1.1. Examples of problems with solutions
Task 1. Write down the thermochemical equation of the reaction if it is known that when 1 mole of hydrogen chloride gas HCl is formed from simple substances under standard conditions, 92 kJ of heat is released.
Solution
Thermochemical equations are the equations of chemical reactions written down indicating the enthalpy value DH (kJ) and the state of aggregation of the substances involved in the reaction.
The enthalpy of the reaction is DH 0 = Q p = -92 kJ, the appearance of the sign (-) is due to the fact that the enthalpies of exothermic reactions are considered negative.
Thermochemical reaction equation
1/2H 2 (g) + 1/2Cl 2 (g) = HCl (g), ∆H 0 = – 92 kJ.
Another possible answer is obtained by doubling this equation:
H 2 (g) + Cl 2 (g) = 2HCl (g), ∆H 0 = – 184 kJ.
Task2 . Calculate the standard enthalpy of formation of Al 2 O 3(t) if the thermochemical equation is known
4Al (t) + 3O 2 (g) = 2Al 2 O 3 (t), DH 0 = – 3340 kJ.
Solution
The enthalpy of formation of a substance is the enthalpy of the reaction of formation of 1 mole of a given substance from simple substances that are stable under standard conditions. The equation for the above reaction corresponds to the formation of 2 mol of aluminum oxide from simple substances - aluminum and oxygen. With thermochemical equations, you can carry out simple mathematical procedures: add, subtract, multiply or divide by any number. Let us divide the reaction equation by two so that it corresponds to the formation of 1 mole of substance (we will proportionally reduce the enthalpy value):
2Al (t) + 3/2O 2 (g) = Al 2 O 3 (t), .
Answer: standard enthalpy of formation of aluminum oxide
Task3 . Arrange the formulas of the substances (see table) in order of increasing stability. Motivate your answer.
Solution
The values of the enthalpies of formation allow us to compare stability of similar connections: The lower the enthalpy of formation, the more stable the compound. Arrangement of formulas of substances in order of increasing stability
H 2 Te (g) H 2 Se (g) H 2 S (g) H 2 O (g).
Task4 . Write down what relationship exists between the enthalpies of the reactions DH 1, DH 2 and DH 3 if the thermochemical equations are known.
1) C (graphite) + O 2 (g) = CO 2 (g), DH 1;
2) C (graphite) + 1/2O 2 (g) = CO (g), DH 2;
3) CO (g) + 1/2O 2 (g) = CO 2 (g), DH 3.
Solution
Thermochemical equations can be added and subtracted etc. Equation (1) can be obtained by adding equations (2) and (3), i.e.
Answer:
Task 5. Determine the standard enthalpy of the reaction from reference data
C 2 H 5 OH (l) + 3O 2 (g) = 2CO 2 (g) + 3H 2 O (g).
Solution
We find the enthalpy of the reaction using the first corollary of Hess’s law
Answer:
Task 6. Calculate the amount of heat released (or absorbed) when slaking 1 kg of lime under standard conditions. The values of the standard enthalpies of formation of substances are given in the table.
Solution
Lime slaking reaction equation:
CaO (s) + H 2 O (l) = Ca(OH) 2 (s).
The thermal effect of the reaction is equal to the enthalpy of the reaction, the value of which is found by the first corollary of Hess’s law:
The enthalpy of the reaction is negative, i.e. When lime is slaked, heat is released. The amount of heat Q = H 0 = -66 kJ corresponds to the quenching of 1 mol of CaO. We calculate the amount of substance contained in 1 kg of calcium oxide:
The amount of heat released when quenching 1 kg of lime is
Answer: When 1 kg of lime is slaked under standard conditions, 1175 kJ of heat is released.
In order to compare the energy effects of different processes, thermal effects are determined by standard conditions. The standard pressure is 100 kPa (1 bar), temperature 25 0 C (298 K), concentration - 1 mol/l. If the starting substances and reaction products are in a standard state, then the thermal effect of a chemical reaction is called standard enthalpy of the system and is designated ΔH 0 298 or ΔH 0 .
Equations of chemical reactions indicating the thermal effect are called thermochemical equations.
Thermochemical equations indicate the phase state and polymorphic modification of the reacting and resulting substances: g - gaseous, l - liquid, k - crystalline, m - solid, p - dissolved, etc. If the aggregate states of substances for the reaction conditions are obvious, for example, ABOUT 2 , N 2 , N 2 - gases, Al 2 ABOUT 3 , CaCO 3 - solids, etc. at 298 K, then they may not be indicated.
The thermochemical equation includes the heat effect of the reaction ΔH, which in modern terminology is written next to the equation. For example:
WITH 6 N 6(W) + 7.5О 2 = 6СО 2 + 3H 2 ABOUT (AND) ΔH 0 = - 3267.7 kJ
N 2 + 3H 2 = 2NH 3(G) ΔH 0 = - 92.4 kJ.
Thermochemical equations can be operated in the same way as algebraic equations (added, subtracted from each other, multiplied by a constant value, etc.).
Thermochemical equations are often (but not always) given for one mole of the substance in question (received or consumed). In this case, other participants in the process can enter the equation with fractional coefficients. This is allowed, since thermochemical equations operate not with molecules, but with moles of substances.
Thermochemical calculations
The thermal effects of chemical reactions are determined both experimentally and using thermochemical calculations.
Thermochemical calculations are based on Hess's law(1841):
The thermal effect of a reaction does not depend on the path along which the reaction proceeds (i.e., on the number of intermediate stages), but is determined by the initial and final state of the system.
For example, the combustion reaction of methane can proceed according to the equation:
CH 4 +2О 2 = CO 2 + 2H 2 ABOUT (G) ΔH 0 1 = -802.34 kJ
The same reaction can be carried out through the stage of CO formation:
CH 4 +3/2О 2 = CO + 2H 2 ABOUT (G) ΔH 0 2 = -519.33 kJ
CO +1/2O 2 = CO 2 ΔH 0 3 = -283.01 kJ
It turns out that ΔH 0 1 = ΔН 0 2 + ΔН 0 3 . Consequently, the thermal effect of the reaction proceeding along two paths is the same. Hess's law is well illustrated using enthalpy diagrams (Fig. 2)
A number of consequences follow from Hess’s law:
1. The thermal effect of the forward reaction is equal to the thermal effect of the reverse reaction with the opposite sign.
2. If, as a result of a series of successive chemical reactions, the system reaches a state that completely coincides with the initial one, then the sum of the thermal effects of these reactions is equal to zero ( ΔH= 0). Processes in which a system, after successive transformations, returns to its original state are called circular processes or cycles. The cycle method is widely used in thermochemical calculations. .
3. The enthalpy of a chemical reaction is equal to the sum of the enthalpies of formation of the reaction products minus the sum of the enthalpies of formation of the starting substances, taking into account stoichiometric coefficients.
Here we meet the concept ""enthalpy of formation"".
The enthalpy (heat) of formation of a chemical compound is the thermal effect of the reaction of the formation of 1 mole of this compound from simple substances taken in their stable state under given conditions. Usually the heat of formation is referred to the standard state, i.e. 25 0 C (298 K) and 100 kPa. The standard enthalpies of formation of chemical substances are designated ΔH 0 298 (or ΔH 0 ), are measured in kJ/mol and are given in reference books. The enthalpy of formation of simple substances that are stable at 298 K and a pressure of 100 kPa is taken equal to zero.
In this case, a corollary from Hess’s law for the thermal effect of a chemical reaction ( ΔH (H.R.)) has the form:
ΔH (H.R.) = ∑ΔН 0 reaction products - ∑ΔН 0 starting materials
Using Hess's law, you can calculate the energy of chemical bonds, the energy of crystal lattices, the heat of combustion of fuels, the calorie content of food, etc.
The most common calculations are the calculation of thermal effects (enthalpies) of reactions, which is necessary for technological and scientific purposes.
Example 1. Write the thermochemical equation for the reaction between CO 2(G) and hydrogen, which results in the formation CH 4(G) And N 2 ABOUT (G) , calculating its thermal effect based on the data given in the appendix. How much heat will be released in this reaction when producing 67.2 liters of methane, based on standard conditions?
Solution.
CO 2(G) + 3H 2(G) = CH 4(G) + 2H 2 ABOUT (G)
We find in the reference book (appendix) the standard heats of formation of compounds involved in the process:
ΔH 0 (CO 2(G) ) = -393.51 kJ/mol ΔH 0 (CH 4(G) ) = -74.85 kJ/mol ΔH 0 (N 2(G) ) = 0 kJ/mol ΔH 0 (N 2 ABOUT (G) ) = -241.83 kJ/mol
Please note that the heat of formation of hydrogen, like all simple substances in their stable state under given conditions, is zero. We calculate the thermal effect of the reaction:
ΔH (H.R.) = ∑ΔН 0 (cont.) -∑ΔН 0 (ref.) =
ΔH 0 (CH 4(G) ) + 2ΔH 0 (N 2 ABOUT (G) ) - ΔН 0 (CO 2(G) ) -3ΔH 0 (N 2(G) )) =
74.85 + 2(-241.83) - (-393.51) - 3·0 = -165.00 kJ/mol.
The thermochemical equation is:
CO 2(G) + 3H 2(G) = CH 4(G) + 2H 2 ABOUT (G) ; ΔH= -165.00 kJ
According to this thermochemical equation, 165.00 kJ of heat will be released when 1 mole is received, i.e. 22.4 liters of methane. The amount of heat released when producing 67.2 liters of methane is found from the proportion:
22.4 l -- 165.00 kJ 67.2 165.00
67.2 l -- Q kJ Q = ------ = 22.4
Example 2. When 1 liter of ethylene C 2 H 4 (G) is burned (standard conditions) with the formation of gaseous carbon monoxide (IV) and liquid water, 63.00 kJ of heat is released. Using these data, calculate the molar enthalpy of combustion of ethylene and write down the thermochemical equation of the reaction. Calculate the enthalpy of formation of C 2 H 4 (G) and compare the obtained value with literature data (Appendix).
Solution. We compose and equalize the chemical part of the required thermochemical equation:
WITH 2 N 4(G) + 3О 2(G) = 2СО 2(G) + 2H 2 ABOUT (AND) ; N= ?
The thermochemical equation created describes the combustion of 1 mole, i.e. 22.4 liters of ethylene. The required molar heat of combustion of ethylene is found from the proportion:
1l -- 63.00 kJ 22.4 63.00
22.4 l -- Q kJ Q = ------ =
1410.96 kJ
H = -Q, the thermochemical equation for ethylene combustion has the form: WITH 2 N 4(G) + 3О 2(G) = 2СО 2(G) + 2H 2 ABOUT (AND) ; N= -1410.96 kJ
To calculate the enthalpy of formation WITH 2 N 4(G) we draw a corollary from Hess’s law: ΔH (H.R.) = ∑ΔН 0 (cont.) -∑ΔН 0 (ref.).
We use the enthalpy of combustion of ethylene that we found and the enthalpies of formation of all (except ethylene) participants in the process given in the appendix.
1410.96 = 2·(-393.51) + 2·(-285.84) - ΔH 0 (WITH 2 N 4(G) ) - 3·0
From here ΔH 0 (WITH 2 N 4(G) ) = 52.26 kJ/mol. This coincides with the value given in the appendix and proves the correctness of our calculations.
Example 3. Write a thermochemical equation for the formation of methane from simple substances, calculating the enthalpy of this process from the following thermochemical equations:
CH 4(G) + 2О 2(G) = CO 2(G) + 2H 2 ABOUT (AND) ΔH 1 = -890.31 kJ (1)
WITH (GRAPHITE) + O 2(G) = CO 2(G) N 2 = -393.51 kJ (2)
N 2(G) + ½О 2(G) = N 2 ABOUT (AND) N 3 = -285.84 kJ (3)
Compare the obtained value with the tabular data (Appendix).
Solution. We compose and equalize the chemical part of the required thermochemical equation:
WITH (GRAPHITE) + 2H 2(G) = CH 4(G) N 4 = N 0 (CH 4(G)) ) =? (4)
Thermochemical equations can be handled in the same way as algebraic ones. As a result of algebraic operations with equations 1, 2 and 3, we must obtain equation 4. To do this, equation 3 should be multiplied by 2, the result added to equation 2 and subtracted by equation 1.
2H 2(G) + O 2(G) = 2H 2 ABOUT (AND) N 0 (CH 4(G) ) = 2 N 3 + N 2 - N 1
+ C (GRAPHITE) + O 2(G) + CO 2(G) N 0 (CH 4(G) ) = 2(-285,84)
- CH 4(G) - 2О 2(G) -CO 2(G) - 2H 2 ABOUT (AND) + (-393,51)
WITH (GRAPHITE) + 2H 2(G) = CH 4(G) N 0 (CH 4(G) ) = -74.88 kJ
This matches the value given in the appendix, which proves that our calculations are correct.
Any chemical reaction is accompanied by the release or absorption of energy in the form of heat.
Based on the release or absorption of heat, they distinguish exothermic And endothermic reactions.
Exothermic reactions are reactions during which heat is released (+Q).
Endothermic reactions are reactions during which heat is absorbed (-Q).
Thermal effect of reaction (Q) is the amount of heat that is released or absorbed during the interaction of a certain amount of initial reagents.
A thermochemical equation is an equation that specifies the thermal effect of a chemical reaction. So, for example, the thermochemical equations are:
It should also be noted that thermochemical equations must necessarily include information about the aggregate states of reagents and products, since the value of the thermal effect depends on this.
Calculations of the thermal effect of the reaction
An example of a typical problem to find the thermal effect of a reaction:
When 45 g of glucose reacts with excess oxygen according to the equation
C 6 H 12 O 6 (solid) + 6O 2 (g) = 6CO 2 (g) + 6H 2 O (g) + Q
700 kJ of heat was released. Determine the thermal effect of the reaction. (Write the number to the nearest whole number.)
Solution:
Let's calculate the amount of glucose:
n(C 6 H 12 O 6) = m(C 6 H 12 O 6) / M(C 6 H 12 O 6) = 45 g / 180 g/mol = 0.25 mol
Those. When 0.25 mol of glucose interacts with oxygen, 700 kJ of heat is released. From the thermochemical equation presented in the condition, it follows that when 1 mole of glucose interacts with oxygen, an amount of heat is formed equal to Q (the thermal effect of the reaction). Then the following proportion is correct:
0.25 mol glucose - 700 kJ
1 mole of glucose - Q
From this proportion the corresponding equation follows:
0.25 / 1 = 700 / Q
Solving which, we find that:
Thus, the thermal effect of the reaction is 2800 kJ.
Calculations using thermochemical equations
Much more often in USE tasks in thermochemistry, the value of the thermal effect is already known, because the condition gives the complete thermochemical equation.
In this case, it is necessary to calculate either the amount of heat released/absorbed with a known amount of a reagent or product, or, conversely, from a known value of heat it is necessary to determine the mass, volume or amount of a substance of any participant in the reaction.
Example 1
According to the thermochemical reaction equation
3Fe 3 O 4 (tv.) + 8Al (tv.) = 9Fe (tv.) + 4Al 2 O 3 (tv.) + 3330 kJ
68 g of aluminum oxide were formed. How much heat was released? (Write the number to the nearest whole number.)
Solution
Let's calculate the amount of aluminum oxide substance:
n(Al 2 O 3) = m(Al 2 O 3) / M(Al 2 O 3) = 68 g / 102 g/mol = 0.667 mol
In accordance with the thermochemical equation of the reaction, when 4 moles of aluminum oxide are formed, 3330 kJ are released. In our case, 0.6667 mol of aluminum oxide is formed. Having denoted the amount of heat released in this case by x kJ, we create the proportion:
4 mol Al 2 O 3 - 3330 kJ
0.667 mol Al 2 O 3 - x kJ
This proportion corresponds to the equation:
4 / 0.6667 = 3330 / x
Solving which, we find that x = 555 kJ
Those. when 68 g of aluminum oxide is formed in accordance with the thermochemical equation in the condition, 555 kJ of heat is released.
Example 2
As a result of a reaction, the thermochemical equation of which
4FeS 2 (tv.) + 11O 2 (g) = 8SO 2 (g) + 2Fe 2 O 3 (tv.) + 3310 kJ
1655 kJ of heat was released. Determine the volume (l) of sulfur dioxide released (no.). (Write the number to the nearest whole number.)
Solution
In accordance with the thermochemical equation of the reaction, when 8 moles of SO 2 are formed, 3310 kJ of heat is released. In our case, 1655 kJ of heat was released. Let the amount of SO 2 formed in this case be x mol. Then the following proportion is fair:
8 mol SO 2 - 3310 kJ
x mol SO 2 - 1655 kJ
From which the equation follows:
8 / x = 3310 / 1655
Solving which, we find that:
Thus, the amount of SO 2 substance formed in this case is 4 mol. Therefore, its volume is equal to:
V(SO 2) = V m ∙ n(SO 2) = 22.4 l/mol ∙ 4 mol = 89.6 l ≈ 90 l(rounded to whole numbers, since this is required in the condition.)
More analyzed problems on the thermal effect of a chemical reaction can be found.
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 constructed. 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 establishment /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.
Problem 10.1. Using the thermochemical equation: 2H 2 (g) + O 2 (g) = 2H 2 O (g) + 484 kJ, determine the mass of water formed if 1479 kJ of energy is released.
Solution. We write the reaction equation in the form:
We have
x = (2 mol 1479 kJ) / (484 kJ) = 6.11 mol.
Where
m(H 2 O) = v M = 6.11 mol 18 g/mol = 110 g
If the problem statement does not indicate the amount of the reactant, but only reports a change in a certain quantity (mass or volume), which, as a rule, relates to a mixture of substances, then it is convenient to introduce an additional term into the reaction equation corresponding to this change.
Problem 10.2. To a 10 L (N.O.) mixture of ethane and acetylene, 10 L (N.O.) hydrogen was added. The mixture was passed over a heated platinum catalyst. After bringing the reaction products to the initial conditions, the volume of the mixture became 16 liters. Determine the mass fraction of acetylene in the mixture.
Solution. Hydrogen reacts with acetylene, but not with ethane.
C 2 H 6 + H2 2 ≠
C 2 H 2 + 2 H 2 → C 2 H 6
In this case, the volume of the system decreases by
ΔV = 10 + 10 – 16 = 4 l.
The decrease in volume is due to the fact that the volume of the product (C 2 H 6) is less than the volume of the reagents (C 2 H 2 and H 2).
Let's write the reaction equation by introducing the expression ΔV.
If 1 liter of C 2 H 2 and 2 liters of H 2 react, and 1 liter of C 2 H 6 is formed, then
ΔV = 1 + 2 – 1 = 2 l.
From the equation it is clear that
V(C 2 H 2) = x = 2 l.
Then
V(C 2 H 6) = (10 - x) = 8 l.
From the expression
m / M = V / V M
we have
m = M V / V M
m(C 2 H 2) = M V / V M= (26 g/mol 2l) / (22.4 l/mol) = 2.32 g,
m(C 2 H 6) = M V / V M,
m(mixture) = m(C 2 H 2) + m(C 2 H 6) = 2.32 g + 10.71 g = 13.03 g,
w(C 2 H 2) = m(C 2 H 2) / m(mixture) = 2.32 g / 13.03 g = 0.18.
Problem 10.3. An iron plate weighing 52.8 g was placed in a solution of copper (II) sulfate. Determine the mass of dissolved iron if the mass of the plate becomes 54.4 g.
Solution. The change in mass of the plate is equal to:
Δm = 54.4 - 52.8 = 1.6 g.
Let's write down the reaction equation. It can be seen that if 56 g of iron dissolves from the plate, then 64 g of copper will be deposited on the plate and the plate will become 8 g heavier: 
It is clear that
m(Fe) = x = 56 g 1.6 g / 8 g = 11.2 g.
Problem 10.4. In 100 g of a solution containing a mixture of hydrochloric and nitric acids, a maximum of 24.0 g of copper(II) oxide is dissolved. After evaporation of the solution and calcination of the residue, its mass is 29.5 g. Write the equations for the reactions occurring and determine the mass fraction of hydrochloric acid in the original solution.
Solution. Let's write the reaction equations:
СuО + 2НCl = СuСl 2 + Н 2 O (1)
CuO + 2HNO 3 = Cu(NO 3) 2 + H 2 O (2)
2Сu(NO 3) 2 = 2СuО + 4NO 2 + O 2 (3)
It can be seen that the increase in mass from 24.0 g to 29.5 g is associated only with the first reaction, because copper oxide, dissolved in nitric acid according to reaction (2), during reaction (3) again turned into copper oxide of the same mass. If during reaction (1) 1 mol of CuO weighing 80 g reacts and 1 mol of CuCl 2 weighing 135 g is formed, then the mass will increase by 55 g. Considering that the mass of 2 mol of HCl is 73 g, we will write equation (1) again, adding the expression Δm.
It is clear that
m(HCl) = x = 73 g 5.5 g / 55 g = 7.3 g.
Find the mass fraction of acid:
w(HCl) = m(HCl) / m solution =
= 7.3 g / 100 g = 0.073.
