By themselves, reversible reactions are rarely of practical interest, but in some cases, technological benefits or production profitability require a shift in the equilibrium of one or another reversible reaction. To shift the balance using techniques such as changing concentration of reagents, change in pressure, temperature.
An increase in the concentration of one of the reactants (or both substances) shifts the equilibrium towards the formation of reaction products. Or vice versa, a decrease in the concentration of reaction products also shifts the equilibrium in the direction of their formation. For example for a reaction:
H 2 +Cl 2 ↔2HCl;
An increase in the concentration of H 2 or Cl 2 (as well as simultaneously H 2 and Cl 2) or a decrease in the concentration of HCl will lead to a shift in this equilibrium from left to right, and to shift the equilibrium from right to left, it is necessary either to increase the concentration of HCl or reduce the concentration of H 2, Cl 2 or both substances.
Let us consider the effect of pressure change on a reversible reaction using the reaction as an example:
2N 2 + H 2 ↔2NHz;
With increasing pressure on a given system, the concentration of substances increases. In this case, the equilibrium will shift towards smaller volumes. On the left side of the equation, two volumes of nitrogen react with one volume of hydrogen. There are two volumes of ammonia on the right side of the equation, i.e. the number of volumes on the right side of the equilibrium reaction is less than on the left and, therefore, with increasing pressure, the equilibrium of the reaction will shift to the right. For reaction:
H 2 +Br 2 ↔2HBr
The number of volumes on the right and left sides of the equation are equal (one volume of hydrogen and one volume of bromine on the left and two volumes of hydrogen bromide on the right) and an increase in pressure will not shift the equilibrium either from left to right or from right to left. Given an equilibrium reaction:
Cl 2 (r) + 2HJ (r) ↔2HCl (r) + J 2 (TB)
Indices (g) correspond to gaseous substances, and (s) - to a substance in the solid phase. A change in pressure on this equilibrium system will affect gaseous substances (Сl 2 , HJ, HCl), and substances that are in the solid state (J2) or liquid (H20) are not affected by pressure. Therefore, for the above reaction, an increase in pressure will shift the equilibrium towards smaller volumes, i.e. from left to right.
An increase in temperature increases the kinetic energy of all molecules involved in the reaction. But molecules entering into a reaction (endothermic) begin to interact with each other faster. As the temperature rises, the equilibrium shifts towards an endothermic reaction, and as the temperature decreases, it shifts towards an exometric reaction. Consider the equilibrium reaction:
Q CaCO3 ↔CaO + CO 2 -Q
in which the left side corresponds to an exothermic reaction, and the right side corresponds to an endothermic one. When CaCO3 is heated, decomposition of this substance occurs, therefore, the higher the decomposition temperature of CaCO3, the greater the concentration of CaO and CO 2 becomes, the equilibrium shifts to the endothermic part of the equation, that is, from left to right, and vice versa, when the temperature decreases, the equilibrium will shift towards the exothermic reaction, those. from right to left.
Changes occurring in an equilibrium system as a result of external influences are determined by Le Chatelier's principle
“If an external influence is exerted on a system in chemical equilibrium, then it leads to a shift in the equilibrium in the direction that counteracts this influence.”
The introduction of catalysts into the equilibrium system does not lead to a shift in the equilibrium.
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The chemical equilibrium in the reaction is shifted towards the formation of the reaction product at
1) pressure reduction
2) rise in temperature
3) adding a catalyst
4) adding hydrogen
Solution.
A decrease in pressure (external influence) will lead to an increase in pressure-increasing processes, which means that the equilibrium will shift towards a larger number of gaseous particles (which create pressure), i.e. towards the reagents.
With an increase in temperature (external influence), the system will tend to lower the temperature, which means that the process of absorbing heat intensifies. the equilibrium will shift towards an endothermic reaction, i.e. towards the reagents.
The addition of hydrogen (external influence) will lead to an increase in hydrogen-consuming processes, i.e. the equilibrium will shift towards the product of the reaction
Answer: 4
Source: Yandex: USE training work in chemistry. Option 1.
Equilibrium shifts towards the starting materials when
1) pressure reduction
2) heating
3) the introduction of a catalyst
4) adding hydrogen
Solution.
Le Chatelier's principle - if a system in equilibrium is acted upon from the outside, changing any of the equilibrium conditions (temperature, pressure, concentration), then processes aimed at compensating for external influences intensify in the system.
A decrease in pressure (external influence) will lead to an increase in pressure-increasing processes, which means that the equilibrium will shift towards a larger number of gaseous particles (which create pressure), i.e. towards the reaction products.
With an increase in temperature (external influence), the system will tend to lower the temperature, which means that the process of absorbing heat intensifies. the equilibrium will shift towards an endothermic reaction, i.e. towards the reaction products.
The catalyst does not affect the equilibrium shift
The addition of hydrogen (external influence) will lead to an increase in hydrogen-consuming processes, i.e. equilibrium will shift in the direction of the original substances
Answer: 4
Source: Yandex: USE training work in chemistry. Option 2.
Dmitry Kolomiets 11.12.2016 17:35
4 cannot be correct. When hydrogen is added, the equilibrium will shift in the direction of its consumption - in the direction of the reaction products
Alexander Ivanov
It remains to figure out which part of the equation PRODUCTS
In system
shift of chemical equilibrium to the right will contribute to
1) temperature decrease
2) increase in the concentration of carbon monoxide (II)
3) pressure increase
4) decrease in chlorine concentration
Solution.
It is necessary to analyze the reaction and find out what factors will contribute to the shift of the equilibrium to the right. The reaction is endothermic, goes with an increase in the volume of gaseous products, homogeneous, occurring in the gas phase. According to Le Chatelier's principle, the external action is counteracted by the system. Therefore, the equilibrium can be shifted to the right if the temperature is increased, the pressure is reduced, the concentration of the starting substances is increased, or the amount of reaction products is reduced. Comparing these parameters with the answer options, we choose answer No. 4.
Answer: 4
The shift of chemical equilibrium to the left in the reaction
will contribute
1) decrease in chlorine concentration
2) decrease in the concentration of hydrogen chloride
3) pressure increase
4) temperature decrease
Solution.
The impact on a system that is in equilibrium is accompanied by opposition from its side. With a decrease in the concentration of the starting substances, the equilibrium shifts towards the formation of these substances, i.e. to the left.
Ekaterina Kolobova 15.05.2013 23:04
The answer is incorrect.. It is necessary to reduce the temperature (as the temperature decreases, the equilibrium will shift towards exothermic release)
Alexander Ivanov
As the temperature decreases, the equilibrium will shift towards exothermic release, i.e. to the right.
So the answer is correct
A. When using a catalyst, there is no shift in chemical equilibrium in this system.
B. With an increase in temperature, the chemical equilibrium in a given system will shift towards the starting materials.
1) only A is true
2) only B is true
3) both statements are correct
4) both judgments are wrong
Solution.
When using a catalyst, there is no shift in chemical equilibrium in this system, because A catalyst speeds up both the forward and reverse reactions.
With an increase in temperature, the chemical equilibrium in this system will shift towards the starting materials, because the reverse reaction is endothermic. Increasing the temperature in the system leads to an increase in the rate of the endothermic reaction.
Answer: 3
will shift in the direction of the reverse reaction if
1) increase pressure
2) add catalyst
3) reduce concentration
4) raise the temperature
Solution.
The chemical equilibrium in the system will shift towards the reverse reaction if the rate of the reverse reaction is increased. We argue as follows: a reverse reaction is an exothermic reaction that occurs with a decrease in the volume of gases. If you decrease the temperature and increase the pressure, the equilibrium will shift in the opposite direction.
Answer: 1
Are the following judgments about the shift of chemical equilibrium in the system correct?
A. When the temperature drops, the chemical equilibrium in this system shifts
towards the reaction products.
B. With a decrease in the concentration of methanol, the equilibrium in the system shifts towards the reaction products.
1) only A is true
2) only B is true
3) both statements are correct
4) both judgments are wrong
Solution.
As the temperature decreases, the chemical equilibrium in the system shifts
in the direction of the reaction products, this is true, because the direct reaction is exothermic.
With a decrease in the concentration of methanol, the equilibrium in the system shifts towards the reaction products, this is true because when the concentration of a substance decreases, the reaction that results in the formation of this substance proceeds faster
Answer: 3
In which system is a change in pressure practically no effect on the shift in chemical equilibrium?
Solution.
In order for the equilibrium not to shift to the right when the pressure changes, it is necessary that the pressure in the system does not change. The pressure depends on the amount of gaseous substances in the system. Let's calculate the volumes of gaseous substances in the left and right parts of the equation (by coefficients).
This will be reaction #3
Answer: 3
Are the following judgments about the shift of chemical equilibrium in the system correct?
A. With a decrease in pressure, the chemical equilibrium in this system will shift
towards the reaction product.
B. With an increase in the concentration of carbon dioxide, the chemical equilibrium of the system will shift towards the reaction product.
1) only A is true
2) only B is true
3) both statements are correct
4) both judgments are wrong
Solution.
Le Chatelier's principle - if a system in equilibrium is acted upon from the outside, changing any of the equilibrium conditions (temperature, pressure, concentration), then processes aimed at compensating for external influences intensify in the system.
A decrease in pressure (external influence) will lead to an increase in pressure-increasing processes, which means that the equilibrium will shift towards a larger number of gaseous particles (which create pressure), i.e. towards reagents. Statement A is false.
The addition of carbon dioxide (external influence) will lead to an increase in the processes that consume carbon dioxide, i.e., the equilibrium will shift towards the reactants. Statement B is false.
Answer: Both statements are wrong.
Answer: 4
Chemical equilibrium in the system
is shifted towards the original substances as a result
1) increasing the concentration of hydrogen
2) temperature rise
3) pressure boost
4) using a catalyst
Solution.
The direct reaction is exothermic, the reverse is endothermic, therefore, with an increase in temperature, the equilibrium will shift towards the starting materials.
Answer: 2
Alexander Ivanov
The pressure does.
To shift the equilibrium towards the reactants, it is necessary to reduce the pressure, and this option has not been proposed.
In which system will the chemical equilibrium shift to the right when the pressure is increased?
Solution.
In order for the equilibrium to shift to the right with increasing pressure, it is necessary that the direct reaction proceed with a decrease in the volumes of gases. Let's calculate the volumes of gaseous substances. on the left and right sides of the equation.
This will be reaction #3
Answer: 3
Are the following judgments about the shift of chemical equilibrium in the system correct?
A. With an increase in temperature, the chemical equilibrium in this system will shift
towards the reaction products.
B. With a decrease in the concentration of carbon dioxide, the equilibrium of the system will shift towards the reaction products.
1) only A is true
2) only B is true
3) both statements are correct
4) both judgments are wrong
Solution.
The forward reaction is exothermic, the reverse reaction is endothermic, therefore, as the temperature increases, the equilibrium will shift towards the reverse reaction. (first statement is false)
With an increase in the concentration of the starting substances, the equilibrium will shift towards the forward reaction, with an increase in the concentration of the reaction products, the equilibrium will shift towards the reverse reaction. When the concentration of a substance decreases, the reaction that results in the formation of this substance proceeds faster. (second statement is true)
Answer: 2
Anton Golyshev
No - the explanation is written correctly, read carefully. With a decrease in the concentration of carbon dioxide, the equilibrium will shift in the direction of the reaction of its formation - in the direction of products.
Lisa Korovina 04.06.2013 18:36
The assignment says:
B. With a decrease in the concentration of carbon dioxide, the equilibrium of the system will shift towards the reaction products ... As I understand it, the right side in the reaction is the reaction products. It follows that both options are correct!
Alexander Ivanov
It follows that the second assertion is true.
In system
The shift of chemical equilibrium to the left will occur when
1) pressure reduction
2) lowering the temperature
3) increase in oxygen concentration
4) adding a catalyst
Solution.
Let's calculate the amount of gaseous products in the right and left parts of the reaction (by coefficients).
3 and 2. This shows that if the pressure is lowered, then the equilibrium will shift to the left, because the system seeks to restore equilibrium in the system.
Answer: 1
In system
1) pressure increase
2) increase in the concentration of carbon monoxide (IV)
3) temperature decrease
4) increase in oxygen concentration
Solution.
Le Chatelier's principle - if a system in equilibrium is acted upon from the outside, changing any of the equilibrium conditions (temperature, pressure, concentration), then processes aimed at compensating for external influences intensify in the system.
An increase in pressure (external influence) will lead to an increase in pressure-reducing processes, which means that the equilibrium will shift towards a smaller number of gaseous particles (which create pressure), i.e. towards the reaction products.
The addition of carbon monoxide (IV) (external influence) will lead to an increase in the processes consuming carbon monoxide (IV), i.e. equilibrium will shift in the direction of the original substances
When the temperature drops (external influence), the system will tend to increase the temperature, which means that the process of generating heat intensifies. The equilibrium will shift towards an exothermic reaction, i.e. towards the reaction products.
The addition of oxygen (external influence) will lead to an increase in oxygen-consuming processes, i.e. the equilibrium will shift towards the products of the reaction.
Answer: 2
A. With an increase in temperature in this system, there is no shift in chemical equilibrium,
B. With an increase in the concentration of hydrogen, the equilibrium in the system shifts towards the starting materials.
1) only A is true
2) only B is true
3) both statements are correct
4) both judgments are wrong
Solution.
According to Le Chatelier's rule, since heat is released in a direct reaction, when it increases, the equilibrium will shift to the left; also, since hydrogen is a reactant, as the hydrogen concentration increases, the equilibrium in the system shifts towards the products. Thus, both statements are false.
Answer: 4
In system
the shift of the chemical equilibrium towards the formation of an ester will be facilitated by
1) adding methanol
2) pressure increase
3) increasing the concentration of ether
4) addition of sodium hydroxide
Solution.
With the addition (increase in concentration) of any starting substance, the equilibrium shifts towards the reaction products.
Answer: 1
In which system will the chemical equilibrium shift in the direction of the starting materials as the pressure increases?
Solution.
By increasing or decreasing pressure, it is possible to shift the equilibrium only in processes in which gaseous substances are involved, and which occur with a change in volume.
To shift the equilibrium towards the starting substances with increasing pressure, conditions are necessary for the process to proceed with an increase in volume.
This is process 2. (Starting substances 1 volumes, reaction products - 2)
Answer: 2
In which system does an increase in hydrogen concentration shift the chemical equilibrium to the left?
Solution.
If an increase in hydrogen concentration shifts the chemical equilibrium to the left, then we are talking about hydrogen as a reaction product. The reaction product is hydrogen only in option 3.
Answer: 3
In system
The shift of chemical equilibrium to the right contributes to
1) temperature increase
2) pressure reduction
3) increase in chlorine concentration
4) decrease in the concentration of sulfur oxide (IV)
Solution.
An increase in the concentration of any of the starting substances shifts the chemical equilibrium to the right.
Alexander Ivanov
A decrease in pressure contributes to a shift in the equilibrium towards a larger number of gaseous particles, in this case towards the reagents. So the correct answer is still one - 3
In system
a shift in the chemical equilibrium towards the starting substances will contribute to
1) pressure reduction
2) temperature decrease
3) increase in concentration
4) decrease in concentration
Solution.
This reaction proceeds with a decrease in volume. When the pressure decreases, the volume increases, therefore, the equilibrium shifts towards an increase in volume. In this reaction, towards the starting materials, i.e. to the left.
Answer: 1
Alexander Ivanov
If the concentration of SO 3 decreases, then the equilibrium will shift towards the reaction that increases the concentration of SO 3, that is, to the right (towards the reaction product)
shifts to the right when
1) pressure increase
2) lowering the temperature
3) increasing concentration
4) rise in temperature
Solution.
With an increase in pressure, a decrease in temperature, or an increase in concentration, the equilibrium, according to Le Chatelier's rule, will shift to the left, only with an increase in temperature will the equilibrium shift to the right.
Answer: 4
On the state of chemical equilibrium in the system
does not affect1) pressure increase
2) increase in concentration
3) temperature increase
4) temperature decrease
Solution.
Since this is a homogeneous reaction that is not accompanied by a change in volume, an increase in pressure does not affect the state of chemical equilibrium in this system.
Answer: 1
In which system will the chemical equilibrium shift in the direction of the starting materials as the pressure increases?
Solution.
According to Le Chatelier's rule, with increasing pressure, the chemical equilibrium will shift towards the starting materials in a homogeneous reaction, accompanied by an increase in the number of moles of gaseous products. There is only one such reaction - number two.
Answer: 2
On the state of chemical equilibrium in the system
does not affect
1) pressure increase
2) increase in concentration
3) temperature increase
4) temperature decrease
Solution.
Changes in temperature and concentration of substances will affect the state of chemical equilibrium. At the same time, the amount of gaseous substances on the left and right is the same, therefore, even though the reaction proceeds with the participation of gaseous substances, an increase in pressure will not affect the state of chemical equilibrium.
Answer: 1
Chemical equilibrium in the system
shifts to the right when
1) pressure increase
2) increase in concentration
3) lowering the temperature
4) rise in temperature
Solution.
Since this is not a homogeneous reaction, a change in pressure will not affect it, an increase in the concentration of carbon dioxide will shift the equilibrium to the left. Since heat is absorbed in a direct reaction, its increase will lead to a shift in equilibrium to the right.
Answer: 4
In which system is a change in pressure practically no effect on the shift in chemical equilibrium?
Solution.
In the case of homogeneous reactions, a change in pressure has practically no effect on the shift in chemical equilibrium in systems in which there is no change in the number of moles of gaseous substances during the reaction. In this case, it's reaction number 3.
Answer: 3
In the system, the shift of chemical equilibrium towards the starting substances will be facilitated by
1) pressure reduction
2) temperature decrease
3) decrease in concentration
4) increase in concentration
Solution.
Since this reaction is homogeneous and is accompanied by a decrease in the number of moles of gaseous substances, with a decrease in pressure, the equilibrium in this system will shift to the left.
Answer: 1
Are the following judgments about the shift of chemical equilibrium in the system correct?
A. With increasing pressure, the chemical equilibrium shifts towards the reaction product.
B. When the temperature drops, the chemical equilibrium in this system will shift towards the reaction product.
1) only A is true
2) only B is true
3) both statements are correct
4) both judgments are wrong
Solution.
Since this is a homogeneous reaction, accompanied by a decrease in the number of moles of gases, as the pressure increases, the chemical equilibrium shifts towards the reaction product. In addition, during the passage of a direct reaction, heat is released, therefore, as the temperature decreases, the chemical equilibrium in this system will shift towards the reaction product. Both judgments are correct.
Answer: 3
In system
the shift of the chemical equilibrium to the right will occur when
1) pressure increase
2) rise in temperature
3) increasing the concentration of sulfur oxide (VI)
4) adding a catalyst
Solution.
The amount of gaseous substances in this system is greater on the left than on the right, that is, when a direct reaction occurs, a decrease in pressure occurs, therefore, an increase in pressure will cause a shift in chemical equilibrium to the right.
Answer: 1
Are the following judgments about the shift of chemical equilibrium in the system correct?
A. With an increase in temperature, the chemical equilibrium in a given system will shift towards the starting materials.
B. With an increase in the concentration of nitric oxide (II), the equilibrium of the system will shift towards the starting materials.
1) only A is true
2) only B is true
3) both statements are correct
4) both judgments are wrong
Solution.
Since heat is released in this system, then, according to Le Chatelier's rule, as the temperature rises, the chemical equilibrium in this system will indeed shift towards the starting materials. Since nitric oxide (II) is a reagent, with an increase in its concentration, the equilibrium will shift towards the products.
Answer: 1
Are the following judgments about the shift of chemical equilibrium in the system correct?
A. With a decrease in temperature, the chemical equilibrium in a given system will shift towards the reaction products.
B. With a decrease in the concentration of carbon monoxide, the equilibrium of the system will shift towards the reaction products.
1) only A is true
2) only B is true
3) both statements are correct
4) both judgments are wrong
Solution.
In this reaction, heat is released, therefore, as the temperature decreases, the chemical equilibrium in this system will indeed shift towards the reaction products. Since carbon monoxide is a reagent, a decrease in its concentration will cause a shift in the equilibrium in the direction of its formation - that is, in the direction of the reagents.
Answer: 1
In system
the shift of the chemical equilibrium to the right will occur when
1) pressure increase
2) rise in temperature
3) increasing the concentration of sulfur oxide (VI)
4) adding a catalyst
Solution.
In this homogeneous reaction, there is a decrease in the number of moles of gaseous substances, therefore, a shift in chemical equilibrium to the right will occur with increasing pressure.
Answer: 1
Chemical equilibrium in the system
shifts to the right when
1) pressure increase
2) increase in concentration
3) lowering the temperature
4) rise in temperature
Solution.
With an increase in pressure, an increase in concentration or a decrease in temperature, the equilibrium will shift in the direction of decreasing these effects - that is, to the left. And since the reaction is endothermic, only with an increase in temperature will the equilibrium shift to the right.
Answer: 4
Increasing the pressure will decrease the yield of the product(s) in the reversible reaction
1) N 2 (g) + 3H 2 (g) 2NH 3 (g)
2) C 2 H 4 (g) + H 2 O (g) C 2 H 5 OH (g)
3) C (tv) + CO 2 (g) 2CO (g)
4) 3Fe (tv) + 4H 2 O (g) Fe 3 O 4 (tv) + 4H 2 (g)
Solution.
According to the principle of Le Chatelier - if a system in a state of chemical equilibrium is acted upon from the outside, changing any of the equilibrium conditions (temperature, pressure, concentration), then the equilibrium in the system will shift in the direction that reduces the impact.
Here it is necessary to find a reaction in which the equilibrium will shift to the left with increasing pressure. In this reaction, the number of moles of gaseous substances on the right must be greater than on the left. This is reaction number 3.
Anastasia Serezhenkova 26.11.2013 09:16
C(tv) + CO2(g) 2CO(g) a solid is involved in this reaction. and should be only gases.
Alexander Ivanov
Only gases need to be taken into account. A reaction can contain both solids and liquids.
Chemical equilibrium in the system
shifts towards the reaction products at
1) lowering the temperature
2) pressure reduction
3) using a catalyst
4) rise in temperature
Solution.
According to the principle of Le Chatelier - if a system in a state of chemical equilibrium is acted upon from the outside, changing any of the equilibrium conditions (temperature, pressure, concentration), then the equilibrium in the system will shift in the direction that reduces the impact.
The equilibrium of an endothermic reaction will shift to the right as the temperature increases.
Answer: 4
Source: USE in Chemistry 06/10/2013. main wave. Far East. Option 2.
REACTION EQUATION | ||
2) towards the starting materials 3) practically does not move |
A | B | AT | G |
Solution.
A) 1) towards the reaction products
Answer: 1131
Establish a correspondence between the equation of a chemical reaction and the direction of displacement of chemical equilibrium with increasing pressure in the system:
REACTION EQUATION | DIRECTION OF SHIFT OF CHEMICAL EQUILIBRIUM | |
1) towards the reaction products 2) towards the starting materials 3) practically does not move |
Write down the numbers in response, arranging them in the order corresponding to the letters:
A | B | AT | G |
Solution.
According to the principle of Le Chatelier - if a system in a state of chemical equilibrium is acted upon from the outside, changing any of the equilibrium conditions (temperature, pressure, concentration), then the equilibrium in the system will shift in the direction that reduces the impact.
With increasing pressure, the equilibrium will shift towards a smaller amount of gaseous substances.
A) - towards the reaction products (1)
B) - towards the reaction products (1)
C) - towards the starting materials (2)
D) - towards the reaction products (1)
Answer: 1121
Establish a correspondence between the equation of a chemical reaction and the direction of displacement of chemical equilibrium with increasing pressure in the system:
REACTION EQUATION | DIRECTION OF SHIFT OF CHEMICAL EQUILIBRIUM | |
1) towards the reaction products 2) towards the starting materials 3) practically does not move |
Write down the numbers in response, arranging them in the order corresponding to the letters:
A | B | AT | G |
Solution.
According to the principle of Le Chatelier - if a system in a state of chemical equilibrium is acted upon from the outside, changing any of the equilibrium conditions (temperature, pressure, concentration), then the equilibrium in the system will shift in the direction that reduces the impact.
With increasing pressure, the equilibrium will shift towards the reaction with a smaller amount of gaseous substances.
B) 2) towards the starting materials
C) 3) practically does not move
D) 1) towards the reaction products
Answer: 2231
Establish a correspondence between the equation of a chemical reaction and the direction of displacement of chemical equilibrium with increasing pressure in the system:
REACTION EQUATION | DIRECTION OF SHIFT OF CHEMICAL EQUILIBRIUM | |
1) towards the reaction products 2) towards the starting materials 3) practically does not move |
Write down the numbers in response, arranging them in the order corresponding to the letters:
A | B | AT | G |
Solution.
According to the principle of Le Chatelier - if a system in a state of chemical equilibrium is acted upon from the outside, changing any of the equilibrium conditions (temperature, pressure, concentration), then the equilibrium in the system will shift in the direction that reduces the impact.
With increasing pressure, the equilibrium will shift towards the reaction with a smaller amount of gaseous substances.
A) 2) towards the starting substances
B) 1) towards the reaction products
C) 3) practically does not move
D) 2) towards the starting materials
Answer: 2132
Establish a correspondence between the equation of a chemical reaction and the direction of displacement of chemical equilibrium with a decrease in pressure in the system:
REACTION EQUATION | DIRECTION OF SHIFT OF CHEMICAL EQUILIBRIUM | |
The state of equilibrium for a reversible reaction can last for an indefinitely long time (without outside intervention). But if an external influence is applied to such a system (to change the temperature, pressure or concentration of the final or initial substances), then the state of equilibrium will be disturbed. The rate of one of the reactions will become greater than the rate of the other. Over time, the system will again take an equilibrium state, but the new equilibrium concentrations of the initial and final substances will differ from the initial ones. In this case, one speaks of a shift in the chemical equilibrium in one direction or another.
If, as a result of an external influence, the rate of the forward reaction becomes greater than the rate of the reverse reaction, then this means that the chemical equilibrium has shifted to the right. If, on the contrary, the rate of the reverse reaction becomes greater, this means that the chemical equilibrium has shifted to the left.
When the equilibrium shifts to the right, the equilibrium concentrations of the initial substances decrease and the equilibrium concentrations of the final substances increase in comparison with the initial equilibrium concentrations. Accordingly, the yield of reaction products also increases.
The shift of chemical equilibrium to the left causes an increase in the equilibrium concentrations of the initial substances and a decrease in the equilibrium concentrations of the final products, the yield of which will decrease in this case.
The direction of the chemical equilibrium shift is determined using the Le Chatelier principle: “If an external effect is exerted on a system that is in a state of chemical equilibrium (change the temperature, pressure, concentration of one or more substances participating in the reaction), then this will lead to an increase in the rate of that reaction, the course of which will compensate (reduce) the impact.
For example, with an increase in the concentration of the starting substances, the rate of the direct reaction increases and the equilibrium shifts to the right. With a decrease in the concentration of the starting substances, on the contrary, the rate of the reverse reaction increases, and the chemical equilibrium shifts to the left.
With an increase in temperature (i.e., when the system is heated), the equilibrium shifts towards the occurrence of an endothermic reaction, and when it decreases (i.e., when the system is cooled), it shifts towards the occurrence of an exothermic reaction. (If the forward reaction is exothermic, then the reverse reaction will necessarily be endothermic, and vice versa).
It should be emphasized that an increase in temperature, as a rule, increases the rate of both the forward and reverse reactions, but the rate of the endothermic reaction increases to a greater extent than the rate of the exothermic reaction. Accordingly, when the system is cooled, the rates of forward and reverse reactions decrease, but also not to the same extent: for an exothermic reaction, it is much less than for an endothermic one.
A change in pressure affects the shift in chemical equilibrium only if two conditions are met:
it is necessary that at least one of the substances participating in the reaction be in a gaseous state, for example:
CaCO 3 (t) CaO (t) + CO 2 (g) - a change in pressure affects the displacement of equilibrium.
CH 3 COOH (l.) + C 2 H 5 OH (l.) CH 3 COOS 2 H 5 (l.) + H 2 O (l.) - a change in pressure does not affect the shift in chemical equilibrium, because none of the starting or end substances is in a gaseous state;
if several substances are in the gaseous state, it is necessary that the number of gas molecules on the left side of the equation for such a reaction is not equal to the number of gas molecules on the right side of the equation, for example:
2SO 2 (g) + O 2 (g) 2SO 3 (g) - pressure change affects the equilibrium shift
I 2 (g) + Н 2 (g) 2НI (g) - pressure change does not affect the equilibrium shift
When these two conditions are met, an increase in pressure leads to a shift in the equilibrium towards the reaction, the course of which reduces the number of gas molecules in the system. In our example (catalytic combustion of SO 2), this will be a direct reaction.
A decrease in pressure, on the contrary, shifts the equilibrium in the direction of the reaction proceeding with the formation of a larger number of gas molecules. In our example, this will be the reverse reaction.
An increase in pressure causes a decrease in the volume of the system, and hence an increase in the molar concentrations of gaseous substances. As a result, the rate of forward and reverse reactions increases, but not to the same extent. Lowering the same pressure in a similar way leads to a decrease in the rates of forward and reverse reactions. But at the same time, the reaction rate, towards which the equilibrium shifts, decreases to a lesser extent.
The catalyst does not affect the equilibrium shift, because it speeds up (or slows down) both the forward and reverse reactions equally. In its presence, the chemical equilibrium is only more quickly (or more slowly) established.
If the system is affected by several factors at the same time, then each of them acts independently of the others. For example, in the synthesis of ammonia
N 2 (gas) + 3H 2 (gas) 2NH 3 (gas)
the reaction is carried out with heating and in the presence of a catalyst to increase its rate. But at the same time, the effect of temperature leads to the fact that the reaction equilibrium is shifted to the left, towards the reverse endothermic reaction. This causes a decrease in the output of NH 3 . In order to compensate for this undesirable effect of temperature and increase the ammonia yield, at the same time the pressure in the system is increased, which shifts the reaction equilibrium to the right, i.e. towards the formation of a smaller number of gas molecules.
At the same time, the most optimal conditions for the reaction (temperature, pressure) are selected empirically, under which it would proceed at a sufficiently high rate and give an economically viable yield of the final product.
Le Chatelier's principle is similarly used in the chemical industry in the production of a large number of different substances of great importance for the national economy.
Le Chatelier's principle is applicable not only to reversible chemical reactions, but also to various other equilibrium processes: physical, physicochemical, biological.
The body of an adult is characterized by the relative constancy of many parameters, including various biochemical indicators, including the concentration of biologically active substances. However, such a state cannot be called equilibrium, because it does not apply to open systems.
The human body, like any living system, constantly exchanges various substances with the environment: it consumes food and releases the products of their oxidation and decay. Therefore, the body is characterized steady state, defined as the constancy of its parameters at a constant rate of exchange of matter and energy with the environment. In the first approximation, the stationary state can be considered as a series of equilibrium states interconnected by relaxation processes. In a state of equilibrium, the concentrations of substances participating in the reaction are maintained by replenishing the initial products from the outside and removing the final products to the outside. Changing their content in the body does not lead, in contrast to closed systems, to a new thermodynamic equilibrium. The system returns to its original state. Thus, the relative dynamic constancy of the composition and properties of the internal environment of the body is maintained, which determines the stability of its physiological functions. This property of a living system is called differently homeostasis.
In the course of the life of an organism in a stationary state, in contrast to a closed equilibrium system, there is an increase in entropy. However, along with this, the reverse process simultaneously proceeds - a decrease in entropy due to the consumption of nutrients with a low entropy value from the environment (for example, high-molecular compounds - proteins, polysaccharides, carbohydrates, etc.) and the release of decay products into the environment. According to the position of I.R. Prigozhin, the total production of entropy for an organism in a stationary state tends to a minimum.
A great contribution to the development of nonequilibrium thermodynamics was made by I. R. Prigozhy, winner of the Nobel Prize in 1977, who stated that “in any non-equilibrium system, there are local areas that are in equilibrium. In classical thermodynamics, equilibrium refers to the whole system, and in non-equilibrium - only to its individual parts.
It has been established that entropy in such systems increases during the period of embryogenesis, during the processes of regeneration and the growth of malignant neoplasms.
HCl + NaOH \u003d NaCl + H 2 O, and if the substances were taken in the right proportions, the solution has a neutral reaction and not even traces of hydrochloric acid and sodium hydroxide remain in it. If you try to carry out a reaction in a solution between the resulting substances - sodium chloride and water, then no changes will be found. In such cases, it is said that the reaction of an acid with an alkali is irreversible, i.e. there is no back reaction. Many reactions are practically irreversible at room temperature, for example,
H 2 + Cl 2 \u003d 2HCl, 2H 2 + O 2 \u003d 2H 2 O, etc.
Many reactions are already reversible under ordinary conditions, which means that the reverse reaction proceeds to a noticeable extent. For example, if you try to neutralize with alkali an aqueous solution of a very weak hypochlorous acid, it turns out that the neutralization reaction does not go to the end and the solution has a strongly alkaline environment. This means that the reaction HClO + NaOH
NaClO + H 2 O is reversible, i.e. the products of this reaction, reacting with each other, partially pass into the starting compounds. As a result, the solution has an alkaline reaction. The reaction of ester formation is reversible (the reverse reaction is called saponification): RCOOH + R "OHRCOOR" + H 2 O, many other processes.Like many other concepts in chemistry, the concept of reversibility is largely arbitrary. Usually, a reaction is considered irreversible, after which the concentrations of the starting substances are so low that they cannot be detected (of course, this depends on the sensitivity of the methods of analysis). When external conditions change (primarily temperature and pressure), an irreversible reaction can become reversible and vice versa. So, at atmospheric pressure and temperatures below 1000 ° C, the reaction 2H 2 + O 2 \u003d 2H 2 O can still be considered irreversible, while at a temperature of 2500 ° C and above, water dissociates into hydrogen and oxygen by about 4%, and at a temperature of 3000 ° C already by 20%.
At the end of the 19th century German physical chemist Max Bodenstein (18711942) studied in detail the processes of formation and thermal dissociation of hydrogen iodine: H 2 + I 2
2HI. By varying the temperature, he could achieve a preferential flow of only the forward or only the reverse reaction, but in the general case, both reactions went simultaneously in opposite directions. There are many such examples. One of the most famous ammonia synthesis reaction 3H 2 + N 22NH3; many other reactions are also reversible, for example, the oxidation of sulfur dioxide 2SO 2 + O 22SO 3 , reactions of organic acids with alcohols, etc.Reaction speed and balance. Let there be a reversible reaction A + BC + D. If we assume that the forward and reverse reactions take place in one stage, then the rates of these reactions will be directly proportional to the concentrations of the reactants: the rate of the forward reaction v 1 = k 1 [A][B], reverse reaction rate v 2 = k 2 [C][D] (square brackets indicate the molar concentrations of the reagents). It can be seen that as the direct reaction proceeds, the concentrations of the starting substances A and B decrease, respectively, and the rate of the direct reaction also decreases. The rate of the reverse reaction, which is zero at the initial moment (there are no products C and D), gradually increases. Sooner or later, the moment will come when the rates of the forward and reverse reactions will equalize. After that, the concentrations of all substances A, B, C and D do not change with time. This means that the reaction has reached an equilibrium position, and concentrations of substances that do not change with time are called equilibrium. But, unlike mechanical equilibrium, at which all movement stops, at chemical equilibrium, both reactions and forward and reverse continue to go on, however, their rates are equal and therefore it seems that no changes occur in the system.There are many ways to prove the flow of forward and reverse reactions after reaching equilibrium. For example, if a little hydrogen isotope deuterium D 2 is introduced into a mixture of hydrogen, nitrogen and ammonia, which is in equilibrium, then a sensitive analysis will immediately detect the presence of deuterium atoms in ammonia molecules. And vice versa, if a little deuterated ammonia NH 2 D is introduced into the system, then deuterium will immediately appear in the initial substances in the form of HD and D 2 molecules. Another spectacular experiment was carried out at the Faculty of Chemistry of Moscow State University. The silver plate was placed in a solution of silver nitrate, and no changes were observed. Then an insignificant amount of radioactive silver ions was introduced into the solution, after which the silver plate became radioactive. This radioactivity could not be "washed away" either by rinsing the plate with water or by washing it with hydrochloric acid. Only etching with nitric acid or mechanical processing of the surface with fine sandpaper made it inactive. There is only one way to explain this experiment: there is a continuous exchange of silver atoms between the metal and the solution, i.e. in the system there is a reversible reaction Ag(tv) e = Ag + . Therefore, the addition of radioactive ions Ag + to the solution led to their "embedding" into the plate in the form of electrically neutral, but still radioactive atoms.
Thus, not only chemical reactions between gases or solutions are in equilibrium, but also the processes of dissolution of metals and precipitation. For example, a solid dissolves fastest when placed in a pure solvent when the system is far from equilibrium, in this case from a saturated solution. Gradually, the dissolution rate decreases, and at the same time the rate of the reverse process increases - the transition of a substance from solution to a crystalline precipitate. When the solution becomes saturated, the system reaches a state of equilibrium, while the dissolution and crystallization rates are equal, and the mass of the precipitate does not change with time.
Equilibrium constant. The most important parameter characterizing a reversible chemical reaction is the equilibrium constant To. If we write for the considered reversible reaction A + DC + D the condition of equality of the rates of the forward and reverse reactions in the state of equilibrium k 1 [A] equals [B] equals = k 2 [C] equals [D] equals, whence [C] equals [D] equals /[A] equals [B] equals = k 1 /k 2 = To, then the value To is called the equilibrium constant of a chemical reaction.So, at equilibrium, the ratio of the concentration of reaction products to the product of the concentration of reactants is constant if the temperature is constant (rate constants k 1 and k 2 and hence the equilibrium constant To depend on temperature, but do not depend on the concentration of reagents). If several molecules of the starting substances participate in the reaction and several molecules of the product (or products) are formed, the concentrations of substances in the expression for the equilibrium constant are raised to the powers corresponding to their stoichiometric coefficients. So for the reaction 3H 2 + N 2
2NH 3 the expression for the equilibrium constant is written as K= 2 equal / 3 equal equal. The described method of deriving the equilibrium constant, based on the rates of forward and reverse reactions, cannot be used in the general case, since for complex reactions the dependence of the rate on concentration is usually not expressed by a simple equation or is not known at all. Nevertheless, in thermodynamics it is proved that the final formula for the equilibrium constant turns out to be correct.For gaseous compounds, instead of concentrations, pressure can be used when writing the equilibrium constant; Obviously, the numerical value of the constant can change in this case if the number of gaseous molecules on the right and left sides of the equation is not the same.
Graphs showing how the system approaches equilibrium (such graphs are called kinetic curves) are shown in the figures.
1. Let the reaction be irreversible. Then k 2 \u003d 0. An example is the reaction of hydrogen with bromine at 300 ° C. Kinetic curves show the change in the concentration of substances A, B, C, D (in this case H 2, Br 2 and HBr) depending on time. For simplicity, the initial concentrations of the reagents H 2 and Br 2 are assumed to be equal. It can be seen that the concentrations of the initial substances as a result of the irreversible reaction decrease to zero, while the sum of the concentrations of the products reaches the sum of the concentrations of the reactants. It can also be seen that the reaction rate (the steepness of the kinetic curves) is maximum at the beginning of the reaction, and after the completion of the reaction, the kinetic curves reach a horizontal section (the reaction rate is zero). For irreversible reactions, the equilibrium constant is not introduced, since it is not defined (K
® Ґ ).2. Let k 2 = 0, and k 2k1 and To> 1 (reaction of hydrogen with iodine at 300°C). Initially, the kinetic curves almost do not differ from the previous case, since the rate of the reverse reaction is low (the concentration of products is low). As HI accumulates, the rate of the reverse reaction increases, while the forward decreases. At some point, they will equalize, after which the concentrations of all substances no longer change with time - the reaction rate became zero, although the reaction did not go to the end. In this case ( K> 1) before equilibrium is reached (shaded part), the direct reaction has time to go to a considerable depth, therefore, in the equilibrium mixture there are more products (C and D) than the starting substances A and B the equilibrium is shifted to the right.
3. For the esterification reaction of acetic acid (A) with ethanol (B) at 50 ° C, the rate constant of the forward reaction is less than the reverse: k 1 k 2 , so K 4. In a relatively rare case, when the rate constants of the forward and reverse reactions are equal ( k 1 = k 2 , K= 1), for the reaction A + B = C + D at [A] 0 = [B] 0 in an equilibrium mixture, the concentrations of the starting materials and products will be the same and the kinetic curves will merge. Sometimes such conditions can be created by appropriate selection of temperature. For example, for a reversible reaction CO + H 2 O \u003d H 2 + CO 2 To\u003d 1 at a temperature of about 900 ° C. At higher temperatures, the equilibrium constant for this reaction is less than 1 (for example, at 1000 ° C To\u003d 0.61) and the equilibrium is shifted towards CO and H 2 O. At lower temperatures K> 1 (e.g. at 700°C To\u003d 1.64) and the equilibrium is shifted towards CO 2 and H 2.
Meaning K can serve as a characteristic of the irreversibility of the reaction under given conditions. So if K is very high, which means that the concentrations of the reaction products are much higher than the concentrations of the starting materials at equilibrium, i.e. the reaction was almost completed. For example, for the reaction NiO + H 2
Ni + H 2 O at 523 K (250 ° C) To\u003d [H 2 O] equal / [H 2 ] equal \u003d 800 (solids concentrations are constant and in the expression for To are not included). Consequently, in a closed volume, after reaching equilibrium, the concentration of water vapor will be 800 times greater than that of hydrogen (here, the concentrations can be replaced by pressures proportional to them). So, this reaction at the indicated temperature goes almost to completion. But for the reaction WO 2 + 2H 2W + 2H 2 O at the same temperature To\u003d ([H 2] equal / [H 2 O] equal) 2 \u003d 10 27, therefore, tungsten dioxide is practically not reduced by hydrogen at 500 K.Values To for some reactions are given in the table.
Reaction | Temperature, o С | To |
H 2 + Cl 2 2HCl | 25 | 4 10 31 |
1270 | 5 10 8 | |
H 2 + I 2 (r) valign="TOP"> 25 | 800 | |
1035 | 45 | |
I 2 (r) valign="TOP"> 1275 | 0,003 | |
1475 | 0,07 | |
3H 2 + N 2 valign="TOP"> 25 | 7 10 5 | |
775 | 0,035 | |
CaCO 3 valign="TOP"> 762 | 100 | |
837 | 300 | |
904 | 800 |
Equilibrium constants have been measured or can be calculated for many reactions at different temperatures if the thermal effect of the reaction is known. Quantitatively, the change in the equilibrium constant with temperature is determined by the sign and absolute value of the thermal effect (enthalpy) of the reaction
D H: K= K 0 e D H/ RT, where K 0 constant, independent of temperature, R gas constant, T absolute temperature, e base of natural logarithms. The most important success of chemical thermodynamics was the ability to calculate the equilibrium constants of chemical reactions at different temperatures and, accordingly, to calculate the equilibrium concentrations of starting materials and products without numerous and time-consuming experiments. Examples of such calculations.Hydrogen reduction reaction of iron(II) oxide FeO + H 2
Fe + H 2 O (g) slightly endothermic: D H = +23 kJ/mol (in thermodynamics it is accepted that for exothermic reactions D H D H > 0). For this reaction K\u003d / \u003d 0.004 at 500 K and increases with increasing temperature to 0.85 at 1500 K. Consequently, FeO at a sufficiently high temperature is reduced by hydrogen, although in a closed vessel, if water vapor is not removed, it goes to a small extent.Chromium(III) oxide reduction reaction Cr 2 O 3 + 3H 2
2Cr + 3H 2 O(g) is much more endothermic: D H \u003d +106 kJ (per 1 mol of Cr 2 O 3). For this reaction, the equilibrium constant K= 3 / 3 = 10 23 at 500 K and even at 1500 K it is very small ( K= 10 9). Therefore, this oxide is not reduced by hydrogen at any temperature.Copper oxide reduction reaction CuO + H 2
Cu + H 2 O (g)exothermic:
D H = 80 kJ/mol. The equilibrium constant is very large already at room temperature ( K= 10 12), but the reaction rate is negligible. As the temperature rises, this constant decreases (because D H Calculations of the equilibrium constant are very important for practice. For example, for the synthesis of ammonia, increase To lowering the temperature contributes, but the lower the temperature, the slower the reaction. To speed it up, you need to raise the temperature (while sacrificing the release of ammonia). The introduction of a catalyst also accelerates the reaction. Thus, it is necessary to find the optimal ratio for industrial synthesis between all process parameters, but so far there are no industrial catalysts that allow the reaction to be carried out at temperatures of at least about 100 ° C, when the ammonia concentration in the equilibrium mixture is high enough, so we have to use another method to shift the equilibrium to side of the ammonia increase the pressure while maintaining the high temperature.An important question for practical purposes arises: is it possible to shift the chemical equilibrium in the right direction with the help of a catalyst and in this way increase the yield of the product? It turns out not. The introduction of a catalyst into a system in which a reversible reaction takes place will lead to a decrease in the activation energy of both the forward and reverse reactions by the same value ( cm. CHEMICAL KINETICS). This means that the catalyst accelerates both reactions equally. Thus, for reversible reactions, the role of a catalyst is only to achieve equilibrium more quickly. Kinetic curves for a reversible reaction in the presence of a catalyst are shown in fig. 4 dotted.
Equilibria in electrolyte solutions: solubility product. In solutions of solid electrolytes (most often these are bases and salts), an equilibrium takes place at the boundary of two phases, for example: AgCl (solid) = Ag + + Cl . Both processes direct and reverse occur very quickly: it is enough to add potassium iodide to a solution with AgCl precipitate and mix the mixture, as almost immediately all white silver chloride turns into yellow AgI iodide; if colorless sodium sulfide is added to the solution, black silver sulfide Ag 2 S is immediately formed.For such processes, one can also write the expression for the equilibrium constant. This constant is called the solubility product (PR). In general, for the equilibrium A x B y(TV)
x A y+ + y B x+ RH = [A] x[B] y. Thus, the product of the concentrations of ions in a saturated solution, raised to the appropriate powers, is a constant value at a given temperature. In terms of the PR value of a number of compounds of the same type (indices x and y they have the same) one can judge their relative solubility: the lower the PR, the lower the solubility. For example, you can compare the solubility (in units of mol/l) for FeS and CuS, but you cannot compare the SP values for CuS and Ag 2 S (different X). The value of PR for some sparingly soluble compounds is given in the table.Substance | ETC |
AgCl | 1.8 10 -10 |
AgBr | 5.3 10 -13 |
AgI | 8.3 10 -17 |
Ca(OH)2 | 6.5 10 -6 |
Fe(OH)2 | 7.1 10 -16 |
Cu(OH)2 | 8.3 10 -20 |
Al(OH)3 | 3.2 10 –34 |
Fe(OH)3 | 6.3 10 –38 |
CaSO4 | 2.5 10 -5 |
PbSO4 | 1.6 10 -8 |
BaSO4 | 1.1 10 -10 |
CaCO3 | 3.8 10 -9 |
BaCO3 | 4.0 10 -10 |
PbCO3 | 7.5 10 -14 |
BaF2 | 1.1 10 -6 |
CaF2 | 4.0 10 -11 |
Ca 3 (PO 4) 2 | 2.0 10 -29 |
Ba 3 (PO 4) 2 | 6.0 10 –39 |
FeS | 5 10 -18 |
ZnS | 2.5 10 -22 |
CuS | 6 10 –36 |
Ag2S | 6 10 -50 |
HgS | 1.6 10 -52 |
From the expression for K a it can be seen that for strong acids that dissociate almost completely, the concentrations of H + and A ions will be close to the initial concentration of the acid introduced into the solution, while the concentration of undissociated HA molecules in the solution will be close to zero. As a result, the dissociation constant will be very large. On the other hand, for weak acids the constant will be small. For example, for acetic acid K a= 1.8 10 5 , and for a stronger formic K a= 1.8 10 4 . For the strongest acids, the value K a may exceed 10 10 .
If the acid dissociates to a small extent, the derived formula for calculating the concentration of hydrogen ions is noticeably simplified. Indeed, at weak dissociation, the concentration of acid molecules With 0 practically does not decrease, so With 0
» . In this case, the expression for the equilibrium constant will be written in the form To= 2 /With 0 , whence =. Acetic acid is a weak acid; at room temperature value To for it is equal to 1.75 10 5 , i.e. the equilibrium is shifted to the left towards undissociated acid molecules. At With 0 = 1 mol/l == 0.0042 mol/l, i.e. very few molecules have dissociated. The degree of dissociation (it can be calculated by the formula) in this case is equal to 0.0042 or 0.42%. As the acid is diluted, the degree of dissociation increases. For example, when c 0 = 0.01 mol/l =![](https://i0.wp.com/files.school-collection.edu.ru/dlrstore/bd0d808e-db78-f3d2-e3b2-5388709d7222/image024.gif)
![](https://i0.wp.com/files.school-collection.edu.ru/dlrstore/bd0d808e-db78-f3d2-e3b2-5388709d7222/image026.gif)
For convenience, the value p is usually used instead of the dissociation constant. K a, which is called the acidity index and is determined (by analogy with the hydrogen index) by the expression p To a = lg To a. Here, the smaller the value of p K a the stronger the acid. So, for acetic acid K a= 4.7, for ant p K a= 3.7. For most known acids p K a takes values in the range from 1 to 14. The strongest acids that dissociate in an aqueous solution almost completely have p K a p-values K a for some acids at room temperature are given in the table (for polybasic acids, a value is given that characterizes dissociation only in the first stage). Acids are listed in order of decreasing p value K a, that is, in order of increasing strength of the acid.
Acid | p K a |
H2O2 | 11,6 |
C6H5OH | 10,0 |
H2SiO3 | 9,7 |
H2SnO3 | 9,4 |
H3BO3 | 9,2 |
HCN | 9,1 |
H 2 S | 7,2 |
H2CO3 | 6,4 |
CH3COOH | 4,8 |
C 6 H 8 O 6 (ascorbic) | 4,1 |
UNSD | 3,8 |
HNO 2 | 3,4 |
HF | 3,2 |
C 6 H 8 O 7 (lemon) | 3,1 |
C 4 H 6 O 6 (wine) | 3,0 |
H3PO4 | 2,1 |
H2SO3 | 1,8 |
CCl 3 COOH | 1,7 |
H 2 C 2 O 4 | 1,3 |
HIO 3 | 0,8 |
H2CrO4 | –1 |
HNO3 | –1,6 |
HMnO 4 | –2,3 |
H2SO4 | –3 |
HCl | –7 |
HBr | –9 |
HI | –11 |
For water, which is both a weak acid and a base of equal strength: H 2 O
H + + OH we can also write the equilibrium constant To= /. In pure water, the concentration of [H 2 O] is a constant value equal to 55.6 mol / l. This value also changes little in dilute aqueous solutions of acids and bases. Therefore, for such solutions, the product = To is also a constant and is called the ionic product of water. At 25 ° C, it is equal to 10 14, and, accordingly, [Н + ] = [OH ] = 10 7 mol/l ( cm. HYDROGEN RATE ( P Н) ).Le Chatelier's principle. In 1884 French physical chemist and metallurgist Henri Louis Le Chatelier(18501936) formulated the general law of displacement of chemical equilibrium: “If an external influence is exerted on a system that is in a state of chemical equilibrium (change temperature, pressure, concentrations of substances), then the equilibrium position is shifted in such a direction as to weaken the external influence.” In any case, the equilibrium will shift until a new equilibrium position is reached, which corresponds to the new conditions. This principle makes it easy to predict qualitative changes in an equilibrium system under changing conditions.How can the system "counteract" changes in external conditions? If, for example, the temperature of the equilibrium mixture is increased by heating, the system itself, of course, cannot “weaken” external heating, but the equilibrium in it is shifted in such a way that heating the reaction system to a certain temperature requires more heat than in the case unless the balance shifted. In this case, the equilibrium is shifted so that the heat is absorbed, i.e. towards an endothermic reaction. This can be interpreted as "the desire of the system to weaken external influences." On the other hand, if there is an unequal number of gaseous molecules on the left and right sides of the equation, then the equilibrium in such a system can also be shifted by changing the pressure. With increasing pressure, the equilibrium shifts to the side where the number of gaseous molecules is less (and in this way, as it were, “opposes” external pressure). If the number of gaseous molecules does not change during the reaction
(H 2 + Br 2 (g)
2HBr, CO + H 2 O (g) CO 2 + H 2), then the pressure does not affect the equilibrium position. It should be noted that when the temperature changes, the equilibrium constant of the reaction also changes, while when only the pressure changes, it remains constant.Several examples of the use of Le Chatelier's principle for predicting shifts in chemical equilibrium. Reaction 2SO 2 + O 2
2SO 3 (d) is exothermic. If the temperature is raised, the endothermic decomposition of SO 3 will take precedence and the equilibrium will shift to the left. If the temperature is lowered, the equilibrium will shift to the right. So, a mixture of SO 2 and O 2, taken in a stoichiometric ratio of 2: 1 ( cm. stoichiomerism), at a temperature of 400 ° C and atmospheric pressure turns into SO 3 with a yield of about 95%, i.e. the state of equilibrium under these conditions is almost completely shifted towards SO 3 . At 600° C, the equilibrium mixture already contains 76% SO 3 , and at 800° C, only 25%. That is why when sulfur is burned in air, mainly SO 2 and only about 4% SO 3 are formed. It also follows from the reaction equation that an increase in the total pressure in the system will shift the equilibrium to the right, and with a decrease in pressure, the equilibrium will shift to the left.The reaction of abstraction of hydrogen from cyclohexane with the formation of benzene
C 6 H 6 + 3H 2 is carried out in the gas phase, also in the presence of a catalyst. This reaction goes with the expenditure of energy (endothermic), but with an increase in the number of molecules. Therefore, the effect of temperature and pressure on it will be directly opposite to that observed in the case of ammonia synthesis. Namely: an increase in the equilibrium concentration of benzene in the mixture is facilitated by an increase in temperature and a decrease in pressure, therefore, the reaction is carried out in industry at low pressures (23 atm) and high temperatures (450500 ° C). Here, an increase in temperature is “doubly favorable”: it not only increases the reaction rate, but also contributes to a shift in the equilibrium towards the formation of the target product. Of course, an even greater decrease in pressure (for example, to 0.1 atm) would cause a further shift of the equilibrium to the right, however, in this case, there will be too little substance in the reactor, and the reaction rate will also decrease, so that the overall productivity will not increase, but will decrease. This example shows once again that an economically justified industrial synthesis is a successful maneuver between Scylla and Charybdis.Le Chatelier's principle "works" in the so-called halogen cycle, which is used to produce titanium, nickel, hafnium, vanadium, niobium, tantalum and other high purity metals. The reaction of a metal with a halogen, for example, Ti + 2I 2
TiI 4 goes with the release of heat, and therefore, as the temperature rises, the equilibrium shifts to the left. Thus, at 600°C, titanium easily forms volatile iodide (the equilibrium is shifted to the right), and at 110°C, the iodide decomposes (the equilibrium is shifted to the left) with the release of a very pure metal. Such a cycle also works in halogen lamps, where tungsten evaporated from the spiral and settled on colder walls forms volatile compounds with halogens, which decompose again on a hot spiral, and tungsten is transferred to its original place.In addition to changing temperature and pressure, there is another effective way to influence the equilibrium position. Imagine that from an equilibrium mixture
C + D any substance is excreted. In accordance with the Le Chatelier principle, the system will immediately “respond” to such an impact: the equilibrium will begin to shift in such a way as to compensate for the loss of a given substance. For example, if substance C or D (or both at once) is removed from the reaction zone, the equilibrium will shift to the right, and if substances A or B are removed, it will shift to the left. The introduction of any substance into the system will also shift the equilibrium, but in the other direction.Substances can be removed from the reaction zone in different ways. For example, if there is sulfur dioxide in a tightly closed vessel with water, an equilibrium will be established between gaseous, dissolved and reacted sulfur dioxide:
SO 2 (p) + H 2 O H2SO3. If the vessel is opened, sulfur dioxide will gradually begin to evaporate and will no longer be able to participate in the process the equilibrium will begin to shift to the left, until the sulfurous acid is completely decomposed. A similar process can be observed every time you open a bottle of lemonade or mineral water: CO 2 equilibrium (g) CO 2 (p) + H 2 O H 2 CO 3 as volatilization CO 2 shifts to the left.The removal of a reagent from the system is possible not only with the formation of gaseous substances, but also by binding one or another reagent with the formation of an insoluble compound that precipitates. For example, if an excess of calcium salt is introduced into an aqueous solution of CO 2, then Ca 2+ ions will form a precipitate of CaCO 3, reacting with carbonic acid; equilibrium CO 2 (p) + H 2 O
H 2 CO 3 will shift to the right until there is no dissolved gas left in the water.The equilibrium can also be shifted by adding a reagent. So, when dilute solutions of FeCl 3 and KSCN are drained, a reddish-orange color appears as a result of the formation of iron thiocyanate (thiocyanate):
FeCl 3 + 3KSCN Fe(SCN) 3 + 3KCl. If additional FeCl 3 or KSCN is added to the solution, the color of the solution will increase, which indicates a shift of the equilibrium to the right (as if weakening the external influence). If, however, an excess of KCl is added to the solution, then the equilibrium will shift to the left with a decrease in color to light yellow.In the formulation of Le Chatelier's principle, it is not for nothing that it is indicated that it is possible to predict the results of external influence only for systems that are in a state of equilibrium. If this indication is neglected, it is easy to come to completely wrong conclusions. For example, it is known that solid alkalis (KOH, NaOH) dissolve in water with the release of a large amount of heat - the solution heats up almost as much as when concentrated sulfuric acid is mixed with water. If we forget that the principle applies only to equilibrium systems, we can make the wrong conclusion that as the temperature rises, the solubility of KOH in water should decrease, since it is precisely this shift in the equilibrium between the precipitate and the saturated solution that leads to "weakening of the external influence." However, the process of dissolving KOH in water is not at all equilibrium, since anhydrous alkali is involved in it, while the precipitate in equilibrium with a saturated solution is KOH hydrates (mainly KOH 2H 2 O). The transition of this hydrate from the precipitate to the solution is an endothermic process, i.e. is accompanied not by heating, but by cooling of the solution, so that Le Chatelier's principle for an equilibrium process is also fulfilled in this case. In the same way, when anhydrous salts CaCl 2, CuSO 4, etc. are dissolved in water, the solution heats up, and when crystalline hydrates CuSO 4 5H 2 O, CaCl 2 6H 2 O is dissolved, it cools.
Another interesting and instructive example of the misuse of Le Chatelier's principle can be found in textbooks and popular literature. If an equilibrium mixture of brown nitrogen dioxide NO 2 and colorless N 2 O 4 tetroxide is placed in a transparent gas syringe, and then the gas is quickly compressed with a piston, the color intensity will immediately increase, and after a while (tens of seconds) it will weaken again, although will not reach the original. This experience is usually explained as follows. The rapid compression of the mixture results in an increase in pressure and therefore in the concentration of both components, so the mixture becomes darker. But an increase in pressure, in accordance with the Le Chatelier principle, shifts the equilibrium in the 2NO 2 system
N 2 O 4 towards colorless N 2 O 4 (the number of molecules decreases), so the mixture gradually brightens, approaching a new equilibrium position, which corresponds to increased pressure.The fallacy of this explanation follows from the fact that both reactions dissociation of N 2 O 4 and dimerization of NO 2 occur extremely quickly, so that the equilibrium is established in any case in millionths of a second, so it is impossible to push the piston so fast as to disturb the equilibrium. This experience is explained differently: gas compression causes a significant increase in temperature (everyone who has had to inflate a tire with a bicycle pump is familiar with this phenomenon). And in accordance with the same principle of Le Chatelier, the equilibrium instantly shifts towards an endothermic reaction that goes with the absorption of heat, i.e. towards the dissociation of N 2 O 4 the mixture darkens. Then the gases in the syringe slowly cool to room temperature and the equilibrium shifts again towards the tetroxide the mixture brightens.
Le Chatelier's principle works well in cases that have nothing to do with chemistry. In a normally functioning economy, the total amount of money in circulation is in equilibrium with the goods that this money can buy. What happens if the “outside influence” is the desire of the government to print more money to pay off debts? In strict accordance with Le Chatelier's principle, the balance between commodity and money will be shifted in such a way as to weaken the citizens' pleasure from having more money. Namely, the prices of goods and services will rise, and in this way a new equilibrium will be reached. Another example. In one of the US cities, it was decided to get rid of constant traffic jams by expanding highways and building interchanges. This helped for a while, but then elated residents began to buy more cars, so traffic jams soon reappeared, but with a new "balance" between the roads and the
ó more cars.Equilibrium in redox reactions in solution. In order to determine the thermodynamic possibility of the occurrence of redox reactions in aqueous solutions, the concept of the electric potential E, expressed in volts, is used ( cm. CHEMICAL CURRENT SOURCES). For the reaction aA + bBcC + dD the Nernst equation is used as a criterion:E = E 0 ( RT/nF)ln([C] c [D] d /[A] a [B] b), where E 0 standard potential, n the number of electrons transferred from the oxidizing agent to the reducing agent, F Faraday's constant. If we limit ourselves to a temperature of 25 ° C (T \u003d 298 K), substitute the values R and F, and go from natural logarithms to decimal logarithms, we get: E = E 0 (0.058/ n)lg([C] c [D] d /[A] a [B] b). If for a given reaction the calculated value E> 0, then the reaction will not go, or rather, the equilibrium will be shifted to the left and the stronger, the more E. If a E E.
Using the Nernst equation, we can consider such a "strange" reaction:
Cu + 2H + = Cu 2+ + H 2 and find out what its potential will be E in different conditions. Let, for simplicity, the hydrogen pressure be 1 atm, and the concentration of hydrogen ions in the solution be 1 mol/l, i.e. only the concentration of copper ions changes. Considering that all redox potentials are usually written as reduction reactions (for example, for copper Cu 2+ + 2e
® Cu), the Nernst equation for the reaction under consideration is obtained in the form: E = E 0 + 0.029lg.Let there already be a copper salt in the solution, and = 1 mol / l. Then log = 0 and E = E 0 . Standard electrode potential for copper E o is known and is equal to +0.34 V. This means that under the conditions considered, copper will not go into solution in the form of ions. On the contrary, the reverse reaction Cu 2+ + H 2 = Cu + 2H + is thermodynamically allowed; the possibility of displacing metals (copper, silver, mercury) from solutions of their salts with hydrogen (under pressure) was proved in 1865 by the Russian chemist N.N. Beketov. Is it possible to force the reaction to go in the opposite direction, i.e. dissolve copper in acid? To shift the equilibrium to the right, you need to decrease the value E and make it negative. As can be seen from the Nernst formula, to reduce E it is necessary to reduce the concentration of copper ions in the solution. The formula also shows that this decrease must be very strong. Indeed, even if the solution contains only 10 9 mol/l of copper ions, i.e. a billion times less than the standard concentration, the potential
E = E 0 + 0.029lg = 0.34 + 0.029lg(10 9) = 0.034 + 0.029(9) = +0.08 V and the reaction will not go. But with such an insignificant concentration, 1 liter of the solution contains only 0.000064 mg of copper ions. Even in pure water, the solubility of copper is higher, not to mention hydrochloric or sulfuric acids. That is why copper does not dissolve either in water or in dilute acids. Rather, the dissolution occurs only for the first moments, while there is no copper in the solution at all and E E will become positive and the reaction will stop (we are not talking here about oxidizing acids such as nitric acid or hot concentrated sulfuric acid, in which the mechanism for dissolving copper is completely different.)
0 ,34 + 0.029lg = 0, whence lg = 11.7 or approximately = 10 12 mol/l. Is it possible to practically achieve such small values? Chemists know various ways to reduce the concentration of copper ions (and other metals) to very low levels. One of them is the binding of ions into very strong complexes, which almost do not dissociate with the formation of free ions. Such powerful complexing agents include, for example, cyanide ions. Therefore, in the presence of potassium cyanide, copper will dissolve even in water (in this case, only one-electron oxidation to Cu + occurs): 2 Cu + H 2 O + 4KCN \u003d 2K + 2KOH + H 2.An additional factor that contributes to the reduction of potential E high concentration of hydrogen ions in the solution. But in practice it is not possible to increase this concentration very much; for example, in concentrated hydrochloric acid = 10 mol/l. But under these conditions, Cl anions, which are also abundant in solution, are able to bind singly charged Cu + cations into fairly strong complexes. As a result, copper can also react slowly with concentrated hydrochloric acid: Cu + 4HCl = 2H + H 2 (dissolved oxygen also contributes to the oxidation of copper). In solutions of strong hydroiodic acid, even silver dissolves with the release of hydrogen, since Ag + ions in this case bind into very strong iodide complexes, which almost do not dissociate into I and Ag + ions.
Ilya Leenson
LITERATURE Shelinskiy G.I. Fundamentals of the theory of chemical processes. M., Enlightenment, 1989Leenson I.A. Chemical reactions: Thermal effect, equilibrium, speed. M., Astrel, 2002
A change in pressure in the system shifts the equilibrium only in reactions involving gaseous substances. When it flows, the number of moles of gaseous substances can change. Then:
when the pressure rises, the system tends to lower it, and the equilibrium shifts in the direction of the reaction, which proceeds with a decrease in the number of moles of gaseous substances;
when the pressure decreases, it tends to increase it, and the equilibrium shifts in the direction of the reaction, which proceeds with an increase in the number of moles of gaseous substances;
pressure does not affect the equilibrium in the system if the number of moles of gaseous substances does not change during the reaction.
Example 6. In which direction will the equilibrium shift?
2NO (g) + Cl 2 (g) 2NOCl (g),
if the pressure in the system is doubled. Support your answer with a calculation.
Solution.
1. Since the number of moles of gases decreases during the reaction (reagents - 3 mol, products - 2 mol), then, according to the Le Chatelier principle, with increasing pressure, the equilibrium shifts in the direction of the direct reaction, i.e. to the right "→".
2. Considering that when the pressure doubles, the concentrations of gaseous substances increase by the same number of times, i.e. 2 times, we find the ratio of the rates of forward and reverse reactions before and after the change in pressure:
Table 5
Reverse |
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1) At equilibrium |
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2) At changed concentrations |
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3) Attitude |
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From this it can be seen that with an increase in pressure in the system by a factor of 2, the rate of the direct reaction increases by a factor of 8, and the reverse reaction by a factor of 4, i.e. the condition ; and the equilibrium shifts to the right (→).
2.4. The effect of changing the volume of the system on the shift in equilibrium
Since the molar concentration of a substance
, mol/l,
is inversely proportional to the volume of the system ( V), then an increase (decrease) in the volume of the system in x times leads, respectively, to a decrease (increase) in the molar concentrations of gaseous or dissolved substances in x once.
It follows that the shift in equilibrium can be judged by the change in the rates of reactions (see example 5).
Example 7. In which direction will the equilibrium of the reaction shift?
CS 2 (g) + 2Cl 2 (g) CCl 4 (g) + 2S (tv),
if the volume of the system is doubled. Support your answer with a calculation.
Solution.
Reducing the volume of the system by 2 times leads to an increase in the concentration of gaseous substances by 2 times. Using the technique described in example (6), we find the relationship:
,
.
With a decrease in volume by 2 times, the rate of the direct reaction increased by 8 times, and the reverse - by 2 times, i.e. the condition and the equilibrium shifts to the right (→).
Answer: The equilibrium in the system shifts to the right.
2.5. Effect of temperature change on equilibrium shift
The direction of equilibrium shift with temperature change can be determined in at least two ways.
Iway. The thermal effect of the reaction is known or can be calculated (Δ H r):
With an increase the system, according to the Le Chatelier principle, tries to lower it, and then the equilibrium shifts in the direction of the reaction proceeding with the absorption of heat, i.e. endothermic reactions (
);
Downgrading the system, according to the Le Chatelier principle, tries to increase it, and then the equilibrium shifts in the direction of the reaction proceeding with the release of heat, i.e. exothermic reaction (
).
When predicting the shift in equilibrium, it is necessary to remember:
,
which means: if the forward reaction is exothermic, then the reverse is endothermic, and vice versa.
Example8 . How will an increase in temperature affect the equilibrium state in the reaction:
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Solution. Calculate the thermal effect of the reaction using the tabular data of the standard heats of formation of substances:
kJ.
, which means that the direct reaction is endothermic, that is, it goes with the absorption of heat. Heating the reaction mixture promotes the forward reaction, and the equilibrium will shift in the direction of the direct reaction, i.e. formation of reaction products (→).
Answer: The equilibrium will shift to the right.
IIway. The temperature coefficients are known ( ) forward and backward reactions. In this case, it is possible to determine the direction of the equilibrium shift with a change in temperature using the van't Hoff rule:
,
(9)
where - reaction rates at temperatures t 1
and t 2
, respectively. Temperature can be expressed in both Celsius and Kelvin. is the temperature coefficient of the reaction rate.
Example 9. Determine in which direction the equilibrium of the reaction will shift:
2MgCl 2 (tv) + O 2 (g) 2MgO (tv) + 2Cl 2 (g)
when the temperature rises by 100°C, if , a
? Support your answer with a calculation.
Solution.
Using the van't Hoff rule (9) and taking into account that t 2
–
t 1
= Δ t= 100, we calculate how many times the rates of forward and reverse reactions will change, i.e. attitude .
direct reaction |
Feedback |
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In our case, the rate of the forward reaction will increase by times, for the reverse reaction, the increase in speed will be
time, the condition
and the equilibrium shifts in the direction of the forward reaction, i.e. right ().
Answer: The equilibrium in the system shifts to the right.
Using the Le Chatelier principle for reversible processes, it is possible to predict how the conditions should be changed in order to increase the yield of reaction products, or vice versa, to stop their formation.