What connects each other is the chemical equilibrium constant. Equilibrium constant
If we consider a reversible gas reaction and assume that gases are ideal, then the following relationships hold:
1) Mendeleev–Clapeyron law pV = nRT (or p i V = n i RT).
Where 
,
(19)
From here we can obtain the following relationships showing the relationship between k p and k c
, but because p i = c i RT, then we get:
.
Here is the change in the number of moles of gaseous substances as a result of one run of this reaction. 
(20)
k р = k с (RT) . (21)
If the partial pressures of the components are expressed in atmospheres and the concentrations in 
, then in the relation (21) connecting k р and k с, instead of R one should put the value R = 0.082 
.
k р does not depend on pressure and concentration, k с does not depend on concentration and pressure.
3.5 Equilibrium in heterogeneous reactions.
Until now we have been talking about homogeneous reactions (occurring in one phase). Let us consider heterogeneous reactions in which not all substances are in the gaseous state. In the case of heterogeneous systems, in which liquid or solid substances do not form solutions with each other or with gaseous substances, the chemical potentials of these condensed substances at a constant temperature will be constant, as well as the vapor pressure above each of these substances in the mixture. Therefore, the expressions for equilibrium constants include only the pressures of gaseous substances.
SaO TV. + CO 2 gas CaCO 3 tv. (22)
If solutions take part in a heterogeneous reaction, the expressions for the equilibrium constants will include the activities (concentrations) of these substances.
3.6 Le Chatelier's principle.
As noted earlier, chemical equilibria are dynamic and fluid. When external conditions change, a shift in equilibrium may occur towards the formation of either reaction products or starting materials. The principle of shifting equilibrium was first formulated by A. Le Chatelier.
Le Chatelier's principle: If an external influence is exerted on a system that is in equilibrium, then the equilibrium shifts towards the process that reduces this influence.
Theoretically, this principle was put forward by F. Brown and is now known as the Le Chatelier–Brown principle.
Typically, the influence of temperature, pressure and changes in the concentration of reacting substances are considered as external factors influencing the state of equilibrium. For example, with increasing temperature, the equilibrium shifts towards an endothermic reaction that occurs with the absorption of heat. A decrease in pressure leads to a shift in equilibrium towards a reaction that occurs with an increase in the number of moles of gaseous substances.
Example 3 .2 Consider the reaction
(
)
Let's increase the temperature: since the reaction is exothermic, i.e. it occurs with the release of heat, the equilibrium shifts towards the starting substances (the reverse reaction occurs with the absorption of heat).
Let's increase the pressure: since the direct reaction occurs with a decrease in the number of moles of gaseous substances (i.e., the volume decreases), the equilibrium shifts towards the reaction products.
1 . Determine the chemical equilibrium constant of the dissociation process of phosphorus pentachloride, which occurs according to the equation PCl 5<=>PCl 3 + Cl 2,
by the equilibrium concentrations of K C and by the partial pressures of the components K R. It is known that by the time equilibrium was established at a temperature of 500 K, 54% of PCl 5 had reacted relative to the initial concentration of 1 mol/l.
Solution. According to the law of mass action for chemical equilibrium
K C = :.
Applying this expression to our equation, we get
K C = C μ (PCl 3) C μ (Cl 2)/C μ (PCl 5).
This expression includes the equilibrium concentrations of the reactant and products. Let's calculate them.
If by the time equilibrium is established, 54% of phosphorus pentachloride has reacted, then 46% still remains in the system, which means that the equilibrium concentration of PCl 5 will be equal to: C μ (PCl 5) = C 0 (PCl 5) 0.46 = 1 0, 46 = 0.46 (mol/l).
Before the start of the reaction, there were no products in the reaction space; therefore, their concentration at equilibrium is determined by the proportion of the consumed reagent (see reaction equation), i.e. C μ (PCl 3) = C μ (Cl 2) = 0.54 mol/l.
Let's substitute the found values into the expression for KC and make calculations:
K C = (0.54)·(0.54)/(0.46) = 0.63 (mol/l).
The equilibrium constant in relation to the partial pressures of the gas system is determined by the equation: K P = K C · (RT) Δ n, in this case, Δn = Σn cont – Σn react. From the chemical equation it is clear that Δn = 2 – 1 = 1 mol.
Let us substitute the found and known values into the expression for KP and make calculations: KP = (0.63)·(8.31·500) 1 = 2617 (Pa/l·K) = 2.62 (kPa/l·K ).
Answer: K C = 0.63 mol/l; K P = 2.62 kPa/l·K.
2 . In a state of equilibrium of the system
CO 2 (g) + H 2 (g)<=>CO(g) + H 2 O(g)
the reaction mixture had the following volumetric composition: 22% CO 2, 41% H 2, 17% CO and 20% H 2 O. Determine K P and K S for this equilibrium at 1900 K and a pressure of 98,501 Pa.
Solution. From the volume fractions of each gas in the mixture, one can determine their partial pressures, knowing that the total pressure in any gas mixture is equal to the sum of the partial pressures of the gases that make up this mixture:
P = P(CO 2) + P(H 2) + P(CO) + P(H 2 O).
And then Р(СО 2) = 0.22·Р = 0.22·98 501 = 21 670 (Pa);
Р(Н 2) = 0.41·98 501 = 40 385 (Pa);
Р(СО) = 0.17·98 501 = 16 745 (Pa); P(H 2 O) = 0.2 98 501 = 19 700 (Pa).
We introduce the found values of partial gas pressures into the equation that establishes the dependence of the equilibrium constant on the partial pressures of the process components:
К Р = Р(СО)·Р(Н 2 О)/Р(СО 2)·Р(Н 2);
K P = 16745·19700/21670·40385 = 0.38.
We will calculate the equilibrium constant in relation to the equilibrium concentrations of reagents K C using the equation K C = K P · (RT) -Δ n. For our case, Δn = 2 – 2 = 0, therefore, the factor (RT) -Δ n = (RT) 0 = 1. and then K C = K P = 0.38.
Answer: K C = K P = 0.38.
3 . Equilibrium in a homogeneous system
4HCl(g) + O2(g)<=>2H 2 O (g) + 2Cl 2 (g)
established at the following concentrations of reactants (mol/l): C μ (H 2 O) = 0.14; C μ (Cl 2) = 0.14; C μ (HCl) = 0.20; C μ (O 2) = 0.32. Calculate the equilibrium constant of this reaction and determine the initial concentrations of hydrogen chloride and oxygen.
Solution. According to the law of mass action, the chemical equilibrium constant for the concentrations of reagents is determined in relation to our case by the equation:
K C = [C μ 2 (H 2 O) · C μ 2 (Cl 2)]/[C μ 4 (HCl) · C μ (O 2)].
Substituting the values of equilibrium concentrations into it and performing calculations, we obtain:
K C = (0.14) 4 / (0.20) 4 · (0.32) = 0.75 (l/mol).
The chemical equation of the reaction shows in what proportions the substances interact with each other: from 4 mol HCl and 1 mol O 2, 2 mol each of H 2 O and Cl 2 are formed. This means that ΔС μ (HCl) = 2·0.14 = 0.28 mol/l and ΔС μ (О 2) = ½·0.14 = 0.07 mol/l were consumed for the formation of 0.14 mol of reaction products .
The initial concentrations of hydrogen chloride and oxygen will be equal:
C μ 0 (HCl) = C μ (HCl) + ΔC μ (HCl), C μ 0 (HCl) = (0.20 + 0.28) mol/l = 0.48 mol/l;
C μ 0 (O 2) = C μ (O 2) + ΔC μ (O 2), C μ 0 (O 2) = (0.32 + 0.07) mol/l = 0.39 mol/l.
Answer: equilibrium constant of this reaction K C = 0.75 l/mol; initial concentrations of reagents C μ 0 (HCl) = 0.48 mol/l; C μ 0 (O 2) = 0.39 mol/l.
4 . Determine the equilibrium constant of the reaction
2NO(g) + Cl 2 (g)<=>2NOCl(g)
at a temperature of 298 K according to the values of the standard enthalpies of formation and entropies of its participants.
Solution. The dependence of the equilibrium constant of a chemical reaction on temperature is determined according to the Kirchhoff equation:
ΔH˚ equals ΔS˚ equals
ℓnК Р = – – + – .
Let's calculate ΔH˚ equal and ΔS˚ equal under standard conditions, using reference data from the table in Appendix No. 1:
ΔH˚ equals = 2ΔH 0 (NOCl) - 2ΔH 0 (NO),
ΔS˚ is equal to = 2S 0 (NOCl) – 2S 0 (NO) – S 0 (Cl 2), then
ΔH˚ equals = 2·53.55 kJ/mol - 2·90.37 kJ/mol = - 73.64 kJ/mol = - 73640 J/mol;
ΔS˚ equals = 2·263.6(J/mol·K) - 2·210.62(J/mol·K) – 223.0(J/mol·K) = - 117.04 J/mol·K .
Let's substitute the found values into the Kirchhoff equation and perform the calculations:
ℓnК Р = - (- 73640)/8.31·298 + (-117.04/8.31) = 29.74 – 14.08 = 15.66.
Then K P = e 15.66 = 5.7 10 6.
Answer: reaction equilibrium constant K P = 5.7·10 6.
5 . For the reaction H 2 (g) + Br 2 (g)<=>2HBr(g) at a certain temperature the equilibrium constant is equal to unity. Determine the composition (in percent by volume) of the equilibrium gas reaction mixture if the initial mixture consisted of 3 moles of hydrogen and 2 moles of bromine.
Solution. Let us write down the expression for the law of mass action for the equilibrium state of the system:
Kr = C 2 μ (HBr)/C μ (H 2) C μ (Br 2) = 1.
Let us assume that by the time equilibrium is established in the system, the reaction has been X moles of H 2 and Br 2. Then their equilibrium concentrations will be equal: C μ (H 2) = (3 - X) mol, C μ (Br 2) = (2 – X) mole. Accordingly, by this moment 2 X mole of hydrogen bromide, i.e. With μ(HBr) = 2 X mole.
Substituting these values of equilibrium concentrations into the expression for the equilibrium constant, we obtain an equation with one unknown: (2 X) 2 /(3-X)·(2- X) = 1.
Having decided it regarding X, we get: 3 X 2 - 5X - 6 = 0;
X 1.2 = (-5±√25+72)/6 = (-5±10)/6. From here X= 0.75 mol.
Thus, at the moment equilibrium was established, the mixture contained: 2.25 mol H 2, 1.25 mol Br 2 and 1.50 mol HBr, which corresponds to 45% hydrogen, 25% bromine and 30% hydrogen bromide.
Answer: at equilibrium, the gas mixture included 45% H 2, 25% Br 2 and 30% HBr.
An arbitrary reversible chemical reaction can be described by an equation of the form:
aA + bB Û dD + eE
In accordance with the law of mass action, in the simplest case, the rate of a direct reaction is related to the concentrations of the starting substances by the equation
v pr = k pr C A a WITH In b,
and the rate of the reverse reaction - with the concentrations of products by the equation
v arr. = k arr. C D d WITH E e.
When equilibrium is achieved, these speeds are equal to each other:
v pr = v arr.
The ratio of the rate constants of the forward and reverse reactions to each other will be equal to equilibrium constant:
Since this expression is based on taking into account the amount of reactants and reaction products, it is a mathematical representation of the law acting masses for reversible reactions.
The equilibrium constant, expressed in terms of the concentrations of the reacting substances, is called the concentration constant and is denoted K s . For a more rigorous consideration, thermodynamic activities of substances should be used instead of concentrations A = fC (Where f - activity coefficient). In this case we are talking about the so-called thermodynamic equilibrium constant
At low concentrations, when the activity coefficients of the starting substances and products are close to unity, K s And K a almost equal to each other.
The equilibrium constant of a reaction occurring in the gas phase can be expressed in terms of partial pressures R substances involved in the reaction:
Between K r And K s there is a relationship that can be derived this way. Let us express the partial pressures of substances in terms of their concentrations using the Mendeleev-Clapeyron equation:
pV = nRT ,
where p = (n /V )RT = CRT .
Then for the reaction in general form, after replacing partial pressures with concentrations, we obtain
Replacing the expression (d + c) - (a + b) with an equal one Dn , we get the final expression
K r = K s (RT )Dn or K s = K r (RT ) - Dn ,
Where Dn - change in the number of moles of gaseous substances during the reaction:
Dn = ån i prod (g) - ån i ref (g) ).
If Dn = 0, i.e. the process proceeds without changing the number of moles of gaseous substances, and K r = K s .
For example, for the ethylene hydration reaction occurring in the gas phase:
C 2 H 4 (g) + H 2 O (g) Û C 2 H 5 OH (g) ,
In this case Dn = 1 - (1 + 1) = -1. This means that the relationship between the constants can be expressed by the following equation:
K r = K s (RT ) - 1 or K s = K p RT .
Thus, knowing K r of this reaction at any given temperature, we can calculate the value K s and vice versa.
The dimension of equilibrium constants depends on the method of expressing concentration (pressure) and the stoichiometry of the reaction. It can often cause confusion, for example, in the example considered [mol - 1 m 3 ] for K s and [Pa - 1] for K r , but there is nothing wrong with that. If the sums of the stoichiometric coefficients of the products and starting substances are equal, the equilibrium constant will be dimensionless.
Since all chemical reactions are reversible, for the reverse reaction (relative to the one when molecules A react with molecules B)
the corresponding expression for the reaction rate will be
Reversibility is indicated by double arrows:
This expression should be read: molecules A and molecules B are in equilibrium with The proportionality sign can be replaced with an equal sign if we introduce a proportionality coefficient k, characteristic of the reaction under consideration. In general
expressions for the speed of the forward reaction (Speed) and reverse reaction (Speed) take the form
When the rates of forward and reverse reactions are equal, the system is said to be in equilibrium:
The ratio is called the equilibrium constant. Remember the following properties of a system in equilibrium
1. The equilibrium constant is equal to the ratio of the rate constants of the forward and reverse reactions,
2. In equilibrium, the rates of forward and reverse reactions (but not their constants) are equal.
3. Equilibrium is a dynamic state. Although there is no total change in the concentration of reactants and products at equilibrium. A and B constantly turn into and vice versa.
4. If the equilibrium concentrations of A and B are known and the numerical value of the equilibrium constant can be found.
Relationship between the equilibrium constant and the change in the standard free energy of a reaction
The equilibrium constant is related to the relation
Here is the gas constant, T is the absolute temperature. Since their values are known, knowing the numerical value, one can find If the equilibrium constant is greater than one, the reaction proceeds spontaneously, that is, in the direction as it is written (from left to right). If the equilibrium constant is less than unity, then the reverse reaction occurs spontaneously. Note, however, that the equilibrium constant indicates the direction in which the reaction can proceed spontaneously, but does not allow us to judge whether the reaction will proceed quickly. In other words, it says nothing about the height of the energy barrier of the reaction (; see above). This follows from the fact that only A (7°) determines. Reaction rates depend on the height of the energy barrier, but not on the magnitude
Lecture 3
Chemical balance. Law of mass action. Chemical equilibrium constant and methods of expressing it.
Chemical equilibrium
In most cases, chemical reactions do not proceed so deeply that the reactants are completely converted into products. The reactions proceed to equilibrium, at which the system contains both products and unreacted starting substances, and no further tendency to change their concentrations is observed. Sometimes the amount of product in an equilibrium mixture is so much greater than the amount of unreacted starting materials that, from a practical point of view, the reaction is complete. Only those reactions reach almost completion in which at least one of the products is removed from the reaction sphere (for example, it precipitates or is released from solution in the form of a gas). But in many important cases, the reaction mixture at equilibrium contains significant concentrations of both products and starting materials.
Chemical equilibrium is a thermodynamic equilibrium in a system in which direct and reverse chemical reactions are possible.
There are thermodynamic and kinetic criteria for chemical equilibrium. From a kinetic point of view, in chemical equilibrium, the rates of all reactions occurring in two opposite directions are equal to each other, therefore, no changes in macroscopic parameters, including the concentrations of reactants, are observed in the system.
From a thermodynamic point of view, chemical equilibrium is characterized by the achievement of a minimum and time-invariant value of the Gibbs energy (or Helmholtz energy).
Knowledge of the basic laws of the study of chemical equilibrium is absolutely necessary for a chemist-technologist. In industry, for example, in chemical and pharmaceutical plants, it is useless to build complex installations for the production of certain substances if thermodynamic calculations show that the reaction tends to go in the “wrong” direction. In addition, when determining the efficiency and profitability of production, it is necessary to know how to obtain the maximum yield of the target product.
The actual mechanism of both forward and reverse reactions is in many cases complex and often not known in detail or completely. Fortunately for chemists, in order to draw correct conclusions about the occurrence of chemical processes, it is not necessary to know the actual reaction mechanism.
Predicting the direction of a chemical reaction, as well as calculating the theoretical equilibrium yield of its products and the composition of the equilibrium reaction mixture depending on the initial composition, temperature and pressure is the main task of the study of chemical equilibrium.
Equilibrium constant
An arbitrary reversible chemical reaction can be described by an equation of the form:
aA + bB Û dD + eE
In accordance with the law of mass action, in the simplest case, the rate of a direct reaction is related to the concentrations of the starting substances by the equation
vpr = k etc WITH Ahh WITH Bb,
and the rate of the reverse reaction - with the concentrations of products by the equation
vorr = kobr WITH Dd WITH Ee.
When equilibrium is achieved, these speeds are equal to each other:
vpr = vorr
The ratio of the rate constants of the forward and reverse reactions to each other will be equal to equilibrium constant:
Since this expression is based on taking into account the amount of reactants and reaction products, it is a mathematical representation of the law acting masses for reversible reactions.
The equilibrium constant, expressed in terms of the concentrations of reacting substances, is called the concentration constant and is denoted KS . For a more rigorous consideration, the thermodynamic activities of substances should be used instead of concentrations A = fC (Where f - activity coefficient). In this case we are talking about the so-called thermodynamic equilibrium constant
At low concentrations, when the activity coefficients of the starting substances and products are close to unity, KS And Ka almost equal to each other.
The equilibrium constant of a reaction occurring in the gas phase can be expressed in terms of partial pressures R substances involved in the reaction:
Between Kr And KS there is a relationship that can be derived this way. Let us express the partial pressures of substances in terms of their concentrations using the Mendeleev-Clapeyron equation:
pV = nRT ,
where p = (n /V )RT = CRT .
Then for the reaction in general form, after replacing partial pressures with concentrations, we obtain
Replacing the expression (d + c) - (a + b) with an equal one D n , we get the final expression
Kr = KS (RT )D n or KS = Kr (RT )-D n ,
Where D n - change in the number of moles of gaseous substances during the reaction:
D n = å ni prod (g) - å ni ref (g) ).
If D n = 0, i.e. the process proceeds without changing the number of moles of gaseous substances, and Kr = KS .
For example, for the ethylene hydration reaction occurring in the gas phase:
C2H4 (g) + H2O (g) Û C2H5OH (g),
In this case D n = 1 - (1 + 1) = -1. This means that the relationship between the constants can be expressed by the following equation:
Kr = KS (RT )- 1 or KS = Kr RT .
Thus, knowing Kr of this reaction at any given temperature, we can calculate the value KS and vice versa.
Calculations using equilibrium constants
Equilibrium constants are used primarily to answer the following questions:
1. Should the reaction proceed spontaneously under certain conditions?
2. What will be the concentration of products (equilibrium yield) after equilibrium is established in the system?
Determining the direction of reversible reactions
Since the equilibrium constant is the ratio of the rate constants of the forward and reverse reactions, its very value indicates the direction of the process. So, if the equilibrium constant is greater than unity, then under these conditions a direct reaction will spontaneously occur, but if it is less than one, a reverse reaction will occur.
According to Le Chatelier's principle, the equilibrium position can be shifted when the conditions under which the reaction occurs change. Therefore, in the general case, it is possible to estimate the shift in equilibrium when the ratio of the initial amounts of substances participating in the reaction changes. If the ratio of concentrations of reacting substances at the initial moment is denoted P :
then according to the ratio Z And KS it is possible to predict the direction of the reaction under given experimental conditions:
at P < K a direct reaction occurs spontaneously;
at P > K the reverse reaction occurs spontaneously;
at P = K the system is in equilibrium.
The more the value of the equilibrium constant differs from unity, the more the reaction equilibrium is shifted to the corresponding side (to the right when TO > 1 and to the left when TO < 1).
Factors influencing balance. Le Chatelier's principle-
Brown
At equilibrium, the forward and reverse reactions exactly cancel each other out. But how sensitive is this compensation to changes in reaction conditions? How can you change the state of balance? These questions are of great practical importance if it is necessary to increase the yield of a useful reaction product, for example, a medicinal substance, or, conversely, to reduce the yield of an undesirable product.
If it is possible to continuously remove products from the reaction mixture (solution) in the form of a gas or precipitate, as well as with the help of technological operations such as freezing, washing, etc., then the reacting system can thereby be constantly maintained in a non-equilibrium, unbalanced state. Under these conditions, there is a need for ever new quantities of reagents and continuous formation of products occurs. This method of disturbing the equilibrium in the direction of obtaining the desired product is carried out without changing the equilibrium constant. But it is often possible to increase the yield of products by increasing the equilibrium constant.
One way to increase the equilibrium constant is to change the temperature. Since in most cases the rates of forward and reverse reactions depend on T , the equilibrium constant also shows a dependence on temperature. Strictly speaking, a change in temperature simultaneously changes the rate of both forward and reverse reactions. But, if an increase in temperature accelerates the forward reaction to a greater extent than the reverse one, then the equilibrium constant will increase.
The temperature dependence of the equilibrium position is one example of the general principle of mobile chemical equilibrium, called Le Chatelier's principle(or Le Chatelier - Brown):
If an external influence is exerted on a system in a state of chemical equilibrium, the equilibrium position shifts in such a direction as to counteract the effect of this influence .
Le Chatelier's principle also applies to other methods of influencing equilibrium, for example, to changing pressure, but it is of a qualitative nature. The quantitative dependence of the equilibrium constant of a reaction on various factors is expressed by the equations of isotherm, isobar and isochore of a chemical reaction, derived by J. Van't Hoff.
Effect on the equilibrium of the initial composition of the reaction
mixtures. Chemical reaction isotherm equation
The maximum work of a reaction occurring in the gas phase at constant temperature and pressure is the algebraic sum of the work performed by all substances participating in the reaction during the transition from initial partial pressures to equilibrium.
Let us consider a gas reaction expressed in general form by the equation
aA + bB Û dD + eE.
Pressure R in the system using the Mendeleev-Clapeyron equation can be expressed in terms of volume V and temperature T :
p = nRT /V ,
whence, assuming that the total number of moles of all components is equal to 1, we obtain for the expansion work
pdV = (RT /V )dV ,
Since the maximum useful work can be calculated by integrating the expression: V2
A'max = ò pdV ,
we get
and since A'max = -D Gr ,
then we can write:
For processes occurring at constant volume, similar expressions can be obtained, which include the maximum work and the change in Helmholtz energy during the reaction. In this case, partial pressures are replaced by the initial concentrations of substances:
Equations (4.1) - (4.4), derived by J. Van't Hoff, are called chemical reaction isotherm equations. They make it possible to determine in which direction and to what extent the reaction can proceed under the conditions under consideration for a given composition of the reaction mixture at a constant temperature.
For standard conditions, when the initial partial pressures (or initial concentrations or activities) of all substances participating in the reaction are equal to unity, the isotherm equations will look like this:
A 'max = RT ln Kp ; D Gor = - RT ln Kp (4.5)
A max = RT ln K With ; D A o r = - RT ln K With .
It follows that when determining the standard value D Gor or D A o r for a reaction, its equilibrium constant can be easily calculated.
Effect of volume change on equilibrium output
and pressure of the reaction mixture
For reactions occurring in the gas phase, the change in the volume of the reaction mixture can be judged by the change in the number of moles of reactants
D n = å ni prod - å ni ref
Three cases are possible, corresponding to different types of chemical reactions:
A) D n < 0 (реакция идет с уменьшением объёма). Например, реакция синтеза аммиака :
N2 (g) + 3H2 (g) Û 2NH3 (g) ; D n = 2 - (1 + 3) = -2
In accordance with Le Chatelier's principle, a decrease in volume (with an increase in pressure) will shift the equilibrium of this and similar reactions to the right, and an increase in volume (with a decrease in pressure) will shift the equilibrium to the left.
b) D n > 0 (the reaction occurs with an increase in volume). For example, the decomposition reaction of methanol:
CH3OH (g) Û CO (g) + 2H2 (g); D n = (1 + 2) - 1 = 2
In this case, a decrease in volume (or an increase in pressure) will shift the equilibrium to the left, and an increase in volume (with a decrease in pressure) will shift the equilibrium to the right.
V) D n = 0 (reaction proceeds without change in volume). For example, the reaction of chlorine with hydrogen bromide:
Cl2 (g) + 2HBr (g) Û Br2 (g) + 2HCl (g) ; D n = (1 + 2) - (1 + 2) = 0
A change in the volume (pressure) of the reaction mixture does not affect the yield of the products of such reactions.
Chemical equilibrium in heterogeneous systems
The patterns discussed earlier relate mainly to homogeneous reactions, that is, to reactions involving substances that are in the same physical state - in the form of a gas or in the form of a solution. Equilibria in which substances in two or more physical states take part (for example, a gas with a liquid or with a solid) are called heterogeneous equilibria.
As an example, consider the decomposition of calcium carbonate CaCO3, used in pharmacy as an antacid (reducing acidity). This is a convenient model for considering the decomposition of various solids, including medicinal ones, leading to the formation of gaseous products:
CaCO3 (t) Û CaO (t) + CO2 (g)
In accordance with the law of mass action, the expression for the equilibrium constant of this reaction can be written as follows:
The partial pressures of CaO and CaCO3 in the gas phase, firstly, are very small, and secondly, they remain practically constant at any time during the reaction. This means that as long as solid CaCO3 and CaO are in contact with the gas, their effect on the equilibrium will be unchanged. In this case, the equilibrium constant does not depend on the amount of solid phase. We can divide both sides of the expression for the equilibrium constant by the quantity p CaO/ p CaCO3 and assume that
K ’ p = p CO2,
Where K ’ p = Kp p CaC03/ p CaO - modified equilibrium constant; in this case, the partial pressures of CaCO3 and CaO are included in the value K ’ p in an implicit form.
If the partial pressure of CO2 above CaCO3 at a given temperature is maintained less than the value K ’ p , then all CaCO3 will turn into CaO and CO2; if the partial pressure p CO2 more than K ’ p , then all CaO will turn into CaCO3. The equilibrium partial pressure of CO2, equal to K ’ p at a given temperature is called dissociation pressure.
When the CO2 pressure reaches 1 atm, the equilibrium in this reaction shifts towards the dissociation of CaCO3, i.e., the decomposition of calcium carbonate. this happens at a temperature of 897°C:
Similar reasoning and the concept of dissociation pressure can be extended to other heterogeneous reactions involving solids. In the case where a medicinal substance (in powder or tablets) can react with gases in the air (H2O, O2, CO2) or decompose with their release, it is necessary to ensure that the partial pressure of these gases and vapors in the warehouse atmosphere is less than the dissociation pressure (or the corresponding equilibrium constant K ’ p ).
