Calculations of the thermal effect of the reaction. Heat of reaction and thermochemical calculations
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The standard heat of formation (enthalpy of formation) of a substance is called the enthalpy of the reaction of formation of 1 mole of this substance from elements ( simple substances, that is, consisting of atoms of the same type) that are in the most stable standard state. Standard enthalpies of formation of substances (kJ/mol) are given in reference books. When using reference values, it is necessary to pay attention to the phase state of the substances participating in the reaction. The enthalpy of formation of the most stable simple substances is 0.
Corollary from Hess's law on calculating the thermal effects of chemical reactions based on the heats of formation : standard the thermal effect of a chemical reaction is equal to the difference between the heats of formation of the reaction products and the heats of formation of the starting substances, taking into account the stoichiometric coefficients (number of moles) of the reactants:
CH 4 + 2 CO = 3 C ( graphite ) + 2H 2 O.
gas gas tv. gas
The heats of formation of substances in the indicated phase states are given in table. 1.2.
Table 1.2
Heat of formation of substances
Solution
Since the reaction takes place at P= const, then we find the standard thermal effect in the form of a change in enthalpy based on the known heats of formation as a consequence of Hess’s law (formula (1.17):
ΔH O 298 = ( 2 (–241.81) + 3 0) – (–74.85 + 2 (–110.53)) = –187.71 kJ = –187710 J.
ΔH O 298 < 0, реакция является экзотермической, протекает с выделением теплоты.
We find the change in internal energy based on equation (1.16):
ΔU O 298 = ΔH O 298 – Δ ν RT.
For a given reaction, changes in the number of moles of gaseous substances due to the passage of a chemical reaction Δν = 2 – (1 + 2) = –1; T= 298 K, then
Δ U O 298 = –187710 – (–1) · 8.314 · 298 = –185232 J.
Calculation of standard thermal effects of chemical reactions using standard heats of combustion of substances participating in the reaction
The standard heat of combustion (enthalpy of combustion) of a substance
is the thermal effect of the complete oxidation of 1 mole of a given substance (to higher oxides or specially indicated compounds) with oxygen, provided that the initial and final substances have a standard temperature. Standard enthalpies of combustion of substances 
(kJ/mol) are given in reference books. When using a reference value, it is necessary to pay attention to the sign of the enthalpy of the combustion reaction, which is always exothermic ( Δ
H
<0), а в таблицах указаны величины
.
The enthalpy of combustion of higher oxides (for example, water and carbon dioxide) is 0.
Corollary from Hess's law on calculating the thermal effects of chemical reactions based on the heat of combustion : the standard thermal effect of a chemical reaction is equal to the difference between the heats of combustion of the starting substances and the heats of combustion of the reaction products, taking into account the stoichiometric coefficients (number of moles) of the reactants:
C 2 H 4 + H 2 O= C 2 N 5 HE.
Thermochemistry studies the thermal effects of chemical reactions. In many cases, these reactions occur at constant volume or constant pressure. From the first law of thermodynamics it follows that under these conditions heat is a function of state. At constant volume, heat is equal to the change in internal energy:
and at constant pressure - the change in enthalpy:
These equalities, when applied to chemical reactions, constitute the essence Hess's law:
The thermal effect of a chemical reaction occurring at constant pressure or constant volume does not depend on the reaction path, but is determined only by the state of the reactants and reaction products.
In other words, the thermal effect of a chemical reaction is equal to the change in the state function.
In thermochemistry, unlike other applications of thermodynamics, heat is considered positive if it is released in environment, i.e. If H < 0 или U
< 0. Под тепловым эффектом химической реакции
понимают значение H(which is simply called the "enthalpy of reaction") or U reactions.
If the reaction occurs in solution or in the solid phase, where the change in volume is negligible, then
H = U + (pV) U. (3.3)
If ideal gases participate in the reaction, then at constant temperature
H = U + (pV) = U+n. RT, (3.4)
where n is the change in the number of moles of gases in the reaction.
In order to facilitate comparison of enthalpies of different reactions, the concept of a “standard state” is used. The standard state is the state of a pure substance at a pressure of 1 bar (= 10 5 Pa) and a given temperature. For gases, this is a hypothetical state at a pressure of 1 bar, having the properties of an infinitely rarefied gas. Enthalpy of reaction between substances in standard states at temperature T, denote ( r means "reaction"). Thermochemical equations indicate not only the formulas of substances, but also their aggregate states or crystalline modifications.
Important consequences follow from Hess's law, which make it possible to calculate the enthalpies of chemical reactions.
Corollary 1.
equal to the difference between the standard enthalpies of formation of reaction products and reagents (taking into account stoichiometric coefficients):
Standard enthalpy (heat) of formation of a substance (f means "formation") at a given temperature is the enthalpy of the reaction of formation of one mole of this substance from elements, which are in the most stable standard state. According to this definition, the enthalpy of formation of the most stable simple substances in the standard state is 0 at any temperature. Standard enthalpies of formation of substances at a temperature of 298 K are given in reference books.
The concept of “enthalpy of formation” is used not only for ordinary substances, but also for ions in solution. In this case, the H + ion is taken as the reference point, for which the standard enthalpy of formation in an aqueous solution is assumed to be zero: 
Corollary 2. Standard enthalpy of a chemical reaction
equal to the difference between the enthalpies of combustion of the reactants and reaction products (taking into account stoichiometric coefficients):
(c means "combustion"). The standard enthalpy (heat) of combustion of a substance is the enthalpy of the reaction of complete oxidation of one mole of a substance. This consequence is usually used to calculate the thermal effects of organic reactions.
Corollary 3. The enthalpy of a chemical reaction is equal to the difference in the energies of the chemical bonds being broken and those formed.
Energy of communication A-B name the energy required to break a bond and separate the resulting particles over an infinite distance:
AB (g) A (g) + B (g) .
Communication energy is always positive.
Most thermochemical data in reference books are given at a temperature of 298 K. To calculate thermal effects at other temperatures, use Kirchhoff equation:
(differential form) (3.7)
(integral form) (3.8)
Where C p- the difference between the isobaric heat capacities of the reaction products and the starting substances. If the difference T 2 - T 1 is small, then you can accept C p= const. If there is a large temperature difference, it is necessary to use the temperature dependence C p(T) type:
where are the coefficients a, b, c etc. for individual substances they are taken from the reference book, and the sign indicates the difference between the products and reagents (taking into account the coefficients).
EXAMPLES
Example 3-1. The standard enthalpies of formation of liquid and gaseous water at 298 K are -285.8 and -241.8 kJ/mol, respectively. Calculate the enthalpy of vaporization of water at this temperature.
Solution. The enthalpies of formation correspond to the following reactions:
H 2 (g) + SO 2 (g) = H 2 O (l), H 1 0 = -285.8;
H 2 (g) + SO 2 (g) = H 2 O (g), H 2 0 = -241.8.
The second reaction can be carried out in two stages: first, burn hydrogen to form liquid water according to the first reaction, and then evaporate the water:
H 2 O (l) = H 2 O (g), H 0 isp = ?
Then, according to Hess's law,
H 1 0 + H 0 isp = H 2 0 ,
where H 0 isp = -241.8 - (-285.8) = 44.0 kJ/mol.
Answer. 44.0 kJ/mol.
Example 3-2. Calculate enthalpy of reaction
6C (g) + 6H (g) = C 6 H 6 (g)
a) by enthalpies of formation; b) by binding energies, under the assumption that the double bonds in the C 6 H 6 molecule are fixed.
Solution. a) Enthalpies of formation (in kJ/mol) are found in the reference book (for example, P.W. Atkins, Physical Chemistry, 5th edition, pp. C9-C15): f H 0 (C 6 H 6 (g)) = 82.93, f H 0 (C (g)) = 716.68, f H 0 (H (g)) = 217.97. The enthalpy of the reaction is:
r H 0 = 82.93 - 6,716.68 - 6,217.97 = -5525 kJ/mol.
b) In this reaction chemical bonds do not break, but only form. In the approximation of fixed double bonds, the C 6 H 6 molecule contains 6 C-H bonds, 3 C-C bonds and 3 C=C bonds. Bond energies (in kJ/mol) (P.W.Atkins, Physical Chemistry, 5th edition, p. C7): E(C-H) = 412, E(C-C) = 348, E(C=C) = 612. The enthalpy of the reaction is:
r H 0 = -(6,412 + 3,348 + 3,612) = -5352 kJ/mol.
The difference with the exact result -5525 kJ/mol is due to the fact that in the benzene molecule there are no C-C single bonds and C=C double bonds, but there are 6 aromatic C C bonds.
Answer. a) -5525 kJ/mol; b) -5352 kJ/mol.
Example 3-3. Using reference data, calculate the enthalpy of the reaction
3Cu (tv) + 8HNO 3(aq) = 3Cu(NO 3) 2(aq) + 2NO (g) + 4H 2 O (l)
Solution. The abbreviated ionic equation for the reaction is:
3Cu (s) + 8H + (aq) + 2NO 3 - (aq) = 3Cu 2+ (aq) + 2NO (g) + 4H 2 O (l).
According to Hess's law, the enthalpy of the reaction is equal to:
r H 0 = 4f H 0 (H 2 O (l)) + 2 f H 0 (NO (g)) + 3 f H 0 (Cu 2+ (aq)) - 2 f H 0 (NO 3 - (aq))
(the enthalpies of formation of copper and the H + ion are equal, by definition, 0). Substituting the values of enthalpies of formation (P.W.Atkins, Physical Chemistry, 5th edition, pp. C9-C15), we find:
r H 0 = 4 (-285.8) + 2 90.25 + 3 64.77 - 2 (-205.0) = -358.4 kJ
(based on three moles of copper).
Answer. -358.4 kJ.
Example 3-4. Calculate the enthalpy of combustion of methane at 1000 K, if the enthalpy of formation at 298 K is given: f H 0 (CH 4) = -17.9 kcal/mol, f H 0 (CO 2) = -94.1 kcal/mol, f H 0 (H 2 O (g)) = -57.8 kcal/mol. The heat capacities of gases (in cal/(mol. K)) in the range from 298 to 1000 K are equal to:
C p (CH 4) = 3.422 + 0.0178. T, C p(O2) = 6.095 + 0.0033. T,
C p (CO 2) = 6.396 + 0.0102. T, C p(H 2 O (g)) = 7.188 + 0.0024. T.
Solution. Enthalpy of methane combustion reaction
CH 4 (g) + 2O 2 (g) = CO 2 (g) + 2H 2 O (g)
at 298 K is equal to:
94.1 + 2 (-57.8) - (-17.9) = -191.8 kcal/mol.
Let's find the difference in heat capacities as a function of temperature:
C p = C p(CO2) + 2 C p(H 2 O (g)) - C p(CH 4) - 2 C p(O2) =
= 5.16 - 0.0094T(cal/(mol K)).
The enthalpy of the reaction at 1000 K is calculated using the Kirchhoff equation:
= + 
= -191800 + 5.16
(1000-298) - 0.0094 (1000 2 -298 2)/2 = -192500 cal/mol.
Answer. -192.5 kcal/mol.
TASKS
3-1. How much heat is required to transfer 500 g of Al (mp 658 o C, H 0 pl = 92.4 cal/g), taken at room temperature, into a molten state, if C p(Al TV) = 0.183 + 1.096 10 -4 T cal/(g K)?
3-2. The standard enthalpy of the reaction CaCO 3 (s) = CaO (s) + CO 2 (g) occurring in an open vessel at a temperature of 1000 K is 169 kJ/mol. What is the heat of this reaction, occurring at the same temperature, but in a closed vessel?
3-3. Calculate the standard internal energy of formation of liquid benzene at 298 K if the standard enthalpy of its formation is 49.0 kJ/mol.
3-4. Calculate the enthalpy of formation of N 2 O 5 (g) at T= 298 K based on the following data:
2NO(g) + O 2 (g) = 2NO 2 (g), H 1 0 = -114.2 kJ/mol,
4NO 2 (g) + O 2 (g) = 2N 2 O 5 (g), H 2 0 = -110.2 kJ/mol,
N 2 (g) + O 2 (g) = 2NO (g), H 3 0 = 182.6 kJ/mol.
3-5. The enthalpies of combustion of -glucose, -fructose and sucrose at 25 o C are equal to -2802,
-2810 and -5644 kJ/mol, respectively. Calculate the heat of hydrolysis of sucrose.
3-6. Determine the enthalpy of formation of diborane B 2 H 6 (g) at T= 298 K from the following data:
B 2 H 6 (g) + 3O 2 (g) = B 2 O 3 (tv) + 3H 2 O (g), H 1 0 = -2035.6 kJ/mol,
2B(tv) + 3/2 O 2 (g) = B 2 O 3 (tv), H 2 0 = -1273.5 kJ/mol,
H 2 (g) + 1/2 O 2 (g) = H 2 O (g), H 3 0 = -241.8 kJ/mol.
3-7. Calculate the heat of formation of zinc sulfate from simple substances at T= 298 K based on the following data.
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. Consequently, during a 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 large quantity energy will be released (or absorbed), i.e. the thermal effect is directly proportional to the amount of substance. Therefore, attribution 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 a chemical reaction equation that specifies state of aggregation 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 liquid state and 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 called the thermal effect of a chemical reaction of formation of this connection 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 school course chemistry. 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.
The thermal effect of a chemical reaction is equal to the sum of the heats of formation of the reaction products minus the sum of the heats of formation of the reactants (taking into account stoichiometric coefficients).
From a common sense point of view, 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 is 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 view. |
|||||
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). In contrast, chemical thermodynamics is the science of behavior 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 opposite sign: if energy is released from the system, the environment receives it and heats up, and 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“Fundamentals of 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.
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As a result of studying this topic, you will learn:
As a result of studying this topic, you will learn:
Study questions: |
6.1. Energy of chemical processes
6.1.1. Internal energy and enthalpy
In any process the law of conservation of energy is observed:
Q = Δ U + A.
This equality means that if heat Q is supplied to the system, then it is spent on changing the internal energy Δ U and doing work A.
Internal energy system is its total reserve, including the energy of translational and rotational movement molecules, the energy of movement of electrons in atoms, the energy of interaction of nuclei with electrons, nuclei with nuclei, etc., i.e. all types of energy except kinetic and potential energy systems as a whole.
The work done by the system during the transition from state 1, characterized by volume V 1, to state 2 (volume V 2) at constant pressure (expansion work) is equal to:
A = p(V 2 - V 1).
At constant pressure (p=const), taking into account the expression for the work of expansion, the law of conservation of energy will be written as follows:
Q = (U 2 + pV 2) – (U 1 + pV 1).
The sum of the internal energy of a system and the product of its volume and pressure is called enthalpy N:
Because the exact value the internal energy of the system is unknown, the absolute values of the enthalpies also cannot be obtained. They also have scientific significance practical use find enthalpy changes ΔH.
The internal energy U and enthalpy H are state functions systems. State functions are those characteristics of the system, changes in which are determined only by the final and initial state of the system, i.e. do not depend on the process path.
6.1.2. Exo- and endothermic processes
The occurrence of chemical reactions is accompanied by the absorption or release of heat. Exothermic called a reaction that occurs with the release of heat into the environment, and endothermic– with the absorption of heat from the environment.
Many processes in industry and laboratory practice take place at constant pressure and temperature (T=const, p=const). The energy characteristic of these processes is the change in enthalpy:
Q P = -Δ N.
For processes occurring at constant volume and temperature (T=const, V=const) Q V =-Δ U.
For exothermic reactions Δ H< 0, а в случае протекания эндотермической реакции Δ Н >0. For example,
N 2 (g) + SO 2 (g) = N 2 O (g); ΔН 298 = +82 kJ,
CH 4 (g) + 2O 2 (g) = CO 2 (g) + 2H 2 O (g); ΔH 298 = -802 kJ.
Chemical equations, which additionally indicate the thermal effect of the reaction (the value DN of the process), as well as the state of aggregation of substances and temperature, are called thermochemical equations.
In thermochemical equations, the phase state and allotropic modifications of the reactants and resulting substances are noted: g - gaseous, g - liquid, j - crystalline; S (diamond), S (monocle), C (graphite), C (diamond), etc.
6.1.3. Thermochemistry; Hess's law
Studies energy phenomena accompanying physical and chemical processes thermochemistry. The basic law of thermochemistry is the law formulated by the Russian scientist G.I. Hess in 1840.
Hess's law: the change in the enthalpy of a process depends on the type and state of the starting materials and reaction products, but does not depend on the path of the process.
By revising thermochemical effects Often, instead of the concept of “change in the enthalpy of a process,” the expression “enthalpy of a process” is used, meaning by this concept the value Δ H. It is incorrect to use the concept “thermal effect of a process” when formulating Hess’s law, since the value Q in the general case is not a function of state. As stated above, only at constant pressure Q P = -Δ N (at constant volume Q V = -Δ U).
Thus, the formation of PCl 5 can be considered as a result of the interaction of simple substances:
P (k, white) + 5/2Cl 2 (g) = PCl 5 (k) ; ΔH 1,
or as a result of a process occurring in several stages:
P (k, white) + 3/2Cl 2 (g) = PCl 3 (g); ΔH 2,
PCl 3(g) + Cl 2(g) = PCl 5(k); ΔH 3,
or in total:
P (k, white) + 5/2Cl 2 (g) = PCl 5 (k) ; Δ H 1 = Δ H 2 + Δ H 3.
6.1.4. Enthalpies of formation of substances
The enthalpy of formation is the enthalpy of the process of formation of a substance in a given state of aggregation from simple substances that are in stable modifications. The enthalpy of formation of sodium sulfate, for example, is the enthalpy of the reaction:
2Na (k) + S (rhombus) + 2O 2 (g) = Na 2 SO 4 (k).
The enthalpy of formation of simple substances is zero.
Since the thermal effect of the reaction depends on the state of the substances, temperature and pressure, when carrying out thermochemical calculations it was agreed to use standard enthalpies of formation– enthalpy of formation of substances located at a given temperature in standard condition. The real state of the substance at a given temperature and pressure of 101.325 kPa (1 atm) is taken as the standard state for substances in a condensed state. Reference books usually give standard enthalpies of formation of substances at a temperature of 25 o C (298 K), referred to 1 mole of substance (Δ H f o 298). The standard enthalpies of formation of some substances at T = 298 K are given in table. 6.1.
Table 6.1.
Standard enthalpies of formation (Δ H f o 298) of some substances
Substance |
Δ Н f o 298, kJ/mol |
Substance |
Δ Н f o 298, kJ/mol |
The standard enthalpies of formation of most complex substances are negative values. For a small number of unstable substances, Δ Н f o 298 > 0. Such substances, in particular, include nitric oxide (II) and nitric oxide (IV), Table 6.1.
6.1.5. Calculation of thermal effects of chemical reactions
To calculate the enthalpies of processes, a corollary from Hess’s law is used: the enthalpy 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, taking into account stoichiometric coefficients.
Let's calculate the enthalpy of decomposition of calcium carbonate. The process is described by the following equation:
CaCO 3 (k) = CaO (k) + CO 2 (g).
The enthalpy of this reaction will be equal to the sum of the enthalpies of formation of calcium oxide and carbon dioxide minus the enthalpy of formation of calcium carbonate:
Δ H o 298 = Δ H f o 298 (CaO (k)) + Δ H f o 298 (CO 2 (g)) - Δ H f o 298 (CaCO 3 (k)).
Using the data in Table 6.1. we get:
Δ H o 298 = - 635.1 -393.5 + 1206.8 = + 178.2 kJ.
From the data obtained it follows that the reaction under consideration is endothermic, i.e. proceeds with the absorption of heat.
CaO (k) + CO 2 (k) = CaCO 3 (k)
Accompanied by the release of heat. Its enthalpy will be equal
Δ H o 298 = -1206.8 +635.1 + 393.5 = -178.2 kJ.
6.2. Rate of chemical reactions
6.2.1. Reaction speed concept
The branch of chemistry that deals with the rate and mechanisms of chemical reactions is called chemical kinetics. One of the key concepts in chemical kinetics is the rate of a chemical reaction.
The rate of a chemical reaction is determined by the change in the concentration of reactants per unit time with a constant volume of the system.
Consider the following process:
Let at some point in time t 1 the concentration of substance A be equal to the value c 1 , and at the moment t 2 – to the value c 2 . Over the period of time from t 1 to t 2, the change in concentration will be Δ c = c 2 – c 1. average speed reaction is equal to:
The minus sign is placed because as the reaction proceeds (Δ t> 0) the concentration of the substance decreases (Δ s< 0), в то время, как скорость реакции является положительной величиной.
The rate of a chemical reaction depends on the nature of the reactants and on the reaction conditions: concentration, temperature, presence of a catalyst, pressure (for gas reactions) and some other factors. In particular, as the area of contact of substances increases, the reaction rate increases. The reaction rate also increases with increasing speed of mixing of the reactants.
Numerical value the reaction rate also depends on which component the reaction rate is calculated from. For example, the speed of the process
H 2 + I 2 = 2HI,
calculated from the change in the concentration of HI is twice the reaction rate calculated from the change in the concentration of the reagents H 2 or I 2.
6.2.2. Dependence of reaction rate on concentration; order and molecularity of the reaction
The basic law of chemical kinetics is law of mass action– establishes the dependence of the reaction rate on the concentration of reactants.
The reaction rate is proportional to the product of the concentrations of the reactants. For a reaction written in general form as
aA + bB = cC + dD,
The dependence of the reaction rate on concentration has the form:
v = k [A] α [B] β .
In this kinetic equation, k is the proportionality coefficient, called rate constant; [A] and [B] are the concentrations of substances A and B. The reaction rate constant k depends on the nature of the reactants and on temperature, but does not depend on their concentrations. Coefficients α and β are found from experimental data.
The sum of exponents in kinetic equations is called the total in order reactions. There are also private order reactions for one of the components. For example, for the reaction
H 2 + C1 2 = 2 HC1
The kinetic equation looks like this:
v = k 1/2,
those. general order is equal to 1.5 and the reaction orders for the components H 2 and C1 2 are equal to 1 and 0.5, respectively.
Molecularity reaction is determined by the number of particles whose simultaneous collision carries out an elementary act of chemical interaction. Elementary act (elementary stage)– a single act of interaction or transformation of particles (molecules, ions, radicals) into other particles. For elementary reactions, the molecularity and order of the reaction are the same. If the process is multistage and therefore writing the reaction equation does not reveal the mechanism of the process, the order of the reaction does not coincide with its molecularity.
Chemical reactions They are divided into simple (single-stage) and complex, occurring in several stages.
Monomolecular reaction is a reaction in which the elementary act is a chemical transformation of one molecule. For example:
CH 3 CHO (g) = CH 4 (g) + CO (g).
Bimolecular reaction– a reaction in which the elementary act occurs when two particles collide. For example:
H 2 (g) + I 2 (g) = 2 HI (g).
Trimolecular reaction- a simple reaction, the elementary act of which occurs during the simultaneous collision of three molecules. For example:
2NO (g) + O 2 (g) = 2 NO 2 (g).
It has been established that the simultaneous collision of more than three molecules, leading to the formation of reaction products, is practically impossible.
The law of mass action does not apply to reactions involving solids, since their concentrations are constant and they react only on the surface. The rate of such reactions depends on the size of the contact surface between the reacting substances.
6.2.3. Dependence of reaction rate on temperature
The rate of chemical reactions increases with increasing temperature. This increase is caused by an increase kinetic energy molecules. In 1884, the Dutch chemist Van't Hoff formulated the rule: With every 10 degree increase in temperature, the rate of chemical reactions increases by 2-4 times.
Van't Hoff's rule is written as:
,
where V t 1 and V t 2 are the reaction rates at temperatures t 1 and t 2; γ is the temperature coefficient of speed, equal to 2 - 4.
Van't Hoff's rule is used to approximate the effect of temperature on the rate of reaction. A more accurate equation describing the dependence of the reaction rate constant on temperature was proposed in 1889 by the Swedish scientist S. Arrhenius:
.
In the Arrhenius equation, A is a constant, E is the activation energy (J/mol); T – temperature, K.
According to Arrhenius, not all collisions of molecules lead to chemical transformations. Only molecules with some excess energy are able to react. This excess energy that colliding particles must have in order for a reaction to occur between them is called activation energy.
6.3. Concept of catalysis and catalysts
A catalyst is a substance that changes the rate of a chemical reaction, but remains chemically unchanged after the reaction is complete.
Some catalysts speed up the reaction, others, called inhibitors, slow down its progress. For example, adding a small amount of MnO 2 to hydrogen peroxide H2O2 as a catalyst causes rapid decomposition:
2 H 2 O 2 –(MnO 2) 2 H 2 O + O 2.
In the presence of small amounts of sulfuric acid, a decrease in the rate of decomposition of H 2 O 2 is observed. In this reaction, sulfuric acid acts as an inhibitor.
Depending on whether the catalyst is in the same phase as the reactants or forms an independent phase, they distinguish homogeneous And heterogeneous catalysis.
Homogeneous catalysis
In the case of homogeneous catalysis, the reactants and the catalyst are in the same phase, for example, gaseous. The mechanism of action of the catalyst is based on the fact that it interacts with reacting substances to form intermediate compounds.
Let us consider the mechanism of action of the catalyst. In the absence of a catalyst, the reaction
It proceeds very slowly. The catalyst forms a reactive intermediate product with the starting substances (for example, with substance B):
which reacts vigorously with another starting material to form the final reaction product:
VK + A = AB + K.
Homogeneous catalysis occurs, for example, in the process of oxidation of sulfur(IV) oxide to sulfur(VI) oxide, which occurs in the presence of nitrogen oxides.
Homogeneous reaction
2 SO 2 + O 2 = 2 SO 3
in the absence of a catalyst it goes very slowly. But when a catalyst (NO) is introduced, an intermediate compound (NO2) is formed:
O 2 + 2 NO = 2 NO 2,
which easily oxidizes SO 2:
NO 2 + SO 2 = SO 3 + NO.
The activation energy of the latter process is very small, so the reaction proceeds with high speed. Thus, the effect of catalysts is reduced to reducing the activation energy of the reaction.
Heterogeneous catalysis
In heterogeneous catalysis, the catalyst and reactants are in different phases. The catalyst is usually in a solid state, and the reactants are in liquid or gaseous states. In heterogeneous catalysis, the acceleration of the process is usually associated with the catalytic effect of the catalyst surface.
Catalysts are distinguished by their selectivity of action. So, for example, in the presence of an aluminum oxide catalyst Al 2 O 3 at 300 o C, water and ethylene are obtained from ethyl alcohol:
C 2 H 5 OH – (Al 2 O 3) C 2 H 4 + H 2 O.
At the same temperature, but in the presence of copper Cu as a catalyst, dehydrogenation of ethyl alcohol occurs:
C 2 H 5 OH – (Cu) CH 3 CHO + H 2 .
Small amounts of some substances reduce or even completely destroy the activity of catalysts (catalyst poisoning). Such substances are called catalytic poisons. For example, oxygen causes reversible poisoning of the iron catalyst during the synthesis of NH 3. The activity of the catalyst can be restored by passing a fresh mixture of nitrogen and hydrogen purified from oxygen. Sulfur causes irreversible poisoning of the catalyst during the synthesis of NH 3. Its activity cannot be restored by passing a fresh N 2 + H 2 mixture.
Substances that enhance the action of reaction catalysts are called promoters, or activators(platinum catalysts are promoted, for example, by adding iron or aluminum).
The mechanism of heterogeneous catalysis is more complex. The adsorption theory of catalysis is used to explain it. The surface of the catalyst is heterogeneous, so there are so-called active centers on it. Adsorption of reacting substances occurs at active centers. The latter process brings the reacting molecules closer together and increases their chemical activity, since the bonds between atoms in adsorbed molecules are weakened and the distance between atoms increases.
On the other hand, it is believed that the accelerating effect of the catalyst in heterogeneous catalysis is due to the fact that the reactants form intermediate compounds (as in the case of homogeneous catalysis), which leads to a decrease in the activation energy.
6.4. Chemical equilibrium
Irreversible and reversible reactions
Reactions that proceed in only one direction and are completed complete transformation initial substances into final substances are called irreversible.
Irreversible, i.e. proceeding to completion are reactions in which
Chemical reactions that can go in opposite directions are called reversible. Typical reversible reactions are the synthesis of ammonia and the oxidation of sulfur(IV) oxide to sulfur(VI) oxide:
N 2 + 3 H 2 2 NH 3,
2 SO 2 + O 2 2 SO 3 .
When writing equations for reversible reactions, instead of an equal sign, use two arrows pointing in opposite directions.
In reversible reactions, the rate of the direct reaction at the initial moment of time has a maximum value, which decreases as the concentration of the initial reagents decreases. On the contrary, the reverse reaction initially has a minimum rate, increasing as the concentration of products increases. As a result, a moment comes when the rates of the forward and reverse reactions become equal and chemical equilibrium is established in the system.
Chemical equilibrium
The state of a system of reacting substances in which the rate of the forward reaction becomes equal to the rate of the reverse reaction is called chemical equilibrium.
Chemical equilibrium is also called true equilibrium. In addition to the equality of the rates of forward and reverse reactions, true (chemical) equilibrium is characterized by the following features:
- the state of the system is the same regardless of which side the system approaches equilibrium from - from the side of the starting substances or from the side of the reaction products.
the invariability of the state of the system is caused by the occurrence of direct and reverse reactions, that is, the equilibrium state is dynamic;
the state of the system remains unchanged over time if there is no external influence on the system;
any external influence causes a shift in the equilibrium of the system; however, if the external influence is removed, the system returns to its original state;
It should be distinguished from the true apparent equilibrium. For example, a mixture of oxygen and hydrogen in a closed vessel at room temperature can be stored for an indefinitely long time. However, the initiation of the reaction (electrical discharge, ultraviolet irradiation, increased temperature) causes the irreversible reaction of water formation.
6.5. Le Chatelier's principle
The influence of changes in external conditions on the equilibrium position is determined Le Châtel's principle e (France, 1884): if any external influence is applied to a system in a state of equilibrium, then the equilibrium in the system will shift towards weakening this influence.
Le Chatelier's principle applies not only to chemical processes, but also to physical ones, such as boiling, crystallization, dissolution, etc.
Consider the impact various factors for chemical equilibrium using the example of ammonia synthesis reaction:
N 2 + 3 H 2 2 NH 3 ; ΔH = -91.8 kJ.
Effect of concentration on chemical equilibrium.
In accordance with Le Chatelier's principle, an increase in the concentration of starting substances shifts the equilibrium towards the formation of reaction products. An increase in the concentration of reaction products shifts the equilibrium towards the formation of the starting substances.
In the process of ammonia synthesis discussed above, the introduction of additional amounts of N 2 or H 2 into the equilibrium system causes a shift in the equilibrium in the direction in which the concentration of these substances decreases; therefore, the equilibrium shifts towards the formation of NH3. An increase in ammonia concentration shifts the equilibrium towards the parent substances.
The catalyst accelerates both forward and reverse reactions equally, therefore the introduction of a catalyst does not affect the chemical equilibrium.
Effect of temperature on chemical equilibrium
As the temperature increases, the equilibrium shifts towards the endothermic reaction, and as the temperature decreases, towards the exothermic reaction.
The degree of equilibrium shift is determined absolute value thermal effect: the larger the ΔH value of the reaction, the greater the influence of temperature.
In the ammonia synthesis reaction under consideration, an increase in temperature will shift the equilibrium towards the starting substances.
Effect of pressure on chemical equilibrium
Changes in pressure affect the chemical equilibrium involving gaseous substances. According to Le Chatelier's principle, an increase in pressure shifts the equilibrium towards the reaction that occurs with a decrease in the volume of gaseous substances, and a decrease in pressure shifts the equilibrium in the opposite direction. The reaction of ammonia synthesis proceeds with a decrease in the volume of the system (there are four volumes on the left side of the equation, two on the right). Therefore, an increase in pressure shifts the equilibrium towards the formation of ammonia. A decrease in pressure will shift the equilibrium to reverse side. If in the equation of a reversible reaction the number of molecules of gaseous substances on the right and left sides are equal (the reaction proceeds without changing the volume of gaseous substances), then pressure does not affect the equilibrium position in this system.
Exercise 81.
Calculate the amount of heat that will be released during the reduction of Fe 2 O 3 metallic aluminum if 335.1 g of iron was obtained. Answer: 2543.1 kJ.
Solution:
Reaction equation:
= (Al 2 O 3) - (Fe 2 O 3) = -1669.8 -(-822.1) = -847.7 kJ
Calculation of the amount of heat that is released when receiving 335.1 g of iron is made from the proportion:
(2 . 55,85) : -847,7 = 335,1 : X; x = (0847.7 . 335,1)/ (2 . 55.85) = 2543.1 kJ,
where 55.85 atomic mass gland.
Answer: 2543.1 kJ.
Thermal effect of reaction
Task 82.
Gaseous ethyl alcohol C2H5OH can be obtained by the interaction of ethylene C 2 H 4 (g) and water vapor. Write the thermochemical equation for this reaction, having first calculated its thermal effect. Answer: -45.76 kJ.
Solution:
The reaction equation is:
C 2 H 4 (g) + H 2 O (g) = C2H 5 OH (g); = ?
The values of standard heats of formation of substances are given in special tables. Considering that the heats of formation of simple substances are conventionally assumed to be zero. Let's calculate the thermal effect of the reaction using a consequence of Hess's law, we get:
= (C 2 H 5 OH) – [ (C 2 H 4) + (H 2 O)] =
= -235.1 -[(52.28) + (-241.83)] = - 45.76 kJ
Reaction equations in which about the symbols chemical compounds their states of aggregation or crystalline modification are indicated, as well as the numerical value of thermal effects, called thermochemical. In thermochemical equations, unless specifically stated, the values of thermal effects at constant pressure Q p are indicated equal to the change in enthalpy of the system. The value is usually given on the right side of the equation, separated by a comma or semicolon. The following abbreviated designations for the state of aggregation of a substance are accepted: G- gaseous, and- liquid, To
If heat is released as a result of a reaction, then< О. Учитывая сказанное, составляем термохимическое уравнение данной в примере реакции:
C 2 H 4 (g) + H 2 O (g) = C 2 H 5 OH (g); = - 45.76 kJ.
Answer:- 45.76 kJ.
Task 83.
Calculate the thermal effect of the reduction reaction of iron (II) oxide with hydrogen based on the following thermochemical equations:
a) EO (k) + CO (g) = Fe (k) + CO 2 (g); = -13.18 kJ;
b) CO (g) + 1/2O 2 (g) = CO 2 (g); = -283.0 kJ;
c) H 2 (g) + 1/2O 2 (g) = H 2 O (g); = -241.83 kJ.
Answer: +27.99 kJ.
Solution:
The reaction equation for the reduction of iron (II) oxide with hydrogen has the form:
EeO (k) + H 2 (g) = Fe (k) + H 2 O (g); = ?
= (H2O) – [ (FeO)
The heat of formation of water is given by the equation
H 2 (g) + 1/2O 2 (g) = H 2 O (g); = -241.83 kJ,
and the heat of formation of iron (II) oxide can be calculated by subtracting equation (a) from equation (b).
=(c) - (b) - (a) = -241.83 – [-283.o – (-13.18)] = +27.99 kJ.
Answer:+27.99 kJ.
Task 84.
When gaseous hydrogen sulfide and carbon dioxide interact, water vapor and carbon disulfide CS 2 (g) are formed. Write the thermochemical equation for this reaction and first calculate its thermal effect. Answer: +65.43 kJ.
Solution:
G- gaseous, and- liquid, To-- crystalline. These symbols are omitted if the aggregative state of the substances is obvious, for example, O 2, H 2, etc.
The reaction equation is:
2H 2 S (g) + CO 2 (g) = 2H 2 O (g) + CS 2 (g); = ?
The values of standard heats of formation of substances are given in special tables. Considering that the heats of formation of simple substances are conventionally assumed to be zero. The thermal effect of a reaction can be calculated using a corollary of Hess's law:
= (H 2 O) + (СS 2) – [(H 2 S) + (СO 2)];
= 2(-241.83) + 115.28 – = +65.43 kJ.
2H 2 S (g) + CO 2 (g) = 2H 2 O (g) + CS 2 (g); = +65.43 kJ.
Answer:+65.43 kJ.
Thermochemical reaction equation
Task 85.
Write the thermochemical equation for the reaction between CO (g) and hydrogen, as a result of which CH 4 (g) and H 2 O (g) are formed. How much heat will be released during this reaction if 67.2 liters of methane were produced in terms of normal conditions? Answer: 618.48 kJ.
Solution:
Reaction equations in which their state of aggregation or crystal modification, as well as the numerical value of thermal effects are indicated next to the symbols of chemical compounds, are called thermochemical. In thermochemical equations, unless specifically stated, the values of thermal effects at constant pressure Q p equal to the change in enthalpy of the system are indicated. The value is usually given on the right side of the equation, separated by a comma or semicolon. The following abbreviated designations for the state of aggregation of a substance are accepted: G- gaseous, and- something, To- crystalline. These symbols are omitted if the aggregative state of the substances is obvious, for example, O 2, H 2, etc.
The reaction equation is:
CO (g) + 3H 2 (g) = CH 4 (g) + H 2 O (g); = ?
The values of standard heats of formation of substances are given in special tables. Considering that the heats of formation of simple substances are conventionally assumed to be zero. The thermal effect of a reaction can be calculated using a corollary of Hess's law:
= (H 2 O) + (CH 4) – (CO)];
= (-241.83) + (-74.84) – (-110.52) = -206.16 kJ.
The thermochemical equation will be:
22,4 : -206,16 = 67,2 : X; x = 67.2 (-206.16)/22?4 = -618.48 kJ; Q = 618.48 kJ.
Answer: 618.48 kJ.
Heat of formation
Task 86.
The thermal effect of which reaction is equal to the heat of formation. Calculate the heat of formation of NO based on the following thermochemical equations:
a) 4NH 3 (g) + 5O 2 (g) = 4NO (g) + 6H 2 O (l); = -1168.80 kJ;
b) 4NH 3 (g) + 3O 2 (g) = 2N 2 (g) + 6H 2 O (l); = -1530.28 kJ
Answer: 90.37 kJ.
Solution:
The standard heat of formation is equal to the heat of reaction of the formation of 1 mole of this substance from simple substances under standard conditions (T = 298 K; p = 1.0325.105 Pa). The formation of NO from simple substances can be represented as follows:
1/2N 2 + 1/2O 2 = NO
Given is reaction (a), which produces 4 mol of NO, and given reaction (b), which produces 2 mol of N2. Oxygen is involved in both reactions. Therefore, to determine the standard heat of formation of NO, we compose the following Hess cycle, i.e., we need to subtract equation (a) from equation (b):
Thus, 1/2N 2 + 1/2O 2 = NO; = +90.37 kJ.
Answer: 618.48 kJ.
Task 87.
Crystalline ammonium chloride is formed by the reaction of ammonia and hydrogen chloride gases. Write the thermochemical equation for this reaction, having first calculated its thermal effect. How much heat will be released if 10 liters of ammonia were consumed in the reaction, calculated under normal conditions? Answer: 78.97 kJ.
Solution:
Reaction equations in which their state of aggregation or crystal modification, as well as the numerical value of thermal effects are indicated next to the symbols of chemical compounds, are called thermochemical. In thermochemical equations, unless specifically stated, the values of thermal effects at constant pressure Q p equal to the change in enthalpy of the system are indicated. The value is usually given on the right side of the equation, separated by a comma or semicolon. The following have been accepted: To-- crystalline. These symbols are omitted if the aggregative state of the substances is obvious, for example, O 2, H 2, etc.
The reaction equation is:
NH 3 (g) + HCl (g) = NH 4 Cl (k). ;
The values of standard heats of formation of substances are given in special tables. Considering that the heats of formation of simple substances are conventionally assumed to be zero. The thermal effect of a reaction can be calculated using a corollary of Hess's law:
= ?
= (NH4Cl) – [(NH 3) + (HCl)];
The thermochemical equation will be:
= -315.39 – [-46.19 + (-92.31) = -176.85 kJ.
22,4 : -176,85 = 10 : The heat released during the reaction of 10 liters of ammonia in this reaction is determined from the proportion:
Answer: X; x = 10 (-176.85)/22.4 = -78.97 kJ; Q = 78.97 kJ.