Self cleaning. The main processes of water self-purification in a water body

One of the most valuable properties of natural waters is their ability to self-purify. Self-purification of waters is their restoration natural properties in rivers, lakes and other water bodies, occurring naturally as a result of interrelated physicochemical, biochemical and other processes (turbulent diffusion, oxidation, sorption, adsorption, etc.). The ability of rivers and lakes to self-purify is closely related to many other natural factors, in particular, physical and geographical conditions, solar radiation, the activity of microorganisms in water, the influence of aquatic vegetation, and especially the hydrometeorological regime. The most intensive self-purification of water in reservoirs and streams is carried out in warm period years when biological activity in aquatic ecosystems is greatest. It flows faster on rivers with fast current and dense thickets of reeds, reeds and cattail along their banks, especially in the forest-steppe and steppe zones country. A complete change of water in rivers takes an average of 16 days, swamps - 5 years, lakes - 17 years.

Reducing the concentration of polluting water bodies is not organic matter occurs by neutralizing acids and alkalis due to the natural buffering of natural waters, the formation of sparingly soluble compounds, hydrolysis, sorption and precipitation. The concentration of organic substances and their toxicity are reduced due to chemical and biochemical oxidation. These natural methods of self-purification are reflected in the accepted methods of purification of polluted waters in industry and agriculture.

To maintain in reservoirs and watercourses the necessary natural quality waters of great importance is the distribution of aquatic vegetation, which plays the role of a kind of biofilter in them. The high cleansing power of aquatic plants is widely used in many industrial enterprises both in our country and abroad. For this, various artificial sedimentation tanks are created, in which lake and marsh vegetation is planted, which cleans polluted water well.

In recent years, artificial aeration has become widespread - one of the effective ways to purify polluted waters, when the self-purification process is sharply reduced when oxygen dissolved in water is deficient. To do this, special aerators are installed in reservoirs and streams or at aeration stations before the discharge of polluted water.

Protection of water resources from pollution.

Security water resources is to prohibit the discharge of untreated water into water bodies and streams, create water protection zones, promote self-purification processes in water bodies, preserve and improve the conditions for the formation of surface and groundwater runoff in watersheds.

Several decades ago, rivers, thanks to their self-purifying function, coped with water purification. Now, in the most populated areas of the country, as a result of the construction of new cities and industrial enterprises, water use sites are located so densely that often places of wastewater discharge and water intakes are practically nearby. Therefore, the development and implementation of effective methods of purification and post-treatment of wastewater, purification and neutralization of tap water is receiving more and more attention. In some enterprises, water related operations are playing an increasingly important role. Particularly high are the costs of water supply, treatment and disposal of wastewater in the pulp and paper, mining and petrochemical industries.

Sequential wastewater treatment at modern enterprises involves primary, mechanical treatment (easily settling and floating substances are removed) and secondary, biological (biologically degradable organic substances are removed). In this case, coagulation is carried out - to precipitate suspended and colloidal substances, as well as phosphorus, adsorption - to remove dissolved organic substances and electrolysis - to reduce the content of dissolved substances of organic and mineral origin. Disinfection of wastewater is carried out by means of their chlorination and ozonation. An important element of the technological process of cleaning is the removal and disinfection of the formed sludge. In some cases, the final operation is the distillation of water.

The most advanced modern treatment facilities ensure the release of wastewater from organic pollution only by 85-90%, and only in some cases - by 95%. Therefore, even after cleaning it is necessary to dilute them 6-12-fold, and often even more. clean water to preserve the normal functioning of aquatic ecosystems. The fact is that the natural self-cleaning capacity of reservoirs and streams is very small. Self-purification occurs only if the discharged waters have been completely purified, and in the water body they have been diluted with water in a ratio of 1:12-15. If, however, large volumes of wastewater enter reservoirs and watercourses, and even more so untreated, the stable natural balance of aquatic ecosystems is gradually lost, and their normal functioning is disrupted.

Recently, more and more effective methods of purification and post-treatment of wastewater after their biological treatment using the latest methods of wastewater treatment: radiation, electrochemical, sorption, magnetic, etc. Improving the technology of wastewater treatment, further increasing the degree of purification are the most important tasks in the field of water protection from pollution.

Much more extensive use should be made of post-treatment of treated wastewater in agricultural irrigation fields. In the post-treatment of wastewater at the ZPO, funds are not spent on their industrial post-treatment, it creates the opportunity to obtain additional agricultural products, water is significantly saved, since the intake of fresh water for irrigation is reduced and there is no need to spend water to dilute wastewater. When used at the WPO, urban wastewater, the nutrients and microelements contained in them are absorbed by plants faster and more completely than artificial mineral fertilizers.

Prevention of pollution of water bodies with pesticides and pesticides is also one of the important tasks. This requires speeding up the implementation of anti-erosion measures, creating pesticides that would decompose within 1-3 weeks without preserving toxic residues in the culture. Until these issues are resolved, it is necessary to limit the agricultural use of coastal areas along watercourses or not to use pesticides in them. The creation of water protection zones also requires more attention.

In protecting water sources from pollution, the introduction of a fee for wastewater discharge, the creation of integrated regional schemes for water consumption, water disposal and wastewater treatment, and automation of water quality control in water sources are of great importance. It should be noted that integrated district schemes make it possible to switch to the reuse and reuse of water, the operation of treatment facilities common to the district, as well as to automate the processes of managing the operation of water supply and sewerage.

In preventing pollution of natural waters, the role of protecting the hydrosphere is important, since the negative properties acquired by the hydrosphere not only modify the aquatic ecosystem and have a depressing effect on its hydrobiological resources, but also destroy land ecosystems, its biological systems, as well as the lithosphere.

It should be emphasized that one of the radical measures to combat pollution is to overcome the ingrained tradition of considering water bodies as wastewater receivers. Where possible, either water abstraction or wastewater discharge should be avoided in the same streams and reservoirs.

Security atmospheric air and soil.

specially protected natural areas. Animal protection and flora.

effective form protection of natural ecosystems, as well as biotic communities are specially protected natural areas. They allow you to save standards (samples) of untouched biogeocenoses, and not only in some exotic, rare places, but also in all typical natural zones of the Earth.

TO specially protected natural areas(SPNA) includes areas of land or water surface, which, due to their environmental and other significance, are completely or partially withdrawn from economic use by decisions of the Government.

The Law on Protected Areas, adopted in February 1995, established the following categories of these territories: a) state nature reserves, incl. biospheric; b) national parks; in) natural parks; d) state nature reserves; e) monuments of nature; f) dendrological parks and botanical gardens.

Reserve- this is a space (territory or water area) specially protected by law, which is completely withdrawn from normal economic use in order to preserve the natural complex in its natural state. Only scientific, security and control activities are allowed in the reserves.

Today in Russia there are 95 nature reserves with a total area of ​​310 thousand square meters. km, which is about 1.5% of the entire territory of Russia. In order to neutralize the technogenic impact of the adjacent territories, especially in areas with developed industry, protected areas are created around the reserves.

Biosphere reserves (BR) perform four functions: the preservation of the genetic diversity of our planet; conducting scientific research; tracking the background state of the biosphere (environmental monitoring); environmental education and international cooperation.

Obviously, the functions of the BR are wider than the functions of any other type of protected natural areas. They serve as a kind of international standards, standards of the environment.

A unified global network of more than 300 biosphere reserves has now been created on Earth (11 in Russia). All of them work according to the coordinated program of UNESCO, conducting constant monitoring of changes in the natural environment under the influence of anthropogenic activities.

national park- a vast territory (from several thousand to several million hectares), which includes both fully protected areas and areas intended for certain types of economic activity.

The goals of creation national parks are: 1) ecological (preservation of natural ecosystems); 2) scientific (development and implementation of methods for preserving the natural complex in conditions of mass admission of visitors) and 3) recreational (regulated tourism and recreation for people).

There are 33 national parks in Russia with a total area of ​​about 66.5 thousand square meters. km.

Nature Park- a territory that has a special ecological and aesthetic value and is used for organized recreation of the population.

Reserve- a natural complex, which is intended for the conservation of one or more species of animals or plants with limited use of others. There are landscape, forest, ichthyological (fish), ornithological (birds) and other types of reserves. Usually, after the restoration of the density of the population of protected species of animals or plants, the reserve is closed and one or another type of economic activity is allowed. In Russia today there are more than 1,600 state natural reserves with a total area of ​​over 600 thousand square meters. km.

natural monument- separate natural objects, which are unique and irreproducible, having scientific, aesthetic, cultural or educational value. These can be very old trees that were “witnesses” to some historical events, caves, rocks, waterfalls, etc. There are about 8 thousand of them in Russia, while on the territory where the monument is located, any activity that can destroy them is prohibited .

Dendrological parks and botanical gardens are collections of trees and shrubs created by man in order to both preserve biodiversity and enrich the flora, and in the interests of science, study, and cultural and educational work. They often carry out work related to the introduction and acclimatization of new plants.

For violation of the regime of specially protected natural areas, Russian legislation establishes administrative and criminal liability. At the same time, scientists and experts strongly recommend a significant increase in the area of ​​specially protected areas. So, for example, in the United States, the area of ​​the latter is more than 7% of the country's territory.

The solution of environmental problems, and, consequently, the prospects for the sustainable development of civilization, is largely associated with the competent use of renewable resources and various functions of ecosystems, and their management. This direction is the most important way of a sufficiently long and relatively inexhaustible use of nature, combined with the preservation and maintenance of the stability of the biosphere, and, consequently, the human environment.

Each species is unique. It contains information about the development of flora and fauna, which is of great scientific and applied importance. Since all the possibilities of using a given organism in the long term are often unpredictable, the entire gene pool of our planet (with the possible exception of some pathogenic organisms dangerous to humans) is subject to strict protection. The need to protect the gene pool from the standpoint of the concept of sustainable development (“co-evolution”) is dictated not so much by economic as by moral and ethical considerations. Humanity alone will not survive.

It is useful to recall one of B. Commoner's environmental laws: "Nature knows best!" Until recently, the possibilities of using the gene pool of animals that were unforeseen are now being demonstrated by bionics, thanks to which there are numerous improvements in engineering structures based on the study of the structure and functions of the organs of wild animals. It has been established that some invertebrates (mollusks, sponges) have the ability to accumulate a large amount of radioactive elements and pesticides. As a result, they can be bioindicators of environmental pollution and help humans solve this important problem.

Protection of the plant gene pool. Being integral part of the general problem of protecting the OPS, the protection of the plant gene pool is a set of measures to preserve the entire species diversity of plants - carriers of the hereditary heritage of productive or scientifically or practically valuable properties.

It is known that under the influence of natural selection and through sexual reproduction of individuals in the gene pool of each species or population, the most useful properties for the species are accumulated; they are in gene combinations. Therefore, the tasks of using natural flora are of great importance. Our modern grain, fruit, vegetable, berry, fodder, industrial, ornamental crops, the centers of origin of which were established by our outstanding compatriot N.I. Vavilov, lead their genealogy either from wild ancestors, or are creations of science, but based on natural gene structures. By using the hereditary properties of wild plants, completely new types of useful plants have been obtained. Through hybrid selection, perennial wheat and grain fodder hybrids were created. According to scientists, about 600 species of wild plants can be used in the selection of agricultural crops from the flora of Russia.

The protection of the gene pool of plants is carried out by creating reserves, natural parks, botanical gardens; formation of a gene pool of local and introduced species; study of biology, ecological needs and competitiveness of plants; ecological assessment of the plant habitat, forecasts of its changes in the future. Thanks to the reserves, Pitsunda and Eldar pines, pistachio, yew, boxwood, rhododendron, ginseng, etc. have been preserved.

Protection of the gene pool of animals. The change in living conditions under the influence of human activity, accompanied by direct persecution and extermination of animals, leads to their depletion. species composition and decline in the abundance of many species. In 1600 there were approximately 4230 species of mammals on the planet, by our time 36 species have disappeared, and 120 species are in danger of extinction. Of the 8684 bird species, 94 have disappeared and 187 are endangered. The situation with subspecies is no better: since 1600, 64 subspecies of mammals and 164 subspecies of birds have disappeared, 223 subspecies of mammals and 287 subspecies of birds are endangered.

Protection of the human gene pool. For this, various scientific directions have been created, such as:

1) ecotoxicology- a section of toxicology (the science of poisons), which studies the ingredient composition, distribution features, biological effects, activation, deactivation of harmful substances in environment;

2) medical genetic counseling in special medical institutions to determine the nature and consequences of the action of ecotoxicants on the human genetic apparatus in order to give birth to healthy offspring;

3) screening- selection and testing for mutagenicity and carcinogenicity of environmental factors (human environment).

Environmental pathology- the doctrine of human diseases, in the occurrence and development of which the leading role is played by unfavorable environmental factors in combination with other pathogenic factors.

Principal directions of environmental protection.

Regulation of environmental quality. Protection of the atmosphere, hydrosphere, lithosphere, biotic communities. Eco-protection equipment and technologies.

Self-purification of water bodies

Between the components of the aquatic ecosystem in the process of its functioning, there is a continuous exchange of matter and energy. This exchange is cyclical varying degrees isolation, accompanied by the transformation of organic matter, in particular phenols under the influence of physical, chemical and biological factors. In the course of transformation, complex organic substances can be gradually decomposed into simple ones, and simple substances can be synthesized into complex ones. Depending on the intensity of the external impact on the aquatic ecosystem and the nature of the processes, either the aquatic ecosystem is restored to the background conditions (self-purification), or the aquatic ecosystem passes to another steady state, which will be characterized by different quantitative and qualitative indicators of biotic and abiotic components. If the external impact exceeds the self-regulating capabilities of the aquatic ecosystem, it may be destroyed.

Self-purification of natural waters is carried out due to the involvement of water coming from external sources substances into continuously ongoing transformation processes, as a result of which the substances received are returned to their reserve fund.

The transformation of substances is the result of various simultaneously operating processes, among which physical, chemical and biological mechanisms can be distinguished. The value of the contribution of each of the mechanisms depends on the properties of the impurity and the characteristics of a particular ecosystem.

Biochemical self-purification.

Biochemical self-purification is a consequence of the transformation of substances carried out by hydrobionts. As a rule, biochemical mechanisms make the main contribution to the process of self-purification, and only when aquatic organisms are inhibited (for example, under the influence of toxicants), do physicochemical processes begin to play a more significant role. Biochemical transformation of organic substances occurs as a result of their inclusion in food webs and is carried out during the processes of production and destruction.

Especially important role plays the role of primary production, since it determines the majority of intra-water processes. The main mechanism of new formation of organic matter is photosynthesis. In most aquatic ecosystems, phytoplankton is a key primary producer. In the process of photosynthesis, the energy of the Sun is directly transformed into biomass. The by-product of this reaction is free oxygen formed by the photolysis of water. Along with photosynthesis in plants, there are processes of respiration with the consumption of oxygen.

Chemical mechanisms of self-purification.

Photolysis is the transformation of molecules of a substance under the action of the light they absorb. Particular cases of photolysis are photochemical dissociation - the decay of particles into several simpler ones and photoionization - the transformation of molecules into ions. From total solar radiation about 1% is used in photosynthesis, from 5% to 30% is reflected by the water surface. The main part of solar energy is converted into heat and participates in photochemical reactions. The most effective part of sunlight is ultraviolet radiation. Ultraviolet radiation is absorbed in a water layer about 10 cm thick, however, due to turbulent mixing, it can also penetrate into deeper layers of water bodies. The amount of a substance subjected to photolysis depends on the type of substance and its concentration in water. Of the substances entering water bodies, humus substances are subject to relatively rapid photochemical decomposition.

Hydrolysis is an ion exchange reaction between various substances and water. Hydrolysis is one of the leading factors in the chemical transformation of organic substances in water bodies. The quantitative characteristic of this process is the degree of hydrolysis, which is understood as the ratio of the hydrolyzed part of the molecules to the total salt concentration. For most salts, it is a few percent and increases with increasing dilution and water temperature. Organic substances are also subject to hydrolysis. In this case, hydrolytic cleavage most often occurs through the bond of a carbon atom with other atoms.

One of the effective ways of self-purification is the transformation of a pollutant due to redox reactions when interacting with redox components. aquatic environment.

The possibility of Red-Ox transformations in the system is characterized by the value of its redox potential (Eh). The E h value of natural waters is affected by free O 2 , H 2 O 2 , Fe 2+ , Fe 3+ , Mn 2+ , Mn 4+ , ​​H + , organic compounds and other "potential-setting components". In natural waters, E h usually ranges from +0.7 to -0.5V. Surface and ground waters saturated with oxygen are most often characterized by an E h interval from +0.150 to +0.700V. Studies show that in the processes of self-purification of natural water bodies from phenols, redox transformations with the participation of H 2 O 2 of natural origin and metal ions of variable valence present in water bodies play an important role. In natural water, the stationary concentration of H 2 O 2 is in the range of 10 -6 - 10 -4 mol/l. Hydrogen peroxide is formed due to photochemical and oxidative processes involving molecular oxygen in a homogeneous medium. Since the decay of H 2 O 2 mainly determines the catalytic amounts of metal ions and sunlight, its rate is almost independent of the initial concentration.

Physical mechanisms of self-purification.

Gas exchange at the "atmosphere-water" interface. Thanks to this process, substances that have a reserve fund in the atmosphere enter the water body and return these substances from the water body to the reserve fund. One of the important special cases of gas exchange is the process of atmospheric reaeration, due to which a significant part of oxygen enters the water body. The intensity and direction of gas exchange are determined by the deviation of the gas concentration in water from the saturation concentration C. The saturation concentration depends on the nature of the substance and the physical conditions in water body- temperature and pressure. At concentrations greater than C, the gas escapes into the atmosphere, and at concentrations less than C s , the gas is absorbed by the water mass.

Sorption is the absorption of impurities by suspended matter, bottom sediments and surfaces of hydrobiont bodies. Colloidal particles and organic substances, such as phenols, which are in the undissociated molecular state, are sorbed most vigorously. The process is based on the phenomenon of adsorption. The rate of accumulation of a substance per unit mass of the sorbent is proportional to its unsaturation with respect to the given substance and the concentration of the substance in water, and is inversely proportional to the content of the substance in the sorbent.

Sedimentation and resuspension. Water bodies always contain a certain amount of suspended solids of inorganic and organic origin. Sedimentation is characterized by the ability of suspended particles to fall to the bottom under the action of gravity. The process of transition of particles from bottom sediments to a suspended state is called resuspension. It occurs under the action of the vertical component of the turbulent flow velocity.

Thus, sorption and redox processes play an important role in the self-purification of natural reservoirs.

Self-purification of water in reservoirs is a set of interrelated hydrodynamic, physicochemical, microbiological and hydrobiological processes leading to the restoration of the original state of a water body.

Among the physical factors, the dilution, dissolution and mixing of incoming contaminants is of paramount importance. Good mixing and reduction of suspended solids concentrations is ensured by the rapid flow of the rivers. It contributes to the self-purification of water bodies by settling to the bottom of insoluble sediments, as well as settling polluted waters. In areas with temperate climate the river is self-cleaning after 200-300 km from the place of pollution, and on Far North- after 2 thousand km.

Disinfection of water occurs under the influence of ultraviolet radiation from the sun. The effect of disinfection is achieved by the direct destructive effect of ultraviolet rays on protein colloids and enzymes of the protoplasm of microbial cells, as well as spore organisms and viruses.

Of the chemical factors of self-purification of water bodies, it should be noted the oxidation of organic and inorganic substances. Self-purification of a water body is often assessed in relation to easily oxidized organic matter or in terms of the total content of organic substances.

The sanitary regime of a reservoir is characterized primarily by the amount of oxygen dissolved in it. It should beat at least 4 mg per 1 liter of water at any time of the year for reservoirs for reservoirs of the first and second types. The first type includes water bodies used for drinking water supply of enterprises, the second - used for swimming, sporting events, as well as those located within the boundaries of settlements.

TO biological factors self-purification of the reservoir include algae, molds and yeasts. However, phytoplankton does not always have a positive effect on self-purification processes: in some cases, the mass development of blue-green algae in artificial reservoirs can be seen as a process of self-pollution.

Representatives of the animal world can also contribute to the self-purification of water bodies from bacteria and viruses. Thus, the oyster and some other amoeba adsorb intestinal and other viruses. Each mollusk filters more than 30 liters of water per day.

The purity of reservoirs is unthinkable without the protection of their vegetation. Only on the basis of a deep knowledge of the ecology of each reservoir, effective control The development of various living organisms inhabiting it can achieve positive results, ensure transparency and high biological productivity of rivers, lakes and reservoirs.

Other factors also adversely affect the processes of self-purification of water bodies. Chemical pollution of water bodies with industrial wastewater, biogenic elements (nitrogen, phosphorus, etc.) inhibits natural oxidative processes and kills microorganisms. The same applies to the discharge of thermal wastewater from thermal power plants.

multi-stage process, sometimes stretching for long time- self-cleaning from oil. IN natural conditions the complex of physical processes of self-purification of water from oil consists of a number of components: evaporation; settling of lumps, especially those overloaded with sediment and dust; adhesion of lumps suspended in the water column; floating lumps forming a film with inclusions of water and air; reducing the concentration of suspended and dissolved oil due to settling, floating and mixing with clean water. The intensity of these processes depends on the properties of a particular type of oil (density, viscosity, coefficient thermal expansion), the presence of colloids in the water, suspended and entrained plankton particles, etc., air temperature and from sunlight.

Pollution entering the reservoir causes a violation of the natural balance in it. The ability of a reservoir to resist this disturbance, to get rid of the pollution introduced, is the essence of the self-purification process.

Self-purification of water systems is due to many natural and sometimes man-made factors. These factors include various hydrological, hydrochemical and hydrobiological processes. Conventionally, three types of self-purification can be distinguished: physical, chemical, biological.

Among physical processes, dilution (mixing) is of paramount importance. Good mixing and a reduction in the concentration of suspended particles is ensured by the intensive flow of rivers. Contributes to the self-purification of water bodies by settling polluted waters and settling to the bottom of insoluble sediments, sorption of pollutants by suspended particles and bottom sediments. For volatile substances, evaporation is an important process.

Among chemical factors self-purification of reservoirs leading role plays the oxidation of organic and inorganic substances. Oxidation occurs in water with the participation of oxygen dissolved in it, therefore, the higher its content, the faster and better the process of mineralization of organic residues and self-purification of the reservoir proceeds. With severe pollution of the reservoir, the reserves of dissolved oxygen are quickly consumed, and its accumulation due to the physical processes of gas exchange with the atmosphere proceeds slowly, which slows down self-purification. Self-purification of water can also occur as a result of some other reactions in which hardly soluble, volatile or non-toxic substances are formed, for example, hydrolysis of pesticides, neutralization reactions, etc. Calcium and magnesium carbonates and bicarbonates contained in natural water neutralize acids, and carbonic acid dissolved in water neutralizes alkalis.

Under the influence of ultraviolet radiation of the sun in the surface layers of the reservoir, photodecomposition of some chemicals, such as DDT, and water disinfection occur - the death of pathogenic bacteria. The bactericidal action of ultraviolet rays is explained by their influence on the protoplasm and enzymes of microbial cells, which causes their death. Ultraviolet rays have a detrimental effect on vegetative forms of bacteria, fungal spores, protozoan cysts, and viruses.

Each body of water is complex living system where bacteria, algae, higher aquatic plants, various invertebrates live. The processes of metabolism, bioconcentration, biodegradation lead to a change in the concentration of pollutants. The biological factors of self-purification of the reservoir also include algae, molds and yeast fungi, however, in some cases mass development blue-green algae in artificial reservoirs can be considered as a process of self-pollution. Representatives of the animal world can also contribute to the self-purification of water bodies from bacteria and viruses. So, oysters and some amoeba adsorb intestinal and other viruses. Each mollusk filters more than 30 liters of water per day. Common reed, narrow-leaved cattail, lake reed and other macrophytes are able to absorb from water not only relatively inert compounds, but also physiologically active substances such as phenols, poisonous salts of heavy metals.

The process of biological purification of water is associated with the content of oxygen in it. With a sufficient amount of oxygen, the activity of aerobic microorganisms that feed on organic substances is manifested. When organic matter is broken down, carbon dioxide and water are formed, as well as nitrates, sulfates, and phosphates. Biological self-purification is the main link in the process and is considered as one of the manifestations of the biotic cycle in a reservoir.

The contribution of individual processes to the ability of the natural aquatic environment to self-purify depends on the nature of the pollutant. For the so-called conservative substances that do not decompose or decompose very slowly (metal ions, mineral salts, persistent organochlorine pesticides, radionuclides, etc.), self-purification has an apparent character, since only the redistribution and dispersion of the pollutant in the environment occurs, pollution adjacent objects to them. The decrease in their concentration in water occurs due to dilution, removal, sorption, bioaccumulation. With regard to biogenic substances, biochemical processes are most important. For water-soluble substances not involved in biological cycle, the reactions of their chemical and microbiological transformation are important.

For most organic compounds and some inorganic substances, microbiological transformation is considered one of the main ways of self-purification of the natural aquatic environment. Microbiological biochemical processes include reactions of several types. These are reactions involving redox and hydrolytic enzymes (oxidases, oxygenases, dehydrogenases, hydrolases, etc.). Biochemical self-purification of water bodies depends on many factors, among which the most important are temperature, active reaction of the environment (pH), and the content of nitrogen and phosphorus. Optimum temperature for the course of biodegradation processes is 25-30ºС. Great importance for the vital activity of microorganisms has a reaction of the environment, which affects the course of enzymatic processes in the cell, as well as a change in the degree of penetration of nutrients into the cell. For most bacteria, a neutral or slightly alkaline reaction of the medium is favorable. At pH<6 развитие и жизнедеятельность микробов чаще всего снижается, при рН <4 в некоторых случаях их жизнедеятельность прекращается. То же самое наблюдается при повышении щелочности среды до рН>9,5.

Task number 6

SELF-PURIFICATION PROCESSES OF NATURAL WATERS

1 TYPES OF POLLUTION AND THEIR EFFECTS

(CHANNELS FOR SELF-CLEANING WATER ENVIRONMENT)

Under the self-purification of the aquatic environment understand the totality of physical, biological and chemical inland processes aimed at reducing the content of pollutants (pollutants).

The contribution of individual processes to the ability of the natural aquatic environment to self-purify depends on the nature of pollutants. In accordance with this, pollutants are conditionally divided into three groups.

one). Preservative substances - non-degradable or biodegradable very slowly . These are mineral salts, hydrophobic compounds such as organochlorine pesticides, oil and oil products. The decrease in the concentration of conservative substances in water damage occurs only due to dilution, physical processes of mass transfer, physicochemical processes of complexation, sorption and bioaccumulation. Self-purification has an apparent character, since there is only a redistribution and dispersion of pollutants in the environment, pollution of adjacent objects by it.

2). Biogenic substances - substances involved in the biological cycle. These are mineral forms of nitrogen and phosphorus, easily digestible organic compounds.

In this case, self-purification of the aquatic environment occurs due to biochemical processes.

3). Water-soluble substances that are not involved in the biological cycle, entering water bodies and streams from anthropogenic sources, are often toxic. Self-purification of the aquatic environment from these substances is carried out mainly due to their chemical and microbiological transformation.

The most significant processes for self-purification of the aquatic environment are the following processes:

– physical transfer processes: dilution (mixing), removal of pollutants to neighboring water bodies (downstream), sedimentation of suspended particles, evaporation, sorption (by suspended particles and bottom sediments), bioaccumulation;

– microbiological transformation;

–chemical transformation: sedimentation, hydrolysis, photolysis, redox reactions, etc.

2 DILUTION OF SAT AT WASTEWATER RELEASE

FROM WATER PURIFICATION FACILITIES

The mass of pollutants in wastewater is equal to the mass of pollutants in the mixed flow (wastewater + watercourse water). Material balance equation for pollutants:

Cct q + γ Q Cf = Cv (q + γ Q),

where Cst is the concentration of pollutants in waste water, g/m3 (mg/dm3);

q is the maximum flow rate of wastewater to be discharged into the watercourse, m3/s

γ - mixing ratio

Q is the average monthly flow rate of the watercourse, m3/s;

Cf is the background concentration of pollutants in the watercourse (established according to long-term observations), g/m3 (mg/dm3);

Cv - concentration of pollutants in the watercourse after mixing (dilution), g/m3 (mg/dm3);

From the material balance equation, one can find the concentration of pollutants in the watercourse after dilution:

Cv = https://pandia.ru/text/80/127/images/image002_20.png" width="117" height="73 src=">

L is the distance along the fairway of the watercourse (fairway is the deepest strip of a given body of water) from the point of release to the control point, m;

α is a coefficient depending on the hydraulic conditions of the flow. Coefficient α is calculated according to the equation:

where ξ is a coefficient depending on the location of the wastewater outlet into the watercourse: ξ = 1 for outlet near the shore, ξ = 1.5 when released into the fairway;

φ is the coefficient of the tortuosity of the watercourse, i.e. the ratio of the distance between the considered sections of the watercourse along the fairway to the distance along the straight line; D is the turbulent diffusion coefficient .

For lowland rivers and simplified calculations, the turbulent diffusion coefficient is found by the formula:

https://pandia.ru/text/80/127/images/image005_9.png" width="59 height=47" height="47">= X-in,

where ac, aw are the activities of substance A in the sorption layer and in the aqueous phase;

γc, γw are the activity coefficients of substance A in the sorption layer and in the aqueous phase;

Cs, Sv are the concentrations of substance A in the sorption layer and in the aqueous phase;

Кс-в - distribution coefficient of substance A (equilibrium constant

AB ↔ AC expressed in terms of concentrations).

Then for relatively constant factor activity of substance A in the sorption layer (organic phase):

X-in = Ka s-in DIV_ADBLOCK4">

This, in particular, determines the existence of a correlation between the distribution coefficients of substances in the system octanol - water and solid organic matter - water:

Ks-in ≈ 0.4 Ko-in ,

where Ko-v is the distribution coefficient of the substance in the octanol-water system.

The value of Ko-in is related to the solubility of a substance in water by a simple empirical relationship:

lg Ko-in = (4.5 ÷ 0.75) lg S,

where S is the solubility of the substance, expressed in mg/dm3.

This ratio holds for many classes of organic compounds, including hydrocarbons, halogenated hydrocarbons, aromatic acids, organochlorine pesticides, chlorinated biphenyls.

In natural sorbents, organic matter makes up only a certain fraction of the mass of the sorbent. Therefore, the distribution coefficient in the sorbent-water system Ks-v is normalized to the content of organic carbon in the sorbent Ks-v*:

Ks-in * \u003d Ks-in ω (C),

where ω(С) – mass fraction organic matter in the sorbent.

In this case, the proportion of the substance sorbed from the aqueous medium ωsorb is equal to:

ωsorb = https://pandia.ru/text/80/127/images/image009_9.png" width="103" height="59">,

where Csorb is the concentration of the sorbent suspended in water.

In bottom sediments, the Csorb value is significant; therefore, for many pollutants Ks-v*· Csorb >> 1, and the unit in the denominator can be neglected. The value of ωsorb tends to unity, i.e., all substance A will be in the sorbed state.

In open water bodies, the situation is different: the concentration of the suspended sorbent is extremely low. Therefore, sorption processes make a significant contribution to the self-purification of the reservoir only for compounds with Ks-v ≥ 105.

Sorption of many pollutants with a water solubility of 10-3 mol/l is one of the main removal processes chemical from the aqueous phase. These substances include organochlorine pesticides, polychlorinated biphenyls, PAHs. These compounds are slightly soluble in water and have big values Co-in (104 - 107). Sorption is the most effective way of self-purification of the aquatic environment from such substances.

4 MICROBIOLOGICAL SELF-CLEANING

Microbiological transformation of pollutants is considered one of the main channels of self-purification of the aquatic environment. . Microbiological biochemical processes include reactions of several types. These are reactions involving redox and hydrolytic enzymes. The optimal temperature for the processes of pollutant biodegradation is 25-30ºС.

The rate of microbiological transformation of a substance depends not only on its properties and structure, but also on the metabolic capacity of the microbial community..png" width="113" height="44 src=">,

where CS is the concentration of the substrate (pollutant), . Here keff is the rate constant of biolysis, .m is the biomass of microorganisms or the population size.

The kinetics of the pseudo-first order transformation of some pollutants at a fixed population size and the directly proportional growth of the rate constant with an increase in the number of bacteria have been experimentally proven in many cases. Moreover, in some cases, kef does not depend on the phase of population growth, on the locality and species composition of the microbial community.

When integrating the kinetic equation of the first order reaction, we obtain:

https://pandia.ru/text/80/127/images/image013_7.png" width="29" height="25 src="> – initial concentration of the substrate (or biochemically oxidizable substances, corresponding to BODtotal);

– current concentration of the substrate (or biochemically oxidizable substances, corresponding to BODtotal – BODτ).

When replacing https://pandia.ru/text/80/127/images/image014_8.png" width="29" height="25"> with the corresponding BOD value in the equation, we get:

.

Let us denote kB/2.303 = k*, where k* is the biochemical oxidation constant (has the dimension of the first-order reaction constant - day-1). When potentiating the equation, we have an equation relating BODtot. and BODτ, in exponential form:

Using this equation, one can determine the time of complete oxidation of biochemically oxidized substances - the time during which 99% of the substance is oxidized .

Under natural conditions of middle latitudes, as a result of microbiological processes, alkanes of a normal structure decompose most quickly (by 60-90% in three weeks). Branched alkanes and cycloalkanes decompose more slowly than n-alkanes - by 40% in a week, by 80% in three weeks. Low molecular weight benzene derivatives mineralize faster than saturated hydrocarbons (for example, phenols and cresols) . Substituted di - and trichlorophenols decompose completely in bottom sediments within a week, nitrophenols - within two to three weeks. However, PAHs are slowly degraded.

Biodegradation processes are influenced by many factors: lighting, dissolved oxygen content, pH , nutrient content, presence of toxicants, etc. . Even if microorganisms have a set of enzymes necessary for the destruction of pollutants, they may not show activity due to the lack of additional substrates or factors.

5 HYDROLYSIS

Many pollutants are weak acids or bases and are involved in acid-base transformations. Salts formed from weak bases or weak acids undergo hydrolysis . Salts formed by weak bases are hydrolyzed by the cation, salts formed by weak acids by the anion. HM, Fe3+, Al3+ cations undergo hydrolysis:

Fe3+ + HOH ↔ FeOH2+ + H+

Al3+ + HOH ↔ AlOH2+ + H+

Cu2+ + HOH ↔ CuOH+ + H+

Pb2+ + HOH ↔ PbOH+ + H+.

These processes cause acidification of the environment.

Anions of weak acids are hydrolyzed:

CO32- + HOH ↔ HCO3- + OH-

SiO32- + HOH ↔ HSiO3- + OH-

PO43- + HOH ↔ HPO42- + OH-

S2- + HOH ↔ HS- + OH-,

which contributes to the alkalization of the environment.

The simultaneous presence of hydrolyzable cations and anions in some cases causes complete irreversible hydrolysis, which can lead to the formation of precipitates of poorly soluble hydroxides Fe(OH)3, Al(OH)3, etc.

Hydrolysis of cations and anions proceeds rapidly, as it refers to ion exchange reactions.

Among organic compounds, esters and amides undergo hydrolysis. carboxylic acids and various phosphoric acids. In this case, water participates in the reaction not only as a solvent, but also as a reagent:

R1–COO–R2 + HOH ↔ R1–COOH + R2OH

R1–COO–NH2 + HOH ↔ R1–COOH + NH3

(R1O)(R2O)–P=O(OR3) + HOH ↔ H3PO4 + R1OH + R2OH + R3OH

As an example, dichlorvos (o,o-diethyl-2,2-dichlorovinyl phosphate) can be mentioned.

(C2H5O)2–P=O(O–CH=CCl2) + 2HOH ↔ (HO)2–P=O(O–CH=CCl2) + 2C2H5OH

Various organohalogen compounds are also hydrolyzed:

R–Cl + HOH ↔ R–OH + HCl;

R–C–Cl2 + 2HOH ↔ R–C–(OH)2 + 2HCl ↔ R–C=O + H2O + 2HCl;

R–C–Cl3 + 3HOH ↔ R–C–(OH)3 + 3HCl ↔ R–COOH + 2H2O + 3HCl.

These hydrolytic processes take place on a different time scale. Hydrolysis reactions can be carried out both without a catalyst and with the participation of acids and bases dissolved in natural waters as catalysts. Accordingly, the hydrolysis rate constant can be represented as:

where https://pandia.ru/text/80/127/images/image020_5.png" width="12" height="19"> – rate constants of acid hydrolysis, hydrolysis in neutral medium and alkaline hydrolysis;

In this case, hydrolysis can be considered a pseudo-first order reaction, since pollutants are present in natural waters in trace amounts. The concentration of water in comparison with their concentrations is much higher and is practically considered unchanged.

To determine the concentration of a pollutant that changes over time, a first-order kinetic reaction equation is used:

where C0 – initial concentration of the pollutant;

FROM – current concentration of the pollutant;

τ – the time elapsed from the start of the reaction;

k – reaction (hydrolysis) rate constant.

The degree of conversion of the pollutant (the proportion of the substance that entered into the reaction) can be calculated by the equation:

β = (С0 – С)/С0 = 1– e-kτ.

6 EXAMPLES OF SOLVING PROBLEMS

Example 1 Calculate the concentration of Fe3+ iron ions in river water at a distance of 500 m from the wastewater outlet, if its concentration in the wastewater at the outlet to the reservoir is 0.75 mg/dm3. The speed of the river flow is 0.18 m/s, the volume flow is 62 m3/s, the depth of the river is 1.8 m, the coefficient of the river sinuosity is 1.0. Wastewater is discharged from the shore. The volume flow of wastewater is 0.005 m3/s. The background concentration of Fe3+ is 0.3 mg/dm3.

Solution:

The turbulent diffusion coefficient is

https://pandia.ru/text/80/127/images/image025_3.png" width="147" height="43">.

The coefficient α according to the condition of the problem (the coefficient taking into account the conditions for the discharge of wastewater ξ = 1 when discharged near the coast; the coefficient of river meandering φ = 1) is calculated by the equation:

= 1.0 1.0https://pandia.ru/text/80/127/images/image028_2.png" width="44" height="28 src="> and find its numerical value

β = https://pandia.ru/text/80/127/images/image030_2.png" width="107" height="73">.png" width="145" height="51 src="> .= 0.302 ≈ 0.3 mg/dm3.

Answer: The concentration of Fe3+ at a distance of 500 m from the place of wastewater discharge is 0.302 mg/dm3, i.e., it is practically equal to the background concentration

Example 2 Calculate the biooxidation rate constant k* if it is experimentally established that BODtotal is observed on the 13th day of sample incubation. What proportion of BODtotal is BOD5 in this case?

Solution:

To determine BODtotal, it is assumed that BODtotal: (BODtotal - BODτ) = 100: 1, i.e. 99% of organic substances are oxidized.

k* = https://pandia.ru/text/80/127/images/image035_1.png" width="72" height="47"> = 1 – 10-k*5 = 1 – 10-0.15 ∙5 = 0.822 or 82.2%.

Answer : Biooxidation rate constant is 0.15 day-1. BOD5 of BODtotal is 82.2%.

Example 3 Calculate the half-life, the degree of hydrolysis and the concentration of methylchoracetate (ClCH2COOCH3) at T = 298K in a stagnant water body with pH = 6.9 after: a) 1 hour; b) 1 day after its entry into the reservoir, if its initial concentration was 0.001 mg/l. The rate constants of hydrolysis of methyl chloroacetate are given in the table.

Solution:

In accordance with the law of mass action, the rate of hydrolysis is

where kHYDR is the hydrolysis rate constant, s-1;

SZV - concentration of pollutants.

Hydrolysis can be considered a pseudo-first order reaction, since pollutants are present in natural waters in trace amounts. The concentration of water in comparison with their concentrations is much higher and is practically considered unchanged.

The hydrolysis constant is calculated by the equation

where https://pandia.ru/text/80/127/images/image020_5.png" width="12" height="19"> – rate constants of acid hydrolysis, hydrolysis in a neutral medium and alkaline hydrolysis (see table in the appendix);

СH+.– concentration of hydrogen ions, mol/l;

СOH is the concentration of hydroxide ions, mol/l.

Since, according to the condition of the problem, pH \u003d 6.9, it is possible to find the concentration of hydrogen ions and the concentration of hydroxide ions.

The concentration of hydrogen ions (mol / l) is equal to:

CH+. \u003d 10 - pH \u003d 10-6.9 \u003d 1.26 10-7.

The sum of the hydrogen and hydroxyl exponents is always constant

Therefore, knowing the pH, you can find the hydroxyl index and the concentration of hydroxide ions.

pOH = 14 - pH = 14 - 6.9 = 7.1

The concentration of hydroxide ions (mol/l) is equal to:

COH - \u003d 10–pOH \u003d 10-7.1 \u003d 7.9 10-8.

The hydrolysis constant of methyl chloroacetate is:

2.1 10-7 1.26 10-7+8.5 10-5+140 7.9 10-8=.

8.5 10-5 + 1.1 10-5 = 9.6 10-5s-1.

The half-life of a substance τ0.5 in a first-order reaction is:

https://pandia.ru/text/80/127/images/image037_1.png" width="155" height="47">s = 2 hours.

The degree of conversion (degree of hydrolysis) of the pollutant can be calculated by the equation:

β = (С0 – С)/С0 = 1– e-kτ.

An hour after the entry of methyl chloroacetate into the reservoir, its degree of hydrolysis is equal to:

β = 1– e-0.000096 3600 = 1– 0.708 = 0.292 (or 29.2%).

After a day, the degree of hydrolysis of pollutants is equal to:

β = 1– e-0.000096 24 3600 = 1– 0.00025 = 0.99975 (or 99.98%).

The current concentration of methyl chloroacetate can be determined by knowing its degree of conversion С = С0(1 – β).

An hour after the entry of methyl chloroacetate into the reservoir, its concentration will be:

C \u003d C0 (1 - β) \u003d 0.001 (1 - 0.292) \u003d 0.001 0.708 \u003d 7.08 10-4 mg / l.

In a day, the concentration of pollutants will be equal to:

C \u003d C0 (1 - β) \u003d 0.001 (1 - 0.99975) \u003d 0.001 0.00025 \u003d 2.5 10-7 mg / l.

Answer: The half-life of methyl chloroacetate is 2 hours. An hour after the pollutant enters the reservoir, its conversion rate will be 29.2%, the concentration will be 7.08 10-4 mg/l. A day after the pollutant enters the reservoir, its conversion rate will be 99.98%, the concentration will be 2.5 10-7 mg/l.

7 TASKS FOR INDEPENDENT SOLUTION

1. Calculate the concentration of Cu2+ ions in river water at a distance of 500m from the wastewater outlet, if the concentration of Cu2+ in wastewater is 0.015 mg/l. The speed of the river flow is 0.25 m/s, the volumetric flow is 70 m3/s, the depth of the river is 3 m, the coefficient of river sinuosity is 1.2. Wastewater is discharged from the shore. The volume flow of wastewater is 0.05 m3/s. The background concentration of Cu2+ is 0.010 mg/l.

2. Calculate the concentration of NH4+ ions in the river water at a distance of 800m from the wastewater outlet, if the concentration of NH4+ in the wastewater is 0.25 mg/l. The speed of the river flow is 0.18 m/s, the volume flow is 50 m3/s, the depth of the river is 1.8 m, the coefficient of river meandering is 1.2. Wastewater is discharged from the shore. The volume flow of wastewater is 0.04 m3/s. The background concentration of NH4+ is 0.045 mg/l.

3. Calculate the concentration of Al3+ ions in river water at a distance of 500m from the wastewater outlet, if the concentration of Al3+ in wastewater is 0.06 mg/l. The speed of the river flow is 0.25 m/s, the volume flow is 70 m3/s, the depth of the river is 3 m, the coefficient of the river sinuosity is 1.0. Wastewater is discharged from the shore. The volume flow of wastewater is 0.05 m3/s. The background concentration of Al3+ is 0.06 mg/l.

4. Calculate the concentration of Fe3+ ions in river water at a distance of 300m from the wastewater outlet, if the concentration of Fe3+ in wastewater is 0.55 mg/l. The speed of the river flow is 0.20 m/s, the volume flow is 65 m3/s, the depth of the river is 2.5 m, the coefficient of river meandering is 1.1. Wastewater is discharged from the shore. The volume flow of wastewater is 0.45 m3/s. The background concentration of Fe3+ is 0.5 mg/l.

5. Calculate the concentration of sulfate ions in the river water at a distance of 500m from the wastewater outlet, if the concentration of SO42- in the wastewater is 105.0 mg/l. The speed of the river flow is 0.25 m/s, the volumetric flow is 70 m3/s, the depth of the river is 3 m, the coefficient of river sinuosity is 1.2. Wastewater is discharged from the shore. The volume flow of wastewater is 0.05 m3/s. The background concentration of SO42- is 29.3 mg/L.

6. Calculate the concentration of chloride ions in river water at a distance of 500m from the wastewater outlet, if the concentration of Cl - in wastewater is 35.0 mg/l. The speed of the river flow is 0.25 m/s, the volume flow is 70 m3/s, the depth of the river is 3 m, the coefficient of the river sinuosity is 1.0. Wastewater is discharged from the shore. The volume flow of wastewater is 0.5 m3/s. The background concentration of SO42- is 22.1 mg/l.

7. The concentration of Cu2+ copper ions in wastewater is 0.02 mg/l. At what distance from the place of wastewater discharge will the concentration of Cu2+ exceed the background by 10% if the volumetric flow rate of wastewater is 0.05 m3/s? The speed of the river flow is 0.15 m/s, the volume flow is 70 m3/s, the depth of the river is 3 m, the coefficient of river meandering is 1.2. Wastewater is discharged from the shore. The background concentration of Cu2+ is 0.010 mg/L.

8. As a result of dry deposition from the atmosphere, aerosol particles with a diameter of 50 µm and a density of 2500 kg/m3 entered a flowing reservoir 1.5 m deep. Water flow rate is 0.8 m/s, water viscosity is 1 10-3 Pa s, water density is 1000 kg/m3. What distance will these particles, carried away by the current, overcome before settling to the bottom?

9. As a result of wet deposition from the atmosphere, aerosol particles with a diameter of 20 µm and a density of 2700 kg/m3 entered a flowing reservoir with a depth of 3.0 m. Water flow rate is 0.2 m/s, water viscosity is 1 10-3 Pa s, water density is 1000 kg/m3. What distance will these particles, carried away by the current, overcome before settling to the bottom?

10. As a result of dry deposition from the atmosphere, aerosol particles with a diameter of 40 μm and a density of 2700 kg/m3 entered a flowing reservoir with a depth of 2.0 m. Water flow velocity is 0.25 m/s, water viscosity is 1 10-3 Pa s, water density is 1000 kg/m3. The length of the reservoir in the direction of the current is 5000 m. Will these particles settle to the bottom of the reservoir or will they be carried out by the current?

11. Calculate the diameter of suspended particles entering the flowing pond with wastewater, which will settle to the bottom of the reservoir 200m from the wastewater outlet, if the particle density is 2600 kg/m3. The water flow rate is 0.6 m/s, the viscosity of water is 1 10-3 Pa s, the density of water is 1000 kg/m3. The depth of the reservoir is 1.8m.

12. As a result of the accident, hexane spread over the surface of the reservoir. The saturation vapor pressure of hexane at 20°C, 30°C and 40°C is 15998.6 Pa, 24798.0 Pa and 37063.6 Pa, respectively. Determine the saturation vapor pressure of hexane at 15°C graphically. Calculate the evaporation rate of hexane at 15°C using the formula if the wind speed is 1m/s. The density of air at 0°C is 1.29 kg/m3, the viscosity of air at 15°C is 18∙10−6 Pa∙s, the diameter of the spot formed by hexane on the water surface is 100m.

13. As a result of the accident, toluene spread over the surface of the reservoir. The saturation vapor pressure of toluene at 20°C, 30°C and 40°C is 3399.7 Pa, 5266.2 Pa and 8532.6 Pa, respectively. Determine the saturation vapor pressure of toluene at 25°C graphically. Calculate the evaporation rate of toluene at 25°C using the formula if the wind speed is 2m/s. The density of air at 0°С is 1.29 kg/m3, the viscosity of air at 25°С is 20∙10−6 Pa∙s, the diameter of the spot formed by toluene on the water surface is 200m.

14. As a result of the accident, the surface of the reservoir spread m-xylene. Saturated steam pressure m-xylene at 20°C and 30°C is equal to 813.3 and 1466.5 Pa, respectively. Determine the saturation vapor pressure m-xylene at 25°C using the integral form of the isobar equation chemical reaction. Calculate Evaporation Rate m-xylene at 25°C according to the formula, if the wind speed is 5m/s. The density of air at 0°C is 1.29 kg/m3, the viscosity of air at 25°C is 20∙10−6 Pa∙s, the diameter of the spot formed m-xylene on the water surface is equal to 500m.

15. Benzene is accidentally spilled on the laboratory table. The saturation vapor pressure of benzene at 20°C and 30°C is 9959.2 and 15732.0 Pa, respectively. Determine the saturation vapor pressure of benzene at 25°C using the integral form of the chemical reaction isobar equation. Calculate the evaporation rate of benzene at 25°C using the emission method harmful substances in atmosphere. The diameter of the spot formed by benzene on the surface of the table is 0.5 m. Will the MPC value be exceeded. h.(С6Н6) = 5 mg/m3 15 minutes after the spill of benzene, if the volume of the room is 200 m3?

16. Chlorobenzene is accidentally spilled on the laboratory table. The saturation vapor pressure of chlorobenzene at 20°C and 30°C is 1173.2 and 199.8 Pa, respectively. Determine the saturation vapor pressure of chlorobenzene at 25°C using the integral form of the chemical reaction isobar equation. Calculate the evaporation rate of chlorobenzene at 25°C using the atmospheric emission method. The diameter of the spot formed by chlorobenzene on the surface of the table is 0.3 m. Will the MPC value be exceeded. z.(С6Н5Cl) = 50mg/m3 10 minutes after the spill of chlorobenzene, if the volume of the room is 150m3?

17. As a result of the accident, a mixture of octane, toluene and m- xylene weighing 1000 kg. The composition of the mixture (mass fractions): octane - 0.3; toluene - 0.4; m-xylene - 0.3. Saturated vapor pressure of octane, toluene and m-xylene at 20°C is equal to 1386.6; 3399.7 Pa and 813.3 Pa, respectively. Calculate the evaporation rates of hydrocarbons at 20°C using the method for determining emissions of harmful substances into the atmosphere. Determine the composition of the mixture (mass fraction) after an hour, if the diameter of the spot formed by the mixture of hydrocarbons on the water surface is 10 m. The wind speed is 1m/s.

18. As a result of the accident, a mixture of benzene, toluene and m- xylene weighing 1000 kg. The composition of the mixture (mass fractions): benzene - 0.5; toluene - 0.3; m-xylene - 0.2. Saturated vapor pressure of benzene, toluene and m-xylene at 20°C is equal to 9959.2; 3399.7 Pa and 813.3 Pa, respectively. Calculate the evaporation rates of hydrocarbons at 20°C using the method for determining emissions of harmful substances into the atmosphere. Determine the composition of the mixture (wt. fraction) after an hour, if the diameter of the spot formed by the mixture of hydrocarbons on the surface of the water is 12m. The wind speed is 0.5m/s.

19. Calculate the proportion of 2,3,7,8-Cl4-dibenzodioxin adsorbed by suspended particles containing 3.5% (wt.) organic carbon. The concentration of suspended particles in the bottom layers of the reservoir is 12000 ppm. The distribution coefficient of 2,3,7,8-Cl4-dibenzodioxin in the octanol-water KO-B system is 1.047 107.

20. Calculate the proportion of 1,2,3,4-Cl4-dibenzodioxin adsorbed by particulate matter containing 4% (wt.) organic carbon. The concentration of suspended particles in the bottom layers of the reservoir is 10,000 ppm. The distribution coefficient of 1,2,3,4-Cl4-dibenzodioxin in the octanol-water KO-B system is 5.888 105.

21. Calculate the proportion of phenol adsorbed by suspended particles containing 10% (wt.) organic carbon. The concentration of suspended particles in the bottom layers of the reservoir is 50,000 ppm. The distribution coefficient of phenol in the system octanol-water KO-B is 31.

22. Will PbSO4 precipitate form when it enters a flowing reservoir with a volume flow of 50m3/s waste water containing 0.01 mg/l of Pb2+ ions? The volume flow rate of waste water is 0.05 m3/s. The background concentration of SO42- is 30 mg/l. Take the mixing ratio γ equal to 1∙10−4. PR(PbSO4) = 1.6 10−8.

23. Will Fe(OH)3 precipitate form when sewage containing 0.7 mg/l of Fe3+ ions enters a flowing reservoir with a volume flow of 60m3/s? The volume flow rate of waste water is 0.06 m3/s. pH = 7.5. Take the mixing ratio γ equal to 4∙10−4. PR(Fe(OH)3) = 6.3 10−38.

24. Calculate the degree of hydrolysis and the concentration of chloroform (CHCl3) at T=298K in a stagnant reservoir with pH=7.5 after: a) 1 day; b) 1 month; c) 1 year after its entry into the reservoir, if its initial concentration was 0.001 mg/l. The rate constants of hydrolysis of chloroform are given in the table.

25. Calculate the degree of hydrolysis (degree of conversion) and the concentration of dichloromethane (CH2Cl2) at T=298K in a stagnant reservoir with pH=8.0 after: a) 1 day; b) 1 month; c) 1 year after its entry into the reservoir, if its initial concentration was 0.001 mg/l. The rate constants of hydrolysis of dichloromethane are given in the table.

26. Calculate the degree of hydrolysis (degree of conversion) and the concentration of bromomethane (CH3Br) at T=298K in a stagnant reservoir with pH=8.0 after: a) 1 day; b) 1 month; c) six months after its entry into the reservoir, if its initial concentration was 0.005 mg/l. The rate constants of hydrolysis, bromine are given in the table.

27. After what time will the concentration of ethyl acetate in a stagnant reservoir become equal to: a) half of the initial concentration; b) 10% of the initial concentration; c) 1% of the initial concentration? T = 298K. pH = 6.5. The rate constants for the hydrolysis of ethyl acetate are given in the table.

28. After what time will the concentration of phenylacetate in a stagnant reservoir become equal to: a) half of the initial concentration; b) 10% of the initial concentration; c) 1% of the initial concentration? T = 298K. pH = 7.8. The rate constants of hydrolysis of phenylacetate are given in the table.

29. After what time will the concentration of phenyl benzoate in a stagnant reservoir become equal to: a) half of the initial concentration; b) 10% of the initial concentration; c) 1% of the initial concentration? T = 298K. pH = 7.5. The rate constants of hydrolysis of phenyl benzoate are given in the table.

30. Calculate the biooxidation constant k* in natural water and the time for removing half of the pollution, if the values ​​of BOD5 and BODtot are experimentally determined, which are equal to 3.0 and 10.0 mgO2/dm3, respectively.

31. Calculate the biooxidation constant k* in natural water and the time for removing half of the pollution, if the values ​​of BOD5 and BODtot are experimentally determined, which are equal to 1.8 and 8.0 mgO2/dm3, respectively.

32. Calculate the biooxidation rate constant k* in natural water, if it is experimentally established that BODtotal is observed on the 13th day of incubation of a sample of this water. What proportion of BODtotal is BOD5 in this case?

33. Calculate the biooxidation rate constant k* in natural water, if it is experimentally established that BODtotal is observed on the 18th day of incubation of a sample of this water. What proportion of BODtotal is BOD5 in this case?

34. The time for complete oxidation of phenol in a pond with natural aeration was 50 days. Calculate the rate constant of biooxidation k* of phenol in this pond, as well as its concentration after 10 days, if the initial concentration of phenol is 20 µg/l.

35. The time of complete oxidation of toluene in a pond with natural aeration was 80 days. Calculate the biooxidation rate constant k* of toluene in this pond, as well as its concentration after 30 days, if the initial concentration of toluene is 50 µg/l.

36. Calculate COD. acetic acid. Calculate COD natural water, which contains 1∙10−4 mol/l of acetic acid. Calculate BODtot. of this water if BODtot: COD = 0.8: 1. Calculate

37. Determine the concentration of phenol in the water of a stagnant reservoir one day after its arrival, if the initial concentration of phenol was 0.010 mg/l. Consider that the transformation of phenol occurs mainly as a result of oxidation by the RO2 radical. The stationary concentration of RO2 is 10-9 mol/l. The reaction rate constant is 104 mol l-1 s-1.

38. Determine the concentration of formaldehyde in the water of a stagnant reservoir 2 days after its arrival, if the initial concentration of formaldehyde was 0.05 mg/l. Consider that the transformation of formaldehyde occurs mainly as a result of oxidation by the RO2 radical. The stationary concentration of RO2 is 10-9 mol/l. The reaction rate constant is 0.1 mol l-1 s-1.

APPENDIX

Table - Rate constants of hydrolysis of some organic substances at T = 298K

Substance	Products hydrolysis	Hydrolysis constants
l mol-1 s-1		l mol-1 s-1
ethyl acetate	CH3COOH + C2H5OH
Methyl chloroacetate	СlCH2COOH + CH3OH
Phenyl acetate	CH3COOH + C6H5OH
Phenyl benzoate	C6H5COOH + C6H5OH
Chloromethane CH3Cl
Bromomethane CH3Br
Dichloromethane CH2Cl2
Trichloromethane CHCl3