Tasks for the section on the basics of thermodynamics with solutions. Thermodynamic calculations in chemical technology
Ministry of Education and Science of the Russian Federation
St. Petersburg State Polytechnic University
Faculty of Technology and Materials Research
Department of "Physical Chemistry, Micro- and Nanotechnologies"
COURSE WORK
"Thermodynamic evaluation of the possibility of flow
chemical process"
Option number 18
in the discipline "Physical Chemistry"
The work was done by a student of group 2068/2
______________ / Dmitrieva A.V.
Work checked
______________ / Art. teacher Elizarova E.P.
The calculation is carried out using the following approximations:
- Take the heat capacities of all participants in the reaction.
- To accept.
- To accept.
Tabular data for all participants in the reaction are given below.
Substance |
kJ/mol∙K |
||||||
At the end, present all the calculated data in a table and, based on the analysis of the obtained values, answer the following questions:
- Determine the thermodynamic possibility of a chemical reaction to proceed at a given temperature.
- Determine the type of this reaction from the point of view of thermochemistry.
- Evaluate the effect of temperature and pressure on the magnitude and shift of equilibrium.
Entropy method for calculating the change in the Gibbs energy and the equilibrium constant of a chemical reaction
This method uses the entropy values of the substances involved in the reaction. It is based on the ratio
(where is the change in the Gibbs energy at temperature Т;
thermal effect of the reaction at temperature T;
change in the entropy of the reaction at temperature T),
derived from the equation G = H – TS for a reaction proceeding at a constant temperature. Since the standard values of the entropies and heats of formation of substances were determined under standard conditions (p = 1 atm, T = 298 K), it is possible to calculate the standard change in the Gibbs energy using the formula:
First, at a temperature of 298 K, the thermal effect of the reaction and the algebraic sum of the entropies of the reaction participants are determined, taking into account the stoichiometric coefficients:
The thermal effect of a reaction at a given temperature is calculated according to the Kirchhoff law: the derivative of the thermal effect with respect to temperature is equal to the algebraic sum of the heat capacities of the substances involved in the reaction
If, then the thermal effect increases with increasing temperature; if so, it decreases.
The algebraic sum of the reaction entropies is found from the equations
Finally, to calculate the change in the Gibbs energy, we obtain
If the reaction participants undergo phase transformations in the interval under study, then the changes in entolpy and entropy must be found by dividing the integration interval into sections:
heat capacities corresponding to the phase in a given temperature range;
if the heat of formation refers to the product of the reaction, then the sign “+” is put; if to the original substance, then the sign “”.
In the first approximation, equation (*) is simplified by equating the sum of heat capacities to zero. That is, we neglect the temperature dependence of the entolpies and entropies of substances:
In the second approximation, the heat capacity is taken as a constant value equal to the heat capacity of substances at T = 298 K and their algebraic sum is found taking into account the stoichiometric coefficients:
Then from the formula (*) we obtain an approximate formula:
The most accurate third approximation takes into account the temperature dependence of the entolpy and entropy of substances, and the calculation is carried out according to the formula (*).
The standard change in the Gibbs energy allows you to determine the main characteristic of a chemical reaction - the constant of chemical equilibrium.
Each chemical reaction after some time after its start comes to an equilibrium state. Equilibrium is a state in which the composition of the system does not change with time. The equilibrium of the reaction will be characterized by the equilibrium constant. The constants expressed in terms of partial pressures are of greatest practical importance.
If all substances involved in the reaction are in standard states, then in this case
By calculating the numerical value of the equilibrium constant, one can calculate the yield of any reaction product and estimate the optimal conditions for the reaction (pressure and temperature).
Also, knowing the sign of the standard change in the Gibbs energy, one can estimate the thermodynamic probability of the reaction. If, then the reaction can proceed spontaneously under the given conditions. If, then the reaction does not proceed under the given conditions.
Settlement part
Thermal effect of the reaction at T=298 K:
The change in the entropy of the reaction at T=298 K:
First approach (:
The heat capacities of the substances involved in the reaction at T=298 K:
Algebraic sum of heat capacities at T=298 K:
Then the change in the thermal effect and the entropy of the reaction at T=1800 K:
second approximation (
In the third approximation, we will take into account phase transitions, in this reaction, the melting of manganese. Then we divide the entire temperature interval 298-1800K into two segments: before the melting point and after it, and we consider the heat capacity of substances as a function depending on temperature.
For the temperature range 298 - 1517 K:
For the interval 1517 - 1800 K:
The values of the change in the thermal effect of the reaction and the change in the entropy of the reaction, taking into account the phase transition:
Third approximation (
Let us determine the reaction equilibrium constant for three approximations:
Table of calculated data.
- In all approximations, the calculated value of the change in the Gibbs energy is positive. This means that the reaction at a temperature of 1800K cannot proceed.
- The change in the thermal effect of the reaction is also positive in all approximations, which means that the reaction is endothermic and proceeds with the absorption of heat.
- a) Effect of temperature on the equilibrium constant:
whence it can be seen that as the temperature rises, the equilibrium constant will increase, and, accordingly, the equilibrium will shift towards the reaction products.
b) Effect of pressure on the equilibrium constant:
where Const is some value; change in molar volume as a result of a reaction.
moreover, that is, with an increase in pressure in the system, the equilibrium constant will increase, and the equilibrium will shift towards the reaction products.
The considered factors generalize the principle of equilibrium displacement, also called Le Chatelier's principle: if an external influence is exerted on a system in a state of true equilibrium, then a spontaneous process occurs in the system that compensates for this influence.
Literature:
- A.G. Morachevsky, I.B. Sladkov. Guide to performing thermodynamic calculations. – L.: LPI, 1975.
- A.P. Ruzinov, B.S. Gulnitsky. Equilibrium transformations of chemical reactions. - M .: Metallurgy, 1976.
Introduction. Thermodynamic calculations make it possible to conclude that this process is possible, to choose the conditions for conducting a chemical reaction, to determine the equilibrium composition of the products, to calculate the theoretically achievable degrees of conversion of the initial substances and the yields of the products, as well as energy effects (heat of reaction, heat of change in the state of aggregation), which is necessary for compiling energy balances and determination of energy costs.
The most important concepts of thermodynamics are “heat of process” and “work”. The quantities characterizing the state of a thermodynamic system are called thermodynamic parameters. These include: temperature, pressure, specific volume, density, molar volume, specific internal energy. Quantities proportional to the mass (or amount of matter) of the considered thermodynamic system are called extensive; these are volume, internal energy, enthalpy, entropy. Intensive quantities do not depend on the mass of the thermodynamic system, and only they serve as thermodynamic parameters as states. These are temperature, pressure, and extensive quantities related to a unit of mass, volume or amount of a substance. Changing intensive parameters in order to accelerate chemical-technological processes is called intensification.
In exothermic reactions, the stock of internal energy of the initial substances (U 1) is greater than that of the resulting products (U 2). The difference ∆U = U 1 - U 2 is converted into the form of heat. On the contrary, in endothermic reactions, due to the absorption of a certain amount of heat, the internal energy of substances increases (U 2 > U 1). ∆U is expressed in J / mol or in technical calculations they are referred to 1 kg or 1 m 3 (for gases). The study of the thermal effects of reactions or states of aggregation, or mixing, dissolution is dealt with by the section of physical chemistry or chemical thermodynamics - thermochemistry. In thermochemical equations, the heat effect of the reaction is indicated. For example: C (graphite) + O 2 \u003d CO 2 + 393.77 kJ / mol. The heats of decomposition have the opposite sign. Tables are used to define them. According to D.P. Konovalov, the heat of combustion is determined from the ratio: Q burn = 204.2n + 44.4m + ∑x (kJ / mol), where n is the number of moles of oxygen required for the complete combustion of 1 mole of a given substance, m is the number of moles of water formed during the combustion of 1 mole of a substance, ∑x is a constant correction for a given homologous series. The more unlimiting, the more ∑x.
For acetylene hydrocarbons ∑x=213 kJ/mol. For ethylene hydrocarbons ∑x=87.9 kJ/mol. For saturated hydrocarbons ∑x=0. If the molecule of the compound has different functional groups and types of bonds, then the thermal characteristic is found by summation.
The thermal effect of a 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 initial substances, taking into account the number of moles of all substances participating in the reaction. For example, for a general reaction: n 1 A + n 2 B \u003d n 3 C + n 4 D + Q x thermal effect: Q x \u003d (n 3 Q C arr + n 4 Q D arr) - (n 1 Q A arr + n 2 Q B arr)
The thermal effect of a reaction is equal to the sum of the heats of combustion of the starting substances minus the sum of the heats of combustion of the reaction products, taking into account the number of moles of all reactants. For the same general reaction:
Q x \u003d (n 1 Q A burn + n 2 Q B burn) - (n 3 Q C burn + n 4 Q D burn)
Probability the course of equilibrium reactions is determined by the thermodynamic equilibrium constant, which is determined by:
К р = e - ∆ G º/(RT) = e - ∆ H º/ RT ∙ e ∆ S º/ R Analysis of this expression shows that for endothermic reactions (Q< 0, ∆Hº > 0) with a decrease in entropy (∆Sº< 0) самопроизвольное протекание реакции невозможно так как – ∆G > 0. In the following, the thermodynamic approach to chemical reactions will be considered in more detail.
Lecture 4
Basic laws of thermodynamics. First law of thermodynamics. Heat capacity and enthalpy. Enthalpy of reaction. Enthalpy of formation of the compound. Enthalpy of combustion. Hess' law and the enthalpy of reaction.
First law of thermodynamics: the change in internal energy (∆E) of the system is equal to the work of external forces (A′) plus the amount of transferred heat (Q): 1)∆E=A′+Q; or (2nd type) 2)Q=∆E+A – the amount of heat transferred to the system (Q) is spent on changing its internal energy (∆E) and work (A) done by the system. This is one of the types of the law of conservation of energy. If the change in the state of the system is very small, then: dQ=dE+δA - such a record for small (δ) changes. For gas (ideal) δА=pdV. In the isochoric process δA=0, then δQ V =dE, since dE=C V dT, then δQ V =C V dT, where C V is the heat capacity at constant volume. In a small temperature range, the heat capacity is constant, therefore Q V =C V ∆T. From this equation, it is possible to determine the heat capacity of the system and the heat of the processes. C V - according to the Joule-Lenz law. In an isobaric process proceeding without useful work, given that p is constant and can be taken out of the bracket under the differential sign, i.e. δQ P =dE+pdV=d(E+pV)=dH, here H is the enthalpy of the system. Enthalpy is the sum of the internal energy (E) of the system and the product of pressure and volume. The amount of heat can be expressed in terms of isobaric heat capacity (С Р): δQ P =С Р dT, Q V =∆E(V = const) and Q P =∆H(p = const) - after generalization. It follows that the amount of heat received by the system is uniquely determined by a change in some state function (enthalpy) and depends only on the initial and final states of the system and does not depend on the form of the path along which the process developed. This provision underlies the consideration of the problem of the thermal effects of chemical reactions.
Thermal effect of the reaction is related to the change in the chemical variable quantity of heat, obtained by the system in which the chemical reaction took place and the reaction products took the temperature of the initial reagents (as a rule, Q V and Q P).
Reactions with negative thermal effect, i.e., with the release of heat into the environment, is called exothermic. Reactions with positive thermal effect, i.e., going with the absorption of heat from the environment, are called endothermic.
The stoichiometric reaction equation will be: (1) ∆H=∑b J H J - ∑a i H i or ∆H=∑y i H i ; j are symbols of products, i are symbols of reagents.
This position is called Hess' law: quantities Е i , H i are functions of the state of the system and, consequently, ∆H and ∆Е, and thus the thermal effects Q V and Q р (Q V =∆Е, Q р =∆H) depend only on what substances react under given conditions and what products are obtained, but do not depend on the path along which the chemical process took place (reaction mechanism).
In other words, the enthalpy of a chemical reaction is equal to the sum of the enthalpies of formation of the reaction components multiplied by the stoichiometric coefficients of the corresponding components, taken with a plus sign for products and with a minus sign for starting substances. Let's find as an example∆H for the reaction PCl 5 +4H 2 O=H 3 PO 4 +5HCl (2)
The tabular values of the enthalpies of formation of the reaction component are, respectively, for PCl 5 - 463 kJ / mol, for water (liquid) - 286.2 kJ / mol, for H 3 PO 4 - 1288 kJ / mol, for HCl (gas) - 92.4 kJ /mol. Substituting these values into the formula: Q V =∆E, we get:
∆H=-1288+5(-92.4)–(-463)–4(-286.2)=-142kJ/mol
For organic compounds, as well as for CO, it is easy to carry out the combustion process to CO 2 and H 2 O. The stoichiometric equation for the combustion of an organic compound with the composition C m H n O p can be written as:
(3) C m H n O p + (p-m-n / 4) O 2 \u003d mCO 2 + n / 2 H 2 O
Therefore, the enthalpy of combustion according to (1) can be expressed in terms of the enthalpies of its formation and the formation of CO 2 and H 2 O:
∆H sg =m∆H CO 2 +n/2 ∆H H 2 O -∆H CmHnOp
Having determined the heat of combustion of the studied compound with the help of a calorimeter and knowing ∆H CO 2 and ∆H H 2 O , one can find the enthalpy of its formation.
Hess' law allows you to calculate the enthalpies of any reactions, if for each component of the reaction one of its thermodynamic characteristics is known - the enthalpy of formation of a compound from simple substances. The enthalpy of formation of a compound from simple substances is understood as ∆H of a reaction leading to the formation of one mole of a compound from elements taken in their typical states of aggregation and allotropic modifications.
Lecture 5
The second law of thermodynamics. Entropy. Gibbs function. Changes in the Gibbs function during chemical reactions. Equilibrium constant and Gibbs function. Thermodynamic estimation of the probability of a reaction.
The second law of thermodynamics called the statement that it is impossible to build a perpetual motion machine of the second kind. The law was obtained empirically and has two formulations equivalent to each other:
a) a process is impossible, the only result of which is the transformation of all the heat received from a certain body into work equivalent to it;
b) a process is impossible, the only result of which is the transfer of energy in the form of heat from a body that is less heated to a body that is hotter.
The function δQ/T is the total differential of some function S: dS=(δQ/T) arr (1) – this function S is called the entropy of the body.
Here Q and S are proportional to each other, that is, with an increase in (Q) (S) - increases, and vice versa. Equation (1) corresponds to an equilibrium (reversible) process. If the process is non-equilibrium, then the entropy increases, then (1) is transformed:
dS≥(δQ/T)(2) Thus, when nonequilibrium processes occur, the entropy of the system increases. If (2) is substituted into the first law of thermodynamics, we get: dE≤TdS-δA. It is customary to write it in the form: dE≤TdS-δA'-pdV, hence: δA'≤-dE+TdS-pdV, here pdV is the equilibrium expansion work, δA' is the useful work. Integration of both parts of this inequality for an isochoric-isothermal process leads to the inequality: A'V≤-∆E+T∆S(3). And integration for the isobaric-isothermal process (Т=const, p=const) leads to the inequality:
A’ P ≤ - ∆E+T∆S – p∆V=-∆H + T∆S (4)
The right parts (3 and 4) can be written as changes to some functions, respectively:
F=E-TS(5) and G=E-TS+pV; or G=H-TS (6)
F is the Helmholtz energy and G is the Gibbs energy, then (3 and 4) can be written as A’ V ≤-∆F (7) and A’ P ≤-∆G (8). The law of equality corresponds to an equilibrium process. In this case, the most useful work is done, that is, (A’ V) MAX = -∆F, and (A’ P) MAX = -∆G. F and G are respectively called isochoric-isothermal and isobaric-isothermal potentials.
Equilibrium of chemical reactions characterized by a process (thermodynamic) in which the system goes through a continuous series of equilibrium states. Each of these states is characterized by the invariability (in time) of thermodynamic parameters and the absence of matter and heat flows in the system. The equilibrium state is characterized by the dynamic nature of the equilibrium, that is, the equality of direct and reverse processes, the minimum value of the Gibbs energy and the Helmholtz energy (that is, dG=0 and d 2 G>0; dF=0 and d 2 F>0). In dynamic equilibrium, the rates of the forward and reverse reactions are the same. Equation must also be observed:
µ J dn J =0, where µ J =(ðG/ðn J) T , P , h =G J is the chemical potential of component J; n J is the amount of component J (mol). A large value of µ J indicates a high reactivity of the particles.
∆Gº=-RTLnK p(9)
Equation (9) is called the van't Haff isotherm equation. The value of ∆Gº in tables in the reference literature for many thousands of chemical compounds.
K p \u003d e - ∆ G º / (RT) \u003d e - ∆ H º / RT ∙ e ∆ S º / R (11). From (11) one can give a thermodynamic estimate of the probability of the reaction occurring. So, for exothermic reactions (∆Нº<0), протекающих с возрастанием энтропии, К р >1, and ∆G<0, то есть реакция протекает самопроизвольно. Для экзотермических реакций (∆Нº>0) with a decrease in entropy (∆Sº>0), the spontaneous flow of the process is impossible.
If ∆Hº and ∆Sº have the same sign, the thermodynamic probability of the process proceeding is determined by the specific values of ∆Hº, ∆Sº and Tº.
Consider, using the example of the ammonia synthesis reaction, the combined effect of ∆Н o and ∆S o on the possibility of implementing the process:
For this reaction, ∆Н o 298 = -92.2 kJ / mol, ∆S o 298 = -198 J / (mol * K), T∆S o 298 = -59 kJ / mol, ∆G o 298 = -33, 2kJ/mol.
It can be seen from the given data that the change in entropy is negative and does not favor the reaction, but at the same time, the process is characterized by a large negative enthalpy effect ∆Hº, due to which the process is possible. With an increase in temperature, the reaction, as shown by calorimetric data, becomes even more exothermic (at T=725K, ∆H=-113kJ/mol), but with a negative value of ∆Sо, an increase in temperature greatly reduces the likelihood of the process occurring.
One of the most important questions in chemistry is the question of the possibility of a chemical reaction. A quantitative criterion for the fundamental feasibility of a chemical reaction is, in particular, the characteristic function of the state of the system, called the Gibbs energy (G). Before proceeding to the consideration of this criterion, let us dwell on a number of definitions.
spontaneous processes. Spontaneous processes are processes that occur without energy supply from an external source. Many chemical processes are spontaneous, such as the dissolution of sugar in water, the oxidation of metals in air (corrosion), etc.
Reversible and irreversible processes. Many chemical reactions proceed in one direction until the reactants are completely exhausted. Such reactions are called chemically irreversible. An example is the interaction of sodium and water.
Other reactions proceed first in the forward direction, and then in the forward and reverse direction due to the interaction of the reaction products. As a result, a mixture is formed containing both the starting materials and the reaction products. Such reactions are called chemically reversible. As a result of a chemically reversible process, true (stable) chemical equilibrium, which is characterized by the following features:
1) in the absence of external influences, the state of the system remains unchanged indefinitely;
2) any change in external conditions leads to a change in the state of the system;
3) the state of equilibrium does not depend on which side it is reached from.
An example of a system in true equilibrium is an equimolecular mixture
CO (g) + H 2 O (g) CO 2 (g) + H 2 (g).
Any change in temperature or other conditions causes a shift in equilibrium, i.e. change in the composition of the system.
In addition to true equilibria, apparent (false, hindered) equilibria are very often encountered, when the state of the system persists in time for a very long time, but a small impact on the system can lead to a strong change in its state. An example is a mixture of hydrogen and oxygen, which at room temperature in the absence of external influences can remain unchanged indefinitely. However, it is enough to introduce platinized asbestos (catalyst) into this mixture, as an energetic reaction will begin.
H 2 (g) + O 2 (g) \u003d H 2 O (g),
leading to complete exhaustion of the starting materials.
If the same catalyst is introduced under the same conditions into liquid water, then it is impossible to obtain the initial mixture.
Entropy. The state of any system can be characterized by the values of directly measured parameters (p, T, etc.). it characteristic of the macrostate of the system. The state of the system can also be described by the characteristics of each particle of the system (atom, molecule): coordinate, vibration frequency, rotation frequency, etc. it characteristic of the microstate of the system. Systems consist of a very large number of particles, so one macrostate will correspond to a huge number of different microstates. This number is called the thermodynamic probability of the state and is denoted as W.
Thermodynamic probability is associated with another property of matter - entropy (S, J / (mol. K)) - Boltzmann formula
where R is the universal gas constant and N A is the Avogadro constant.
The physical meaning of entropy can be explained by the following thought experiment. Let an ideal crystal of some substance, such as sodium chloride, be cooled to absolute zero temperature. Under these conditions, the sodium and chlorine ions that make up the crystal become practically immobile, and this macroscopic state is characterized by a single microstate, i.e. W=1, and in accordance with (3.13) S=0. As the temperature rises, the ions begin to oscillate around the equilibrium positions in the crystal lattice, the number of microstates corresponding to one macrostate increases, and, consequently, S>0.
In this way, entropy is a measure of the disorder of the state of a system. The entropy of the system increases in all processes accompanied by a decrease in order (heating, dissolution, evaporation, decomposition reactions, etc.). Processes that occur with an increase in order (cooling, crystallization, compression, etc.) lead to a decrease in entropy.
Entropy is a function of state, but unlike most other thermodynamic functions, it is possible to experimentally determine the absolute value of the entropy of a substance. This possibility is based on the postulate of M. Planck, according to which at absolute zero, the entropy of an ideal crystal is zero(third law of thermodynamics).
The temperature dependence of the entropy of a substance is presented qualitatively in Fig. . 3.1.
On fig. 3.1 it can be seen that at a temperature equal to 0 K, the entropy of a substance is zero. With an increase in temperature, the entropy increases smoothly, and at the points of phase transitions, an abrupt increase in entropy takes place, determined by the relation
(3.14)
where Δ f.p S, Δ f.p H and T f.p are the changes in entropy, enthalpy and phase transition temperature, respectively.
The entropy of a substance B in the standard state is denoted as . For many substances, the absolute values of the standard entropies are determined and are given in reference books.
Entropy, like internal energy and enthalpy, is a function of state, so the change in the entropy of a system in a process does not depend on its path and is determined only by the initial and final states of the system. The change in entropy during a chemical reaction (3.10) can be found as the difference between the sum of the entropies of the reaction products and the sum of the entropies of the starting materials:
The concept of entropy is used in one of the formulations second law of thermodynamics: in isolated systems, only processes occurring with an increase in entropy (ΔS>0) can spontaneously proceed. Isolated systems are understood as systems that do not exchange either matter or energy with the environment. Systems in which chemical processes take place do not belong to isolated systems, because they exchange energy with the environment (the thermal effect of the reaction), and in such systems processes can also occur with a decrease in entropy.
SO 2 (g) + 2H 2 S (g) \u003d 3S (t) + 2H 2 O (l), if the standard entropies of sulfur oxide (IV), hydrogen sulfide, sulfur and water are 248.1; 205.64; 31.88 and 69.96 J/(mol K), respectively.
Solution. Based on equation (3.15), we can write:
The entropy in this reaction decreases, which is associated with the formation of solid and liquid products from gaseous substances.
Example 3.8. Without making calculations, determine the sign of the change in entropy in the following reactions:
1) NH 4 NO 3 (c) \u003d N 2 O (g) + 2H 2 O (g),
2) 2H 2 (g) + O 2 (g) \u003d 2H 2 O (g),
3) 2H 2 (g) + O 2 (g) \u003d 2H 2 O (g).
Solution. In reaction (1), 1 mol of NH 4 NO 3 in the crystalline state forms 3 mol of gases, therefore, D r S 1 >0.
In reactions (2) and (3), both the total number of moles and the number of moles of gaseous substances decrease. Therefore, D r S 2<0 и D r S 3 <0. При этом уменьшение энтропии в реакции (3) больше, чем в реакции (2) , так как S о (H 2 O (ж)) < S о (H 2 O (г)).
Gibbs energy(isobaric-isothermal potential). In many cases, spontaneous processes in nature occur in the presence of a potential difference, for example, the difference in electrical potentials causes charge transfer, and the difference in gravitational potentials causes the body to fall. These processes end when the minimum potential is reached. The driving force of chemical processes occurring at constant pressure and temperature is the isobaric-isothermal potential, called Gibbs energy and denoted G. The change in the Gibbs energy in a chemical process is determined by the relation
ΔG = ΔH –TΔS, (3.16)
where ΔG is the change in the Gibbs energy of the chemical process; ΔH is the change in the enthalpy of the chemical process; ΔS is the change in the entropy of the chemical process; T is temperature, K.
Equation (3.16) can be represented in the following form:
∆H = ∆G + T∆S. (3.17)
The meaning of equation (3.17) is that part of the heat effect of the reaction is spent on doing work (ΔG), and part is dissipated into the environment (TΔS).
The Gibbs energy is a criterion for the fundamental possibility of a spontaneous reaction. If the Gibbs energy decreases during the reaction, then the process can proceed spontaneously under these conditions:
ΔG< 0. (3.18)
The process under these conditions is not feasible if
ΔG > 0. (3.19)
Expressions (3.18) and (3.19) simultaneously mean that the reverse reaction cannot (3.18) or can (3.19) proceed spontaneously.
The reaction is reversible, i.e. can flow in both forward and reverse directions, if
Equation (3.20) is a thermodynamic condition for chemical equilibrium.
Relations (3.18) – (3.20) are also applicable to phase equilibria, i.e. to cases when two phases (aggregate states) of the same substance are in equilibrium, for example, ice and liquid water.
Enthalpy and entropy factors. It follows from equations (3.16) and (3.18) that processes can proceed spontaneously (ΔG<0), если они сопровождаются уменьшением энтальпии (ΔH<0) и увеличением энтропии системы (ΔS>0). If the enthalpy of the system increases (ΔH>0), and the entropy decreases (ΔS<0), то такой процесс протекать не может (ΔG>0). With other signs of ΔS and ΔН, the fundamental possibility of the process proceeding is determined by the ratio of the enthalpy (ΔH) and entropy (ТΔS) factors.
If ΔН>0 and ΔS>0, i.e. Since the enthalpy component counteracts, and the entropy component favors the course of the process, the reaction can proceed spontaneously due to the entropy component, provided that |ΔH|<|TΔS|.
If the enthalpy component favors and the entropy counteracts the process, then the reaction can proceed spontaneously due to the enthalpy component, provided that |ΔH|>|TΔS|.
Effect of temperature on the direction of the reaction. Temperature affects the enthalpy and entropy components of the Gibbs energy, which may be accompanied by a change in the sign of the Gibbs energy of these reactions, and hence the direction of the reactions. For a rough estimate of the temperature at which the sign of the Gibbs energy changes, we can neglect the dependence of ΔН and ΔS on temperature. Then it follows from Eq. (3.16) that the sign of the Gibbs energy will change at the temperature
It is obvious that the sign change of the Gibbs energy with temperature change is possible only in two cases: 1) ΔН>0 and ΔS>0 and 2) ΔН<0 и ΔS<0.
The standard Gibbs energy of formation is the change in the Gibbs energy of the reaction of formation of 1 mol of a compound from simple substances that are stable under standard conditions. The Gibbs energy of the formation of simple substances is assumed to be zero. The standard Gibbs energies of the formation of substances can be found in the relevant reference books.
Gibbs energy of a chemical reaction. The Gibbs energy is a state function, i.e. its change in the process does not depend on the path of its flow, but is determined by the initial and final states of the system. Therefore, the Gibbs energy of the chemical reaction (3.10) can be calculated from the formula
Note that the conclusions about the fundamental possibility of the reaction proceeding in terms of Δ r G are applicable only to those conditions for which the change in the Gibbs energy of the reaction is calculated. If the conditions differ from the standard, then the equation can be used to find Δ r G van't Hoff isotherms, which for the reaction (3.10) between gases is written as
(3.23)
and between solutes
(3.24)
where are the partial pressures of the corresponding substances; c A, c B, c D , c E are the concentrations of the corresponding dissolved substances; a, b, d, e are the corresponding stoichiometric coefficients.
If the reactants are in the standard state, then equations (3.23) and (3.24) become the equation
Example 3.9. Determine the possibility of the reaction NH 3 (g) + HCl (g) \u003d NH 4 Cl (k) under standard conditions at a temperature of 298.15 K, using data on standard enthalpies of formation and entropies.
Solution. Based on the first corollary of the Hess law, we find the standard enthalpy of the reaction:
; the reaction is exothermic, therefore, the enthalpy component favors the reaction.
We calculate the change in the entropy of the reaction according to the equation
The reaction is accompanied by a decrease in entropy, which means that the entropy component counteracts the reaction.
We find the change in the Gibbs energy of the process according to equation (3.16):
Thus, this reaction can proceed spontaneously under standard conditions.
Example 3.10. Using data on standard enthalpies of formation and entropies, determine at what temperature equilibrium will occur in the system N 2 (g) + 3H 2 (g) \u003d 2NH 3 (g).
Solution. The equilibrium condition for the system is ΔG=0. To do this, using relation (3.21), we find the temperature at which ΔG=0. Calculate the standard enthalpy and entropy of the reaction:
The enthalpy component favors, and the entropy component opposes the reaction, which means that at a certain temperature it is possible to change the sign of the Gibbs energy, i.e. change the direction of the reaction.
The equilibrium condition will be written as follows:
∆G = ∆H –T∆S,
or, substituting numerical values, we get
0 \u003d - 92.38 - T (-198.3) 10 -3.
Therefore, the reaction will be in equilibrium at a temperature
TO.
Below this temperature, the reaction will proceed in the forward direction, and above this temperature, in the opposite direction.
Example 3.11. At a certain temperature T, the endothermic reaction A®B practically goes to completion. Determine: a) the sign D r S of the reaction; b) sign DG of the reaction B ® A at temperature T; c) the possibility of the reaction B ® A at low temperatures.
Solution. a) The spontaneous occurrence of the reaction A ® B indicates that DG<0. Поскольку DН>0, then from the equation
DG = DH - TDS implies that DS>0; for reverse reaction B ® A DS<0.
b) For the reaction A ® B DG<0. Следовательно, для обратной реакции при той же температуре DG>0.
c) The reaction A ® B is endothermic (DH<0), следовательно, обратная реакция В ® А экзотермическая. При низких температурах абсолютная величина члена TDS мала, так что знак DG определяется знаком DН. Следовательно, при достаточно низких температурах протекание реакции В ® А возможно.
Example 3.12. Calculate the value of the Gibbs energy and determine whether the reaction CO + Cl 2 ÛCOCl 2 is possible at a temperature of 700 K, if the equilibrium constant of the reaction at this temperature is 10.83 atm -1 and the partial pressures of all components are the same and equal to one.
Solution. The relationship D r G 0 and K r of the reaction A + B Û C + D is given by the isotherm equation (3.22)
Under standard conditions, when the partial pressure of each reactant is 1 atm, this ratio will take the form
Consequently, the reaction at T=700 K can proceed spontaneously in the forward direction.
Questions and tasks for self-study
1. Give the numerical values of pressure and temperature in the international system of units, as well as in atmospheres, millimeters of mercury and degrees Celsius, corresponding to standard and normal conditions.
2. What condition do the state functions satisfy? What determines the change in the value of the state function in the process?
3. The constancy of what parameters characterizes isobaric-isothermal and isochoric-isothermal processes?
4. Formulate the first law of thermodynamics.
5. Under what conditions will the thermal effect of the process be: a) equal to the change in the enthalpy of this process; b) is equal to the change in the internal energy of the process?
6. The chemical reaction takes place in a sealed reactor. The change in which state function will determine the thermal effect of the reaction?
7. During a chemical reaction, the temperature of the system rises. Is this process exothermic or endothermic? What sign (+) or (-) does the enthalpy change of this process have?
8. Formulate Hess' law.
9. Define the term "standard enthalpy of formation of a substance."
10. What are the standard enthalpies of formation of molecular chlorine and stable at a temperature of 298 K modification of iron α-Fe?
11. The standard enthalpy of formation of white phosphorus is zero, and red - (-18.41) kJ / mol. Which of the allotropic modifications is more stable at a temperature of 25 o C?
12. Formulate the 1st corollary of Hess' law.
13. Define the concept of "standard enthalpy of combustion of a substance."
14. How are the standard enthalpy of formation of carbon dioxide and the standard enthalpy of combustion stable at T = 298 K modification of carbon - graphite?
15. Give 3 examples of spontaneous chemical processes.
16. List the signs of chemical (true) equilibrium.
17. Give examples of processes accompanied by: a) an increase in entropy; b) a decrease in entropy.
18. What sign should the change in the entropy of a spontaneously occurring reaction have if Δ r Н=0?
19. What sign should the change in the entropy of the reaction of thermal decomposition of calcium carbonate have? Why? Write the reaction equation.
20. What thermodynamic properties of the participants in the reaction do you need to know in order to resolve the issue of the possibility of a reaction?
21. An exothermic reaction between gases is accompanied by an increase in volume. What can be said about the possibility of such a reaction?
22. In which of the following cases is it possible to change the direction of the reaction with a change in temperature: a) DH<0, DS<0; б) DH>0, DS>0; c) DH<0, DS>0; d) DH>0, DS<0?
23. Find the standard enthalpy for the oxidation of gaseous sulfur(IV) oxide with oxygen to gaseous sulfur(VI) oxide. Standard enthalpies of formation SO 2 - (-297 kJ / mol) and SO 3 - (-395 kJ / mol).
Answer: -196 kJ.
24. Indicate the sign of the change in entropy in the following reactions:
a) CO (G) + H 2 (G) \u003d C (T) + H 2 O (G);
b) CO 2 (G) + C (T) \u003d 2CO (G);
c) FeO (T) + CO (G) \u003d Fe (T) + CO 2 (G);
d) H 2 O (W) \u003d H 2 O (G);
Answer: a) (-); b) (+); c)(~0); d) (+); e) (-).
25. Find the standard entropy of the reaction of oxidation of gaseous sulfur(IV) oxide with oxygen to gaseous sulfur(VI) oxide. Standard entropy of formation of SO 2 - (248 J / (mol K), SO 3 - (256 J / (mol K)), O 2 - (205 J / (mol K)).
Answer: -189 J/K.
26. Find the enthalpy of the reaction for the synthesis of benzene from acetylene, if the enthalpy of combustion of benzene is (-3302 kJ / mol), and acetylene - (-1300 kJ / mol).
Answer: - 598 kJ.
27. Find the standard Gibbs energy of the decomposition reaction of sodium bicarbonate. Is it possible for the reaction to proceed spontaneously under these conditions?
Answer: 30.88 kJ.
28. Find the standard Gibbs energy of the reaction 2Fe (T) + 3H 2 O (G) \u003d Fe 2 O 3 (T) + 3H 2 (G) (corrosion reactions of carbon steel with water vapor). Is it possible for the reaction to proceed spontaneously under these conditions?
Answer: -54.45 kJ.
29. At what temperature will chemical equilibrium occur in the system 2NO (g) + O 2 (g) Û 2NO 2 (g)?
Answer: 777 K.
30. Find the thermal effect of the evaporation process of 1 g of water (specific heat of evaporation) at a temperature of 298 K, if the standard enthalpy of formation of H 2 O (l) is (-285.84 kJ / mol), and gaseous - (-241.84 kJ /mol).
Answer: 2.44 kJ / g.
3.4. Tasks for current and intermediate controls
Section I
1. The process of formation of carbon dioxide during the combustion of graphite in oxygen can proceed in two ways:
I. 2C (g) + O 2 (g) \u003d 2CO (g); 2CO (g) + O 2 \u003d 2CO 2 (g), D r H ° \u003d -566 kJ.
II. C (gr) + O 2 (g) \u003d CO 2 (g), D r H ° \u003d -393 kJ.
Find D f H°(CO).
Answer: -110 kJ / mol.
2. Calculate the enthalpy of formation and enthalpy of combustion of carbon monoxide (CO) based on the following reactions:
I. 2C (g) + O 2 (g) \u003d 2CO (g), D r H ° \u003d -220 kJ.
II. 2CO (g) + O 2 (g) \u003d 2CO 2 (g), D r H ° \u003d -566 kJ.
Answer: -110 kJ/mol; -283 kJ/mol.
3. Find the standard enthalpy of formation of sodium sulfite from the thermochemical equation
4Na 2 SO 3 (cr) \u003d 3Na 2 SO 3 (cr) + Na 2 S (cr) - 181.1 kJ,
if 
kJ/mol and 
kJ/mol.
Answer: -1090 kJ / mol.
4. Find the standard enthalpy of combustion of methane based on the reaction CH 4 (g) + 2O 2 (g) \u003d CO 2 (g) + 2H 2 O (g), D r H ° \u003d -802 kJ.
Answer: -802 kJ / mol.
5. Predict whether it will be positive or negative
change in the entropy of the system in the reactions:
a) H 2 O (g) ® H 2 O (g) (at a temperature of 25 ° C);
b) CaCO 3 (t) ® CaO (t) + CO 2 (g);
c) N 2 (g) + 3H 2 (g) \u003d 2NH 3 (g);
d) N 2 (g) + O 2 (g) \u003d 2NO (g);
e) Ag + (solution) + Cl - (solution) = AgCl (t).
Give explanations without making calculations.
Answer: a) +; b) +; in) -; d) ~0; e) -.
6. Predict the sign of the DS system in each of the following
processes:
a) evaporation of 1 mol CCl 4(g);
b) Br 2(g) → Br 2(g);
c) precipitation of AgCl(t) by mixing NaCl(aq.) and AgNO 3 (aq.).
Give explanations.
Answer: a) +; b) -; in)-.
7. Using the tabular values of the absolute values of the entropies of substances under standard conditions (S °), compare the values of the absolute entropies of substances at a temperature of 298 K in each of the following pairs:
a) O 2 (g) and O 3 (g);
b) C (diamond) and C (graphite);
c) NaCl (t) and MgCl 2(t).
Explain the reason for the difference in S° in each case.
8. Calculate D r S° for reactions
a) N 2 (g) + 3H 2 (g) \u003d 2NH 3 (g); b) 2SO 2 (g) + O 2 (g) \u003d 2SO 3 (g),
using tabular values of the absolute entropies of substances under standard conditions.
Answer: a) -197.74 J/K; b) -188.06 J/K.
9. Using the tabular values of the absolute en-
tropium (S°), calculate D r S° for the following processes:
a) CO (g) + 2H 2 (g) \u003d CH 3 OH (g);
b) 2HCl (g) + Br 2 (g) \u003d 2HBr (g) + Cl 2 (g);
c) 2NO 2 (g) = N 2 O 4 (g).
Does the sign of D r S° agree in each case with that which should be expected on the basis of qualitative representations? Explain answers.
Answer: a) -218.83 J/K; b) 94.15 J/K; c) -175.77 J/K.
10. The standard enthalpy of formation of CO (g) is -110.5 kJ/mol. The combustion of 2 mol CO (g) released 566 kJ of heat. Calculate
Answer: -393.5 kJ / mol.
11. Determine the amount of heat released when quenching 100 kg of lime with water: CaO (k) + H 2 O (l) \u003d Ca (OH) 2 (k), if the standard heats of formation CaO (k), H 2 O (l) , Ca(OH) 2(k) are respectively -635.14; -285.84; -986.2 kJ/mol.
Answer: -1165357.2 kJ.
12. Determine the enthalpy of decomposition of hydrogen peroxide (H 2 O 2) into water and oxygen using the data below:
SnCl 2 (p) + 2HCl (p) + H 2 O 2 (p) \u003d SnCl 4 (p) + 2H 2 O (l), D r H ° \u003d -393.3 kJ;
SnCl 2 (p) + 2HCl (p) + 1 / 2O 2 (g) \u003d SnCl 4 (p) + H 2 O (l), D r H ° \u003d -296.6 kJ.
Answer: - 96.7 kJ.
13. Calculate the amount of heat that is released in the production of 10 6 kg of ammonia per day, if
Answer: -2.7. 10 9 kJ.
14. Determine based on the following data:
P 4 (cr) + 6Cl 2 (g) \u003d 4PCl 3 (l), D r H ° \u003d -1272.0 kJ;
PCl 3 (g) + Cl 2 (g) \u003d PCl 5 (cr), D r H ° \u003d -137.2 kJ.
Answer: -455.2 kJ / mol.
15. Calculate the change in the enthalpy of reaction under standard conditions: H 2 (g) + 1 / 3O 3 (g) \u003d H 2 O (g), based on the following data:
2O 3 (g) \u003d 3O 2 (g), D r H ° \u003d -288.9 kJ,
kJ/mol.
Answer: -289.95 kJ.
16. Calculate the standard enthalpy of PbO formation reaction using the following data:
1) 2Pb (cr) + O 2 (g) \u003d 2PbO 2 (cr) - 553.2 kJ;
2) 2PbO 2 (cr) \u003d 2PbO (cr)) + O 2 (g) + 117.48 kJ.
Answer: -217.86 kJ / mol.
17. Calculate the standard enthalpy of the CuCl formation reaction using the following data:
1) CuCl 2 (cr) + Cu (cr) = 2 CuCl (cr) - 63.5 kJ;
2) Cu (cr) + Cl 2 (g) = CuCl 2 (cr) - 205.9 kJ.
Answer: 134.7 kJ / mol.
18. Calculate Δ f H° of methyl alcohol in the liquid state, knowing the following data:
H 2 (g) + 1 / 2O 2 (g) \u003d H 2 O (g), D r H ° \u003d -285.8 kJ;
C (gr) + O 2 (g) \u003d CO 2 (g), D r H ° \u003d -393.7 kJ;
CH 3 OH (l) + 3 / 2O 2 (g) \u003d CO 2 (g) + 2H 2 O (l), D r H ° \u003d -715.0 kJ.
Answer: -250.3 kJ / mol.
19. The standard enthalpies of combustion of benzene and acetylene are -3270 and -1302 kJ / mol, respectively. Determine D r H ° of the conversion of acetylene to benzene: 3C 2 H 2 (g) \u003d C 6 H 6 (g).
Answer: -636 kJ.
20. Determine the standard enthalpy of formation of iron oxide (III), if 146.8 kJ of heat was released during the oxidation of 20 g of iron.
Answer: -822 kJ / mol.
21. Calculate the amount of heat that is released when receiving 22.4 liters of ammonia (n.o.) if
N 2 (g) + 3H 2 (g) \u003d 2NH 3 (g), D r H ° \u003d -92 kJ.
Answer: -46 kJ.
22. Determine Δ f H° of ethylene using the following data.
C 2 H 4 (g) + 3O 2 (g) \u003d 2CO 2 (g) + 2H 2 O (g) -1323 kJ;
C (gr) + O 2 (g) \u003d CO 2 (g) -393.7 kJ;
H 2 (g) + 1 / 2O 2 (g) \u003d H 2 O (g) -241.8 kJ.
Answer: 52 kJ / mol.
23. Calculate the enthalpy of reaction F (g) + Li (g) \u003d F - (g) + Li + (g),
if F (g) + e \u003d F - (g) -322 kJ / mol;
Li (g) \u003d Li + (g) + e + 520 kJ / mol.
Answer: 198 kJ.
24. Calculate the standard enthalpy of the formation of Hg 2 Br 2 using the following data:
1) HgBr 2 (cr) + Hg (g) = Hg 2 Br 2 (cr) - 37.32 kJ;
2) HgBr 2 (cr) \u003d Hg (l) + Br 2 (l) + 169.45 kJ.
Answer: -206.77 kJ / mol.
25. Calculate the standard enthalpy of sodium bicarbonate formation using the following data:
2NaHCO 3 (cr) \u003d Na 2 CO 3 (cr) + CO 2 (g) + H 2 O (g) + 130.3 kJ,
if 
kJ/mol;
C (gr) + O 2 (g) \u003d CO 2 (g) - 393.7 kJ; H 2 (g) + 1 / 2O 2 (g) \u003d H 2 O (g) -241.8 kJ.
Answer: -947.4 kJ / mol.
26. Calculate the standard enthalpy of the formation of CaCO 3 (cr) using the following data:
Ca (OH) 2 (c) + CO 2 (g) \u003d CaCO 3 (cr) + 173.9 kJ;
C (gr) + O 2 (g) \u003d CO 2 (g) - 393.7 kJ;
kJ/mol.
Answer: -1206 kJ / mol.
27. Determine the standard enthalpy of formation of iron oxide (III), if during the reaction
2Fe + Al 2 O 3 \u003d Fe 2 O 3 + 2Al
for every 80 g of Fe 2 O 3, 426.5 kJ of heat is absorbed, 
kJ/mol.
Answer: -823 kJ / mol.
28. How much heat must be spent to obtain 11.2 kg of iron, if, in accordance with the thermochemical equation, FeO (t) + H 2 (g) \u003d Fe (t) + H 2 O (g) + 23 kJ.
Answer: 4600 kJ.
29. Find the heat of combustion of diamond if the standard heat of combustion of graphite is -393.51 kJ / mol, and the heat is
and the phase transition С(graphite) ® С(diamond) is
1.88 kJ/mol.
Answer: -395.39 kJ / mol.
30. How much heat is released when 1 kg of red phosphorus is converted to black phosphorus, if known,
that the standard enthalpies of formation of red and black phosphorus are -18.41 and -43.20 kJ/mol, respectively.
Answer: -800 kJ.
Section II
Calculate the standard change in the Gibbs energy of a chemical reaction at a temperature of 25 °C from the values of the standard enthalpies of formation and absolute entropies of chemical compounds and establish the possibility of a spontaneous reaction:
1. 4NH 3g + 5O 2g = 4NO g + 6H 2 O g.
Answer: -955.24 kJ; reaction is possible.
2. SO 2g + 2H 2 S g \u003d 3S to + 2H 2 O well.
Answer: -107.25 kJ; reaction is possible.
3. 2H 2 S g + 3O 2g = 2H 2 O g + 2SO 2g.
Answer: -990.48 kJ; reaction is possible.
4. 2NO g + O 3g + H 2 O well \u003d 2HNO 3g.
Answer: - 260.94 kJ; reaction is possible.
5. 3Fe 2 O 3k + CO g \u003d 2Fe 3 O 4k + CO 2g.
Answer: - 64.51 kJ; reaction is possible.
6. 2CH 3 OH w + 3O 2g \u003d 4H 2 Og + 2CO 2g.
Answer: - 1370.46 kJ; reaction is possible.
7. CH 4g + 3CO 2g \u003d 4CO g + 2H 2 O g.
Answer: 228.13 kJ; reaction is not possible.
8. Fe 2 O 3k + 3CO g \u003d 2Fe k + 3CO 2g.
Answer: -31.3 kJ; reaction is possible.
9. C 2 H 4g + 3O 2g \u003d 2CO 2g + 2H 2 O g.
Answer: -1313.9 kJ; reaction is possible.
10. 4NH 3g + 3O 2g = 6H 2 O g + 2N 2g.
Answer: -1305.69 kJ; reaction is possible.
11. 4NO 2g + O 2g + 2H 2 O x = 4HNO 3g.
Answer: -55.08 kJ; reaction is possible.
12. 2HNO 3l + NO g = 3NO 2g + H 2 O l.
Answer: -7.71 kJ; reaction is possible.
13. 2C 2 H 2g + 5O 2g \u003d 4CO 2g + 2H 2 O g.
Answer: -2452.81 kJ; reaction is possible.
14. Fe 3 O 4k + 4H 2g \u003d 3Fe to + 4H 2 O g.
Answer: 99.7 kJ; reaction is not possible.
15. 2Fe 2 O 3k + 3C k \u003d 4Fe k + 3CO 2g.
Answer: 297.7 kJ; reaction is not possible.
16. Fe 3 O 4k + 4CO g \u003d 3Fe k + 4CO 2g.
Answer: -14.88 kJ; reaction is possible.
17. 2H 2 S g + O 2g \u003d 2H 2 O well + 2S c.
Answer: -407.4 kJ; reaction is possible.
18. Fe 2 O 3k + 3H 2g \u003d 2Fe to + 3H 2 O g.
Answer: 54.47 kJ; reaction is not possible.
Calculate the standard change in the Gibbs energy of a chemical reaction at a temperature of 25 °C from the values of the standard enthalpies of formation and absolute entropies of chemical compounds and determine at what temperature equilibrium will occur in the system.
19. 4HCl g + O 2g ↔ 2Cl 2g + 2H 2 O f.
Answer: -93.1 kJ; ~552 K.
20. Cl 2g + 2HI g ↔ I 2c + 2HCl g.
Answer: -194.0 kJ; ~1632 K.
21. SO 2g + 2CO g ↔ 2CO 2g + S c.
Answer: -214.24 kJ; ~1462 K.
22. CH 4g + 2H 2 Og ↔ CO 2g + 4H 2g.
Answer: 113.8 kJ; ~959 K.
23. CO g + 3H 2g ↔ CH 4g + H 2 O g.
Answer: -142.36 kJ; ~ 963 K.
Calculate the change in the Gibbs energy of a chemical reaction at a temperature of 350 °C from the standard enthalpies of formation and absolute entropies of chemical compounds. Ignore the temperature dependence of D f H° and S°. Set the possibility of spontaneous reactions:
24. 2РН 3g + 4O 2g \u003d P 2 O 5k + 3H 2 O g.
Answer: 1910.47 kJ; reaction is possible.
25. Cl 2 g + SO 2 g + 2H 2 O w = H 2 SO 4 w + 2HCl g.
Answer: -80.0 kJ; reaction is possible.
26. P 2 O 5k + 5C k \u003d 2P k + 5CO g.
Answer: 860.0 kJ; reaction is not possible.
27. 2CO g + SO 2g \u003d S to + 2CO 2g.
Answer: -154.4 kJ; reaction is possible.
28. CO 2g + 4H 2g \u003d CH 4g + 2H 2 O g.
Answer: -57.9 kJ; reaction is possible.
29. NO g + O 3 g = O 2 g + NO 2 g.
Answer: -196.83 kJ; reaction is possible.
30. CH 4g + 2O 2g \u003d CO 2g + 2H 2 O g.
Answer: -798.8 kJ; reaction is possible.
The possibility of a spontaneous reaction;
Entropy;
Isobaric - isothermal potential or Gibbs free energy.
Spontaneous processes are called, as a result of which useful work can be obtained, non-spontaneous are the processes that need to spend work.
Consider two spontaneous processes - the dissolution of sodium hydroxide and the dissolution of ammonium nitrate:
These are spontaneous processes, but one of them is accompanied by the release of heat, and the other by the absorption of heat. As we can see, the sign of the thermal effect of the process (enthalpy factor) does not unambiguously determine the possibility of a spontaneous process. There is a 2nd factor in the spontaneity of the process - entropy factor.
What is entropy?
The state of any system can be described, on the one hand, by the value of the measured parameters of the system (macrostates of the system), on the other hand, the state of the system can be described by a set of instantaneous microstates, which correspond to different energy levels of the microparticles that make up our system.
The number of microstates that corresponds to a given macrostate of matter is called thermodynamic probability its states (W), i.e. W is the number of ways in which molecules can be distributed among different energy levels.
Associated with the thermodynamic state probability is the state function of the system, called the entropy(S) .
S = k ln W, where k is the Boltzmann constant, k ≈ 1.38∙10 -23 J/K,
W is the thermodynamic probability of the state of the system.
For 1 mole of a substance:
S = R ln W, where R is the universal gas constant, here S is measured in .
State Probability maximum at the maximum disorder of the system, i.e., the entropy is maximum when the system is in the most disordered state. This is what the system strives for spontaneously.
Any system tends to move into the state of the greatest disorder, that is, spontaneously, any system tends to increase entropy. And entropy is a measure of the disorder in the system. It increases in such physical processes as melting, boiling, expansion of gases. In chemical processes, entropy increases if gaseous reaction products are obtained from the initial substances taken in solid or liquid states, or if the number of molecules increases during the reaction.
That is, the entropy grows, because the number of moving particles increases.
D.S.< 0 , - энтропия уменьшается т. к. уменьшается количество частиц (из 3-х в 2) и система переходит из газообразного состояния в жидкое.
Consider the change in entropy in the system during the transition from one state characterized by volume V 1 to another - with volume V 2:
If V 2 > V 1, then DS > 0, if V 2< V 1 , то DS < 0, т.е. при увеличении объема энтропия увеличивается.
The entropy of an ideal crystal at absolute zero is zero, so you can calculate the absolute value of the entropy for each substance. The tables show the standard value of entropy (S°) under standard conditions.
Entropy is a function of the state of matter, which means that it does not depend on the path of the system's transition from one state to another. For reversible isothermal processes ( phase transitions), the change in entropy is equal to the change in enthalpy divided by the temperature:
Entropy depends on temperature:
Where C P is the molar heat capacity at constant pressure.
Theoretical information
The chemical process can be considered as the first step in the ascent from chemical objects - electron, proton, atom - to a living system.
The doctrine of chemical processes- this is the field of science in which there is the deepest interpenetration of physics, chemistry, biology. This theory is based on chemical thermodynamics and kinetics.
The ability of a substance to undergo chemical transformations is determined by their reactivity, i.e. the nature of the reacting substances - the composition, structure, nature of the chemical bond; energy factors that determine the possibility of the process and kinetic factors that determine the speed of its course.
Almost all chemical processes are accompanied by the release or absorption of energy, most often in the form of heat and work.
Warmth - a quantitative measure of the random, chaotic movement of particles that form a given system.
Work - a quantitative measure of the ordered motion of particles in a directed force field.
The section of chemistry that studies the transitions of energy from one form to another during chemical reactions and establishes the direction and limits of their spontaneous flow under given conditions is called chemical thermodynamics .
The object of study of chemical thermodynamics is a chemical system.
System - this is the body under study or a group of bodies that interact with each other and are mentally or actually separated from the environment by boundaries that conduct or do not conduct heat.
Depending on the nature of the interaction of the system with the environment, there are open, closed and insulated systems.
open systems can exchange energy and matter with the environment. For example, an aqueous solution of sodium chloride in an open vessel. When water evaporates from the solution and during heat exchange, the mass of the system and its temperature will change, and, consequently, the energy.
Closed systems do not exchange matter with the environment. For example, a solution of sodium chloride in a closed vessel. If the solution and the environment have different temperatures, then the solution will be heated or cooled, and, consequently, its energy will change.
Isolated systems cannot exchange matter or energy with the environment. An isolated system is an idealization. There are no such systems in nature. But, despite the impossibility of practical implementation, isolated systems make it possible to determine the maximum theoretical energy differences between the system and its environment.
The state of the system is determined by a set of properties and is characterized by thermodynamic parameters : temperature (), pressure (), volume (), density (), amount of substance (), work done (), heat (). A change in at least one thermodynamic parameter leads to a change in the state of the system as a whole. If all parameters are constant in time and space, then this state of the system is called equilibrium .
In thermodynamics, the properties of a system are considered in its equilibrium states: initial and final, regardless of the path of the system's transition from one state to another. The transition of the system from one state to another at , = const called isobaric-isothermal, at , = const– isochoric-isothermal.
The most important tasks of chemical thermodynamics are to elucidate the possibility or impossibility of a spontaneous process of a particular chemical reaction under given conditions and in a given direction; setting the value of thermodynamic parameters at which the maximum output of the process is achieved; determination of the characteristics of the energy change occurring in the system. Find it with thermodynamic functions ().
The state function characterizes internal energy of the system – the sum of the potential energy of the interaction of all particles of the body with each other and the kinetic energy of their movement. It depends on the state of matter - type, mass, state of aggregation. The absolute value of internal energy cannot be measured. To study chemical processes, it is important to know only the change in internal energy during the transition of a system from one state to another.
(27)
Wherein 
the internal energy of the system decreases when 
- increases. All changes in internal energy occur due to the chaotic collision of molecules (the measure of energy transferred in this way is heat) and the movement of masses consisting of a large number of particles under the action of any forces (the measure of energy transferred in this way is work). Thus, the transfer of internal energy can be carried out partly in the form of heat and partly in the form of work:
(28)
The above equation is a mathematical expression I law of thermodynamics : If heat is supplied to the system, then the supplied heat is spent on increasing the internal energy of the system and on doing work for it.
In an isochoric-isothermal process, all the heat supplied to the system is spent on changing the internal energy:
(29)
In an isobaric-isothermal process, the only kind of work done by the system is the expansion work:
(30)
where is the pressure in the system, is the change in volume
Then the mathematical expression of the first law of thermodynamics takes the form: 
(31)
Denoting 
, we get 
System state function H - enthalpy is the total energy reserve of the system, i.e. is the energy content of the system. The enthalpy of the system is greater than the internal energy by the amount of work.
To characterize the energy manifestations in the course of a reaction, the concept thermal effect.
thermal effect- this is the amount of heat that is released or absorbed during the irreversible course of the reaction, when the only work is the work of expansion. In this case, the temperatures of the starting materials and reaction products should be the same. thermal effect endothermic reaction(flows with heat absorption) will be positive: 
, 
. thermal effect exothermic reaction(flows with the release of heat) will be negative: 
, 
.
The branch of chemistry devoted to the study of the thermal effects of chemical reactions is called thermochemistry .
Any chemical reaction is accompanied by changes in the energy reserve of the reacting substances. The more energy released during the formation of a chemical compound, the more stable this compound is, and, conversely, the substance obtained as a result of an endothermic reaction is unstable.
In chemical equations in which the heat of reaction is indicated, they are called thermochemical. They are compiled on the basis of the laws of conservation of mass and energy.
To compare the thermal effects of various processes, the conditions for their occurrence are standardized. Standard conditions - T 0 \u003d 298 K, p 0 \u003d 101.313 kPa, n - 1 mol of pure substance, enthalpy change ( 
) refers to the unit amount of the substance, kJ/mol. All standard thermodynamic functions are table values, which depend on the state of aggregation of the substance.
The quantitative laws of thermochemistry follow from the I law of thermodynamics.
Lavoisier-Laplace law(1780 - 1784) - for each chemical compound, the heat of decomposition is equal to the heat of its formation, but has the opposite sign.
Law G.I. Hess(1840) - the thermal effect of a chemical reaction depends on the nature and physical state of the initial substances and final products, but does not depend on the nature and path of the reaction, i.e. from the sequence of individual intermediate stages. This law is the theoretical basis of thermochemistry. A number of consequences follow from it:
In thermochemical calculations, the heat of formation (enthalpy) of simple substances under standard conditions is assumed to be zero.
(a simple substance) = 0
The amount of energy that is released or absorbed during the formation of 1 mol of a complex substance from simple ones under standard conditions is called the standard enthalpy of formation ( 
, kJ/mol).
The amount of energy that is released or absorbed by 1 mole of organic matter decomposing to carbon dioxide and water under standard conditions is called the standard enthalpy of combustion ( 
, kJ/mol).
The thermal effect of a chemical reaction is equal to the difference between the sum of the heats of formation of the reaction products and the sum of the heats of formation of the initial substances, taking into account the stoichiometric coefficients:
where is the thermal effect of a chemical reaction under standard conditions; 
- the sum of the standard heats of formation of the reaction products; 
- the sum of the standard heats of formation of the starting substances; 
, 
- stoichiometric coefficients, respectively, of the reaction products and starting materials.
Hess's law allows you to calculate the thermal effects of various reactions. But the sign and magnitude of the thermal effect does not allow one to judge the ability of the processes to proceed spontaneously and does not contain information about the direction and completeness of the processes.
Spontaneous processes(natural or positive) - flow in the system without interference from the external environment and are accompanied by a decrease in the internal energy of the system and the transfer of energy to the environment in the form of heat and work. Endothermic spontaneous processes do not contradict this definition, since they can take place in a non-isolated system and produce work due to the heat of the environment.
Processes that cannot take place by themselves (without external influence) are called non-spontaneous , unnatural or negative. Such processes are carried out by transferring energy from the external environment to the system in the form of heat or work.
According to the second law of thermodynamics, spontaneous processes go in the direction of reducing the stock of internal energy or the enthalpy of the system. However, such processes are known that proceed spontaneously without changing the internal energy of the system. The driving force of such processes is the entropy of the system.
Entropy(bound energy) (
S)
- this is a measure of the irreversibility of the process, a measure of the transition of energy into a form from which it cannot independently into another energy. Entropy characterizes the disorder in the system, the higher the disorder, the higher the entropy. It increases with increasing particle motion. In systems isolated from the external environment, processes proceed spontaneously in the direction of increasing entropy (). Processes for which the entropy decreases ( 
) are not feasible in isolated systems. If the process is possible in the forward and reverse directions, then in an isolated system it will proceed in the direction of increasing entropy. The course of a spontaneous process in an isolated system ends with a state of equilibrium. Therefore, in equilibrium system entropy maximum
.
Boltzmann derived an equation according to which
(34) where is the Boltzmann constant, W is the probability of a state, determines the number of microstates corresponding to a given microstate.
This relationship shows that entropy can be viewed as a measure of the molecular disorder of a system.
According to the II law of thermodynamics for an isothermal process, the change in entropy is:
; [J/(mol K] (35)
The entropy of simple substances is not equal to zero. Unlike enthalpy, the absolute value of entropy can be measured. “At absolute zero, the entropy of an ideal crystal is zero” - this postulate by M. Planck (1911) is called III law of thermodynamics.
The change in the entropy of a chemical process is determined by the balance equation:
Any system is characterized by order () and disorder (). Their ratio determines the direction of the reaction.
Thus, during the spontaneous movement of the system to a stable state, two tendencies appear: a decrease in the enthalpy 
and an increase in entropy. The combined effect of the two trends at constant temperature and pressure reflects isobaric-isothermal potential or Gibbs energy ()
.
State function
characterizes the general driving force of the process, the maximum possible useful work ("free energy") performed by the system; 
- part of the energy that cannot be converted into useful work ("bound energy").
Chemical reactions proceed in an open vessel with a change in volume, so the possibility (spontaneity) and the direction of the process characterizes the function determined by the balance equation under standard conditions:
; 
(38)
The spontaneous course of the process corresponds to a decrease in the Gibbs energy, 
. The more it decreases, the more irreversibly the process proceeds towards the formation of the final reaction products. Increase in isobaric potential 
is a sign of the impracticability of the process under given conditions. Meaning 
characterizes the state of equilibrium, i.e. a state in which the system is not producing useful work.
An analysis of the values and in the Gibbs equation showed that the possibility of a reversible process is due to the same signs and . At a certain temperature, the values and become equal. Therefore, from the Gibbs equation, one can determine the "equilibrium" temperature or the temperature of the beginning of the process ():
; = 0
; 
; 
(39)
Thus, reactions proceed spontaneously in which the change in free energy is negative. reactions in which , flow only under the condition that work is done on the system by external forces or energy is transferred to the system from outside. The conditions for the spontaneous flow of the process are shown in fig. 3.
Chemical reactions, Chemical reactions,
flowing spontaneously flowing not spontaneously
exothermic reactions, exothermic reactions,
accompanying accompanying
increase in entropy decrease in entropy
at any temperatures at high temperatures
endothermic reactions
accompanied
entropy increase
at low temperatures
Rice. 3. Conditions for the spontaneous flow of the process.
3.2. Control questions and tasks
1. What is called a system? What are the parameters of the system?
2. Describe the internal energy of the system, the concept of isochoric and isobaric processes.
3. What is called enthalpy?
4. Describe the enthalpy of formation of compounds, standard enthalpies of combustion and formation of substances.
5. Hess' law and its consequences, its application in thermochemical calculations.
6. Determination of heats (enthalpies) of neutralization, dissolution, hydration.
7. Entropy. Boltzmann equation. How does entropy change with temperature?
8. Gibbs energy. Criteria for the spontaneous flow of the process.
9. Using the reference data in Appendix 3, calculate the change in the standard enthalpy of reaction ():
10. Using the reference data in Appendix 3, calculate the change in the standard entropy of the reaction ( 
):
11. Calculate reactions at 846 0 С, if = 230 kJ, = 593 J/K.
Examples of problem solving
Example 1 The combustion reaction of ethyl alcohol is expressed by the thermochemical equation C 2 H 5 OH (L) + 3O 2 (G) \u003d 2CO 2 (G) + 3H 2 O (L). Calculate the thermal effect of the reaction if it is known that the molar heat of vaporization of C 2 H 5 OH (L) is +42.36 kJ, and the heat of formation of C 2 H 5 OH (G) = -235.31 kJ, CO 2 (G) \u003d -393.51 kJ, H 2 O (L) \u003d -285.84 kJ.
Solution. To determine ΔΗ of the reaction, it is necessary to know the heat of formation C 2 H 5 OH (L). We find the latter from the data:
C 2 H 5 OH (F) \u003d C 2 H 5 OH (G); ΔΗ = +42.36 kJ
42.36 \u003d -235.31 - ΔΗ (C 2 H 5 OH (W))
ΔΗ (C 2 H 5 OH (W)) \u003d -235.31-42.36 \u003d -277.67 kJ
We calculate ΔΗ of the reaction, applying the consequences from the Hess law:
ΔΗ H.R. \u003d 2 (-393.51) + 3 (-285.84) + 277.67 \u003d -1366.87 kJ.
The thermal effect of the reaction is 1366.87 kJ.
a) Fe 2 O 3 (K) + 3H 2 (G) \u003d 2Fe (K) + 3H 2 O (G)
b) Fe 2 O 3 (K) + 3CO (G) \u003d 2Fe (K) + 3CO 2 (G)
Which process requires more energy?
Solution. To calculate ΔΗ XP, we use the formula of the consequence from the Hess law and the standard enthalpies of formation of each substance [Appendix 3]:
a) ΔΗ XP \u003d 2ΔΗ (Fe) + 3ΔΗ (H 2 O) - (ΔΗ (Fe 2 O 3) + 3ΔΗ (H 2)) \u003d 2 (0) + 3 (-241.8) - ((-822 .2) + 3(0)) = -725.4 + 822.2 = 96.8 kJ.
b) ΔΗ XP \u003d 2ΔΗ (Fe) + 3ΔΗ (CO 2) - (ΔΗ (Fe 2 O 3) + 3ΔΗ (CO)) \u003d 2 (0) + 3 (-393.5) - ((-822.2 ) + 3(-110.5)) = -1180.5 + 822.2 + 331.5 = -26.5 kJ.
According to the calculations, process a) - the reduction of iron oxide (III) with hydrogen, requires more energy than process b). In process b), the reaction is even exothermic (energy is released in the form of heat).
Example 3 Water gas is a mixture of equal volumes of hydrogen and carbon monoxide (II). Find the amount of heat released during the combustion of 112 liters of water gas, (n.o.).
Solution. Let us compose the thermochemical equation of the process:
H 2 (G) + CO (G) + O 2 (G) \u003d H 2 O (G) + CO 2 (G) ΔΗ XP \u003d - Q.
Let us calculate ΔΗ XP when 2 mol of water gas burns (1 mol of H 2 and 1 mol of CO), i.e. 22.4 l / mol 2 mol \u003d 44.8 l. The calculation is carried out according to the formula of the consequence of the Hess law and the standard enthalpies of formation of each substance [Add. 3]:
ΔΗ XP \u003d ΔΗ (H 2 O) + ΔΗ (CO 2) - (ΔΗ (H 2) + ΔΗ (CO) + ΔΗ (O 2)) \u003d -241.8 - 393.5 - (0 - 110.5 + 0) = - 635.3 + 110.5 = - 524.8 kJ
We make a proportion:
44.8 liters of water gas burns - 524.8 kJ of heat is released
112 l - X kJ
X \u003d 112 524.8 / 44.8 \u003d 1312 kJ
When burning 112 liters of water gas, 1312 kJ of heat is released.
Example 4 Give a thermodynamic characteristic of the process Ga + HCl) ↔ GaCl 3 (t) + H 2 (g) according to the plan:
1. Write down the stoichiometric equation.
2. Write down the thermodynamic functions of the substances involved.
3. Calculate the change in the standard enthalpy of a chemical reaction and plot the enthalpy diagram.
4. Determine if the reaction is exothermic or endothermic; the temperature in the system increases or decreases as a result of this reaction.
5. Calculate the change in the standard entropy of the reaction, explain the change in entropy during the reaction.
6. Calculate the standard change in the Gibbs energy from the balance equation and the Gibbs equation. Give an analysis of the obtained data.
7. Match the signs of the quantities 
. and 
Conclude that the reaction is reversible.
8. For a reversible reaction, calculate the equilibrium temperature according to the Gibbs equation, assuming that the maximum allowable temperature is 3000 K. Make a conclusion: Tp - realizable or not realizable.
9. Calculate the value 
at three temperatures (500, 1000 and 1500 K). Plot Graphical Dependency 
..
10. Draw a conclusion about the spontaneity of the chemical reaction. Determine the conditions under which the reaction is possible
Solution.
1 Write down the stoichiometric equation.
2. We write out the standard thermodynamic functions for the formation of reaction components (Table 21) (thermodynamic parameters of substances from [Appendix 3]).
