 1.00. Calculate the equilibrium concentrations of all reactants if the initial concentrations are equal (mol/l): [CO] ref = 0.10; [H 2 O] ref = 0.40.
Answer: [CO 2] P \u003d [H 2] P \u003d 0.08; [CO]P = 0.02; [H 2 O] P = 0.32.
Solution:
The reaction equation is:
CO (g) + H 2 O (g) ↔ CO 2 (g) + H 2 (g)
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 equilibrium constant given system:
We denote by x mol / l the equilibrium concentration of one of the reaction products, then the equilibrium concentration of the other will also be x mol / l, since they are both formed in the same amount. The equilibrium concentrations of the starting substances will be:
[CO] ref = 0.10 – x mol/l; [H 2 O] ref = 0.40 - x mol / l. (since the formation of x mol / l of the reaction product consumes, respectively, x mol / l of CO and H 2 O. At the moment of equilibrium, the concentration of all substances will be (mol / l): [CO 2] P \u003d [H 2] P \u003d x ; [CO] P \u003d 0.10 - x; [H 2 O] P \u003d 0.4 - x.
We substitute these values into the expression for the equilibrium constant:
Solving the equation, we find x = 0.08. From here concentration equilibrium(mol/l):
[CO 2 ] P = [H 2 ] P = x = 0.08 mol/l;
[H 2 O] P \u003d 0.4 - x \u003d 0.4 - 0.08 \u003d 0.32 mol / l;
[CO] P \u003d 0.10 - x \u003d 0.10 - 0.08 \u003d 0.02 mol / l.
Task 137.
The equilibrium constant of a homogeneous system N 2 + 3H 2 \u003d 2NH 3 at a certain temperature is 0.1. The equilibrium concentrations of hydrogen and ammonia are 0.2 and 0.08 mol/l, respectively. Calculate the equilibrium and initial concentrations of nitrogen. Answer: P = 8 moles/l; ref = 8.04 mol/l.
Solution:
The reaction equation is:
N 2 + ZN 2 ↔ 2NH 3
Let us denote the equilibrium concentration of N2 as x mol/L. The expression for the equilibrium constant of this reaction is:
Let us substitute the data of the problem into the expression of the equilibrium constant and find the concentration N 2
To find the initial concentration of N 2, we take into account that, according to the reaction equation for the formation of 1 mol of NH 3, ½ mol of N 2 is spent. Since, according to the condition of the problem, 0.08 mol of NH 3 was formed in each liter of the system, 0.08 .
1/2 \u003d 0.04 mol N 2. Thus, the desired initial concentration of N 2 is equal to:
Ref \u003d P + 0.04 \u003d 8 + 0.04 \u003d 8.04 mol / l.
Answer: P = 8 moles/l; ref = 8.04 mol/l.
Task 138.
At some temperature, the equilibrium of a homogeneous system
2NO + O 2 ↔ 2NO 2 was established at the following concentrations of reactants (moles/l): p = 0.2; [O 2 ] p = 0.1; p = 0.1. Calculate the equilibrium constant and the initial concentration of NO and O 2 . Answer: K = 2.5; ref = 0.3 mol/l; [O 2 ] ex x = 0.15 mol/L.
Solution:
Reaction equation:
2NO + O 2 ↔ 2NO 2
Equilibrium constant
To find the initial concentrations of NO and O 2, we take into account that, according to the reaction equation, 2 mol NO 2 is formed from 2 mol NO and 1 mol O2, then 0.1 mol NO and 0.05 mol O 2 were spent. Thus, the initial concentrations of NO and O 2 are equal:
Ref = NO] p + 0.1 = 0.2 + 0.1 = 0.3 mol/l;
[O 2] ref \u003d [O 2] p + 0.05 \u003d 0.1 + 0.05 \u003d 0.15 mol / l.
Answer: Kp = 2.5; ref = 0.3 mol/l; [O 2] ref = 0.15 mol / l.
Shift in the equilibrium of a chemical system
Task 139.
Why does the equilibrium of the system shift when pressure changes?
N 2 + 3Н 2 ↔ 2NH 3 and, the equilibrium of the N 2 + O 2 2NO system does not shift? Justify your answer based on the calculation of the rate of forward and reverse reactions in these systems before and after the pressure change. Write expressions for the equilibrium constants of each of these systems.
Solution:
a) Reaction equation:
N 2 + 3H 2 ↔ 2NH 3.
It follows from the reaction equation that the reaction proceeds with a decrease in volume in the system (from 4 moles of gaseous substances, 2 moles of a gaseous substance are formed). Therefore, with a change in pressure in the system, a shift in equilibrium will be observed. If the pressure in this system is increased, then, according to Le Chatelier's principle, the equilibrium will shift to the right, towards a decrease in volume. When the equilibrium in the system shifts to the right, the rate of the forward reaction will be greater than the rate of the reverse reaction:
pr>arr or pr \u003d k 3\u003e o br \u003d k 2.
If the pressure in the system is reduced, then the equilibrium of the system will shift to the left, towards an increase in volume, then when the equilibrium shifts to the left, the rate of the direct reaction will be less than the rate of the direct one:
etc<
обр
или (пр =
k
3 )< (обр
= k
2).
b) Reaction equation:
N2 + O2) ↔ 2NO. .
It follows from the reaction equation that during the course of the reaction it is not accompanied by a change in volume, the reaction proceeds without changing the number of moles of gaseous substances. Therefore, a change in pressure in the system will not lead to a shift in equilibrium, so the rates of the forward and reverse reactions will be equal:
pr \u003d arr \u003d or (pr k [O 2]) \u003d (arr \u003d k 2) .
Task 140.
Initial concentrations ref and [C1 2 ]ref in a homogeneous system
2NO + Cl 2 ↔ 2NOС1 are 0.5 and 0.2 mol/l, respectively. Calculate the equilibrium constant if 20% NO has reacted by the time equilibrium is reached. Answer: 0.417.
Solution:
The reaction equation is: 2NO + Cl 2 ↔ 2NOС1
According to the condition of the problem, 20% NO entered the reaction, which is 0.5 .
0.2 = 0.1 mol, but 0.5 - 0.1 = 0.4 mol NO did not react. It follows from the reaction equation that for every 2 moles of NO, 1 mole of Cl2 is consumed, and 2 moles of NOCl are formed. Therefore, 0.05 mol Cl 2 reacted with 0.1 mol NO and 0.1 mol NOCl was formed. 0.15 mol Cl 2 remained unused (0.2 - 0.05 = 0.15). Thus, the equilibrium concentrations of the substances involved are equal (mol / l):
P = 0.4; p=0.15; p = 0.1.
Equilibrium constant this reaction is expressed by the equation:
Substituting in this expression the equilibrium concentrations of substances, we obtain.
Let us derive the equilibrium constant for reversible chemical reactions (in general form)
back reaction rate:
We transfer the constants (rate constants) to the left side of the equation, and the variables (concentrations) to the right side of the equation, i.e. We write this equality in the form of a proportion:
The expression of the constant includes the equilibrium concentrations of substances, taken in powers equal to the coefficients in front of the substance in the reaction equation.
The equilibrium constant reflects the depth of the process. The greater the value of the equilibrium constant, the higher the concentration of reaction products at the moment of equilibrium, i.e. the more complete the reaction.
The equilibrium constant depends on the nature of the reactants, but does not depend on the presence of a catalyst, since it equally accelerates both the forward and reverse reactions. The influence of other factors (concentration of substances, gas pressure and temperature) on the value of the equilibrium constant will be analyzed below using specific examples.
Consider the derivation of the expression for the equilibrium constant on specific examples.
Example 2 for reaction: N 2(g) +3H 2(g) Û 2NH 3(g)
V pr \u003d k 1 3; V arr. = k 2 2 . If V pr \u003d V arr. , That k 1 [ H 2 ] 3 = k 2 2 ,
.
If solid substances participate in the reaction (heterogeneous system), then their concentration is not included in the expression of the reaction rate (because it remains constant per unit surface per unit time), and hence the equilibrium constants.
Example 3 for reaction: C (tv.) + O 2 (g) Û CO 2 (g) the chemical equilibrium constant will be equal to 
.
Example 4 In a reversible chemical reaction A + 2B Û C equilibrium occurred at the following equilibrium concentrations: [A] = 0.6 mol/l; [B] = 1.2 mol/l; [C] = 2.16 mol/l. Determine the equilibrium constant and the initial concentrations of the substance A And IN.
A quantitative characteristic showing the direction of the reaction and the shift in the concentration of substances is called the equilibrium constant of a chemical reaction. The equilibrium constant depends on the temperature and the nature of the reactants.
Reversible and irreversible reactions
All reactions can be divided into two types:
- reversible, simultaneously flowing in two mutually opposite directions;
- irreversible flowing in the same direction with a total consumption of at least one initial substance.
In irreversible reactions, insoluble substances are usually formed in the form of a precipitate or gas. These reactions include:
- combustion:
C 2 H 5 OH + 3O 2 → 2CO 2 + H 2 O;
- decomposition:
2KMnO 4 → K 2 MnO 4 + MnO 2 + H 2 O;
- connection with the formation of a precipitate or gas:
BaCl 2 + Na 2 SO 4 → BaSO 4 ↓ + 2NaCl.
Rice. 1. Precipitation of BaSO 4 .
Reversible reactions are possible only under certain constant conditions. The original substances give a new substance, which immediately breaks down into its constituent parts and is collected again. For example, as a result of the reaction 2NO + O 2 ↔ 2NO 2 nitric oxide (IV) easily decomposes into nitric oxide (II) and oxygen.
Equilibrium
After a certain time, the rate of the reversible reaction slows down. Chemical equilibrium is achieved - a state in which there is no change in the concentration of the starting substances and reaction products over time, since the rates of the forward and reverse reactions are equalized. Equilibrium is possible only in homogeneous systems, that is, all reacting substances are either liquids or gases.
Consider the chemical equilibrium on the example of the reaction of the interaction of hydrogen with iodine:
- direct reaction -
H 2 + I 2 ↔ 2HI;
- back reaction -
2HI ↔ H 2 + I 2 .
As soon as two reagents are mixed - hydrogen and iodine - hydrogen iodine does not yet exist, since simple substances only react. A large number of starting substances actively react with each other, so the rate of the direct reaction will be maximum. In this case, the reverse reaction does not proceed, and its rate is zero.
The rate of a direct reaction can be expressed graphically:
ν pr = k pr ∙ ∙ ,
where k pr is the rate constant of the direct reaction.
Over time, the reagents are consumed, their concentration decreases. Accordingly, the rate of the forward reaction decreases. At the same time, the concentration of a new substance, hydrogen iodide, increases. When accumulated, it begins to decompose, and the rate of the reverse reaction increases. It can be expressed as
ν arr = k arr ∙ 2 .
Hydrogen iodide is squared, since the coefficient of the molecule is two.
At some point, the rates of the forward and reverse reactions equalize. There is a state of chemical equilibrium.
Rice. 2. Graph of reaction rate versus time.
The equilibrium can be shifted either towards the starting materials or towards the products of the reaction. The displacement under the influence of external factors is called Le Chatelier's principle. Equilibrium is affected by temperature, pressure, concentration of one of the substances.
Constant calculation
In a state of equilibrium, both reactions proceed, but at the same time, the concentrations of substances are in equilibrium (equilibrium concentrations are formed), since the rates are balanced (ν pr \u003d ν arr).
Chemical equilibrium is characterized by the chemical equilibrium constant, which is expressed by the summary formula:
K p \u003d k pr / k arr \u003d const.
The reaction rate constants can be expressed in terms of the reaction rate ratio. Let's take the conditional equation of the reverse reaction:
aA + bB ↔ cC + dD.
Then the rates of the forward and reverse reactions will be equal:
- ν inc = k inc ∙ [A] p a ∙ [B] p b
- ν arr = k arr ∙ [C] p c ∙ [D] p d .
Accordingly, if
ν pr \u003d ν arr,
k ex ∙ [A] p a ∙ [B] p b = k arr ∙ [C] p c ∙ [D] p d .
From here we can express the ratio of constants:
k arr / k inc = [C] p c ∙ [D] p d / [A] p a ∙ [B] p b .
This ratio is equal to the equilibrium constant:
K p = [C] p c ∙ [D] p d / [A] p a ∙ [B] p b .
Rice. 3. The formula for the equilibrium constant.
The value shows how many times the rate of the forward reaction is greater than the rate of the reverse reaction.
What have we learned?
Reactions depending on the final products are classified into reversible and irreversible. Reversible reactions proceed in both directions: the starting materials form final products, which decompose into starting substances. During a reaction, the rates of the forward and reverse reactions are balanced. This state is called chemical equilibrium. It can be expressed as the ratio of the product of the equilibrium concentrations of the reaction products to the product of the equilibrium concentrations of the starting materials.
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In chemistry, the equilibrium state characterizes most gaseous and liquid systems, as well as a large group of hard alloys. Therefore, the laws of chemical equilibrium are of great practical importance. When analyzing the equation for the Gibbs energy, it is found that states can arise in thermodynamic systems when oppositely directed processes occur simultaneously, but the state of the system as a whole remains in equilibrium, i.e. its parameters are unchanged (A= 0). The constancy of the parameters of the system in time, however, is a necessary but not sufficient condition for true chemical equilibrium. Under certain conditions, the parameters of systems in which reactions of the type
For example, a mixture of gaseous ammonia with air is unchanged over time. However, it is enough to add a heated catalyst - chromium oxide Cr 2 Oe to this mixture, as the reaction begins, leading to the formation of nitrogen dioxide M0 2: 
The calculation according to the Gibbs-Helmholtz equation shows that for this reaction AG 2 ° 98 0 and it belongs to reactions of the type (11.1). So, in this case, we are not dealing with true balance in the system, but only inhibited (metastable) state.
The model reaction describing truly equilibrium systems will be
A specific example of a truly equilibrium homogeneous system is an aqueous solution prepared from two salts: iron(III) chloride FeCl 3 and potassium thiocyanate KCNS and containing four substances. In addition to the indicated reagents, it also contains two reaction products - iron thiocyanate) Fe (CNS) 3 and potassium chloride KSL:
This type of reaction is called kinetically inverted, since they flow both in the forward and in the opposite direction at any state of the system. In the range of parameters, when the rates of the forward and reverse reactions become equal, the system also becomes thermodynamically reversible. Therefore, one often speaks of the reversibility of such reactions without specifying which type of reversibility is meant. The change in the Gibbs energy in systems where reactions of the type (11.1) and (11.2) take place can be represented by a diagram (Fig. 11.2).
Rice. 11.2.
The abscissa axis of the diagram shows the composition of the system (it changes from pure reagents A; B in the initial (initial) state, to pure reaction products P; R in the final (con) state) in mole fractions of X, each component, and on the ordinate axis - the value Gibbs energy for a mixture of the current composition. As follows from the diagram, for a reaction of the type (11.1), the dependence of the Gibbs energy on the composition is monotonic. The value C of the system decreases as the process proceeds, and the value D, C (index G emphasizes that this change in the Gibbs energy for a chemical reaction), equal to the difference between the Gibbs energies in the initial and final states, is always negative.
For a reaction of the type (11.2), the picture is different. The value C of the system in this diagram has a minimum, and the entire diagram is divided into two areas: to the left of the point Z as the process proceeds, the value ts.fi(change in the Gibbs energy for a direct (pr) reaction) is negative, and to the right of the point Z as the process proceeds, the value of AG is negative (? arr (change in the Gibbs energy for the reverse (arr) reaction). Point Z is the equilibrium point - in it and A g C pr, and A; .C oG)p are equal to zero. Composition of the system at a point Z called balanced composition. The state of equilibrium at a point Z in chemical thermodynamics it is characterized by a special value - equilibrium constant
Siya Yaravn-
The equilibrium constant for the state of the system at constant temperature There is constant value. The equilibrium constant can be expressed in terms of various parameters of the system, which is reflected by the indices: K s(via molar concentrations), K x(through mole fractions), K r(through partial pressures), etc. The algorithm for constructing an expression for the equilibrium constant is simple: it is equal to a fraction, in the numerator of which is the product of the equilibrium parameters (concentrations, partial pressures, mole fractions, etc.) of the reaction products in powers equal to the stoichiometric coefficients of the corresponding substances, and in the denominator - a similar product for the reagents.
Consider the expression 
for model reaction (11.2):
The formula of a substance enclosed in square brackets indicates the molar concentration of the substance in the system. When it is clear from the context that we are talking about a state of equilibrium, the index "equal" is omitted. The features of the form of writing the equilibrium constants in the case of heterogeneous systems will be considered later in this chapter. If some component in the system is a solid, then its concentration is a constant value, it is transferred to the left side and included in the equilibrium constant. In this case, such a component is not represented in the analytical expression.
Analytically, the equilibrium constants of various types are related to each other as follows:
At the same time, no matter how K is expressed, it is a parameter of the system and does not depend on the position of the point Z. This point is mobile and depends on the method of preparation of the system, which can be verified experimentally. If you change the ratio of components in the initial state, for example, add components O and E (reaction products) to the system, then the position of the point Z will shift to the left, but if you add components A and B (reagents), then to the right. Below we consider such a shift in the position of the equilibrium point of the system based on the Le Chatelier-Brown principle.
Equilibrium constants can be both dimensional and dimensionless quantities. The dimension of the equilibrium constant is determined by the dimension of the quantity through which it is expressed (concentration, partial pressure, etc.), and is equal to this dimension to a degree equal to the difference between the coefficients of the final and initial components of the reaction: 
The equilibrium constant is written as a fraction, the numerator of which includes the parameters of the reaction products, and the denominator - the parameters of the initial substances. It is most convenient to express the equilibrium constant in terms of mole fractions (in this case, it has no dimension):
For processes involving gaseous components, it is convenient to express the equilibrium constant in terms of partial pressures of gaseous components:
The expression of the equilibrium constant in terms of molar concentrations is convenient for processes at a constant volume or in solutions (if the volume of solutions remains practically constant):
If the process goes without changing the number of moles (Dy = 0), then all expressions of the equilibrium constants coincide: 
In this case, all equilibrium constants become dimensionless.
