What is the constant of chemical equilibrium. Chemical equilibrium: chemical equilibrium constant and ways of expressing it
Concept chemical equilibrium
The state of the system, which remains unchanged, is considered to be in equilibrium, and this state is not caused by the action of any external forces. The state of a system of reactants, at which the rate of the direct reaction becomes equal to the rate of the reverse reaction, is called chemical equilibrium... This balance is also called mobile m or dynamic balance.
Signs of chemical equilibrium
1. The state of the system remains unchanged over time while maintaining external conditions.
2. Equilibrium is dynamic, that is, it is caused by the flow of direct and reverse reactions at the same rates.
3. Any external influence causes a change in the equilibrium of the system; if the external influence is removed, then the system returns to its original state.
4. The state of equilibrium can be approached from two sides - both from the side of the initial substances and from the side of the reaction products.
5. In a state of equilibrium, the Gibbs energy reaches its minimum value.
Le Chatelier's principle
The influence of changes in external conditions on the equilibrium position is determined Le Chatelier principle (principle of moving balance): if an external influence is made on a system in a state of equilibrium, then in the system one of the directions of the process will intensify, which weakens the effect of this influence, and the equilibrium position will shift in the same direction.
Le Chatelier's principle applies not only to chemical processes, but also to physical ones such as boiling, crystallization, dissolution, etc.
Consider the impact various factors on chemical equilibrium using the example of the NO oxidation reaction:
2 NO (d) + O 2 (d) 2 NO 2 (d); H about 298 = - 113.4 kJ / mol.
Effect of temperature on chemical equilibrium
As the temperature rises, the equilibrium shifts towards the endothermic reaction, and as the temperature decreases, towards the exothermic reaction.
The degree of displacement of equilibrium is determined by the absolute value of the heat effect: the greater in absolute value is the enthalpy of reaction H, the more significant is the effect of temperature on the state of equilibrium.
In the considered reaction for the synthesis of nitric oxide (IV ) an increase in temperature will shift the equilibrium towards the starting substances.
Effect of pressure on chemical equilibrium
Compression shifts the equilibrium in the direction of the process, which is accompanied by a decrease in the volume of gaseous substances, and a decrease in pressure shifts the equilibrium in the opposite direction. In this example, there are three volumes on the left side of the equation, and two on the right. Since an increase in pressure favors a process that occurs with a decrease in volume, then with an increase in pressure the equilibrium will shift to the right, i.e. towards the reaction product - NO 2 ... A decrease in pressure will shift the balance in the opposite direction. It should be noted that if in the equation reversible reaction the number of molecules of gaseous substances in the right and left parts are equal, then the change in pressure does not affect the position of equilibrium.
Effect of concentration on chemical equilibrium
For the reaction under consideration, the introduction into the equilibrium system additional quantities NO or O 2 causes a shift in the equilibrium in the direction at which the concentration of these substances decreases, therefore, there is a shift in the equilibrium towards the formation NO 2 ... Increased concentration NO 2 shifts the equilibrium towards the starting materials.
The catalyst equally accelerates both forward and reverse reactions and therefore does not affect the shift in chemical equilibrium.
When introduced into an equilibrium system (at Р = const ) inert gas, the reagent concentrations (partial pressures) decrease. Since the oxidation process under consideration NO goes with a decrease in volume, then when adding in
Chemical equilibrium constant
For a chemical reaction:
2 NO (g) + O 2 (g) 2 NO 2 (d)
the constant of the chemical reaction K c is the ratio:
(12.1)
In this equation in square brackets are the concentrations of reactants that are established at chemical equilibrium, i.e. equilibrium concentrations of substances.
The chemical equilibrium constant is related to the change in the Gibbs energy by the equation:
G T о = - RTlnK. (12.2).
Examples of problem solving
At a certain temperature, the equilibrium concentrations in the system 2CO (g) + O 2 (d) 2CO 2 (d) were: = 0.2 mol / L, = 0.32 mol / L, = 0.16 mol / l. Determine the equilibrium constant at this temperature and the initial concentrations of CO and O 2 if the initial mixture did not contain CO 2 .
.
2CO (g) + O 2 (g) 2CO 2 (d).
In the second line, c proreagir means the concentration of the reacted starting materials and the concentration of the resulting CO 2 , moreover, with initial = with proreagir + with equal .
Using the reference data, calculate the process equilibrium constant3 H 2 (D) + N 2 (G) 2 NH 3 (G) at 298 K.
G 298 o = 2 ( - 16.71) kJ = -33.42 10 3 J.
G T about = - RTlnK.
lnK = 33.42 · 10 3 / (8.314 × 298) = 13.489. K = 7.21 × 10 5.
Determine the equilibrium concentration of HI in the systemH 2 (d) + I 2 (d) 2HI (G) ,
if at a certain temperature the equilibrium constant is equal to 4, and the initial concentrations of H 2, I 2 and HI are equal, respectively, 1, 2 and 0 mol / l.
Solution. Let x mol / L H 2 have reacted at some point in time.
.
Solving this equation, we get x = 0.67.
This means that the equilibrium concentration of HI is 2 × 0.67 = 1.34 mol / L.
Using the reference data, determine the temperature at which the equilibrium constant of the process: H 2 (g) + HCOH (d) CH 3 OH (d) becomes equal to 1. Accept that H o T »H o 298, and S o T »S about 298.If K = 1, then G о T = - RTlnK = 0;
G o T »N about 298 - T D S about 298 ... Then ;
N about 298 = -202 - (- 115.9) = -86.1 kJ = - 86.1 x 10 3 J;
S about 298 = 239.7 - 218.7 - 130.52 = -109.52 J / K;
TO.
Solution. Let x mol / L SO 2 have reacted at some point in time.
SO 2 (G) + Cl 2 (G) SO 2 Cl 2 (D)
Then we get:
.
Solving this equation, we find: x 1 = 3 and x 2 = 1.25. But x 1 = 3 does not satisfy the condition of the problem.
Therefore, = 1.25 + 1 = 2.25 mol / l.
Tasks for independent solution
12.1. In which of the following reactions will an increase in pressure shift the equilibrium to the right? Justify the answer.
1) 2 NH 3 (g) 3 H 2 (g) + N 2 (d)
2) ZnCO 3 (q) ZnO (q) + CO 2 (d)
3) 2HBr (g) H 2 (g) + Br 2 (g)
4) CO 2 (d) + C (graphite) 2CO (g)
12.2.At a certain temperature, the equilibrium concentrations in the system
2HBr (g) H 2 (g) + Br 2 (d)
were: = 0.3 mol / L, = 0.6 mol / L, = 0.6 mol / L. Determine the equilibrium constant and the initial concentration of HBr.
12.3.For the reaction H 2 (g)+ S (d) H 2 S (d) at a certain temperature, the equilibrium constant is 2. Determine the equilibrium concentrations of H 2 and S if the initial concentrations of H 2, S and H 2 S are equal, respectively, 2, 3 and 0 mol / l.
Study questions
State of equilibrium
Equilibrium constant
Calculation of equilibrium concentrations
Chemical equilibrium shift. Le Chatelier's principle
State of equilibrium
Reactions proceeding under the same conditions simultaneously in opposite directions are called reversible.
Consider a reversible reaction that occurs in closed system
The direct reaction rate is described by the equation:
pr = k pr [A] [B],
where 
pr is the speed of the direct reaction;
k pr is the rate constant of the direct reaction.
Over time, the concentration of reagents BUT and IN decrease, the reaction rate decreases (Fig. 1, curve 
etc).
Reaction between BUT and IN leads to the formation of substances C and D, molecules of which in collisions can again give substances BUT and IN.
The back reaction rate is described by the equation:
arr = k
arr [C] [D],
where 
obr - the speed of the reverse reaction;
k obr - rate constant of the reverse reaction.
As the concentration of substances C and D increase, the rate of the reverse reaction increases (Fig. 1, curve 
arr).
Fig. 1. Change in the rates of forward and backward reactions in time
Over time the rates of forward and backward reactions become equal:
pr = 
arr
This state of the system is called state of equilibrium .
In a state of equilibrium, the concentrations of all its participants stop changing over time. . Such concentrations are called equilibrium .
Chemical equilibrium – This dynamic balance. The invariability of the concentrations of substances present in a closed system is a consequence of continuously running chemical processes. The rates of the forward and reverse reactions are not equal to zero, and the observed rate of the process is equal to zero.
The equality of the rates of forward and reverse reactions is the kinetic condition of chemical equilibrium.
2. Equilibrium constant
When the speeds of the forward and reverse reactions are equal
pr = 
arr
fair equality
k pr [A] [B] = k arr [C] [D],
where [ A], [B], [WITH], [D] - equilibrium concentrations of substances.
Since the rate constants do not depend on the concentration, the equality can be written differently:
The ratio of the rate constants of the forward and reverse reactions ( k etc / k arr ) 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, in a state of equilibrium they should provide a stoichiometric transition of the initial substances to the reaction products and vice versa. This means that the algebraic sum of all stages of the process is equal to the stoichiometric equation of the reaction, i.e. stoichiometric coefficients are the sum of the molecularities of all stages of the mechanism.
For a complex reaction
aA + bB cC + dD
|
K c = |
For the same temperature, the ratio of the product of the equilibrium concentrations of the reaction products in powers equal to the stoichiometric coefficients to the product of the equilibrium concentrations of the starting substances in powers equal to the stoichiometric coefficients is a constant value.
This is the second formulation of the law of the masses at work.
The expression for the equilibrium constant of a heterogeneous reaction includes only the concentrations of substances in the liquid or gaseous phase, since the concentrations of solids, as a rule, remain constant.
For example, the expression for the equilibrium constant of the following reaction
CO 2 (g) + C (tv) 2CO (g)
is written like this:
TO c = 
.
The equation of the equilibrium constant shows that under equilibrium conditions, the concentrations of all substances participating in the reaction are related to each other. The numerical value of the equilibrium constant determines what the ratio of the concentrations of all reacting substances should be at equilibrium.
A change in the concentration of any of these substances entails changes in the concentration of all other substances. As a result, new concentrations are established, but the ratio between them again corresponds to the equilibrium constant.
The value of the equilibrium constant depends on the nature of the reactants and the temperature.
Equilibrium constant expressed in terms of molar concentrations of reactants ( TOwith) and the equilibrium constant, expressed in terms of the equilibrium partial pressures ( TOR) (see "Fundamentals of Chemical Thermodynamics"), are related by the relations:
TOR= KwithRT , Kc = KR / (RT) ,
where is the change in the number of gaseous moles in the reaction.
The standard change in Gibbs energy is
G Т = - RT ln Kp,
G T = H – T S.
After equating the right-hand sides of the equations:
- RT ln Kp = H – T S
ln K R = - H / ( RT) + S/ R .
The equation not only establishes the form of the dependence of the constant on temperature, but also shows that the constant is determined by the nature of the reacting substances.
The equilibrium constant does not depend on the concentration (as well as the reaction rate constant), the reaction mechanism, activation energy, and the presence of catalysts. 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.
All chemical reactions can be divided into reversible and irreversible. Reversible reactions include those reactions that, at a certain temperature, with an appreciable rate proceed in two opposite directions - forward and backward. Reversible reactions do not proceed completely, none of the reactants is completely consumed. An example is the reaction
In a certain temperature range, this reaction is reversible. Sign " » is a sign of reversibility.
Irreversible reactions are reactions that proceed only in one direction to the end, i.e. until one of the reactants is completely consumed. An example of an irreversible reaction is the decomposition reaction of potassium chlorate:
Formation of potassium chlorate from potassium chloride and oxygen in normal conditions impossible.
Chemical equilibrium state. Chemical equilibrium constant
Let's write the equation of some reversible reaction in general form:
By the time the reaction started, the concentrations of the starting substances A and B were at their maximum. In the course of the reaction, they are consumed, and their concentration decreases. Moreover, in accordance with the law of mass action, the rate of the direct reaction is
will decrease. (Hereinafter, the arrow at the top denotes the direction of the process.) At the initial moment, the concentrations of the reaction products D and E were equal to zero. In the course of the reaction, they increase, the speed of the reverse reaction increases from zero according to the equation:
In fig. 4.5 shows the change in the speeds of forward and backward
reactions over time. After the expiration of time t, these speeds are equal to - " 
Rice. 4.5. Change in the rate of forward (1) and reverse (2) reaction in time: - in the absence of a catalyst: .......... - in the presence of a catalyst
This state is called chemical equilibrium. Chemical equilibrium is the most stable, limiting state of spontaneous processes. It can last as long as you want, if you do not change the external conditions. In isolated systems in a state of equilibrium, the entropy of the system reaches a maximum and remains constant, i.e. dS = 0. Under isobaric-isothermal conditions, the driving force of the process, the Gibbs energy, at equilibrium takes on a minimum value and does not change further; dG = 0.
The concentrations of the participants in the reaction in a state of equilibrium are called equilibrium. As a rule, they are denoted by the formulas of the corresponding substances, enclosed in square brackets, for example, the equilibrium concentration of ammonia is denoted in contrast to the initial, nonequilibrium concentration of C ^ NH ^.
Since the rates of the forward and reverse processes in the equilibrium state are equal, we equate the right-hand sides of equations (4.44) and
- -^ i-
- (4.45), replacing the concentration designation: A: [A] "" [B] "=? [D] /; )
