The composition of the initial mixture. Tasks for mixtures and alloys on the exam in chemistry
dissolving water. The solution obtained after passing gases through water had an acidic reaction. When this solution was treated with silver nitrate, 14.35 g of a white precipitate precipitated. Determine the quantitative and qualitative composition of the initial mixture of gases. Decision.
The gas that burns to form water is hydrogen, which is slightly soluble in water. React in sunlight with an explosion hydrogen with oxygen, hydrogen with chlorine. Obviously, there was chlorine in the mixture with hydrogen, because. the resulting HC1 is highly soluble in water and gives a white precipitate with AgNO3.
Thus, the mixture consists of gases H2 and C1:
1 mol 1 mol
HC1 + AgN03 -» AgCl 4- HN03.
x mol 14.35
When processing 1 mol of HC1, 1 mol of AgCl is formed, and when processing x mol, 14.35 g or 0.1 mol. Mr(AgCl) = 108 + 2 4- 35.5 = 143.5, M(AgCl) = 143.5 g/mol,
v= - = = 0.1 mol,
x = 0.1 mol of HC1 was contained in the solution. 1 mol 1 mol 2 mol H2 4-C12 2HC1 x mol y mol 0.1 mol
x \u003d y \u003d 0.05 mol (1.12 l) of hydrogen and chlorine reacted to form 0.1 mol
HC1. The mixture contained 1.12 liters of chlorine, and hydrogen 1.12 liters + 1.12 liters (excess) = 2.24 liters.
Example 6 A laboratory has a mixture of sodium chloride and iodide. 104.25 g of this mixture was dissolved in water and an excess of chlorine was passed through the resulting solution, then the solution was evaporated to dryness and the residue was calcined to constant weight at 300 °C.
The mass of dry matter turned out to be 58.5 g. Determine the composition of the initial mixture in percent.
Mr(NaCl) = 23 + 35.5 = 58.5, M(NaCl) = 58.5 g/mol, Mr(Nal) = 127 + 23 = 150 M(Nal) = 150 g/mol.
In the initial mixture: the mass of NaCl - x g, the mass of Nal - (104.25 - x) g.
When passing through a solution of chloride and sodium iodide, iodine is displaced by them. When passing the dry residue, the iodine evaporated. Thus, only NaCl can be a dry matter.
In the resulting substance: the mass of NaCl of the original x g, the mass of the obtained (58.5-x):
2 150 g 2 58.5 g
2NaI + C12 -> 2NaCl + 12
(104.25 - x)g (58.5 - x)g
2 150 (58.5 - x) = 2 58.5 (104.25 x)
x = - = 29.25 (g),
those. NaCl in the mixture was 29.25 g, and Nal - 104.25 - 29.25 = 75 (g).
Find the composition of the mixture (in percent):
w(Nal) = 100% = 71.9%,
©(NaCl) = 100% - 71.9% = 28.1%.
Example 7 68.3 g of a mixture of nitrate, iodide and potassium chloride are dissolved in water and treated with chlorine water. As a result, 25.4 g of iodine was released (neglect the solubility of which in water). The same solution was treated with silver nitrate. 75.7 g of sediment fell out. Determine the composition of the initial mixture.
Chlorine does not interact with potassium nitrate and potassium chloride:
2KI + C12 -» 2KS1 + 12,
2 mol - 332 g 1 mol - 254 g
Mg (K1) \u003d 127 + 39 - 166,
x = = 33.2 g (KI was in the mixture).
v(KI) - - = = 0.2 mol.
1 mol 1 mol
KI + AgN03 = Agl + KN03.
0.2 mol x mol
x = = 0.2 mol.
Mr(Agl) = 108 + 127 = 235,
m(Agl) = Mv = 235 0.2 = 47 (r),
then AgCl will be
75.7 g - 47 g = 28.7 g.
74.5 g 143.5 g
KCl + AgN03 = AgCl + KN03
X \u003d 1 L_ \u003d 14.9 (KCl).
Therefore, the mixture contained: 68.3 - 33.2 - 14.9 = 20.2 g KN03.
Example 8. To neutralize 34.5 g of oleum, 74.5 ml of a 40% potassium hydroxide solution are consumed. How many moles of sulfur oxide (VI) account for 1 mole of sulfuric acid?
100% sulfuric acid dissolves sulfur oxide (VI) in any ratio. The composition expressed by the formula H2S04*xS03 is called oleum. Let's calculate how much potassium hydroxide is needed to neutralize H2SO4:
1 mol 2 mol
H2S04 + 2KOH -> K2S04 + 2H20 xl mol y mol
y - 2*x1 mole of KOH is used to neutralize SO3 in oleum. Let's calculate how much KOH is needed to neutralize 1 mol of SO3:
1 mol 2 mol
S03 4- 2KOH -> K2SO4 + H20 x2 mol z mol
z - 2 x2 mol of KOH goes to neutralize SOg in oleum. 74.5 ml of a 40% KOH solution is used to neutralize the oleum, i.e. 42 g or 0.75 mol KOH.
Therefore, 2 xl + 2x 2 \u003d 0.75,
98 xl + 80 x2 = 34.5 g,
xl = 0.25 mol H2SO4,
x2 = 0.125 mol SO3.
Example 9 There is a mixture of calcium carbonate, zinc sulfide and sodium chloride. If 40 g of this mixture is treated with an excess of hydrochloric acid, 6.72 liters of gases will be released, the interaction of which with an excess of sulfur oxide (IV) releases 9.6 g of sediment. Determine the composition of the mixture.
When exposed to a mixture of excess hydrochloric acid, carbon monoxide (IV) and hydrogen sulfide could be released. Only hydrogen sulfide interacts with sulfur oxide (IV), therefore, according to the amount of precipitate, its volume can be calculated:
CaC03 + 2HC1 -> CaC12 + H20 + C02t(l)
100 g - 1 mol 22.4 l - 1 mol
ZnS + 2HC1 -> ZnCl2 + H2St (2)
97 g - 1 mol 22.4 l - 1 mol
44.8 l - 2 mol 3 mol
2H2S + S02 -» 3S + 2H20 (3)
xl l 9.6 g (0.3 mol)
xl = 4.48 L (0.2 mol) H2S; from equations (2 - 3) it can be seen that ZnS was 0.2 mol (19.4 g):
2H2S + S02 -> 3S + 2H20.
Obviously, the carbon monoxide (IV) in the mixture was:
6.72 l - 4.48 l \u003d 2.24 l (CO2).
Introduction
Rectification is a mass transfer process, which is carried out in most cases in countercurrent column apparatuses with contact elements (packings, plates) similar to those used in the absorption process. Therefore, the methods of approach to the calculation and design of distillation and absorption plants have much in common. Nevertheless, a number of features of the distillation process (different ratio of liquid and steam loads in the lower and upper parts of the column, physical properties of the phases and distribution coefficient that are variable along the height of the column, the joint flow of mass and heat transfer processes) complicates its calculation.
One of the difficulties is the lack of generalized regularities for calculating the kinetic coefficients of the rectification process. To the greatest extent, this applies to columns with a diameter of more than 800 mm with packings and trays, which are widely used in chemical industries. Most of the recommendations come down to the use of kinetic dependences obtained in the study of absorption processes for the calculation of distillation columns.
In the process of distillation, there is a continuous exchange between the liquid and vapor phases. The liquid phase is enriched with a higher-boiling component, and the vapor phase is enriched with a lower-boiling one. The process of mass transfer occurs along the entire height of the column between the phlegm flowing down and the steam rising up. To intensify the mass transfer process, contact elements are used, which allows increasing the mass transfer surface. In the case of using a packing, the liquid flows down in a thin film over its surface; in the case of using trays, the vapor passes through the liquid layer on the surface of the trays. This paper presents the calculation of a tray distillation column for the separation of a binary mixture of acetone - benzene
Principal diagram of a distillation plant
Schematic diagram of the distillation unit is presented on. The initial mixture from the intermediate tank 1 is fed by a centrifugal pump 2 to the heat exchanger 3, where it is heated to the boiling point. The heated mixture is fed to the separation in the distillation column 5 on the feed plate, where the composition of the liquid is equal to the composition of the initial mixture X F .
Flowing down the column, the liquid interacts with the steam rising upwards, which is formed during the boiling of the bottom liquid in the boiler 4. The initial composition of the vapor is approximately equal to the composition of the bottom residue X w, i.e. depleted in the volatile component. As a result of mass exchange with the liquid, the vapor is enriched with a highly volatile component. For a more complete enrichment, the upper part of the column is irrigated in accordance with a given reflux ratio with a liquid (phlegm) of composition X p obtained in a reflux condenser 6 by condensing the steam leaving the column. Part of the condensate is removed from the dephlegmator in the form of a finished separation product - distillate, which is cooled in the heat exchanger 7 and sent to the intermediate tank 8.
From the bottom part of the column, pump 9 continuously removes bottom liquid - a product enriched with a low-volatile component, which is cooled in heat exchanger 10 and sent to tank 11.
Thus, in the distillation column, a continuous non-equilibrium process of separation of the initial binary mixture into a distillate (with a high content of a volatile component) and a distillation residue (enriched with a non-volatile component) is carried out.
Rice. 1 Schematic diagram of the distillation unit:
1 - container for the initial mixture; 2.9 – pumps; 3 – heat exchanger-heater; 4 - boiler; 5 - distillation column; 6 - dephlegmator; 7 – distillate cooler; 8 - container for collecting distillate; 10 – bottom liquid cooler; 11 - container for bottom liquid.
Technological calculation of a distillation column of continuous action
The task
Design a distillation plant for mixture separation.
Mixture: acetone - benzene.
The amount of the original mixture: 
t/h=15000 kg/h
The composition of the initial mixture: 
% wt.
The composition of the VAT residue: 
% wt.
Distillate composition: 
% wt.
Heating steam pressure: 5 atm
Pressure in the column: 1 ata
Type of contact devices: valve discs
Building the steps of the rectification process
Conversion of mass fractions to molar
,
where M A and M B are the molar masses of acetone and benzene, respectively, kg/mol.
M A = 58 kg/mol; M B = 78 kg/mol
Material balance of the column
Molar mass of the initial mixture
Mole second consumption of the mixture
Distillate consumption
Let us substitute this expression into , where F, D, W are the costs of the initial mixture, distillate, distillation residue, kmol/s.
kmol/s
Equilibrium between vapor and liquid
Table 1. Equilibrium compositions of the liquid ( x) and pair ( y) in the mol. and boiling point ( t) in °C binary mixtures at 760 mm. rt. Art.
Rice. 2 Equilibrium curve and position of the operating line at R min
Rice. 3 Diagram t – x, y.
1 - liquid line; 2 - steam line.
Minimum Reflux Number
bmax = 0.35 (Fig. 2)
Working Reflux Number
1.
;
– coefficient of excess phlegm
2.
3.
4.
5.
Rice. 4 Graphical determination of the number of theoretical plates at 
Rice. 5 Graphical determination of the number of theoretical plates at
Rice. 6 Graphical determination of the number of theoretical plates at
Rice. 7 Graphical determination of the number of theoretical plates at
Rice. 8 Graphical determination of the number of theoretical plates at
Optimal Reflux Number
Table 2. The number of theoretical steps for different reflux ratios
|
|
Rice. 9 Dependence of the number of theoretical steps on the reflux number
Rice. 10 Determining the optimal reflux ratio
The graph (Fig. 10), built on the basis of the data (Table 2), shows that the minimum column volume will occur at R=2.655. We accept this phlegm number for further calculations and the corresponding number of theoretical steps n tc in = 19; n ts n = 5
Molar flow rate of liquid at the top and bottom of the column
Molar flow rate of steam in the column
Physico-chemical properties of the vapor and liquid phases for the top and bottom of the column
Average molar concentrations of liquid and vapor
mol. USD
mol. USD
According to the diagram t – x, y (Fig. 3), at medium liquid concentrations 
and , we determine the average temperatures of the liquid: ° С and 
°С
mol. USD
mol. dollars,
where y D = x D ; y W = x W ; y F is the vapor composition corresponding to the composition of the initial mixture x F(Fig. 6)
According to the t – x, y diagram (Fig. 3), at average vapor concentrations 
and , we determine the average steam temperatures: 
°C (334K) and 
°С (347К);
Average molar masses of liquid and vapor
Average liquid and vapor densities
Conversion of molar concentrations to mass:
wt. USD
wt. USD
kg / m 3,
where 
and 
are the densities of acetone and benzene, respectively, at a temperature of °C, 
kg / m 3, 
kg / m 3
kg / m 3,
where and are the densities of acetone and benzene, respectively, at a temperature of °C, 
kg / m 3, kg / m 3
where T 0 is the absolute temperature equal to 273K
Average viscosities of liquid and steam
,
where 
and are the viscosities of acetone and benzene, respectively, at a temperature of °C, 
,
,
,
,
,
where and are the viscosities of acetone and benzene, respectively, at a temperature of °C, 
,
Mass and volume flow rates of liquid and steam
Average mass costs:
Volume costs:
Table 3. Parameters of vapor and liquid flows in the column
|
Stream name |
Parameter dimension |
|||||
|
Liquid at the top of the column |
|
|
|
|
||
|
Liquid at the bottom of the column |
|
|
|
|
||
|
Steam at the top of the column |
|
|
||||
|
Steam at the bottom of the column |
|
|
|
|||
=63
=767,5
=5,12
=6,67∙10 -3
=72,2
=802,6
=10,31
=1,29∙10 -2
=62
=1,25
=1,4
=7,95
=5,68Hydraulic calculation of the column
Load increase factor
Estimated steam speed
for the top of the column:
for the bottom:
Diameter
top of the column:
the bottom of the column:
Actual steam speed
Because 
we accept a column with a diameter 
at the top of the column:
at the bottom of the column:
Relative active tray area
Drain perimeter 
%; overflow section 0.3m 2
load factor
for the top of the column:
for the bottom of the column:
Surface tension coefficient for the top of the column:
where 
and are the surface tension of the mixture for the upper and lower parts of the column, respectively, 
N/m, N/m
Let's take the minimum distance between the plates 
m 
;
m/s for the bottom of the column:
Checking the conditions for the admissibility of steam speeds for the upper and lower parts of the column:
>
>
It can be seen that the condition is not satisfied for either the upper or lower parts of the column. By successively increasing the distance between the plates, as well as the diameter of the column, we find that the condition will be satisfied only with a diameter of m; 
m
Actual steam velocity at the top of the column:
at the bottom of the column:
Drain perimeter 
m; relative free section 
%; overflow section 0.52m 2
Relative overflow section:
Relative active tray area:
Distance between plates m 
;
Permissible steam velocity in the working section of the column for the upper part:
for the bottom of the column:
Conditions for the admissibility of steam speeds for the upper and lower parts of the column:
>
>
>
>
The conditions are met.
The specific load of the liquid on the drain partition
at the top of the column:
at the bottom of the column:
Steam load factor
for the top of the column:
for the bottom of the column:
Liquid back-up over the drain threshold in the upper part of the column:
for the bottom:
Bubbling depth
The height of the vapor-liquid layer on the plates of the upper part of the column:
on plates at the bottom of the column:
Drain threshold height
at the top of the column:
at the bottom of the column:
Dynamic bubbling depth
m
The minimum allowable steam velocity in the free section of the column
where 
- valve thickness equal to 0.001 m; 
- material density (steel) equal to 7700 kg / m 3
The minimum allowable steam speed in the free section of the upper plates:
m/s
in the free section of the lower plates:
where 
– drag coefficient
Section safety factor:
Since > 1 and 
> 1, the selected free section of the plates ensures their uniform operation, we accept 
aeration factor
for top plates:
for bottom plates:
on top plates:
on bottom plates:
Liquid layer height
on top plates:
on bottom plates:
Hydraulic resistance of plates
at the top of the column:
at the bottom of the column:
The height of the separation space between the plates
at the top of the column:
at the bottom of the column:
where K 5 \u003d 1 - foaming coefficient of the mixture
Intertray liquid entrainment
at the top of the column:
at the bottom of the column:
Cross-sectional area of the column:
Liquid velocity in the overflow devices of the upper plates:
in the overflow devices of the lower plates:
Permissible liquid velocity in the overflow devices of the upper plates:
m / s in the overflow devices of the lower plates:
The actual velocity of the liquid in the overflows is less than the permissible ones.
Local contact efficiency
Vapor diffusion coefficient
,
where ; specific volume of acetone and benzene, respectively
Vapor diffusion coefficient at the top of the column:
where 
- steam temperature at the top of the column
Steam diffusion coefficient at the bottom of the column:
where 
- steam temperature at the bottom of the column
for the top of the column:
where and are the viscosities of acetone and benzene, respectively, at a temperature of °С, ,
where Ф = 1 is a dimensionless parameter, takes into account the association of solvent molecules
Liquid diffusion coefficient for the bottom of the column:
,
where and are the viscosities of acetone and benzene, respectively, at a temperature of °С, ,
Number of transfer units
in the gas phase for the top of the column:
for the bottom of the column:
Number of liquid transfer units for the top of the column:
for the bottom:
Diameter of fitting for reflux inlet, 
The diameter of the fitting for the output of the VAT residue,
Bibliography
Ulyanov. B.A., Badenikov V.Ya., Likuchev V.G., processes and apparatuses of chemical technology. Textbook - Angarsk: Publishing House of the Angarsk State Technical Academy, 2005 - 903 p.
Dytnersky Yu.I. Basic Processes and Apparatuses of Chemical Technology: Design Guide / Ed. Yu.I. Dytnersky. M.: chemistry, 1991.-496 p.
Guidelines for the course design of processes and apparatuses of chemical technology - Ed. 2nd, rev. And extra. - Angarsk, AGTA, 2005 - 64 p.
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The composition of the initial mixture for the production of artificial stone. (Photo gallery "Our technologies" on the page of the same name. What is included in the composition of artificial facing stone produced on flexible elastic molds. In essence, the decorative facing stone we are talking about is a typical Portland cement-based sand concrete, made by vibrocasting into special flexible elastic matrices - shaped and specially colored.Consider the main components of the concrete mixture for the production of artificial facing stone by vibrocasting.The binder is the basis of any artificial facing stone.In this case, it is Portland cement grade M-400 or M-500.To ensure that the quality of concrete always remains consistently high, we recommend using only "fresh" cement (as you know, it quickly loses its properties over time and from improper storage) of the same manufacturer with a good reputation.For the production of decorative facing stone, both ordinary, gray cement, and and white cement. In nature, there are a number of colors and shades that can only be replicated on white cement. In other cases, gray portland is used (for reasons of economic feasibility).
Many domestic manufacturers of artificial facing stone have recently been actively using gypsum as a binder. At the same time, they claim that their products are expanded clay concrete. And, as a rule, expanded clay concrete is really presented at the stands of companies. But there is one point that determines the behavior of manufacturers of artificial facing stone. The cost of flexible elastic injection molds that allow you to accurately repeat the texture of the stone is very high.
And if the technology is followed, the turnover of injection molds, that is, the time from the moment the concrete is poured to the moment the product is stripped, is 10-12 hours, versus 30 minutes on plaster. This is what drives companies to use gypsum as a binder. And the price of gypsum is at least five times lower than the price of white cement. All this provides companies with super profits. But the issue price for the end user is very high! The extremely low frost resistance and strength of such products will not allow you to enjoy the view of the facades for a long time.
In the presented photographs, the plaster product is one year after installation. Multiple cracks and fractures are clearly visible. Therefore, the use of this material on an industrial scale is difficult. Based on the tasks we are facing, we prefer to produce artificial facing stone - a material close to natural stone in terms of hardness and abrasion properties, suitable for both external and internal cladding, rather than decorations that are fragile and whimsical to water. Filler. Depending on the type of fillers used, cement-based artificial facing stone can be "heavy" (2-2.4 g/cm3) or "light" (about 1.6 g/cm3). Ideally, heavy concrete is used for the production of paving stone, decorative paving slabs, curbs, plinth frames, and interior stone. For the production of artificial facing stone used for exterior decoration, lightweight concrete is used.
Approximately this is what manufacturers working on American technology do. In the regions, unfortunately, heavy concrete is used predominantly. Of course, it is much easier to make a decorative stone on sand, but a light stone will always be preferable for the consumer. It's just a matter of choice. For the production of heavy artificial facing stone, coarse quartz sand of a fraction of 0.63-1.5 mm is used (the use of fine sand worsens the strength characteristics of concrete) and, when appropriate, fine crushed stone, for example, marble, of a fraction of 5-10 mm. "Light" facing stone is made using expanded clay sand. But in the production of artificial facing stone on expanded clay, the following factor should be taken into account. In July 2001, we received information from customers about the appearance of "shots" on the surface of products (lightweight concrete) (dotted swelling of the white material). As a result of consultations with specialists, it was found that the "shots" appear as a result of the decay of lime inclusions in expanded clay.
When free calcium interacts with moisture (water or its vapor), a chemical reaction occurs accompanied by an increase in the volume of free calcium grains, resulting in the so-called “shot” effect. CaO + H2O \u003d Ca (OH) 2 + CO2 \u003d CaCO3 The peculiarity of this chemical reaction is that it takes a very long time - up to 6 months. Expanded clay manufacturers produce products in accordance with GOST, which allows the presence of lime grains up to 3% of the total mass. The effect of "shots" reduces the consumer properties of products, so the task was to find a new filler for the production of lightweight concrete.
It has been observed that the lime disintegration reaction causes degradation of the surface of the product ONLY during interior finishing. When using products for finishing plinths and facades of buildings, visible damage to the finishing material is not observed. According to the statements of an employee of NIIZhB, lime decay is leveled when using products for exterior decoration of buildings. In connection with the identification of this pattern, since August 2001, products for interior work have been produced not on expanded clay, but on another (heavier) aggregate. To switch to a single filler, we offer the following solutions to this problem: 1. Use crushed expanded clay with a fraction of at least 2 cm as a filler. 2. Create expanded clay dumps with exposure in an open area for at least 6-9 months.
3. Creation of a non-uniform filler from quartz sand and a lighter artificial filler. 4. Use of slag pumice. however, the bulk density of the finished product will increase to 1800-2000 kg/m3. Lightweight aggregate must meet the following requirements. bulk weight is about 600 kg/m3. sand of a fraction of 0-0.5 cm or 0-1 cm (the presence of a fine fraction of 15% by volume. Compressive strength 18 kg / cm (expanded clay index. Water absorption up to 25% (expanded clay index. In the production of artificial facing stone, decorative paving slabs) , small architectural products on flexible elastic molds, the following fillers can be used: Slag pumice, Granulated slag, Crushed stone and slag sand, Foam glass, Expanded perlite sand, Rigidly expanded perlite, Expanded vermoculite, Expanded polystyrene, Enriched quartz sand, Marble chips, Building sand (white ), Molding sand, Volcanic pumice. Pigments and dyes. The most important component of a decorative facing stone is the pigments (dyes) used. Skillful or inept use of colorants directly affects the appearance of the final product. In experienced hands, ordinary concrete turns into something completely indistinguishable from natural "wild" stone. How to achieve this? Mineral inorganic pigments (titanium, iron, chromium oxides) and special light and weather resistant dyes are used for coloring cement. Experienced manufacturers usually choose dyes from companies such as Bayer, Du Pont, Kemira and others no less reputable. This is due not only to the consistently high quality of their products, but also to their wide range of products. So, Bayer offers several dozen iron oxide pigments. By combining them with each other, you can choose almost any desired shade of color. So, Portland cement, expanded clay sand and pigments are the main composition of artificial facing stone. Many manufacturers of architectural concrete products limit themselves to this, despite the fact that there are a huge number of various additives in cements to improve certain characteristics. In any major city, you can find suppliers of domestic and imported additives for concrete. These are various superplasticizers that improve workability and increase the strength of concrete; polymer-latex additives that have a beneficial effect on the durability of concrete; concrete hardening accelerators and air-entraining additives; volumetric water repellents, many times reducing water absorption (useful for facade, basement and paving stone); chemical fibers for dispersed reinforcement, which dramatically increases crack resistance and much more. To use any of these additives or not - decide for yourself, we just want to recommend the use of protective impregnating compounds for surface treatment of decorative facing stone. Properly selected water repellent for concrete will achieve the following results. will increase the aesthetics of the perception of the stone and eliminate the "dustiness" - a characteristic feature of any cement concrete. increase the service life of the facade stone (the point here is that the process of destruction of decorative concrete primarily affects the color saturation long before the first signs of destruction appear, the reason for which is the exposure of aggregate particles on the front surface of the stone. It will sharply reduce the risk of efflorescence on the surface of the stone, which are a real disaster for cementitious decorative concretes, which is why they should be given the closest attention.
The composition of an equilibrium mixture can be expressed using:
a) degree of dissociation ()
b) degree of transformation ()
c) product yield (x)
Let's take a look at each of these cases with examples:
a) according to the degree of dissociation
Degree of dissociation () is the fraction of dissociated molecules from the initial number of molecules. It can be expressed in terms of the amount of matter
where n diss is the number of decomposed moles of the starting material; n ref is the number of moles of the starting material before the reaction.
Let there be, for example, 5 mol NO 2 before the reaction, and α is the degree of dissociation of NO 2.
By equation (1.20) 
, unreacted NO 2 will remain (5 - 5).
According to the reaction equation, when 2 moles of NO 2 dissociate, 2 moles of NO and 1 mole of O 2 are obtained, and from 5, respectively, 5 moles of NO and 
mole O 2 . The balance line will be:
b ) according to the degree of transformation
The degree of transformation of a substance () is the proportion of reacted molecules of a given substance to the initial number of molecules of this substance. We express through the amount of substance in moles
(1.21)
Let 2 moles of CO and 2 moles of H 2 be taken, is the degree of hydrogen conversion in the reaction
Let us explain the equilibrium line. We proceed from a substance for which the degree of transformation is known, i.e. H 2 . From equation (1.21) we get n reag = n ref · = 2 .
It can be seen from the stoichiometric equation that CO is consumed 3 times less than H 2, that is, if H 2 reacts 2, then CO will react 
, and the rest will remain unreacted by the time of equilibrium. We also talk about products using the stoichiometric equation.
in) by product yield.
Product yield (x) is the amount of the final substance in moles. Let "x" be the yield of methanol in the reaction
in all three cases, the reasoning is similar and proceeds from a substance for which something is known (in the examples this value is underlined).
Knowing the composition of the equilibrium mixture, we can express the equilibrium constant. So, for the case "in"
and from equation (1.19)
The yield of the substance in fractions(or %) - the ratio of the amount of the product formed to the total amount of the substance in the equilibrium mixture:
In this example:
1.3.4 Influence of various factors on the equilibrium shift (on the composition of the equilibrium mixture)
Influence of pressure (or volume) at T=const
If the system is ideal, then the equilibrium constant K p does not depend on pressure (or volume). If the reaction proceeds at high pressures, then the following equation should be used:
, (1.22)
where f- fugacity.
K f does not depend on pressure, while the value of K p depends on pressure, but as the pressure decreases, it approaches the value of K f, since the real gas mixture approaches the ideal state, f p. So for the reaction:
at 350 atm K f = 0.00011 K R = 0,00037
At low pressures, one can consider To R independent of pressure, that is 
. In what follows, we will consider this particular case.
It can be seen from relation (1.12) that the quantities 
,
will depend on pressure, therefore, without affecting the equilibrium constant 
, a change in pressure can affect the composition of the equilibrium mixture and the yield of products.
(1.23)
Equation (1.23) shows that the effect of pressure on 
due to the value n:
n 0, the reaction proceeds with an increase in the number of moles of gaseous products, for example:
, that is, with an increase in the total pressure To X decreases, decreases and the number of products in the equilibrium mixture, that is, the equilibrium shifts to the left, towards the formation of COCl 2 .
n = 0-2-1= -3
, that is, with increasing pressure, K x (and the product yield) increases.
K
=
K
= const. In this case, the composition of the equilibrium mixture does not depend on pressure.
Addition of inert gas at P = const, it affects the equilibrium shift similarly to a decrease in pressure. Inert gases in chemical equilibrium are gases that do not interact with reactants or reaction products.
Volume increase at constant pressure affects the equilibrium shift similarly to a decrease in pressure.
Influence of the ratio between the components
The composition of the equilibrium mixture is also affected by the ratio of the reagents taken for the reaction.
The highest yield of products will be at a stoichiometric ratio. So for reaction
the ratio of hydrogen and nitrogen 3:1 will give the highest yield of ammonia.
In some cases, a high degree of conversion of one of the reactants is required even to the detriment of the product yield.
For example, in the formation of hydrogen chloride by the reaction
a more complete conversion of chlorine is necessary so that the equilibrium mixture contains as little Cl 2 as possible. The equilibrium mixture dissolves in water and thus hydrochloric acid is obtained. In this case, hydrogen is almost insoluble in water and is not contained in acid, while free chlorine dissolves and the quality of hydrochloric acid deteriorates.
To achieve the maximum degree of conversion of Cl 2 take the second reagent, H 2 , in large excess.
An increase in the degree of conversion of both components can be achieved if the reaction products are removed from the reaction zone by binding them into poorly dissociating, hardly soluble or non-volatile substances.
Effect of Temperature on Equilibrium
Experience shows that temperature has a great influence on the composition of an equilibrium mixture, increasing the content of reaction products in some reactions and decreasing it in others. Quantitatively, this dependence is reflected equations isobars(1.24) and isochores (1.25) Van't Hoff:
(1.24) 
(1.25)
It can be seen from these equations that the change in the equilibrium constant with increasing temperature (and hence the change in the yield of the reaction product) is determined by the sign of the thermal effect H and U:
H0 or U0 - endothermic reaction (with heat absorption). The right sides of the equations are greater than zero, which means that the derivatives are also greater than zero:
> 0;
> 0
Thus, the functions lnK p and lnK c (as well as K p and K c) increase with increasing temperature.
H0 or U0 - exothermic reaction (with heat release).
< 0;
< 0
The equilibrium constant decreases with increasing temperature, i.e. the content of the reaction products in the equilibrium mixture decreases, and the content of the initial substances increases.
Thus, an increase in temperature contributes to a more complete flow endothermic processes. We integrate the isobar equation.
Let Hf(Т) separate the variables and integrate,
;
(1.26)
As you can see, the equilibrium constant depends on temperature according to the exponential law: 
, and in coordinates ln K = f( 
) linear dependence (equation 1.26, figure 1.7)
|
|
|
Figure 1.7 - Temperature dependence of the equilibrium constant
Certain integration of the isobar equation gives:
(1.27)
Knowing the value of the equilibrium constant at any one temperature, you can find K p at any other with a known value of H.
Learning to solve problems on a mixture of organic substances
Generalization of the experience of teaching organic chemistry in specialized biological and chemical classes
One of the main criteria for mastering chemistry as an academic discipline is the ability of students to solve computational and qualitative problems. In the process of teaching in specialized classes with in-depth study of chemistry, this is of particular relevance, since all entrance exams in chemistry offer tasks of an increased level of complexity. The greatest difficulty in the study of organic chemistry is caused by the tasks of determining the quantitative composition of a multicomponent mixture of substances, the qualitative recognition of a mixture of substances, and the separation of mixtures. This is due to the fact that in order to solve such problems, it is necessary to deeply understand the chemical properties of the substances under study, be able to analyze, compare the properties of substances of different classes, and also have a good mathematical background. A very important point in learning is the generalization of information about the classes of organic substances. Let us consider the methodological methods of developing the ability of students to solve problems on a mixture of organic compounds.
hydrocarbons
- Where is what substance (qualitative composition)?
- How much substance is in the solution (quantitative composition)?
- How to separate the mixture?
STAGE 1. Summarizing knowledge about the chemical properties of hydrocarbons using a table(Table 1).
STEP 2. Solving quality problems.
Task 1. The gas mixture contains ethane, ethylene and acetylene. How to prove the presence of each of the gases in a given mixture? Write the equations for the necessary reactions.
Decision
Of the remaining gases, only ethylene will decolorize bromine water:
C 2 H 4 + Br 2 \u003d C 2 H 4 Br 2.
The third gas, ethane, burns:
2C 2 H 6 + 7O 2 4CO 2 + 6H 2 O.
Table 1
Chemical properties of hydrocarbons
| Reagent | Representatives of hydrocarbons | ||||||
|---|---|---|---|---|---|---|---|
| CH 3 CH 3 ethane | CH 2 \u003d CH 2 ethylene | CHCH acetylene | C 6 H 6 benzene | C 6 H 5 CH 3 toluene | C 6 H 5 CH \u003d CH 2 styrene | C 6 H 10 cyclohexene | |
| Br 2 (water) | – | + | + | – | – | + | + |
| KMnO 4 | – | + | + | – | + | + | + |
| Ag2O (solution in NH 3 aq.) |
– | – | + | – | – | – | – |
| Na | – | – | + | – | – | – | – |
| O2 | + | + | + | + | + | + | + |
Task 2. Isolate in pure form the components of the mixture consisting of acetylene, propene and propane. Write the equations for the necessary reactions.
Decision
When the mixture is passed through an ammonia solution of silver oxide, only acetylene is absorbed:
C 2 H 2 + Ag 2 O \u003d C 2 Ag 2 + HOH.
To regenerate acetylene, the resulting silver acetylenide is treated with hydrochloric acid:
C 2 Ag 2 + 2HCl \u003d C 2 H 2 + 2AgCl.
When the remaining gases are passed through bromine water, propene will be absorbed:
C 3 H 6 + Br 2 \u003d C 3 H 6 Br 2.
For propene regeneration, the resulting dibromopropane is treated with zinc dust:
C 3 H 6 Br 2 + Zn \u003d C 3 H 6 + ZnBr 2.
STAGE 3. Solution of calculation problems.
Task 3. It is known that 1.12 l (n.o.) of a mixture of acetylene with ethylene in the dark completely binds with 3.82 ml of bromine (= 3.14 g/ml). How many times will the volume of the mixture decrease after passing it through an ammonia solution of silver oxide?
Decision
Both components of the mixture react with bromine. Let's compose the reaction equations:
C 2 H 4 + Br 2 \u003d C 2 H 4 Br 2,
C 2 H 2 + 2Br 2 \u003d C 2 H 2 Br 4.
Let us denote the amount of ethylene substance through X mol, and the amount of acetylene substance through
y mol. It can be seen from the chemical equations that the amount of the reacting bromine substance will be in the first case X mole, and in the second - 2 y mol. Amount of gas mixture substance:
= V/V M \u003d 1.12 / 22.4 \u003d 0.05 mol,
and the amount of bromine substance:
(Br2) = V/M\u003d 3.82 3.14 / 160 \u003d 0.075 mol.
Let us compose a system of equations with two unknowns:
Solving the system, we get that the amount of ethylene substance in the mixture is equal to the amount of acetylene substance (0.025 mol each). Only acetylene reacts with an ammonia solution of silver, therefore, when the gas mixture is passed through a solution of Ag 2 O, the volume of the gas will decrease exactly by a factor of two.
Task 4. The gas released during the combustion of a mixture of benzene and cyclohexene was passed through an excess of barite water. This gave 35.5 g of sediment. Find the percentage composition of the initial mixture if the same amount of it can discolor 50 g of a solution of bromine in carbon tetrachloride with a mass fraction of bromine of 3.2%.
Decision
C 6 H 10 + Br 2 \u003d C 6 H 10 Br 2.
The amount of cyclohexene substance is equal to the amount of bromine substance:
(Br2) = m/M= 0.032 50/160 = 0.01 mol.
The mass of cyclohexene is 0.82 g.
Let us write the equations for the reactions of hydrocarbon combustion:
C 6 H 6 + 7.5O 2 \u003d 6CO 2 + 3H 2 O,
C 6 H 10 + 8.5O 2 \u003d 6CO 2 + 5H 2 O.
0.01 mol of cyclohexene forms 0.06 mol of carbon dioxide when burned. The released carbon dioxide forms a precipitate with barite water according to the equation:
CO 2 + Ba (OH) 2 \u003d BaCO 3 + H 2 O.
The amount of substance of the sediment of barium carbonate (BaCO 3) \u003d m/M\u003d 35.5 / 197 \u003d 0.18 mol is equal to the amount of substance of all carbon dioxide.
The amount of carbon dioxide substance formed during the combustion of benzene is:
0.18 - 0.06 \u003d 0.12 mol.
Using the benzene combustion reaction equation, we calculate the amount of benzene substance - 0.02 mol. The mass of benzene is 1.56 g.
Weight of the whole mixture:
0.82 + 1.56 = 2.38 g
The mass fractions of benzene and cyclohexene are 65.5% and 34.5%, respectively.
Oxygen-containing
organic compounds
Solving problems on mixtures in the topic "Oxygen-containing organic compounds" occurs in a similar way.
STAGE 4. Compilation of a comparative summary table(Table 2).
STEP 5. Recognition of substances.
Task 5. Use qualitative reactions to prove the presence of phenol, formic acid and acetic acid in this mixture. Write the reaction equations, indicate the signs of their occurrence.
Decision
Of the components of the mixture, phenol reacts with bromine water to form a white precipitate:
C 6 H 5 OH + 3Br 2 \u003d C 6 H 2 Br 3 OH + 3HBr.
The presence of formic acid can be established using an ammonia solution of silver oxide:
HCOOH + 2Ag (NH 3) 2 OH \u003d 2Ag + NH 4 HCO 3 + 3NH 3 + HOH.
Silver is released in the form of a precipitate or a mirror coating on the walls of the test tube.
If, after adding an excess of ammonia solution of silver oxide, the mixture boils with a solution of baking soda, then it can be argued that acetic acid is present in the mixture:
CH 3 COOH + NaHCO 3 \u003d CH 3 COOHa + CO 2 + H 2 O.
table 2
Chemical properties of oxygen-containing
organic matter
| Reagent | Representatives of oxygen-containing compounds | ||||||
|---|---|---|---|---|---|---|---|
| CH 3 OH methanol | C 6 H 5 OH phenol | HCHO methanal | HCOOH formic acid | CH 3 CHO acet- aldehyde |
HCOCH 3 methyl- formate |
C 6 H 12 O 6 glucose | |
| Na | + | + | – | + | – | – | + |
| NaOH | – | + | – | + | – | + | – |
| NaHCO3 | – | – | – | + | – | – | – |
| Ba 2 (water) | – | + | + | + | + | + | + |
| Ag2O (solution in NH 3 aq.) |
– | – | + | + | + | + | + |
Task 6. Four unlabeled tubes contain ethanol, acetaldehyde, acetic acid, and formic acid. What reactions can be used to distinguish substances in test tubes? Write reaction equations.
Decision
Analyzing the features of the chemical properties of these substances, we come to the conclusion that to solve the problem, one should use a solution of sodium bicarbonate and an ammonia solution of silver oxide. Acetaldehyde reacts only with silver oxide, acetic acid only with sodium bicarbonate, and formic acid with both. A substance that does not react with any of the reagents is ethanol.
Reaction equations:
CH 3 CHO + 2Ag (NH 3) 2 OH \u003d CH 3 COOHNH 4 + 2Ag + 3NH 3 + HOH,
CH 3 COOH + NaHCO 3 \u003d CH 3 COOHa + CO 2 + HOH,
HCOOH + 2Ag (NH 3) 2 OH \u003d 2Ag + NH 4 HCO 3 + 3NH 3 + HOH,
HCOOH + NaHCO 3 \u003d HCOOHa + CO 2 + HOH.
STEP 6. Determination of the quantitative composition of the mixture.
Task 7. To neutralize 26.6 g of a mixture of acetic acid, acetaldehyde and ethanol, 44.8 g of a 25% potassium hydroxide solution was used. When the same amount of the mixture interacted with an excess of metallic sodium, 3.36 liters of gas were released at n.o. Calculate the mass fractions of substances in this mixture.
Decision
Acetic acid and ethanol will react with metallic Na, and only acetic acid will react with KOH. Let's compose the reaction equations:
CH 3 COOH + Na \u003d CH 3 COONa + 1 / 2H 2, (1)
C 2 H 5 OH + Na \u003d C 2 H 5 ONa + 1/2H 2, (2)
Task 8. A mixture of pyridine and aniline weighing 16.5 g was treated with 66.8 ml of 14% hydrochloric acid (= 1.07 g/ml). To neutralize the mixture, it was necessary to add 7.5 g of triethylamine. Calculate the mass fractions of salts in the resulting solution.
Decision
Let's compose the reaction equations:
C 5 H 5 N + HCl \u003d (C 5 H 5 NH) Cl,
C 6 H 5 NH 2 + HCl \u003d (C 6 H 5 NH 3) Cl,
(C 2 H 5) 3 N + Hcl \u003d ((C 2 H 5) 3 NH) Cl.
Calculate the amount of substances - participants in the reactions:
(HCl) = 0.274 mol,
((C 2 H 5) 3 N) = 0.074 mol.
0.074 mol of acid was also spent on the neutralization of triethylamine, and for the reaction with the mixture: 0.274 - 0.074 = 0.2 mol.
We use the same technique as in Problem 3. Denote X is the number of moles of pyridine and y is the number of aniline in the mixture. Let's make a system of equations:
Solving the system, we get that the amount of pyridine is 0.15 mol, and aniline is 0.05 mol. Let's calculate the amounts of substances of hydrochloric salts of pyridine, aniline and triethylamine, their masses and mass fractions. They are respectively 0.15 mol, 0.05 mol, 0.074 mol; 17.33 g, 6.48 g, 10.18 g; 18.15%, 6.79%, 10.66%.
LITERATURE
Kuzmenko N.E., Eremin V.V. Chemistry. 2400 tasks for schoolchildren and university applicants. Moscow: Bustard, 1999;
Ushkalova V.N., Ioanidis N.V.. Chemistry: competitive tasks and answers. Allowance for entering universities. M.: Education, 2000.
