Stoichiometric calculations in chemistry. Selection of stoichiometric coefficients svr The sum of stoichiometric coefficients in the equation is equal to
When drawing up an equation for a redox reaction (ORR), it is necessary to determine the reducing agent, oxidizing agent, and the number of given and received electrons. OVR stoichiometric coefficients are selected using either the electron balance method or the electron-ion balance method (the latter is also called the half-reaction method). Let's look at a few examples. As an example of compiling OVR equations and selecting stoichiometric coefficients, we analyze the process of oxidation of iron (II) disulfide (pyrite) with concentrated nitric acid: First of all, we determine the possible reaction products. Nitric acid is a strong oxidizing agent, so the sulfide ion can be oxidized either to the maximum oxidation state S (H2S04) or to S (SO2), and Fe to Fe, while HN03 can be reduced to N0 or N02 (the set of specific products is determined concentrations of reagents, temperature, etc.). Let's choose the following possible option: H20 will be on the left or right side of the equation, we don't know yet. There are two main methods for selecting coefficients. Let us first apply the method of electron-ion balance. The essence of this method is in two very simple and very important statements. First, this method considers the transition of electrons from one particle to another, with the obligatory consideration of the nature of the medium (acidic, alkaline, or neutral). Secondly, when compiling the equation of the electron-ion balance, only those particles are recorded that actually exist during the course of a given OVR - only really existing cations or annones are recorded in the form of ions; Substances that are poorly disociated, insoluble or liberated in the form of a gas are written in molecular form. When compiling the equation for the processes of oxidation and reduction, to equalize the number of hydrogen and oxygen atoms, one introduces (depending on the medium) either water molecules and hydrogen ions (if the medium is acidic), or water molecules and hydroxide ions (if the medium is alkaline). Consider for our case the oxidation half-reaction. Molecules of FeS2 (a poorly soluble substance) are converted into Fe3+ ions (iron nitrate (II) completely dissociates into ions) and sulfate ions S042 "(dissociation of H2SO4): Consider now the reduction half-reaction of the nitrate ion: To equalize oxygen, add 2 to the right side water molecules, and to the left - 4 H + ions: To equalize the charge to the left side (charge +3), add 3 electrons: Finally, we have: Reducing both parts by 16H + and 8H20, we get the final, reduced ionic equation of the redox reaction: Adding the corresponding number of NOJ nH+ ions to both sides of the equation, we find the molecular reaction equation: Please note that to determine the number of given and received electrons, we never had to determine the oxidation state of the elements. In addition, we took into account the influence of the environment and “automatically” determined that H20 is on the right side of the equation. There is no doubt that this method has a great chemical meaning. Empirical balance method. The essence of the method of finding the stoichiometric coefficients in the equations of the OVR is the obligatory determination of the oxidation states of the atoms of the elements involved in the OVR. Using this approach, we again equalize the reaction (11.1) (above we applied the method of half-reactions to this reaction). The reduction process is described simply: It is more difficult to draw up an oxidation scheme, since two elements are oxidized at once - Fe and S. You can assign iron an oxidation state of +2, sulfur - 1 and take into account that there are two S atoms per Fe atom: You can, however, do without determination of oxidation states and write down a scheme resembling scheme (11.2): The right side has a charge of +15, the left side has a charge of 0, so FeS2 must give up 15 electrons. We write down the overall balance: We still need to “figure out” the resulting balance equation - it shows that 5 HN03 molecules are used to oxidize FeS2 and another 3 HNO molecules are needed to form Fe(N03)j: To equalize hydrogen and oxygen, to the right part you need to add 2 molecules of H20: The electron-ion balance method is more versatile than the electron balance method and has an undeniable advantage in the selection of coefficients in many OTS, in particular, with the participation of organic compounds, in which even the procedure for determining oxidation states is very complicated . - Consider, for example, the process of ethylene oxidation, which occurs when it is passed through an aqueous solution of potassium permanganate. As a result, ethylene is oxidized to ethylene glycol HO - CH2 - CH2 - OH, and permanganate is reduced to manganese oxide (TV), in addition, as will be obvious from the final balance equation, potassium hydroxide is also formed on the right: After making the necessary reductions of such terms, we write the equation in the final molecular form * Influence of the medium on the nature of the OVR flow. The examples (11.1) - (11.4) clearly illustrate the "technique" of using the electron-ion balance method in the case of OVR flow in an acidic or alkaline medium. The nature of the environment! influences the course of one or another OVR; in order to “feel” this influence, let us consider the behavior of one and the same oxidizing agent (KMnO4) in different environments. , recovering up to Mn+4(Mn0j), and the minimum - in the strength of the last one, in which the risen Shaiyaaapsya up to (mvnganat-nOn Mn042"). This is explained as follows. The acids of the dissociation line form hydroxide ions ffjO +, which strongly polarize 4 "MoOH ions Weaken the bonds of manganese with oxygen (thereby enhancing the action of the reducing agent) .. In a neutral medium, the polarizing effect of water molecules is significantly c-aafep. >"MnO ions; much less polarized. In a strongly alkaline medium, hydroxide ions “even strengthen the Mn-O bond, as a result of which the effectiveness of the reducing agent decreases and MnO^ accepts only one electron. An example of the behavior of potassium permanganate in a neutral medium is represented by the reaction (11.4). Let us also give one example of reactions involving KMnOA in acidic and alkaline media
Which studies the quantitative relationships between the substances that entered into the reaction and formed during it (from other Greek "stechion" - "elemental composition", "meitren" - "I measure").
Stoichiometry is the most important for material and energy calculations, without which it is impossible to organize any chemical production. Chemical stoichiometry allows you to calculate the amount of raw materials needed for a particular production, taking into account the desired performance and possible losses. No enterprise can be opened without preliminary calculations.
A bit of history
The very word "stoichiometry" is an invention of the German chemist Jeremy Benjamin Richter, proposed by him in his book, in which the idea of the possibility of calculations using chemical equations was first described. Later, Richter's ideas received theoretical justification with the discovery of the laws of Avogadro (1811), Gay-Lussac (1802), the law of constancy of composition (J.L. Proust, 1808), multiple ratios (J. Dalton, 1803), and the development of atomic and molecular theory. Now these laws, as well as the law of equivalents, formulated by Richter himself, are called the laws of stoichiometry.
The concept of "stoichiometry" is used in relation to both substances and chemical reactions.
Stoichiometric Equations
Stoichiometric reactions - reactions in which the starting substances interact in certain ratios, and the amount of products corresponds to theoretical calculations.
Stoichiometric equations are equations that describe stoichiometric reactions.
Stoichiometric equations) show the quantitative relationships between all participants in the reaction, expressed in moles.
Most inorganic reactions are stoichiometric. For example, three successive reactions to produce sulfuric acid from sulfur are stoichiometric.
S + O 2 → SO 2
SO 2 + ½O 2 → SO 3
SO 3 + H 2 O → H 2 SO 4
Calculations using these reaction equations can determine how much each substance needs to be taken in order to obtain a certain amount of sulfuric acid.
Most organic reactions are non-stoichiometric. For example, the reaction equation for cracking ethane looks like this:
C 2 H 6 → C 2 H 4 + H 2 .
However, in reality, during the reaction, different amounts of by-products will always be obtained - acetylene, methane and others, which cannot be calculated theoretically. Some inorganic reactions also defy calculations. For example, ammonium nitrate:
NH 4 NO 3 → N 2 O + 2H 2 O.
It goes in several directions, so it is impossible to determine how much starting material needs to be taken in order to obtain a certain amount of nitric oxide (I).
Stoichiometry is the theoretical basis of chemical production
All reactions that are used in or in production must be stoichiometric, that is, subject to accurate calculations. Will the plant or factory be profitable? Stoichiometry allows you to find out.
On the basis of stoichiometric equations, a theoretical balance is made. It is necessary to determine how much of the starting materials will be required to obtain the desired amount of the product of interest. Further, operational experiments are carried out, which will show the real consumption of the starting materials and the yield of products. The difference between theoretical calculations and practical data allows you to optimize production and evaluate the future economic efficiency of the enterprise. Stoichiometric calculations also make it possible to compile the heat balance of the process in order to select equipment, determine the masses of by-products formed that will need to be removed, and so on.
Stoichiometric substances
According to the law of composition constancy proposed by J.L. Proust, any chemical has a constant composition, regardless of the method of preparation. This means that, for example, in a molecule of sulfuric acid H 2 SO 4, regardless of the method by which it was obtained, there will always be one sulfur atom and four oxygen atoms per two hydrogen atoms. All substances that have a molecular structure are stoichiometric.
However, substances are widespread in nature, the composition of which may differ depending on the method of preparation or source of origin. The vast majority of them are crystalline substances. One could even say that for solids, stoichiometry is the exception rather than the rule.
For example, consider the composition of well-studied titanium carbide and oxide. In titanium oxide TiO x X=0.7-1.3, that is, from 0.7 to 1.3 oxygen atoms per titanium atom, in carbide TiC x X=0.6-1.0.
The nonstoichiometric nature of solids is explained by an interstitial defect at the nodes of the crystal lattice or, conversely, by the appearance of vacancies at the nodes. Such substances include oxides, silicides, borides, carbides, phosphides, nitrides and other inorganic substances, as well as high-molecular organic ones.
And although evidence for the existence of compounds with a variable composition was presented only at the beginning of the 20th century by I.S. Kurnakov, such substances are often called berthollides by the name of the scientist K.L. Berthollet, who suggested that the composition of any substance changes.
All quantitative ratios in the calculation of chemical processes are based on the stoichiometry of reactions. It is more convenient to express the amount of a substance in such calculations in moles, or derived units (kmol, mmol, etc.). The mole is one of the basic SI units. One mole of any substance corresponds to its quantity, numerically equal to the molecular weight. Therefore, the molecular weight in this case should be considered as a dimensional value with units: g/mol, kg/kmol, kg/mol. So, for example, the molecular weight of nitrogen is 28 g/mol, 28 kg/kmol, but 0.028 kg/mol.
Mass and molar amounts of a substance are related by known relationships
N A \u003d m A / M A; m A = N A M A,
where N A is the amount of component A, mol; m A is the mass of this component, kg;
M A - molecular weight of component A, kg/mol.
In continuous processes, the flow of substance A can be expressed by its mol-
quantity per unit of time
where W A is the molar flow of component A, mol/s; τ - time, s.
For a simple reaction that proceeds almost irreversibly, usually a stoichiomet
ric equation is written in the form
v A A + v B B = v R R + v S S.
However, it is more convenient to write the stoichiometric equation in the form of an algebraic
th, assuming that the stoichiometric coefficients of the reactants are negative, and the reaction products are positive:
Then for each simple reaction we can write the following equalities:
Index "0" refers to the initial amount of the component.
These equalities give grounds to obtain the following material balance equations for the component for a simple reaction:
Example 7.1. The hydrogenation reaction of phenol to cyclohexanol proceeds according to the equation
C 6 H 5 OH + ZN 2 \u003d C 6 H 11 OH, or A + 3B \u003d R.
Calculate the amount of product formed if the initial amount of component A was 235 kg and the final amount was 18.8 kg
Solution: We write the reaction as
R - A - ZV \u003d 0.
The molecular weights of the components are: M A = 94 kg/kmol, M B = 2 kg/kmol and
M R = 100 kg/kmol. Then the molar amounts of phenol at the beginning and at the end of the reaction will be:
N A 0 \u003d 235/94 \u003d 2.5; N A 0 \u003d 18.8 / 94 \u003d 0.2; n \u003d (0.2 - 2.5) / (-1) \u003d 2.3.
The amount of cyclohexanol formed will be equal to
N R \u003d 0 + 1 ∙ 2.3 \u003d 2.3 kmol or m R \u003d 100 2.3 \u003d 230 kg.
The determination of stoichiometrically independent reactions in their system in the material and thermal calculations of reaction apparatuses is necessary to exclude reactions that are the sum or difference of some of them. Such an assessment can be most easily carried out using the Gram criterion.
In order not to carry out unnecessary calculations, it should be assessed whether the system is stoichiometrically dependent. For these purposes it is necessary:
Transpose the original matrix of the reaction system;
Multiply the original matrix by the transposed one;
Calculate the determinant of the resulting square matrix.
If this determinant is equal to zero, then the reaction system is stoichiometrically dependent.
Example 7.2. We have a reaction system:
FeO + H 2 \u003d Fe + H 2 O;
Fe 2 O 3 + 3H 2 \u003d 2Fe + 3H 2 O;
FeO + Fe 2 O 3 + 4H 2 \u003d 3Fe + 4H 2 O.
This system is stoichiometrically dependent since the third reaction is the sum of the other two. Let's make a matrix
When compiling the equations of redox reactions, the following two important rules must be observed:
Rule 1: In any ionic equation, charge conservation must be observed. This means that the sum of all charges on the left side of the equation ("left") must match the sum of all charges on the right side of the equation ("right"). This rule applies to any ionic equation, both for complete reactions and for half-reactions.
Charges from left to right
Rule 2: The number of electrons lost in the oxidation half-reaction must be equal to the number of electrons gained in the reduction half-reaction. For example, in the first example given at the beginning of this section (the reaction between iron and hydrated cuprous ions), the number of electrons lost in the oxidative half-reaction is two:
Therefore, the number of electrons acquired in the reduction half-reaction must also be equal to two:
The following procedure can be used to derive the full redox equation from the equations of the two half-reactions:
1. The equations of each of the two half-reactions are balanced separately, and to fulfill the above rule 1, the corresponding number of electrons is added to the left or right side of each equation.
2. The equations of both half-reactions are balanced with respect to each other so that the number of electrons lost in one reaction becomes equal to the number of electrons gained in the other half-reaction, as required by rule 2.
3. The equations for both half-reactions are summed to obtain the complete equation for the redox reaction. For example, summing the equations of the two half-reactions above and removing from the left and right sides of the resulting equation
equal number of electrons, we find
We balance the equations of the half-reactions given below and compose an equation for the redox reaction of the oxidation of an aqueous solution of any ferrous salt into a ferric salt with an acidic potassium solution.
Stage 1. First, we balance the equation of each of the two half-reactions separately. For equation (5) we have
To balance both sides of this equation, you need to add five electrons to its left side, or subtract the same number of electrons from the right side. After that we get
This allows us to write the following balanced equation:
Since electrons had to be added to the left side of the equation, it describes a reduction half-reaction.
For equation (6), we can write
To balance this equation, you can add one electron to its right side. Then
