Problem 10.1. Using the thermochemical equation: 2H 2 (g) + O 2 (g) \u003d 2H 2 O (g) + 484 kJ, determine the mass of water formed if 1479 kJ of energy was 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) \u003d v M \u003d 6.11 mol 18 g / mol \u003d 110 g
If the condition of the problem does not indicate the amount of the reactant, but only reports a change in a certain quantity (mass or volume), which, as a rule, refers 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 mixture of ethane and acetylene with a volume of 10 l (n.o.) was added 10 l (n.o.) of hydrogen. The mixture was passed over a heated platinum catalyst. After bringing the reaction products to the initial conditions, the volume of the mixture became equal to 16 l. 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 + 2H 2 → C 2 H 6
In this case, the volume of the system is reduced by
ΔV \u003d 10 + 10 - 16 \u003d 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).
We write the reaction equation by introducing the expression ΔV.
If 1 l C 2 H 2 and 2 l H 2 enter into the reaction, and 1 l C 2 H 6 is formed, then
ΔV \u003d 1 + 2 - 1 \u003d 2 l.
It can be seen from the equation that
V (C 2 H 2) \u003d x \u003d 2 l.
Then
V (C 2 H 6) \u003d (10 - x) \u003d 8 l.
From expression
m / M = V / V M
we have
m = M V / V M
m (C 2 H 2) \u003d M V / V M\u003d (26 g / mol 2l) / (22.4 l / mol) \u003d 2.32 g,
m (C 2 H 6) \u003d M V / V M,
m (mixtures) \u003d m (C 2 H 2) + m (C 2 H 6) \u003d 2.32 g + 10.71 g \u003d 13.03 g,
w (C 2 H 2) \u003d m (C 2 H 2) / m (mixtures) \u003d 2.32 g / 13.03 g \u003d 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 the mass of the plate is:
Δm = 54.4 - 52.8 = 1.6 g.
Let's write the reaction equation. It can be seen that if 56 g of iron is dissolved from the plate, then 64 g of copper will be deposited on the plate and the plate will become 8 g heavier: 
It's clear that
m(Fe) \u003d x \u003d 56 g 1.6 g / 8 g \u003d 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 taking place and determine the mass fraction of hydrochloric acid in the initial solution.
Solution. Let's write the reaction equations:
CuO + 2HCl \u003d CuCl 2 + H 2 O (1)
CuO + 2HNO 3 \u003d Cu (NO 3) 2 + H 2 O (2)
2Cu (NO 3) 2 \u003d 2CuO + 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 in the course of reaction (1) 1 mol of CuO with a mass of 80 g reacts and 1 mol of CuCl 2 with a mass of 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 write equation (1) again, by adding the expression Δm.
It's clear that
m (HCl) \u003d x \u003d 73 g 5.5 g / 55 g \u003d 7.3 g.
Find the mass fraction of the acid:
w(HCl) = m(HCl) / m solution =
= 7.3 g / 100 g = 0.073.
From the lesson materials, you will learn which equation of a chemical reaction is called thermochemical. The lesson is devoted to the study of the calculation algorithm for the thermochemical equation of reactions.
Topic: Substances and their transformations
Lesson: Calculations using thermochemical equations
Almost all reactions proceed with the release or absorption of heat. The amount of heat 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, in contrast to conventional chemical equations, the state of aggregation of a substance (solid, liquid, gaseous) is necessarily indicated.
For example, the thermochemical equation for the reaction between calcium oxide and water looks like this:
CaO (t) + H 2 O (l) \u003d Ca (OH) 2 (t) + 64 kJ
The amount of heat Q released or absorbed during a chemical reaction is proportional to the amount of the substance of the reactant or product. Therefore, using thermochemical equations, various calculations can be made.
Consider 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 reaction of water decomposition:
You can solve this problem using the proportion:
during the decomposition of 36 g of water, 484 kJ were absorbed
in the decomposition of 3.6 g of water absorbed x kJ
Thus, the equation for the reaction can be drawn up. The complete solution of the problem is shown in Fig.1.
Rice. 1. Formulation of the solution of problem 1
The problem can be formulated in such a way that you will need to write a thermochemical reaction equation. Let's consider an example of such a task.
Task 2: The interaction of 7 g of iron with sulfur released 12.15 kJ of heat. Based on these data, make a thermochemical equation for the reaction.
I draw your attention to the fact that the answer to this problem is the thermochemical reaction equation itself.
Rice. 2. Formulation of the solution of problem 2
1. Collection of tasks and exercises in chemistry: 8th grade: to textbook. P.A. Orzhekovsky and others. “Chemistry. Grade 8 / P.A. Orzhekovsky, N.A. Titov, F.F. Hegel. - M.: AST: Astrel, 2006. (p. 80-84)
2. Chemistry: inorganic. chemistry: textbook. for 8kl. general inst. /G.E. Rudzitis, F.G. Feldman. - M.: Enlightenment, JSC "Moscow textbooks", 2009. (§23)
3. Encyclopedia for children. Volume 17. Chemistry / Chapter. edited by V.A. Volodin, leading. scientific ed. I. Leenson. - M.: Avanta +, 2003.
Additional web resources
1. Problem solving: calculations according to thermochemical equations ().
2. Thermochemical equations ().
Homework
1) with. 69 tasks №№ 1,2 from the textbook "Chemistry: inorgan. chemistry: textbook. for 8kl. general inst.» /G.E. Rudzitis, F.G. Feldman. - M .: Education, JSC "Moscow textbooks", 2009.
2) p.80-84 nos. 241, 245 from the Collection of tasks and exercises in chemistry: 8th grade: to textbook. P.A. Orzhekovsky and others. “Chemistry. Grade 8 / P.A. Orzhekovsky, N.A. Titov, F.F. Hegel. - M.: AST: Astrel, 2006.
Thermochemical equations. Quantity of heat. which is released or absorbed as a result of the reaction between certain amounts of reagents, given by stoichiometric coefficients, is called the thermal effect of a chemical reaction and is usually denoted by the symbol Q. Exothermic and endothermic reactions. Hesse's thermochemical law Reactions that proceed with the release of energy in the form of heat are called exothermic; reactions proceeding 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, respectively, increase) of the enthalpy of the reaction LA. Thus, in exothermic reactions, when heat is released, AH is negative. In endothermic reactions (heat is absorbed), AH is positive. The magnitude of the thermal effect of a chemical reaction depends on the nature of the initial substances and products of the reaction, their state of aggregation and temperature. The reaction equation, on the right side of which, along with the reaction products, the change in the enthalpy AH 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 bonds in H2 and 02 molecules. These amounts of energy are respectively 435 and 494 kJ / mol . On the other hand, the formation of an O - H bond releases 462 kJ/mol of energy. 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 bonds H - H and O \u003d O, therefore the reaction is exothermic, i.e. during the formation two moles of vaporous water will release 484 kJ of energy. The reaction equation for the formation of water, written taking into account the change in enthalpy Exothermic and endothermic reactions. Hesse's thermochemical law will already be a thermochemical reaction equation. 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 the bonds in the molecules of the starting substances is 1439 kJ and more than the released energy of bond formation 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. Thermochemical Equations Exothermic and endothermic reactions. Hesse's thermochemical law This explains why nitric oxide (II) is formed only at high temperatures, for example, in car exhaust gases, in lightning discharges and is not formed under normal conditions.
In order to compare the energy effects of different processes, thermal effects are determined at standard conditions. The standard pressure is 100 kPa (1 bar), temperature 25 0 C (298 K), concentration - 1 mol / l. If the initial substances and products of the reaction are in the standard state, then the heat effect of a chemical reaction is called system standard enthalpy and denoted ΔN 0 298 or ΔN 0 .
The equations of chemical reactions indicating the thermal effect are called thermochemical equations.
In thermochemical equations, the phase state and the polymorphic modification of the reacting and formed substances are indicated: g - gas, g - liquid, k - crystalline, m - solid, p - dissolved, etc. If the aggregate states of substances for the reaction conditions are obvious, for example, O 2 , N 2 , N 2 - gases, Al 2 O 3 , CaCO 3 - solids, etc. at 298 K, they may not be indicated.
The thermochemical equation includes the thermal effect of the reaction ΔN, which in modern terminology is written next to the equation. For example:
FROM 6 H 6(W) + 7.5O 2 = 6CO 2 + 3H 2 O (AND) ΔN 0 = - 3267.7 kJ
N 2 + 3H 2 = 2NH 3(G) ΔN 0 = - 92.4 kJ.
It is possible to operate with thermochemical equations, as well as with algebraic equations (add, subtract from each other, multiply by a constant value, etc.).
Thermochemical equations are often (but not always) given for one mole of the substance in question (produced or consumed). At the same time, 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 by thermochemical calculations.
Thermochemical calculations are based on Hess' law(1841):
The thermal effect of the reaction does not depend on the path along which the reaction proceeds (ie, 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 +2O 2 = CO 2 + 2H 2 O (G) ΔN 0 1 = -802.34 kJ
The same reaction can be carried out through the stage of formation of CO:
CH 4 +3/2O 2 = CO + 2H 2 O (G) ΔN 0 2 = -519.33 kJ
CO +1/2O 2 = CO 2 ΔN 0 3 = -283.01 kJ
In doing so, it turns out that ΔN 0 1 = ΔH 0 2 + ΔН 0 3 . Therefore, the thermal effect of the reaction proceeding along the 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 heat effect of the forward reaction is equal to the heat effect of the reverse reaction with the opposite sign.
2. If, as a result of a series of successive chemical reactions, the system comes to a state that completely coincides with the initial one, then the sum of the thermal effects of these reactions is equal to zero ( ΔN= 0). The processes in which the 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 initial 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 heat effect of the reaction of formation of 1 mole of this compound from simple substances taken in their stable state under given conditions. Usually, the heats of formation are referred to the standard state, i.e. 25 0 C (298 K) and 100 kPa. The standard enthalpies of formation of chemicals are denoted ΔN 0 298 (or ΔN 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, the consequence from the Hess law for the thermal effect of a chemical reaction ( ΔN (H.R.)) looks like:
ΔN (H.R.) = ∑ΔН 0 reaction products - ∑ΔN 0 starting materials
Using Hess's law, one can calculate the energy of a chemical bond, 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 SO 2(G) and hydrogen, resulting in the formation CH 4(G) and H 2 O (G) , calculating its thermal effect based on the data given in the appendix. How much heat will be released in this reaction upon receipt of 67.2 liters of methane in terms of standard conditions?
Solution.
SO 2(G) + 3H 2(G) = CH 4(G) + 2H 2 O (G)
We find in the directory (application) the standard heats of formation of the compounds involved in the process:
ΔN 0 (SO 2(G) ) \u003d -393.51 kJ / mol ΔN 0 (CH 4(G) ) = -74.85 kJ/mol ΔN 0 (H 2(G) ) = 0 kJ/mol ΔN 0 (H 2 O (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:
ΔN (H.R.) = ∑ΔN 0 (prod.) -∑ΔN 0 (ref.) =
ΔN 0 (CH 4(G) ) + 2ΔН 0 (H 2 O (G) ) - ΔН 0 (SO 2(G) ) -3ΔН 0 (H 2(G) )) =
74.85 + 2 (-241.83) - (-393.51) - 3 0 \u003d -165.00 kJ / mol.
The thermochemical equation has the form:
SO 2(G) + 3H 2(G) = CH 4(G) + 2H 2 O (G) ; ΔN= -165.00 kJ
According to this thermochemical equation, 165.00 kJ of heat will be released upon receipt of 1 mol, i.e. 22.4 liters of methane. The amount of heat released upon receipt of 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 During the combustion of 1 l of ethylene C 2 H 4 (G) (standard conditions) with the formation of gaseous carbon monoxide (IV) and liquid water, 63.00 kJ of heat is released. Calculate the molar enthalpy of combustion of ethylene from these data and write down the thermochemical reaction equation. Calculate the enthalpy of formation of C 2 H 4 (G) and compare the obtained value with the literature data (Appendix).
Solution. We compose and equalize the chemical part of the required thermochemical equation:
FROM 2 H 4(G) + 3O 2(G) = 2СО 2(G) + 2H 2 O (AND) ; H= ?
The created thermochemical equation describes the combustion of 1 mol, i.e. 22.4 liters of ethylene. The molar heat of combustion of ethylene required for it 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 the combustion of ethylene has the form: FROM 2 H 4(G) + 3O 2(G) = 2СО 2(G) + 2H 2 O (AND) ; H= -1410.96 kJ
To calculate the enthalpy of formation FROM 2 H 4(G) we draw a corollary from the Hess law: ΔN (H.R.) = ∑ΔN 0 (prod.) -∑ΔN 0 (ref.).
We use the ethylene combustion enthalpy found by us 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) - ΔN 0 (FROM 2 H 4(G) ) - thirty
From here ΔN 0 (FROM 2 H 4(G) ) = 52.26 kJ/mol. This matches the value given in the appendix and proves the correctness of our calculations.
Example 3 Write the thermochemical equation for the formation of methane from simple substances by calculating the enthalpy of this process from the following thermochemical equations:
CH 4(G) + 2O 2(G) = CO 2(G) + 2H 2 O (AND) ΔN 1 = -890.31 kJ (1)
FROM (GRAPHITE) + O 2(G) = CO 2(G) H 2 = -393.51 kJ (2)
H 2(G) + ½O 2(G) = H 2 O (AND) H 3 = -285.84 kJ (3)
Compare the obtained value with the tabular data (application).
Solution. We compose and equalize the chemical part of the required thermochemical equation:
FROM (GRAPHITE) + 2H 2(G) = CH 4(G) H 4 = H 0 (CH 4(G)) ) =? (4)
You can operate with thermochemical equations in the same way as with algebraic ones. As a result of algebraic operations with equations 1, 2 and 3, we must obtain equation 4. To do this, multiply equation 3 by 2, add the result to equation 2 and subtract equation 1.
2H 2(G) + O 2(G) = 2H 2 O (AND) H 0 (CH 4(G) ) = 2 H 3 + H 2 - H 1
+ C (GRAPHITE) + O 2(G) + CO 2(G) H 0 (CH 4(G) ) = 2(-285,84)
- CH 4(G) - 2O 2(G) -CO 2(G) - 2H 2 O (AND) + (-393,51)
FROM (GRAPHITE) + 2H 2(G) = CH 4(G) H 0 (CH 4(G) ) = -74.88 kJ
This matches the value given in the appendix, which proves the correctness of our calculations.
Task 1.
When burning 560 ml (N.O.) of acetylene according to the thermochemical equation:
2C 2 H 2 (G) + 5O 2 (g) \u003d 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) \u003d 560 ml.
Find: the amount of released heat.
Solution:
To select the correct answer, it is most convenient to calculate the value sought in the problem and compare it with the proposed options. The calculation according to the thermochemical equation is no different from the calculation according to the usual reaction equation. Above the reaction, we indicate the data in the condition and the desired values, under the reaction - their ratios according to the coefficients. Heat is one of the products, so we consider its numerical value as a coefficient.
Comparing the received answer with the proposed options, we see that answer No. 2 is suitable.
A little trick that led inattentive students to the wrong answer No. 3 was the units of acetylene volume. The volume indicated in the condition in milliliters must have been converted to liters, since the molar volume is measured in (l / mol).
Occasionally, there are problems in which the thermochemical equation must be compiled independently from the value of the heat of formation of a complex substance.
Task 1.2.
The heat of formation of aluminum oxide is 1676 kJ/mol. Determine the thermal effect of the reaction in which the interaction of aluminum with oxygen yields
25.5 g A1 2 O 3 .
1) 140kJ; 2) 209.5 kJ; 3) 419kJ; 4) 838kJ.
Given:
heat of formation of aluminum oxide: Qobr (A1 2 O 3) = = 1676 kJ/mol;
mass of the obtained aluminum oxide: m (A1 2 O 3) \u003d 25.5 g.
Find: thermal effect.
Solution:
This type of problem can be solved in two ways:
I way
According to the definition, the heat of formation of a complex substance is the heat 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 O 2. When arranging the coefficients in the resulting equation, we take into account that before A1 2 O 3 there should be a coefficient "one"
, which corresponds to the amount of substance in 1 mol. In this case, we can use the heat of formation given in the condition:
2A1 (TV) + 3 / 2O 2 (g) -----> A1 2 O 3 (TV) + 1676 kJ
We have obtained a thermochemical equation.
In order for the coefficient in front of A1 2 O 3 to remain equal to "1", the coefficient in front 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:
We make a proportion:
upon receipt of 25.5 g of A1 2 O 3 x kJ is released (according to the condition)
upon receipt of 102 g of A1 2 O 3, 1676 kJ is released (according to the equation)
Suitable answer is #3.
When solving the last problem in the conditions of the Unified State Examination, it was possible not to draw up a thermochemical equation. Let's consider this method.
II way
According to the definition of the heat of formation, 1676 kJ is released during the formation of 1 mol of Al 2 O 3 . The mass of 1 mol of A1 2 O 3 is 102 g, therefore, it is possible to make a proportion:
1676 kJ is released during the formation of 102 g of A1 2 O 3
x kJ is released during the formation of 25.5 g of A1 2 O 3
Suitable answer is #3.
Answer: Q = 419kJ.
Task 1.3.
With the formation of 2 mol CuS from simple substances, 106.2 kJ of heat is released. During the formation of 288 g of CuS, heat is released by the amount:
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(CuS) = n(CuS) . M(CuS) = 2. 96 = 192
In the text of the condition, instead of the value of the amount of the substance CuS, we substitute the mass of 2 mol of this substance and get the finished proportion:
in the formation of 192 g of CuS, 106.2 kJ of heat is released
in the formation of 288 g of CuS, heat is released by the amount X kJ.
Appropriate answer number 2.
The second type of problems can be solved both according to the law of volumetric relations, and without its use. Let's look at both solutions with an example.
Tasks for applying the law of volumetric relations:
Task 1.4.
Determine the volume of oxygen (n.o.s.) required to burn 5 liters of carbon monoxide (o.s.).
1) 5 l; 2) 10 l; 3) 2.5 l; 4) 1.5 l.
Given:
carbon monoxide volume (n.o.): VCO) = 5 l.
Find: oxygen volume (n.o.): V (O 2) \u003d?
Solution:
First of all, you need to write an equation for the reaction:
2CO + O 2 \u003d 2CO
n = 2 mol n = 1 mol
We apply the law of volumetric ratios:
We find the ratio by the reaction equation, and
V(CO) will be taken from the condition. Substituting all these values into the law of volumetric ratios, we get:
Hence: V (O 2) \u003d 5/2 \u003d 2.5 l.
Suitable answer is #3.
Without using the law of volumetric ratios, the problem is solved by calculating according to the equation:
We make a proportion:
5 l CO2 interact with chl O2 (according to the condition) 44.8 l CO2 interact with 22.4 l O2 (according to the equation):
We received the same answer option number 3.
