Task 127.
How will the rate of a reaction occurring in the gas phase change with an increase in temperature by 60°C, if the temperature coefficient of the rate of this reaction is 2?
Solution:
Consequently, the reaction rate with an increase in temperature by 600 C 0 is 64 times greater than the initial reaction rate.
Task 121.
The oxidation of sulfur and its dioxide proceeds according to the equations:
a) S (k) + O 2 = SO 2 (d); b) 2SO 2 (d) + O 2 = 2SO 3 (d).
How will the rate of these reactions change if the volumes of each system are reduced by four times?
Solution:
a) S (k) + O 2 = SO 2 (d)
Let us denote the concentrations of gaseous reactants: = a, = b. According to law active masses
, the rates of forward and reverse reactions before the volume change are respectively equal:
V pr = k. a; V arr = k. b.
After reducing the volume of the heterogeneous system by four times, the concentration gaseous substances will increase fourfold: = 4a, = 4b. At new concentrations, the rates of forward and reverse reactions will be equal
Consequently, after reducing the volume in the system, the rates of forward and reverse reactions increased fourfold. The equilibrium of the system has not shifted.
b) 2SO 2 (g) + O 2 = 2SO 3 (g)
Let us denote the concentrations of reactants: = a, = b, = With. According to the law of mass action, the rates of forward and reverse reactions before the change in volume are respectively equal:
V pr = ka 2 b; Vo b r = kc 2 .
After reducing the volume of a homogeneous system by four times, the concentration of reactants will increase four times: = 4 a, = 4b, = 4 s At new concentrations, the rates of forward and reverse reactions will be equal:
Consequently, after reducing the volume in the system, the rate of the forward reaction increased by 64 times, and the reverse reaction by 16. The equilibrium of the system shifted to the right, towards a decrease in the formation of gaseous substances.
Equilibrium constants of a homogeneous system
Task 122.
Write an expression for the equilibrium constant of a homogeneous system:
N 2 + ZN 2 = 2NH 3. How will the rate of the direct reaction of ammonia formation change if the hydrogen concentration is increased threefold?
Solution:
Reaction equation:
N 2 + ZN 2 = 2NH 3
The expression for the equilibrium constant of this reaction has the form:
Let us denote the concentrations of gaseous reactants: = a, = b. According to the law of mass action, the rate of direct reactions before increasing the hydrogen concentration is equal to: V pr = kab 3. After increasing the hydrogen concentration to three times the concentration starting materials will be equal: = a, = 3b. At new concentrations, the rate of direct reactions will be equal to:
Consequently, after increasing the hydrogen concentration three times, the reaction rate increased 27 times. Equilibrium, according to Le Chatelier’s principle, shifted towards decreasing hydrogen concentration, i.e. to the right.
Z assignment 123.
The reaction follows the equation N 2 + O 2 = 2NO. The concentrations of the starting substances before the start of the reaction were = 0.049 mol/L, = 0.01 mol/L. Calculate the concentration of these substances when = 0.005 mol/l. Answer: 0.0465 mol/l; = 0.0075 mol/l.
Solution:
The reaction equation is:
From the reaction equation it follows that the formation of 2 moles of NO requires 1 mole of N2 and O2, i.e., the formation of NO requires half as much N2 and O2. Based on the above, it can be assumed that the formation of 0.005 mol of NO requires 0.0025 mol of N 2 and O 2. Then the final concentrations of the starting substances will be equal to:
Final = ref. – 0.0025 = 0.049 – 0.0025 = 0.0465 mol/l;
finite = ref. - 0.0025 = 0.01 – 0.0025 = 0.0075 mol/l.
Answer: finite = 0.0465 mol/l; finite = 0.0075 mol/l.
Task 124.
The reaction proceeds according to the equation N 2 + ZH 2 = 2NH 3. Concentrations of substances involved (mol/l): = 0.80; = 1.5; = 0.10. Calculate the concentration of hydrogen and ammonia = 0.5 mol/l. Answer: = 0.70 mol/l; [H 2) = 0.60 mol/l.
Solution:
The reaction equation is:
N2 + ZH2 = 2NH3
From the equation it follows that from 1 mol N 2 2 mol NH 3 are formed and 3 mol H 2 are consumed. Thus, with the participation of a certain amount of nitrogen in the reaction, twice the amount of nitrogen is formed large quantity ammonia and will react with three times more hydrogen. Let's calculate the amount of nitrogen that reacted: 0.80 – 0.50 = 0.30 mol. Let's calculate the amount of ammonia that was formed: 0.3 . 2 = 0.6 mol. Let's calculate the amount of reacted hydrogen: 0.3. 3 = 0.9 mol. Now let's calculate the final concentrations of the reactants:
finite = 0.10 + 0.60 = 0.70 mol;
[H 2 ]final = 1.5 - 0.90 = 0.60 mol;
finite = 0.80 - 0.50 = 0.30 mol.
Answer:= 0.70 mol/l; [H 2) = 0.60 mol/l.
Speed, temperature coefficient of reaction rate
Task 125.
The reaction proceeds according to the equation H 2 + I 2 = 2HI. The rate constant of this reaction at a certain temperature is 0.16. Initial concentrations of reactants (mol/l): [H 2 ] = 0.04:
= 0.05. Calculate initial speed reaction and its rate at = 0.03 mol/l. Answer: 3.2 .
10 -4 , 1,92 .
10 -4
Solution:
The reaction equation is:
H 2 + I 2 = 2HI
At the initial concentrations of the reacting substances, according to the law of mass action, the reaction rate will be equal when denoting the concentrations of the starting substances: [H 2 ] = a, = b.
V pr = k ab = 0,16 . 0,04 . 0,05 = 3,2 . 10 -4 .
Let's calculate the amount of hydrogen that reacted if its concentration changed and became 0.03 mol/l, we get: 0.04 - 0.03 = 0.01 mol. From the reaction equation it follows that hydrogen and iodine react with each other in a ratio of 1: 1, which means that 0.01 mol of iodine also entered into the reaction. Hence, the final concentration of iodine is: 0.05 -0.01 = 0.04 mol. At new concentrations, the rate of the direct reaction will be equal to:
Answer: 3.2 . 10 -4 , 1,92 . 10 -4 .
Task 126.
Calculate by how many times the rate of the reaction occurring in the gas phase will decrease if the temperature is lowered from 120 to 80 ° C. Temperature coefficient reaction speed Z.
Solution:
Speed dependence chemical reaction on temperature is determined by the empirical Van't Hoff rule according to the formula:
Therefore, the reaction rate; at 800 C 0 the reaction rate at 1200 C 0 is 81 times less.
Chemical reaction rate- change in the amount of one of the reacting substances per unit of time in a unit of reaction space.
The speed of a chemical reaction is influenced by the following factors:
- the nature of the reacting substances;
- concentration of reactants;
- contact surface of reacting substances (in heterogeneous reactions);
- temperature;
- action of catalysts.
Active collision theory allows us to explain the influence of certain factors on the rate of a chemical reaction. The main provisions of this theory:
- Reactions occur when particles of reactants that have a certain energy collide.
- The more particles of reagents, the closer they are to each other, the more chances they have to collide and react.
- Only effective collisions lead to a reaction, i.e. those in which “old connections” are destroyed or weakened and therefore “new ones” can be formed. To do this, the particles must have sufficient energy.
- The minimum excess energy required for effective collision of reactant particles is called activation energy Ea.
- Activity chemical substances manifests itself in the low activation energy of reactions involving them. The lower the activation energy, the higher the reaction rate. For example, in reactions between cations and anions, the activation energy is very low, so such reactions occur almost instantly
The influence of the concentration of reactants on the reaction rate
As the concentration of reactants increases, the reaction rate increases. In order for a reaction to occur, two chemical particles must come together, so the rate of the reaction depends on the number of collisions between them. An increase in the number of particles in a given volume leads to more frequent collisions and an increase in the reaction rate.
An increase in the rate of reaction occurring in the gas phase will result from an increase in pressure or a decrease in the volume occupied by the mixture.
Based on experimental data in 1867, Norwegian scientists K. Guldberg and P. Waage, and independently of them in 1865, Russian scientist N.I. Beketov formulated the basic law of chemical kinetics, establishing dependence of the reaction rate on the concentrations of the reactants -
Law of mass action (LMA):
The rate of a chemical reaction is proportional to the product of the concentrations of the reacting substances, taken in powers equal to their coefficients in the reaction equation. (“effective mass” is a synonym modern concept"concentration")
aA +bB =cС +dD, Where k– reaction rate constant
ZDM is performed only for elementary chemical reactions occurring in one stage. If a reaction proceeds sequentially through several stages, then the total speed of the entire process is determined by its slowest part.
Expressions for speeds various types reactions
ZDM refers to homogeneous reactions. If the reaction is heterogeneous (reagents are in different states of aggregation), then the ZDM equation includes only liquid or only gaseous reagents, and solid ones are excluded, affecting only the rate constant k.
Molecularity of the reaction is the minimum number of molecules participating in the elementary chemical process. Based on molecularity, elementary chemical reactions are divided into molecular (A →) and bimolecular (A + B →); trimolecular reactions are extremely rare.
Rate of heterogeneous reactions
- Depends on surface area of contact between substances, i.e. on the degree of grinding of substances and the completeness of mixing of reagents.
- An example is wood burning. A whole log burns relatively slowly in air. If you increase the surface of contact of wood with air, splitting the log into chips, the burning rate will increase.
- Pyrophoric iron is poured onto a sheet of filter paper. During the fall, the iron particles become hot and set fire to the paper.
Effect of temperature on reaction rate
In the 19th century, the Dutch scientist Van't Hoff experimentally discovered that with an increase in temperature by 10 o C, the rates of many reactions increase by 2-4 times.
Van't Hoff's rule
For every 10 ◦ C increase in temperature, the reaction rate increases by 2-4 times.
Here γ ( greek letter"gamma") - the so-called temperature coefficient or Van't Hoff coefficient, takes values from 2 to 4.
For each specific reaction, the temperature coefficient is determined experimentally. It shows exactly how many times the rate of a given chemical reaction (and its rate constant) increases with every 10 degree increase in temperature.
Van't Hoff's rule is used to approximate the change in the reaction rate constant with increasing or decreasing temperature. More exact ratio Swedish chemist Svante Arrhenius established between the rate constant and temperature:
How more E a specific reaction, so less(at a given temperature) will be the rate constant k (and rate) of this reaction. An increase in T leads to an increase in the rate constant, this is explained by the fact that an increase in temperature leads to a rapid increase in the number of “energetic” molecules capable of overcoming the activation barrier Ea.
Effect of catalyst on reaction rate
You can change the rate of a reaction by using special substances that change the reaction mechanism and direct it along an energetically more favorable path with a lower activation energy.
Catalysts- these are substances that participate in a chemical reaction and increase its speed, but at the end of the reaction they remain unchanged qualitatively and quantitatively.
Inhibitors– substances that slow down chemical reactions.
Changing the rate of a chemical reaction or its direction using a catalyst is called catalysis .
Example 1
How many times will the reaction rate increase?
A) C + 2 H 2 = CH 4
b) 2 NO + Cl 2 = 2 NOCl
when the pressure in the system increases three times?
Solution:
Increasing the pressure in the system by three times is equivalent to increasing the concentration of each of the gaseous components by three times.
In accordance with the law of mass action, we write down the kinetic equations for each reaction.
a) Carbon is a solid phase, and hydrogen is a gas phase. Speed heterogeneous reaction does not depend on the concentration of the solid phase, so it is not included in the kinetic equation. The rate of the first reaction is described by the equation
Let the initial concentration of hydrogen be equal to X, Then v 1 = kh 2 . After increasing the pressure three times, the concentration of hydrogen became 3 X, and the reaction rate v 2 = k(3x) 2 = 9kx 2. Next we find the speed ratio:
v 1:v 2 = 9kx 2:kx 2 = 9.
So, the reaction rate will increase 9 times.
b) Kinetic equation the second reaction, which is homogeneous, will be written in the form 
. Let the initial concentration NO equal to X, and the initial concentration Cl 2 equal to at, Then v 1 = kx 2 y; v 2 = k(3x) 2 3y = 27kx 2 y;
v 2:v 1 = 27.
The reaction speed will increase by 27 times.
Example 2
The reaction between substances A and B proceeds according to the equation 2A + B = C. The concentration of substance A is 6 mol/l, and substance B is 5 mol/l. The reaction rate constant is 0.5 (l 2 ∙mol -2 ∙s –1). Calculate the rate of chemical reaction in starting moment and at the moment when 45% of substance B remains in the reaction mixture.
Solution:
Based on the law of mass action, the rate of a chemical reaction at the initial moment is equal to:
= 0.5∙6 2 ∙5 = 90.0 mol∙s -1 ∙l -1
After some time, 45% of substance B will remain in the reaction mixture, that is, the concentration of substance B will become equal to 5. 0.45= 2.25 mol/l. This means that the concentration of substance B decreased by 5.0 - 2.25 = 2.75 mol/l.
Since substances A and B interact with each other in a ratio of 2:1, the concentration of substance A decreased by 5.5 mol/l (2.75∙2=5.5) and became equal to 0.5 mol/l (6. 0 - 5.5=0.5).
= 0.5(0.5) 2 ∙2.25 = 0.28 mol∙s -1 ∙l -1 .
Answer: 0.28 mol∙s -1 ∙l -1
Example 3
Temperature coefficient of reaction rate g equals 2.8. By how many degrees was the temperature increased if the reaction time was reduced by 124 times?
Solution:
According to van't Hoff's rule v 1 = v 2 ×. Reaction time t there is a quantity, inverse proportional to speed, Then v 2 /v 1 = t 1 /t 2 = 124.
t 1 /t 2 = = 124
Let's take a logarithm of the last expression:
lg( )= log 124;
DT/ 10×lgg=lg 124;
DT = 10×lg124/ lg2.8 » 47 0 .
The temperature was increased by 47 0.
Example 4
When the temperature increased from 10 0 C to 40 0 C, the reaction rate increased 8 times. What is the activation energy of the reaction?
Solution:
Reaction rate ratio at different temperatures equal to the ratio of rate constants at the same temperatures and equal to 8. In accordance with the Arrhenius equation
k 2 / k 1 = A× /A = 8
Since the pre-exponential factor and activation energy are practically independent of temperature, then
Example 5
At a temperature of 973 TO reaction equilibrium constant
NiO+H 2 = Ni+H 2 O (g)
Solution:
We assume that the initial concentration of water vapor was zero. The expression for the equilibrium constant of this heterogeneous reaction has the following form: 
.
Let the concentration of water vapor become equal to the moment of equilibrium x mol/l. Then, in accordance with the stoichiometry of the reaction, the hydrogen concentration decreased by x mol/l and became equal (3 – x) mol/l.
Let us substitute the equilibrium concentrations into the expression for the equilibrium constant and find X:
K = x / (3 – x); x / (3 – x) = 0.32; x=0.73 mol/l.
So, the equilibrium concentration of water vapor is 0.73 mol/l, the equilibrium concentration of hydrogen is 3 – 0.73 = 2.27 mol/l.
Example 6
How will the reaction balance be affected? 2SO 2 +O 2 ⇄2SO 3 ; DH= -172.38 kJ:
1) increase in concentration SO 2, 2) increase in pressure in the system,
3) cooling the system, 4) introducing a catalyst into the system?
Solution:
According to Le Chatelier's principle, with increasing concentration SO 2 the equilibrium will shift towards the process leading to consumption SO 2, that is, towards the direct reaction of formation SO 3.
The reaction comes with a change in number mole gaseous substances, so a change in pressure will shift the equilibrium. With increasing pressure, the equilibrium will shift towards a process that counteracts this change, that is, proceeding with a decrease in the number mole gaseous substances, and, consequently, with a decrease in pressure. According to the reaction equation, the number mole gaseous starting substances is three, and the number mole products of the direct reaction is equal to two. Therefore, with increasing pressure, the equilibrium will shift towards the direct reaction of formation SO 3.
Because DH< 0, then straight reaction is underway with the release of heat (exothermic reaction). The reverse reaction will occur with the absorption of heat (endothermic reaction). In accordance with Le Chatelier's principle, cooling will cause a shift in equilibrium towards the reaction that releases heat, that is, towards the direct reaction.
The introduction of a catalyst into the system does not cause a shift in the chemical equilibrium.
Example 7
At 10 0 C the reaction ends in 95 s, and at 20 0 C in 60 s. Calculate the activation energy for this reaction.
Solution:
Reaction time is inversely proportional to its speed. Then 
.
The relationship between the reaction rate constant and the activation energy is determined by the Arrhenius equation:
= 1,58.
ln1.58 = 
;
Answer: 31.49 kJ/mol.
Example 8
During the synthesis of ammonia N 2 + 3H 2 2NH 3, equilibrium was established at the following concentrations of reactants (mol/l):
Calculate the equilibrium constant for this reaction and initial concentrations nitrogen and hydrogen.
Solution:
We determine the equilibrium constant K C of this reaction:
K C= 
= (3,6) 2 / 2,5 (1,8) 3 = 0,89
We find the initial concentrations of nitrogen and hydrogen based on the reaction equation. The formation of 2 moles of NH 3 requires 1 mole of nitrogen, and the formation of 3.6 moles of ammonia required 3.6/2 = 1.8 moles of nitrogen. Taking into account the equilibrium concentration of nitrogen, we find its initial concentration:
C out (H 2) = 2.5 + 1.8 = 4.3 mol/l
To form 2 moles of NH 3, it is necessary to consume 3 moles of hydrogen, and to obtain 3.6 moles of ammonia, 3 ∙ 3.6 is required: 2 = 5.4 moles.
C out (H 2) = 1.8 + 5.4 = 7.2 mol/l.
Thus, the reaction began at concentrations (mol/l): C(N 2) = 4.3 mol/l; C(H2) = 7.2 mol/l
List of tasks for topic 3
1. The reaction proceeds according to the scheme 2A + 3B = C. The concentration of A decreased by 0.1 mol/l. How did the concentrations of substances B and C change?
2. The initial concentrations of substances involved in the reaction CO + H 2 O = CO 2 + H 2 were equal (mol/l, from left to right): 0.3; 0.4; 0.4; 0.05. What are the concentrations of all substances at the moment when ½ of the initial concentration of CO has reacted?
3. How many times will the reaction rate 2A + B change? C, if the concentration of substance A is increased by 2 times, and the concentration of substance B is decreased by 3?
4. Some time after the start of reaction 3A + B 2C + D concentrations of substances were (mol/l, from left to right): 0.03; 0.01; 0.008. What are the initial concentrations of substances A and B?
5. In the system CO + Cl 2 COCl 2 CO concentration was increased from 0.03 to 0.12 mol/l, and chlorine from 0.02 to 0.06 mol/l. How many times did the rate of the forward reaction increase?
6. How many times should the concentration of substance B be increased in the system 2A + B A 2 B, so that when the concentration of substance A decreases by 4 times, the rate of the direct reaction does not change?
7. How many times should the concentration of carbon monoxide (II) in the 2CO system be increased? CO 2 + C, so that the reaction rate increases 100 times? How will the reaction rate change when the pressure increases by 5 times?
8. How long will it take to complete the reaction at 18 0 C, if at 90 0 C it is completed in 20 seconds, and the temperature coefficient of the reaction rate is γ = 3.2?
9. At 10 0 C the reaction ends in 95 s, and at 20 0 C in 60 s. Calculate the activation energy.
10. How many times will the reaction rate increase when the temperature increases from 30 0 to 50 0 C if the activation energy is 125.5 kJ/mol?
11. What is the activation energy of a reaction whose rate at 300 K is 10 times greater than at 280 K?
12. What is the activation energy of the reaction if, with an increase in temperature from 290 to 300 K, its speed doubles?
13. The activation energy of a certain reaction is 100 kJ/mol. How many times will the reaction rate change when the temperature increases from 27 to 37 0 C?
14. The initial concentrations of the substances involved in the reaction N 2 +3H 2 =2NH 3 are equal (mol/l, from left to right): 0.2; 0.3; 0. What are the concentrations of nitrogen and hydrogen at the moment when the ammonia concentration becomes 0.1 mol/l.
15. How many times will the reaction rate 2A + B change? C, if the concentration of substance A is increased by 3 times, and the concentration of substance B is decreased by 2 times?
16. Initial concentrations of substances A and B in the reaction A+2B C were 0.03 and 0.05 mol/L, respectively. The reaction rate constant is 0.4. Find the initial rate of the reaction and the rate after some time, when the concentration of substance A decreases by 0.01 mol/l.
17. How will the reaction rate of 2NO+ O 2 change? 2NO 2 if: a) increase the pressure in the system by 3 times; b) reduce the volume of the system by 3 times?
18. How many times will the rate of a reaction occurring at 298 K increase if its activation energy is reduced by 4 kJ/mol?
19. At what temperature will the reaction complete in 45 minutes, if at 293 K it takes 3 hours? Temperature coefficient of reaction is 3.2.
20. The activation energy of the reaction NO 2 = NO + 1/2O 2 is 103.5 kJ/mol. The rate constant of this reaction at 298K is 2.03∙10 4 s -1. Calculate the rate constant for this reaction at 288 K.
21. The reaction CO + Cl 2 COCl 2 occurs in a volume of 10 liters. Composition of the equilibrium mixture: 14 g CO; 35.6 g Cl 2 and 49.5 g COCl 2. Calculate the equilibrium constant of the reaction.
22. Find the equilibrium constant of the reaction N 2 O 4 2NO 2 if the initial concentration of N 2 O 4 is 0.08 mol/l, and by the time equilibrium occurs, 50% of N 2 O 4 has dissociated.
23. The equilibrium constant of the reaction A + B C + D is equal to unity. Initial concentration [A] o =0.02 mol/l. What percentage of A is converted if the initial concentrations of B, C and D are 0.02; 0.01 and 0.02 mol/l respectively?
24. For the reaction H 2 + Br 2 2HBr at a certain temperature K = 1. Determine the composition of the equilibrium mixture if the initial mixture consisted of 3 mol H 2 and 2 mol bromine.
25. After mixing gases A and B in the system A + B C + D, equilibrium is established at the following concentrations (mol/l): [B] = 0.05; [C] = 0.02. The equilibrium constant of the reaction is 4∙10 3. Find the initial concentrations of A and B.
26. The equilibrium constant of the reaction A + B C + D is equal to unity. Initial concentration [A] = 0.02 mol/l. What percentage of A is converted if the initial concentrations [B] are 0.02; 0.1 and 0.2 mol/l?
27. At the initial moment of the reaction, ammonia synthesis concentrations were (mol/l): = 1.5; = 2.5; = 0. What is the concentration of nitrogen and hydrogen when the ammonia concentration is 0.15 mol/l?
28. Equilibrium in the H 2 +I 2 2HI system was established at the following concentrations (mol/l): =0.025; =0.005; =0.09. Determine the initial concentrations of iodine and hydrogen if there was no HI at the initial moment of the reaction.
29. When a mixture of carbon dioxide and hydrogen was heated in a closed vessel, the equilibrium CO 2 + H 2 CO + H 2 O was established. The equilibrium constant at a certain temperature is 1. What percent of CO 2 will turn into CO if you mix 2 moles of CO 2 and 1 mole H 2 at the same temperature.
30. The equilibrium constant of the reaction FeO + CO Fe + CO 2 at a certain temperature is 0.5. Find the equilibrium concentrations of CO and CO 2 if the initial concentrations of these substances were 0.05 and 0.01 mol/l, respectively.
Solutions
Theoretical explanations
The concentration of a solution is the relative amount of solute in a solution. There are two ways to express the concentration of solutions - fractional and concentration.
Share method
Mass fraction substances ω – dimensionless quantity or expressed as a percentage, calculated using the formula
%, (4.1.1)
Where m(in-va)- mass of substance, G;
m(size)- mass of solution, G.
%, (4.1.2)
Where ν(in-va)– amount of substance, mole;
ν 1+ν 2+… - the sum of the quantities of all substances in the solution, including the solvent, mole.
Volume fraction φ – dimensionless value or expressed as a percentage, calculated using the formula
%, (4.1.3)
Where V(v-va)- volume of substance, l;
V(mixtures)- volume of the mixture, l.
Concentration method
Molar concentration C M , mol/l, calculated by the formula
, (4.1.4)
Where ν(in-va)- amount of substance, mole;
V(r-ra)- volume of solution, l.
The abbreviation 0.1 M means a 0.1 molar solution (concentration 0.1 mol/L).
Normal concentration C N , mol/l, calculated by the formula
or 
, (4.1.5)
Where ν(eq)- amount of equivalent substance, mole;
V(r-ra)- volume of solution, l;
Z– equivalent number.
Abbreviated designation 0.1n. means 0.1 normal solution (concentration 0.1 mol eq/l).
Molal concentration C b , mol/kg, calculated by the formula
(4.1.6)
Where ν(in-va)- amount of substance, mole;
m(r-la)- mass of solvent, kg.
Titer T , g/ml, calculated by the formula
(4.1.7)
Where m(in-va)- mass of substance, G;
V(r-ra)- volume of solution, ml.
Let us consider the properties of dilute solutions, which depend on the number of particles of the solute and on the amount of solvent, but practically do not depend on the nature of the dissolved particles (colligative properties ) .
These properties include: pressure reduction saturated steam solvent above the solution, increase in boiling point, decrease in the freezing point of the solution compared to a pure solvent, osmosis.
Osmosis- this is the one-way diffusion of substances from solutions through a semi-permeable membrane that separates the solution and a pure solvent or two solutions of different concentrations.
In a solvent-solution system, solvent molecules can move through the partition in both directions. But the number of solvent molecules passing into solution per unit time is more number molecules moving from solution to solvent. As a result, the solvent passes through the semi-permeable membrane into a more concentrated solution, diluting it.
The pressure that must be applied to a more concentrated solution to stop the flow of solvent into it is called osmotic pressure .
Solutions characterized by the same osmotic pressure are called isotonic .
Osmotic pressure calculated using the Van't Hoff formula
Where ν - amount of substance, mole;
R- gas constant equal to 8.314 J/(mol K);
T - absolute temperature, TO;
V- volume of solution, m 3;
WITH- molar concentration, mol/l.
According to Raoult's law, the relative decrease in saturated vapor pressure above the solution is equal to the mole fraction of the dissolved nonvolatile substance:
(4.1.9)
The increase in boiling point and decrease in the freezing point of solutions compared to a pure solvent, as a consequence of Raoult’s law, are directly proportional to the molal concentration of the dissolved substance:
(4.1.10)
where is the temperature change;
Molal concentration, mol/kg;
TO- proportionality coefficient, in the case of an increase in the boiling point is called an ebullioscopic constant, and in the case of a decrease in the freezing point - cryoscopic.
These constants, numerically different for the same solvent, characterize an increase in the boiling point and a decrease in the freezing point of a one-molal solution, i.e. when 1 mole of non-volatile electrolyte is dissolved in 1 kg of solvent. Therefore, they are often called the molal increase in boiling point and decrease in the freezing point of a solution.
Cryoscopic and ebullioscopic constants do not depend on the nature of the solute, but depend on the nature of the solvent and are characterized by dimension 
.
Table 4.1.1 - Cryoscopic K K and ebullioscopic K E constants for some solvents
Cryoscopy and ebullioscopy– methods for determining certain characteristics of substances, for example, molecular weights of dissolved substances. These methods allow you to determine molecular weight substances that do not dissociate when dissolved by a decrease in the freezing point and an increase in the boiling point of solutions of known concentration:
(4.1.11)
where is the mass of the dissolved substance in grams;
Solvent mass in grams;
Molar mass of solute in g/mol;
1000 is the conversion factor from grams of solvent to kilograms.
Then molar mass nonelectrolyte is determined by the formula
(4.1.12)
Solubility S shows how many grams of a substance can dissolve in 100 g of water at a given temperature. The solubility of solid substances, as a rule, increases with increasing temperature, and for gaseous substances it decreases.
Solids characterized by very different solubility. Along with soluble substances, there are slightly soluble and practically insoluble in water. However, there are no absolutely insoluble substances in nature.
IN saturated solution a slightly soluble electrolyte, a heterogeneous equilibrium is established between the precipitate and the ions in solution:
A m B n 
mA n + +nB m - .
sediment 
saturated solution
In a saturated solution, the rates of dissolution and crystallization processes are the same , and the ion concentrations above the solid phase are equilibrium at a given temperature.
The equilibrium constant of this heterogeneous process is determined only by the product of the activities of the ions in solution and does not depend on the activity of the solid component. She got the name solubility product PR .
(4.1.13)
Thus, the product of ion activities in a saturated solution of a slightly soluble electrolyte at a given temperature is a constant value.
If an electrolyte has very low solubility, then the ion concentrations in its solution are negligible. In this case, the interionic interaction can be neglected and the concentrations of ions can be considered equal to their activities. Then the solubility product can be expressed in terms of the equilibrium molar concentrations of electrolyte ions:
. (4.1.14)
The solubility product, like any equilibrium constant, depends on the nature of the electrolyte and the temperature, but does not depend on the concentration of ions in the solution.
When the concentration of one of the ions increases in a saturated solution of a sparingly soluble electrolyte, for example, as a result of the introduction of another electrolyte containing the same ion, the product of the ion concentrations becomes greater than the value solubility products. In this case, the equilibrium between the solid phase and the solution shifts towards the formation of a precipitate. A precipitate will form until a new equilibrium is established, at which condition (4.1.14) is again satisfied, but at different ratios of ion concentrations. As the concentration of one of the ions in a saturated solution above the solid phase increases, the concentration of the other ion decreases so that the solubility product remains constant under constant conditions.
So, the condition for precipitation is:
. (4.1.15)
If in a saturated solution of a sparingly soluble electrolyte we reduce the concentration of any of its ions, then ETC will become more work ion concentrations. The equilibrium will shift towards the dissolution of the precipitate. Dissolution will continue until condition (4.1.14) is met again.
reaction is proportional to the product of the concentrations of the starting substances in powers equal to their stoicheometric coefficients.
O = K-s[A]t. c [B]p, where c [A] and c [B] are the molar concentrations of substances A and B, K is the proportionality coefficient, called the reaction rate constant.
Effect of temperature
The dependence of the reaction rate on temperature is determined by Van't Hoff's rule, according to which, with every 10 C increase in temperature, the rate of most reactions increases by 2-4 times. Mathematically, this dependence is expressed by the relation:
where and i)t, i>t are the reaction rates, respectively, at the initial (t:) and final (t2) temperatures, and y is the temperature coefficient of the reaction rate, which shows how many times the reaction rate increases with an increase in the temperature of the reactants by 10 °C.
Example 1. Write an expression for the dependence of the rate of a chemical reaction on the concentration of reactants for the processes:
a) H2 4- J2 -» 2HJ (in the gas phase);
b) Ba2+ 4- S02-= BaS04 (in solution);
c) CaO 4- C02 -» CaC03 (with the participation of solid
substances).
Solution. v = K-c(H2)c(J2); v = K-c(Ba2+)-c(S02); v = Kc(C02).
Example 2. How will the rate of the reaction 2A + B2^± 2AB, occurring directly between molecules in a closed vessel, change if the pressure is increased by 4 times?
According to the law of action of molecules, the rate of a chemical reaction is directly proportional to the product of the molar concentrations of the reacting substances: v = K-c[A]m.c[B]n. By increasing the pressure in the vessel, we thereby increase the concentration of reactants.
Let the initial concentrations of A and B be equal to c[A] = a, c[B] = b. Then = Ka2b. Due to an increase in pressure by 4 times, the concentration of each of the reagents also increased 4 times and steel c[A] = 4a, c[B] = 4b.
At these concentrations:
vt = K(4a)2-4b = K64a2b.
The value of K is the same in both cases. The rate constant for a given reaction is a constant value, numerically equal to speed reactions at molar concentrations of reactants equal to 1. Comparing v and vl9 we see that the reaction rate has increased by 64 times.
Example 3. How many times will the rate of a chemical reaction increase when the temperature increases from 0°C to 50°C, taking the temperature coefficient of the rate equal to three?
The rate of a chemical reaction depends on the temperature at which it occurs. When the temperature increases by 10 °C, the reaction rate will increase by 2-4 times. If the temperature decreases, it decreases by the same amount. The number showing how many times the reaction rate increases when the temperature increases by 10 °C is called the temperature coefficient of the reaction.
IN mathematical form The dependence of the change in reaction rate on temperature is expressed by the equation:
The temperature increases by 50 °C, and y = 3. Substitute these values
^5о°с = ^о°с "3у = "00оС? 3 = v0oC ? 243. The speed increases by 243 times.
Example 4. The reaction at a temperature of 50 °C proceeds in 3 minutes 20 s. The temperature coefficient of the reaction rate is 3. How long will it take for this reaction to complete at 30 and 100 °C?
When the temperature increases from 50 to 100 °C, the reaction rate increases in accordance with Van't Hoffe's rule by the following number of times:
H _ 10 „O 10 - Q3
U yu = z yu = z* = 243 times.
If at 50°C the reaction ends in 200 s (3 min 20 s), then at 100°C it will end in 200/
243 = 0.82 s. At 30 °C the reaction rate decreases
sews 3 10 = 32 = 9 times and the reaction ends in 200 * 9 = 1800 s, i.e. in 30 min.
Example 5. The initial concentrations of nitrogen and hydrogen are respectively 2 and 3 *mol/l. What will be the concentrations of these substances at the moment when 0.5 mol/L of nitrogen has reacted?
Let's write the reaction equation:
N2 + ZH2 2NH3, the coefficients show that nitrogen reacts with hydrogen in a molar ratio of 1:3. Based on this, we create the ratio:
1 mole of nitrogen reacts with 3 moles of hydrogen.
0.5 mol of nitrogen reacts with x mol of hydrogen.
From - = - ; x =-- = 1.5 mol.
1.5 mol/l (2 - 0.5) nitrogen and 1.5 mol/l (3 - 1.5) hydrogen did not react.
Example 6. How many times will the speed of a chemical reaction increase when one molecule of substance A and two molecules of substance B collide:
A(2) + 2B -» C(2) + D(2), with an increase in the concentration of substance B by 3 times?
Let us write an expression for the dependence of the rate of this reaction on the concentration of substances:
v = K-c(A)-c2(B),
where K is the rate constant.
Let us take the initial concentrations of substances c(A) = a mol/l, c(B) = b mol/l. At these concentrations, the reaction rate is u1 = Kab2. When the concentration of substance B increases by 3 times, c(B) = 3b mol/l. The reaction rate will be equal to v2 = Ka(3b)2 = 9Kab2.
Speed increase v2: ig = 9Kab2: Kab2 = 9.
Example 7. Nitric oxide and chlorine react according to the reaction equation: 2NO + C12 2NOC1.
How many times should the pressure of each source be increased?
