THERMAL EFFECT, heat released or absorbed thermodynamically. system when chemical flows through it. districts. It is determined under the condition that the system does not perform any work (except for possible expansion work), and the t-ry and products are equal. Since heat is not a state function, i.e. during the transition between states depends on the transition path, then in the general case the thermal effect cannot serve as a characteristic of a specific district. In two cases, an infinitesimal amount of heat (elementary heat) d Q coincides with the total differential of the state function: with constant volume d Q = = dU (U is the internal energy of the system), and with constant d Q = dH (H - enthalpy of the system).
Two types of thermal effects are practically important: isothermal-isobaric (at constant temperatures T and p) and isothermal-isochoric (at constant T and volume V). There are differential and integral thermal effects. The differential thermal effect is determined by the expressions:
where u i, h i -resp. partial molar extr. energy and ; v i -stoichiometric coefficient (v i > 0 for products, v i<0 для ); x
= (n i - n i 0)/v i ,-хим.
переменная, определяющая состав системы в любой момент протекания р-ции (n i
и n i0 - числа i-го компонента
в данный момент времени и в начале хим. превращения соотв.). Размерность дифференциального
thermal effect reaction-kJ/ . If u T,V , h T,p > 0, the r-tion is called. endothermic, with the opposite sign of the effect - exothermic. The two types of effects are related by:
The temperature dependence of the thermal effect is given, the application of which, strictly speaking, requires knowledge of the partial molars of all substances involved in the solution, but in most cases these quantities are unknown. Since for rivers flowing in real solutions and other thermodynamically non-ideal environments, thermal effects, like others, significantly depend on the composition of the system and experiment. conditions, an approach has been developed that facilitates the comparison of different districts and the taxonomy of thermal effects. This purpose is served by the concept of standard thermal effect (denoted). By standard we mean a thermal effect carried out (often hypothetically) under conditions when all the entities participating in the district are in the given conditions. Differential and integral standard thermal effects are always numerically the same. The standard thermal effect can be easily calculated using tables of standard heats of formation or heats of combustion (see below). For non-ideal media, there is a large discrepancy between the actually measured and standard thermal effects, which must be kept in mind when using thermal effects in thermodynamic calculations. For example, for alkaline diacetimide [(CH 3 CO) 2 NH (sol) + H 2 O (l) = CH 3 SOKH 2 (sol) + CH 3 COOH (l) +] in 0.8 n. NaOH solution in aqueous (58% by weight) at 298 K measured thermal effect DH 1 = - 52.3 kJ/. For the same district under standard conditions, = - 18.11 kJ/ was obtained. It means so much. the difference is explained by the thermal effects accompanying the substance in the specified solution (heat). For solid, liquid acetic acid and heat are equal, respectively: D H 2 = 13.60; D H 3 = - 48.62; D H 4 = - 0.83 kJ/, so = D H 1 - D H 2 - D H 3 + D H 4. From the example viewbut that when studying thermal effects, it is important to measure the thermal effects of the accompanying physical-chemical. processes.
The study of thermal effects is a very important task. Basic let's experiment method - calorimetry. Modern The equipment makes it possible to study thermal effects in gas, liquid and solid phases, at the interface, as well as in complex ones. systems. The range of typical values of measured thermal effects is from hundreds of J/ to hundreds of kJ/. In table calorimetric data are given. measurements of thermal effects of certain districts. Measuring thermal effects, dilution, and heat allows one to move from actually measured thermal effects to standard ones.
An important role belongs to thermal effects of two types - the heat of formation of the compound. from simple ones and the heat of combustion of the substance in pure form with the formation of higher elements, of which the substance consists. These thermal effects are brought to standard conditions and tabulated. With their help it is easy to calculate any thermal effect; it is equal to algebraic. the sum of the heats of formation or heats of combustion of all the substances involved in the process:
Application of tabular values 
allowscalculate thermal effects plural. thousand rubles, although these values themselves are known only for several. thousand connections. This method of calculation is unsuitable, however, for districts with small thermal effects, since the calculated small value obtained as an algebraic amount several large values, characterized by an error, edges in abs. may exceed the thermal effect. Calculation of thermal effects using quantities 
based on the fact that there is a state function. This makes it possible to compose thermochemical systems. equations to determine the thermal effect of the required solution (see). Standard thermal effects are almost always calculated. In addition to the method discussed above, the calculation of thermal effects is carried out using the temperature dependence of -eq.
Just as one of the physical characteristics of a person is physical strength, the most important characteristic of any chemical bond is the strength of the bond, i.e. her energy.
Let us recall that the energy of a chemical bond is the energy that is released during the formation of a chemical bond or the energy that must be spent in order to destroy this bond.
A chemical reaction in general is the transformation of one substance into another. Therefore, during chemical reaction some bonds are broken and others are formed, i.e. energy conversion.
The fundamental law of physics states that energy does not appear from nothing and does not disappear without a trace, but only passes from one form to another. Due to its universality, this principle is obviously applicable to chemical reactions.
Thermal effect of a chemical reaction is called the amount of heat
released (or absorbed) during a reaction and referred to 1 mole of reacted (or formed) substance.
Thermal effect is denoted by the letter Q and is usually measured in kJ/mol or kcal/mol.
If a reaction occurs with the release of heat (Q > 0), it is called exothermic, and if with the absorption of heat (Q< 0) – эндотермической.
If we schematically depict the energy profile of a reaction, then for endothermic reactions the products are higher in energy than the reactants, and for exothermic reactions, on the contrary, the reaction products are lower in energy (more stable) than the reactants.
It is clear that the more substance reacts, the greater the amount of energy released (or absorbed), i.e. the thermal effect is directly proportional to the amount of substance. Therefore, attributing the thermal effect to 1 mole of a substance is due to our desire to compare the thermal effects of various reactions.
Lecture 6. Thermochemistry. Thermal effect of a chemical reaction Example 1. When 8.0 g of copper(II) oxide was reduced with hydrogen, metallic copper and water vapor were formed and 7.9 kJ of heat was released. Calculate the thermal effect of the reduction reaction of copper(II) oxide.
Solution . Reaction equation: CuO (solid) + H2 (g) = Cu (solv) + H2 O (g) + Q (*)
Let's make a proportion: when reducing 0.1 mol - 7.9 kJ is released; when reducing 1 mol - x kJ is released
Where does x = + 79 kJ/mol. Equation (*) takes the form
CuO (solid) + H2 (g) = Cu (solid) + H2 O (g) +79 kJ
Thermochemical equation is the equation of a chemical reaction, which indicates the state of aggregation of the components of the reaction mixture (reagents and products) and the thermal effect of the reaction.
Thus, in order to melt ice or evaporate water, certain amounts of heat are required, whereas when liquid water freezes or water vapor condenses, the same amounts of heat are released. This is why we feel cold when we get out of the water (evaporation of water from the surface of the body requires energy), and sweating is biological defense mechanism from overheating of the body. On the contrary, a freezer freezes water and heats the surrounding room, releasing excess heat to it.
This example shows the thermal effects of changes in the state of aggregation of water. Heat of fusion (at 0o C) λ = 3.34×105 J/kg (physics), or Qpl. = - 6.02 kJ/mol (chemistry), heat of evaporation (vaporization) (at 100o C) q = 2.26×106 J/kg (physics) or Qex. = - 40.68 kJ/mol (chemistry).
melting
evaporation |
|||||
arr. 298.
Lecture 6. Thermochemistry. Thermal effect of a chemical reaction Of course, sublimation processes are possible when a solid
passes into the gas phase, bypassing the liquid state and the reverse processes of deposition (crystallization) from the gas phase; for them it is also possible to calculate or measure the thermal effect.
It is clear that every substance has chemical bonds, therefore, every substance has a certain amount of energy. However, not all substances can be converted into each other by one chemical reaction. Therefore, we agreed to introduce a standard state.
Standard state of matter– this is the state of aggregation of a substance at a temperature of 298 K, a pressure of 1 atmosphere in the most stable allotropic modification under these conditions.
Standard terms– this is a temperature of 298 K and a pressure of 1 atmosphere. Standard conditions (standard condition) are indicated by the index 0.
Standard heat of formation of a compound is the thermal effect of the chemical reaction of the formation of a given compound from simple substances taken in their standard state. The heat of formation of a compound is indicated by the symbol Q 0 For many compounds, standard heats of formation are given in reference books of physicochemical quantities.
The standard heats of formation of simple substances are equal to 0. For example, Q0 sample, 298 (O2, gas) = 0, Q0 sample, 298 (C, solid, graphite) = 0.
For example . Write down the thermochemical equation for the formation of copper(II) sulfate. From the reference book Q0 sample 298 (CuSO4) = 770 kJ/mol.
Cu (solid) + S (solid) + 2O2 (g) = CuSO4 (solid) + 770 kJ.
Note: the thermochemical equation can be written for any substance, but one must understand that in real life the reaction occurs in a completely different way: from the listed reagents, copper(II) and sulfur(IV) oxides are formed when heated, but copper(II) sulfate is not formed. An important conclusion: the thermochemical equation is a model that allows calculations; it agrees well with other thermochemical data, but does not stand up to practical testing (i.e., it is unable to correctly predict the possibility or impossibility of a reaction).
(B j ) - ∑ a i × Q arr 0 .298 i
Lecture 6. Thermochemistry. Thermal effect of a chemical reaction
Clarification . In order not to mislead you, I will immediately add that chemical thermodynamics can predict the possibility/impossibility of a reaction, however, this requires more serious “tools” that go beyond the scope of a school chemistry course. The thermochemical equation in comparison with these techniques is the first step against the background of the Cheops pyramid - you cannot do without it, but you cannot rise high.
Example 2. Calculate the thermal effect of condensation of water weighing 5.8 g. Solution. The condensation process is described by the thermochemical equation H2 O (g.) = H2 O (l.) + Q – condensation is usually an exothermic process. The heat of condensation of water at 25o C is 37 kJ/mol (reference book).
Therefore, Q = 37 × 0.32 = 11.84 kJ.
In the 19th century, the Russian chemist Hess, who studied the thermal effects of reactions, experimentally established the law of conservation of energy in relation to chemical reactions - Hess's law.
The thermal effect of a chemical reaction does not depend on the process path and is determined only by the difference between the final and initial states.
From the point of view of chemistry and mathematics, this law means that we are free to choose any “calculation trajectory” to calculate the process, because the result does not depend on it. For this reason, the very important Hess law has an incredibly important corollary of Hess's law.
Thermal effect of a chemical reaction equal to the sum heats of formation of reaction products minus the sum of heats of formation of reactants (taking into account stoichiometric coefficients).
From point of view common sense This consequence corresponds to a process in which all the reactants were first converted into simple substances, which then reassembled to form reaction products.
In equation form, the consequence of Hess’s law looks like this: Reaction equation: a 1 A 1 + a 2 A 2 + … + a n A n = b 1 B 1 + b 2 B 2 + … b
In this case, a i and b j are stoichiometric coefficients, A i are reagents, B j are reaction products.
Then the consequence of Hess’s law has the form Q = ∑ b j × Q arr 0 .298
k Bk + Q
(Ai)
Lecture 6. Thermochemistry. Thermal effect of a chemical reaction Since the standard heats of formation of many substances
a) are summarized in special tables or b) can be determined experimentally, then it becomes possible to predict (calculate) the thermal effect very large quantity reactions with fairly high accuracy.
Example 3. (Corollary of Hess's law). Calculate the thermal effect of steam reforming of methane occurring in the gas phase under standard conditions:
CH4 (g) + H2 O (g) = CO (g) + 3 H2 (g)
Determine whether this reaction exothermic or endothermic?
Solution: Corollary of Hess's Law
Q = 3 Q0 | G ) +Q 0 | (CO ,g ) −Q 0 | G ) −Q 0 | O, g) - in general terms. |
|||||
arr. 298 | arr. 298 | arr. 298 | arr. 298 | ||||||
Q rev0 | 298 (H 2,g) = 0 | Simple substance in standard state |
|||||||
From the reference book we find the heats of formation of the remaining components of the mixture.
O,g) = 241.8 | (CO,g) = 110.5 | Г) = 74.6 | |||||||||
arr. 298 | arr. 298 | arr. 298 | |||||||||
Substituting the values into the equation
Q = 0 + 110.5 – 74.6 – 241.8 = -205.9 kJ/mol, the reaction is highly endothermic.
Answer: Q = -205.9 kJ/mol, endothermic
Example 4. (Application of Hess's law). Known heats of reactions
C (solid) + ½ O (g) = CO (g) + 110.5 kJ
C (solid) + O2 (g) = CO2 (g) + 393.5 kJ Find the thermal effect of the reaction 2CO (g) + O2 (g) = 2CO2 (g). Solution Multiply the first and second equation on 2
2C (sol.) + O2 (g.) = 2CO (g.) + 221 kJ 2C (solv.) + 2O2 (g.) = 2CO2 (g.) + 787 kJ
Subtract the first from the second equation
O2 (g) = 2CO2 (g) + 787 kJ – 2CO (g) – 221 kJ,
2CO (g) + O2 (g) = 2CO2 (g) + 566 kJ Answer: 566 kJ/mol.
Note: When studying thermochemistry, we consider a chemical reaction from the outside (outside). On the contrary, chemical thermodynamics - the science of the behavior of chemical systems - considers the system from the inside and operates with the concept of “enthalpy” H as the thermal energy of the system. Enthalpy, so
Lecture 6. Thermochemistry. The thermal effect of a chemical reaction has the same meaning as the amount of heat, but has the opposite sign: if energy is released from the system, environment it receives it and heats up, but the system loses energy.
Literature:
1. textbook, V.V. Eremin, N.E. Kuzmenko et al., Chemistry 9th grade, paragraph 19,
2. Educational and methodological manual"Basics general chemistry" Part 1.
Compiled by S.G. Baram, I.N. Mironov. - take with you! for the next seminar
3. A.V. Manuilov. Basics of chemistry. http://hemi.nsu.ru/index.htm
§9.1 Thermal effect of a chemical reaction. Basic laws of thermochemistry.
§9.2** Thermochemistry (continued). The heat of formation of a substance from elements.
Standard enthalpy of formation.
Attention!
We are moving on to solving calculation problems, so a calculator is now desirable for chemistry seminars.
Any chemical reaction is accompanied by the release or absorption of energy in the form of heat.
Based on the release or absorption of heat, they distinguish exothermic And endothermic reactions.
Exothermic reactions are reactions during which heat is released (+Q).
Endothermic reactions are reactions during which heat is absorbed (-Q).
Thermal effect of reaction (Q) is the amount of heat that is released or absorbed during the interaction of a certain amount of initial reagents.
A thermochemical equation is an equation that specifies the thermal effect of a chemical reaction. So, for example, the thermochemical equations are:
It should also be noted that thermochemical equations must necessarily include information about the aggregate states of reagents and products, since the value of the thermal effect depends on this.
Calculations of the thermal effect of the reaction
An example of a typical problem to find the thermal effect of a reaction:
When 45 g of glucose reacts with excess oxygen according to the equation
C 6 H 12 O 6 (solid) + 6O 2 (g) = 6CO 2 (g) + 6H 2 O (g) + Q
700 kJ of heat was released. Determine the thermal effect of the reaction. (Write the number to the nearest whole number.)
Solution:
Let's calculate the amount of glucose:
n(C 6 H 12 O 6) = m(C 6 H 12 O 6) / M(C 6 H 12 O 6) = 45 g / 180 g/mol = 0.25 mol
Those. When 0.25 mol of glucose interacts with oxygen, 700 kJ of heat is released. From the thermochemical equation presented in the condition, it follows that the interaction of 1 mole of glucose with oxygen produces an amount of heat equal to Q (thermal effect of the reaction). Then the following proportion is correct:
0.25 mol glucose - 700 kJ
1 mole of glucose - Q
From this proportion the corresponding equation follows:
0.25 / 1 = 700 / Q
Solving which, we find that:
Thus, the thermal effect of the reaction is 2800 kJ.
Calculations using thermochemical equations
Much more often in Unified State Exam assignments in thermochemistry, the value of the thermal effect is already known, because the condition gives the complete thermochemical equation.
In this case, it is necessary to calculate either the amount of heat released/absorbed with a known amount of reagent or product, or, conversely, by known value heat, it is required to determine the mass, volume or quantity of a substance of any participant in the reaction.
Example 1
According to the thermochemical reaction equation
3Fe 3 O 4 (tv.) + 8Al (tv.) = 9Fe (tv.) + 4Al 2 O 3 (tv.) + 3330 kJ
68 g of aluminum oxide were formed. How much heat was released? (Write the number to the nearest whole number.)
Solution
Let's calculate the amount of aluminum oxide substance:
n(Al 2 O 3) = m(Al 2 O 3) / M(Al 2 O 3) = 68 g / 102 g/mol = 0.667 mol
In accordance with the thermochemical equation of the reaction, when 4 moles of aluminum oxide are formed, 3330 kJ are released. In our case, 0.6667 mol of aluminum oxide is formed. Having denoted the amount of heat released in this case by x kJ, we create the proportion:
4 mol Al 2 O 3 - 3330 kJ
0.667 mol Al 2 O 3 - x kJ
This proportion corresponds to the equation:
4 / 0.6667 = 3330 / x
Solving which, we find that x = 555 kJ
Those. when 68 g of aluminum oxide is formed in accordance with the thermochemical equation in the condition, 555 kJ of heat is released.
Example 2
As a result of a reaction, the thermochemical equation of which
4FeS 2 (tv.) + 11O 2 (g) = 8SO 2 (g) + 2Fe 2 O 3 (tv.) + 3310 kJ
1655 kJ of heat was released. Determine the volume (l) of sulfur dioxide released (no.). (Write the number to the nearest whole number.)
Solution
In accordance with the thermochemical equation of the reaction, when 8 moles of SO 2 are formed, 3310 kJ of heat is released. In our case, 1655 kJ of heat was released. Let the amount of SO 2 formed in this case be x mol. Then the following proportion is fair:
8 mol SO 2 - 3310 kJ
x mol SO 2 - 1655 kJ
From which the equation follows:
8 / x = 3310 / 1655
Solving which, we find that:
Thus, the amount of SO 2 substance formed in this case is 4 mol. Therefore, its volume is equal to:
V(SO 2) = V m ∙ n(SO 2) = 22.4 l/mol ∙ 4 mol = 89.6 l ≈ 90 l(rounded to whole numbers, since this is required in the condition.)
More analyzed problems on the thermal effect of a chemical reaction can be found.
Thermal effect of reaction the amount of heat that is released or absorbed by a system as a result of a chemical reaction. This can be H (P,T = const) or U (V,T = const).
If heat is released as a result of a reaction, i.e. the enthalpy of the system decreases ( N 0 ), then the reaction is called exothermic.
Reactions accompanied by the absorption of heat, i.e. with increasing enthalpy of the system ( N 0), are called e ndothermic.
Like other state functions, enthalpy depends on the amount of substance, so it is scaled ( N) usually referred to 1 mol of substance and expressed in kJ/mol.
Typically, system functions are determined by standard conditions, which, in addition to the standard state parameters, includes the standard temperature T = 298.15 K (25C). Temperature is often indicated as a subscript ().
5.3. Thermochemical equations
Thermochemical reaction equations equations that indicate the thermal effect, reaction conditions and aggregative states of substances. Usually the enthalpy of the reaction is indicated as the thermal effect. For example,
C (graphite) + O 2 (gas) = CO 2 (gas), H 0 298 = -396 kJ.
The thermal effect can be written in the reaction equation:
C (graphite) + O 2 (gas) = CO 2 (gas) + 396 kJ.
In chemical thermodynamics, the first form of notation is used more often.
Features of thermochemical equations.
1. The thermal effect depends on the mass of the reactant, po-
Therefore, it is usually calculated per mole of substance. In this regard, in thermochemical equations one can use fractional odds. For example, for the case of the formation of one mole of hydrogen chloride, the thermochemical equation is written as follows:
½H 2 + ½Cl 2 = HCl, H 0 298 = 92 kJ
or H 2 + Cl 2 = 2HCl, H 0 298 = –184 kJ.
2. Thermal effects depend on the state of aggregation of the reagents; it is indicated in thermochemical equations by indices: and liquid, G- gaseous, T hard or To - crystalline, R– dissolved.
For example: H 2 + ½ O 2 = H 2 O (l), H 0 298 = -285.8 kJ.
H 2 + ½ O 2 = H 2 O (g), H 0 298 = 241.8 kJ.
3. You can perform algebraic operations with thermochemical equations (they can be added, subtracted, multiplied by any coefficients along with the thermal effect).
Thermochemical equations more fully than ordinary ones reflect the changes occurring during a reaction - they show not only the qualitative and quantitative composition of reagents and products, but also the quantitative energy transformations that accompany this reaction.
5.4. Hess's law and its consequences
Thermochemical calculations are based on the law discovered by the Russian scientist G. I. Hess (1841). Its essence is as follows: the thermal effect of a chemical reaction depends only on the initial and final state of the system, but does not depend on the speed and path of the process, that is, on the number of intermediate stages. This means, in particular, that thermochemical reactions can be added together with their thermal effects. For example, the formation of CO 2 from carbon and oxygen can be represented by the following scheme:
C+O 2 H 1 CO 2 1. C (graph.) + O 2 (g) = CO 2 (g), H 0 1 = -396 kJ.
2. C (graph.) + 1/2O 2 (g) = CO (g), H 0 2 = X kJ.
Н 2 Н 3
3. CO (g) + 1/2O 2 (g) = CO 2 (g), H 0 3 = 285.5 kJ.
CO +½ ABOUT 2
All three of these processes are widely used in practice. As is known, the thermal effects of CO 2 formation (H 1) and CO combustion (H 3) are determined experimentally. The thermal effect of the formation of CO (H 2) cannot be measured experimentally, since when carbon burns under conditions of lack of oxygen, a mixture of CO and CO 2 is formed. But the enthalpy of the reaction of formation of CO from simple substances can be calculated.
From Hess’s law it follows that H 0 1 = H 0 2 + H 0 3. Hence,
H 0 2 = H 0 1 H 0 3 = 396 (285.5) = 110.5 (kJ) - this is the true value
Thus, using Hess's law, it is possible to find the heat of reactions that cannot be determined experimentally.
In thermochemical calculations, two consequences of Hess's law are widely used. According to the first, the thermal effect of a reaction is equal to the sum of the enthalpies of formation of the reaction products minus the sum of the enthalpies of formation of the starting substances (reagents).
N 0 h.r. = n prod · H 0 ƒ prod - n ref · N 0 ƒ reagents ,
where n is the amount of substance; Н 0 ƒ standard enthalpy (heat) of formation of a substance.
The thermal effect of the reaction of the formation of 1 mole of a complex substance from simple substances, determined under standard conditions, is called the standard enthalpy of formation of this substance (H 0 image or H 0 ƒ kJ/mol).
Since the absolute enthalpy of a substance cannot be determined, for measurements and calculations it is necessary to determine the origin, that is, the system and conditions for which the value is taken : H = 0. In thermodynamics, the states of simple substances in their most stable forms at normal conditions- in standard condition.
For example: H 0 ƒ (O 2) = 0, but H 0 ƒ (O 3) = 142.3 kJ/mol. Standard enthalpies of formation are determined for many substances and listed in reference books (Table 5.1).
In general, for the reaction aA + bB = cC + dD, enthalpy, according to the first corollary, is determined by the equation:
H 0 298 h.r. = (cН 0 ƒ, C + dН 0 ƒ, E) (аH 0 ƒ, A + вH 0 ƒ, B).
The second consequence of Hess's law applies to organic substances. The thermal effect of a reaction involving organic substances is equal to the sum of the heats of combustion of the reactants minus the heats of combustion of the products.
In this case, the heat of combustion is determined under the assumption of complete
combustion: carbon is oxidized to CO 2, hydrogen to H 2 O, nitrogen to N 2.
The thermal effect of the reaction of oxidation of elements that make up a substance with oxygen to the formation of higher oxides is called heat of combustion of this substance(H 0 sg.). In this case, it is obvious that the heats of combustion of O 2, CO 2, H 2 O, N 2 are taken equal to zero.
Table 5.1
Thermodynamic constants of some substances
|
Substance |
Н 0 f, 298, kJ/mol |
S 0 298 , J/molK |
G 0 f, 298, kJ/mol |
Substance |
Н 0 f , 298, kJ/mol |
J/molK |
G 0 f, 298, |
|
C(graphite) | |||||||
For example, the heat of combustion of ethanol
C 2 H 5 OH (l) + 3O 2 = 2CO 2 + 3H 2 O (g)
H 0 h.r. = Н 0 сг (C 2 H 5 OH) = 2Н 0 ƒ, (CO 2)+3Н 0 ƒ, (H 2 O) Н 0 ƒ, (C 2 H 5 OH).
Н 0 сг (C 2 H 5 OH) = 2(393.5) + 3(241.8) – (277.7) = 1234.7 kJ/mol.
Calorific values are also given in reference books.
Example 1. Determine the thermal effect of the ethanol dehydration reaction if
H 0 sg (C 2 H 4) =1422.8; H 0 sg (H 2 O) = 0; Н 0 сг (C 2 H 5 OH) = -1234.7 (kJ/mol).
Solution. Let's write the reaction: C 2 H 5 OH (l) = C 2 H 4 + H 2 O.
According to the second corollary, we determine the thermal effect of the reaction from the heats of combustion, which are given in the reference book:
H 0 298 x.p = H 0 sg (C 2 H 5 OH) H 0 sg (C 2 H 4) H 0 sg (H 2 O) =
1234.7 + 1422.8 = 188.1 kJ/mol.
In technology for characterizing thermal qualities individual species fuels usually use them calorific value.
Calorific value fuel is called the thermal effect, which corresponds to the combustion of a unit of mass (1 kg) for solid and liquid fuels or a unit of volume (1 m 3) for gaseous fuels (Table 5.2).
Table 5.2
Calorific value and composition of some
common fuels
|
Calorific value, |
||||
|
oxygen | ||||
|
Anthracite* | ||||
|
Ancient coal | ||||
|
Nature gas | ||||
|
Raw oil | ||||
*Anthracite is hard coal with the maximum carbon content (94-96%).
Hydrogen is the most effective chemical energy carrier for energy, transport and technology of the future, since it has a very high calorific value (Table 4.2), it is relatively easy to transport, and its combustion produces only water, i.e. It is a “clean” fuel and does not cause air pollution. However, its widespread use as an energy source is hampered by the too low content of hydrogen in nature in a free state. Most Hydrogen is produced by the decomposition of water or hydrocarbons. However, such decomposition requires a lot of energy, and in practice, due to thermal losses, it is necessary to expend more energy to produce hydrogen than can then be obtained. In the future, if it is possible to create large and cheap energy sources (for example, as a result of the development of technology for producing nuclear or solar energy), part of it will be used to produce hydrogen. Many scientists are convinced that the energy of the future is hydrogen energy.
Using Hess's law and its consequences, it is possible to determine many quantities, including those not determined experimentally, if the reaction corresponding to the unknown quantity can be obtained by adding other reactions with known characteristics.
Example 2. Based on the calorific value of CH 4 (H 0 сг =890 kJ/mol) and Н 2 (Н 0 сг =286 kJ/mol), calculate the calorific value of a gas containing 60% hydrogen and 40% methane CH 4 .
Solution. Let us write the thermochemical equations of combustion reactions:
1) H 2 +½O 2 = H 2 O (l);H 0 f (H 2 O) = -286 kJ/mol;
CH 4 + 2O 2 = CO 2 + 2H 2 O (l);H 0 2
H 0 2 = H 0 ƒ, (CO 2) + 2H 0 ƒ, (H 2 0)H 0 ƒ, (CH 4) =3932. 286 + 75 =890 kJ/mol.
1 m 3 of gas contains 600 liters of H 2 and 400 liters of CH 4, which is H 2 and CH 4. The calorific value of the gas will be:
kJ/m3.
Example 3. Using the data in Table 5.1, calculate the thermal effect of the ethylene combustion reaction: C 2 H 4 + 3O 2 = 2CO 2 + 2H 2 O (g).
Solution. From Table 5.1 we write down the values of the enthalpies of formation of the substances participating in the reaction (in kJ/mol):
H 0 ƒ, co 2 =393.5;H 0 ƒ, s 2 n 4 = 52.3;H 0 ƒ, n 2 o =241.8.
(Recall that the enthalpy of formation of simple substances is zero.)
According to the corollary of Hess’s law (4.4):
H 0 298 x.r =n cont · Н 0 ƒ , cont n ref · N 0 ƒ , ref = 2N 0 ƒ , с 2 + 2Н 0 ƒ , р 2 оН 0 ƒ , с 2 Н 4 =
2. (393.5) + 2 . (241.8)52.3 =1322.9 kJ.
Example 4. Based on the thermal effect of the reaction
3CaO (t) + P 2 O 5 (t) = Ca 3 (PO 4) 2 (t),H 0 =739 kJ,
determine the enthalpy of formation of calcium orthophosphate.
Solution. According to the corollary of Hess's law:
H 0 298 х.р =Н 0 ƒ, Ca 3 (PO 4) 2 (3Н 0 ƒ, CaO + Н 0 ƒ, P 2 O 5).
From the table 4.1: H 0 ƒ, (CaO) =635.5;H 0 ƒ, (P 2 O 5) =1492 (kJ/mol).
H 0 ƒ, Ca 3 (PO 4) 2 =739 + 3. (635.5)1492 =4137.5 kJ/mol.
Example 5. Write a thermochemical equation for the combustion reaction of solid sulfur in N 2 O, if it is known that the combustion of 16 g of sulfur releases 66.9 kJ of heat (it is assumed that when measuring heat, the temperature of the products decreases to the temperature of the reactants, equal to 298 K).
Solution. To write a thermochemical equation, you need to calculate the thermal effect of the reaction:
S (t) + 2N 2 O (g) = SO 2 (g) + 2N 2 (g) ;H 0 = X kJ.
According to the conditions of the problem, it is known that when 16 g of sulfur is burned, 66.9 kJ is released, and 32 g of sulfur is involved in the reaction. Let's make a proportion:
16g 66.9 kJ
32g X kJ X = 133.8 kJ.
Thus, the thermochemical equation is written as follows:
S (t) + 2N 2 O (g) = SO 2 (g) + 2N 2 (g) ,H 0 x..r. =133.8 kJ.
(Since heat is released, the reaction is exothermic, H 0 0).
Example 6. What amount of heat will be released when 5.6 liters of hydrogen combine with chlorine (n.o.), if the enthalpy of formation of hydrogen chloride is equal to -91.8 kJ/mol (the temperature of the products and reagents is 25C).
Solution.H 0 ƒ , (HCl) = -91.8 kJ/mol, this means that when one mole of HCl is formed from simple substances, 91.8 kJ of heat is released, which corresponds to the thermochemical equation:
½Cl 2 +½ H 2 =HCl,H 0 ƒ =91.8 kJ.
From the equation it is clear that to obtain 1 mol of HCl, 0.5 mol of H 2 is consumed, i.e. 0.5·22.4 l = 11.2 l. Let's make a proportion:
11.2 l 91.8 kJ
5.6 l XX= 45.19 kJ.
Answer: 45.19 kJ of heat will be released.
Example 7. Determine the enthalpy of formation of iron (III) oxide based on three thermochemical equations (do not use a reference book):
Fe 2 O 3 + 3CO = 2Fe + 3CO 2, H 0 1 = 26.5 kJ;
C (graphite) +½O 2 = CO,H 0 2 =110.4 kJ;
CO 2 = C (graphite) + O 2,H 0 3 = + 393.3 kJ.
Solution: Let's write down the equation whose thermal effect needs to be determined:
4Fe + 3O 2 = 2Fe 2 O 3; H 0 4 = 2X kJ.
To get the fourth from the first three equations, you need to multiply equation 1) by (2), and equations 2) and 3) by (6) and add:
1) 4Fe + 6CO 2 = 2Fe 2 O 3 + 6CO, H 0 1 = 2·(+26.5) kJ;
2) 6CO = 6C (graphite) + 3O 2,H 0 2 = 6·(+110.4) kJ;
3) 6C (graphite) + 6O 2 = 6CO 2 ,H 0 3 = 6·(393.3) kJ;
H 0 4 = 2H 0 1 + 6H 0 2 + 6H 0 3 = +53 + 662.42359.8 =1644.4 kJ.
Hence H 0 ƒ (Fe 2 O 3) =822.2 kJ/mol.
Introduction
The thermal effects of chemical reactions are necessary for many technical calculations. They find wide application in many industries, as well as in military developments.
The purpose of this course work is the study of the practical application of the thermal effect. We will look at some options for its use, and find out how important it is to use the thermal effects of chemical reactions in the context of the development of modern technologies.
Thermal effect of a chemical reaction
Each substance stores a certain amount of energy. We encounter this property of substances already at breakfast, lunch or dinner, since food allows our body to use the energy of a wide variety of chemical compounds contained in food. In the body, this energy is converted into movement, work, and is used to maintain a constant (and quite high!) body temperature.
One of the most famous scientists working in the field of thermochemistry is Berthelot. Berthelot - professor of chemistry at the Higher Pharmaceutical School in Paris (1859). Minister of Education and Foreign Affairs.
Beginning in 1865, Berthelot was actively involved in thermochemistry and conducted extensive calorimetric research, which led, in particular, to the invention of the “calorimetric bomb” (1881); he owns the concepts of “exothermic” and “endothermic” reactions. Berthelot obtained extensive data on the thermal effects of a huge number of reactions, on the heat of decomposition and formation of many substances.
Berthelot studied the effect of explosives: explosion temperature, combustion speed and blast wave propagation, etc.
The energy of chemical compounds is concentrated mainly in chemical bonds. It takes energy to break a bond between two atoms. When a chemical bond is formed, energy is released.
Any chemical reaction consists of breaking some chemical bonds and forming others.
When, as a result of a chemical reaction during the formation of new bonds, more energy is released than was required to destroy the “old” bonds in the starting substances, the excess energy is released in the form of heat. An example is combustion reactions. For example, natural gas (methane CH 4) burns in oxygen in the air, releasing a large amount of heat (Fig. 1a). Such reactions are exothermic.
Reactions that occur with the release of heat exhibit a positive thermal effect (Q>0, DH<0) и называются экзотермическими.
In other cases, the destruction of bonds in the original substances requires more energy than can be released during the formation of new bonds. Such reactions occur only when energy is supplied from outside and are called endothermic.
Reactions that occur with the absorption of heat from the environment (Q<0, DH>0), i.e. with a negative thermal effect, are endothermic.
An example is the formation of carbon monoxide (II) CO and hydrogen H2 from coal and water, which occurs only when heated (Fig. 1b).
Rice. 1a,b. Depiction of chemical reactions using molecular models: a) exothermic reaction, b) endothermic reaction. The models clearly show how, with a constant number of atoms between them, old chemical bonds are destroyed and new chemical bonds arise.
Thus, any chemical reaction is accompanied by the release or absorption of energy. Most often, energy is released or absorbed in the form of heat (less often in the form of light or mechanical energy). This heat can be measured. The measurement result is expressed in kilojoules (kJ) for one mole of reactant or (less commonly) for one mole of reaction product. This quantity is called the thermal effect of the reaction.
Thermal effect is the amount of heat released or absorbed by a chemical system when a chemical reaction occurs in it.
Thermal effect is indicated by the symbols Q or DH (Q = -DH). Its value corresponds to the difference between the energies of the initial and final states of the reaction:
DH = Hfin. - Nish. = Efin. - Eout.
Icons (d), (g) indicate the gaseous and liquid states of substances. There are also designations (tv) or (k) - solid, crystalline substance, (aq) - substance dissolved in water, etc.
The designation of the state of aggregation of a substance is important. For example, in the combustion reaction of hydrogen, water is initially formed in the form of steam (gaseous state), upon condensation of which some more energy can be released. Consequently, for the formation of water in the form of a liquid, the measured thermal effect of the reaction will be slightly greater than for the formation of only steam, since when the steam condenses, another portion of heat will be released.
A special case of the thermal effect of the reaction is also used - the heat of combustion. From the name itself it is clear that the heat of combustion serves to characterize the substance used as fuel. The heat of combustion is referred to 1 mole of a substance that is a fuel (a reducing agent in an oxidation reaction), for example:
|
acetylene |
heat of combustion of acetylene |
The energy (E) stored in molecules can be plotted on the energy scale. In this case, the thermal effect of the reaction ( E) can be shown graphically (Fig. 2).
Rice. 2. Graphic representation of the thermal effect (Q = E): a) exothermic reaction of hydrogen combustion; b) endothermic reaction of water decomposition under the influence of electric current. The reaction coordinate (horizontal axis of the graph) can be considered, for example, as the degree of conversion of substances (100% is the complete conversion of the starting substances).
Chemical Reaction Equations
Equations of chemical reactions in which the thermal effect of the reaction is written along with the reagents and products are called thermochemical equations.
The peculiarity of thermochemical equations is that when working with them, you can transfer the formulas of substances and the magnitude of thermal effects from one part of the equation to another. As a rule, this cannot be done with ordinary equations of chemical reactions.
Term-by-term addition and subtraction of thermochemical equations is also allowed. This may be necessary to determine the thermal effects of reactions that are difficult or impossible to measure experimentally.
Let's give an example. In the laboratory it is extremely difficult to carry out pure form"reaction of producing CH4 methane by direct combination of carbon with hydrogen:
C + 2H 2 = CH 4
But you can learn a lot about this reaction through calculations. For example, find out whether this reaction will be exo- or endothermic, and even quantitatively calculate the magnitude of the thermal effect.
The thermal effects of the combustion reactions of methane, carbon and hydrogen are known (these reactions occur easily):
a) CH 4 (g) + 2O 2 (g) = CO 2 (g) + 2H 2 O (l) + 890 kJ
b) C(tv) + O 2 (g) = CO 2 (g) + 394 kJ
c) 2H 2 (g) + O 2 (g) = 2H 2 O (l) + 572 kJ
Let us subtract the last two equations (b) and (c) from equation (a). We will subtract the left sides of the equations from the left, and the right sides from the right. In this case, all molecules O 2, CO 2 and H 2 O will contract. We get:
CH 4 (g) - C (tv) - 2H 2 (g) = (890 - 394 - 572) kJ = -76 kJ
This equation looks somewhat unusual. Let's multiply both sides of the equation by (-1) and move CH 4 to the right side with the opposite sign. We get the equation we need for the formation of methane from coal and hydrogen:
C(tv) + 2H 2 (g) = CH 4 (g) + 76 kJ/mol
So, our calculations showed that the thermal effect of the formation of methane from carbon and hydrogen is 76 kJ (per mole of methane), and this process must be exothermic (energy will be released in this reaction).
It is important to pay attention to the fact that only substances that are in identical states of aggregation can be added, subtracted and reduced term by term in thermochemical equations, otherwise we will make a mistake in determining the thermal effect on the value of the heat of transition from one state of aggregation to another.
Basic laws of thermochemistry
The branch of chemistry that studies the transformation of energy in chemical reactions is called thermochemistry.
There are two most important laws of thermochemistry. The first of them, the Lavoisier–Laplace law, is formulated as follows:
The thermal effect of a forward reaction is always equal to the thermal effect of a reverse reaction with the opposite sign.
This means that during the formation of any compound, the same amount of energy is released (absorbed) as is absorbed (released) during its decomposition into the original substances. For example:
2H 2 (g) + O 2 (g) = 2H 2 O (l) + 572 kJ (combustion of hydrogen in oxygen)
2 H 2 O (l) + 572 kJ = 2H 2 (g) + O 2 (g) (decomposition of water by electric current)
Lavoisier–Laplace's law is a consequence of the law of conservation of energy.
The second law of thermochemistry was formulated in 1840 by Russian academician G. I. Hess:
The thermal effect of a reaction depends only on the initial and final states of the substances and does not depend on the intermediate stages of the process.
This means that the total thermal effect of a series of successive reactions will be the same as that of any other series of reactions if the starting and ending substances are the same at the beginning and at the end of these series. These two basic laws of thermochemistry give thermochemical equations some similarity to mathematical ones, when in reaction equations it is possible to transfer terms from one part to another, to add, subtract and reduce the formulas of chemical compounds term by term. In this case, it is necessary to take into account the coefficients in the reaction equations and not to forget that the substances being added, subtracted or reduced by moles must be in the same state of aggregation.
Application of the thermal effect in practice
The thermal effects of chemical reactions are needed for many technical calculations. For example, consider the powerful Russian Energia rocket, capable of launching spacecraft and other payloads into orbit. The engines of one of its stages operate on liquefied gases - hydrogen and oxygen.
Suppose we know the work (in kJ) that will have to be spent to deliver a rocket with cargo from the surface of the Earth to orbit, we also know the work to overcome air resistance and other energy costs during the flight. How to calculate the required supply of hydrogen and oxygen, which (in a liquefied state) are used in this rocket as fuel and oxidizer?
Without the help of the thermal effect of the reaction of the formation of water from hydrogen and oxygen, this is difficult to do. After all, the thermal effect is the very energy that should launch the rocket into orbit. In the combustion chambers of a rocket, this heat is converted into the kinetic energy of molecules of hot gas (steam), which escapes from the nozzles and creates jet thrust.
In the chemical industry, thermal effects are needed to calculate the amount of heat to heat reactors in which endothermic reactions occur. In the energy sector, thermal energy production is calculated using the heat of combustion of fuel.
Dietitians use the thermal effects of oxidation food products in the body to create proper diets not only for patients, but also for healthy people - athletes, workers of various professions. Traditionally, calculations here use not joules, but other energy units - calories (1 cal = 4.1868 J). The energy content of food is referred to any mass of food products: 1 g, 100 g, or even standard packaging of the product. For example, on the label of a jar of condensed milk you can read the following inscription: “calorie content 320 kcal/100 g.”
The thermal effect is calculated when producing monomethylaniline, which belongs to the class of substituted aromatic amines. The main area of application of monomethylaniline is as an anti-knock additive for gasoline. It is possible to use monomethylaniline in the production of dyes. Commercial monomethylaniline (N-methylaniline) is isolated from the catalyzate by periodic or continuous rectification. Thermal effect of the reaction ∆Н= -14±5 kJ/mol.
Heat-resistant coatings
The development of high-temperature technology necessitates the creation of particularly heat-resistant materials. This problem can be solved by using refractory and heat-resistant metals. Intermetallic coatings are attracting increasing attention because they have many valuable qualities: resistance to oxidation, aggressive melts, heat resistance, etc. Of interest is also the significant exothermicity of the formation of these compounds from their constituent elements. There are two possible ways to use the exothermicity of the reaction for the formation of intermetallic compounds. The first is the production of composite, two-layer powders. When heated, the components of the powder interact, and the heat of the exothermic reaction compensates for the cooling of the particles, reaching the protected surface in a completely molten state and forming a low-porosity coating firmly adhered to the base. Another option would be to apply a mechanical mixture of powders. When the particles are heated sufficiently, they interact already in the coating layer. If the magnitude of the thermal effect is significant, then this can lead to self-melting of the coating layer, the formation of an intermediate diffusion layer that increases the adhesion strength, and obtaining a dense, low-porosity coating structure. When choosing a composition that forms an intermetallic coating with a great thermal effect and has many valuable qualities - corrosion resistance, sufficient heat resistance and wear resistance, nickel aluminides, in particular NiAl and Ni 3 Al, attract attention. The formation of NiAl is accompanied by a maximum thermal effect.
Thermochemical method of diamond processing
The “thermochemical” method received its name due to the fact that it occurs at elevated temperatures, and is based on the use of the chemical properties of diamond. The method is carried out as follows: the diamond is brought into contact with a metal capable of dissolving carbon, and in order for the dissolution or processing process to proceed continuously, it is carried out in a gas atmosphere that interacts with carbon dissolved in the metal, but does not react directly with the diamond. During the process, the magnitude of the thermal effect takes on a high value.
To determine the optimal conditions for thermochemical processing of diamond and identify the capabilities of the method, it was necessary to study the mechanisms of certain chemical processes, which, as shown by an analysis of the literature, have not been studied at all. A more specific study of the thermochemical processing of diamond was hampered, first of all, by insufficient knowledge of the properties of the diamond itself. They were afraid of ruining it with heat. Research on the thermal stability of diamond has only been carried out in recent decades. It has been established that diamonds that do not contain inclusions can be heated to 1850 “C” in a neutral atmosphere or in a vacuum without any harm to them, and only higher.
Diamond is the best blade material due to its unique hardness, elasticity and low friction against biological tissue. Operating with diamond knives facilitates operations and reduces the healing time of incisions by 2-3 times. According to the microsurgeons of the MNTK for eye microsurgery, knives sharpened by thermochemical method are not only not inferior, but also superior in quality to the best foreign samples. Thousands of operations have already been performed with thermochemically sharpened knives. Diamond knives of different configurations and sizes can be used in other areas of medicine and biology. Thus, microtomes are used to make preparations in electron microscopy. The high resolution of the electron microscope places special demands on the thickness and quality of the section of specimens. Diamond microtomes, sharpened by thermochemical method, make it possible to produce sections of the required quality.
Technogenic raw materials for cement production
Further intensification of cement production involves the widespread introduction of energy and resource-saving technologies using waste from various industries.
When processing skarn-magnetite ores, dry magnetic separation (DMS) tailings are released, which are crushed stone material with a grain size of up to 25 mm. SMS tailings have a fairly stable chemical composition, wt.%:
SiO 2 40…45,
Al 2 O 3 10…12,
Fe 2 O 3 15…17,
CaO 12…13,
MgO 5…6,
The possibility of using SMS tailings in the production of Portland cement clinker has been proven. The resulting cements are characterized by high strength properties.
The thermal effect of clinker formation (TEC) is defined as the algebraic sum of the heats of endothermic processes (decarbonization of limestone, dehydration of clay minerals, formation of a liquid phase) and exothermic reactions (oxidation of pyrite introduced by CMS tailings, formation of clinker phases).
The main advantages of using skarn-magnetite ore enrichment waste in cement production are:
Expansion of the raw material base due to man-made sources;
Saving natural raw materials while maintaining cement quality;
Reducing fuel and energy costs for clinker firing;
Possibility of producing low-energy active low-basic clinkers;
Solving environmental problems through rational waste disposal and reducing gas emissions into the atmosphere during clinker firing.
Biosensors
Biosensors are sensors based on immobilized enzymes. They allow you to quickly and efficiently analyze complex, multicomponent mixtures of substances. Currently, they are increasingly used in a number of branches of science, industry, agriculture and healthcare. The basis for the creation of automatic enzymatic analysis systems was the latest advances in the field of enzymology and engineering enzymology. The unique qualities of enzymes - specificity of action and high catalytic activity - contribute to the simplicity and high sensitivity of this analytical method, and the large number of enzymes known and studied to date makes it possible to constantly expand the list of analyzed substances.
Enzyme microcalorimetric sensors - use the thermal effect of an enzymatic reaction. It consists of two columns (measuring and control), filled with a carrier with an immobilized enzyme and equipped with thermistors. When the analyzed sample is passed through the measuring column, a chemical reaction occurs, which is accompanied by a recorded thermal effect. This type of sensor is interesting for its versatility.
Conclusion
So, after analyzing the practical application of the thermal effect of chemical reactions, we can conclude: the thermal effect is closely related to our everyday life, it is constantly being studied and is finding new applications in practice.
With the development of modern technologies, the warm effect has found its application in various industries. Chemical, military, construction, food, mining and many other industries use the thermal effect in their developments. It is used in internal combustion engines, refrigeration units and various combustion devices, as well as in the production of surgical instruments, heat-resistant coatings, new types of building materials and so on.
In modern conditions of constantly developing science, we are seeing the emergence of more and more new developments and discoveries in the field of production. This entails more and more new areas of application of the thermal effect of chemical reactions.
Chernykh E. A.
Bibliography
Musabekov Yu. S., Marcelin Berthelot, M., 1965; Centenaire de Marcelin Berthelot, 1827-1927, P., 1929.
Patent 852586 Russian Federation. MKI V 28 D 5/00. Method for dimensional processing of diamond / A.P.Grigoriev, S.H.Lifshits, P.P.Shamaev (Russian Federation). - 2 s.
