Qualitative problems for calculations using the thermochemical equation. Study guide

1.1. Examples of tasks with solutions

Objective 1. Write down the thermochemical equation of the reaction if it is known that during the formation of 1 mol of gaseous hydrogen chloride HCl from simple substances under standard conditions, 92 kJ of heat is released.

Solution

Thermochemical equations are called equations of chemical reactions, written with an indication of the value of the enthalpy DH (kJ) and the state of aggregation of the substances involved in the reaction.

The enthalpy of reaction DH 0 = Q p = ‑92 kJ, the appearance of the sign (-) is due to the fact that the enthalpies of exothermic reactions are considered to be negative.

Thermochemical reaction equation

1 / 2H 2 (g) + 1 / 2Cl 2 (g) = HCl (g), ∆Н 0 = - 92 kJ.

Another answer is possible, 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 (s) + 3O 2 (g) = 2Al 2 O 3 (s), DH 0 = - 3340 kJ.

Solution

The enthalpy of formation of a substance is the enthalpy of the reaction of formation of 1 mol of a given substance from simple substances that are stable under standard conditions. The equation of the given reaction corresponds to the formation of 2 mol of aluminum oxide from simple substances - aluminum and oxygen. With thermochemical equations, you can carry out the simplest mathematical procedures: add, subtract, multiply, or divide by any number. Let's divide the reaction equation by two so that it corresponds to the formation of 1 mol of substance (we will proportionally decrease 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 resistance. Motivate your answer.

Solution

The values ​​of the enthalpies of formation make it possible to compare stability of the same type of connections: the lower the value of 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 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 / 2O2 (g) = CO 2 (g), DH 3.

Solution

Thermochemical equations can be added, subtracted etc. Equation (1) can be obtained by adding equations (2) and (3), i.e.

Answer:

Task 5. Determine the standard enthalpy of reaction from the reference data

C 2 H 5 OH (g) + 3O 2 (g) = 2CO 2 (g) + 3H 2 O (g).

Solution

The value of the enthalpy of reaction is found by the first consequence 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

The reaction equation for slaking lime:

CaO (s) + H 2 O (l) = Ca (OH) 2 (s).

The heat effect of the reaction is equal to the enthalpy of the reaction, the value of which is found by the first consequence of Hess's law:

The enthalpy of reaction is negative, i.e. when lime is slaked, heat is released. The amount of heat Q = H 0 = ‑66 kJ corresponds to quenching 1 mol of CaO. We calculate the amount of substance contained in 1 kg of calcium oxide:

The amount of heat released during the slaking of 1 kg of lime,

Answer: when slaking 1 kg of lime under standard conditions, 1175 kJ of heat is released.

From the lesson materials you will learn which equation chemical reaction 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 by thermochemical equations

Almost all reactions proceed with the release or absorption of heat. The amount of heat that is released or absorbed during the reaction is called thermal effect of 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 state of aggregation of matter (solid, liquid, gaseous) must be indicated.

For example, the thermochemical equation for the reaction between calcium oxide and water looks like this:

CaO (t) + H 2 O (l) = Ca (OH) 2 (t) + 64 kJ

The amount of heat Q released or absorbed during a chemical reaction is proportional to the amount of reagent substance or product. Therefore, using thermochemical equations, one can produce various calculations.

Let's consider examples of solving problems.

Objective 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:

upon decomposition of 36 g of water, 484 kJ was absorbed

on decomposition 3.6 g of water absorbed x kJ

Thus, the reaction equation can be drawn up. Complete solution tasks are shown in Fig. 1.

Rice. 1. Registration of the solution to problem 1

The problem can be formulated in such a way that you will need to draw up a thermochemical equation for the reaction. Let's consider an example of such a task.

Task 2: When 7 g of iron interacted with sulfur, 12.15 kJ of heat was released. Based on these data, draw up the thermochemical equation of the reaction.

I draw your attention to the fact that the answer in this problem is the thermochemical equation of the reaction itself.

Rice. 2. Registration of the solution to problem 2

1. Collection of tasks and exercises in chemistry: 8th grade .: for textbook. P.A. Orzhekovsky and others. "Chemistry. Grade 8 "/ P.А. Orzhekovsky, N.A. Titov, F.F. Hegel. - M .: AST: Astrel, 2006. (p.80-84)

2. Chemistry: inorganic. chemistry: textbook. for 8kl. general institutions / G.E. Rudzitis, F.G. Feldman. - M .: Education, JSC "Moscow textbooks", 2009. (§23)

3. Encyclopedia for children. Volume 17. Chemistry / Chap. ed. by V.A. Volodin, led. scientific. ed. I. Leenson. - M .: Avanta +, 2003.

Additional web resources

1. Solving problems: calculations by thermochemical equations ().

2. Thermochemical equations ().

Homework

1) p. 69 problems No. 1,2 from the textbook "Chemistry: Inorgan. chemistry: textbook. for 8kl. general institution. " / G.E. Rudzitis, F.G. Feldman. - M .: Education, JSC "Moscow textbooks", 2009.

2) p.80-84 No. 241, 245 from the Collection of tasks and exercises in chemistry: 8th grade .: for textbook. P.A. Orzhekovsky and others. "Chemistry. Grade 8 "/ P.А. Orzhekovsky, N.A. Titov, F.F. Hegel. - M .: AST: Astrel, 2006.

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Lesson number 17

Lesson topic: Calculations by thermochemical equations

Occupation type: a lesson in learning new material

The purpose of the lesson:

consider chemical processes from the point of view of their energy component, update the concepts of "thermochemical reactions", "thermal effect of a chemical reaction", "exothermic and endothermic processes";

to reveal the concept of "heat of formation of compounds", standard enthalpy, Hess's law;

to acquaint with the concept of entropy and the assessment of the possibility of spontaneous reactions;

develop the ability to solve computational problems using thermochemical equations, calculate the thermal effect of a chemical reaction using the concepts of "heat of formation", draw up thermochemical equations of reactions, determine the heats of formation of substances by thermochemical equations of reactions

Means of education:

Computer, projection equipment

Course of the lesson:

I. Introductory word of the teacher, introduction to the topic of the lesson Slide 1

II. Knowledge update. Slide 2

Thermal effects of chemical reactions. A chemical reaction consists in the breaking of some and the formation of other bonds, therefore it is accompanied by the release or absorption of energy in the form of heat, light, the work of expansion of the formed gases.

Introductory assignment (classification of chemistry by thermal effect) Independent activity of students to actualize the concepts of “thermochemical reactions”, “thermal effect of a chemical reaction”, “exothermic and endothermic reactions”.

III. Explanation of the new material.

Chemical reactions proceed with the release or absorption of energy, most often in the form of heat. Reactions in which heat is released are called exothermic, absorbed - endothermic. The amount of heat released or absorbed in a chemical reaction that occurs at a constant temperature is called the heat effect of the reaction. At constant pressure, the thermal effect of the reaction is equal to the change in enthalpy (ΔН).

The heat effect of a reaction is expressed in units of energy - kilojoules (kJ) or kilocalories (kcal) (1 kcal = 4.1868 kJ).

The science that studies the thermal effects of chemical reactions is called thermochemistry, and the equations of chemical reactions, in which the thermal effect is indicated, are called thermochemical equations.

The heat effect of the reaction (ΔН) depends on the nature of the reacting substances, on the amount of these substances and their state of aggregation, and on temperature.

To compare the energy effects of various reactions and to carry out thermochemical calculations, standard thermal effects are used (indicated).

Slide 3 The standard refers to the heat effect of the reaction carried out under conditions when all the substances participating in the reaction are in the specified standard states (pressure 101 kPa).

Slide 4 In thermochemical equations, it is necessary to indicate the aggregate states of substances using letter indices, and the heat effect of the reaction (ΔН) should be written separately, separated by commas.

For example, the thermochemical equation

shows that this chemical reaction is accompanied by the release of 1531 kJ of heat, if the pressure is 101 kPa, and refers to the number of moles of each of the substances, which corresponds to the stoichiometric coefficient in the reaction equation.

In exothermic reactions, when heat is released, ∆H is negative. In endothermic reactions (heat is absorbed) and ∆H is positive.

H 2 + Cl 2 = 2Сl 2 + Q,

where Q is the amount of heat released. If you use the enthalpy (characteristic of the energy content of the system), then this equation should be written differently:

H 2 + Cl 2 = 2Сl 2, ∆Н

The most important quantity in thermochemistry is the standard heat of formation (standard enthalpy of formation). Standard heat (enthalpy) of formation a complex substance is called the thermal effect (change in the standard enthalpy) of the reaction of the formation of one mole of this substance from simple substances in the standard state. The standard enthalpy of formation of simple substances in this case is taken equal to zero.

In thermochemistry, equations are often used in which the thermal effect is attributed to one mole of the formed substance, using fractional coefficients, if necessary.

For example, kJ.

The heat effect of this chemical reaction is equal to the enthalpy of formation of HCl (g), i.e.

Slide 5 Where did the minus sign in front of the heat effect come from? It is accepted to represent the energy lost by any system with a minus sign. Consider, for example, the familiar system of methane and oxygen molecules. As a result of an exothermic reaction between them, heat is released:

CH 4 (g) + 2 O 2 (g) = CO2 (g) + 2 H 2 O (g) + 890 kJ

You can write this reaction and another equation, where the released ("lost") heat has a "minus" sign:

CH 4 (g) + 2 O 2 (g) - 890 kJ = CO 2 (g) + 2 H 2 O (g)

By tradition, the enthalpy of this and other exothermic reactions in thermodynamics is usually written with a minus sign:

∆H about 298 = –890 kJ / mol (energy is released).

Slide 6 On the contrary, if as a result of an endothermic reaction the system has absorbed energy, then the enthalpy of such an endothermic reaction is written with a plus sign. For example, for the reaction of obtaining CO and hydrogen from coal and water (when heated):

C (TV) + H 2 O (g) + 131.3 kJ = CO (g) + H 2 (g)

(∆H about 298 = +131.3 kJ / mol)

The heat effect of the exothermic reaction is considered negative (ΔH 0).

Slide 7 Thermochemical calculations are based on law Hess... The thermal effect (∆H) of a chemical reaction (at constant P and T) does not depend on the path of its course, but depends on the nature and physical state of the initial substances and reaction products.

ΔН c.r. = ∑ ΔН prod arr - ∑ ΔН ref arr

Consequences from Hess's Law

The heat effects of direct and reverse reactions are equal in magnitude and opposite in sign.

The heat effect of a chemical reaction (∆Н) is equal to the difference between the sum of the enthalpies of formation of the reaction products and the sum of the enthalpies of formation of the starting materials, taken with account of the coefficients in the reaction equation.

IV... Explanation of the algorithm for solving computational problems using thermochemical calculations using equations.

Slide 8 Example With the formation of 1.8 g of water (H 2 O (l)), 28.6 kJ of heat was released from gaseous hydrogen and oxygen. Calculate the enthalpy of formation of H 2 O (g) and write the equation for the reaction, the thermal effect of which is
.

Solution. 1st way .

Since 1 mole of water is equal to 18 g, the enthalpy of formation

1 mol H 2 O (g) can be calculated

kJ / mol,

which corresponds to the equation

kJ / mol.

2nd way: From the condition: H = -28.6 kJ.

A-priory:
;

Hence,

kJ / mol.

V... Primary control of the assimilation of theoretical material, the solution of computational problems.

Slide 9 Example 1. How much heat will be released upon receipt of 1 kg of iron by reaction

Fe 2 O 3 (k) + 3CO (g) = 2Fe (k) + 3CO 2 (g), if the enthalpies of formation of Fe 2 O 3 (k), CO (g), and CO 2 (g) are, respectively, (kJ / mol): -822.7; -110.6 and -394.0.

Solution

1.Calculate the thermal effect of the reaction (H) using a corollary from Hess's law.

Since the enthalpy of formation of a simple substance is taken to be zero, then
.

Let's carry out the calculation according to the thermochemical equation:

if 256g Fe is formed, then 27.2 kJ is released;

if 1000 g of Fe is formed, then NS kJ.

We solve the proportion and get

kJ, i.e.

242.9 kJ of heat will be released.

Slide 10 Example 2. The ethane combustion reaction is expressed by the thermochemical equation

C 2 H 6 (g) + 3½O 2 = 2 CO 2 (g) + 3H 2 O (g); ΔHx.r. = -1559.87 kJ. Calculate the heat of formation of ethane, if the heats of formation of CO 2 (g) and H 2 O (l) are known

Solution.

Based on the following data:

a) C 2 H 6 (g) + 3 ½O 2 (g) = 2CO 2 (g) + 3H 2 O (g); ΔН = -1559.87 kJ

b) C (graphite) + O 2 (g) = CO2 (g); ΔН = -393.51 kJ

c) H 2 (g) + ½O 2 = H 2 O (g); ΔН = -285.84 kJ

Based on Hess's law

С 2 Н 6 = 3 ½О 2 - 2С - 2О 2 - 3Н 2 - 3/2 О 2 = 2СО 2 + 3Н 2 О - 2СО 2 - 3Н 2 О

ΔH = -1559.87 - 2 (-393.51) - 3 (-285.84) = +84.67 kJ;

ΔH = -1559.87 + 787.02 + 857.52; C 2 H 2 = 2C + 3H 2;

ΔН = +84.67 kJ

Hence

∆H sample С 2 Н 6 = -84.67 kJ

Slide 11 Example 3. The combustion reaction of ethyl alcohol is expressed by the thermochemical equation:

C 2 H 5 OH (g) + 3O 2 (g) = 2CO 2 (g) + 3H 2 O (g); ΔН =?

Calculate the heat of reaction if it is known that the molar (molar) heat of vaporization of C 2 H 5 OH (l) is +42.36 kJ and the heats of formation are known: C 2 H 5 OH (g); CO 2 (g); H 2 O (g).

Solution. To determine the ΔН reaction, it is necessary to know the heat of formation of С 2 Н 5 ОН (l). We find the latter from the data:

C 2 H 5 OH (g) = C 2 H 5 OH (g); ΔH = + 42.36 kJ.

42.36 = -235.31 - ∆HC 2 H 5 OH (l);

∆HC 2 H 5 OH (l) = -235.31 - 42.36 = -277.67 kJ.

We calculate the ΔН of the reaction, applying the consequences of the Hess law:

ΔН c.r. = 2 (-393.51) + 3 (-285.84) + 277.67 = -1366.87 kJ.

Slide 12 Example 4. Calculate the enthalpy of formation of N 2 O 5 (cr), if the heat effect of the reaction N 2 O 5 (k) + 2KOH (k) = 2KNO 3 (k) + H 2 O (g) is known; kJ, as well as the enthalpies of formation of KOH (k), KNO 3 (k) and H 2 O (g), which are respectively -425.0; -493.2 and –286.0 (kJ / mol).

Solution.

Using a corollary from Hess's law, we write

Substitute the data from the condition and get

380,6=(2-493,2-286)-(
+2-425)

We carry out arithmetic calculations:

380,6=-422,4-
.

KJ / mol

VI... Reflection Slide 13

What is the thermal effect of a chemical reaction (DH)?

List the factors that influence the thermal effect of a chemical reaction (DH).

What reactions are called exothermic and endothermic? Give examples.

What is the sign of the thermal effect (DH) for exothermic and endothermic reactions?

Give the definition of the standard enthalpy of formation of a complex substance.

Give the formulation of Hess's law.

Formulate the consequences of Hess's law.

Vii... Homework Slide 14

1. When 1 mol of hydrogen and 1 mol of chlorine interact, 184 kJ is released. What is the enthalpy of formation of hydrogen chloride?

2. The decomposition of 1 mole of hydrogen bromide into simple substances requires 72 kJ of heat. What is the enthalpy of formation of HBr?

3. How much heat will be released during combustion of 1 kg of aluminum, if
kJ / mol.

4. When burning, how much magnesium is released 1000 kJ, if
kJ / mol?

Table 1

Standard heats (enthalpies) of formation ΔH O 298 of some substances

Substance	State	ΔH about 298, kJ / mol	Substance	State	ΔH about 298, kJ / mol
		92,31 equations... 1. When combining 4.2 g of iron with sulfur, ... kJ of heat was released. Make up thermochemical the equation phosphorus combustion reactions. 7. By thermochemical equation combustion of hydrogen 2H2 + ... Calculations using the thermochemical equation (thx) of the reaction Document TASKS Grade 11 Calculations on equation reaction What mass ... with a mass fraction of bromine 3.2%? Calculations on thermochemical equation(TCA) reactions What is the amount of ... ethylene (n.o.) and 5 g of water. Calculations on equations consecutive reactions Burned 12 liters (at ... Ib calculation of the heat effect of the reaction, drawing up a thermochemical equation Solution Computational tasks " Calculations on thermochemical equations»IA Payment on thermochemical equation Sample problem solution Thermochemical the equation limestone decomposition reactions: CaCO3 = CaO ... Natural science work program for 5 Akclass Compiled by Working programm Skills and skills: problem solving on equations chemical reactions. 51 8 Calculations on thermochemical equations... - combined lesson - explanation ... Explanatory note, the work program in chemistry in grade 8 was compiled on the basis of: Federal component of the educational standard of basic general education in chemistry Explanatory note Oxygen production and properties. Computational tasks. Calculations on thermochemical equations... Topic 3. Hydrogen (3 hours) Hydrogen ... air Be able to: write down equations oxidation reactions; lead calculations on thermochemical equations; receive and collect ...

In order to compare energetic effects different processes, thermal effects are determined at standard conditions... A pressure of 100 kPa (1 bar), a temperature of 25 0 C (298 K), and a concentration of 1 mol / l are taken as standard. If the starting materials and reaction products are in a standard state, then the thermal effect of the chemical reaction is called standard enthalpy of the system and denoted ΔH 0 298 or ΔH 0 .

The equations of chemical reactions indicating the thermal effect are called thermochemical equations.

The thermochemical equations indicate the phase state and polymorphic modification of the reacting and formed substances: r - gas, g - liquid, k - crystalline, t - solid, p - dissolved, etc. If the aggregate states of substances for the reaction conditions are obvious, for example, O 2 , N 2 , H 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 ΔH, which in modern terminology is written next to the equation. For example:

WITH 6 N 6 (F) + 7.5O 2 = 6CO 2 + 3H 2 O (F) ΔH 0 = - 3267.7 kJ

N 2 + 3H 2 = 2NH 3 (D) ΔH 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 under consideration (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 heat effect of the 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 reaction of methane combustion can proceed according to the equation:

CH 4 + 2O 2 = CO 2 + 2H 2 O (G) ΔH 0 1 = -802.34 kJ

The same reaction can be carried out through the stage of CO formation:

CH 4 + 3 / 2O 2 = CO + 2H 2 O (G) ΔH 0 2 = -519.33 kJ

CO + 1 / 2O 2 = CO 2 ΔH 0 3 = -283.01 kJ

Moreover, it turns out that ΔH 0 1 = ΔН 0 2 + ΔH 0 3 ... Consequently, the heat 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 thermal effect of the direct 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 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 ( ΔH= 0). 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 the stoichiometric coefficients.

Here we meet with 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 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 ΔH 0 298 (or ΔH 0 ), are measured in kJ / mol and are given in reference books. The enthalpy of formation of simple substances stable at 298 K and a pressure of 100 kPa is taken to be zero.

In this case, a consequence of 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, it is possible to calculate the energy of a chemical bond, the energy of crystal lattices, the heats of combustion of fuels, the caloric 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 (D) and hydrogen, as a result of which CH 4 (D) and N 2 O (G) by calculating its thermal effect based on the data provided in the appendix. How much heat will be released in this reaction when 67.2 liters of methane are obtained in terms of standard conditions?

Solution.

CO 2 (D) + 3H 2 (D) = CH 4 (D) + 2H 2 O (G)

We find in the reference book (appendix) the standard heats of formation of the compounds participating in the process:

ΔH 0 (CO 2 (D) ) = -393.51 kJ / mol ΔH 0 (CH 4 (D) ) = -74.85 kJ / mol ΔH 0 (N 2 (D) ) = 0 kJ / mol ΔH 0 (N 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 these conditions, is zero. We calculate the thermal effect of the reaction:

ΔH (H.R.) = ∑ΔH 0 (cont.) -∑ΔH 0 (out.) =

ΔH 0 (CH 4 (D) ) + 2ΔH 0 (N 2 O (G) ) - ΔН 0 (CO 2 (D) ) -3ΔН 0 (N 2 (D) )) =

74.85 + 2 (-241.83) - (-393.51) - 30 = -165.00 kJ / mol.

The thermochemical equation is:

CO 2 (D) + 3H 2 (D) = CH 4 (D) + 2H 2 O (G) ; ΔH= -165.00 kJ

According to this thermochemical equation, 165.00 kJ of heat will be released when 1 mol is obtained, i.e. 22.4 liters of methane. The amount of heat released during the production 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 liter 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 from these data 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 the literature data (appendix).

Solution. We compose and equalize the chemical part of the required thermochemical equation:

WITH 2 N 4 (D) + 3O 2 (D) = 2CO 2 (D) + 2H 2 O (F) ;  N= ?

The created thermochemical equation describes 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: WITH 2 N 4 (D) + 3O 2 (D) = 2CO 2 (D) + 2H 2 O (F) ;  N= -1410.96 kJ

To calculate the enthalpy of formation WITH 2 N 4 (D) we draw a consequence from Hess's law: ΔH (H.R.) = ∑ΔH 0 (cont.) -∑ΔH 0 (out.).

We use the enthalpy of combustion of ethylene found by us and the enthalpies of formation of all (except for ethylene) participants in the process given in the appendix.

1410.96 = 2 (-393.51) + 2 (-285.84) - ΔH 0 (WITH 2 N 4 (D) ) - thirty

From here ΔH 0 (WITH 2 N 4 (D) ) = 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 (D) + 2O 2 (D) = CO 2 (D) + 2H 2 O (F) ΔH 1 = -890.31 kJ (1)

WITH (GRAPHITE) + O 2 (D) = CO 2 (D)  N 2 = -393.51 kJ (2)

N 2 (D) + ½O 2 (D) = H 2 O (F)  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 (D) = CH 4 (D)  N 4 =  N 0 (CH 4 (D)) ) =? (4)

You can operate with thermochemical equations in the same way as with algebraic ones. We should, as a result of algebraic operations with equations 1, 2 and 3, obtain equation 4. To do this, multiply equation 3 by 2, add the result with equation 2 and subtract equation 1.

2H 2 (D) + O 2 (D) = 2H 2 O (F)  N 0 (CH 4 (D) ) = 2  N 3 +  N 2 -  N 1

+ C (GRAPHITE) + O 2 (D) + CO 2 (D)  N 0 (CH 4 (D) ) = 2(-285,84)

- CH 4 (D) - 2O 2 (D) -CO 2 (D) - 2H 2 O (F) + (-393,51)

WITH (GRAPHITE) + 2H 2 (D) = CH 4 (D)  N 0 (CH 4 (D) ) = -74.88 kJ

This matches the value given in the appendix, which proves the correctness of our calculations.