Write the thermochemical equation for the reaction between CO (g) and hydrogen, resulting in the formation of CH4 (g) and H2O (g). How much heat will be released during this reaction if 67.2 liters of methane were obtained in terms of normal conditions

Answer: 618.48 kJ

Let's write the reaction equation:

CO (g) + 3H 2 (g) > CH 4 (g) + H 2 O (g)

Let's calculate the change in enthalpy of this reaction:

Thus, the equation becomes:

CO(g) + 3H2(g) > CH4(g) + H2O(g) + 206.16 kJ

This equation is valid for the formation of 1 mole or 22.4 liters (n.s.) of methane. When 67.2 liters or 3 mol of methane are formed, the equation takes the form:

• 3CO (g) + 9H 2 (g) > 3CH 4 (g) + 3H 2 O (g) + 618.48 kJ
• 3. Entropy decreases or increases during transitions: a) water into steam; b) graphite into diamond? Why? Calculate ?S°298 for each transformation. Draw a conclusion about the quantitative change in entropy during phase and allotropic transformations

Answer: a) 118.78 J/(mol K); b) - 3.25 J/(mol K)

a) When water transforms into steam, the entropy of the system increases.

In 1911, Max Planck proposed the following postulate: the entropy of a properly formed crystal of a pure substance at absolute zero is zero. This postulate can be explained by statistical thermodynamics, according to which entropy is a measure of the disorder of a system at the micro level:

where W is the number of different states of the system available to it under given conditions, or the thermodynamic probability of the macrostate of the system; R = 1.38.10-16 erg/deg - Boltzmann constant.

It is obvious that the entropy of the gas significantly exceeds the entropy of the liquid. This is confirmed by calculations:

H2O(l)< H2O(г)

• ?S°prot. = 188.72 - 69.94 = 118.78 J/mol*K
• b) When graphite transforms into diamond, the entropy of the system decreases, because the number of different states of the system decreases. This is confirmed by calculations:

Cgraph. > Salm.

S°prot. = 2.44 - 5.69 = -3.25 J/mol*K

Conclusion about the quantitative change in entropy during phase and allotropic transformations, since entropy characterizes the disorder of the system, then during allotropic transformations, if the system becomes more ordered (in this case, diamond is harder and stronger than graphite), then the entropy of the system decreases. During phase transformations: when a substance passes from a solid, liquid phase into a gaseous phase, the system becomes less ordered and entropy increases and vice versa.

Video tutorial 2: Calculations using thermochemical equations

Lecture: Thermal effect of a chemical reaction. Thermochemical equations

Thermal effect of a chemical reaction

Thermochemistry is a branch of chemistry that studies thermal, i.e. thermal effects of reactions.

As you know, each chemical element has n-amount of energy. We face this every day, because... Every meal our body stores energy from chemical compounds. Without this, we will not have the strength to move or work. This energy maintains a constant t of 36.6 in our body.

At the time of reactions, the energy of the elements is spent either on destruction or on the formation of chemical bonds between atoms. To break a bond, energy must be expended and energy must be released to form it. And when the energy released is greater than the energy expended, the resulting excess energy turns into heat. Thus:

The release and absorption of heat during chemical reactions is called thermal effect of reaction, and is designated by the letters Q.

Exothermic reactions– in the process of such reactions, heat is released and it is transferred to the environment.

This type of reaction has a positive thermal effect +Q. As an example, take the combustion reaction of methane:

Endothermic reactions– in the process of such reactions, heat is absorbed.

This type of reaction has a negative thermal effect -Q. For example, consider the reaction of coal and water at high t:

Thermochemical equations

The thermal effect of a reaction is determined using a thermochemical equation. How is it different? In this equation, next to the symbol of an element, its state of aggregation (solid, liquid, gaseous) is indicated. This must be done because The thermal effect of chemical reactions is influenced by the mass of the substance in the aggregate state. At the end of the equation, after the = sign, the numerical value of thermal effects in J or kJ is indicated.

As an example, a reaction equation is presented showing the combustion of hydrogen in oxygen: H 2 (g) + ½O 2 (g) → H 2 O (l) + 286 kJ.

The equation shows that 286 kJ of heat is released per 1 mole of oxygen and 1 mole of water formed. The reaction is exothermic. This reaction has a significant thermal effect.

When any compound is formed, the same amount of energy will be released or absorbed as is absorbed or released during its decomposition into primary substances.

Almost all thermochemical calculations are based on the law of thermochemistry - Hess's law. The law was derived in 1840 by the famous Russian scientist G. I. Hess.

Basic law of thermochemistry: the thermal effect of a reaction depends on the nature and physical state of the starting and final substances, but does not depend on the path of the reaction.

By applying this law, it will be possible to calculate the thermal effect of an intermediate stage of a reaction if the overall thermal effect of the reaction and the thermal effects of other intermediate stages are known.

Knowledge of the thermal effect of a reaction is of great practical importance. For example, nutritionists use them when preparing a proper diet; in the chemical industry, this knowledge is necessary when heating reactors and, finally, without calculating the thermal effect it is impossible to launch a rocket into orbit.

The concept of thermochemical reaction equations

Equations of chemical reactions in which the thermal effect is indicated are called thermochemical equations. The thermal effect is given as the value of the change in the enthalpy of the reaction AN. In thermochemical equations, unlike ordinary chemical equations, the aggregate states of substances must be indicated (liquid “liquid,” solid “solid,” or gaseous “g”). This is due to the fact that the same substance in different states of aggregation has different enthalpy. Therefore, a chemical reaction involving the same substances, but in a different state of aggregation, is characterized by a different thermal effect.

The thermal effect of a reaction in thermochemical equations is denoted in two ways:

1) indicate only the sign AN - if you just need to note whether the reaction is exo- or endothermic:

The change in enthalpy given in the thermochemical equation is the same part of the chemical equation as the formulas of substances, and therefore obeys the same relationships. For example, for the ethane combustion equation:

For other quantities of reactants or products, the amount of heat will change proportionally.

Often, to facilitate the use of thermochemical equations, the coefficients in them are reduced so that the formulas of the substances used for calculations are preceded by a coefficient of 1. Of course, in this case, other coefficients may turn out to be fractional, and it is necessary to proportionally reduce the value of the change in enthalpy. Thus, for the reaction of sodium with water given above, we can write the thermochemical equation:

Drawing up thermochemical reaction equations Example 1. When nitrogen reacts with 1 mole of a substance with oxygen to form nitrogen(N) oxide, 181.8 kJ of energy is absorbed. Write a thermochemical equation for the reaction.

Solution. Since energy is absorbed, AH is a positive number. The thermochemical equation would look like this:

Example 2. For the synthesis reaction of hydrogen iodide from simple gaseous substances AN = +52 kJ/mol. Write a thermochemical equation for the decomposition of hydrogen iodide to simple substances.

Solution. The reactions of hydrogen iodide synthesis and its decomposition are opposite reactions. Analyzing Figure 18.4, we can conclude that in this case the substances, and therefore their enthalpies, are the same. The only difference is which substance is the product of the reaction and which is the reactant. Based on this, we conclude that in opposite processes the ANs are identical in value, but different in sign. So, for the synthesis reaction of hydrogen iodide:

Since in practice the mass or volume of substances is measured, there is a need to compose thermochemical equations using precisely these data. Example. When liquid water weighing 18 g was formed, 241.8 kJ of heat was released from simple substances. Write a thermochemical equation for this reaction. Solution. Water weighing 18 g corresponds to the amount of substance n(H 2 O) = m / M = 18 g / 18 g / mol = 1 mol. And in the equation for the reaction of the formation of water from simple substances, the formula of water is preceded by a coefficient of 2. This means that in the thermochemical equation it is necessary to note the change in enthalpy when water is formed with an amount of substance of 2 mol, i.e. 241.8. 2 = 483.6:

Food labels must include data on their energy value, which is often called calorie content. For most people, information about the calorie content of foods makes them think: “How much weight will I gain if I eat this?” In fact, the numbers indicated on the label are the thermal effect of the reaction of complete combustion of 100 g of this product to carbon dioxide and water. This thermal effect is often given in outdated units of heat measurement - calories or kilocalories (1 cal = 4.18 J, 1 kcal = 4.18 kJ), which is where the term “calorie” comes from.

Key idea

Enthalpy change is a quantitative characteristic of the heat released or absorbed during a chemical reaction.

Assignments for mastering the material

210. What reaction equations are called thermochemical?

211. Determine which of the given thermochemical equations correspond to exothermic processes? endothermic processes?

212. Using the thermochemical equation for the synthesis of ammonia, calculate how much heat will be released: a) when nitrogen is consumed in an amount of 1 mole of substance; b) the formation of ammonia with an amount of substance of 2 mol. 1\1 2 (g) + 3H 2 (g) = 2NH 3 (n); DN = -92 kJ/mol.

213. The change in enthalpy of the combustion reaction of coal is 393.5 kJ/mol. Write a thermochemical equation for this reaction.

214. When methane was burned, 1 mole of the substance released 890 kJ of energy. Write a thermochemical equation for this reaction.

215. Ferrum(11) oxide is reduced by carbon(11) oxide to iron. This reaction is accompanied by the release of 1318 kJ of heat when 1 mole of iron is produced. Write a thermochemical equation for this reaction.

216. When hydrogen interacts with iodine, hydrogen iodide is formed with an amount of substance of 2 mol. In this case, 101.6 kJ of energy was absorbed. Write a thermochemical equation for this reaction.

217. Using the thermochemical equations in task 211, create thermochemical equations for the reactions: a) the formation of mercury(II) oxide from simple substances; b) decomposition of hydrogen chloride; c) the formation of glucose during photosynthesis.

218. During the combustion of carbon(I) oxide, 2 mol of a substance released 566 kJ of energy. Write a thermochemical equation for the reaction.

219. The decomposition of barium carbonate weighing 197 g requires 272 kJ of heat. Write a thermochemical equation for this reaction.

220. When iron weighing 56 g interacts with sulfur, 95 kJ of heat is released. Write a thermochemical equation for this reaction.

221. Compare the thermochemical equations given and explain the differences in enthalpy change:

222*. The change in the enthalpy of the reaction of neutralization of chloride acid with sodium hydroxide is -56.1 kJ/mol, and with potassium hydroxide - -56.3 kJ/mol. When nitrate acid reacts with lithium hydroxide, the enthalpy change is -55.8 kJ/mol. Why do you think the thermal effects of these reactions are almost the same?

This is textbook material

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 reaction equation 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.

Task 1.Thermochemical reaction equation

Gaseous ethyl alcohol can be obtained by the interaction of ethylene and water vapor. Write the thermochemical equation for this reaction, calculating its thermal effect. How much heat will be released if 10 liters of ethylene react at ambient conditions?

Solution: Let's create a thermochemical equation for the reaction:

C 2 H 4 (r) + H 2 O (r) = C 2 H 5 OH (r) DHhr = ?

According to the corollary of Hess's law:

DHhr = DH C2H5OH(r) - DH C 2 H 4(r) - DH H 2 O (r)

We substitute the values ​​of DN from the table:

DНхр = -235.31 – 52.28 – (-241.84) = -45.76 kJ

One mole of ethylene (no.) occupies a volume of 22.4 liters. Based on the corollary of Avogardo's law, we can create the proportion:

22.4 l C 2 H 4 ¾ 45.76 kJ

10 l C 2 H 4 ¾DНхр DНхр =20.43 kJ

If 10 liters of C 2 H 4 react, then 20.43 kJ of heat is released.

Answer: 20.43 kJ of heat.

Problem 2. Determination of reaction enthalpy
Determine the change in enthalpy of a chemical reaction and its thermal effect.
2NaOH + H 2 SO 4 = Na 2 SO 4 + 2H 2 O
Solution:
Using the reference book, we determine the enthalpies of formation of the components.
ΔH 0 (NaOH) = -426 kJ/mol.
ΔH 0 (H 2 SO 4) = -813 kJ/mol.
ΔH 0 (H 2 O) = -285 kJ/mol.
ΔH 0 (Na 2 SO 4) = -1387 kJ/mol.
Based on the corollary of Hess’s law, we determine the change in the enthalpy of the reaction:
ΔHх.р. = [ΔH(Na 2 SO 4) + 2ΔH(H 2 O)] - [ΔH(H 2 SO 4) + 2ΔH(NaOH)] =
= [-1387 + 2(-285)] - [-813 + 2(-426)] = - 1957 - (-1665) = - 292 kJ/mol.
Let's define the thermal effect:
Q = - ΔHх.р. = 292 kJ.
Answer: 292 kJ.
Task 3.Lime slaking is described by the equation: CaO + H 2 O = Ca (OH) 2.
ΔHх.р. = - 65 kJ/mol. Calculate the heat of formation of calcium oxide if ΔH 0 (H 2 O) = -285 kJ/mol,
ΔH 0 (Ca(OH) 2) = -986 kJ/mol.
Solution:
Let us write according to Hess's law:
ΔHх.р. = ΔH 0 (Ca(OH) 2) - ΔH 0 (H 2 O) - ΔH 0 (CaO)
From here,
ΔH0(CaO) = ΔH 0 (Ca(OH) 2) - ΔH 0 (H 2 O) - ΔHх.р. = - 986 - (-285) - (-65) = - 636 kJ/mol.

Answer: - 636 kJ/mol.

Task 4.Calculate the enthalpy of formation of zinc sulfate from simple substances at T = 298 K based on the following data:
ZnS = Zn + S ΔH 1 = 200.5 kJ
2ZnS + 3O 2 = 2ZnO + 2SO 2 ΔH 2 = - 893.5 kJ
2SO 2 + O 2 = 2SO 3 ΔH 3 = - 198.2 kJ
ZnSO 4 = ZnO + SO 3 ΔH 4 = 235.0 kJ
Solution:
It follows from Hess's law that since the path of transition is not important, the calculations follow the algebraic rules of working with ordinary equations. In other words, they can be “shuffled” however you like. Let's try to use this opportunity.
We need to arrive at the equation:
Zn + S + 2O 2 = ZnSO 4.
To do this, we will arrange the available “material” so that Zn, S, O 2 are on the left, and zinc sulfate is on the right. Let's flip the first and fourth equations from left to right, and in the second and third we divide the coefficients by 2.
We get:
Zn + S = ZnS
ZnS + 1.5O 2 = ZnO + SO 2
SO 2 + 0.5O 2 = SO 3
ZnO + SO 3 = ZnSO 4.
Now let's simply add up the right parts and the left parts.
Zn + S + ZnS + 1.5O 2 + SO 2 + 0.5O 2 + ZnO + SO 3 = ZnS + ZnO + SO 2 + SO 3 + ZnSO 4
That it will be equal
Zn + S + 2O 2 + ZnS + SO 2 + SO 3 + ZnO = ZnS + SO 2 + SO 3 + ZnO+ ZnSO 4

Apparently, yes, what happens? All underlined reduce (again, pure arithmetic!)
And in the end we have
Zn + S + 2O 2 = ZnSO 4 - as required.
Now we apply the same principle to enthalpies. The first and fourth reactions were reversed, which means that the enthalpies will receive the opposite sign. We divide the second and third in half (since we divided the coefficients).
ΔH = - 200.5 + (-893.5/2) + (-198.2/2) + (-235.0) = - 981.35 kJ/mol.
Answer:- 981.35 kJ/mol.

Task 5.Calculate the enthalpy of the reaction of complete oxidation of ethyl alcohol to acetic acid if the enthalpy of formation of all substances participating in the reaction is equal to:

∆Нº arr. C 2 H 5 OH l = - 277 kJ/mol;

∆Нº arr. CH 3 COOH w = - 487 kJ/mol;

∆Нº arr. H 2 O w = - 285.9 kJ/mol;

∆Нº arr. O 2 = 0

Solution: Ethyl alcohol oxidation reaction:

C 2 H 5 OH + O 2 = CH 3 COOH + H 2 O

From Hess’s law it follows that ∆Н r-tion = (∆Нº sample CH 3 COOH + ∆Нº sample H 2 O) –

(∆Hº sample C 2 H 5 OH + ∆Hº sample O 2) = - 487 – 285.9 + 277.6 = - 495.3 kJ.

Task 6.Calorific value determination

Calculate the heat of combustion of ethylene C 2 H 4 (g) + 3O 2 = 2CO 2 (g) + 2H 2 O (g) if its heat of formation is 52.3 kJ/mol. What is the thermal effect of combustion of 5 liters. ethylene?
Solution:
Let us determine the change in the enthalpy of the reaction according to Hess's law.
Using the reference book, we determine the enthalpies of formation of the components, kJ/mol:
ΔH 0 (C 2 H 4 (g)) = 52.
ΔH 0 (CO 2 (g)) = - 393.
ΔH 0 (H 2 O (g)) = - 241.
ΔHх.р. = - = -1320 kJ/mol.
The amount of heat released during the combustion of 1 mole of ethylene Q = - ΔHх.р. = 1320 kJ
The amount of heat released during combustion of 5 liters. ethylene:
Q1 = Q * V / Vm = 1320 * 5 / 22.4 = 294.6 kJ.
Answer: 294.6 kJ.

Task 7.Equilibrium temperature
Determine the temperature at which equilibrium of the system occurs:
ΔHх.р. = + 247.37 kJ.
Solution:
The criterion for the possibility of a chemical reaction occurring is the Gibbs energy, ΔG.
ΔG< 0, реакция возможна.
ΔG = 0, threshold of possibility.
ΔG > 0, reaction is impossible.
The Gibbs energy is related to enthalpy and entropy by the relation:
ΔG = ΔH - TΔS.
Hence, for equilibrium to occur (to reach the threshold), the following relation must be satisfied:
T = ΔH/ΔS
Let us determine the change in entropy by a consequence of Hess’s law.
CH 4 (g) + CO 2 (g) = 2CO (g) + 2H 2 (g)
ΔS 0 h.r. = -
Having written out from the reference book accordingly. values, we solve:
ΔS 0 h.r. = (2*198 + 2*130) - (186 + 213) = 656 - 399 = 257 J/mol*K = 0.257 kJ/mol*K.
T = ΔH/ΔS = 247.37/0.257 = 963 o K.
Answer: 963 about K.

Task 8.Entropy change sign

Without making calculations, determine the sign of the change in the entropy of the processes:
1. H 2 O(g) ---> H 2 O(l)
2. 2H 2 S + O 2 = 2S(solid) + 2H 2 O(l)
3. (NH 4) 2 CO 3 (solid) = 2NH 3 + CO 2 + H 2 O (all products are gaseous).

Solution:
Since entropy is a measure of the disorder of a system, the general rule holds:
S(TV)< S(жидкость) < S(газ).
In light of this, let us analyze the problem.
1. Liquid condenses from gas.
Since S(liquid)< S(газ), ΔS < 0.
2. From 3 moles of gases 2 moles of solids are obtained. substance and 2 moles of liquid.
It is obvious that ΔS< 0.
3. Gases are obtained from solid matter.
Since S(tv.)< S(газ), ΔS > 0.

Task 9.Process capability

Conditions specified:
1. ΔS< 0, ΔH < 0
2. ΔS< 0, ΔH > 0
3. ΔS > 0, ΔH< 0
4. ΔS > 0, ΔH > 0
Analyze the possibility of the reaction occurring.
Solution:
In the solution we will rely on the formula: ΔG = ΔH - TΔS. (For more details, see task No. 7).
1. At ΔS< 0, ΔH < 0.
The first term of the formula (ΔH) is less than zero, and the second, due to the negative sign of entropy, is greater than zero
(-T(-ΔS) = +TΔS) . The possibility of a reaction will be determined by the ratio of the values ​​of the first and second terms. If the enthalpy value (modulo) is greater than the product TΔS, (|ΔH| > |TΔS|), i.e. in general, the Gibbs energy will be less than zero, the reaction is possible.
2. ΔS< 0, ΔH > 0.
Both the first and second terms are greater than zero. Gibbs energy is greater than zero. No reaction possible.
3. ΔS > 0, ΔH< 0.
The first term is less than zero, the second is also less. The Gibbs energy is less than zero, the reaction is possible.
4. ΔS > 0, ΔH > 0
The first term of the formula (ΔH) is greater than zero, and the second, due to the positive sign of entropy, is greater than zero
(-T(+ΔS) = - TΔS) . The possibility of a reaction will be determined by the ratio of the values ​​of the first and second terms. If the enthalpy value (modulo) is greater than the product TΔS, (|ΔH| > |TΔS|), i.e. in general, the Gibbs energy will be greater than zero, the reaction is impossible. However, with increasing temperature, the second term will increase (in absolute value), and beyond a certain temperature limit the reaction will become possible.
Answer: 1 – possible; 2 - impossible.; 3 – possible; 4 – possible.
Problem 10.Based on the standard heats of formation and absolute standard entropies of the corresponding substances, calculate DG o 298 of the reaction CO (g) + H 2 O (l) = CO 2 (g) + H 2 (g) Is this reaction possible under standard conditions?

Solution: DG o is determined from the equation DG o =DH o -TDS ​​o

DHхр = DH CO2 - DH CO - DH H2O (l) == -393.51 – (110.52) – (-285.84) = -218.19 kJ.

DSхр = S CO2 + S H2 - S CO – S H2O (l) = = 213.65+130.59–197.91–69.94=76.39 J/mol×K

or 0.07639 kJ.

DG = -218.19 – 298 × 0.07639 = -240.8 kJ

DG<0, значит реакция возможна.

Answer: a reaction is possible.

Options for test tasks

Option 1

1. How to calculate the change in the Gibbs energy in a reaction based on the thermodynamic characteristics of the starting materials and reaction products?

2. Calculate the thermal effect of the reduction reaction of iron (II) oxide with hydrogen based on the following thermochemical equations:

FeO(k) + CO(g) = Fe(k) + CO 2 (g); ∆Н 1 = -13.18 kJ;

CO (g) + O 2 (g) = CO 2 (g); ∆H 2 = -283.0 kJ;

H 2 (g) + O 2 (g) = H 2 O (g); ∆H 3 = -241.83 kJ.

Answer: +27.99 kJ.

Option 2

1. What are the thermodynamic conditions for the spontaneous occurrence of a chemical reaction?

2. Gaseous ethyl alcohol C 2 H 5 OH can be obtained by the interaction of ethylene C 2 H 4 (g) and water vapor. Write the thermochemical equation for this reaction, having first calculated its thermal effect. Answer:-45.76 kJ.

Option 3

1. What is called a thermochemical equation? Why is it necessary to indicate the state of aggregation of substances and their polymorphic modifications?

2. Crystalline ammonium chloride is formed by the interaction of gaseous ammonia and hydrogen chloride. Write the thermochemical equation for this reaction, having first calculated its thermal effect. How much heat will be released if 10 liters of ammonia were consumed in the reaction, calculated under normal conditions? Answer: 78.97 kJ.

Option 4

1. What are the two systems of symbols for thermal effects?

2. The thermal effect of the combustion reaction of liquid benzene 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.

Option 5

1. What is the standard heat (enthalpy) of formation of a compound? What conditions are called standard?

2. Write the thermochemical equation for the reaction between CO(g) and hydrogen, as a result of which CH 4 (g) and H 2 O(g) are formed. How much heat will be released during this reaction if 67.2 liters of methane were obtained in terms of normal conditions? Answer: 618.48 kJ.

Option 6

1. Formulate Hess’s law and a consequence of this law. What is the relationship between Hess's law and the law of conservation of energy?

2. The reduction of Fe 3 O 4 with carbon monoxide follows the equation

Fe 3 O 4 (k) + CO (g) = 3FeO (k) + CO 2 (g).

Calculate ∆G 0 298 and draw a conclusion about the possibility of spontaneous occurrence of this reaction under standard conditions. What is ∆S 0 298 equal to in this process? Answer:+24.19 kJ; +31.34 J/K.

Option 7

1. In what direction do chemical reactions spontaneously occur? What is the driving force of a chemical process?

2. The combustion of 11.5 g of liquid ethyl alcohol released 308.71 kJ of heat. Write the thermochemical equation for the reaction that results in the formation of water vapor and carbon dioxide. Calculate the heat of formation of C 2 H 5 OH (l). Answer: -277.67 kJ.

Option 8

1. What is the isobaric-isothermal potential of a chemical reaction and how is it related to the change in enthalpy and entropy of the reaction?

2. The thermal effect of the reaction is –560.0 kJ. Calculate Standard Heat of Formation .Answer: 83.24 kJ/mol.

Option 9

1. What is the entropy of a reaction?

2. Based on the values ​​of the standard heats of formation and the absolute standard entropies of the corresponding substances, calculate ∆G 0 298 of the reaction occurring according to the equation NH 3 (g) + HCl (g) = NH 4 Cl (k). Can this reaction occur spontaneously under standard conditions? Answer: -92.08 kJ.

Option 10

1. How does entropy change with increasing movement of particles in the system?

2. Using values reactants, calculate reaction and determine whether it can occur under standard conditions.

Option 11

1. Basic Concepts of Thermodynamics: system, phase, types of systems, parameters of the state of systems, types of processes.

2. Determine the enthalpy of the reaction of alcoholic fermentation of glucose

C 6 H 12 O 6 2C 2 H 5 OH + 2CO 2

enzymes

∆Hº 298 (C 6 H 12 O 6) = - 1273.0 kJ/mol

∆Hº 298 (C 2 H 5 OH) = - 1366.91 kJ/mol

∆Hº 298 (CO 2) = - 393.5 kJ/mol

Option 12

1. The first law of thermodynamics for isochoric and isobaric processes. Enthalpy.

2. Determine the enthalpy of the reaction: NH 3 (g) + HCl (g) = NH 4 Cl (T)

∆Нº 298 (НCl) = - 92.3 kJ/mol

∆Нº (NН 3) = - 46.2 kJ/mol

∆Нº (NH 4 Cl) = - 313.6 kJ/mol

Option 13

1. Thermochemistry: exo- and endothermic reactions. Thermochemical equations, their features.

2. Determine which of these reactions is exo- and which is endothermic? Justify your answer.

N 2 + O 2 D 2NO ∆H = + 80 kJ

N 2 + 3H 2 D 2NO 3 ∆Н = - 88 kJ

Option 14

1.What are the system parameters? What parameters do you know?

2. Calculate the enthalpy of formation of gaseous sulfuric anhydride if the combustion of 16 g of sulfur released 197.6 kJ of heat.

Option 15

1. List the functions of the system state.

4HCl (g) + O 2 (g) ↔ 2H 2 O (g) + 2Cl 2 (g); ∆H = -114.42 J.

Is chlorine or oxygen the stronger oxidizing agent in this system and at what temperature? Answer: 891K.

Option 16

1. What types of thermodynamic processes do you know?

2. How can we explain that under standard conditions the exothermic reaction H 2 (g) + CO 2 (g) = CO (g) + H 2 O (l); ∆H = -2.85 kJ. Knowing the thermal effect of the reaction and the absolute standard entropies of the corresponding substances, determine ∆G 0 298 of this reaction. Answer: -19.91 kJ.

Option 17

1. Hess's law and consequences arising from it.

2. Identify systems. Answer: 160.4 J/(mol K).

Option 18

1. How does the enthalpy of formation of a substance differ from the enthalpy of reaction?

2. Calculate ∆H 0 ,∆S 0 ,∆G 0 T of the reaction proceeding according to the equation Fe 2 O 3 (k) + 3H 2 (g) = 2Fe (k) + 2H 2 O (g). Is the reduction reaction of Fe 2 O 3 with hydrogen possible at 500 and 2000 K? Answer: +96.61 kJ; 138.83 J/K; 27.2 kJ; -181.05 kJ.

Option 19

2. The thermal effect of which reaction is equal to the heat of methane formation? Calculate the heat of formation of methane based on the following thermochemical equations:

H 2 (g) + O 2 (g) = H 2 O; ∆H 1 = -285.84 kJ;

C(k) + O 2 (g) = CO 2 (g); ∆H 2 = -393.51 kJ;

CH 4 (g) + 2O 2 (g) = 2H 2 O (l) + CO 2 (g); ∆H 3 = -890.31 kJ.

Answer: -74.88 kJ.

Option 20

1. What processes are accompanied by an increase in entropy?

2. After counting the reactions, determine which of the two reactions is thermodynamically possible: ; .

Option 21

1. What is the standard enthalpy of formation?

2. Based on the standard heats of formation and absolute standard entropies of the corresponding substances, calculate ∆G 0 298 of the reaction proceeding according to the equation CO 2 (g) + 4H 2 (g) = CH 4 (g) + 2H 2 O (l). Is this reaction possible under standard conditions? Answer: -130.89 kJ.

Option 22

1. What is the sign of ∆ G of the ice melting process at 263 K?

2. Entropy decreases or increases during the transition of a) water into steam; b) graphite into diamond? Why? Calculate ∆S 0 298 for each transformation. Draw a conclusion about the quantitative change in entropy during phase and allotropic transformations. Answer: a) 118.78 J/(mol∙K); b) -3.25 J/(mol∙K).

Option 23

1. What is the sign of ∆ H of the coal combustion process?

2. Under standard conditions, the reaction proceeds spontaneously. Determine the signs of ∆Nor ∆S in this system.

Option 24

1. What is the sign of ∆ S of the “dry ice” sublimation process?

2. Calculate ∆H O, ∆S O, ∆G O T of the reaction proceeding according to the equation TiO 2 (k) + 2C (k) = Ti (k) + 2CO (g). Is the reduction reaction of TiO 2 with carbon possible at 1000 and 3000 K? Answer:+722.86 kJ; 364.84 J/K; +358.02 kJ; -371.66 kJ.

Option 25

1. What is the sign of the change in entropy during the boiling process of water?

2. Find the change in internal energy during the evaporation of 75 g of ethyl alcohol at the boiling point, if the specific heat of evaporation is 857.7 J/g, and the specific volume of steam at the boiling point is 607 cm 3 /g. Neglect the volume of liquid. Answer: 58.39 kJ.

Option 26

1. II law of thermodynamics. Carnot-Clausius theorem.

2. Calculate the thermal energy consumption during the reaction if 336 g of iron were obtained. Answer: –2561.0 kJ.

Option 27

1. III law of thermodynamics.

2. The combustion reaction of acetylene proceeds according to the equation

C 2 H 2 (g) + O 2 (g) = 2CO 2 (g) + H 2 O (l)

Calculate ∆G 0 298 and ∆S 0 298. Explain the decrease in entropy as a result of this reaction. Answer: -1235.15 kJ; -216.15 J/(mol∙K).

Option 28

1. Nernst's theorem.

2. When ammonia gas burns, water vapor and nitrogen oxide are formed. 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.

Option 29

1. Planck's postulate.

2. At what temperature will the system reach equilibrium?

CH 4 (g) + CO 2 (g) ↔ 2CO (g) + 2H 2 (g); ∆Н = +247.37 kJ?

Option 30

1. Basics of thermodynamic calculations

2. Having calculated the thermal effect and the change in Gibbs energy at 25ºC for the reaction, determine for this reaction. Answer: –412.4 J/(mol K).

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