Task 1.
When 560 ml (n.s.) of acetylene is burned according to the thermochemical equation:
2C 2 H 2 (G) + 5O 2 (G) = 4CO 2 (G) + 2H 2 O (G) + 2602.4 kJ
stood out:
1) 16.256 kJ; 2) 32.53 kJ; 3) 32530 kJ; 4) 16265kJ
Given:
volume of acetylene: V(C 2 H 2) = 560 ml.
Find: the amount of heat released.
Solution:
To select the correct answer, it is most convenient to calculate the quantity sought in the problem and compare it with the proposed options. Calculation using a thermochemical equation is no different from calculation using a conventional reaction equation. Above the reaction we indicate the data in the condition and the required quantities, under the reaction - their relationships according to the coefficients. Heat is one of the products, so we consider its numerical value as a coefficient.
Comparing the answer received with the proposed options, we see that answer No. 2 is suitable.
A small trick that led inattentive students to incorrect answer No. 3 was the units of measurement for the volume of acetylene. The volume indicated in the condition in milliliters had to be converted into liters, since the molar volume is measured in (l/mol).
Occasionally there are problems in which a thermochemical equation must be compiled independently based on the value of the heat of formation of a complex substance.
Problem 1.2.
The heat of formation of aluminum oxide is 1676 kJ/mol. Determine the thermal effect of the reaction in which, when aluminum interacts with oxygen,
25.5 g A1 2 O 3.
1) 140 kJ; 2) 209.5 kJ; 3) 419 kJ; 4) 838kJ.
Given:
heat of formation of aluminum oxide: Qrev (A1 2 O 3) = = 1676 kJ/mol;
mass of the resulting aluminum oxide: m(A1 2 O 3) = 25.5 g.
Find: thermal effect.
Solution:
This type of problem can be solved in two ways:
Method I
According to the definition, the heat of formation of a complex substance is the thermal effect of the chemical reaction of the formation of 1 mole of this complex substance from simple substances.
We write down the reaction of the formation of aluminum oxide from A1 and O2. When arranging the coefficients in the resulting equation, we take into account that before A1 2 O 3 there must be a coefficient "1"
, which corresponds to the amount of substance in 1 mole. In this case, we can use the heat of formation specified in the condition:
2A1 (TV) + 3/2O 2(g) -----> A1 2 O 3(TV) + 1676 kJ
We obtained a thermochemical equation.
In order for the coefficient of A1 2 O 3 to remain equal to “1”, the coefficient of oxygen must be fractional.
When writing thermochemical equations, fractional coefficients are allowed.
We calculate the amount of heat that will be released during the formation of 25.5 g of A1 2 O 3:
Let's make a proportion:
upon receipt of 25.5 g of A1 2 O 3, x kJ is released (according to the condition)
when receiving 102 g of A1 2 O 3, 1676 kJ is released (according to the equation)
Answer number 3 is suitable.
When solving the last problem under the conditions of the Unified State Exam, it was possible not to create a thermochemical equation. Let's consider this method.
II method
According to the definition of the heat of formation, 1676 kJ is released when 1 mol of A1 2 O 3 is formed. The mass of 1 mole of A1 2 O 3 is 102 g, therefore, the proportion can be made:
1676 kJ is released when 102 g of A1 2 O 3 are formed
x kJ is released when 25.5 g of A1 2 O 3 are formed
Answer number 3 is suitable.
Answer: Q = 419 kJ.
Problem 1.3.
When 2 moles of CuS are formed from simple substances, 106.2 kJ of heat is released. When 288 g of CuS is formed, heat is released in the amount of:
1) 53.1 kJ; 2) 159.3 kJ; 3) 212.4 kJ; 4) 26.6 kJ
Solution:
Find the mass of 2 mol CuS:
m(СuS) = n(СuS) . M(CuS) = 2. 96 = 192 g.
In the text of the condition, instead of the value of the amount of substance CuS, we substitute the mass of 2 moles of this substance and get the finished proportion:
when 192 g of CuS is formed, 106.2 kJ of heat is released
when 288 g of CuS is formed, heat is released in the amount X kJ.
Answer number 2 is suitable.
The second type of problem can be solved both using the law of volumetric relations and without using it. Let's look at both solutions using an example.
Tasks for applying the law of volumetric relations:
Problem 1.4.
Determine the volume of oxygen (n.o.) that will be required to burn 5 liters of carbon monoxide (n.o.).
1) 5 l; 2) 10 l; 3) 2.5 l; 4) 1.5 l.
Given:
volume of carbon monoxide (n.s.): VCO) = 5 l.
Find: volume of oxygen (no): V(O 2) = ?
Solution:
First of all, you need to create an equation for the reaction:
2CO + O 2 = 2CO
n = 2 mol n =1 mol
We apply the law of volumetric relations:
We find the relationship from the reaction equation, and
We take V(CO) from the condition. Substituting all these values into the law of volumetric relations, we get:
Hence: V(O 2) = 5/2 = 2.5 l.
Answer number 3 is suitable.
Without using the law of volumetric relations, the problem is solved using calculation using the equation:
Let's make a proportion:
5 l of CO2 interact with x l of O2 (according to the condition) 44.8 l of CO2 interact with 22.4 l of O2 (according to the equation):
We received the same answer option No. 3.
Task 88.
The thermal effect of which reaction is equal to the heat of formation of methane? Calculate the heat of formation of methane based on the following thermochemical equations:
A) H 2 (g) + 1/2O 2 (g) = H 2 O (l); = -285.84 kJ;
b) C (k) + O 2 (g) = CO 2 (g); = -393.51 kJ;
c) CH 4 (g) + 2O 2 (g) = 2H 2 O (l) + CO 2 (g); = -890.31 kJ.
Answer: -74.88 kJ.
Solution:
.
105 Pa). The formation of methane from hydrogen and carbon can be represented as follows:
C (graphite) + 2H 2 (g) = CH 4 (g); = ?
Based on these equations according to the conditions of the problem, taking into account that hydrogen burns to water, carbon to carbon dioxide, methane to carbon dioxide and water and, based on Hess’s law, thermochemical equations can be operated in the same way as algebraic ones. To obtain the desired result, you need to multiply the hydrogen combustion equation (a) by 2, and then subtract the sum of the hydrogen (a) and carbon (b) combustion equations from the methane combustion equation (c):
CH 4 (g) + 2O 2 (g) - 2 H 2 (g) + O 2 (g) - C (k) + O 2 (g) =
= 2H 2 O (l) + CO 2 - 2H 2 O - CO 2;
= -890,31 – [-393,51 + 2(-285,84).
CH 4 (g) = C (k) + 2H 2 (k); = +74.88 kJ.2
Since the heat of formation is equal to the heat of decomposition with the opposite sign, then
(CH 4) = -74.88 kJ.
Answer: -74.88 kJ.
Task 89.
The thermal effect of which reaction is equal to the heat of formation of calcium hydroxide? Calculate the heat of formation of calcium hydroxide based on the following thermochemical equations:
Ca (k) + 1/2O (g) = CaO (k); = -635.60 kJ;
H 2 (g) + 1/2O 2 (g) = H 2 O (l); = -285.84 kJ;
CaO (k) + H 2 O (l) = Ca (OH) 2 (k); = -65.06 kJ.
Answer: -986.50 kJ.
Solution:
The standard heat of formation is equal to the heat of reaction of the formation of 1 mole of this substance from simple substances under standard conditions (T = 298 K; p = 1.0325 .
105 Pa). The formation of calcium hydroxide from simple substances can be represented as follows:
Ca (k) + O 2 (g) + H 2 (g) = Ca (OH) 2 (k); = ?
Based on the equations that are given according to the conditions of the problem and taking into account that hydrogen burns to water, and calcium, reacting with oxygen, forms CaO, then, based on Hess’s law, thermochemical equations can be operated in the same way as algebraic ones. To obtain the desired result, you need to add all three equations together:
CaO (k) + H 2 O (l) + Ca (k) + 1/2O (g) + H 2 (g) + 1/2O 2 (g = (OH) 2 (k) + CaO (k) + H 2 O (l);
= -65.06 + (-635.60) + (-285.84) = -986.50 kJ.
Since the standard heats of formation of simple substances are conventionally assumed to be zero, the heat of formation of calcium hydroxide will be equal to the thermal effect of the reaction of its formation from simple substances (calcium, hydrogen and oxygen):
== (Ca(OH) 2 = -986.50 kJ.2
Answer: -986.50 kJ.
Task 90.
The thermal effect of the combustion reaction of liquid gasoline with the formation of water vapor and carbon dioxide is equal to -3135.58 kJ. Make up a thermochemical equation for this reaction and calculate the heat of formation of C 6 H 6 (l). Answer: +49.03 kJ.
Solution:
Reaction equations in which their state of aggregation or crystal modification, as well as the numerical value of thermal effects are indicated next to the symbols of chemical compounds, are called thermochemical. In thermochemical equations, unless specifically stated, the values of thermal effects at constant pressure Qp are indicated equal to the change in enthalpy of the system. The value is usually given on the right side of the equation, separated by a comma or semicolon. The following abbreviated designations for the state of aggregation of a substance are accepted: g - gaseous, g - liquid, j - crystalline. These symbols are omitted if the aggregative state of the substances is obvious, for example, O 2, H 2, etc.
The thermochemical equation of the reaction is:
C 6 H 6 (l) + 7/2O 2 = 6CO 2 (g) + 3H 2 O (g); = -3135.58 kJ.
The values of standard heats of formation of substances are given in special tables. Considering that the heats of formation of simple substances are conventionally assumed to be zero. The thermal effect of a reaction can be calculated using a corollary of Hess's law:
6 (CO 2) + 3 =0 (H 2 O) – (C 6 H 6)
(C 6 H 6) = -;
(C 6 H 6) = - (-3135.58) = +49.03 kJ.
Answer:+49.03 kJ.
Heat of formation
Task 91.
Calculate how much heat will be released during the combustion of 165 liters (n.s.) of acetylene C 2 H 2 if the combustion products are carbon dioxide and water vapor? Answer: 924.88 kJ.
Solution:
Reaction equations in which their state of aggregation or crystal modification, as well as the numerical value of thermal effects are indicated next to the symbols of chemical compounds, are called thermochemical. In thermochemical equations, unless specifically stated, the values of thermal effects at constant pressure Qp are indicated equal to the change in enthalpy of the system. The value is usually given on the right side of the equation, separated by a comma or semicolon. The following abbreviated designations for the state of aggregation of a substance are accepted: G- gaseous, and- liquid, To-- crystalline. These symbols are omitted if the aggregative state of the substances is obvious, for example, O 2, H 2, etc.
The reaction equation is:
C 2 H 2 (g) + 5/2O 2 (g) = 2CO 2 (g) + H 2 O (g); = ?
2(CO 2) + (H 2 O) – (C 2 H 2);
= 2(-393.51) + (-241.83) – (+226.75) = -802.1 kJ.
The heat released during the combustion of 165 liters of acetylene by this reaction is determined from the proportion:
22.4: -802.1 = 165: x; x = 165 (-802.1)/22.4 = -5908.35 kJ; Q = 5908.35 kJ.
Answer: 5908.35 kJ.
Task 92.
When ammonia gas burns, it produces water vapor and nitrogen oxide. How much heat will be released during this reaction if 44.8 liters of NO were obtained, based on normal conditions? Answer: 452.37 kJ.
Solution:
The reaction equation is:
NH 3 (g) + 5/4O 2 = NO (g) + 3/2H 2 O (g)
The values of standard heats of formation of substances are given in special tables. Considering that the heats of formation of simple substances are conventionally assumed to be zero. The thermal effect of a reaction can be calculated using a corollary of Hess's law:
= (NO) + 3/2 (H 2 O) – (NH 3);
= +90.37 +3/2 (-241.83) – (-46.19) = -226.185 kJ.
The thermochemical equation will be:
We calculate the heat released during the combustion of 44.8 liters of ammonia from the proportion:
22.4: -226.185 = 44.8: x; x = 44.8 (-226.185)/22.4 = -452.37 kJ; Q = 452.37 kJ.
Answer: 452.37 kJ
Thermochemical equations. Quantity of heat. which is released or absorbed as a result of the reaction between certain quantities of reagents, specified by stoichiometric coefficients, is called the thermal effect of a chemical reaction and is usually denoted by the symbol Q. Exothermic and endothermic reactions. Hessian thermochemical law Reactions that occur with the release of energy in the form of heat are called exothermic; reactions that occur with the absorption of energy in the form of heat are endothermic. It has been proven that in isobaric chemical processes, the released (or absorbed) heat is a measure of the decrease (or, accordingly, increase) in the enthalpy of the reaction. Thus, in exothermic reactions, when heat is released, AN is negative. In endothermic reactions (heat is absorbed), AN is positive. The magnitude of the thermal effect of a chemical reaction depends on the nature of the starting substances and reaction products, their state of aggregation and temperature. A reaction equation, on the right side of which, along with the reaction products, the change in enthalpy AN or the thermal effect of the reaction Qp is indicated, is called thermochemical. An example of an exothermic reaction is the reaction of water formation: 2H2(G) + 02(g) = 2H20(G) To carry out this reaction, it is necessary to expend energy to break the bonds in the H2 and 02 molecules. These amounts of energy are respectively 435 and 494 kJ/mol . On the other hand, when an O - H bond is formed, 462 kJ/mol of energy is released. The total amount of energy (1848 kJ) released during the formation of O - H bonds is greater than the total amount of energy (1364 kJ) expended on breaking the H - H and O = O bonds, therefore the reaction is exothermic, i.e., upon formation two moles of vaporous water will release 484 kJ of energy. The equation for the reaction of water formation, written taking into account the change in enthalpy. Exothermic and endothermic reactions. Hessian thermochemical law will already be a thermochemical equation of the reaction. An example of an endothermic reaction is the formation of nitric oxide (II). To carry out this reaction, it is necessary to expend energy to break the N = N and 0 = 0 bonds in the molecules of the starting substances. They are respectively equal to 945 and 494 kJ/mol. When the N = O bond is formed, energy is released in the amount of 628.5 kJ/mol. The total amount of energy required to break bonds in the molecules of the starting substances is 1439 kJ and is greater than the released energy for the formation of bonds in the molecules of the reaction product (1257 kJ). Therefore, the reaction is endothermic and requires the absorption of energy in the amount of 182 kJ from the environment for it to occur. Thermochemical equations Exothermic and endothermic reactions. Hessian thermochemical law This explains that nitrogen oxide (II) is formed only at high temperatures, for example, in car exhaust gases, in lightning discharges and is not formed under normal conditions.
From the lesson materials you will learn which chemical reaction equation is called thermochemical. The lesson is devoted to studying the calculation algorithm for the thermochemical reaction equation.
Topic: Substances and their transformations
Lesson: Calculations using thermochemical equations
Almost all reactions occur with the release or absorption of heat. The amount of heat that is released or absorbed during a reaction is called thermal effect of a chemical reaction.
If the thermal effect is written in the equation of a chemical reaction, then such an equation is called thermochemical.
In thermochemical equations, unlike ordinary chemical ones, the aggregate state of the substance (solid, liquid, gaseous) must be indicated.
For example, the thermochemical equation for the reaction between calcium oxide and water looks like this:
CaO (s) + H 2 O (l) = Ca (OH) 2 (s) + 64 kJ
The amount of heat Q released or absorbed during a chemical reaction is proportional to the amount of substance of the reactant or product. Therefore, using thermochemical equations, various calculations can be made.
Let's look at examples of problem solving.
Task 1:Determine the amount of heat spent on the decomposition of 3.6 g of water in accordance with the TCA of the water decomposition reaction:
You can solve this problem using the proportion:
during the decomposition of 36 g of water, 484 kJ were absorbed
during decomposition 3.6 g of water was absorbed x kJ
In this way, an equation for the reaction can be written. The complete solution to the problem is shown in Fig. 1.
Rice. 1. Formulation of the solution to problem 1
The problem can be formulated in such a way that you will need to create a thermochemical equation for the reaction. Let's look at an example of such a task.
Problem 2: When 7 g of iron interacts with sulfur, 12.15 kJ of heat is released. Based on these data, create a thermochemical equation for the reaction.
I draw your attention to the fact that the answer to this problem is the thermochemical equation of the reaction itself.
Rice. 2. Formalization of the solution to problem 2
1. Collection of problems and exercises in chemistry: 8th grade: for textbooks. P.A. Orzhekovsky and others. “Chemistry. 8th grade” / P.A. Orzhekovsky, N.A. Titov, F.F. Hegel. - M.: AST: Astrel, 2006. (p.80-84)
2. Chemistry: inorganic. chemistry: textbook. for 8th grade general education inst. /G.E. Rudzitis, F.G. Feldman. - M.: Education, OJSC “Moscow Textbooks”, 2009. (§23)
3. Encyclopedia for children. Volume 17. Chemistry / Chapter. ed.V.A. Volodin, Ved. scientific ed. I. Leenson. - M.: Avanta+, 2003.
Additional web resources
1. Solving problems: calculations using thermochemical equations ().
2. Thermochemical equations ().
Homework
1) p. 69 problems No. 1,2 from the textbook “Chemistry: inorganic.” chemistry: textbook. for 8th grade general education institution." /G.E. Rudzitis, F.G. Feldman. - M.: Education, OJSC “Moscow Textbooks”, 2009.
2) pp. 80-84 No. 241, 245 from the Collection of problems and exercises in chemistry: 8th grade: for textbooks. P.A. Orzhekovsky and others. “Chemistry. 8th grade” / P.A. Orzhekovsky, N.A. Titov, F.F. Hegel. - M.: AST: Astrel, 2006.
Equations of chemical reactions, which indicate their thermal
effects are called thermochemical equations.
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Thermochemical equations have a number of features:
a) Since the state of the system depends on the aggregate states of substances
in general, in thermochemical equations using letter subscripts
(j), (g), (p) and (d) indicate the states of substances (crystalline, liquid, dissolved and gaseous). For example,
b) So that the thermal effect of the reaction is expressed in kJ/mol of one of the starting substances or reaction products, in thermochemical equations
Fractional odds are allowed. For example,
=−46.2 kJ/mol.
c) Often the heat of reaction (thermal effect) is written as ∆H
The upper index 0 means the standard value of the thermal effect (the value obtained under standard conditions, i.e., at a pressure of 101 kPa), and the lower index means the temperature at which the interaction occurs.
The peculiarity of thermochemical equations is that when working with them, you can transfer the formulas of substances and the magnitude of thermal effects from one part of the equation to another. As a rule, this cannot be done with ordinary equations of chemical reactions.
Term-by-term addition and subtraction of thermochemical equations is also allowed. This may be necessary to determine the thermal effects of reactions that are difficult or impossible to measure experimentally.
11.Formulate Hess’s law and a corollary to Hess’s law.
Hess's law is formulated as follows: 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 (enthalpy) of the starting substances and reaction products.
Corollary 1. The thermal effect of the reaction is equal to the difference between the sums of the heats of formation of the reaction products and the heats of formation of the starting substances, taking into account their stoichiometric coefficients.
Corollary 2. If the thermal effects of a number of reactions are known, then it is possible to determine the thermal effect of another reaction, which includes substances and compounds included in the equations for which the thermal effect is known. At the same time, with thermochemical equations you can perform a variety of arithmetic operations (addition, subtraction, multiplication, division) as with algebraic equations.
12.What is the standard enthalpy of formation of a substance?
The standard enthalpy of formation of a substance is the thermal effect of the reaction of formation of 1 mole of a given substance from the corresponding amount of simple substances under standard conditions.
13.What is entropy? How is it measured?
Entropy is a thermodynamic function of the state of the system, and its value depends on the amount of the substance (mass) under consideration, temperature, and state of aggregation.
Units J/C
14.Formulate the 2nd and 3rd laws of thermodynamics.
Second law of thermodynamics
In isolated systems (Q= 0, A= 0, U= const) spontaneously occur
only those processes that are accompanied by an increase in the entropy of the system, i.e. S>0.
The spontaneous process ends when the maximum at
given entropy conditions S max, i.e. when ∆S= 0.
Thus, in isolated systems, the criterion for a spontaneous process is an increase in entropy, and the limit of such a process is -∆S = 0.
Third law of thermodynamics
The entropy of each chemical element in an ideal crystalline state at a temperature close to absolute zero is close to zero.
The entropy of nonideal crystals is greater than zero, since they can be considered
as mixtures with entropy of mixing. This is also true for crystals that have defects in the crystal structure. This leads to the principle
the unattainability of absolute zero temperature. Currently achieved
lowest temperature 0.00001 K.
