Le Chatelier's principle formula. Le Chatelier principle

It remains unchanged as long as the parameters at which it was established are constant. When conditions change, the equilibrium is disturbed. After some time, equilibrium again occurs in the system, characterized by a new equality of velocities and new equilibrium concentrations of all substances.

The process of transition of a system from one equilibrium state to another is called displacement or shift of chemical equilibrium.

The equilibrium shifts in one direction or another because changing conditions affect the rates of the forward and reverse reactions in different ways. The equilibrium shifts in the direction of the reaction, the rate of which becomes greater when the equilibrium is disturbed. For example, if, when external conditions change, the equilibrium is disturbed so that the rate of the forward reaction becomes greater than the rate of the reverse reaction (V ® > V ¬), then the equilibrium shifts to the right.

In general, the direction of equilibrium shift is determined by Le Chatelier's principle: if an external influence is exerted on a system that is in equilibrium, then the equilibrium is shifted in the direction that weakens the effect of external influence.

Equilibrium shift can be caused by:

temperature change;

Change in the concentration of one of the reagents;

Change in pressure.

Let us dwell on the influence of each of these factors on the state of chemical equilibrium in more detail.

Temperature change. An increase in temperature causes an increase in the rate constant of the endothermic process (DH 0 T > 0 and DU 0 T > 0) and a decrease in the rate constant of the exothermic process (DH 0 T< 0 и DU 0 Т < 0), следовательно, when the temperature rises, the equilibrium shifts towards an endothermic reaction, and when the temperature decreases, an exothermic reaction.

for instance:

N 2 (g) + 3H 2 (g) Û 2NH 3 (g) DH 0 T \u003d -92.4 kJ / mol, i.e. the direct process is exothermic, therefore, with increasing temperature, the equilibrium will shift to the left (in the direction of the reverse reaction).

Change in concentration. With an increase in the concentration of any of the substances, the equilibrium shifts towards the consumption of this substance, and a decrease in the concentration of any substance shifts the equilibrium towards its formation.

For example, for the reaction 2HCl (g) Û H 2(g) + Cl 2(g), an increase in the concentration of hydrogen chloride leads to a shift of the equilibrium to the right (in the direction of the direct reaction). The same result can be obtained by decreasing the concentration of hydrogen or chlorine.

Change in pressure. If several gaseous substances are involved in the reaction, then with increasing pressure, the equilibrium shifts towards the formation of a smaller number of moles of gaseous substances in the gas mixture and, accordingly, towards a decrease in pressure in the system. Conversely, when the pressure decreases, the equilibrium shifts towards the formation of more moles of gas, which causes an increase in pressure in the system.

Example:

N 2(g) + 3H 2(g) Û 2NH 3(g) .

1 mol + 3 mol Û 2 mol

With an increase in pressure in the system, the equilibrium of this reaction shifts to the right (in the direction of the direct reaction).

If the same number of moles of gaseous substances participate in the forward and reverse reactions, then a change in pressure does not cause a shift in chemical equilibrium.

The catalyst does not affect the equilibrium shift, it only accelerates the onset of chemical equilibrium.

The principle is applicable to equilibrium of any nature: mechanical, thermal, chemical, electrical (Lenz effect, Peltier phenomenon).

If external conditions change, this leads to a change in the equilibrium concentrations of substances. In this case, one speaks of a violation or shift in chemical equilibrium.

The chemical equilibrium shifts in one direction or another when any of the following parameters changes:

1. temperature of the system, that is, when it is heated or cooled
2. pressure in the system, that is, when it is compressed or expanded
3. concentration of one of the participants in the reversible reaction

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✪ Le Chatelier principle

✪ 84. Le Chatelier's principle. Shift in balance (part 1)

✪ Chemistry. Grade 11, 2014. Displacement of chemical equilibrium. Foxford Online Learning Center

Subtitles

Let's say we have a reaction. Molecule A plus molecule B is in dynamic equilibrium with molecules C plus D... plus D. This means that the rate of the forward reaction is equal to the rate of the reverse reaction. There will be some equilibrium concentrations of A, B, C and D here, and if we want we can calculate the equilibrium constant. And I will repeat again. I've said this four times before. Just because the rate of the forward reaction is equal to the rate of the reverse reaction does not mean that all concentrations are equal. The concentrations of molecules can be very different. They just don't change because the reaction rates are the same. Given that there is an equilibrium, what happens if I add more A to the system? Let me remind you that she was in balance. The concentrations were constant. Now I am adding more A to the system. Now the chances that particles A and B (even though I don't add more B molecules) will collide are slightly higher, so the direct reaction is more likely to occur. As the number of molecules A increases, there will be more collisions with B, as a result there will be a little less of them. Because they will be consumed. At the same time, the number of C and D will increase markedly, which is important. That's what would happen if A was added. They would collide more with B, and the rate of the forward reaction would be faster than the rate of the reverse. The reaction would have gone in that direction. Then there would be more C and D, they would also collide more often, and the reaction would go in the opposite direction. Eventually, a new equilibrium would emerge. The bottom line is that you will have more A on the left, but slightly less B, because you did not add B. More B will be spent reacting with these A that you added. And then you will get more C and D at equilibrium. And if you added more A and more B... Let's say if you added more B, then the reaction would go forward even more intense. This, of course, is understandable. Obviously, if you affect this reaction by adding on this side, then naturally it will go in the direction that removes the effect. If you add more A, then you will have more A hitting B, and it will go in that direction and probably use up a little more B. If you add both kinds of molecules, then the overall reaction will go in that direction. Likewise... We need to rewrite the reaction. Another color. A plus B, C plus D. If I add more C (I think you get the point here), what happens? The amount of A and B will rise, and perhaps a little more D will be consumed. Then if you add C and D, then of course there will be a lot more A and B. This derivation seems pretty obvious, but it has a nice name, it's called... and it's called Le Chatelier's principle. Le Chatelier. So, Le Chatelier. I need to be careful with how I write. It says that when you act on a reaction that is in equilibrium, it will prefer the direction that weakens this effect. "Affecting a reaction" is, for example, adding more A, and the reaction will go in the forward direction to reduce the effect of this increased A. The effect here is any change. You are changing one in relation to the other. And before that, all the elements were balanced. Let us analyze some situations taking into account Le Chatelier's principle. Given A plus B... A plus B plus heat, the output is C plus D. And plus some E. Let's add heat to this system, see what happens. Heat is required for the reaction to proceed in the forward direction. The more heat, the more likely progress in the forward direction. Le Chatelier's principle states that when we influence this reaction by adding heat, the reaction will prefer the direction that removes this influence. To remove the effect (you have more of this at the entrance), you will increase the consumption of A. The stable concentration of A will decrease when equilibrium is reached. The amount of B will decrease because these molecules will be more actively consumed. The direct reaction is faster. And the number of C, D and E is increasing. What if you do the opposite? Okay, now erase... Instead of adding heat, you're removing heat. Lower the temperature. So, if you take heat away, what happens? There will be predominance in the other direction, because there will be less heat. There is less heat for the reaction to proceed, and this rate will begin to dominate this rate. With a decrease in temperature, the rate of this reaction will decrease, and this one will increase, the concentration will change in this direction, that is, the reverse reaction will be predominant. Now consider pressure. We have previously mentioned the Haber process. And here is the reaction for the Haber process. Nitrogen gas plus 3 moles of hydrogen gas in equilibrium with 2 moles of ammonia gas. What will happen if I apply pressure to this system? I will apply pressure. What happens in this case? A contraction occurs, although the volume does not necessarily decrease, but it does cause all the molecules to tend to be closer to each other. Now that the molecules are close together, the effect of pressure can be lifted if we get fewer molecules at the output. Now I will explain this point to you. PV is equal to nRT. We've seen this many times, right? We can write P equal to nRT / V. If we increase the pressure, how can we remove this effect? Let me remind you that Le Chatelier's principle says that whatever happens, everything will strive to reduce the impact. The reaction will go in the direction that reduces the impact. If we reduce the number of molecules, then this will reduce the pressure, right? There will be fewer molecules colliding with each other. If we reduce the number of molecules here. It's not the best way to write it, it's not an exact equality, but I want you to reason like that. So, I'd better erase it. This was probably not entirely clear. So let's continue. I have a container... No, it's too bright... No, it's the same... So here's the container, and I put pressure on it. Let me have 2 molecules in one container, no, 4 is better. And here let there be only 2 molecules. In both containers, the reaction can go between these molecules. These 4 can combine and form 2 molecules. I am using our example. The nitrogen molecule is this blue molecule. I'll highlight it with a different color. This brown molecule can combine with 3 hydrogens. And this is what happens. This is another way of writing this reaction, perhaps more visually. Now, if I apply pressure to this system... So, I just think of pressure as a kind of force acting on the area from all sides. Which of these situations is more likely to be removed? A situation where we have fewer molecules colliding with each other because it's easier to compress them than when you have a lot of molecules colliding with each other. It's all very conditional, but gives you an understanding. If you apply pressure to the system... By the way, this arrow does not mean that the pressure is decreasing. It means that pressure is applied to the system. But when the pressure increases, which side of the reaction will prevail? The reaction will favor the side with fewer molecules. There are 2 molecules on this side, although they will obviously be large molecules, because of course no mass is lost. And there are 4 molecules on this side, right? 1 mole of nitrogen gas and 3 moles of hydrogen. And just to bring it back to the idea we saw earlier with kinetic equilibrium, let's just imagine a reaction like this. To show that it obeys Le Chatelier's principle, is consistent with everything we have learned about equilibrium constants. So here's the reaction. 2 moles, or just a factor of two, 2 A in gaseous form plus B in gaseous form are in equilibrium with C in gaseous form. Let's say that initially the molar concentration or molarity of A is 2. And the molar concentration of B is 6, and then our molar concentration of C is 8. Equal to 8's. What is the equilibrium constant here? The equilibrium constant is the product (concentration C, which is 8) divided by 2 squared due to this, multiplied by 6. This is equal to 8/24, which is equal to 1/3. Let's say we add more A, no matter how much more, so as not to get confused with the math. But after adding A, our concentration has changed. Now, the concentration of A is molarity 3. You may be asking yourself if I added molarity 1. No. I added, probably more than 1 molarity. It's just that whatever I add, the reaction will shift to the right, that is, in the forward direction. So, some of this here will be swallowed up and go in this direction, but the rest will be here. I could add even more A's to this system. But everything above 1 is absorbed, and this equilibrium concentration of 3 remains. I didn't have to add 1. You can add more. Let's say our new equilibrium is with molarity 12 for C, which agrees with what we're talking about. If we add some A, then the concentration of C must increase, and it is clear that the concentration of B must decrease a little, because a little more B will be consumed, because these molecules will be more likely to collide with more A molecules. Let's see what the new concentration B. Let me remind you that the equilibrium constant remains constant. Our equilibrium constant will now be equal to the concentration C. Here is our reaction. So the molarity is 12, I won't write units divided by our new A concentration of 3. But let's remember the reaction. The factor for A is 2. So that's 3 times the new concentration for B. There's no factor here, so I don't have to worry about any exponents. Now let's just count. So you end up with 1/3 which is 12/9 divided by B. If we simply multiply, we get 9 times the concentration of B, which is 3 times 12, which is 36. Divide both sides of the equation by 9. The new concentration B is 4, or the molarity is 4. So the molarity of B is 4. We've added more A to the reaction. We started with 2 mol for A, 6 mol for B, and 8 for C. We added more A, the reaction went in that direction, maybe it went back and forth a little. But stabilized at molarity 3 for A, molarity 12 for C. So there was an increase in C. Notice that our stable equilibrium concentration of B decreased, which is consistent with our statement that the reaction proceeds in the direction that produces more C, consumes more B. I hope you now have a good understanding of the whole theoretical scheme for influencing the reaction and Le Chatelier's principle.

Temperature effect

Symbol +Q or −Q, written at the end of the thermochemical equation, characterizes the thermal effect of the direct reaction. It is equal in magnitude to the thermal effect of the reverse reaction, but opposite in sign.

The effect of temperature depends on the sign of the thermal effect of the reaction. As the temperature rises, the chemical equilibrium shifts in the direction of the endothermic reaction, and as the temperature drops, in the direction of the exothermic reaction. In the general case, when the temperature changes, the chemical equilibrium shifts towards the process, the sign of the change in entropy in which coincides with the sign of the change in temperature.

The temperature dependence of the equilibrium constant in condensed systems is described by the van't Hoff isobar equation:

(d ln ⁡ KP d T) p = Δ H 0 RT 2 , (\displaystyle \left((\frac (d\ln K_(P))(dT))\right)_(p)=(\frac ( \Delta H^(0))(RT^(2))),)

in systems with a gas phase - the van't Hoff isochore equation

(d ln ⁡ K C d T) v = Δ U 0 R T 2 . (\displaystyle \left((\frac (d\ln K_(C))(dT))\right)_(v)=(\frac (\Delta U^(0))(RT^(2))) .)

In a small range of temperatures in condensed systems, the relationship between the equilibrium constant and temperature is expressed by the following equation:

Ln ⁡ K P = − Δ H 0 R T + Δ S 0 R . (\displaystyle \ln K_(P)=-(\frac (\Delta H^(0))(RT))+(\frac (\Delta S^(0))(R)).)

For example, in the ammonia synthesis reaction

the thermal effect under standard conditions is −92 kJ/mol, the reaction is exothermic, therefore, an increase in temperature leads to a shift in the equilibrium towards the starting materials and a decrease in the product yield.

Pressure influence

Pressure significantly affects the equilibrium position in reactions involving gaseous substances, accompanied by a change in volume due to a change in the amount of substance in the transition from starting substances to products:

With increasing pressure, the equilibrium shifts in the direction in which the total number of moles of gases decreases and vice versa.

In the ammonia synthesis reaction, the amount of gases is halved: N 2 + 3H 2 ↔ 2NH 3

This means that with increasing pressure, the equilibrium shifts towards the formation of NH 3, as evidenced by the following data for the ammonia synthesis reaction at 400 ° C:

Influence of inert gases

The introduction of inert gases into the reaction mixture or the formation of inert gases during the reaction has the same effect as lowering the pressure, since the partial pressure of the reactants is lowered. It should be noted that in this case, a gas not participating in the reaction is considered as an inert gas. In systems with a decrease in the number of moles of gases, inert gases shift the equilibrium towards the starting materials, therefore, in production processes in which inert gases can form or accumulate, periodic blowing of gas pipelines is required.

Influence of concentration

The influence of concentration on the state of equilibrium obeys the following rules:

• With an increase in the concentration of one of the starting substances, the equilibrium shifts in the direction of the formation of reaction products (to the right);
• With a decrease in the concentration of one of the reaction products, the equilibrium shifts in the direction of the formation of the starting substances (to the left).

The main milestones of the biography

Le Chatelier was born in Paris in the family of a mining engineer. From an early age, his father instilled in his son an interest in science. Mother brought up in severity and discipline under the motto "Order is one of the most perfect forms of civilization." Le Chatelier received his primary and secondary education at Rolland College, at the same time he studied at the Military Academy.

Educated at the Polytechnic School, later - at the Higher School of Mines in Paris. During his studies, Le Chatelier worked for A.E. St. Clair Deville in the laboratory, attended lectures at the College de France. He was fond of natural sciences, ancient languages, religious issues.

He worked as a mining engineer in Besançon and in Paris.

In 1875 he married.

From 1878 to 1919 - professor at the Higher Mining School and almost simultaneously (1898-1907) - professor at the College de France.

1886 - Chevalier of the Order of the Legion of Honor.

Between 1907 and 1925 He worked at the University of Paris as an assistant professor and head of the department of chemistry.

In 1898 he succeeded Paul Schützenberg at the Collège de France, where he taught inorganic chemistry.

1907 - chief inspector of mines.

Since 1907 he was a member of the Paris Academy of Sciences.

In 1916, the Royal Society of London honored Le Chatelier with the Davy Medal.

Since 1931 - President of the French Chemical Society. He was a member of many academies of sciences and scientific societies, including a foreign corresponding member of the St. Petersburg Academy of Sciences and an honorary member of the USSR Academy of Sciences.

Le Chatelier died in 1936 at the age of 85.

Scientific activity

The main scientific achievements include:

1. He studied the processes of combustion, ignition, explosions, detonation of firedamp (together with F. Mallar and P.E.M. Berthelot).
2. He proposed a method for determining the heat capacities of gases at high temperatures.
3. He studied chemical and technological processes in metallurgy.
4. He formulated the law of displacement of chemical equilibrium, according to which the equilibrium in an equilibrium system under external influence will shift in the direction opposite to this action (Le Chatelier's principle).
5. He designed a thermoelectric pyrometer, which makes it possible to determine the temperature of bodies by their color; created a metallographic microscope, which helps to study opaque bodies, improved the methodology for studying the structure of metals and alloys.
6. He confirmed the analogy between solutions and alloys by examining the temperature regime of crystallization of systems consisting of two metals and two salts.
7. He studied the methods of preparation and properties of cements, investigated the problems of firing cement and its hardening. He created the theory of "crystallization" - the theory of cement hardening.
8. He derived a thermodynamic equation that establishes the relationship between the temperature of the dissolution process, the solubility and heat of fusion of a substance.
9. Invented the platinum-rhodium thermocouple.
10. He discovered the conditions for the synthesis of ammonia.

A change in external conditions can lead to a change in the thermodynamic parameters and functions that characterize the system, while the state of equilibrium is disturbed. Processes begin in the system, leading to a new state of equilibrium with other equilibrium parameters. Let's show this with an example. The reactor contains a mixture of gases N 2, H 2 and NH 3 in a state of equilibrium:

Let us introduce an additional amount of N 2 into the reactor under isothermal conditions, i.e. increase its concentration. The constant is - 2

weighty TO=---^ will remain unchanged, since it does not depend on

[M 2 PN 2] 3

from concentration. This is possible only as a result of a change in the values ​​of equilibrium concentrations: an increase will lead to a decrease in [H 2 ] due to the additional interaction of a part of the introduced hydrogen with nitrogen, while increasing accordingly. A change in the parameters of a system that brings it to a new state of equilibrium through the predominant flow of direct or reverse processes is called chemical equilibrium shift respectively in the forward or reverse direction. In the example under consideration, the equilibrium shifted in the forward direction.

Qualitative problems of shifting chemical equilibrium can be solved without thermodynamic or kinetic calculations using the rule formulated in 1884 by Le Chatelier.

It was called Le Chatelier's principle (independently of Le Chatelier, this principle was formulated in 1887 by Brown): If any external influence is exerted on a system that is in a state of equilibrium, then as a result of the processes in the system, the equilibrium will shift in the direction leading to a decrease in the impact.

At increase concentration of any substance in equilibrium (for example, NH 3 in the system discussed above), the equilibrium shifts towards expense this substance (in the opposite direction). At decrease concentration of any substance (for example, H 2), the equilibrium shifts towards education of this substance (i.e. in this case also in the opposite direction).

Let us consider the effect of pressure on the process of ammonia synthesis (4.51). Let the pressure in the reactor be increased by 2 times by compression. Under isothermal conditions, the volume will decrease by half, therefore, the concentrations of all components will double. Before the change in pressure, the rate of the forward reaction was

After shrinking, she became

those. increased 16 times. The rate of feedback also increased:

but only 4 times. Therefore, the equilibrium has shifted in the forward direction.

In accordance with the Le Chatelier principle, as the pressure increases by compressing the system, the equilibrium shifts towards a decrease in the number of gas molecules, i.e. in the direction of decreasing pressure (in the above example in the forward direction); when the pressure decreases, the equilibrium shifts towards an increase in the number of gas molecules, i.e. in the direction of increasing pressure (in the above example, in the opposite direction). If the reaction proceeds without changing the number of gas molecules, the equilibrium is not disturbed when the system is compressed or expanded. So, for example, in the system

H 2 (g) + 1 2 (g) 2H1 (g) when the pressure changes, the equilibrium is not disturbed; HI output is independent of pressure.

Pressure has practically no effect on the equilibrium of reactions occurring without the participation of the gas phase, since liquids and solids are almost incompressible. However, at ultrahigh pressures, the equilibrium shifts towards a denser packing of particles in the crystal lattice. For example, graphite, one of the allotropic modifications of carbon (density p \u003d 2.22 g / cm 3), at a pressure of the order of 10 u Pa (10 5 atm) and a temperature of about 2000 ° C, passes into diamond, another modification of carbon with a denser packing atoms (p \u003d 3.51 g / cm 3).

When the temperature rises, the equilibrium shifts in the direction of the endothermic reaction, and when it decreases, it shifts in the direction of the exothermic reaction. For example, the synthesis of ammonia (Equation 4.51) is an exothermic reaction (DN^ 98 = -92.4 kJ). Therefore, as the temperature rises, the equilibrium in the H 2 - N 2 - NH 3 system shifts to the left - towards the decomposition of ammonia, since this process proceeds with the absorption of heat. Conversely, the synthesis of nitric oxide (II) is an endothermic reaction:

Therefore, as the temperature rises, the equilibrium in the N 2 - About 2- NO shifts to the right - towards the formation of N0.

The nature of the displacement under the influence of external influences can be predicted by applying the Le Chatelier principle: if a system in equilibrium is affected from outside, then the equilibrium in the system is shifted in such a way as to weaken the external influence.

1. Influence of concentrations.

An increase in the concentration of one of the reactants shifts the equilibrium of the reaction in the direction of spending the substance.

A decrease in concentration is in the direction of the formation of a substance.

2. Effect of temperature.

An increase in temperature shifts the equilibrium towards a reaction proceeding with the absorption of heat (endothermic), and a decrease in temperature shifts the equilibrium towards a reaction proceeding with the release of heat (exothermic).

3. Influence of pressure.

An increase in pressure shifts the equilibrium towards a reaction proceeding with a decrease in volume, and, conversely, a decrease in pressure shifts towards a reaction proceeding with an increase in volume.

3.1. Examples of problem solving.

Example 1 How will the rate of a reaction proceeding in a closed vessel change if the pressure is increased by 4 times?

2NO (g.) + O 2 (g.) \u003d 2NO 2

Solution: to increase the pressure by 4 times means to increase the concentration of gases by the same amount.

Determine the reaction rate before increasing the pressure.

V 1 \u003d K * C 2 NO * CO 2

Determine the rate of reaction after increasing the pressure.

V 2 \u003d K * (4C NO) 2 * (4CO 2) \u003d 64 K * C 2 NO * CO 2

Determine how many times the reaction rate has increased

V2 = 64*K*C 2 NO *CO 2 = 64

V1 K*C2 NO*CO2

Answer: the reaction rate increased 64 times.

Example 2 How many times will the reaction rate increase with an increase in temperature from 20 C to 50 C0. The temperature coefficient is 3.

Solution: according to the Van't - Hoff rule Vt 2 \u003d Vt 1 *γ T 2 -T 1 / 10

According to the condition of the problem, it is required to determine VT 2

Substitute the data in the formula:

VT 2 \u003d γ T 2 - T 1 / 10 \u003d 3 (50-20) / 10 \u003d 3 3 \u003d 27

Answer: The reaction rate increased 27 times.

Example 3 Calculation of the equilibrium constant of the reaction from the equilibrium concentrations of the reactants and determination of their initial concentrations.

In the synthesis of ammonia N 2 + 3H 2 == 2NH 3, the equilibrium was established at the following concentrations of the reactants (mol / l): C N 2 = 2.5; C H 2 = 1.8; CNH 3 = 3.6. Calculate the equilibrium constant of this reaction and the concentrations of nitrogen and hydrogen.

Solution: we determine the equilibrium constant of this reaction:

K*C= C 2 NH 3 = (3,6) 2 = 0,89

C N 2 *C 3 H 3 2.5*(1.8) 3

The initial concentrations of nitrogen and hydrogen are found on the basis of the reaction equation. The formation of two moles of NH 3 consumes one mole of nitrogen, and the formation of 3.6 moles of ammonia required 3.6/2=1.8 moles of nitrogen. Considering the equilibrium concentration of nitrogen,

find its initial concentration:

C exN 2 \u003d 2.5 + 1.8 \u003d 4.3 mol / l

For the formation of two moles of NH3, it is necessary to spend 3 moles of hydrogen, and the share of obtaining 3.6 moles of ammonia is required

3 * 3.6 / 2 \u003d 5.4 mol.

C refH 2 \u003d 1.8 + 5.4 \u003d 7.2 mol / l

Answer: C N 2 \u003d 4.3

Example 4 The equilibrium constant of a homogeneous system

CO (g) + H 2 O (g) \u003d\u003d CO 2 (g) + H 2 (g)

at 850 0 C it is equal to 1. Calculate the concentrations of all substances at equilibrium if the initial concentrations are: ref = 3 mol/l, ref = 2 mol/l.

Solution: at equilibrium, the rates of direct and reverse reactions are equal, and the ratio of the constants of these rates is also a constant value and is called the equilibrium constant of the given system:

V pr \u003d K 1;

V arr \u003d K 2;

K equals = K 1 =

K2

In the condition of the problem, the initial concentrations are given, while the expression K equals includes only the equilibrium concentrations of all substances in the system. Let us assume that by the moment of equilibrium the concentration is equal to = x mol/l. According to the equation of the system, the number of moles of hydrogen formed in this case will also be x mol / l. By the same number of moles (x mol / l) CO and H 2 O are consumed to form x moles of CO 2 and H 2. Therefore, the equilibrium concentrations of all four substances will be:

Equal \u003d [H 2] equal \u003d x mol / l,

Equal \u003d (3 - x) mol / l,

[H 2 O] equals = (2 - x) mol / l.

Knowing the equilibrium constant, we find the value of x, and then the initial concentrations of all substances:

1 = X 2

x 2 \u003d 6 - 2x - 3x + x 2; 5x = 6, x = 1.2 mol/l

Thus, the desired equilibrium concentrations are:

Equal = 1.2 mol/l.

[H 2] equals = 1.2 mol / l.

Equal \u003d 3 - 1.2 \u003d 1.8 mol / l.

[H 2 O] equals \u003d 2 - 1.2 \u003d 0.8 mol / l.

Example 5 The endothermic reaction of decomposition of phosphorus pentachloride proceeds according to the equation:

PCl 5 (g) == PCl 3 (g) + Cl 2 (g); ΔН = + 129.7 kJ.

How to change: a) temperature, b) pressure; c) concentration in order to shift the equilibrium in the direction of a direct reaction - decomposition of PCl 5?

Solution: a shift or a shift in chemical equilibrium is a change in the equilibrium concentrations of reactants as a result of a change in one of the reaction conditions. The direction in which the equilibrium has shifted is determined by the Le Chatelier principle: a) since the decomposition reaction of PC1 5 is endothermic (ΔН> 0), to shift the equilibrium towards the direct reaction, it is necessary to increase the temperature; 6) since the decomposition of PCl 5 in this system leads to an increase in volume (two gaseous molecules are formed from one gas molecule), then to shift the equilibrium towards a direct reaction, it is necessary to reduce the pressure; c) shifting the equilibrium in the indicated direction can be achieved both by increasing the concentration of PCl 5 and by decreasing the concentration of PCl 3 or Cl 2 .