Equilibrium constant. Chemical equilibrium: chemical equilibrium constant and ways of expressing it Factors affecting chemical equilibrium

Let's return to the ammonia production process, expressed by the equation:

N 2 (g) + 3H 2 (g) → 2NH 3 (g)

Being in a closed volume, nitrogen and hydrogen combine and form ammonia. How long will this process last? It is logical to assume that until any of the reagents runs out. However, in real life this is not entirely true. The fact is that some time after the reaction has begun, the resulting ammonia will begin to decompose into nitrogen and hydrogen, i.e., a reverse reaction will begin:

2NH 3 (g) → N 2 (g) + 3H 2 (g)

In fact, in a closed volume, two reactions, directly opposite to each other, will take place at once. Therefore, this process is written by the following equation:

N 2 (g) + 3H 2 (g) ↔ 2NH 3 (g)

A double arrow indicates that the reaction proceeds in two directions. The reaction of nitrogen and hydrogen combining is called direct reaction. Ammonia decomposition reaction - backlash.

At the very beginning of the process, the rate of direct reaction is very high. But over time, the concentrations of the reagents decrease, and the amount of ammonia increases - as a result, the rate of the forward reaction decreases, and the rate of the reverse reaction increases. There comes a time when the rates of forward and reverse reactions are compared - chemical equilibrium or dynamic equilibrium occurs. At equilibrium, both forward and reverse reactions occur, but their rates are the same, so no changes are noticeable.

Equilibrium constant

Different reactions proceed in different ways. In some reactions, a fairly large number of reaction products are formed before equilibrium occurs; in others - much less. Thus, we can say that a particular equation has its own equilibrium constant. Knowing the equilibrium constant of a reaction, it is possible to determine the relative amounts of reactants and reaction products at which chemical equilibrium occurs.

Let some reaction be described by the equation: aA + bB = cC + dD

• a, b, c, d - coefficients of the reaction equation;
• A, B, C, D - chemical formulas of substances.

Equilibrium constant:

[C] c [D] d K = ———————— [A] a [B] b

Square brackets indicate that the formula involves molar concentrations of substances.

What does the equilibrium constant say?

For the synthesis of ammonia at room temperature K = 3.5·10 8. This is a fairly large number, indicating that chemical equilibrium will occur when the ammonia concentration is much greater than the remaining starting materials.

In actual ammonia production, the technologist’s task is to obtain the highest possible equilibrium coefficient, i.e., so that the direct reaction proceeds to completion. How can this be achieved?

Le Chatelier's principle

Le Chatelier's principle reads:

How to understand this? Everything is very simple. There are three ways to upset the balance:

• changing the concentration of the substance;
• changing the temperature;
• changing the pressure.

When the ammonia synthesis reaction is in equilibrium, it can be depicted as follows (the reaction is exothermic):

N 2 (g) + 3H 2 (g) → 2NH 3 (g) + Heat

Changing concentration

Let's introduce additional nitrogen into a balanced system. This will upset the balance:

The forward reaction will begin to proceed faster because the amount of nitrogen has increased and more of it reacts. After some time, chemical equilibrium will occur again, but the nitrogen concentration will be greater than the hydrogen concentration:

But, it is possible to “skew” the system to the left side in another way - by “lightening” the right side, for example, by removing ammonia from the system as it forms. Thus, the direct reaction of ammonia formation will again predominate.

Changing the temperature

The right side of our “scales” can be changed by changing the temperature. In order for the left side to “outweigh”, it is necessary to “lighten” the right side - reduce the temperature:

Changing the pressure

It is possible to disturb the equilibrium in a system using pressure only in reactions with gases. There are two ways to increase pressure:

• reducing the volume of the system;
• introduction of inert gas.

As pressure increases, the number of molecular collisions increases. At the same time, the concentration of gases in the system increases and the rates of forward and reverse reactions change - the equilibrium is disturbed. To restore balance, the system “tries” to reduce the pressure.

During the synthesis of ammonia, two molecules of ammonia are formed from 4 molecules of nitrogen and hydrogen. As a result, the number of gas molecules decreases - the pressure drops. As a consequence, in order to reach equilibrium after increasing pressure, the rate of the forward reaction increases.

Summarize. According to Le Chatelier's principle, ammonia production can be increased by:

• increasing the concentration of reagents;
• reducing the concentration of reaction products;
• reducing the reaction temperature;
• increasing the pressure at which the reaction occurs.

At a certain temperature, the enthalpy and entropy factors of the reaction can be balanced, then state of balance, which corresponds to the equality ∆ r G T= 0. In this state, the free energy of the system is minimal, and the possibility of forward and reverse reactions occurring is equally probable, while per unit time the same number of reaction products are obtained as they are consumed in the reverse reaction of the formation of the starting substances. Under such conditions, the partial pressures and concentrations of all reaction components will be constant in time and at all points of the system and are called equilibrium pressures and concentrations.

If the reaction proceeds under isochoric-isothermal conditions, then the condition for chemical equilibrium is the equality Δ r F T= 0. From equations (1.12) and (1.15) it follows that when the chemical reaction is in equilibrium a A(g)+ b B(g)+ d D(k) ↔ e E(g)+ f F(g)

∆r G 0 T= - RT ln( p e E equals pf F equals / p a A is equal p b B equal) . (2.1)

If this heterogeneous reaction with the participation of gaseous components occurs at a constant volume, then

∆r F 0 T= - RT ln( c e E equals c f F equals / c a A is equal c b B equal) . (2.2)

If the reaction a A(p)+ b B(p)+ d D(k)= e E(p)+ f F(p) occurs in an ideal solution, then from (1.12a) it follows:

∆r G 0 T=∆r F 0 T = - RT ln( c e E equals c f F equals / c a A is equal c b B equal) . (2.3)

Since the values ​​∆ r F 0 T and ∆ r G 0 T for a given temperature there are constant values, then these equations are valid if under the sign of the logarithm there are constant expressions for a given temperature, called equilibrium constants K c And K r:

K s = (c e E equals c f F equals / c a A is equal c b B equal) (2.4)

K r = (p e E equals pf F equals / p a A is equal p b B equal) . (2.5)

Equations (2.4) and (2.5) are the mathematical expression law of mass action.

For reactions with gaseous components, the relationship between K r And K s expressed by the equation: K r = K s(RT) ∆ν , (2.6) where ∆ν =( e+f-a-b) is the change in the number of moles of gases as a result of the reaction, and R= 0.082 atm . l . mole -1 . K -1. It should be noted that in the expression for K s And K r components in a more condensed state (for example, substance D in a crystalline state) are not included.

Equilibrium constant K r can also be expressed in terms of the equilibrium amounts of moles of gaseous components n i equals the total pressure P 0, at which an isobaric-isothermal reaction is carried out. Considering that the partial pressure i th component is proportional to the mole fraction of this component p i = (n i/Σ n i)P0, from equation (2.5) we obtain:

K r=(p e E equals pf F equals / p a A is equal p b B equal)=( n e E equals nf F equals / n a A is equal n b B equals)( P0/Σ n i) ∆ν (2.6)

where Σ n i =(n E equals + n F equals + n A equals + n B equal) is the sum of the equilibrium moles of all gaseous components.

Combining equations (2.1), (2.2), (2.3) with equations (2.4) and (2.5) we obtain expressions often used for calculations:

∆r G 0 T= - RT ln K r and(2.7)

∆r F 0 T= - RT ln K s for gas-phase reactions. (2.8)

∆r G 0 T =- RT ln K s for reactions in condensed systems. (2.7a)

Thus, having calculated the Gibbs energy of the reaction for a given temperature, we can use these formulas to calculate K s And K r at this temperature. The greater the value of the equilibrium constant under given conditions, the greater the values ​​of the equilibrium concentrations of reaction products, and therefore, the higher the yield of reaction products. The yield of a reaction product is understood as the ratio of the amount (or mass) of a reaction product that is formed under given conditions to the maximum possible (theoretically) quantity (or mass) of this product, provided that any starting substance is completely converted into a reaction product. It is obvious that the complete (100%) conversion of the starting substance into the product is impossible from a thermodynamic point of view, since in this case the equilibrium constant becomes infinitely large.

The degree of conversion of the starting substance is understood as the ratio of the amount (or mass) of the starting substance that reacted under given conditions to the initial amount (or mass) of this substance. If the product yield approaches one (100%), then the degree of conversion of the starting substance also approaches one (100%).

Values K r And K s at a given temperature do not depend on the values ​​of partial pressures and concentrations of components, as well as the total pressure in the system, but depend on temperature. The dependence of the equilibrium constant on temperature can be expressed in differential form:

(d ln Kp/ dT) = ∆r H 0/(RT 2) , (2.9) where ∆ r H 0 is the standard enthalpy of the reaction, which, to a first approximation, is considered independent of temperature. As can be seen from (2.9), with increasing temperature, the equilibrium constant of the exothermic reaction decreases, and the equilibrium constant of the endothermic reaction increases.

When integrating expression (2.9) taking into account the indicated approximation, we obtain (with T 2 > T 1) formula

ln( K 2 /K 1) = (∆r H 0/R)(1/T 1 – 1/T 2) , (2.10)

from which it follows that the greater the absolute value of the thermal effect of the reaction, the more the value of the equilibrium constant changes with temperature. This formula can also be used to calculate the value K equal for any T 3 if values ​​are known TO 2 and TO 1 at temperatures T 2 and T 1 .

Example 10. Write down the expression for K s And K r and calculate K r And K s reactions C(k) + CO 2 (g) = 2CO(g) at 298 K and at 1000 K. Draw a conclusion from the obtained values ​​about the yield of the reaction product at given temperatures and about the effect of temperature on the value of the equilibrium constant.

Solution. Let us write down the expressions for the equilibrium constants of this reaction, taking into account that the reaction is heterogeneous and the substance graphite C(k) is in the solid state:

K p = p 2 CO equals / p CO 2equal; K s = With 2 CO equals / With CO 2equal

From equation (2.7) we have Kp=exp(- ∆G 0 T/RT). Using the results of Example 5, we calculate K r for 298 K and 1000 K:

K r 298 = exp(-120 . 10 3 /8,31 . 298)= exp(-48.5)<< 1;

Kp 1000 =exp(+316/8.31 . 1000) = exp(0.038) = 1.039.

Using formula (2.6) we find K s = K r/(RT) ∆ν = 1.039/0.082. 1000 = 0.013, since ∆ν = 2-1=1. Based on the data obtained, we can conclude that at 298 K the equilibrium constant is K r tends to zero, which indicates that there are practically no reaction products in the equilibrium mixture and the equilibrium of the reaction is strongly shifted towards the starting substances. With increasing temperature, the value of the equilibrium constant increases (the reaction is endothermic) and at 1000 K K r is already greater than 1, that is, reaction products begin to predominate in the equilibrium mixture, their yield increases with increasing T.

Example 11. For some reaction A(g) = 2B(g), occurring at constant pressure and temperature, the equilibrium constant K r equal to 0.02 at 400 K and 4.0 at 600 K. Determine ∆ from these data rH 0 298 , ∆r S 0 298 and ∆ r G 0 298 this reaction, as well as K r at 800 K.

Solution. Neglecting the dependence ∆ rH 0 and ∆ r S 0 on temperature and using expressions (1.14) and (2.7) we compose a system of two equations with two unknowns ( T 1 =400 K, T 2 =600 K):

∆r G 0 T 1 =∆r H 0 298 –T 1 ∆r S 0 298= -RT 1 ln K r 1 or x – 400y= -8.31.400 ln2 . 10 -2

∆r G 0 T 2 =∆r H 0 298 –T 2 ∆r S 0 298= -RT 2 ln K r 2 or x – 600y= -8.31. 600ln4

Where X = ∆rH 0 298 = 52833(J) = 52.833 kJ; y =∆r S 0 298 =99.575 J/K.

Meaning K r at 800 K we calculate it using formula (2.10). We have:

ln( K 800 /K 400) = log( K 800 /0.02)= (52833/8.31)(1/400 -1/800) = 7.95. Where TO 800 = 56,55.

Example 10. Determine the temperature at which in the reaction CaCO 3 (k) = CaO (k) + CO 2 (g) the equilibrium partial pressure of CO 2 R CO2 = 10 4 Pa.

Solution. For this heterogeneous reaction, we write the expression for the equilibrium constant: K r = R CO2, that is, the equilibrium constant is equal to the relative partial pressure of CO2 at a given temperature. For the desired temperature K r =R CO2 = 10 4 /10 5 =0.1. Neglecting the dependence ∆ rH 0 and ∆ r S 0 on temperature, we use formulas (1.14) and (2.7) and equate two expressions for ∆ r G 0 T : ∆r G 0 T= ∆r H 0 298 –T∆r S 0 298= -RT ln K r. ∆ values rH 0 298 and ∆ r S 0 298 is determined as discussed above, using tabular data: ∆r H 0 298 =178.1 kJ; ∆r S 0 298 =160.5 J. We have:

178,1 . 10 3 –T . 160,5

∆r G 0 T= -8,31T ln0,1

Solving the resulting system of equations for T, we find T=991K

K p = ∏ p i ν i (\displaystyle K_(p)=\prod p_(i)^((\nu )_(i)))

For example, for the oxidation reaction of carbon monoxide:

2CO + O 2 = 2CO 2

the equilibrium constant can be calculated using the equation:

Where Δn- change in the number of moles of substances during the reaction. It's clear that K x depends on pressure. If the number of moles of reaction products is equal to the number of moles of starting substances ( Δ n = 0 (\displaystyle \Delta n=0)), That K p = K x (\displaystyle K_(p)=K_(x)).

Standard equilibrium constant

The standard equilibrium constant of a reaction in a mixture of ideal gases (when the initial partial pressures of the reaction participants are equal to their values ​​in the standard state = 0.1013 MPa or 1 atm) can be calculated by the expression:

The standard equilibrium constant is a dimensionless quantity. She is connected with Kp ratio:

It is clear that if p i 0 (\displaystyle p_(i)^(0)) expressed in atmospheres, then (p i 0) Δ n = 1 (\displaystyle (p_(i)^(0))^(\Delta n)=1) And K p = K 0 (\displaystyle K_(p)=K^(0)).

For a reaction in a mixture of real gases in the standard initial state, the partial fugacity of gases is taken equal to their partial pressures f i 0 = p i 0 (\displaystyle f_(i)^(0)=p_(i)^(0))= 0.1013 MPa or 1 atm. Kf associated with K 0 ratio:

Equilibrium constant for reactions in heterogeneous systems

the equilibrium constant (assuming that the gas phase is ideal) has the form:

Thermodynamic description of equilibrium

Along with the designation Q for the ratio of the activities of substances at an arbitrary moment of reaction t (“reaction coefficient”)

in chemical thermodynamics the notation is used K eq for the same form of relationship between the equilibrium activities of substances

In a system where a chemical reaction occurs

equilibrium can be described by the condition

or, the same equilibrium condition can be written using chemical potentials as

where are the chemical potentials

Equilibrium composition of the mixture and direction of reaction

The "reaction coefficient" mentioned above Q(other designations found in the literature are Ω (\displaystyle \Omega ) or π (\displaystyle \pi ), "reaction product")

reflects the ratio of the current activities of all participants in the reaction and can be used to determine the direction of the reaction at the moment for which Q is known

If at moment t the coefficient Q > K, then the current activities of the products are greater than the equilibrium ones, and therefore they must decrease by the time equilibrium is established, that is, a reverse reaction is currently occurring;< K, то If Q = K, then the equilibrium state has been achieved and the rates of forward and reverse reactions are equal;

If Q v 1 > v − 1 (\displaystyle v_(1)>v_(-1)) Using the value Q r (\displaystyle Q_(r))

chemical reaction isotherms Δ G p , T = R T ln ⁡ Q r − R T ln ⁡ K e q = R T ln ⁡ Q r K e q = ∑ ν i μ i (\displaystyle \Delta G_(p,T)=RT\ln Q_(r) -RT\ln K_(eq)=RT\ln (\frac (Q_(r))(K_(eq)))=\sum \nu _(i)\mu _(i)) Where ν (\displaystyle \nu )- stoichiometric coefficients (for products - with a “+” sign, for starting substances - with a “-” sign; the same as in the expressions for Q and K), and

chemical reaction isotherms - chemical potentials and the standard Gibbs energy and the standard constant areΔ G p , T o = − R T ln ⁡ K e q o = ∑ ν i μ i o (\displaystyle \Delta G_(p,T)^(o)=-RT\ln K_(eq)^(o)=\sum \nu _(i)\mu _(i)^(o))

μ o (\displaystyle \mu ^(o))

- standard chemical potentials The isotherm equation shows how the value of Q is related to the change in the free energy of the reaction: At Q > K (\displaystyle Q>K) for direct reaction Δ G > 0 (\displaystyle \Delta G>0), that is ∑ ν j μ j (\displaystyle \sum \nu _(j)\mu _(j)) At for the products of the direct reaction is greater than for the starting substances - this means that the direct reaction is prohibited (which means the reverse reaction is not prohibited); at Q = K (\displaystyle Q=K)< K {\displaystyle Q At Δ G = 0 (\displaystyle \Delta G=0)< 0 {\displaystyle \Delta G<0} , that is, the reaction has reached an equilibrium state;

at Q And Δ G, that is, this spontaneous occurrence of this reaction is allowed The value, by definition, makes sense only for the equilibrium state, that is, for the state with does not say anything about the rates of reactions, but it describes the composition of the system in a state of equilibrium.

If K >> 1, then the system is dominated by (direct) reaction products If K<< 1, то в системе преобладают исходные вещества (продукты обратной реакции)

Standard states

The standard Gibbs energy of a reaction in a gas mixture is the Gibbs energy of a reaction at standard partial pressures of all components equal to 0.1013 MPa (1 atm). The standard Gibbs energy of a reaction in solution is the Gibbs energy at the standard state of the solution, which is taken to be hypothetical a solution with the properties of an extremely dilute solution, but with a concentration of all reagents equal to unity. For a pure substance and a liquid, the standard Gibbs energy coincides with the Gibbs energy of formation of these substances. The value of the standard Gibbs energy of a reaction can be used to approximate the thermodynamic possibility of a reaction proceeding in a given direction, if the initial conditions do not differ much from the standard ones. In addition, by comparing the values ​​of the standard Gibbs energy of several reactions, it is possible to select the most preferable ones, for which it has the largest modulus negative size.

Kinetic description

For a reversible chemical reaction, the equilibrium constant is K eq can be expressed through the rate constants of direct and reverse reactions. Let us consider an elementary reversible chemical reaction of the first order

By definition, equilibrium is given by the condition v 1 = v − 1 (\displaystyle v_(1)=v_(-1)), that is, the equality of the rates of forward and reverse reactions.

In accordance with the law of mass action v = k ∏ a j n j (\displaystyle v=k(\prod )(a_(j))^(n_(j)))

chemical reaction isotherms k is the rate constant of the corresponding reaction, and a j n j (\displaystyle (a_(j))^(n_(j)))- equilibrium activities of the reactants of this reaction, raised to powers equal to their stoichiometric coefficients

we can write the equilibrium condition in the form

(see thermodynamic description of the equilibrium constant), which is only possible if

This important relationship provides one of the “points of contact” between chemical kinetics and chemical thermodynamics.

Multiple equilibria

In the case when several equilibria are established in a system at once (that is, several processes occur simultaneously or sequentially), each of them can be characterized by its own equilibrium constant, from which the general equilibrium constant for the entire set of processes can be expressed. We can consider this situation using the example of stepwise dissociation of dibasic acid H 2 A. Its aqueous solution will contain particles (solvated) H +, H 2 A, HA - and A 2-. The dissociation process occurs in two stages:

K 1 and K 2 - constants of the first and second stages of dissociation, respectively. From these we can express the “total” equilibrium constant for the process of complete dissociation:

Another example of multiple equilibrium is the analysis of the precipitate/soluble complex system. Suppose there is an equilibrium

The reaction can be represented in the form of two successive equilibria - the equilibrium of the decomposition of a complex ion into its constituent ions, which is characterized by an “instability constant” (the reciprocal of the “stability constant” β):

and equilibrium of the transition of ions from the bulk of the solvent to the crystal lattice

taking into account that for solid substances the activity is assumed to be equal to 1 , and in dilute solutions the activities can be replaced by molar concentrations, we get

Where K s p (\displaystyle K_(sp))- solubility product

Then the total equilibrium will be described by the constant

And the value of this constant will be the condition for the predominance of a complex compound or solid salt in the equilibrium mixture: as above, if K<< 1, то в равновесной смеси большая часть ионов связана в комплексное соединение, если K >> 1, then in the equilibrium state in the system most of the ions are bound in the crystalline phase. reactions occurring, respectively, at constant pressure or constant volume. IfΔ H > 0 (\displaystyle \Delta H>0) (the thermal effect is positive, the reaction is endothermic), then the temperature coefficient of the equilibrium constant d ln ⁡ K p d T (\displaystyle (\frac (d\ln K_(p))(dT)))

is also positive, that is, with increasing temperature, the equilibrium constant of the endothermic reaction increases, the equilibrium shifts to the right (which is quite consistent with Le Chatelier’s principle).

Methods for calculating the equilibrium constant Calculation methods for determining the equilibrium constant of a reaction usually come down to calculating in one way or another the standard change in the Gibbs energy during the reaction (ΔG 0

- universal gas constant. Calculation methods for determining the equilibrium constant of a reaction usually come down to calculating in one way or another the standard change in the Gibbs energy during the reaction ( It should be remembered that the Gibbs energy is a function of the state of the system, that is, it does not depend on the path of the process, on the reaction mechanism, but is determined only by the initial and final states of the system. Therefore, if direct determination or calculation Calculation methods for determining the equilibrium constant of a reaction usually come down to calculating in one way or another the standard change in the Gibbs energy during the reaction ( known or can be easily determined, and the summation of which will give the reaction in question (see Hess's Law). In particular, reactions of the formation of compounds from elements are often used as such intermediate reactions.

Entropy calculation of the change in the Gibbs energy and the equilibrium constant of the reaction

Entropy calculation method ΔG reaction is one of the most common and convenient. It is based on the relationship:

or, accordingly, for standard Gibbs energy changes:

Here ΔH 0 at constant pressure and temperature is equal to the thermal effect of the reaction, the methods of calculation and experimental determination of which are known - see, for example, the Kirchhoff equation:

It is necessary to obtain the change in entropy during the reaction. This problem can be solved in several ways, for example:

• According to thermal data - based on Nernst’s thermal theorem and using information about the temperature dependence of the heat capacity of reaction participants. For example, for substances that are in a solid state under normal conditions:

where A and B are tabular constants depending on the type of compound in question, M is the molecular weight. So, if known, Δ H 298 0 (\displaystyle \Delta H_(298)^(0))Δ S 298 0 (\displaystyle \Delta S_(298)^(0)) Δ G T 0 (\displaystyle \Delta G_(T)^(0)) can be calculated using the formula:

A somewhat simplified version of this formula is obtained by considering the sum of the heat capacities of substances to be independent of temperature and equal to the sum of the heat capacities at 298 K:

And an even more simplified calculation is carried out by equating the sum of the heat capacities to zero:

Transfer from Δ G T 0 (\displaystyle \Delta G_(T)^(0)) to the equilibrium constant is carried out according to the above formula.

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 such technological operations 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 ).

Many chemical reactions are reversible, i.e. can simultaneously flow in both directions - forward and reverse. If a reversible reaction is carried out in a closed system, then after some time the system will reach a state of chemical equilibrium - the concentrations of all reacting substances will cease to change over time. It should be noted that the achievement of a state of equilibrium by the system does not mean the cessation of the process; chemical equilibrium is dynamic, i.e. corresponds to the simultaneous occurrence of a process in opposite directions at the same speed. Chemical equilibrium is mobile - any infinitesimal external influence on an equilibrium system causes an infinitesimal change in the state of the system; upon termination of the external influence, the system returns to its original state. Another important property of chemical equilibrium is that a system can spontaneously reach a state of equilibrium from two opposite sides. In other words, any state adjacent to the equilibrium state is less stable, and the transition to it from the equilibrium state is always associated with the need to expend work from outside

Chemically reversible reactions initially proceed in one direction due to the interaction of the starting substances with each other. As reaction products accumulate, they begin to interact with each other to form the starting substances.

Let us consider a reversible reaction in which the reaction order for each substance coincides with the stoichiometric coefficients.

A(A) + b(B) = c(C) + d(D)

Graphical dependence of the rates of direct (V 1) and reverse (V 2) reactions of a reversible chemical process on time

As a result, chemical equilibrium is established in the system, and the concentrations of the initial and final substances cease to change. Invariance of concentrations of substances over time can also be observed in the case of very slowly occurring reactions. But such a state is not a true chemical equilibrium. Signs of true chemical balance :

1. the state of the system remains unchanged over time in the absence of external influences;

2. the state of the system changes under the influence of external influences, no matter how small they are;

3. the state of the system does not depend on which side it approaches equilibrium.

Chemical equilibrium position- this is the ratio of concentrations of reacting substances achieved at equilibrium, which is unchanged for a given state. A quantitative characteristic of chemical equilibrium is the equilibrium constant, which can be expressed in terms of equilibrium concentrations C and partial pressures P.

The state of chemical equilibrium is described by the law active masses .

When equilibrium is achieved, the ratio of the product of concentrations (partial pressures) of reaction products in degrees corresponding to stoichiometric coefficients to the product of concentrations of starting substances also in degrees corresponding to stoichiometric coefficients is a constant. It depends only on the nature of the reactants and temperature. Constant expressed in terms of equilibrium concentrations C i denoted by K s, and expressed in terms of partial pressures P i – K R.

For a homogeneous process

A(A) + b(B) = d(D) + k(K)

K c = or K p = .

For a heterogeneous process

A(A) + b(B) = With[C] + d(D)

K c = or K p = .

The concentration of substance C does not change during the interaction, therefore it is not included in the expression of the equilibrium constant (C is not an effective mass).

The expression for K c and K p is a mathematical expression of the law of mass action as applied to reversible processes.

Using the Mendeleev-Clapeyron equation РV= (m / M) RT, the transformation of which

gives the connection between Kc and Kr:

K p = K s (RT) D n,

Dn is the difference between the amount of gaseous products formed and the amount of initial gaseous substances.

Based on the value of the chemical equilibrium constant, one can judge the depth of the process at the time equilibrium is reached.

In calculations involving the use of equilibrium constants K with, in the proposed problems the concepts of equilibrium and initial concentrations, changes in concentrations are used. Equilibrium concentrations of reactants in a state of chemical equilibrium are called ( WITH), the initial concentrations of substances specified before the start of the reaction are considered ( WITH 0). Equilibrium concentrations of reagents ( WITH) are related to their initial concentration ( WITH 0) equation WITH = WITH 0–D WITH, where D WITH- the amount of the starting substance that reacted before reaching equilibrium.

Example 1. The equilibrium of the reaction 2(NO) + (O 2) = 2(NO 2) was established at the following concentrations of the reactants: C NO = 0.02 mol/l; WITH O = 0.01 mol/l; WITH NO = 0.01 mol/l. Calculate the equilibrium constant and the initial concentrations of NO and O 2.

Solution. To calculate the chemical equilibrium constant, we substitute the values ​​of the equilibrium concentrations of all reacting substances into the expression K with:

K with = = = 0,25.

Let's find the initial concentrations of NO and O 2 from the relationship WITH 0 = WITH+D WITH. To determine the initial concentrations of each substance, it is necessary to calculate the values ​​of D WITH. From the reaction equation we see that from 2 moles of NO, 2 moles of NO 2 are formed as a result of the reaction, therefore, 0.01 moles of NO were consumed to form 0.01 mol of NO 2 by the time equilibrium was established. The equilibrium concentration of NO is 0.02 mol/l, which means C 0(NO) = 0.02 + 0.01 = 0.03 mol/l. Reasoning similarly, we obtain the value D WITH O = 0.005 mol/l, since according to the reaction equation, 1 mole of O 2 is consumed when 2 moles of NO 2 are formed. C 0(O) = 0.01 + 0.005 = 0.015 mol/l.

Example 2. The initial concentrations of carbon monoxide (IV) and hydrogen are 6 mol/L and 4 mol/L, respectively. The equilibrium constant of the process (CO 2) + (H 2) = (CO) + (H 2 O) is 0.5. Calculate the equilibrium concentrations of all substances.

Solution

K with = .

Let us denote D WITH CO, concentration of reacted CO 2 at the moment of equilibrium for X. Then WITH CO = 6 – X. 1 mole of H2 interacts with 1 mole of CO 2, so the amount of reacted hydrogen in the reaction will also be x, WITH H = 4 – X.

From the reaction equation it follows that from 1 mole of CO 2 1 mole of CO is formed, and from 1 mole of hydrogen - 1 mole of H2O, therefore, the concentrations of the formed substances will be equal in X(mol/l).

Substituting the values ​​of equilibrium concentrations into the expression of the constant, we solve it for X:

0,5 = , where X= 2 mol/l.

This means that the amount of reacted CO 2 and H 2 and the resulting CO and H 2 O is 2 mol/l.

The equilibrium concentrations of CO 2 and H 2 are respectively:

6 – 2 = 4 mol/l and 4 – 2 = 2 mol/l.

Example 3. The initial concentration of substance A in the system (A) = 2(B) is 2 mol/l. Equilibrium was established when 20% of the starting material reacted. Calculate the equilibrium constant of the process.

Solution. Let us write down the expression for the equilibrium constant:

K with = .

Let us calculate the values ​​of the equilibrium concentrations of substances A and B. From the conditions of the problem it follows that the amount of reacted substance A is 20% of the original amount, i.e. D WITH A = = 0.2 × 2 = 0.4 mol/l. The equilibrium concentration of substance A is defined as the difference WITH 0(A) – D WITH A = 2 – 0.4 = 1.6 mol/l. From 1 mole of A, 2 moles of B are formed. Therefore, if by the time equilibrium is established, 0.4 mol/L of substance A is consumed, then 0.8 mol/L of substance B is formed. Then WITH B = 0.8 mol/l.

Let us substitute the values ​​of equilibrium concentrations into the expression of the equilibrium constant and obtain the value of the constant:

K with = = 1,6.

ENTROPY

The possibility of spontaneous occurrence of a chemical process is determined by two factors: the desire of the system to reduce internal energy due to an exothermic reaction (-D H) and the tendency of the system to increase disorder in the arrangement of particles due to thermal motion. At low temperatures, the tendency to minimize energy prevails. At high temperatures, due to the intensification of the chaotic movement of particles of matter, the main role begins to be played by the factor of increasing disorder, the measure of which is the state function called entropy.

Entropy ( S) is a thermodynamic function that uniquely characterizes the state of the system regardless of its history. It is a quantitative measure of disorder systems, a measure of chaotic motion and relative position of particles . Entropy depends on the nature of the substance, its quantity (concentration), pressure and temperature. Unlike internal energy and enthalpy, the absolute value of entropy can be determined.

In isolated systems (where exchange of energy and mass with the environment is impossible), processes spontaneously proceed only in the direction of increasing entropy (second law of thermodynamics). Those. in isolated systems, the change in entropy can be used to determine the direction of the spontaneous occurrence of the process.

The entropy of a substance in a certain state (with given parameters) in a certain state of aggregation (gas, liquid, solid) is proportional to the logarithm of the thermodynamic probability of finding a substance in this state (Boltzmann equation):

S = k log W,

Where k = R / N A , W is the thermodynamic probability of the state, which is determined by the number of microstates with the help of which a given macrostate can be realized.

In an ideal crystal at absolute zero T=0K, the movement stops, the position of the particles remains unchanged, and there can be only one probability of the state of the system. Consequently, the entropy of an ideal crystal at absolute zero is S=0.

Third law of thermodynamics states that the entropy of a pure ideal crystalline body at a temperature of 0 K is zero.

Standard entropy is considered per 1 mole of a substance under standard conditions (pressure 1 atm (101.32 kPa) and temperature 298 K (25 O C)). Standard entropy values ​​for most chemical compounds are presented in reference literature. The unit of measurement for entropy is usually Joule per mole kelvin(J/(mol K)).