Heat of formation of a simple substance under standard conditions. Enthalpy of formation. Thermochemical reaction equation
The standard heat of formation is understood as the heat effect of the reaction of formation of one mole of a substance from simple substances, its constituents, which are in stable standard states.
For example, the standard enthalpy of formation of 1 mole of methane from carbon and hydrogen is equal to the heat of the reaction:
C (tv) + 2H 2 (g) \u003d CH 4 (g) + 76 kJ / mol.
The standard enthalpy of formation is denoted Δ H fO . Here the index f means formation (education), and the crossed out circle, resembling a Plimsol disk, means that the value refers to the standard state of matter. In the literature, another designation for the standard enthalpy is often found - ΔH 298.15 0, where 0 indicates pressure equal to one atmosphere (or, somewhat more accurately, standard conditions), and 298.15 is temperature. Sometimes index 0 is used for quantities related to pure substance, stipulating that it is possible to designate standard thermodynamic quantities with it only when it is a pure substance that is chosen as the standard state. The standard can also be taken, for example, the state of a substance in an extremely dilute solution. "Plimsol disk" in this case means the actual standard state of matter, regardless of its choice.
The energy of chemical processes is part of chemical thermodynamics (part of general thermodynamics).
Energy state chemical reaction how the system is described using the following characteristics: U-internal energy, H-enthalpy, S-entropy, G-Gibbs energy.
The heat received by the system is used to increase internal energy and do work: Q=D U+A. If the system does no work other than expansion work, then Q=D U+pD V. The value D H=D U+pD V at p=const is called the enthalpy of reaction. Because it is impossible to measure the internal energy of the body (only the change D U can be measured), then it is also impossible to measure the enthalpy of the body - the change in enthalpy D H is used in the calculations.
The standard enthalpy of formation is the isobaric heat effect of the reaction to produce one mole complex substance from simple substances taken in their most stable form under standard conditions (T=298K, p=1 atm., C=1 mol/l). The enthalpy of formation of simple substances in their steady state under standard conditions is assumed to be 0.
The laws of thermochemistry:
1. Lavoisier-Laplace: the thermal effect of the formation of chemical compounds is equal, but opposite in sign, to the thermal effect of its decomposition.
2. Hess: the thermal effect of a reaction at constant pressure or volume depends only on the initial and final state of the system and does not depend on the transition path.
Entropy is a quantitative measure of the disorder of the system. It has a statistical meaning and is a characteristic of systems consisting of a sufficiently large but limited number of particles. Entropy is expressed in terms of the thermodynamic probability of the system - the number of microstates corresponding to a given microstate. It is assumed that at absolute zero the entropy of an ideal crystal is 0. It is also assumed that for a hydrated proton H + the absolute value of entropy in an aqueous solution is 0. The entropy depends on: the number of particles in the system, the nature of the substance, and the state of aggregation. For chemical reactions, the change in entropy is calculated from the absolute values of the entropy of the components. For reactions taking place in an aqueous solution, the calculation is made using a short ionic formula. For gaseous substances, the sign of D S is determined by the change in volume. If the volume does not change, then the sign cannot be determined. In isolated systems, processes are possible that go with an increase in entropy. This means that the sign D S can be taken as a criterion for a possible spontaneous reaction (only in isolated systems!). In general, this criterion cannot be applied in open systems.
The total influence of energy and entropy factors at constant pressure and temperature reflects the change in the isothermal potential, which is called the change in Gibbs free energy: D G=D H-TD S. free energy Gibbs is called the energy made up of the energies of chemical bonds. Sign D G is a criterion of thermodynamic probability of spontaneous flow of the process under given conditions (p,T=const). Under these conditions, only those processes D G for which is less than 0 can spontaneously proceed. not taken into account temperature dependence enthalpy and entropy. At low temperatures, mainly exothermic reactions take place. At high temperatures, the main role is played by the entropy term of the equation, which can be seen from the example of the fact that the reactions of decomposition of complex substances into simple ones mainly proceed at high temperature.
The standard Gibbs energy of a substance is the energy of obtaining a given substance under standard conditions. Standard conditions do not exist in practice, so all calculations using standard values are approximate.
Standard Conditions
The thermal effects of reactions depend on the conditions under which they occur. Therefore, in order to be able to compare the obtained values of the thermal effects of reactions, the enthalpies of formation of substances, we agreed to determine or bring them to certain, identical, so-called standard conditions. The standard conditions are considered to be the state of 1 mol of a pure substance at a pressure of 101,325 Pa (1 atm or 760 mm Hg) and a temperature of 25 ° C or 298 K. For substances in solution, a concentration equal to one mol in liter (C \u003d 1 mol / l). Moreover, it is assumed that the solution behaves at this concentration in exactly the same way as at infinite dilution, i.e. is ideal. The same assumption applies to substances that are in a gaseous state (a gas, as it were, is ideal both at a pressure of 1 atmosphere and at a pressure much lower).
Therefore, the change in the enthalpy of the reaction system during the transition from one state to another under standard conditions will also have a standard character. Therefore, the enthalpy of formation of one mole of a complex substance from simple substances under standard conditions will also be called standard enthalpy (warmth ) education.
Standard changes in the enthalpy of formation are denoted by DYa (^ p. In what follows, we will simply call them standard enthalpies of formation of substances or enthalpies of reaction (omitting the word change). For example, the standard enthalpy of formation of water in a liquid state is denoted as follows:
This entry means that under standard conditions, the formation of one mole of water in a liquid state from simple substances is accompanied by a loss of 285.85 kJ by the reacting system. The thermochemical equation for this reaction looks like this:
The standard enthalpies of formation for most known substances have been determined empirically or calculated and summarized in reference tables of the thermodynamic properties of substances.
The standard values of the enthalpies of formation of simple substances (for example, H 2 (g), O 2 (g), Cu (cr) and other substances) for those aggregate states in which these substances are stable are taken equal to zero, i.e.
The standard enthalpy of formation of a compound is a measure of its thermodynamic stability, strength, and is of a periodic nature for one class, a group of substances of the same type.
Sometimes there are exceptions to the choice of the standard state, for example, when we talk about the standard heat of formation of vaporous water, we mean that water vapor is formed, the pressure of which is 101.3 kPa and the temperature is 25 ° C. But at 25°C, water vapor has a much lower equilibrium pressure. This means that the heat of formation of water in the vapor state Dc 2 o(n) is a purely conditional state.
Thermochemical laws
Hess' law
The independence of the heat of a chemical reaction from the path of the process at R = const and T = const was established in the first half of the 19th century. Russian scientist G. I. Hess. Hess formulated the law that now bears his name: the thermal effect of a chemical reaction does not depend on the path of its occurrence, but depends only on the nature and physical state of the initial substances and reaction products.
This law is valid for those interactions that occur in isobaric-isothermal (or isochoric-isothermal) conditions, despite the fact that the only type of work performed is work against external pressure forces.
Imagine that there is a reaction system in which substances A And IN turn into products D And E, according to the thermochemical equation:
The change in the enthalpy of this reaction АH^ eacci. reaction products D And E can be obtained directly and directly from the starting materials A And IN , as shown schematically in Fig. 2.2, but along the path 1-2, bypassing any intermediate stages. The thermal effect in this method of transformation (Fig. 2.2, 6) will be equal to:
Get the same products D And E it is possible by carrying out the process through the formation of any intermediate substances, for example, along the path 1-3 4-5-2 or 1-6-7-2 (Fig. 2.2, A). Moreover, each stage of education
intermediate substances will be characterized by their thermal effect or enthalpy change: D H 1, DN 2, DN 3, DN 4, DN 5, DN 6, and DN 7, respectively, for each section of the process path (Fig. 2.2, b).
Rice. 2.2. :
A - possible ways of carrying out the process; b - schemes of changing the enthalpies of intermediate stages depending on the reaction path
If we consider the final result of the energy changes of the process through the intermediate stages, then it turns out that it is equal to the algebraic sum of the change in the enthalpies of the intermediate stages:
That is, the thermal effect of the reaction does not depend on the method of carrying out the process, but depends only on the initial state of the starting materials and the final state of the reaction products (Fig. 2.2, b).
On a specific reaction, for example, the oxidation of iron with oxygen, we will check the feasibility of Hess's law. The thermochemical equation for this process is:
Let's go through this process step by step. First, we oxidize iron to iron oxide (I) according to the equation:
I stage :
with a thermal effect of 2,263.7 kJ, and then we oxidize iron (I) oxide in the second stage to iron (III) oxide according to the equation:
II stage-.
in which 293.9 kJ will be released. Adding the equations of the first and second stages of the reactions, we get:
The total thermal effect of these stages is also equal to 821.3 kJ, as if the process was carried out without intermediate stages. That is, Hess's law is fulfilled.
Thermochemical equations can be added and subtracted like ordinary algebraic equations.
Consider an illustration of Hess's law with another example.
Known:
Find DH° for the following reactions:
Based on the initial data, it is convenient to draw up a diagram of possible pathways for the formation of CO 2 (Fig. 2.3).
Rice. 2.3.
According to Hess' law
The same result can be reached, given that the reaction equation (3) can be obtained by subtracting equation (2) from equation (1). A similar operation with thermal effects will give
To obtain equation (4), we must subtract equation (2) from equation (1) multiplied by 2. Therefore,
For practical use, the consequences of Hess's law are important. Let's consider two of them.
The first corollary of Hess' law
This consequence is associated with the heats of formation of compounds. Heat (enthalpy) of formation compounds is called the amount of heat,
released or absorbed during the formation of 1 mol of this compound from simple substances that are in the most stable state under given conditions. (Simple substances consist of atoms of the same type, for example, N 2, H 2, 0 2, C, S, Fe, etc.) In this case, the reaction may turn out to be hypothetical, i.e. not really flow. For example, the heat of formation of calcium carbonate is equal to the heat of the reaction for the formation of 1 mole of crystalline calcium carbonate from metallic calcium, carbon in the form of graphite, and gaseous oxygen:
The heats (enthalpies) of formation of stable simple substances (N 2 , H 2 , 0 2 , Fe, etc.) are equal to zero.
Let us designate the heat of formation of the substance as DY oG)p
In accordance with the first corollary of the Hess law, the heat effect of any reaction can be calculated from the tenlots (enthalpies) of formation: the heat effect of the reaction is equal to the difference between the heats (enthalpies) of formation of the reaction products and starting materials, taking into account stoichiometric coefficients.
(2.11)
Subscript icons here j And і refer to reaction products and starting materials, respectively; v- stoichiometric coefficients.
The scheme in fig. 2.4 illustrates the proof of this corollary. Equation (2.11) follows from the vector addition rule.
Rice. 2.4.
As stated in Section 2.4, the heats of formation are usually referred to as standard conditions and are called standard heat (enthalpy) of compound formation and denote AHob r. The Anob values of the most common compounds are given in thermodynamic reference tables. With their help, the standard thermal effects of chemical reactions AN 0 are calculated:
The second corollary of Hess' law
Note that in all the above examples, standard enthalpies (heats) of formation of individual substances were used. But for some compounds it is not possible to determine them directly experimentally if we proceed only from simple substances. In such cases, the law of G. I. Hess is used to calculate the standard enthalpies (heats) of formation according to known enthalpies (heats) of combustion these substances, since in most of these cases it is possible to carry out the reaction of complete combustion of simple and complex substances.
At the same time, under calorific value understand thermal effect of combustion of 1 mol of a complex substance (or 1 mol of atoms of a simple substance) until stable oxides are formed.
Standard calorific values referred to 25°C (298 K) and pressure
- 101.3 kPa. The heats of combustion of oxygen and combustion products in their steady state under standard conditions (25°С,
- 101.3 kPa), i.e. consider the energy content of gaseous oxygen, nitrogen, carbon dioxide, sulfur dioxide, liquid water and other non-combustible substances to be conditionally equal to zero.
The practical significance of knowing the heats of combustion of substances is that their values can be used to calculate the thermal effects of chemical reactions in the same way as is done when using the enthalpies (heats) of formation of substances. After all, the thermal effect of the reaction does not depend on the method of its implementation, intermediate stages, but is determined only by the initial and final state of the initial substances and reaction products according to the Hess law. especially large practical value calorific values are used to determine the thermal effects of reactions involving organic compounds. For example, the heat of formation of methane from simple substances
cannot be measured directly. To determine the heat of formation of organic matter, it is burned and, based on the heat of combustion of complex organic matter and the heat of combustion of simple substances, its heat of formation is found. The relationship between the heat of formation of methane and the heat of combustion of the reaction products is visible in the diagram (Fig. 2.5).
According to Hess's law, the thermal effects of the first and second paths must be equal
The heat of combustion of a simple substance, for example, graphite and hydrogen to a stable oxide, i.e. before the formation of carbon dioxide or water, is identical to the heat of formation of carbon dioxide or water:
Rice. 2.5.
Taking this into account, we get:
Substituting the numerical values of the corresponding heats of formation into the equation, we obtain:
Some thermodynamic handbooks contain tables with isobaric heats of combustion - A//J rop of many organic substances, which can be used in calculations. However, if non-combustible substances are involved in the reaction, then the thermal effect can only be determined through the heats of formation. For example:
under standard conditions, the thermal effect is:
those. this reaction is exothermic Q= +168.07 kJ/mol.
Hess's law and its consequences serve as the basis for all thermochemical calculations, while it is necessary that all heats of combustion or formation refer to the same conditions - isobaric or isochoric. The thermodynamic tables give the values AN formation or combustion under standard conditions (/? = 101.3 kPa and T = 298 K), i.e. for isobaric-isothermal process.
To go from Qp to Qn
you need to use the equation: 
Chemical transformations of food substances in the body, like any chemical reactions outside the body, obey the laws of thermochemistry. Consequently, Hess's law gives reason to use the heat of combustion of food substances to represent the energy of their oxidation in the body. Although nutrients, introduced into the body, pass until their final transformation hard way and participate in a large number of reactions, the total energy effect of all these reactions, according to the Hess law, is equal to the thermal effect of the direct combustion of the introduced substances.
For example, when burning one mole of glucose (to carbon dioxide and water) in a calorimetric bomb, 2816 kJ is released, which means that with complete oxidation and in the body of one mole of glucose, an amount of energy equivalent to 2816 kJ is released. The pathways of glucose oxidation in a calorimetric bomb and in an organism are different, but the energy effect is the same in both cases, since the initial and final states of the substances involved in the reaction are the same.
Thermochemical calculations
Thermochemical calculations associated with the determination of the thermal effects of reactions, the heats of formation of compounds, make it possible to some extent predict the probable direction of the process, and approximately characterize the strength of the joint. All calculations are based on two laws of thermochemistry and on its basic concepts and definitions.
Consider a few concrete examples thermochemical calculations.
Example 2.1. Find the standard thermal effect А// 0 of the reaction of obtaining crystalline Al2(SO4)3 at 298 K from crystalline Al 2 0 3 and gaseous S0 3:
The standard enthalpies of formation of the substances involved in this reaction at 298 K are:
Then by equation (2.12) we find
Solution. We write the thermochemical equation for the combustion of methane
From the handbook of thermodynamic properties of substances, we write out the standard values of the enthalpies of formation (heats of formation) of the starting materials and reaction products:
Since carbon dioxide (1 mol) and water (2 mol) are formed in the liquid state during the combustion of methane, we will compose the thermochemical equations for the formation of these substances from simple substances:
And since during combustion, methane CH 4 (g) decomposes, turning into water in a liquid state and carbon dioxide, we write the thermochemical equation for the decomposition of methane into simple substances:
Adding these last three equations, we obtain the thermochemical equation for the reaction of methane combustion:
Thus, the thermal effect of this reaction under standard conditions is equal to Q °„ \u003d 890.94 kJ / mol or the change in the enthalpy of the reaction is DH ° ktsnn \u003d - 890.94 kJ / mol.
If you carefully look at how this numerical value was obtained, it turns out that the sum of the heats of formation of the initial substances was subtracted from the sum of the heats of formation of the reaction products. This consequence from Hess's law, which can be written as follows:
Or, in relation to the concept of a change in the enthalpy of a reaction:
As applied to our problem, the thermal effect of the reaction can be calculated without compiling the equations for the formation and decomposition of substances:
Or, substituting numerical data, we get:
A similar calculation can be carried out using not the heats of formation, but enthalpies:
Example 2.3. Calculate the thermal effect of the reaction:
The enthalpies of combustion are:
for acetylene (g) DH a = -1298.3 kJ/mol; for benzene (l) AN" = -3264.2 kJ/mol.
By equation (2.13) we find
Knowing the heat of combustion, it is easy to determine the heat of formation, and vice versa. If, for example, the heat of combustion of methyl alcohol is -729 kJ / mol, then, using the values of the heat of formation of CO 2 and H 2 0, the following thermochemical equations can be drawn up:
)
Multiplying the equation (V) by 2, adding with equation (b) and subtracting equation (a), we obtain after transformations the reaction of formation of methyl alcohol
Having carried out similar transformations with the thermal effects of reactions, we obtain the thermal effect of the formation of methyl alcohol AN
Hess's law is also valid for complex biochemical processes. Thus, the amount of heat obtained during the oxidation of carbohydrates and fats in a living organism, where these processes proceed in several stages, and the amount of heat released during the combustion of these substances in oxygen, turned out to be equal. For proteins, this is not the case, since the end product of protein oxidation in the body is urea, while in oxygen, protein oxidation is complete.
The content of the article
CHEMICAL THERMODYNAMICS, considers the relationship between work and energy in relation to chemical transformations. Since a chemical transformation is usually accompanied by the release or absorption of a certain amount of heat, it, like other natural phenomena (including electrical and magnetic), accompanied by thermal effects, obeys the fundamental principles (beginnings) of thermodynamics. Chemical thermodynamics determines, first of all, the conditions (such as temperature and pressure) for the occurrence of chemical reactions and the equilibrium states they reach. The analysis of thermal phenomena is based on three fundamental principles, confirmed by numerous observations.
First law of thermodynamics.
The first law of thermodynamics essentially expresses the law of conservation of energy. For a system surrounded by a closed boundary through which there is no transfer of matter, the relation
Where U 1 and U 2 – energies of the system in states 1 and 2; Q– heat received from external sources; W- the work done by the system on external bodies in the process by which the system passes from state 1 to state 2. If the process is a chemical reaction, then it is usually carried out under such conditions that it is possible to separate the energy of chemical transformation from the energy associated with simultaneous changes temperature or pressure. Therefore, the energy (heat) of a chemical reaction is usually determined under conditions in which the products are at the same temperature and pressure as the reactants. The energy of a chemical reaction is then determined by the heat Q received from or transferred to the environment. Measurement Q can be done with a calorimeter suitable type. The reaction could be carried out, for example, in a metal vessel immersed in a thermally insulated volume of water, the change in temperature of which (usually by several degrees) corresponds to the heat of reaction. For quantitative measurements, the calorimeter is usually calibrated using an independent electric heater or by conducting a chemical reaction in a vessel, the heat of which is known.
Slow reactions are especially difficult for calorimetric measurements because complex precautions are needed to protect the calorimeter from heat exchange with the environment. The so-called adiabatic calorimeter is completely immersed in an isothermal shell with independent heating, the temperature of which during the experiment is kept as close as possible to the temperature inside the calorimeter. Reactions that release heat (negative Q in equation (1)), are called exothermic, and reactions in which heat is absorbed are called endothermic.
As equation (1) shows, the internal energy of a reacting system is determined not only by the amount of released or absorbed heat. It also depends on how much energy the system expends or acquires through the work done. At constant pressure p the total work done by the system is described by the expression p (V 2 – V 1) +We, where the first term is the work of expansion associated with a change in volume from V 1 to V 2 , and We- additional, or so-called. "useful", the work done by the system in addition to the work of expansion. If work is done on the system, then both terms have a negative sign. Therefore, equation (1) can be transformed into the form
An auxiliary measure of the energy of the system is introduced H, determined by the general relation
If the pressure is constant (usually a pressure of 1 atm is taken as the standard), then the change in the function H, called the enthalpy of the system, differs from the change in its internal energy by the value of the work of expansion:
With the exception of gas-phase systems, this difference is negligible compared to the typical thermal effects of reactions. However, for the general case, it follows from formula (2) that the heat Q measured at constant pressure and We= 0 (the usual condition for a chemical reaction to occur if it does not occur, for example, in a battery or a galvanic cell), is equal to the change in the enthalpy of the system:
In any case, since the difference H 2 – H 1 , as well as U 2 – U 1 , is determined, according to the first law of thermodynamics, exclusively by the initial and final states of the system and does not depend on the method of transition from the initial state to the final state, the total amount of heat absorbed in the process of chemical transformation at constant temperature and pressure (at We= 0) depends only on the initial reagents and final products and does not depend on the intermediate stages through which the reaction proceeds.
Here, the letters in brackets denote the aggregate states of substances (gas or liquid). Symbol D H° denotes the enthalpy change in a chemical transformation at a standard pressure of 1 atm and a temperature of 298 K (25 ° C) (the degree sign in the superscript H indicates that this value refers to substances in standard states (at p= 1 atm and T= 298 K)). Chemical formula each substance in such an equation denotes a well-defined amount of a substance, namely its molecular weight, expressed in grams. The molecular weight is obtained by adding the atomic masses of all the elements included in the formula with coefficients equal to the number of atoms of a given element in the molecule. The molecular weight of methane is 16.042, and, according to the previous equation, the combustion of 16.042 g (1 mol) of methane produces products whose enthalpy is 212.798 kcal less than the enthalpy of the reactants. In accordance with equation (5), this amount of heat is released when 1 mole of methane burns in oxygen at a constant pressure of 1 atm. The corresponding decrease in the internal energy of the system during the reaction is 211.615 kcal. Difference between D H° and D U° is equal to - 1.183 kcal and represents the work p (V 2 – V 1), performed when 3 moles of gaseous reagents are compressed at a pressure of 1 atm to 1 mole of gaseous carbon dioxide and 2 moles of liquid water.
Standard heat of formation.
It follows from the law of conservation of energy that when a substance is formed from atoms and (or) simpler substances, the internal energy or enthalpy of the system changes by a certain amount, called the heat of formation of this substance. The heat of formation can be determined different ways, including direct calorimetric measurements and by indirect calculation (based on the Hess law) from the heat of reaction in which a given substance participates. When making calculations, they use standard (with p= 1 atm and T= 298 K) heats of formation of substances included in the reaction equation. For example, the standard heat (enthalpy) of formation of methane can be calculated using the thermochemical equation
Although this reaction is not practical at 25°C, the standard heat of formation of methane is indirectly calculated from the measured heats of combustion of methane, hydrogen, and graphite. Based on the Hess law, it is established that the heat of reaction is equal to the difference between the heats of combustion of the substances indicated on the left side of the equation and the heats of combustion of the substances indicated on the right side of the reaction equation (taken with the appropriate signs and stoichiometric coefficients).
In addition to using thermochemical data to solve problems of the practical use of thermal energy, they are widely used in the theoretical evaluation of chemical bond energies. This issue is considered in detail by L. Pauling in the book Nature chemical bond (The Nature of the Chemical Bond, 1960).
The second law of thermodynamics.
The second law of thermodynamics essentially determines the unidirectionality of heat transfer in various processes occurring spontaneously under certain conditions, namely, the direction of heat transfer from bodies with high temperature to bodies with low temperatures. The second law of thermodynamics can be formulated as follows: there can be no spontaneous general transfer of heat from less heated bodies to hotter bodies.
Heat transfer Q from a source with a temperature T can be characterized by Q/T. For any spontaneous heat transfer process in which a source with temperature T 1 gives off the amount of heat Q 1 , and as a result of the transfer, the system with temperature T 2 receives the amount of heat Q 2 , the Clausius inequality Q 1 /T 1 Ј Q 2 /T 2. Thus, in order for heat transfer to take place, T 1 should be more T 2. For the transition of a system from one state to another, a more general formulation of the second law of thermodynamics states that the direction of heat transfer is determined by the condition
Where S 2 – S 1 is the difference between the entropies of the system in two states. If we combine this condition with equations (2) and (3), we get a relation that is important for describing a chemical reaction at constant temperature and pressure:
If we introduce the system state function
then the formulation of the second law of thermodynamics will take the following form:
This means that for a system at constant temperature and pressure, only such transitions from one state to another can occur for which the useful work We does not exceed a certain maximum value equal to the difference D G two values G. If G 1 > G 2 , then the transition from state 1 to state 2 (say, from reactants to products) can occur spontaneously even when We= 0. If G 2 > G 1 , then the transition from state 1 to state 2 can be carried out only due to the external useful work; it means that work We must be a negative value, as, for example, the electrical energy expended in the electrolytic decomposition of water. If G 1 = G 2, then the system is in equilibrium.
Function G is called the Gibbs energy, or isobaric-isothermal potential. Various methods have shown that the value of D G° , the "standard Gibbs energy of formation", by analogy with the standard enthalpy of formation can be determined for chemical compounds relative to elements from chemical equilibrium and chemical process data. Standard Gibbs energy of formation D G° , which characterizes any chemical reaction, can be established using tables of standard Gibbs energies of formation by subtracting the sum of their values for the reactants from the sum of the values for the products. D values G° for clean chemical elements at 25°C and a pressure of 1 atm are taken equal to zero.
The standard Gibbs energy of a chemical reaction is essentially a measure of how far the reactants and products are from equilibrium with each other at a given temperature and a standard pressure of 1 atm. According to the second law of thermodynamics, all spontaneous changes in the system and its environment tend to bring them to the final state of equilibrium. Therefore, it is the change in Gibbs energy, and not the change in enthalpy or internal energy, that determines the possibility of a chemical reaction occurring. In particular, the potential difference between the electrodes of chemical current sources depends on the change in the Gibbs energy during a chemical reaction.
The standard change in Gibbs energy is related to standard change enthalpy, according to (7), by the relation
Exercise 81.
Calculate the amount of heat that will be released during the reduction of Fe 2O3 metallic aluminum if 335.1 g of iron was obtained. Answer: 2543.1 kJ.
Solution:
Reaction equation:
\u003d (Al 2 O 3) - (Fe 2 O 3) \u003d -1669.8 - (-822.1) \u003d -847.7 kJ
Calculation of the amount of heat that is released upon receipt of 335.1 g of iron, we produce 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 is the atomic mass of iron.
Answer: 2543.1 kJ.
Thermal effect of the 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 previously calculated its thermal effect. Answer: -45.76 kJ.
Solution:
The reaction equation is:
C 2 H 4 (g) + H 2 O (g) \u003d C2H 5 OH (g); = ?
The values of the standard heats of formation of substances are given in special tables. Considering that the heats of formation of simple substances are conditionally taken equal to zero. Calculate the thermal effect of the reaction, using the consequence of the Hess law, we get:
\u003d (C 2 H 5 OH) - [ (C 2 H 4) + (H 2 O)] \u003d
= -235.1 -[(52.28) + (-241.83)] = - 45.76 kJ
Reaction equations in which their states of aggregation or crystalline modification are indicated near the symbols of chemical compounds, as well as the numerical value of thermal effects, are called thermochemical. In thermochemical equations, unless otherwise specified, the values of thermal effects at a constant pressure Q p are indicated equal to the change in the enthalpy of the system. The value is usually given on the right side of the equation, separated by a comma or semicolon. The following abbreviations for the aggregate state of matter 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) \u003d 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) EEO (c) + CO (g) \u003d Fe (c) + 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 oxide (II) with hydrogen has the form:
EeO (k) + H 2 (g) \u003d Fe (k) + H 2 O (g); = ?
\u003d (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 oxide (II) can be calculated if equation (a) is subtracted from equation (b).
\u003d (c) - (b) - (a) \u003d -241.83 - [-283.o - (-13.18)] \u003d + 27.99 kJ.
Answer:+27.99 kJ.
Task 84.
During the interaction of gaseous hydrogen sulfide and carbon dioxide, water vapor and carbon disulfide СS 2 (g) are formed. Write the thermochemical equation for this reaction, preliminarily calculate its thermal effect. Answer: +65.43 kJ.
Solution:
G- gaseous, and- liquid, To- crystalline. These symbols are omitted if the aggregate state of substances is obvious, for example, O 2, H 2, etc.
The reaction equation is:
2H 2 S (g) + CO 2 (g) \u003d 2H 2 O (g) + CS 2 (g); = ?
The values of the standard heats of formation of substances are given in special tables. Considering that the heats of formation of simple substances are conditionally taken equal to zero. The thermal effect of the reaction can be calculated using the corollary e from the Hess law:
\u003d (H 2 O) + (CS 2) - [(H 2 S) + (CO 2)];
= 2(-241.83) + 115.28 – = +65.43 kJ.
2H 2 S (g) + CO 2 (g) \u003d 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 obtained in terms of normal conditions? Answer: 618.48 kJ.
Solution:
Reaction equations in which their states of aggregation or crystalline modification are indicated near the symbols of chemical compounds, as well as the numerical value of thermal effects, are called thermochemical. In thermochemical equations, unless it is specifically stated, the values of thermal effects at constant pressure Q p are indicated equal to the change in the enthalpy of the system. The value is usually given on the right side of the equation, separated by a comma or semicolon. The following abbreviations for the aggregate state of matter are accepted: G- gaseous, and- something To- crystalline. These symbols are omitted if the aggregate state of substances is obvious, for example, O 2, H 2, etc.
The reaction equation is:
CO (g) + 3H 2 (g) \u003d CH 4 (g) + H 2 O (g); = ?
The values of the standard heats of formation of substances are given in special tables. Considering that the heats of formation of simple substances are conditionally taken equal to zero. The thermal effect of the reaction can be calculated using the corollary e from the Hess law:
\u003d (H 2 O) + (CH 4) - (CO)];
\u003d (-241.83) + (-74.84) - (-110.52) \u003d -206.16 kJ.
The thermochemical equation will look like:
22,4 : -206,16 = 67,2 : X; x \u003d 67.2 (-206.16) / 22? 4 \u003d -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 from the following thermochemical equations:
a) 4NH 3 (g) + 5O 2 (g) \u003d 4NO (g) + 6H 2 O (g); = -1168.80 kJ;
b) 4NH 3 (g) + 3O 2 (g) \u003d 2N 2 (g) + 6H 2 O (g); = -1530.28 kJ
Answer: 90.37 kJ.
Solution:
The standard heat of formation is equal to the heat of formation of 1 mol 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 the reaction (a) in which 4 moles of NO are formed and the reaction (b) is given in which 2 moles of N2 are formed. Both reactions involve oxygen. 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 interaction of gaseous ammonia and hydrogen chloride. Write the thermochemical equation for this reaction, having previously calculated its thermal effect. How much heat will be released if 10 liters of ammonia were consumed in the reaction in terms of normal conditions? Answer: 78.97 kJ.
Solution:
Reaction equations in which their states of aggregation or crystalline modification are indicated near the symbols of chemical compounds, as well as the numerical value of thermal effects, are called thermochemical. In thermochemical equations, unless it is specifically stated, the values of thermal effects at constant pressure Q p are indicated equal to the change in the enthalpy of the system. The value is usually given on the right side of the equation, separated by a comma or semicolon. The following are accepted To- crystalline. These symbols are omitted if the aggregate state of substances is obvious, for example, O 2, H 2, etc.
The reaction equation is:
NH 3 (g) + HCl (g) \u003d NH 4 Cl (k). ; = ?
The values of the standard heats of formation of substances are given in special tables. Considering that the heats of formation of simple substances are conditionally taken equal to zero. The thermal effect of the reaction can be calculated using the corollary e from the Hess law:
\u003d (NH4Cl) - [(NH 3) + (HCl)];
= -315.39 - [-46.19 + (-92.31) = -176.85 kJ.
The thermochemical equation will look like:
The heat released during the reaction of 10 liters of ammonia in this reaction is determined from the proportion:
22,4 : -176,85 = 10 : X; x \u003d 10 (-176.85) / 22.4 \u003d -78.97 kJ; Q = 78.97 kJ.
Answer: 78.97 kJ.
thermal effect reaction is the amount of heat that is released or absorbed by the system during the reaction.
where , - stoichiometric coefficients of the reaction products and starting materials; , - standard enthalpies of formation of reaction products and starting materials. The heat of formation. Here the index means formation(formation), and zero that the value refers to the standard state of matter.
Heat of Formation substances is determined from reference books or calculated based on the structure of the substance.
Heat of combustion called the amount of heat released during the complete combustion of a unit amount of a substance, provided that the initial and final products are under standard conditions.
Distinguish:
· molar- for one mole (kJ/mol),
· mass− for one kilogram (kJ/kg),
· volumetric− for one cubic meter of substance (kJ/m³) the heat of combustion.
Depending on the state of aggregation of the water formed during the combustion process, there are higher and lower calorific values.
Higher calorific value called the amount of heat that is released during the complete combustion of a unit amount of combustible substance, including the heat of condensation of water vapor.
lower calorific value called the amount of heat that is released during the complete combustion of a unit amount of a combustible substance, provided that the water in the combustion products is in a gaseous state.
The molar heat of combustion is calculated in accordance with the law Hess. To convert the molar heat of combustion to mass heat, you can use the formula:
where is the molar mass of the combustible substance, .
For substances in the gaseous state, when converted from standard heat of combustion to volumetric heat, the formula is used:
where is the molar volume of the gas, which under standard conditions is .
Sufficiently accurate results for complex combustible substances or mixtures are given by the Mendeleev formula for the higher calorific value:
Where , ; , , , , - the content in the combustible substance, respectively, of carbon, hydrogen, sulfur, oxygen and nitrogen in mass. percent.
For lower calorific value
Where , ; - moisture content in the combustible substance in mass. percent.
The calculation of the heat of combustion of combustible mixtures is performed according to the formula
where is the lowest calorific value of the combustible mixture, ; - volume fraction of the i-th fuel in the mixture; is the lowest calorific value of the i-th fuel in the mixture, .
The calculation of the heat of combustion of gas-air mixtures is carried out using the formula
where is the lowest calorific value of a combustible substance, ; - concentration of combustible substance in the gas-air mixture, volume fraction; is the heat of combustion of the gas-air mixture, .
Heat capacity body is called physical quantity, which determines the ratio of an infinitesimal amount of heat received by the body to the corresponding increment in its temperature
The amount of heat supplied to or removed from a body is always proportional to the amount of matter.
specific heat capacity is the heat capacity per unit quantity of a substance. The amount of a substance can be measured in kilograms, cubic meters and moles. Therefore, a distinction is made between mass, volume and molar heat capacity.
Denote:
· - molar heat capacity, . This is the amount of heat that needs to be suspended on 1 mole of a substance so that its temperature rises by 1 Kelvin;
· - mass heat capacity, . This is the amount of heat that needs to be suspended on 1 kilogram of a substance that its temperature has risen by 1 Kelvin;
· - volumetric heat capacity, . This is the amount of heat that needs to be suspended from 1 cubic meter substance that its temperature has risen by 1 Kelvin.
The relationship between molar and mass heat capacities is expressed by the formula
where is the molar mass of the substance. Volumetric heat capacity is expressed in terms of molar as follows
where is the molar volume of the gas under normal conditions.
The heat capacity of a body depends on the process during which heat is supplied.
The heat capacity of a body at constant pressure called the ratio of the specific (per 1 mol of substance) amount of heat supplied in the isobaric process to the change in body temperature.
The heat capacity of a body at constant volume called the ratio of the specific (per 1 mole of a substance) amount of heat supplied in an isochoric process to a change in body temperature.
The heat capacity of ideal gases is
where is the number of degrees of freedom of the molecule. The relationship between the isobaric and isochoric heat capacities of ideal gases is determined by the Mayer equation
where is the universal gas constant.
The heat capacity of substances in the solid phase for conditions close to normal according to the Dulong-Petit law is
Due to the fact that the heat capacity depends on temperature, the heat consumption for the same temperature increase varies (Fig. 3.1).
True heat capacity called the heat capacity, which, under a certain thermodynamic process, is expressed by the following formula
where - denotes the process in which the heat capacity is measured. The parameter can take values , etc.
Rice. 3.1. Dependence of heat capacity on temperature
Average heat capacity is the ratio of the amount of heat imparted to the body in a given process to the change in temperature, provided that the temperature difference is a finite value. With a known dependence of the true heat capacity on temperature, the average heat capacity in the temperature range from to can be found using the mean value theorem
where is the average heat capacity, is the true heat capacity.
In experimental studies of the heat capacity of substances, the average heat capacity is often found as a function of the upper limit, with a fixed value of the lower limit, which is taken equal to
The dependences of the average heat capacities of gases on the temperature of the upper limit are given in Table 3.1.
Heat capacity gas mixture depends on the composition of the mixture and the heat capacities of the components. Let us denote: - the molar fraction of the component in the mixture; - volume fraction; - mass fraction. Here - the amount of the -th component in moles, m 3, kg, respectively. The heat capacity of a gas mixture can be determined by the formulas
where , , are the average molar, mass and volumetric heat capacities of the th component of the mixture.
Table 3.1.
| Gas name | Formulas for determining the average molar heat capacities of individual gases at constant volume, J/(mol deg), for temperatures, 0 С | |
| from 0 to 1500 | from 1501 to 2800 | |
| Air | ||
| Oxygen | ||
| Nitrogen | ||
| Hydrogen | ||
| carbon monoxide | ||
| Carbon dioxide | ||
| water vapor |
In heat engines and engines, at the beginning of each cycle, a portion of a fresh mixture is fed into the combustion chamber, which is called fresh charge. However, as a rule, exhaust gases from the previous cycle remain in the combustion chamber.
Residual gas ratio is called the ratio
where is the number of moles of residual gases, is the number of moles of fresh charge. The mixture in the combustion chamber of residual gases with a fresh charge is called working mixture. The heat capacity of the working mixture is calculated by the formula
where , are the average heat capacities of the fresh charge and residual gases at the temperature of the working mixture; - coefficient of residual gases.
The heat released in the combustion zone is spent on heating the combustion products and heat loss (the latter include preheating of the combustible substance and radiation from the combustion zone to environment). The maximum temperature to which the products of combustion are heated is called combustion temperature.
Depending on the conditions in which the combustion process takes place, there are calorimetric, adiabatic, theoretical, And valid combustion temperature.
Under calorimetric combustion temperature understand the temperature to which the products of combustion are heated under the following conditions:
all the heat released during the reaction goes to heating the combustion products;
complete combustion of the stoichiometric combustible mixture ();
In the process of formation of combustion products, their dissociation does not occur;
The combustible mixture is at an initial temperature of 273K and a pressure of 101.3 kPa.
Adiabatic combustion temperature determined for a non-stoichiometric combustible mixture ().
Theoretical combustion temperature differs from the calorimetric one in that the calculations take into account heat losses due to the dissociation of combustion products.
Actual combustion temperature is the temperature to which the combustion products are heated in real conditions.
Let us consider the calculation of only the calorimetric and adiabatic combustion temperatures with a small correction. We assume that the initial temperature of the initial mixture differs from . We denote and the number of moles of the working mixture and the mixture of combustion products. Then the heat balance of combustion at constant pressure can be written as
where , are the average heat capacities of the initial mixture and combustion products; is the heat released during the combustion of 1 mole of the working mixture, ; and are the temperatures of the working mixture and combustion products, respectively. With respect to one mole of the working mixture, formula (3.20) can be represented as
where is the coefficient of molecular change in the composition of the mixture. Calorimetric and adiabatic combustion temperatures are found from the heat balance equation.
The explosion pressure can be found using the Klaiperon-Mendeleev equation, given that the volume does not change during the process.
"Calculation of the heat of combustion of substances"
Target: Learn the basic concepts of the energy balance of combustion processes. Learn how to calculate the heat of combustion for different type combustible substance (individual substances and mixtures; complex substances represented by elemental composition).
Calculation formulas and algorithms
1. To calculate the calorific value individual substances formula (3.1) is used. First, a combustion reaction equation is drawn up, with the help of which stoichiometric coefficients and products are determined. Then, according to the table (see table 3.1), the standard enthalpies of formation of the starting materials and reaction products are found. The parameters found are substituted into formula (3.1) and the heat of combustion of the combustible substance is calculated.
2. Calorific value complex substances are found according to the formulas of D. I. Mendeleev (3.4) and (3.5). To perform the calculation, it is necessary to know only the mass fractions of elements in percent. The heat of combustion is calculated in kJ/kg.
3. For calculation combustible mixtures formulas (3.1) - (3.6) are used. First, the lower calorific value of each combustible gas is found as an individual substance according to formula (3.2) or as a complex substance according to formulas (3.4), (3.5). Formulas (3.2), (3.3) are used to pass to the volumetric heat of combustion. The calculation is completed by calculating the net calorific value of the combustible mixture according to the formula (3.6).
4. To determine the calorific value of 1 m 3 gas-air mixture calculate the volume fraction of combustible gases in the presence of air, the amount of which depends on . Then, using formula (3.7), the heat of combustion of the gas-air mixture is calculated.
Example 3.1. Determine the net calorific value of acetylene.
Solution. Let us write the equation for the combustion of acetylene.
In accordance with the equation, the stoichiometric coefficients are , , , . Using Appendix 3.1, we find the standard enthalpies of formation of reaction substances: , , , . Using formula (3.1), we calculate the net calorific value of acetylene
To calculate the amount of heat released during the combustion of 1 m 3 of acetylene, it is necessary to divide the resulting value by the molar volume under standard conditions (3.3):
Answer: ;
Solution. By the Mendeleev formulas (3.4) and (3.5) we find
Answer: .
Example 3.3. Determine the heat of combustion of a gas mixture consisting of - 40%, - 20%, - 15%, - 5%, - 10%, - 10%.
Solution. Of these gases, , , , are combustible. For each fuel, we write the equation for the reaction with oxygen:
We find the standard enthalpies of formation of substances using tabular data in Table 3.2.
; ; ; ; ; ; ; .
According to the formula (3.1), in accordance with the combustion equations (1) - (4), we find the heat of combustion, :
For a mixture of combustible gases, we use formula (3.6), taking into account that the molar and volume fractions are the same. As a result of calculations, we obtain the lower calorific value of a mixture of gases
During the combustion of 1 m 3 of such a mixture of gases, heat is released equal to
Answer: ; .
Solution. We write the propane combustion equation
In accordance with the reaction equation, 1 m 3 of propane should account for m 3 of air for a stoichiometric mixture. Considering that m 3 of air is actually consumed per 1 m 3 of propane. Thus, in 1 m 3 in a propane-air mixture, the volume fraction of propane will be
We find the lower calorific value of propane using formula (3.1). The standard enthalpy of formation of propane can be determined from Table 3.2.
The heat of combustion of propane is
The lower calorific value of the propane-air mixture can be determined by the formula (3.7)
1536,21
Table 3.3. Parameters for control task No. 3.1
| Option | Condition | Option | Condition | Option | Condition |
| 1. | CH3OH | 11. | C 4 H 8 | 21. | C 8 H 18 |
| 2. | C2H5OH | 12. | C 4 H 10 | 22. | C 10 H 8 |
| 3. | NH3 | 13. | C 3 H 8 | 23. | C 12 H 10 |
| 4. | SO 3 | 14. | C 7 H 8 | 24. | CH4O |
| 5. | HNO3 | 15. | C 7 H 16 | 25. | C2H4O2 |
| 6. | C 3 H 4 | 16. | C 5 H 12 | 26. | C2H6O |
| 7. | H 2 S | 17. | C 6 H 12 | 27. | C3H6O |
| 8. | C 5 H 5 N | 18. | C 6 H 14 | 28. | C4H10O |
| 9. | C 2 H 5 O | 19. | C 8 H 6 | 29. | CH 6 N 2 |
| 10. | C 3 H 6 | 20. | C 8 H 10 | 30. | C6H7N |
Table 3.4. Parameters for control task No. 3.2 ( W - moisture)
