Determination of equilibrium constants of chemical reactions and calculation of chemical equilibrium. The equilibrium constant of a chemical reaction
Study questions
Balance state
Equilibrium constant
Calculation of equilibrium concentrations
Bias chemical equilibrium. Le Chatelier's principle
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:
|
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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 – T S.
After equating the right parts of the equations:
- RT ln Kp = H – T S
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.
Task 135.
Calculate Equilibrium Constant for Homogeneous System
if the equilibrium concentration of the reactants (moles/l):
[CO] 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: K = 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) + H2 (g)
The equation constant for this reaction has the expression:
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.
The equilibrium constant of a homogeneous system
at a certain temperature it is equal to 1. Calculate the equilibrium concentrations of all reacting substances if the initial concentrations are equal (moles/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 the equilibrium constant of the 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. Hence the 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 \u003d 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
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.
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 when the reaction 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 [С1 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.
The equilibrium constant of this reaction is expressed by the equation:
Substituting in this expression the equilibrium concentrations of substances, we obtain.
Chemical equilibrium constant
All chemical reactions can be divided into 2 groups: irreversible reactions, i.e. reactions proceeding until the complete consumption of one of the reacting substances, and reversible reactions in which none of the reacting substances is completely consumed. This is due to the fact that an irreversible reaction proceeds in only one direction. A reversible reaction can proceed both in the forward and in the reverse direction. For example, the reaction
Zn + H 2 SO 4 ® ZnSO 4 + H 2
proceeds until the complete disappearance of either sulfuric acid or zinc and does not proceed in the opposite direction: metallic zinc and sulfuric acid cannot be obtained by passing hydrogen into an aqueous solution of zinc sulfate. Therefore, this reaction is irreversible.
A classic example of a reversible reaction is the synthesis of ammonia from nitrogen and hydrogen: N 2 + 3 H 2 ⇆ 2 NH 3.
If 1 mol of nitrogen and 3 mol of hydrogen are mixed at a high temperature, then even after a sufficiently long reaction time, not only the reaction product (NH 3) will be present in the reactor, but also unreacted starting materials(N 2 and H 2). If, under the same conditions, not a mixture of nitrogen and hydrogen, but pure ammonia is introduced into the reactor, then after a while it will turn out that part of the ammonia has decomposed into nitrogen and hydrogen, i.e. the reaction proceeds in the opposite direction.
To understand the nature of chemical equilibrium, it is necessary to consider the question of the rates of forward and reverse reactions. Under speed chemical reaction understand the change in the concentration of the starting substance or reaction product per unit of time. When studying issues of chemical equilibrium, the concentrations of substances are expressed in mol / l; these concentrations indicate how many moles of a given reactant are contained in 1 liter of the vessel. For example, the statement “ammonia concentration is 3 mol/l” means that each liter of the volume under consideration contains 3 mol of ammonia.
Chemical reactions are carried out as a result of collisions between molecules, therefore, the more molecules are in a unit volume, the more often collisions occur between them, and the greater the reaction rate. Thus, the greater the concentration of the reactants, the greater the rate of the reaction.
equilibrium in the system is not observed
there is no visible change.
So, for example, the concentrations of all substances can remain unchanged for an arbitrarily long time if no external influence is exerted on the system. This constancy of concentrations in a system in a state of chemical equilibrium does not at all mean the absence of interaction and is explained by the fact that the forward and reverse reactions proceed at the same rate. This state is also called true chemical equilibrium. Thus, true chemical equilibrium is dynamic equilibrium.
False equilibrium must be distinguished from true equilibrium. The constancy of the parameters of the system (concentrations of substances, pressure, temperature) is a necessary but not sufficient sign of true chemical equilibrium. This can be illustrated by the following example. The interaction of nitrogen and hydrogen with the formation of ammonia, as well as the decomposition of ammonia, proceeds at a noticeable rate at a high temperature (about 500 ° C). If hydrogen, nitrogen and ammonia are mixed at room temperature in any ratio, then the reaction N 2 + 3 H 2 ⇆ 2 NH 3
will not leak, and all system parameters will remain constant. However, in this case, the equilibrium is false, not true, because it is not dynamic; there is no chemical interaction in the system: the rate of both the forward and reverse reactions is zero.
In the further presentation of the material, the term "chemical equilibrium" will be used in relation to the true chemical equilibrium.
The quantitative characteristic of a system in a state of chemical equilibrium is equilibrium constant K .
For the general case of a reversible reaction a A + b B + ... ⇆ p P + q Q + ...
The equilibrium constant is expressed the following formula:
In formula 5.1 C(A), C(B), C(P) C(Q) are the equilibrium concentrations (mol/l) of all substances participating in the reaction, i.e. concentrations that are established in the system at the moment of chemical equilibrium; a, b, p, q are stoichiometric coefficients in the reaction equation.
The expression for the equilibrium constant for the ammonia synthesis reaction N 2 +3H 2 ⇆2NH 3 is as follows: . (5.2)
Thus, numerical value the chemical equilibrium constant is equal to the ratio of the product of the equilibrium concentrations of the reaction products to the product of the equilibrium concentrations of the starting substances, and the concentration of each substance must be raised to a power equal to the stoichiometric coefficient in the reaction equation.
It is important to understand that the equilibrium constant is expressed in terms of equilibrium concentrations, but does not depend on them ; on the contrary, the ratio of the equilibrium concentrations of the substances participating in the reaction will be such as to correspond to the equilibrium constant. The equilibrium constant depends on the nature of the reacting substances and temperature and is a constant (at a constant temperature) value .
If K >> 1, then the numerator of the fraction of the expression of the equilibrium constant is many times greater than the denominator, therefore, at the moment of equilibrium, the reaction products predominate in the system, i.e. the reaction proceeds largely in the forward direction.
If K<< 1, то знаменатель во много раз превышает числитель, следовательно, в момент равновесия в системе преобладают исходные вещества, т.е. реакция лишь в незначительной степени протекает в прямом направлении.
If K ≈ 1, then the equilibrium concentrations of the initial substances and reaction products are comparable; the reaction proceeds to a significant extent both in the forward and in the reverse direction.
It should be borne in mind that the expression of the equilibrium constant includes the concentrations of only those substances that are in the gas phase or in the dissolved state (if the reaction proceeds in solution). If a solid substance is involved in the reaction, then the interaction occurs on its surface, so the concentration of the solid substance is assumed to be constant and is not written in the expression of the equilibrium constant.
CO 2 (gas) + C (solid) ⇆ 2 CO (gas)
CaCO 3 (solid) ⇆ CaO (solid) + CO 2 (gas) K = C (CO 2)
Ca 3 (PO 4) 2 (solid) ⇆ 3Ca 2+ (solution) + 2PO 4 3– (solution) K = C 3 (Ca 2+) C 2 (PO 4 3–)
Chemical equilibrium constant
The quantitative characteristic of chemical equilibrium is equilibrium constant , which can be expressed in terms of equilibrium concentrations C i , partial pressures P i or mole fractions X i of the reactants. For some reaction
the corresponding equilibrium constants are expressed as follows:
The equilibrium constant is a characteristic quantity for every reversible chemical reaction; the value of the equilibrium constant depends only on the nature of the reacting substances and temperature. Based on the equation of state of an ideal gas, written as the relation P i = C i RT, where С i = n i /V, and Dalton's law for an ideal gas mixture, expressed by the equation P = ΣP i , we can derive the relationship between the partial pressure P i , molar concentration C i and mole fraction X i of the i-th component:
From here we get the relation between K c , K p and K x:
Here Δν is the change in the number of moles of gaseous substances during the reaction:
Δν = – ν 1 – ν 2 – ... + ν" 1 + ν" 2 + ...
The value of the equilibrium constant K x, in contrast to the equilibrium constants K c and K p , depends on the total pressure Р.
The expression for the equilibrium constant of an elementary reversible reaction can be derived from kinetic concepts. Consider the process of establishing equilibrium in a system in which at the initial moment of time only the initial substances are present. The rate of the forward reaction V 1 at this moment is maximum, and the rate of the reverse reaction V 2 is zero:
As the concentration of the starting substances decreases, the concentration of the reaction products increases; accordingly, the rate of the forward reaction decreases, the rate of the reverse reaction increases. Obviously, after some time, the rates of the forward and reverse reactions will become equal, after which the concentrations of the reactants will stop changing, i.e. chemical equilibrium is established.
Assuming that V 1 \u003d V 2, we can write:
Thus, the equilibrium constant is the ratio of the rate constants of the forward and reverse reactions. This implies the physical meaning of the equilibrium constant: it shows how many times the rate of the forward reaction is greater than the rate of the reverse at a given temperature and concentrations of all reacting substances equal to 1 mol / l. The above derivation of the expression for the equilibrium constant, however, proceeds from a false general case the assumption that the rate of a chemical reaction is directly proportional to the product of the concentrations of the reactants, taken in powers equal to stoichiometric coefficients. As is known, in the general case, the exponents at concentrations of reagents in the kinetic equation of a chemical reaction do not coincide with stoichiometric coefficients.
11. Redox reactions: definition, basic concepts, the essence of oxidation and reduction, the most important oxidizing and reducing agents of the reaction.
Redox is called processes that are accompanied by the displacement of electrons from one free or bound atoms to others. Since in such cases it is not the degree of displacement that matters, but only the number of displaced electrons, it is customary to conditionally consider the displacement to be always complete and speak of the recoil or displacement of electrons.
If an atom or ion of an element donates or accepts electrons, then in the first case, the oxidation state of the element rises, and it goes into the oxidized form (OF), and in the second case, it decreases, and the element goes into the reduced form (WF). Both forms form a conjugated redox pair. Each redox reaction involves two conjugated pairs. One of them corresponds to the transition of an oxidizing agent that accepts electrons to its reduced form (OF 1 → VF 1), and the other corresponds to the transition of a reducing agent that donates electrons to its oxidized form (VF 2 → OF 2), for example:
Cl 2 + 2 I - → 2 Cl - + I 2
OF 1 WF 1 WF 2 OF 2
(here Cl 2 is an oxidizing agent, I is a reducing agent)
Thus, the same reaction is always both the process of oxidation of the reducing agent and the process of reduction of the oxidizing agent.
The coefficients in the equations of redox reactions can be found electronic balance methods and electron-ion balance. In the first case, the number of received or donated electrons is determined by the difference in the oxidation states of the elements in the initial and final states. Example:
HN 5+ O 3 + H 2 S 2– → N 2+ O + S + H 2 O
In this reaction, the oxidation state is changed by two elements: nitrogen and sulfur. Electronic balance equations:
The fraction of dissociated H 2 S molecules is insignificant, therefore, not the S 2– ion, but the H 2 S molecule is substituted into the equation. First, the balance of particles is equalized. At the same time, in an acidic medium, hydrogen ions added to the oxidized form and water molecules added to the reduced form are used for equalization. Then the balance of charges is equalized, and the coefficients equalizing the number of given and received electrons are indicated to the right of the line. After that, the summary equation is written below, taking into account the coefficients:
We have obtained a reduced ion-molecular equation. Adding Na + and K + ions to it, we obtain a similar equation in full form, as well as a molecular equation:
NaNO 2 + 2 KMnO 4 + 2 KOH → NaNO 3 + 2 K 2 MnO 4 + H 2 O
In a neutral medium, the balance of particles is equalized by adding water molecules to the left side of the half-reactions, and H + or OH - ions are added to the right side:
I 2 + Cl 2 + H 2 O → HIO 3 + HCl
The starting materials are not acids or bases, therefore, in the initial period of the reaction, the medium in the solution is close to neutral. Half reaction equations:
| I 2 + 6 H 2 O + 10e → 2 IO 3 – + 12 H + | |
| Cl 2 + 2e → 2 Cl - | |
| I 2 + 5 Cl 2 + 6 H 2 O → 2 IO 3 - + 12 H + + 10 Cl - |
Reaction equation in molecular form:
I 2 + 5 Cl 2 + 6 H 2 O → 2 HIO 3 + 10 HCl.
THE MOST IMPORTANT OXIDIZERS AND REDUCERS. CLASSIFICATION OF REDOX REACTIONS
The limits of oxidation and reduction of an element are expressed by the maximum and minimum values of the oxidation states *. In these extreme states, determined by the position in the periodic table, the element has the ability to show only one function - an oxidizing or reducing agent. Accordingly, substances containing elements in these oxidation states are only oxidizing agents (HNO 3, H 2 SO 4, HClO 4, KMnO 4, K 2 Cr 2 O 7, etc.) or only reducing agents (NH 3, H 2 S, hydrogen halides, Na 2 S 2 O 3, etc.). Substances containing elements in intermediate oxidation states can be both oxidizing and reducing agents (HClO, H 2 O 2 , H 2 SO 3, etc.).
Redox reactions are divided into three main types: intermolecular, intramolecular and disproportionation reactions.
The first type includes processes in which the atoms of the oxidizing element and the reducing element are part of different molecules.
Intramolecular reactions are called reactions in which the oxidizing agent and reducing agent in the form of atoms of different elements are part of the same molecule. For example, thermal decomposition of potassium chlorate according to the equation:
2 KClO 3 → 2 KCl + 3 O 2
Disproportionation reactions are processes in which the oxidizing and reducing agents are the same element in the same oxidation state, which both decreases and increases in the reaction, for example:
3 HClO → HClO 3 + 2 HCl
Reverse disproportionation reactions are also possible. These include intramolecular processes in which the same element is the oxidizing and reducing agent, but in the form of atoms that are in different degrees of oxidation and equalize it as a result of the reaction, for example.
SEI VPO "Ural State Technical University - UPI"
Determination of chemical equilibrium constants
reactions and calculation of chemical equilibrium
at the rate physical chemistry
for full-time students
Yekaterinburg 2007
UDC 544(076)С79
Compiler
Scientific editor, Ph.D., associate professor
Determination of equilibrium constants of chemical reactions and calculation of chemical equilibrium: guidelines for laboratory work No. 4 at the course of physical chemistry / comp. - Yekaterinburg: GOU VPO USTU-UPI, 20s.
The guidelines are intended for additional in-depth study of the material on chemical equilibrium as part of the calculation and analytical laboratory work. They contain 15 options for individual tasks, which contributes to the achievement of the goal.
Bibliography: 5 titles. Rice. Tab.
© GOU VPO "Ural State
Technical University - UPI", 2007
Introduction
This work, although carried out within the framework of a laboratory workshop, belongs to the calculation and analytical and consists in mastering the 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, the consideration of the isobar equation and the isotherm of a chemical reaction, etc., is presented in lectures and studied in practical classes (therefore, this material is not presented in this work). This manual considers in detail the 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.
So, the implementation of this work by students will solve 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 the mixture, based on a variety of experimental data.
1. THEORETICAL INFORMATION ABOUT METHODS
DEFINITIONS OF EQUILIBRIUM CONSTANTS FOR CHEMICAL REACTIONS
Let us dwell briefly on the main 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">- standard Gibbs molar energy of the reaction r.
Equation (1) is the defining 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 law of chemical equilibrium is written as follows
, (2)
where https://pandia.ru/text/78/005/images/image005_99.gif" width="23" height="25">- activity k- reaction participant; - dimension of activity; stoichiometric coefficient k- reaction participant r.
Experimental determination of the equilibrium constants is a rather difficult task. First of all, it is necessary to be sure that equilibrium is reached at a given temperature, i.e., the composition of the reaction mixture corresponds to the equilibrium state - a state with a minimum Gibbs energy, zero reaction affinity, and equal 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 quantitative analysis methods with characteristic chemical reactions. However, the introduction of a foreign 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 sufficiently low. That is why very often, when studying equilibrium, various physical methods are also used to determine the composition of the system.
1.1 Chemical methods
There are static chemical methods and dynamic chemical methods. Consider the 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 reaction under study must be slow enough so that the introduction of a foreign reagent does not practically disturb the equilibrium state. To slow down the process, it is possible to cool the reaction flask sufficiently quickly. A classic example of such a study is the reaction between iodine and hydrogen
H2(g) + I2(g) = 2HI(g) (3)
Lemoyne placed either a mixture of iodine with hydrogen or hydrogen iodide in glass cylinders. At 200 °C, the reaction practically does not proceed; at 265 °C, the duration of equilibrium 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 vessel was rapidly cooled by lowering it into water, then a 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 quickly cooled for subsequent analysis. These methods are most applicable to fairly fast reactions. Reactions are usually accelerated either by carrying them out 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 based primarily 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
Every reaction that is accompanied by a change in the number of moles of gaseous reactants is accompanied by a change in pressure at 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 on the basis of one molecule of the starting material
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 relationships 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, by measuring the equilibrium pressure, the degree of dissociation can be determined using formula (13), and then, using formula (14), the equilibrium constant can also be calculated.
1.2.2 Measurement of mass density
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 mass density 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">- 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"> - the mass density of the system at the initial moment and at the moment of equilibrium, respectively. By measuring the mass density of the system, we can use formula (16) to calculate the degree of dissociation, and then the equilibrium constant.
1.2.3 Direct partial pressure measurement
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 the general case, 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 to be permeable to hydrogen at high temperatures is used. 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 capable of passing through the walls of the iridium tank.
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 like:
https://pandia.ru/text/78/005/images/image033_14.gif" width="89 height=23" height="23">. (17)
1.2.4 Optical methods
There are equilibrium methods based on adsorption measurements that are particularly effective for colored gases. It is also possible to determine the composition of a binary gas mixture by measuring the refractive index (refractometrically). For example, Chadron (1921) studied the reduction of metal oxides with carbon monoxide by measuring the composition of a gas mixture of oxide and carbon dioxide refractometrically.
1.2.5 Measurement of thermal conductivity
This method has been used in the study of dissociation reactions in the gas phase, for example
Let us assume that a mixture of N2O4 and NO2 is placed in a vessel, the right wall of which has a temperature T2, and the left one T1, with T2>T1 (Fig. 2). The dissociation of N2O4 will be to a greater extent in that part of the vessel that has more high temperature. Consequently, the concentration of NO2 in the right side of the vessel will be greater than in the left, and diffusion of NO2 molecules from right to left and N2O4 from left to right will be observed. However, reaching the right side of the reaction vessel, the N2O4 molecules again dissociate with the absorption of energy in the form of heat, and the 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 makes it possible to determine the composition of the equilibrium mixture.
1.2.6 Measurement of the electromotive force (EMF) of a galvanic cell
Measurement of 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 one under study, the equilibrium constant of which must be determined; 2) measure the EMF of a galvanic cell in a thermodynamically equilibrium process. To do this, it is necessary that the corresponding current-generating process be performed infinitely slowly, that is, that the element work at an infinitely small current strength, which is why the compensation method is used to measure the EMF of a galvanic cell, which is based on the fact that the galvanic cell under study is switched on in series against an external potential difference , and the latter was chosen in such a way that there was no current in the circuit. EMF value, measured by the compensation method, corresponds to the thermodynamically equilibrium process occurring in the element and the useful work of the process is maximum and is equal to the loss of the Gibbs energy
https://pandia.ru/text/78/005/images/image035_12.gif" width="181" height="29 src="> (20)
for p, T=const, where F– Faraday number = 96500 C/mol, n is the smallest common multiple of the number of electrons involved in electrode reactions, Eo- standard EMF, V.
The value of the equilibrium constant can be found from relation (21)
(21)
2. EXAMPLE OF LABORATORY WORK ON DETERMINING THE VALUE OF THE EQUILIBRIUM CONSTANT
In physical chemistry workshops, one often encounters laboratory work related to the study of the reaction of dissociation of metal carbonates. Let's bring summary similar work.
Goal of the work – determination of the equilibrium constant and calculation of the main thermodynamic quantities 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 undissociated calcium carbonate remains.
The equilibrium constant of reaction (22) is written as:
, (23)
where https://pandia.ru/text/78/005/images/image041_11.gif" width="68" height="51"> in general view or ; 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 elasticity of carbonate dissociation, if the latter is expressed in atmospheres. Thus, having determined experimentally the elasticity of dissociation of calcium carbonate, it is possible to determine the value of the equilibrium constant of this reaction.
experimental part
A static method is used to determine the elasticity of dissociation of calcium carbonate. Its essence lies in the direct measurement at a given temperature of the pressure of carbon dioxide in the installation.
Equipment. The main components 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 tap (4) to a manual vacuum pump (5). The temperature in the furnace is maintained by a regulator (6), the temperature is controlled by 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 oven and set the required initial temperature of the reaction vessel with the help of a regulator. Record the first readings of the thermocouple and pressure gauge. After that, using the regulator (6) increase the temperature in the furnace by 10-20 degrees, wait for the establishment of a new constant temperature value and record the pressure value corresponding to this temperature. 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 valve (4). Then turn off the oven and the 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 THE EQUILIBRIUM CONSTANTS
WITHOUT EXPERIMENT
3.1 Calculation of the equilibrium constant of a chemical reaction from
the value of the standard Gibbs molar function of the reaction
This method does not involve experimentation at all. If the standard molar enthalpy and entropy of the reaction at a given temperature are known, then using the corresponding equations, it is possible to calculate the standard Gibbs molar 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 entropies and enthalpies at a given temperature are unknown, then you can use the Temkin and Schwartzman method, that is, by the value of the standard molar enthalpies and entropies at a temperature of 298 K and the values of the temperature dependence coefficients 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)
According to 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 PROBLEM SOLVING
Find the equilibrium constant for 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 moles of nitrogen is equal to the stoichiometric, that is, 0.5 mol (https://pandia.ru/text/78/005/images/image069_3.gif " width="247" height="57 src=">
We write the reaction, under the symbols of the elements we indicate the initial and equilibrium numbers 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 of the direct problem of chemical equilibrium is the calculation of the equilibrium composition of the system in which a given reaction occurs (several reactions). Obviously, the basis of the solution is 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 600 K and pressure expressed 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 the starting substances 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 of the number of moles of each participant at the initial moment and at the moment of equilibrium established by 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.
Task 3
Its condition differs from problem 2 only in that the initial amounts of hydrogen and iodine moles are 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; mol. 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 in terms of 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 determine from it the reaction depth that corresponds to the 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 complete the job, you need to complete the following tasks
Exercise 1
1. Describe a method for experimentally determining the elasticity of carbon dioxide when studying the reaction of dissociation СaCO3⇄CaO+CO2
(options 1 - 15, Table 3);
2. Write down the law of chemical equilibrium for the reaction under study; determine the values of the equilibrium constants of the dissociation reaction of calcium carbonate according to experimental data (Table 3) at different temperatures; tasks from section B (according to the indicated option) and tasks 1-3, p;
3. Write down the defining expression for the equilibrium constant and theoretically calculate the equilibrium constant of the reaction under study at the last temperature indicated in the table.
Task 2
1. Prepare an answer to question 1 (options 1-15, Table 4)
2. Solve problems 2 and 3.
Reference data required to complete the job
Quantity for calculating the standard molar change in the Gibbs energy by the method of Temkin and Schwartzman
Table 1
Thermodynamic data for calculating the Gibbs standard molar energy
table 2
Experimental data for task 1
Table 3
Option | Experimental data |
|||
t, oC | ||||
p, mmHg | ||||
p, mmHg | ||||
p, mmHg | ||||
p, mmHg | ||||
p, mmHg | ||||
p, mmHg | ||||
p, mmHg | ||||
p, mmHg | ||||
p, mmHg | ||||
p, mmHg | ||||
p, mmHg | ||||
p, mmHg | ||||
p, mmHg | ||||
p, mmHg | ||||
p, mmHg |
Task conditions for completing task 2
Table 4
1 option | 1. Tell us about chemical methods for determining the values of chemical equilibrium constants. 2. There is a mixture of gaseous substances A and B, which can enter into a chemical reaction with the formation of a reaction product C, according to the stoichiometric equation 0.5 A + 2B = C. At the initial time, there is no reaction product in the system, and the starting substances are taken in stoichiometric quantities. After equilibrium is established, the equilibrium mixture contains the number of moles of product C equal to 0.4, and the total pressure is 2 atm. Find the equilibrium constant in the p-scale. 3 . At 1273 K and a total pressure of 30 atm, the equilibrium mixture with the assumed reaction CO2(g) + C(s) = 2CO(g) contains 17% (by volume) CO2. What percentage of CO2 will be contained in the gas at a total pressure of 20 atm?. At what pressure will the gas contain 25% CO2? |
Option 2 | 1 . Tell us about the physical method for determining the value of the chemical equilibrium constant by measuring pressure. 2. There is a mixture of gaseous substances A and B, which can enter into a chemical reaction with the formation of a reaction product C, according to the stoichiometric equation 2A + B = C. At the initial time, there is no reaction product in the system, and the initial substances are taken in stoichiometric quantities. After equilibrium is established, the equilibrium mixture contains the number of moles of product C equal to 0.5, and the total pressure is 2 atm. Find the equilibrium constant in the p-scale. 3 . At 2000 °C and a total pressure of 1 atm, 2% of water dissociates into hydrogen and oxygen according to the reaction H2O(g)= H2(g) + 0.5 O2(g). Calculate the equilibrium constant of the reaction under these conditions. |
3 option | 1 . Describe the method for determining the value of the equilibrium constant by measuring the density. What methods does this method refer to? 2. There is a mixture of gaseous substances A and B, which can enter into a chemical reaction with the formation of a reaction product C, according to the stoichiometric equation A + 2B = C. At the initial time, there is no reaction product in the system, and the initial substances are taken in stoichiometric quantities. After equilibrium is established, the equilibrium mixture contains the number of moles of product C equal to 0.6, and the total pressure is 2 atm. Find the equilibrium constant in the p-scale. 3 . The equilibrium constant of the reaction CO(g) + H2O(g) = H2(g) + CO2(g) at 500 °C is 5.5 ([p]=1 atm). A mixture consisting of 1 mol of CO and 5 mol of H2O was heated to this temperature. Calculate the mole fraction of water in the equilibrium mixture. |
4 option | 1 . Describe a method for determining the value of the equilibrium constant by direct measurement of partial pressure. 2. There is a mixture of gaseous substances A and B, which can enter into a chemical reaction with the formation of a reaction product C, according to the stoichiometric equation 0.5 A + B \u003d C. At the initial time, there is no reaction product in the system, and the starting substances are taken in stoichiometric quantities. After equilibrium is established, the equilibrium mixture contains the number of moles of product C equal to 0.3, and the total pressure is 1.5 atm. Find the equilibrium constant in the p-scale. 3 .The equilibrium constant of the reaction N2O4 (g) \u003d 2NO2 (g) at 25 ° C is 0.143 ([p] \u003d 1 atm). Calculate the pressure that will be established in a 1 liter vessel containing 1 g of N2O4 at this temperature. |
5 option | 1 . How can you determine the value of the equilibrium constant of a reaction without resorting to experiment. 2. There is a mixture of gaseous substances A and B, which can enter into a chemical reaction with the formation of a reaction product C, according to the stoichiometric equation 0.5 A + 3B = C. At the initial time, there is no reaction product in the system, and the starting materials are taken in stoichiometric quantities. After equilibrium is established, the equilibrium mixture contains the number of moles of product C equal to 0.3, and the total pressure is 2 atm. Find the equilibrium constant in the p-scale. 3 . A 3-liter vessel containing 1.79·10 -2 mol I2 was heated to 973 K. The pressure in the vessel at equilibrium turned out to be 0.49 atm. Assuming ideal gases, calculate the equilibrium constant at 973 K for the reaction I2(r) = 2I(r). |
6 option | 1. Using the reaction isobar equation to determine the value of the chemical equilibrium constant at a previously unexplored temperature. 2. There is a mixture of gaseous substances A and B, which can enter into a chemical reaction with the formation of a reaction product C, according to the stoichiometric equation 3A + B = C. At the initial time, there is no reaction product in the system, and the initial substances are taken in stoichiometric quantities. After equilibrium is established, the equilibrium mixture contains the number of moles of product C equal to 0.4, and the total pressure is 2 atm. Find the equilibrium constant in the p-scale. 3 . For the reaction PCl5(g) =PCl3(g) +Cl2(g) at 250 °C, the standard molar change in the Gibbs energy = - 2508 J/mol. At what total pressure will the degree of conversion of PCl5 into PCl3 and Cl2 be 30% at 250 °C? |
7 option | 1. The system in which the endothermic gas-phase reaction A + 3B = 2C takes place is in equilibrium at 400 K and 5 atm. If the gases are ideal, how will the yield of the product be affected by the addition of an inert gas at constant volume? 2. There is a mixture of gaseous substances A and B, which can enter into a chemical reaction with the formation of the reaction product C, according to the stoichiometric equation 2A + B = 2C. At the initial moment of time, there is no reaction product in the system, and the initial substances are taken in stoichiometric quantities. After equilibrium is established, the equilibrium mixture contains the number of moles of product C equal to 0.3, and the total pressure is 2 atm. Find the equilibrium constant in the p-scale. 3 . For the reaction 2HI(g) = H2 + I2(g), the equilibrium constant is Kp\u003d 0.0183 ([p] \u003d 1 atm) at 698.6 K. How many grams of HI are formed when 10 g of I2 and 0.2 g of H2 are heated to this temperature in a three-liter vessel? What are the partial pressures of H2, I2 and HI? |
8 option | 1. The system in which the endothermic gas-phase reaction A + 3B = 2C takes place is in equilibrium at 400 K and 5 atm. If the gases are ideal, how will the product yield be affected by increasing the temperature? 2. There is a mixture of gaseous substances A and B, which can enter into a chemical reaction with the formation of the reaction product C, according to the stoichiometric equation 0.5A + 2B = 2C. At the initial moment of time, there is no reaction product in the system, and the initial substances are taken in stoichiometric quantities. After equilibrium is established, the equilibrium mixture contains the number of moles of product C equal to 0.3, and the total pressure is 2 atm. Find the equilibrium constant in the p-scale. 3 . A 1-liter vessel containing 0.341 mol of PCl5 and 0.233 mol of N2 was heated to 250°C. The total pressure in the vessel at equilibrium turned out to be 29.33 atm. Considering all gases to be ideal, calculate the equilibrium constant at 250 °C for the reaction PCl5(g) = PCl3(g) + Cl2(g) taking place in the vessel. |
9 option | 1 . The system in which the endothermic gas-phase reaction A+3B=2C proceeds is in equilibrium at 400 K and 5 atm. If the gases are ideal, how will the pressure increase affect the yield of the product? 2. There is a mixture of gaseous substances A and B, which can enter into a chemical reaction with the formation of the reaction product C, according to the stoichiometric equation 0.5A + B = 2C. At the initial moment of time, there is no reaction product in the system, and the initial substances are taken in stoichiometric quantities. After equilibrium is established, the equilibrium mixture contains the number of moles of product C equal to 0.5, and the total pressure is 2 atm. Find the equilibrium constant in the p-scale. 3 . The equilibrium constant of the reaction CO(g) + 2H2 = CH3OH(g) at 500 K is kr= 0.00609 ([p]=1 atm). Calculate the total pressure required to produce methanol with 90% yield if CO and H2 are taken in a 1:2 ratio. |
10 option | 1. Describe the method for determining equilibrium constants by measuring partial pressure. 2. There is a mixture of gaseous substances A and B, which can enter into a chemical reaction with the formation of the reaction product C, according to the stoichiometric equation 0.5A + 1.5B = 2C. At the initial moment of time, there is no reaction product in the system, and the initial substances are taken in stoichiometric quantities. After equilibrium is established, the equilibrium mixture contains the number of moles of product C equal to 0.4, and the total pressure is 2 atm. Find the equilibrium constant in the p-scale. 3 . Equilibrium in the reaction 2NOCl (g)=2NO(g)+Cl2(g) is established at 227°C and a total pressure of 1.0 bar, when the partial pressure of NOCl is 0.64 bar (initially only NOCl was present). Calculate this reaction at the given temperature. |
11 option | 1 . Describe the chemical methods for determining equilibrium constants. 2. There is a mixture of gaseous substances A and B, which can enter into a chemical reaction with the formation of the reaction product C, according to the stoichiometric equation 2A + 0.5B = 2C. At the initial moment of time, there is no reaction product in the system, and the initial substances are taken in stoichiometric quantities. After equilibrium is established, the equilibrium mixture contains the number of moles of product C equal to 0.2, and the total pressure is 2 atm. Find the equilibrium constant in the p-scale. 3 . Calculate the total pressure that must be applied to a mixture of 3 parts H2 and 1 part N2 to obtain an equilibrium mixture containing 10 vol% NH3 at 400°C. Equilibrium constant for the reaction N2(g) + 3 H2(g)= 2NH3(g) at 400°C and the expression of pressure in atm is 1.6 10-4. |
12 option | 1 . The system in which the endothermic gas-phase reaction A+3B=2C proceeds is in equilibrium at 400 K and 5 atm. If the gases are ideal, how will the product yield be affected by a decrease in pressure? 2. There is a mixture of gaseous substances A and B, which can enter into a chemical reaction with the formation of the reaction product C, according to the stoichiometric equation 2A + B = 0.5C. At the initial moment of time, there is no reaction product in the system, and the initial substances are taken in stoichiometric quantities. After equilibrium is established, the equilibrium mixture contains the number of moles of product C equal to 0.4, and the total pressure is 2 atm. Find the equilibrium constant in the p-scale. 3 . At 250 °C and a total pressure of 1 atm, PCl5 is dissociated by 80% according to the reaction PCl5(g) = PCl3(g) + Cl2(g). What will be the degree of dissociation of PCl5 if nitrogen is added to the system so that the partial pressure of nitrogen is 0.9 atm? The total pressure is maintained at 1 atm. |
13 option | 1 . System in which an exothermic reaction occurs CO(g) + 2H2 = CH3OH(g) is in equilibrium at 500 K and 10 bar. If the gases are ideal, how will the methanol yield be affected by a decrease in pressure? 2. There is a mixture of gaseous substances A and B, which can enter into a chemical reaction with the formation of the reaction product C, according to the stoichiometric equation 1.5A + 3B = 2C. At the initial moment of time, there is no reaction product in the system, and the initial substances are taken in stoichiometric quantities. After equilibrium is established, the equilibrium mixture contains the number of moles of product C equal to 0.5, and the total pressure is 2 atm. Find the equilibrium constant in the p-scale. 3 . The equilibrium constant of the reaction CO(g) + 2H2 = CH3OH(g) at 500 K is 6.09 × 10 5 ([p] = 1 atm). The reaction mixture, consisting of 1 mol of CO, 2 mol of H2 and 1 mol of inert gas (nitrogen) are heated to 500 K and a total pressure of 100 atm. Calculate the composition of the reaction mixture. |
14 option | 1 . Describe a method for determining equilibrium constants from electrochemical data. 2. There is a mixture of gaseous substances A and B, which can enter into a chemical reaction with the formation of a reaction product C, according to the stoichiometric equation 2A + 0.5B = C. At the initial time, there is no reaction product in the system, and the starting substances are taken in stoichiometric quantities . After equilibrium is established, the equilibrium mixture contains the number of moles of product C equal to 0.4, and the total pressure is 2 atm. Find the equilibrium constant in the p-scale. 3. For the reaction N2 (g) + 3 H2 (g) \u003d 2NH3 (g) at 298 K, the equilibrium constant when expressed in atm pressure is 6.0 × 10 5, and the standard molar enthalpy of ammonia formation is = - 46.1 kJ / mol . Find the value of the equilibrium constant at 500 K. |
15 option | 1 . The system with the exothermic reaction CO(g) + 2H2 = СH3OH(g) is in equilibrium at 500 K and 10 bar. If the gases are ideal, how will the methanol yield be affected by lowering the temperature. 2. There is a mixture of gaseous substances A and B, which can enter into a chemical reaction with the formation of the reaction product C, according to the stoichiometric equation 2A + B = 1.5C. At the initial moment of time, there is no reaction product in the system, and the initial substances are taken in stoichiometric quantities. After equilibrium is established, the equilibrium mixture contains the number of moles of product C equal to 0.5, and the total pressure is 2 atm. Find the equilibrium constant in the p-scale. 3. The equilibrium constant of the reaction N2(g) + 3 H2(g)= 2NH3(g) at 400 °C and expressed in atm is 1.6 10-4. What total pressure must be applied to an equimolar mixture of nitrogen and hydrogen to convert 10% of the nitrogen to ammonia? The gases are assumed to be ideal. |
In a report on laboratory work it seems appropriate to include the following sections: introduction, part 1, part 2, conclusions.
1. Introduction you can briefly present theoretical information on one of the following issues: either on the law of mass action, the history of its discovery and its authors; or about the basic concepts and defining relationships of the section "Chemical Equilibrium"; or to derive the law of chemical equilibrium in its modern formulation; or talk about the factors that affect the value of the equilibrium constant, etc.
The section "Introduction" should end with a statement of the objectives of the work.
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 . Give the results of the calculation of the equilibrium constant according to the experimental data
2.3. Give the calculation of the equilibrium constant according to thermodynamic data
Part 2 necessary
3.1 . Give a full justified answer to question 1 of task 2.
3.2 . Give the solution of tasks 2 and 3 of task 2. The condition of the tasks must be written in symbolic notation.
In the conclusions it is advisable to reflect the fulfillment of the goals set in the work, as well as to compare the values of the equilibrium constant calculated in 2.2 and 2.3.
Bibliographic list
1. Karjakin of chemical thermodynamics: Proc. allowance for universities. M.: Academy., 20s.
2. Prigozhin I., Kondepudi D. Modern thermodynamics. From heat engines to dissipative structures. M.: Mir, 20s.
3., Cherepanov on physical chemistry. Toolkit. Yekaterinburg: publishing house of the Ural State University, 2003.
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