Calculation of thermal effects of reactions in an indirect way. The rate of chemical reactions
In thermochemistry, the amount of heat Q that is released or absorbed as a result of a chemical reaction is called thermal effect. Reactions that release heat are called exothermic (Q>0), and with the absorption of heat - endothermic (Q<0 ).
In thermodynamics, respectively, the processes in which heat is released are called exothermic, and the processes in which heat is absorbed - endothermic.
According to the corollary of the first law of thermodynamics for isochoric-isothermal processes, the thermal effect is equal to the change in the internal energy of the system .
Since in thermochemistry the opposite sign is used with respect to thermodynamics, then .
For isobaric-isothermal processes, the thermal effect is equal to the change in the enthalpy of the system .
If D H > 0- the process proceeds with the absorption of heat and is endothermic.
If D H< 0 - the process is accompanied by the release of heat and is exothermic.
From the first law of thermodynamics it follows Hess' law:
the thermal effect of chemical reactions depends only on the type and state of the initial substances and final products, but does not depend on the path of transition from the initial state to the final one.
A consequence of this law is the rule that with thermochemical equations, you can perform the usual algebraic operations.
As an example, consider the reaction of coal oxidation to CO 2 .
The transition from the initial substances to the final one can be carried out by directly burning coal to CO 2:
C (t) + O 2 (g) \u003d CO 2 (g).
The thermal effect of this reaction Δ H 1.
This process can be carried out in two stages (Fig. 4). At the first stage, carbon burns to CO by the reaction
C (t) + O 2 (g) \u003d CO (g),
on the second CO burns down to CO 2
CO (t) + O 2 (g) \u003d CO 2 (g).
The thermal effects of these reactions, respectively, Δ H 2 and Δ H 3.
Rice. 4. Scheme of the combustion process of coal to CO 2
All three processes are widely used in practice. Hess's law allows you to relate the thermal effects of these three processes by the equation:
Δ H 1=Δ H 2 + Δ H 3.
The thermal effects of the first and third processes can be measured relatively easily, but the combustion of coal to carbon monoxide at high temperatures is difficult. Its thermal effect can be calculated:
Δ H 2=Δ H 1 - Δ H 3.
Values H 1 and Δ H 2 depend on the type of coal used. Value Δ H 3 not related to this. During the combustion of one mole of CO at constant pressure at 298K, the amount of heat is Δ H 3= -283.395 kJ/mol. Δ H 1\u003d -393.86 kJ / mol at 298K. Then at 298K Δ H 2\u003d -393.86 + 283.395 \u003d -110.465 kJ / mol.
Hess's law makes it possible to calculate the thermal effects of processes for which there are no experimental data or for which they cannot be measured under the required conditions. This also applies to chemical reactions, and to the processes of dissolution, evaporation, crystallization, adsorption, etc.
When applying Hess's law, the following conditions must be strictly observed:
Both processes must have really the same start states and really the same end states;
Not only the chemical compositions of the products should be the same, but also the conditions for their existence (temperature, pressure, etc.) and the state of aggregation, and for crystalline substances, the crystalline modification.
When calculating the thermal effects of chemical reactions based on the Hess law, two types of thermal effects are usually used - the heat of combustion and the heat of formation.
The heat of education called the thermal effect of the reaction of formation of a given compound from simple substances.
Heat of combustion called the thermal effect of the reaction of oxidation of a given compound with oxygen with the formation of higher oxides of the corresponding elements or compounds of these oxides.
Reference values of thermal effects and other quantities are usually referred to the standard state of matter.
As standard condition individual liquid and solid substances take on their state at a given temperature and at a pressure equal to one atmosphere, and for individual gases, their state is such that at a given temperature and pressure equal to 1.01 10 5 Pa (1 atm.), They have properties of an ideal gas. To facilitate calculations, reference data refer to standard temperature 298 K.
If any element can exist in several modifications, then such a modification is accepted as standard, which is stable at 298 K and atmospheric pressure equal to 1.01 10 5 Pa (1 atm.)
All quantities related to the standard state of substances are marked with a superscript in the form of a circle: 
. In metallurgical processes, most compounds are formed with the release of heat, so for them the enthalpy increment. For elements in the standard state, the value .
Using the reference data of the standard heats of formation of the substances involved in the reaction, one can easily calculate the heat effect of the reaction.
From the law of Hess it follows:the thermal effect of the reaction is equal to the difference between the heats of formation of all substances indicated on the right side of the equation(final substances or reaction products) , and the heats of formation of all substances indicated on the left side of the equation(starting materials) , taken with coefficients equal to the coefficients in front of the formulas of these substances in the reaction equation:
where n- the number of moles of the substance involved in the reaction.
Example. Let us calculate the thermal effect of the reaction Fe 3 O 4 + CO = 3FeO + CO 2 . The heats of formation of the substances involved in the reaction are: for Fe 3 O 4, for CO, for FeO, for CO 2.
Thermal effect of the reaction:
Since , the reaction at 298K is endothermic, i.e. goes with the absorption of heat.
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.
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 needs to be spent in order to destroy this bond.
In general, a chemical reaction is the transformation of one substance into another. Consequently, in the course of a chemical reaction, some bonds are broken and others are formed, i.e. energy conversion.
The fundamental law of physics says that energy does not arise from nothing and does not disappear without a trace, but only passes from one form to another. Due to its universality, this principle obviously applies to a chemical reaction.
Thermal effect of a chemical reaction called the amount of heat
released (or absorbed) during the reaction and referred to 1 mol of the reacted (or formed) substance.
The thermal effect is denoted by the letter Q and is usually measured in kJ/mol or kcal/mol.
If the 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 the 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 located lower in energy (more stable) than the reactants.
It is clear that the more matter reacts, the more energy is released (or absorbed), i.e. the thermal effect is directly proportional to the amount of substance. Therefore, the assignment of the thermal effect to 1 mol of a substance is due to our desire to compare the thermal effects of various reactions with each other.
Lecture 6. Thermochemistry. Thermal effect of a chemical reaction Example 1 . During the reduction of 8.0 g of copper(II) oxide with hydrogen, metallic copper and water vapor were formed and 7.9 kJ of heat was released. Calculate the thermal effect of the copper(II) oxide reduction reaction.
Decision . Reaction equation CuO (solid) + H2 (g) = Cu (solid) + H2 O (g) + Q (*)
Let's make a proportion for the reduction of 0.1 mol - 7.9 kJ is released; for the restoration of 1 mol - x kJ is released
Where x = + 79 kJ/mol. Equation (*) becomes
CuO (solid) + H2 (g) = Cu (solid) + H2 O (g) +79 kJ
Thermochemical equation- this is an equation of a chemical reaction, in which the state of aggregation of the components of the reaction mixture (reagents and products) and the thermal effect of the reaction are indicated.
So, in order to melt ice or evaporate water, it is required to expend certain amounts of heat, while when liquid water freezes or water vapor condenses, the same amount of heat is released. That is why we are cold when we get out of the water (evaporation of water from the surface of the body requires energy), and sweating is a biological defense mechanism against overheating of the body. On the contrary, the freezer freezes water and heats the surrounding room, giving it excess heat.
This example shows the thermal effects of a change in the state of aggregation of water. Heat of fusion (at 0o C) λ = 3.34×105 J/kg (physics), or Qpl. \u003d - 6.02 kJ / mol (chemistry), heat of evaporation (vaporization) (at 100o C) q \u003d 2.26 × 106 J / kg (physics) or Qisp. \u003d - 40.68 kJ / mol (chemistry).
melting
evaporation |
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mod 298.
Lecture 6. Thermochemistry. The 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 precipitation (crystallization) from the gas phase, it is also possible to calculate or measure the thermal effect for them.
It is clear that in every substance there are chemical bonds, therefore, every substance has a certain amount of energy. However, not all substances can be converted into each other by a single chemical reaction. Therefore, we agreed to introduce a standard state.
standard state of matter is the state of aggregation of a substance at a temperature of 298 K and a pressure of 1 atmosphere in the most stable allotropic modification under these conditions.
Standard Conditions is a temperature of 298 K and a pressure of 1 atmosphere. Standard conditions (standard state) is denoted by index 0 .
The standard heat of formation of the compound called 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 denoted by the symbol Q 0 For many compounds, the standard heats of formation are given in reference books of physicochemical quantities.
The standard heats of formation of simple substances are 0. For example, Q0 arr.298 (O2, gas) = 0, Q0 arr.298 (C, solid, graphite) = 0.
For example . Write down the thermochemical equation for the formation of copper(II) sulfate. From the reference book Q0 arr. 298 (CuSO4 ) = 770 kJ/mol.
Cu (s.) + S (s.) + 2O2 (g.) = CuSO4 (s.) + 770 kJ.
Note: the thermochemical equation can be written for any substance, but it must be understood that in real life the reaction occurs in a completely different way: copper (II) and sulfur (IV) oxides are formed from the listed reagents when heated, but copper (II) sulfate is not formed . An important conclusion: the thermochemical equation is a model that allows calculations, it is in good agreement with other thermochemical data, but does not withstand 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 methods is the first step against the background of the pyramid of Cheops - one cannot do without it, but one cannot rise high.
Example 2 . Calculate the thermal effect of condensation of water with a mass of 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 water condensation 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 path of the process 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 Hessian law has an incredibly important corollary of Hess' 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 the point of view of common sense, this consequence corresponds to a process in which all the reactants were first converted into simple substances, which were then assembled in a new way, so that the reaction products were obtained.
In the form of an equation, 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 the Hess law has the form Q = ∑ b j × Q arr 0 .298
k Bk + Q
(A i )
Lecture 6. Thermochemistry. The 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 of a very large number of reactions with a sufficiently high accuracy.
Example 3 . (Consequence of Hess' 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 if this reaction is exothermic or endothermic?
Solution: Consequence of Hess' law
Q = 3 Q0 | D ) +Q 0 | (CO ,g ) −Q 0 | D ) −Q 0 | O, d) - in general terms. |
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mod ,298 | mod ,298 | mod ,298 | mod ,298 | ||||||
Q arr0 | 298 (H 2, g) \u003d 0 | A simple substance in its standard state |
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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 | D) = 74.6 | |||||||||
mod ,298 | mod ,298 | mod ,298 | |||||||||
Plugging the values into the equation
Q \u003d 0 + 110.5 - 74.6 - 241.8 \u003d -205.9 kJ / mol, the reaction is strongly endothermic.
Answer: Q \u003d -205.9 kJ / mol, endothermic
Example 4. (Application of Hess' law). Known heats of reactions
C (solid) + ½ O (g) \u003d CO (g) + 110.5 kJ
C (s.) + O2 (g.) = CO2 (g.) + 393.5 kJ Find the heat effect of the reaction 2CO (g.) + O2 (g.) = 2CO2 (g.). Solution We multiply the first and second equations on 2
2C (s.) + O2 (g.) \u003d 2CO (g.) + 221 kJ 2C (s.) + 2O2 (g.) \u003d 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, the environment receives it and heats up, and the system loses energy.
Literature:
1. textbook, V.V. Eremin, N.E. Kuzmenko and others, Chemistry grade 9, paragraph 19,
2. Educational and methodical 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. Fundamentals 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 matter from elements.
Standard enthalpy of formation.
Attention!
We are moving on to solving computational problems, therefore, from now on, a calculator is desirable for seminars in chemistry.
7. Calculate the thermal effect of the reaction under standard conditions: Fe 2 O 3 (t) + 3 CO (g) \u003d 2 Fe (t) + 3 CO 2 (g), if the heat of formation: Fe 2 O 3 (t) \u003d - 821.3 kJ / mol; CO (g ) = – 110.5 kJ/mol;
CO 2 (g) \u003d - 393.5 kJ / mol.
Fe 2 O 3 (t) + 3 CO (g) \u003d 2 Fe (t) + 3 CO 2 (g),
Knowing the standard thermal effects of combustion of the initial substances and reaction products, we calculate the thermal effect of the reaction under standard conditions:
16. Dependence of the rate of a chemical reaction on temperature. Van't Hoff's rule. Temperature coefficient of reaction.
Only collisions between active molecules lead to reactions, the average energy of which exceeds the average energy of the participants in the reaction.
When a certain activation energy E is communicated to molecules (excess energy above the average), the potential energy of interaction of atoms in molecules decreases, bonds within molecules weaken, molecules become reactive.
The activation energy is not necessarily supplied from the outside; it can be imparted to some part of the molecules by redistributing the energy during their collisions. According to Boltzmann, among N molecules there is the following number of active molecules N with increased energy :
N N e – E / RT
where E is the activation energy, showing the necessary excess of energy, compared with the average level, that molecules must have in order for the reaction to become possible; the rest of the designations are well known.
During thermal activation for two temperatures T 1 and T 2 the ratio of the rate constants will be:
, (2)
, (3)
which allows you to determine the activation energy by measuring the reaction rate at two different temperatures T 1 and T 2 .
An increase in temperature by 10 0 increases the reaction rate by 2–4 times (approximate van't Hoff rule). The number showing how many times the reaction rate (and hence the rate constant) increases with an increase in temperature by 10 0 is called the temperature coefficient of the reaction:
(4)
.(5)
This means, for example, that with an increase in temperature by 100 0 for a conditionally accepted increase in the average rate by 2 times ( = 2), the reaction rate increases by 2 10 , i.e. approximately 1000 times, and when = 4 - 4 10 , i.e. 1000000 times. The van't Hoff rule is applicable to reactions occurring at relatively low temperatures in a narrow range. The sharp increase in the reaction rate with increasing temperature is explained by the fact that the number of active molecules increases exponentially.
25. Van't Hoff chemical reaction isotherm equation.
In accordance with the law of mass action for an arbitrary reaction
and A + bB = cC + dD
The equation for the rate of a direct reaction can be written:
,
and for the rate of the reverse reaction:
.
As the reaction proceeds from left to right, the concentrations of substances A and B will decrease and the rate of the forward reaction will decrease. On the other hand, as reaction products C and D accumulate, the reaction rate will increase from right to left. There comes a moment when the speeds υ 1 and υ 2 become the same, the concentrations of all substances remain unchanged, therefore,
,Where K c = k 1 / k 2 =
.
The constant value K c, equal to the ratio of the rate constants of the forward and reverse reactions, quantitatively describes the state of equilibrium through the equilibrium concentrations of the starting substances and the products of their interaction (in terms of their stoichiometric coefficients) and is called the equilibrium constant. The equilibrium constant is constant only for a given temperature, i.e.
K c \u003d f (T). The equilibrium constant of a chemical reaction is usually expressed as a ratio, the numerator of which is the product of the equilibrium molar concentrations of the reaction products, and the denominator is the product of the concentrations of the starting substances.
If the reaction components are a mixture of ideal gases, then the equilibrium constant (K p) is expressed in terms of the partial pressures of the components:
.
For the transition from K p to K with we use the equation of state P · V = n · R · T. Insofar as
, then P = C·R·T. .It follows from the equation that K p = K s, provided that the reaction proceeds without changing the number of moles in the gas phase, i.e. when (c + d) = (a + b).
If the reaction proceeds spontaneously at constant P and T or V and T, then the valuesG and F of this reaction can be obtained from the equations:
,
where C A, C B, C C, C D are the nonequilibrium concentrations of the initial substances and reaction products.
,where P A, P B, P C, P D are the partial pressures of the initial substances and reaction products.
The last two equations are called the van't Hoff chemical reaction isotherm equations. This relation makes it possible to calculate the values of G and F of the reaction, to determine its direction at different concentrations of the initial substances.
It should be noted that both for gas systems and for solutions, with the participation of solids in the reaction (i.e. for heterogeneous systems), the concentration of the solid phase is not included in the expression for the equilibrium constant, since this concentration is practically constant. So for the reaction
2 CO (g) \u003d CO 2 (g) + C (t)
the equilibrium constant is written as
.The dependence of the equilibrium constant on temperature (for temperature T 2 relative to temperature T 1) is expressed by the following van't Hoff equation:
,
where Н 0 is the thermal effect of the reaction.
For an endothermic reaction (the reaction proceeds with the absorption of heat), the equilibrium constant increases with increasing temperature, the system, as it were, resists heating.
34. Osmosis, osmotic pressure. Van't Hoff equation and osmotic coefficient.
Osmosis is the spontaneous movement of solvent molecules through a semipermeable membrane that separates solutions of different concentrations from a solution of a lower concentration to a solution of a higher concentration, which leads to the dilution of the latter. As a semi-permeable membrane, through small holes of which only small solvent molecules can selectively pass and large or solvated molecules or ions are retained, a cellophane film is often used - for high molecular weight substances, and for low molecular weight - a copper ferrocyanide film. The process of solvent transfer (osmosis) can be prevented if an external hydrostatic pressure is applied to a solution with a higher concentration (under equilibrium conditions this will be the so-called osmotic pressure, denoted by the letter ). To calculate the value of in solutions of non-electrolytes, the empirical Van't Hoff equation is used:
where C is the molar concentration of the substance, mol/kg;
R is the universal gas constant, J/mol K.
The value of osmotic pressure is proportional to the number of molecules (in the general case, the number of particles) of one or more substances dissolved in a given volume of solution, and does not depend on their nature and the nature of the solvent. In solutions of strong or weak electrolytes, the total number of individual particles increases due to the dissociation of molecules; therefore, it is necessary to introduce the appropriate proportionality coefficient, called the isotonic coefficient, into the equation for calculating the osmotic pressure.
i C R T,
where i is the isotonic coefficient, calculated as the ratio of the sum of the numbers of ions and undissociated electrolyte molecules to the initial number of molecules of this substance.
So, if the degree of electrolyte dissociation, i.e. the ratio of the number of molecules decomposed into ions to the total number of molecules of the solute is and the electrolyte molecule decomposes into n ions, then the isotonic coefficient is calculated as follows:
i = 1 + (n – 1) ,(i > 1).
For strong electrolytes, we can take = 1, then i = n, and the coefficient i (also greater than 1) is called the osmotic coefficient.
The phenomenon of osmosis is of great importance for plant and animal organisms, since the membranes of their cells in relation to solutions of many substances have the properties of a semipermeable membrane. In pure water, the cell swells strongly, in some cases up to the rupture of the shell, and in solutions with a high salt concentration, on the contrary, it decreases in size and shrinks due to the large loss of water. Therefore, when preserving food, a large amount of salt or sugar is added to them. Cells of microorganisms in such conditions lose a significant amount of water and die.
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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 = ∆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 on doing work A.
Internal energy system is its total reserve, including the energy of the translational and rotational motion of molecules, the energy of the motion of electrons in atoms, the energy of interaction of nuclei with electrons, nuclei with nuclei, etc., i.e. all types of energy, except for the kinetic and potential energy of the system as a whole.
The work performed 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 \u003d p (V 2 - V 1).
At constant pressure (р=const), taking into account the expression for the expansion work, the energy conservation law will be written as follows:
Q \u003d (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 H:
Since the exact value of the internal energy of the system is unknown, the absolute values of the enthalpies cannot be obtained either. Changes in enthalpies Δ H are of scientific importance and have practical applications.
Internal energy U and enthalpy H are state functions systems. State functions are such characteristics of the system, changes in which are determined only by the final and initial state of the system, i.e. are independent of the process path.
6.1.2. Exo- and endothermic processes
The flow of chemical reactions is accompanied by the absorption or release of heat. exothermic called a reaction that proceeds with the release of heat into the environment, and endothermic- with the absorption of heat from the environment.
Many processes in industry and in laboratory practice proceed at constant pressure and temperature (T=const, p=const). The energy characteristic of these processes is the change in enthalpy:
Q P \u003d -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) \u003d N 2 O (g); ΔН 298 = +82kJ,
CH 4 (g) + 2O 2 (g) \u003d CO 2 (g) + 2H 2 O (g); ΔN 298 = -802kJ.
Chemical equations in which the thermal effect of the reaction is additionally indicated (the value of the DH 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 reagents and formed substances are noted: d - gaseous, g - liquid, k - crystalline; S (rhombus), S (monocle), C (graphite), C (diamond), etc.
6.1.3. Thermochemistry; Hess' law
Energy phenomena accompanying physical and chemical processes studies thermochemistry. The basic law of thermochemistry is the law formulated by the Russian scientist G.I. Hess in 1840.
Hess' law: the change in the enthalpy of the process depends on the type and state of the starting materials and reaction products, but does not depend on the path of the process.
When considering thermochemical effects, the expression “process enthalpy” is often used instead of the concept of “change in the enthalpy of the process”, meaning by this concept the value Δ H. It is incorrect to use the concept of “heat effect of the process” when formulating the Hess law, since the value Q in general is not a function of state . As mentioned above, only at a constant pressure Q P =-Δ N (at a constant volume Q V =-Δ U).
So, the formation of PCl 5 can be considered as the result of the interaction of simple substances:
P (c, white) + 5/2Cl 2 (g) = PCl 5 (c); Δ H 1,
or as a result of a process that takes place in several stages:
P (k, white) + 3/2Cl 2 (g) = PCl 3 (g); Δ H 2,
PCl 3 (g) + Cl 2 (g) \u003d PCl 5 (c); Δ H 3,
or in total:
P (c, white) + 5/2Cl 2 (g) = PCl 5 (c); Δ H 1 \u003d 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 reaction:
2Na (c) + S (rhombus) + 2O 2 (g) \u003d Na 2 SO 4 (c).
The enthalpy of formation of simple substances is zero.
Since the thermal effect of a reaction depends on the state of substances, temperature, and pressure, it was agreed to use in thermochemical calculations standard enthalpies of formation are the enthalpies of formation of substances that are at a given temperature in standard condition. As a standard state for substances in a condensed state, the real state of the substance at a given temperature and pressure of 101.325 kPa (1 atm) is taken. Reference books usually give the standard enthalpies of formation of substances at a temperature of 25 o C (298K), referred to 1 mol of a substance (Δ H f o 298). Standard enthalpies of formation of some substances at T=298K are given in Table. 6.1.
Table 6.1.
Standard enthalpies of formation (Δ H f o 298) of some substances
Substance |
Δ H f o 298, kJ/mol |
Substance |
Δ H f o 298, kJ/mol |
The standard enthalpies of formation for most complex substances are negative values. For a small number of unstable substances, Δ H 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 consequence of the Hess law is used: the enthalpy of 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 .
Calculate the enthalpy of decomposition of calcium carbonate. The process is described by the following equation:
CaCO 3 (c) \u003d CaO (c) + 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 \u003d Δ H f o 298 (CaO (c)) + Δ H f o 298 (CO 2 (g)) - Δ H f o 298 (CaCO 3 (c)).
Using the data in Table 6.1. we get:
Δ H o 298 = - 635.1 -393.5 + 1206.8 = + 178.2 kJ.
It follows from the obtained data that the considered reaction is endothermic, i.e. proceeds with the absorption of heat.
CaO (c) + CO 2 (c) \u003d CaCO 3 (c)
Accompanied by the release of heat. Its enthalpy will be equal to
Δ H o 298 = -1206.8 + 635.1 + 393.5 = -178.2 kJ.
6.2. The rate of chemical reactions
6.2.1. The concept of reaction rate
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 the reacting substances per unit time at 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 - the value c 2 . For a period of time from t 1 to t 2, the change in concentration will be Δ c \u003d c 2 - c 1. The average reaction rate is:
The minus sign is put because as the reaction proceeds (Δ t> 0), the concentration of the substance decreases (Δ c< 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, with an increase in the contact area of substances, the reaction rate increases. The reaction rate also increases with an increase in the stirring rate of the reactants.
The numerical value of the reaction rate also depends on which component is used to calculate the reaction rate. For example, the speed of the process
H 2 + I 2 \u003d 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 the 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 factor, 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 reacting substances and on the temperature, but does not depend on their concentrations. Coefficients α and β are found from experimental data.
The sum of the exponents in the kinetic equations is called the total in order reactions. There is also a particular order of the reaction in one of the components. For example, for the reaction
H 2 + C1 2 \u003d 2 HC1
The kinetic equation looks like this:
v = k 1/2,
those. the overall order is 1.5 and the reaction orders for the H 2 and C1 2 components are 1 and 0.5, respectively.
Molecularity reaction is determined by the number of particles, the simultaneous collision of which is the 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 multi-stage and therefore the reaction equation does not reveal the mechanism of the process, the order of the reaction does not coincide with its molecularity.
Chemical reactions 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) \u003d CH 4 (g) + CO (g).
Bimolecular reaction- a reaction in which the elementary act is carried out when two particles collide. For example:
H 2 (g) + I 2 (g) \u003d 2 HI (g).
trimolecular reaction- a simple reaction, the elementary act of which is carried out with the simultaneous collision of three molecules. For example:
2NO (g) + O 2 (g) \u003d 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 reactants.
6.2.3. Temperature dependence of the reaction rate
The rate of chemical reactions increases with increasing temperature. This increase is caused by an increase in the kinetic energy of the molecules. In 1884, the Dutch chemist van't Hoff formulated the rule: for every 10 degrees 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 - temperature coefficient of speed, equal to 2 - 4.
The van't Hoff rule is used to approximate the effect of temperature on the reaction rate. 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 is 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. The concept of catalysis and catalysts
A catalyst is a substance that changes the rate of a chemical reaction but remains chemically unchanged at the end of the reaction.
Some catalysts speed up the reaction, while others, called inhibitors, slow it down. For example, adding a small amount of MnO 2 as a catalyst to hydrogen peroxide H2O2 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, there are 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 reactants to form intermediate compounds.
Consider the mechanism of action of the catalyst. In the absence of a catalyst, the reaction
It flows very slowly. The catalyst forms with the starting materials (for example, with substance B) a reactive intermediate product:
which reacts vigorously with another starting material to form the final reaction product:
VK + A \u003d AB + K.
Homogeneous catalysis takes place, 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 \u003d 2 SO 3
in the absence of a catalyst is very slow. But when a catalyst (NO) is introduced, an intermediate compound (NO2) is formed:
O 2 + 2 NO \u003d 2 NO 2,
which easily oxidizes SO 2:
NO 2 + SO 2 \u003d SO 3 + NO.
The activation energy of the latter process is very low, so the reaction proceeds at a high rate. Thus, the action of catalysts is reduced to a decrease in the activation energy of the reaction.
heterogeneous catalysis
In heterogeneous catalysis, the catalyst and reactants are in different phases. The catalyst is usually in the solid state and the reactants are in the liquid or gaseous state. In heterogeneous catalysis, the acceleration of the process is usually associated with the catalytic effect of the catalyst surface.
Catalysts differ in selectivity (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, ethyl alcohol is dehydrogenated:
C 2 H 5 OH - (Cu) CH 3 CHO + H 2.
Small amounts of certain 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 in 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 in the synthesis of NH 3 . Its activity can no longer be restored by passing a fresh mixture of N 2 +H 2 .
Substances that enhance the action of catalysts are called promoters, or activators(promotion of platinum catalysts, for example, is carried out by adding iron or aluminum).
The mechanism of heterogeneous catalysis is more complex. To explain it, the adsorption theory of catalysis is used. The surface of the catalyst is heterogeneous, so it has the so-called active centers. Reacting substances are adsorbed on active sites. The latter process causes the approach of the reacting molecules and an increase in their chemical activity, since the bond between the atoms of the adsorbed molecules is weakened, the distance between the atoms increases.
On the other hand, it is believed that the accelerating effect of a 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 end with the complete transformation of the starting substances into final substances are called irreversible.
Irreversible, i.e. proceeding to the end are reactions in which
Chemical reactions that can go in opposite directions are called reversible. Typical reversible reactions are the reactions of ammonia synthesis and 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 the equations of reversible reactions, instead of the equal sign, put 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, which increases as the concentration of the products increases. As a result, there comes a moment 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 reactants 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 immutability of the state of the system is caused by the flow of direct and reverse reactions, that is, the equilibrium state is dynamic;
the state of the system remains unchanged in 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, then the system returns to its original state again;
must be distinguished from the real apparent balance. So, for example, a mixture of oxygen and hydrogen in a closed vessel at room temperature can be stored for an arbitrarily long time. However, the initiation of the reaction (electric discharge, ultraviolet irradiation, temperature increase) causes the reaction of water formation to proceed irreversibly.
6.5. Le Chatelier's principle
The influence of changes in external conditions on the equilibrium position is determined by Le Chatel principle e (France, 1884): if any external influence is produced on a system in equilibrium, the equilibrium in the system will shift in the direction of 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 influence of various factors on the chemical equilibrium using the ammonia synthesis reaction as an example:
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 the initial substances shifts the equilibrium towards the formation of reaction products. An increase in the concentration of the reaction products shifts the equilibrium towards the formation of the starting substances.
In the process of ammonia synthesis considered 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. Increasing the concentration of ammonia shifts the equilibrium towards the starting materials.
A catalyst speeds up both the forward and reverse reactions equally, so the introduction of a catalyst does not affect the chemical equilibrium.
Effect of Temperature on Chemical Equilibrium
As the temperature rises, the equilibrium shifts towards an endothermic reaction, and as the temperature decreases, it shifts towards an exothermic reaction.
The degree of equilibrium shift is determined by the absolute value of the thermal effect: the greater the value of ΔH of the reaction, the greater the effect of temperature.
In the ammonia synthesis reaction under consideration, an increase in temperature will shift the equilibrium towards the starting materials.
Effect of pressure on chemical equilibrium
A change in pressure affects the chemical equilibrium with the participation of gaseous substances. According to Le Chatelier's principle, an increase in pressure shifts the equilibrium in the direction of a reaction proceeding with a decrease in the volume of gaseous substances, and a decrease in pressure shifts the equilibrium in the opposite direction. The ammonia synthesis reaction proceeds with a decrease in the volume of the system (there are four volumes on the left side of the equation, and two volumes on the right). Therefore, an increase in pressure shifts the equilibrium towards the formation of ammonia. A decrease in pressure will shift the equilibrium in the opposite direction. If in the equation of a reversible reaction the number of molecules of gaseous substances in the right and left parts are equal (the reaction proceeds without changing the volume of gaseous substances), then pressure does not affect the equilibrium position in this system.
Any chemical reaction is accompanied by the release or absorption of energy in the form of heat.
On the basis of the release or absorption of heat, they distinguish exothermic and endothermic reactions.
exothermic reactions - such reactions during which heat is released (+ Q).
Endothermic reactions - reactions during which heat is absorbed (-Q).
The thermal effect of the 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 in which the heat effect of a chemical reaction is indicated. For example, thermochemical equations are:
It should also be noted that thermochemical equations must necessarily include information about the aggregate states of reactants and products, since the value of the thermal effect depends on this.
Reaction Heat Calculations
An example of a typical problem for finding the heat effect of a reaction:
When interacting 45 g of glucose with an excess of oxygen in accordance with the equation
C 6 H 12 O 6 (solid) + 6O 2 (g) \u003d 6CO 2 (g) + 6H 2 O (g) + Q
700 kJ of heat were released. Determine the thermal effect of the reaction. (Write down the number to the nearest integer.)
Decision:
Calculate the amount of glucose substance:
n (C 6 H 12 O 6) \u003d m (C 6 H 12 O 6) / M (C 6 H 12 O 6) \u003d 45 g / 180 g / mol \u003d 0.25 mol
Those. the interaction of 0.25 mol of glucose with oxygen releases 700 kJ of heat. From the thermochemical equation presented in the condition, it follows that when 1 mol of glucose interacts with oxygen, an amount of heat equal to Q (the heat of the reaction) is formed. Then the following proportion is true:
0.25 mol glucose - 700 kJ
1 mol of glucose - Q
From this proportion follows the corresponding equation:
0.25 / 1 = 700 / Q
Solving which, we find that:
Thus, the thermal effect of the reaction is 2800 kJ.
Calculations according to thermochemical equations
Much more often, in the USE assignments in thermochemistry, the value of the thermal effect is already known, because. the complete thermochemical equation is given in the condition.
In this case, it is required to calculate either the amount of heat released / absorbed with a known amount of the reactant or product, or, conversely, the known value of heat is required to determine the mass, volume or amount of a substance of any involved in the reaction.
Example 1
In accordance with the thermochemical reaction equation
3Fe 3 O 4 (solid) + 8Al (solid) \u003d 9Fe (solid) + 4Al 2 O 3 (solid) + 3330 kJ
formed 68 g of aluminum oxide. How much heat is released in this case? (Write down the number to the nearest integer.)
Decision
Calculate the amount of aluminum oxide substance:
n (Al 2 O 3) \u003d m (Al 2 O 3) / M (Al 2 O 3) \u003d 68 g / 102 g / mol \u003d 0.667 mol
In accordance with the thermochemical equation of the reaction, 3330 kJ are released during the formation of 4 mol of aluminum oxide. In our case, 0.6667 mol of aluminum oxide is formed. Denoting the amount of heat released in this case, through x kJ we will make up 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. in the formation of 68 g of aluminum oxide, in accordance with the thermochemical equation, 555 kJ of heat is released under the condition.
Example 2
As a result of the reaction, the thermochemical equation of which
4FeS 2 (solid) + 11O 2 (g) \u003d 8SO 2 (g) + 2Fe 2 O 3 (solid) + 3310 kJ
1655 kJ of heat were released. Determine the volume (l) of sulfur dioxide released (n.o.s.). (Write down the number to the nearest integer.)
Decision
In accordance with the thermochemical reaction equation, the formation of 8 mol of SO 2 releases 3310 kJ of heat. In our case, 1655 kJ of heat was released. Let the amount of substance SO 2 formed in this case be equal to x mol. Then the following proportion is valid:
8 mol SO 2 - 3310 kJ
x mol SO 2 - 1655 kJ
From which follows the equation:
8 / x = 3310 / 1655
Solving which, we find that:
Thus, the amount of substance SO 2 formed in this case is 4 mol. Therefore, its volume is:
V (SO 2) \u003d V m ∙ n (SO 2) \u003d 22.4 l / mol ∙ 4 mol \u003d 89.6 l ≈ 90 l(round up to integers, because this is required in the condition.)
More analyzed problems on the thermal effect of a chemical reaction can be found.
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