Chemical equilibrium and principles of its displacement (Le Chatelier's principle)
IN reversible reactions under certain conditions, a state of chemical equilibrium may occur. This is a condition in which the rate of the reverse reaction becomes equal to the rate of the forward reaction. But in order to shift the equilibrium in one direction or another, it is necessary to change the conditions for the reaction. The principle of shifting equilibrium is Le Chatelier's principle.
Key points:
1. An external influence on a system that is in a state of equilibrium leads to a shift in this equilibrium in a direction in which the effect of the effect is weakened.
2. When the concentration of one of the reacting substances increases, the equilibrium shifts towards the consumption of this substance; when the concentration decreases, the equilibrium shifts towards the formation of this substance.
3. With an increase in pressure, the equilibrium shifts towards a decrease in the amount of gaseous substances, that is, towards a decrease in pressure; when the pressure decreases, the equilibrium shifts towards increasing amounts of gaseous substances, that is, towards increasing pressure. If the reaction proceeds without changing the number of molecules of gaseous substances, then pressure does not affect the equilibrium position in this system.
4. When the temperature increases, the equilibrium shifts towards the endothermic reaction, and when the temperature decreases, towards the exothermic reaction.
For the principles we thank the manual “Beginnings of Chemistry” Kuzmenko N.E., Eremin V.V., Popkov V.A.
Unified State Exam assignments for chemical equilibrium(formerly A21)
Task No. 1.
H2S(g) ↔ H2(g) + S(g) - Q
1. Increased pressure
2. Rising temperature
3. Decreased pressure
Explanation: First, let's consider the reaction: all substances are gases and on the right side there are two molecules of products, and on the left there is only one, the reaction is also endothermic (-Q). Therefore, let us consider the change in pressure and temperature. We need the equilibrium to shift towards the reaction products. If we increase the pressure, then the equilibrium will shift towards decreasing volume, that is, towards the reactants - this does not suit us. If we increase the temperature, then the equilibrium will shift towards the endothermic reaction, in our case towards the products, which is what was required. The correct answer is 2.
Task No. 2.
Chemical equilibrium in the system
SO3(g) + NO(g) ↔ SO2(g) + NO2(g) - Q
will shift towards the formation of reagents when:
1. Increasing NO concentration
2. Increasing SO2 concentration
3. Temperature rises
4. Increased pressure
Explanation: all substances are gases, but the volumes on the right and left sides of the equation are the same, so pressure will not affect the equilibrium in the system. Consider a change in temperature: as the temperature increases, the equilibrium shifts towards the endothermic reaction, precisely towards the reactants. The correct answer is 3.
Task No. 3.
In system
2NO2(g) ↔ N2O4(g) + Q
a shift of balance to the left will contribute
1. Increase in pressure
2. Increase in N2O4 concentration
3. Temperature drop
4. Introduction of catalyst
Explanation: Let us pay attention to the fact that the volumes of gaseous substances on the right and left sides of the equation are not equal, therefore a change in pressure will affect the equilibrium in this system. Namely, with increasing pressure, the equilibrium shifts towards a decrease in the amount of gaseous substances, that is, to the right. This doesn't suit us. The reaction is exothermic, therefore a change in temperature will affect the equilibrium of the system. As the temperature decreases, the equilibrium will shift towards the exothermic reaction, that is, also to the right. As the concentration of N2O4 increases, the equilibrium shifts towards the consumption of this substance, that is, to the left. The correct answer is 2.
Task No. 4.
In reaction
2Fe(s) + 3H2O(g) ↔ 2Fe2O3(s) + 3H2(g) - Q
the equilibrium will shift towards the reaction products when
1. Increased pressure
2. Adding a catalyst
3. Adding iron
4. Adding water
Explanation: the number of molecules in the right and left parts is the same, so a change in pressure will not affect the equilibrium in this system. Let's consider an increase in the concentration of iron - the equilibrium should shift towards the consumption of this substance, that is, to the right (towards the reaction products). The correct answer is 3.
Task No. 5.
Chemical equilibrium
H2O(l) + C(t) ↔ H2(g) + CO(g) - Q
will shift towards the formation of products in the case
1. Increased pressure
2. Increase in temperature
3. Increasing the process time
4. Catalyst Applications
Explanation: a change in pressure will not affect the equilibrium in a given system, since not all substances are gaseous. As the temperature increases, the equilibrium shifts towards the endothermic reaction, that is, to the right (towards the formation of products). The correct answer is 2.
Task No. 6.
As the pressure increases, the chemical equilibrium will shift towards the products in the system:
1. CH4(g) + 3S(s) ↔ CS2(g) + 2H2S(g) - Q
2. C(t) + CO2(g) ↔ 2CO(g) - Q
3. N2(g) + 3H2(g) ↔ 2NH3(g) + Q
4. Ca(HCO3)2(t) ↔ CaCO3(t) + CO2(g) + H2O(g) - Q
Explanation: reactions 1 and 4 are not affected by changes in pressure, because not all participating substances are gaseous; in equation 2, the number of molecules on the right and left sides is the same, so pressure will not affect. Equation 3 remains. Let's check: with increasing pressure, the equilibrium should shift towards decreasing amounts of gaseous substances (4 molecules on the right, 2 molecules on the left), that is, towards the reaction products. The correct answer is 3.
Task No. 7.
Does not affect balance shift
H2(g) + I2(g) ↔ 2HI(g) - Q
1. Increasing pressure and adding catalyst
2. Raising the temperature and adding hydrogen
3. Lowering the temperature and adding hydrogen iodide
4. Adding iodine and adding hydrogen
Explanation: in the right and left parts the amounts of gaseous substances are the same, so a change in pressure will not affect the equilibrium in the system, and adding a catalyst will also not affect it, because as soon as we add a catalyst, the direct reaction will accelerate, and then immediately the reverse and equilibrium in the system will be restored . The correct answer is 1.
Task No. 8.
To shift the equilibrium in a reaction to the right
2NO(g) + O2(g) ↔ 2NO2(g); ΔH°<0
required
1. Introduction of catalyst
2. Lowering the temperature
3. Lower pressure
4. Reduced oxygen concentration
Explanation: a decrease in oxygen concentration will lead to a shift in equilibrium towards the reactants (to the left). A decrease in pressure will shift the equilibrium towards a decrease in the amount of gaseous substances, that is, to the right. The correct answer is 3.
Task No. 9.
Product yield in an exothermic reaction
2NO(g) + O2(g) ↔ 2NO2(g)
with a simultaneous increase in temperature and decrease in pressure
1. Increase
2. Will decrease
3. Will not change
4. First it will increase, then it will decrease
Explanation: when the temperature increases, the equilibrium shifts towards the endothermic reaction, that is, towards the products, and when the pressure decreases, the equilibrium shifts towards an increase in the amounts of gaseous substances, that is, also to the left. Therefore, the product yield will decrease. The correct answer is 2.
Task No. 10.
Increasing the yield of methanol in the reaction
CO + 2H2 ↔ CH3OH + Q
promotes
1. Increase in temperature
2. Introduction of catalyst
3. Introduction of inhibitor
4. Increased pressure
Explanation: with increasing pressure, the equilibrium shifts towards the endothermic reaction, that is, towards the reactants. An increase in pressure shifts the equilibrium towards decreasing amounts of gaseous substances, that is, towards the formation of methanol. The correct answer is 4.
Tasks for independent solution (answers below)
1. In the system
CO(g) + H2O(g) ↔ CO2(g) + H2(g) + Q
a shift in chemical equilibrium towards reaction products will be facilitated by
1. Reducing pressure
2. Increase in temperature
3. Increase in carbon monoxide concentration
4. Increase in hydrogen concentration
2. In which system, when pressure increases, does the equilibrium shift towards the reaction products?
1. 2СО2(g) ↔ 2СО2(g) + O2(g)
2. C2H4(g) ↔ C2H2(g) + H2(g)
3. PCl3(g) + Cl2(g) ↔ PCl5(g)
4. H2(g) + Cl2(g) ↔ 2HCl(g)
3. Chemical equilibrium in the system
2HBr(g) ↔ H2(g) + Br2(g) - Q
will shift towards the reaction products when
1. Increased pressure
2. Rising temperature
3. Decreased pressure
4. Using a catalyst
4. Chemical equilibrium in the system
C2H5OH + CH3COOH ↔ CH3COOC2H5 + H2O + Q
shifts towards the reaction products when
1. Adding water
2. Reducing the concentration of acetic acid
3. Increasing ether concentration
4. When removing ester
5. Chemical equilibrium in the system
2NO(g) + O2(g) ↔ 2NO2(g) + Q
shifts towards the formation of the reaction product when
1. Increased pressure
2. Rising temperature
3. Decreased pressure
4. Application of catalyst
6. Chemical equilibrium in the system
CO2(g) + C(s) ↔ 2СО(g) - Q
will shift towards the reaction products when
1. Increased pressure
2. Lowering the temperature
3. Increasing CO concentration
4. Temperature rises
7. Changes in pressure will not affect the state of chemical equilibrium in the system
1. 2NO(g) + O2(g) ↔ 2NO2(g)
2. N2(g) + 3H2(g) ↔ 2NH3(g)
3. 2CO(g) + O2(g) ↔ 2CO2(g)
4. N2(g) + O2(g) ↔ 2NO(g)
8. In which system, with increasing pressure, will the chemical equilibrium shift towards the starting substances?
1. N2(g) + 3H2(g) ↔ 2NH3(g) + Q
2. N2O4(g) ↔ 2NO2(g) - Q
3. CO2(g) + H2(g) ↔ CO(g) + H2O(g) - Q
4. 4HCl(g) + O2(g) ↔ 2H2O(g) + 2Cl2(g) + Q
9. Chemical equilibrium in the system
С4Н10(g) ↔ С4Н6(g) + 2Н2(g) - Q
will shift towards the reaction products when
1. Increase in temperature
2. Lowering the temperature
3. Using a catalyst
4. Reducing butane concentration
10. On the state of chemical equilibrium in the system
H2(g) + I2(g) ↔ 2HI(g) -Q
does not affect
1. Increase in pressure
2. Increasing iodine concentration
3. Increase in temperature
4. Reduce temperature
2016 assignments
1. Establish a correspondence between the equation of a chemical reaction and the shift in chemical equilibrium with increasing pressure in the system.
Reaction equation Shift of chemical equilibrium
A) N2(g) + O2(g) ↔ 2NO(g) - Q 1. Shifts towards the direct reaction
B) N2O4(g) ↔ 2NO2(g) - Q 2. Shifts towards the reverse reaction
B) CaCO3(s) ↔ CaO(s) + CO2(g) - Q 3. There is no shift in equilibrium
D) Fe3O4(s) + 4CO(g) ↔ 3Fe(s) + 4CO2(g) + Q
2. Establish a correspondence between external influences on the system:
CO2(g) + C(s) ↔ 2СО(g) - Q
and a shift in chemical equilibrium.
A. Increase in CO concentration 1. Shifts towards the direct reaction
B. Decrease in pressure 3. No shift in equilibrium occurs
3. Establish a correspondence between external influences on the system
HCOOH(l) + C5H5OH(l) ↔ HCOOC2H5(l) + H2O(l) + Q
External influence Shift in chemical equilibrium
A. Addition of HCOOH 1. Shifts towards the direct reaction
B. Dilution with water 3. No shift in equilibrium occurs
D. Increase in temperature
4. Establish a correspondence between external influences on the system
2NO(g) + O2(g) ↔ 2NO2(g) + Q
and a shift in chemical equilibrium.
External influence Shift in chemical equilibrium
A. Decrease in pressure 1. Shifts towards the forward reaction
B. Increase in temperature 2. Shifts towards the reverse reaction
B. Increase in NO2 temperature 3. No equilibrium shift occurs
D. Addition of O2
5. Establish a correspondence between external influences on the system
4NH3(g) + 3O2(g) ↔ 2N2(g) + 6H2O(g) + Q
and a shift in chemical equilibrium.
External influence Shift in chemical equilibrium
A. Decrease in temperature 1. Shift towards direct reaction
B. Increase in pressure 2. Shifts towards the reverse reaction
B. Increase in concentration in ammonia 3. There is no shift in equilibrium
D. Removal of water vapor
6. Establish a correspondence between external influences on the system
WO3(s) + 3H2(g) ↔ W(s) + 3H2O(g) +Q
and a shift in chemical equilibrium.
External influence Shift in chemical equilibrium
A. Increase in temperature 1. Shifts towards a direct reaction
B. Increase in pressure 2. Shifts towards the reverse reaction
B. Use of a catalyst 3. There is no shift in equilibrium
D. Removal of water vapor
7. Establish a correspondence between external influences on the system
С4Н8(g) + Н2(g) ↔ С4Н10(g) + Q
and a shift in chemical equilibrium.
External influence Shift in chemical equilibrium
A. Increase in hydrogen concentration 1. Shifts towards a direct reaction
B. Increase in temperature 2. Shifts towards the reverse reaction
B. Increase in pressure 3. No shift in equilibrium occurs
D. Use of a catalyst
8. Establish a correspondence between the equation of a chemical reaction and a simultaneous change in the parameters of the system, leading to a shift in the chemical equilibrium towards a direct reaction.
Reaction equation Changing system parameters
A. H2(g) + F2(g) ↔ 2HF(g) + Q 1. Increase in temperature and hydrogen concentration
B. H2(g) + I2(s) ↔ 2HI(g) -Q 2. Decrease in temperature and hydrogen concentration
B. CO(g) + H2O(g) ↔ CO2(g) + H2(g) + Q 3. Increasing temperature and decreasing hydrogen concentration
D. C4H10(g) ↔ C4H6(g) + 2H2(g) -Q 4. Decrease in temperature and increase in hydrogen concentration
9. Establish a correspondence between the equation of a chemical reaction and the shift in chemical equilibrium with increasing pressure in the system.
Reaction equation Direction of chemical equilibrium shift
A. 2HI(g) ↔ H2(g) + I2(s) 1. Shifts towards the direct reaction
B. C(g) + 2S(g) ↔ CS2(g) 2. Shifts towards the reverse reaction
B. C3H6(g) + H2(g) ↔ C3H8(g) 3. There is no shift in equilibrium
G. H2(g) + F2(g) ↔ 2HF(g)
10. Establish a correspondence between the equation of a chemical reaction and a simultaneous change in the conditions for its implementation, leading to a shift in the chemical equilibrium towards a direct reaction.
Reaction equation Changing conditions
A. N2(g) + H2(g) ↔ 2NH3(g) + Q 1. Increase in temperature and pressure
B. N2O4(l) ↔ 2NO2(g) -Q 2. Decrease in temperature and pressure
B. CO2(g) + C(s) ↔ 2CO(g) + Q 3. Increase in temperature and decrease in pressure
D. 4HCl(g) + O2(g) ↔ 2H2O(g) + 2Cl2(g) + Q 4. Decrease in temperature and increase in pressure
Answers: 1 - 3, 2 - 3, 3 - 2, 4 - 4, 5 - 1, 6 - 4, 7 - 4, 8 - 2, 9 - 1, 10 - 1
1. 3223
2. 2111
3. 1322
4. 2221
5. 1211
6. 2312
7. 1211
8. 4133
9. 1113
10. 4322
For the assignments, we thank the collections of exercises for 2016, 2015, 2014, 2013, authors:
Kavernina A.A., Dobrotina D.Yu., Snastina M.G., Savinkina E.V., Zhiveinova O.G.
Chemical equilibrium is maintained as long as the conditions in which the system is located remain unchanged. Changing conditions (concentration of substances, temperature, pressure) causes an imbalance. After some time, the chemical equilibrium is restored, but under new, different from previous conditions. Such a transition of a system from one equilibrium state to another is called displacement(shift) of equilibrium. The direction of displacement obeys Le Chatelier's principle.
As the concentration of one of the starting substances increases, the equilibrium shifts towards greater consumption of this substance, and the direct reaction intensifies. A decrease in the concentration of the starting substances shifts the equilibrium towards the formation of these substances, as the reverse reaction intensifies. An increase in temperature shifts the equilibrium towards an endothermic reaction, while a decrease in temperature shifts the equilibrium towards an exothermic reaction. An increase in pressure shifts the equilibrium towards decreasing amounts of gaseous substances, that is, towards smaller volumes occupied by these gases. On the contrary, as pressure decreases, the equilibrium shifts towards increasing amounts of gaseous substances, that is, towards larger volumes formed by gases.
Example 1.
How will an increase in pressure affect the equilibrium state of the following reversible gas reactions:
a) SO 2 + C1 2 =SO 2 CI 2;
b) H 2 + Br 2 = 2НВr.
Solution:
We use Le Chatelier's principle, according to which an increase in pressure in the first case (a) shifts the equilibrium to the right, towards a smaller amount of gaseous substances occupying a smaller volume, which weakens the external influence of the increased pressure. In the second reaction (b), the quantities of gaseous substances, both the starting materials and the reaction products, are equal, as are the volumes they occupy, so pressure has no effect and the equilibrium is not disturbed.
Example 2.
In the reaction of ammonia synthesis (–Q) 3H 2 + N 2 = 2NH 3 + Q, the forward reaction is exothermic, the reverse reaction is endothermic. How should the concentration of reactants, temperature and pressure be changed to increase the yield of ammonia?
Solution:
To shift the balance to the right you need to:
a) increase the concentrations of H 2 and N 2;
b) reduce the concentration (removal from the reaction sphere) of NH 3;
c) lower the temperature;
d) increase the pressure.
Example 3.
The homogeneous reaction between hydrogen chloride and oxygen is reversible:
4HC1 + O 2 = 2C1 2 + 2H 2 O + 116 kJ.
1. What effect will the following have on the equilibrium of the system?
a) increase in pressure;
b) increase in temperature;
c) introduction of a catalyst?
Solution:
a) In accordance with Le Chatelier's principle, an increase in pressure leads to a shift in equilibrium towards the direct reaction.
b) An increase in t° leads to a shift in equilibrium towards the reverse reaction.
c) The introduction of a catalyst does not shift the equilibrium.
2. In what direction will the chemical equilibrium shift if the concentration of reactants is doubled?
Solution:
υ → = k → 0 2 0 2 ; υ 0 ← = k ← 0 2 0 2
After increasing concentrations, the rate of the forward reaction became:
υ → = k → 4 = 32 k → 0 4 0
that is, it increased 32 times compared to the initial speed. Similarly, the rate of the reverse reaction increases 16 times:
υ ← = k ← 2 2 = 16k ← [H 2 O] 0 2 [C1 2 ] 0 2 .
The increase in the rate of the forward reaction is 2 times greater than the increase in the rate of the reverse reaction: the equilibrium shifts to the right.
Example 4.
IN which direction will the equilibrium of a homogeneous reaction shift:
PCl 5 = PC1 3 + Cl 2 + 92 KJ,
if you increase the temperature by 30 °C, knowing that the temperature coefficient of the forward reaction is 2.5, and the reverse reaction is 3.2?
Solution:
Since the temperature coefficients of the forward and reverse reactions are not equal, increasing the temperature will have different effects on the change in the rates of these reactions. Using Van't Hoff's rule (1.3), we find the rates of forward and reverse reactions when the temperature increases by 30 °C:
υ → (t 2) = υ → (t 1)=υ → (t 1)2.5 0.1 30 = 15.6υ → (t 1);
υ ← (t 2) = υ ← (t 1) =υ → (t 1)3.2 0.1 30 = 32.8υ ← (t 1)
An increase in temperature increased the rate of the forward reaction by 15.6 times, and the reverse reaction by 32.8 times. Consequently, the equilibrium will shift to the left, towards the formation of PCl 5.
Example 5.
How will the rates of forward and reverse reactions change in the isolated system C 2 H 4 + H 2 ⇄ C 2 H 6 and where will the equilibrium shift when the volume of the system increases by 3 times?
Solution:
The initial rates of forward and reverse reactions are as follows:
υ 0 = k 0 0 ; υ 0 = k 0 .
An increase in the volume of the system causes a decrease in the concentrations of reactants by 3 times, hence the change in the rate of forward and reverse reactions will be as follows:
υ 0 = k = 1/9υ 0
υ = k = 1/3υ 0
The decrease in the rates of forward and reverse reactions is not the same: the rate of the reverse reaction is 3 times (1/3: 1/9 = 3) higher than the rate of the reverse reaction, therefore the equilibrium will shift to the left, to the side where the system occupies a larger volume, that is, towards the formation of C 2 H 4 and H 2.
Codifier Topics: reversible and irreversible reactions. Chemical balance. Shift in chemical equilibrium under the influence of various factors.
If a reverse reaction is possible, chemical reactions are divided into reversible and irreversible.
Reversible chemical reactions are reactions whose products under given conditions can interact with each other.
Irreversible reactions are reactions whose products cannot interact with each other under given conditions.
More details about classification of chemical reactions can be read.
The likelihood of product interaction depends on the process conditions.
So, if the system open, i.e. exchanges both matter and energy with the environment, then chemical reactions in which, for example, gases are formed, will be irreversible. For example , when calcining solid sodium bicarbonate:
2NaHCO 3 → Na 2 CO 3 + CO 2 + H 2 O
Carbon dioxide gas will be released and evaporate from the reaction zone. Therefore, this reaction will be irreversible under these conditions. If we consider closed system , which can not exchange a substance with the environment (for example, a closed box in which the reaction occurs), then carbon dioxide will not be able to escape from the reaction zone, and will interact with water and sodium carbonate, then the reaction will be reversible under these conditions:
2NaHCO 3 ⇔ Na 2 CO 3 + CO 2 + H 2 O
Let's consider reversible reactions. Let the reversible reaction proceed according to the scheme:
aA + bB = cC + dD
The speed of the forward reaction according to the law of mass action is determined by the expression: v 1 =k 1 ·C A a ·C B b, the speed of the reverse reaction: v 2 =k 2 ·C С с ·C D d. If at the initial moment of the reaction there are no substances C and D in the system, then particles A and B mainly collide and interact, and a predominantly direct reaction occurs. Gradually, the concentration of particles C and D will also begin to increase, therefore, the rate of the reverse reaction will increase. At some point the rate of the forward reaction will be equal to the rate of the reverse reaction. This state is called chemical equilibrium .
Thus, chemical equilibrium is a state of the system in which the rates of forward and reverse reactions are equal .
Because the rates of forward and reverse reactions are equal, the rate of formation of substances is equal to the rate of their consumption, and the current concentrations of substances do not change . Such concentrations are called equilibrium .
Please note that at equilibrium there are both direct and reverse reactions, that is, the reactants interact with each other, but the products also interact at the same speed. At the same time, external factors can influence displace chemical equilibrium in one direction or another. Therefore, chemical equilibrium is called mobile or dynamic.
Research in the field of mobile equilibrium began in the 19th century. The works of Henri Le Chatelier laid the foundations of the theory, which was later generalized by the scientist Karl Brown. The principle of mobile equilibrium, or the Le Chatelier-Brown principle, states:
If a system in a state of equilibrium is influenced by an external factor that changes any of the equilibrium conditions, then processes in the system aimed at compensating for the external influence are intensified.
In other words: when there is an external influence on the system, the equilibrium will shift so as to compensate for this external influence.
This principle, which is very important, works for any equilibrium phenomena (not just chemical reactions). However, we will now consider it in relation to chemical interactions. In the case of chemical reactions, external influences lead to changes in the equilibrium concentrations of substances.
Chemical reactions at equilibrium can be affected by three main factors—temperature, pressure, and concentrations of reactants or products.
1. As is known, chemical reactions are accompanied by a thermal effect. If the direct reaction occurs with the release of heat (exothermic, or +Q), then the reverse reaction occurs with the absorption of heat (endothermic, or -Q), and vice versa. If you raise temperature in the system, the equilibrium will shift so as to compensate for this increase. It is logical that in an exothermic reaction the temperature increase cannot be compensated. Thus, as the temperature increases, the equilibrium in the system shifts towards heat absorption, i.e. towards endothermic reactions (-Q); with decreasing temperature - towards an exothermic reaction (+Q).
2. In the case of equilibrium reactions, when at least one of the substances is in the gas phase, the equilibrium is also significantly affected by a change pressure in system. As pressure increases, the chemical system tries to compensate for this effect and increases the rate of reaction, in which the amount of gaseous substances decreases. As the pressure decreases, the system increases the rate of reaction, which produces more molecules of gaseous substances. Thus: with an increase in pressure, the equilibrium shifts towards a decrease in the number of gas molecules, and with a decrease in pressure - towards an increase in the number of gas molecules.
Note! Systems where the number of molecules of reactant gases and products are the same are not affected by pressure! Also, changes in pressure have virtually no effect on the equilibrium in solutions, i.e. on reactions where there are no gases.
3. Also, equilibrium in chemical systems is affected by changes concentrations reactants and products. As the concentration of reactants increases, the system tries to use them up and increases the rate of the forward reaction. As the concentration of reagents decreases, the system tries to produce them, and the rate of the reverse reaction increases. As the concentration of products increases, the system also tries to consume them and increases the rate of the reverse reaction. When the concentration of products decreases, the chemical system increases the rate of their formation, i.e. rate of forward reaction.
If in a chemical system the rate of forward reaction increases right , towards the formation of products And reagent consumption . If the rate of reverse reaction increases, we say that the balance has shifted left , towards food consumption And increasing the concentration of reagents .
For example, in the ammonia synthesis reaction:
N 2 + 3H 2 = 2NH 3 + Q
An increase in pressure leads to an increase in the rate of reaction, in which fewer gas molecules are formed, i.e. direct reaction (the number of molecules of reactant gases is 4, the number of gas molecules in products is 2). As pressure increases, the equilibrium shifts to the right, towards the products. At temperature rise the balance will shift in the opposite direction of the endothermic reaction, i.e. to the left, towards the reagents. An increase in the concentration of nitrogen or hydrogen will shift the equilibrium towards their consumption, i.e. to the right, towards the products.
Catalyst does not affect balance, because accelerates both forward and reverse reactions.
The equilibrium state for a reversible reaction can last indefinitely (without outside intervention). But if an external influence is exerted on such a system (change the temperature, pressure or concentration of final or initial substances), then the state of equilibrium will be disrupted. The speed of one of the reactions will become greater than the speed of the other. Over time, the system will again occupy an equilibrium state, but the new equilibrium concentrations of the initial and final substances will differ from the original ones. In this case, they talk about a shift in 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, 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 starting substances decrease and the equilibrium concentrations of the final substances increase compared to the initial equilibrium concentrations. Accordingly, the yield of reaction products also increases.
A shift of chemical equilibrium to the left causes an increase in the equilibrium concentrations of the starting substances and a decrease in the equilibrium concentrations of the final products, the yield of which will decrease.
The direction of the shift in chemical equilibrium is determined using Le Chatelier’s principle: “If an external influence is exerted on a system in a state of chemical equilibrium (change temperature, pressure, concentration of one or more substances participating in the reaction), this will lead to an increase in the rate of that reaction, the occurrence of which will compensate (reduce) the impact."
For example, as the concentration of starting substances increases, the rate of the forward reaction increases and the equilibrium shifts to the right. When the concentration of the starting substances decreases, on the contrary, the rate of the reverse reaction increases, and the chemical equilibrium shifts to the left.
When the temperature increases (i.e. when the system is heated), the equilibrium shifts towards the endothermic reaction, and when it decreases (i.e. when the system cools) - towards the 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 forward and reverse reactions, but the rate of an endothermic reaction increases to a greater extent than the rate of an 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 significantly less than for an endothermic one.
A change in pressure affects a 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 (s) CaO (s) + CO 2 (g) - a change in pressure affects the displacement of the equilibrium.
CH 3 COOH (liquid) + C 2 H 5 OH (liquid) CH 3 COOC 2 H 5 (liquid) + H 2 O (liquid) – a change in pressure does not affect the shift in chemical equilibrium, because none of the starting or final 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 changes affect the equilibrium shift
I 2(g) + H 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 equilibrium towards a reaction, the occurrence 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 towards the reaction that occurs with the formation of a larger number of gas molecules. In our example, this will be the opposite reaction.
An increase in pressure causes a decrease in the volume of the system, and therefore 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. A decrease in pressure according to a similar scheme 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 to the same extent. In its presence, chemical equilibrium is only established faster (or slower).
If a system is affected by several factors simultaneously, 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 by heating and in the presence of a catalyst to increase its speed. But the effect of temperature leads to the fact that the equilibrium of the reaction shifts to the left, towards the reverse endothermic reaction. This causes a decrease in the output of NH 3. To compensate for this undesirable effect of temperature and increase the yield of ammonia, the pressure in the system is simultaneously increased, which shifts the equilibrium of the reaction to the right, i.e. towards the formation of fewer gas molecules.
In this case, the most optimal conditions for the reaction (temperature, pressure) are selected experimentally, at which it would proceed at a sufficiently high speed 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 adult human body is characterized by the relative constancy of many parameters, including various biochemical indicators, including the concentrations of biologically active substances. However, such a state cannot be called equilibrium, because it is not applicable to open systems.
The human body, like any living system, constantly exchanges various substances with the environment: it consumes food and releases products of their oxidation and decay. Therefore, it is typical for an organism steady state, defined as the constancy of its parameters at a constant rate of exchange of matter and energy with the environment. To a first approximation, a 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 due to the replenishment of the initial products from the outside and the removal of the final products to the outside. A change in their content in the body does not lead, unlike 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.
During the life of an organism in a stationary state, in contrast to a closed equilibrium system, an increase in entropy occurs. However, along with this, the reverse process also occurs simultaneously - 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 decomposition products into the environment. According to the position of I.R. Prigogine, the total production of entropy for an organism in a stationary state tends to a minimum.
A major contribution to the development of nonequilibrium thermodynamics was made by I. R. Prigozhy, Nobel Prize winner in 1977, who argued that “in any nonequilibrium system there are local areas that are in an equilibrium state. In classical thermodynamics, equilibrium refers to the entire system, but in nonequilibrium, only to its individual parts.”
It has been established that entropy in such systems increases during embryogenesis, during regeneration processes and the growth of malignant neoplasms.
Concept of chemical equilibrium
An equilibrium state is considered to be a state of a system that remains unchanged, and this state is not caused by the action of any external forces. The state of a system of reacting substances in which the rate of the forward reaction becomes equal to the rate of the reverse reaction is called chemical equilibrium. This equilibrium is also called mobile m or dynamic balance.
Signs of chemical balance
1. The state of the system remains unchanged over time while maintaining external conditions.
2. Equilibrium is dynamic, that is, it is caused by the occurrence of forward and reverse reactions at the same rates.
3. Any external influence causes a change in the equilibrium of the system; if the external influence is removed, the system returns to its original state.
4. The state of equilibrium can be approached from two sides - both from the side of the starting substances and from the side of the reaction products.
5. In a state of equilibrium, the Gibbs energy reaches its minimum value.
Le Chatelier's principle
The influence of changes in external conditions on the equilibrium position is determined Le Chatelier's principle (principle of moving equilibrium): If any external influence is applied to a system in a state of equilibrium, then in the system that direction of the process that weakens the effect of this influence will be strengthened, and the equilibrium position will shift in the same direction.
Le Chatelier's principle applies not only to chemical processes, but also to physical ones, such as boiling, crystallization, dissolution, etc.
Let us consider the influence of various factors on chemical equilibrium using the example of the NO oxidation reaction:
2 NO (g) + O 2(g) 2 NO 2(g) ; H o 298 = - 113.4 kJ/mol.
Effect of temperature on chemical equilibrium
As the temperature increases, the equilibrium shifts towards the endothermic reaction, and as the temperature decreases, towards the exothermic reaction.
The degree of equilibrium shift is determined by the absolute value of the thermal effect: the greater the absolute value of the enthalpy of the reaction H, the greater the influence of temperature on the equilibrium state.
In the reaction under consideration for the synthesis of nitric oxide (IV ) an increase in temperature will shift the equilibrium towards the starting substances.
Effect of pressure on chemical equilibrium
Compression shifts the equilibrium in the direction of a process that is accompanied by a decrease in the volume of gaseous substances, and a decrease in pressure shifts the equilibrium in the opposite direction. In the example under consideration, there are three volumes on the left side of the equation, and two on the right. Since an increase in pressure favors the process that occurs with a decrease in volume, then with an increase in pressure the equilibrium will shift to the right, i.e. towards the reaction product – NO 2 . Reducing the pressure will shift the equilibrium in the opposite direction. It should be noted that if in the equation of a reversible reaction the number of molecules of gaseous substances in the right and left sides are equal, then a change in pressure does not affect the equilibrium position.
Effect of concentration on chemical equilibrium
For the reaction under consideration, the introduction of additional amounts of NO or O 2 into the equilibrium system causes a shift in equilibrium in the direction in which the concentration of these substances decreases, therefore, there is a shift in equilibrium towards the formation NO 2 . Increased concentration NO 2 shifts the equilibrium towards the starting substances.
The catalyst equally accelerates both forward and reverse reactions and therefore does not affect the shift in chemical equilibrium.
When introduced into an equilibrium system (at P = const ) of inert gas, the concentrations of reagents (partial pressures) decrease. Since the oxidation process under consideration NO goes with a decrease in volume, then when adding in
Chemical equilibrium constant
For a chemical reaction:
2 NO (g) + O 2 (g) 2 NO 2(g)
chemical reaction constant K c is the ratio:
(12.1)
In this equation, in square brackets are the concentrations of reacting substances that are established at chemical equilibrium, i.e. equilibrium concentrations of substances.
The chemical equilibrium constant is related to the change in Gibbs energy by the equation:
G T o = – RTlnK . (12.2).
Examples of problem solving
At a certain temperature, the equilibrium concentrations in the system 2CO (g) + O 2(g)2CO 2 (g) were: = 0.2 mol/l, = 0.32 mol/l, = 0.16 mol/l. Determine the equilibrium constant at this temperature and the initial concentrations of CO and O 2 , if the original mixture did not contain CO 2 .
.
2CO (g) + O 2(g) 2CO 2(d).
In the second line, “proreact” refers to the concentration of the reacted starting substances and the concentration of the resulting CO 2 , and, with initial = with react + with equal .
Using reference data, calculate the equilibrium constant of the process3 H 2 (G) + N 2 (G) 2 NH 3 (G) at 298 K.
G 298 o = 2·( - 16.71) kJ = -33.42 10 3 J.
G T o = - RTlnK.
lnK = 33.42 10 3 /(8.314 × 298) = 13.489. K = 7.21× 10 5 .
Determine the equilibrium concentration of HI in the systemH 2(g) + I 2(g) 2HI (G) ,
if at a certain temperature the equilibrium constant is 4, and the initial concentrations of H 2, I 2 and HI are equal to 1, 2 and 0 mol/l, respectively.
Solution. Let x mol/l H2 react at some point in time.
.
Solving this equation, we get x = 0.67.
This means that the equilibrium concentration of HI is 2 × 0.67 = 1.34 mol/L.
Using reference data, determine the temperature at which the equilibrium constant of the process is: H 2 (g) + HCOH (d)CH3OH (d) becomes equal to 1. Assume that H o T » H o 298 and S o T "S o 298.If K = 1, then G o T = - RTlnK = 0;
G o T » N about 298 - T D S o 298 . Then ;
N about 298 = -202 – (- 115.9) = -86.1 kJ = - 86.1× 10 3 J;
S o 298 = 239.7 – 218.7 – 130.52 = -109.52 J/K;
TO.
Solution. Let x mol/l SO 2 react at some point in time.
SO 2(G) + Cl 2(G) SO 2 Cl 2(G)
Then we get:
.
Solving this equation, we find: x 1 = 3 and x 2 = 1.25. But x 1 = 3 does not satisfy the conditions of the problem.
Therefore, = 1.25 + 1 = 2.25 mol/l.
Problems to solve independently
12.1. In which of the following reactions will an increase in pressure shift the equilibrium to the right? Justify the answer.
1) 2 NH 3 (g) 3 H 2 (g) + N 2 (d)
2) ZnCO 3 (k) ZnO (k) + CO 2 (d)
3) 2HBr (g) H 2 (g) + Br 2 (w)
4) CO 2 (g) + C (graphite) 2CO (g)
12.2.At a certain temperature, the equilibrium concentrations in the system
2HBr (g) H 2 (g) + Br 2 (d)
were: = 0.3 mol/l, = 0.6 mol/l, = 0.6 mol/l. Determine the equilibrium constant and the initial concentration of HBr.
12.3.For the reaction H 2(g)+S (d) H 2 S (d) at a certain temperature the equilibrium constant is 2. Determine the equilibrium concentrations of H 2 and S, if the initial concentrations of H 2, S and H 2 S are equal to 2, 3 and 0 mol/l, respectively.