Enthalpy calculation. How to find the enthalpy of a reaction. Initial knowledge of students

Example 1. Calculation of enthalpies of formation of substances and thermal effects of chemical processes

1. The standard enthalpy of formation of gaseous ozone is 142.3 kJ/mol. Indicate in which of the following reactions the thermal effect ΔH 0 arr will correspond to the standard enthalpy of formation O 3 (g):

a) 3O (g) \u003d O 3 (g); b) 1.5O 2 (g) \u003d O 3 (g); c) O 2 (g) + O (g) \u003d O 3 (g); d) 2O 2 (g) \u003d O (g) + O 3 (g).

Write the thermochemical equation for this process.

Solution. By definition, ΔH 0 shows the thermal effect of the reaction of formation of 1 mol of substance O 3 (g) by direct synthesis from simple substances that are stable at 298K and a pressure of 101kPa. Case d) does not fit this definition, because does not correspond to the synthesis reaction. Case c) does not meet the requirements of the standard conditions, because at 298 K and a pressure of 101 kPa, oxygen cannot be in a liquid state. Case a) must also be excluded, because atomic oxygen O(g) is not a stable form of existence of a simple oxygen substance. Thus, only reaction b) meets the requirements of the definition. The thermochemical equation will look like this:

O 2 (g) + ½O 2 (g) \u003d O 3 (g); ΔH 0 arr [O 3 (g)] = 142.3 kJ / mol.

Answer: the conditions of the problem correspond to equation b); ΔH 0 arr [O 3 (g)] = 142.3 kJ / mol.

2. During the combustion of calcium weighing 8 g, the amount of energy released was 127 kJ. Write a thermochemical equation for this reaction.

Solution. Let us first compose the chemical equation for the combustion reaction of a metal in oxygen: 2Са + О 2 = 2СаО. The thermochemical equation differs from the chemical one in that it indicates the aggregate states of the reactants and products, as well as the thermal effect of the process. Therefore, our case will correspond to the thermochemical equation of the following form:

2Ca(t) + O 2 (g) = 2CaO(t), ΔH = X kJ.

In this equation, the desired thermal effect corresponds to 2 moles of calcium.

And according to the condition of the problem, 8 g of calcium entered into the reaction, which corresponds to the amount of substance n Ca = m(Ca)/M(Ca); nCa \u003d 8g / 40g / mol \u003d 0.2 mol. We calculate the amount of energy that will be released during the combustion of 2 mol of calcium, using the proportion: 0.2 mol of Ca - -127 kJ

2 mol Ca - X kJ X\u003d 2mol (-127kJ) / 0.2mol \u003d -1270kJ.

Thus, during the combustion of 2 mol of metallic calcium, 1270 kJ of energy is released.

Answer: 2Ca(t) + O 2 (g) = 2CaO(t), ΔH = –1270 kJ.

3. Calculate the standard enthalpy of formation of benzene C 6 H 6 (l), if the enthalpies of combustion of hydrogen, carbon and benzene are equal, respectively (kJ / mol):

285,84; -393,51; -3267,70.

Solution. Let us write the reaction equation, the thermal effect of which must be determined. The formation of benzene from simple substances can be represented by the following thermochemical equation:

6C (t) + 3H 2 (g) \u003d C 6 H 6 (g), ΔH 0 arr [C 6 H 6 (g)] = X kJ/mol .

To determine the enthalpy of formation of benzene, we compose a Hess cycle using these problems:

H 2 (g) + ½O 2 (g) \u003d H 2 O (g), ΔH 0 1 \u003d -285.84 kJ / mol; (1)

C (t) + O 2 (g) \u003d CO 2 (g), ΔH 0 2 \u003d -393.51 kJ / mol; (2)

C 6 H 6 (l) + 15 / 2O 2 (g) \u003d 6CO 2 (g) + 3H 2 O (l), ΔH 0 3 \u003d -3267.70 kJ / mol. (3)

To obtain the desired equation for the formation of benzene from simple substances, it is enough to add equations (1) and (2) in the Hess cycle, multiplying them by the corresponding coefficients 3 and 6, and subtract equation (3) from them:

3H 2 (g) + 3 / 2O 2 (g) + 6C (t) + 6O 2 (g) -C 6 H 6 (l) -15 / 2O 2 (g) \u003d 3H 2 O (l) + 6CO 2 (g) -6CO 2 (g) -3H 2 O (g).

We reduce the homogeneous terms and transfer -C 6 H 6 (g) to the right side of the equality with the opposite sign. We get the desired equation: 6C (t) + 3H 2 (g) \u003d C 6 H 6 (g).

We will do similar actions with thermal effects:

ΔH 0 arr [C 6 H 6 (l)] = 3ΔH 0 1 + 6ΔH 0 2 - ΔH 0 3,

ΔH 0 arr [C 6 H 6 (l)] \u003d 3 (-285.84) kJ / mol + 6 (-393.51) kJ / mol - (-3267.70) kJ / mol \u003d

(-857.52 -2361.06 + 3267.70) kJ/mol = 49.12 kJ/mol.

Answer: ΔH 0 arr [C 6 H 6 (g)] = 49.12 kJ/mol.

4. Using the value of the standard enthalpies of formation of the participants in the chemical process, determine the amount of energy released when 100 kg of lead dioxide is reduced by carbon monoxide to oxide with the formation of carbon dioxide.

Solution. Let us write the thermochemical reaction equation, the thermal effect of which must be determined:

PbO 2 (t) + CO (g) \u003d PbO (t) + CO 2 (g), ΔH 0 \u003d X kJ/mol.

According to the 3rd corollary of the Hess law, the thermal effect of the process is determined by the difference between the sums of the enthalpies of formation of products and reactants. Using the data in Table 1 of the Appendix, we find the values ​​of the desired enthalpies (kJ / mol):

ΔH 0 arr = -276.86; ΔH 0 arr = -110.50; ΔH 0 arr = - 217.86;

ΔH 0 arr = -393.51.

Let's compose a Hess cycle to calculate the heat effect of the reduction of lead dioxide with carbon monoxide:

ΔH 0 \u003d (ΔH 0 arr + ΔH 0 arr) - (ΔH 0 arr + ΔH 0 arr)

and, substituting the known values, calculate it:

ΔH 0 \u003d [(- 217.86) kJ / mol + (-393.51) kJ / mol] - [(-276.86) kJ / mol + (-110.50) kJ / mol] \u003d -224, 01 kJ/mol.

Calculations have shown that 224.01 kJ of energy is released during the reduction of 1 mole of PbO 2 . Let us determine what mass of lead dioxide corresponds to this amount of substance:

m (PbO 2) \u003d n M (PbO 2) \u003d 1 mol (207 + 2 16) g / mol \u003d 239 g.

Now we calculate the amount of energy that will be released during the combustion of 100 kg of PbO 2 using the proportion:

239 10 -3 kg - -224.01 kJ

100kg - X kJ, X= = -93728 kJ.

Answer: when reducing 100 kg of lead dioxide, 93728 kJ of energy is released.

5. The C–H bond energy in methane is 435 kJ/mol, the C–Cl bond energy in chloromethane is 350 kJ/mol. Knowing that E Cl - Cl \u003d 240 kJ / mol, and E H - Cl \u003d 430 kJ / mol, calculate the thermal effect (ΔH 0) of the reaction: CH 4 + Cl 2 → CH 3 Cl + HCl.

Solution. The enthalpies of formation of chemical substances can be calculated from the bond energies that are formed in these substances. For example, 4 C–H covalent bonds are formed in a methane molecule, therefore,

ΔH 0 arr (CH 4) \u003d 4 , and ΔH 0 arr (CH 4) \u003d 4 435 kJ / mol \u003d 1740 kJ / mol. We will carry out similar calculations for all other molecules:

ΔH 0 arr (Cl 2) \u003d 3 [E Cl - Cl] \u003d 3 240 kJ / mol \u003d 720 kJ / mol (see the theory of the formation of a dative bond, in the Cl 2 molecule - a triple bond);

ΔH 0 arr (HCl) = E H - Cl = 430 kJ/mol;

ΔH 0 arr (CH 3 Cl) \u003d 3 + E C - Cl \u003d 3 435 kJ / mol + 350 kJ / mol \u003d 1655 kJ / mol.

Now, according to the 3rd consequence from the Hess law, we calculate the heat effect of the desired reaction: ΔH 0 \u003d [ΔH 0 arr (CH 3 Cl) + ΔH 0 arr (HCl)] - [ΔH 0 arr (CH 4) + ΔH 0 arr (Сl 2)] and

ΔH 0 \u003d (1655 + 430) - (1740 + 720) kJ / mol \u003d -375 kJ / mol.

Answer: the thermal effect of the exothermic process of methane chlorination with the formation of chloromethane is ΔH 0 = -375 kJ / mol.

Example 2. Determination of the relationship between internal energy and enthalpy of thermodynamic processes

1. The change in the internal energy of the Fe(t) + Cl 2 (g) = FeCl 2 (t) system is -334.0 kJ. Determine the thermal effect of this reaction under standard conditions.

Solution. According to the first law of thermodynamics, the heat effect of a chemical reaction at a constant temperature ΔH T is associated with a change in the internal energy of the system ΔU by the equation ΔH T = ΔU ± RTΔn. In this equation, the change in the amount of substance Δn is determined only by substances that are in the least condensed phase, in our case, in the gaseous phase. Since there are no gaseous substances in the reaction products, then Δn \u003d 0 - 1 mol (Cl 2) \u003d -1 mol.

Under standard conditions, T 0 \u003d 298 K, R \u003d 8.31 10 -3 kJ / mol K. Substituting these and the found values ​​into the equation for ΔH 0 T, we find the thermal effect of the reaction for the synthesis of iron (2) chloride:

ΔH 0 x.r. \u003d -334 kJ - (8.31 10 -3 kJ / mol K) 298 K 1 mol \u003d -336.5 kJ.

Answer: ΔH 0 x.r. = -336.5 kJ.

2. Calculate the change in internal energy during the evaporation of 50 g of ethanol at the boiling temperature, if its specific heat of evaporation is 857.7 J/g, and the specific vapor volume at the boiling temperature is 607 10 -3 l/g. The volume of the liquid can be neglected.

Solution. The process of evaporation (transition of a liquid substance into a gaseous state) is physical, it occurs at constant pressure and constant temperature (if the substance is chemically pure). For such a process (occurring, as a rule, at constant pressure), the relationship between the change in the total ΔH P and internal ΔU energy of the thermodynamic system, according to the first law of thermodynamics, obeys the equation ΔH P = ΔU ± PΔV. Since the volume of the system increases in this case, ΔV > 0 and the equation is simplified: ΔH P = ΔU + PΔV.

The change in the volume of the system ΔV will be equal to the volume of the formed steam, taking into account the conditions of the problem. If the specific volume of gaseous alcohol v at the boiling point is equal to 607·10 -3 l/g, then the change in volume during the vaporization of 50 g of alcohol can be easily calculated by the equation ΔV=v·m; ΔV \u003d 607 10 -3 (l / g) 50 (g) \u003d 3035 10 -2 (l) \u003d 30.35 l.

The enthalpy effect ΔH 0 during the phase transition under standard conditions is determined by the formula ΔH 0 =L·m, where L is the specific heat of vaporization. Substituting the values ​​from the condition of the problem, we will make the corresponding calculations ΔH 0:

ΔH 0 \u003d 857.7 (J / g) 50 (g) \u003d 42885 J \u003d 42885 kPa l.

Transforming the thermodynamic equation for ΔH 0 relative to ΔU 0 and solving it, we get: ΔU 0 = ΔH 0 - PΔV; ΔU 0 = 42885kPa l - 101kPa 30.35l = 39820kPa l = 39820J = 39.82kJ.

Answer: the internal energy of the thermodynamic system has increased by 39.82 kJ.

3 . Calculate the thermal effect and change in the internal energy of the reaction system during the reduction of iron oxide (2) with hydrogen, if the thermal effects of the following reactions are known: FeO(t) + CO(g) = Fe(t) + CO 2 (g), ΔH 1 = - 13.18 kJ; (1)

CO (g) + ½O 2 (g) \u003d CO 2 (g), ΔH 2 \u003d -283.00 kJ; (2)

H 2 (g) + ½O 2 (g) \u003d H 2 O (g), ΔH 3 \u003d -241.83 kJ (3).

Solution. According to Hess's law, to obtain the desired reaction equation, one can perform arithmetic operations with the equations given in the problem. The same can be done with thermal effects.

Therefore, to get the equation

FeO (t) + H 2 (g) \u003d Fe (t) + H 2 O (g), ΔH \u003d X kJ,

you need to sum equations (1) and (3) and subtract equation (2) from this sum. We perform the same action with thermal effects. Then the thermal effect of the reduction reaction of iron oxide (2) with hydrogen is determined by the formula:

∆H = ∆H 1 + ∆H 3 - ∆H 2 .

Substituting the known values ​​​​into this formula and making calculations, we get:

ΔH = - 13.18 kJ + (-241.83 kJ) - (-283.00 kJ) = 27.99 kJ.

To determine the change in the internal energy of the system for a given process, we apply the first law of thermodynamics ΔH = ΔU ± RTΔn. The calculation of the change in the amount of substance of gaseous products after (H 2 O) and before (H 2) reactions show that Δn = 0. Then the equation relating ΔU and ΔH is simplified: ΔH = ΔU. This means that the reduction process is endothermic and the internal energy of the system increases by 27.99 kJ.

Answer: ΔH = ΔU = 27.99 kJ.

4. The internal energy during the evaporation of 90 g of water at 100 0 C increased by 188.1 kJ. The specific volume of water vapor is 1.699 l / g, the pressure is 1.01 10 5 Pa. Determine the heat of vaporization of water (kJ/mol).

Solution. For the vaporization process

H 2 O (l)<=>H 2 O (g), ΔH = X kJ/mol,

the relationship between the heat of vaporization ΔH and the change in internal energy ΔU of the system at constant pressure (Р = Const) is expressed by the equation ΔH = ΔU ± PΔV, where ΔV = V Н2О(g) – V Н2О(l) > 0, because V H2O (g) > V H2O (l). Given this conclusion, the equation will be simplified: ΔH = ΔU + PΔV.

Knowing the specific volume of water vapor under given conditions (v) and the mass of water (m), we find: V H2O (g) = vm; V H2O (g) \u003d 1.699 (l / g) 90 (g) \u003d 152.91 l. Since the density of liquid water is also known (ρ Н2О(l) = 1 10 -3 g/l), we find the volume of liquid water using the formula

V H2O (l) \u003d ρm and V H2O (l) \u003d 1 10 -3 (g / l) 90 (g) \u003d 0.09 l.

Taking into account these values, the change in volume during the evaporation of 90 g of water ΔV will be:

ΔV \u003d 152.91l - 0.09l \u003d 152.82l.

Substituting the found value ΔV, as well as these conditions of the problem into the expression for ΔH, we determine the heat of evaporation of 90 g of water:

ΔH P \u003d 188.1 kJ + 1.01 10 5 (10 -3 kPa) 152.82 (10 -3 m 3) \u003d 188.1 kJ + 15.43 kJ \u003d 203.53 kJ.

Based on 1 mole of the resulting steam, this value will be: ΔH = ΔH P ·M/m, where M is the molar mass of water. Then ∆H = = 40.71 kJ/mol.

Answer: The heat of vaporization of water at its boiling point is 40.71 kJ / mol.

5. The dissolution of 130 g of metallic zinc in dilute sulfuric acid at 20 0 C is accompanied by the release of 286.2 kJ of energy. The gaseous hydrogen released in this process does work against external pressure. Determine the change in the internal energy of this process.

Solution. For the chemical reaction Zn + H 2 SO 4 = ZnSO 4 + H 2

the relationship between the thermal effect of the process (ΔH) and the change in the internal energy of the system (ΔU) obeys the equation of the first law of thermodynamics ΔH = ΔU ± RTΔn. Since in this process work is done on the environment, it means that the internal energy of the system decreases, i.e.

ΔH = ΔU - RTΔn or ΔU = ΔH + RTΔn.

In this equation, Δn corresponds to the amount of the released gaseous hydrogen substance n H 2 , determined by the amount of the metal zinc reacted with the acid n Zn . And then n H 2 \u003d n Zn \u003d m Zn / M Zn, where m and M are the mass and molar mass of zinc, respectively. After doing the calculations, we get:

n H 2 \u003d 130 (g) / 65 (g / mol) \u003d 2 mol. Therefore, Δn = 2 mol.

Now we calculate the change in the internal energy of the process, remembering that in exothermic processes ΔH< 0, т.е. ΔH = -286,2 кДж; Т = 273 + 20 = 293 К;

R = 8.31 10 -3 kJ/mol K. And then:

ΔU \u003d -286.2 kJ + 8.31 10 -3 (kJ / mol K) 293 K 2 mol \u003d -281.3 kJ.

Answer: during the reaction, the internal energy of the system will decrease by 281.3 kJ.

Example 3. Calculation of entropy, its relationship with the enthalpy of a chemical process

and temperature

1. The specific heat of evaporation of bromobenzene at a temperature of 156.8 0 C is 241.0 J/g. Determine the change in the entropy of the phase transition during the evaporation of 1.25 mol of bromobenzene.

Solution. The change in entropy in the equilibrium process of the transition of a substance from one state of aggregation to another is determined according to the second law of thermodynamics as

ΔS = , where ΔH is the heat of vaporization (or the enthalpy of the phase transition from liquid to gaseous state), Т is the phase transition temperature.

To determine the thermal effect of the ΔH process, you must first calculate the molar mass of the starting material bromobenzene C 6 H 5 Br, it will be equal to: M (C 6 H 5 Br) \u003d 6 12 + 5 1 + 1 80 \u003d 157 (g / mol ). Knowing the amount of bromobenzene substance n involved in the phase transition, we determine its mass: m(C 6 H 5 Br) = M n;

m (C 6 H 5 Br) \u003d 157 g / mol 1.25 mol \u003d 196.25 g.

For a given mass of substance, taking into account the specific heat of evaporation (L), we calculate the heat effect of the process according to the formula: ΔH = L m, ΔH = 241 (J/g) 196.25 (g) = 47296.25 J.

Phase transition temperature T \u003d t 0 C + 273 \u003d 156.8 + 273 \u003d 429.8 K.

Substituting the obtained values ​​into the equation of the 2nd law of thermodynamics, we obtain:

∆S = = 110,04 .

Answer: upon evaporation of 1.25 mol of bromobenzene, the entropy of the system increases by 110.04 J/K.

2. Determine the change in entropy under standard conditions for the following chemical process: Al (c) + Cr 2 O 3 (c) → Cr (c) + Al 2 O 3 (c).

Solution. According to the 3rd corollary of the Hess law, the change in the entropy of a chemical process (ΔS) is defined as the difference between the sums of the entropies of the reaction products and reactants, taking into account their stoichiometric coefficients in the reaction equation.

Given this circumstance, the process scheme must be reduced to a chemical equation by placing the appropriate coefficients. Then we get:

2Al (c) + Cr 2 O 3 (c) \u003d 2Cr (c) + Al 2 O 3 (c).

For this reaction, we will compose an equation for calculating the change in entropy under standard conditions: ΔS 0 = - .

According to the table of Appendix No. 7, we set the entropy values ​​(S 0) of the participants in the process (J / mol K):

S 0 Al(k) = 28.32; S 0 Cr 2 O 3 (c) = 81.10; S 0 Cr(k) = 23.76; S 0 Al 2 O 3 (c) = 50.94.

Substituting the found entropy values ​​into the desired equation, and making calculations, we get: ΔS 0 = (2 23.76 + 50.94) - (2 28.32 + 81.10) = -39.28 (J / mol TO).

Note that the negative value of the change in entropy (decrease in entropy) indicates the impossibility of carrying out this process spontaneously under standard conditions.

Answer: ΔS 0 \u003d -39.28 J / mol K. Under standard conditions, such a process is impossible.

The decomposition reaction of magnesium nitrate according to the equation

2Mg (NO 3) 2 (t) \u003d 2MgO (t) + 4NO 2 (g) + O 2 (g)

is accompanied by an increase in the entropy of the system by 891 J/K and a change in enthalpy by 510 kJ. Calculate the standard enthalpy of formation and entropy of formation of magnesium nitrate. Determine which of the factors - enthalpy or entropy - contributes to the spontaneous flow of this process.

Solution. We will calculate ΔH 0 arr and S 0 arr on the basis of the 3rd corollary from Hess's law, according to which:

a) ΔH 0 x.r. \u003d 2 ΔH 0 arr + 4 ΔH 0 arr - 2 ΔH 0 arr;

hence ΔH 0 arr = ΔH 0 arr + 2 ΔH 0 arr -½ΔH 0 x.r.

b) ΔS 0 x.r. \u003d 2 S 0 arr + 4 S 0 arr + S 0 arr - 2 S 0 arr; hence S 0 arr = S 0 arr + 2 S 0 arr + ½ S 0 arr - ½ ΔS 0 x.r.

Using the data in the table of Appendix No. 1, we find the values ​​of the enthalpies of formation and the entropies of the reaction products:

ΔH 0 arr \u003d -601.24 kJ / mol; ΔH 0 arr = 33.50 kJ/mol; S 0 arr \u003d 26.94 J / mol K; S 0 arr \u003d 240.45 J / mol K; S 0 arr \u003d 205.04 J / mol K.

Substituting the found values ​​into equations a) and b), we calculate the required values:

ΔH 0 arr \u003d 1 mol (-601.24 kJ / mol) + 2 mol 33.50 kJ / mol -½ (510 kJ) \u003d

789.24 kJ;

S 0 arr \u003d 1 mol 26.94 J / mol K + 2 mol 240.45 J / mol K + ½ mol 205.04 J / mol K - ½ 891 J / K \u003d -164.87 J / K.

As is known, the spontaneous flow of the reaction is facilitated by a decrease in its enthalpy factor (ΔH 0 x.r.< 0) и увеличение энтропийного фактора (Т·ΔS 0 х.р. >0). According to the given conditions of the problem, the entropy during the process increases, and, consequently, the product Т·ΔS 0 x.r. , which contributes to its spontaneous flow. On the other hand, the enthalpy of the reaction also increases, which does not contribute to the spontaneity of the process in the forward direction.

Answer: ΔH 0 arr = - 789.24 kJ; S 0 arr \u003d -164.87 J / K. The entropy factor of this reaction contributes to the spontaneity of the process of decomposition of magnesium nitrate.

4. When 100 g of copper is melted, the entropy of the system increases by 1.28 J/K. Calculate the specific heat of fusion of copper at a temperature of 1083 0 C.

Solution. Between specific heat (L, J/kg) and melting enthalpy (ΔH, J) there is a relation L = ΔH/m. The relationship between the enthalpy of the process and the change in its entropy is expressed by the equation of the 2nd law of thermodynamics ΔH = Т·ΔS. Combining the two expressions, we get:

Let us substitute the data from the condition of the problem into the found ratio, perform the corresponding calculations and get:

L= .

Answer: specific heat of fusion of copper is 17.4.

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

C 2 H 2 (g) + 5 / 2O 2 (g) \u003d 2CO 2 (g) + H 2 O (l).

Calculate the change in the entropy of the system under standard conditions and explain the reasons for its decrease.

Solution. According to the corollary of the Hess law, the change in entropy is defined as the difference between the sums of the entropies of the products and reactants of the process, taking into account the stoichiometric coefficients of the reaction. Then

ΔS 0 x.r. = - .

In the table of Appendix No. 1 we find the values ​​of the required entropies:

S 0 arr CO 2 (g) \u003d 213.65 J / mol K; S 0 arr H 2 O (l) \u003d 69.94 J / mol K; S 0 arr C 2 H 2 (g) \u003d 219.45 J / mol K; S 0 arr O 2 (g) \u003d 205.03 J / mol K.

Substituting these values ​​into the equation for the change in the entropy of the process, and making calculations, we get:

ΔS 0 x.r. \u003d (2 213.65 + 69.94 - 219.45 - (5/2) 205.03) J / mol K \u003d -234.79 J / mol K.

The decrease in the entropy of the process is explained by an increase in the ordering of the system, since the amount of gas substance in the reaction products is 2.7 times less than in the reagents (5.5/2).

Answer: ΔS 0 x.r. = -234.79 J/mol K; ΔS 0 x.r.<0 т.к. Δn(г) < 0.

Example 4. Gibbs energy change calculation, direction determination

chemical process

1. Restoration of the natural mineral magnetite Fe 3 O 4 with carbon monoxide (2) is carried out under production conditions according to the reaction

Fe 3 O 4 (c) + CO (g) \u003d 3FeO (c) + CO 2 (g).

Determine the change in the Gibbs energy and draw a conclusion about the possibility of spontaneous flow of this process under standard conditions.

Solution. The isobaric-isothermal potential of a thermodynamic system or the Gibbs energy G reflects the overall driving force of the process, i.e. denotes that part of the total energy of the system (H), which can completely and without a trace turn into useful work (the actual chemical process). Change in the Gibbs energy ΔG (at T = Const and P = Const) in the direction of its decrease (ΔG< 0) указывает на меру химической активности системы: чем больше |ΔG|, тем сильнее стремление к протеканию процесса и тем дальше он отстоит от состояния равновесия. Энергия Гиббса является функцией состояния и поэтому к ней применим закон Гесса:

ΔG ch.r. \u003d ∑ΔG arr (prod) - ∑ΔG arr (reag).

Applying this expression to the equation for the reduction of double iron oxide Fe 3 O 4 under standard conditions, we obtain:

ΔG 0 cold \u003d [ 3 ΔG 0 arr FeO (c) + ΔG 0 arr CO 2 (g)] - [ΔG 0 arr Fe 3 O 4 (c) + ΔG 0 arr CO (g)].

Using the table of Appendix No. 1, we set the values ​​of ΔG 0 arr of the reaction products and reagents:

ΔG 0 arr FeO(c) = -244.3 kJ/mol; ΔG 0 arr CO 2 (g) = -394.38 kJ / mol; ΔG 0 arr Fe 3 O 4 (k) \u003d -1014.20 kJ / mol; ΔG 0 arr CO (g) \u003d -137.27 kJ / mol.

Substituting the found values ​​into the expression for ΔG 0 x.r. and doing the calculations, we get:

ΔG 0 cold = - [(-1014.20) + (-137.27)] = 24.19 (kJ / mol).

Calculations showed that ΔG 0 x.r.> 0, which means that this process cannot proceed under standard conditions.

Answer: under standard conditions, the process of spontaneous reduction of iron dioxide with carbon monoxide (2) is not feasible, because ΔG 0 cold > 0.

2. Explain why the exothermic reaction H 2 (g) + CO 2 (g) \u003d CO (g) + H 2 O (l), ΔH 1 \u003d -2.85 kJ / mol, does not proceed under standard conditions; but the reaction

2NO (g) + O 2 (g) \u003d 2NO 2 (g), ΔH 2 \u003d -113.74 kJ / mol.

Solution. According to the first law of thermodynamics, the relationship between the enthalpy and the Gibbs energy of a chemical process is expressed by the equation: ΔH = ΔG + TΔS. Hence ΔG = ΔH – ТΔS. Let us calculate the change in the Gibbs energy of both processes, using the data in the table of Appendix No. 1 to calculate the change in entropy ΔS.

For the first reaction we get:

ΔS 0 x.r. (1) \u003d S 0 arr CO (g) + S 0 arr H 2 O (l) - S 0 arr H 2 (g) - S 0 arr CO 2 (g) and

ΔS 0 x.r. (1) = (197.91 + 69.94 - 130.59 - 213.65) J/mol K = -76.39 J/mol K.

For the second reaction, the result will be as follows:

ΔS 0 x.r. (2) \u003d 2 S 0 arr NO 2 (g) - 2 S 0 arr NO (g) - S 0 arr O 2 (g) and

ΔS 0 x.r. (2) = (2 240.46 - 2 210.20 - 205.03) J/mol K = -144.51 J/mol K.

Now we calculate the change in the Gibbs energy at T = 298K for these reactions:

ΔG 0 cold (1) \u003d ΔH 1 - TΔS 0 x.r. (1) and

ΔG 0 cold (1) = -2.85 kJ/mol - 298K (-76.39 10 -3 kJ/mol K) = 19.91 kJ/mol;

ΔG 0 cold (2) \u003d ΔH 2 - TΔS 0 x.r. (2) and

ΔG 0 cold (2) = -113.74 kJ/mol - 298K (-144.51 10 -3 kJ/mol K) = -70.68 kJ/mol.

According to the calculation results, ΔG 0 x.r. (1) > 0 and, therefore, this process will not proceed spontaneously, but ΔG 0 x.r. (2)< 0, что свидетельствует о самопроизвольности процесса при стандартных условиях.

Answer: under standard conditions, the reaction of reduction of carbon dioxide with hydrogen does not proceed, because for it ΔG 0 x.r. > 0, but the oxidation reaction of nitric oxide (2) with oxygen is possible, accompanied by a decrease in the Gibbs energy ΔG 0 x.r. (2)< 0.

3. Determine the possibility of spontaneous flow of the process of aluminothermy

Fe 2 O 3 (c) + 2Al (c) \u003d Al 2 O 3 (c) + 2Fe (c)

at 298K and 500K and the standard state of all substances. Set the minimum temperature above which the specified process proceeds spontaneously.

Solution. To calculate ΔG 0 x.r. We use Hess' law:

ΔG 0 cold \u003d [ΔG 0 arr Al 2 O 3 (c) + 2 ΔG 0 arr Fe (c)] - [ΔG 0 arr Fe 2 O 3 (c) +2 ΔG 0 arr Al (c)].

In this case, we take into account that ΔG 0 arr Fe(c) = ΔG 0 arr Al(c) = 0, and according to the table of Appendix No. 7 ΔG 0 arr Al 2 O 3 (c) = -1580.00 kJ/mol; ΔG 0 arr Fe 2 O 3 (k) = -740.98 kJ / mol. Substituting the found values ​​and making calculations, we get:

ΔG 0 cold \u003d [-1580.00 - (-740.98)] kJ / mol \u003d -839.02 kJ / mol.

To calculate ΔG 500 x.r. use the first law of thermodynamics

ΔG 500 x.r. = ΔH 500 x.r. – ТΔS 500 x.r. At the same time, in accordance with the indication of the condition of the problem (all substances are in the standard state), we use the tabular values ​​\u200b\u200bof ΔH 0 and ΔS 0 of reagents and products at 298K:

ΔH 0 arr Al 2 O 3 (k) = -1676.00 kJ / mol; ΔH 0 arr Fe 2 O 3 (k) = -822.16 kJ / mol; S 0 arr Al 2 O 3 (k) \u003d 50.94 J / mol K; S 0 arr Fe 2 O 3 (k) \u003d 89.96 J / mol K; S 0 arr Al(k) = 42.69 J/mol K; S 0 arr Fe(k) = 27.15 J/mol K.

Let us substitute these values ​​into the expressions for ΔH 500 x.r. and ΔS 500 x.r. and do the calculations:

ΔH 500 x.r. \u003d ΔH 0 arr Al 2 O 3 (c) - ΔH 0 arr Fe 2 O 3 (c); ΔH 500 x.r. \u003d [-1676.00 - (-822.16)] kJ / mol \u003d -853.84 kJ / mol.

ΔS 500 x.r. = - ; ΔS 500 x.r. \u003d (50.94 + 2 27.15) - (89.96 + 2 42.69) J / mol K \u003d -70.10 J / mol K.

Now we find ΔG 500 x.r. , expressing ΔS 500 x.r. in kJ/mol K:

ΔG 500 x.r. \u003d [-853.84 - 500 (-70.10 10 -3)] kJ / mol \u003d -818.79 kJ / mol.

To find the minimum temperature above which the process proceeds spontaneously, we apply the condition T = 0K to the system, and then ΔG 0 x.r. \u003d ΔH 0 x.r.< 0. Таким образом, даже при отрицательных значениях температуры (а такие значения практически недостижимы) реакция будет протекать самопроизвольно.

To set the upper temperature limit at which the process ceases to be spontaneous, we apply the condition of the state of chemical equilibrium: ΔG = 0 and ΔH = ТΔS, hence Т = .

Let us substitute the found values ​​ΔH 500 x.r. into the obtained expression. and ΔS 500 x.r. and, having made calculations, we get: T = = 12180 K.

Thus, only at a very high temperature (T≥12180 K) the process of aluminothermy is impossible.

Answer: at 298K and 500K, the process of reducing iron oxide (3) with aluminum proceeds spontaneously, because ΔG 298 x.r.< 0 и ΔG 500 х.р.< 0. Самопроизвольность процесса обеспечивается уже при температуре 0К и прекращается при температуре выше 12180К.

4. Determine the standard change in the Gibbs energy of the reaction

COCl 2 (g)<=>CO (g) + Cl 2 (g), if at a temperature of 885K 70% of phosgene decomposed, taken at an initial pressure of 100 kPa.

Solution. If, before the start of the reaction, the partial pressure of COCl 2 (g) was equal to P 0 \u003d 100 kPa, 70% of the gas was consumed during the reaction, then at the time of equilibrium, the partial pressure of the remaining phosgene P is equal to COCl 2 (g) \u003d P 0 0.3 = 30 kPa. The partial pressures of the reaction products in the equilibrium state are equal to the fraction of phosgene consumed, which means that P is equal to CO (g) \u003d P is equal to Cl 2 (g) \u003d P 0 0.7 \u003d 70 kPa.

According to the law of mass action for an equilibrium process

K equals = .

Let us substitute the found values ​​of the partial equilibrium pressures of products and reactants into this equation and calculate the value of the equilibrium constant:

K equals = = 163.3.

Now, using the van't Hoff isotherm equation ΔG 0 = –RTlnKp, we calculate the change in the standard Gibbs energy in the equilibrium state at a given temperature:

ΔG 0 \u003d (-8.31 885) J / mol ℓn163.3 \u003d -37434 J / mol \u003d -37.4 kJ / mol.

Answer: in a state of equilibrium under standard conditions, the change in the Gibbs energy of the reaction ΔG 0 x.r. = -37.4 kJ / mol.

Thermodynamics

Discipline: general chemistry.

Full-time education.

Ministry of Health and Social Development of the Russian Federation

State budgetary educational institution of higher

professional education

Saratov State Medical University

named after V.I. Razumovsky" of the Ministry of Health and Social

Development of the Russian Federation

(SBEI HPE Saratov State Medical University named after V.I. Razumovsky of the Ministry of Health and Social Development of Russia)

Guidelines for the laboratory - practical lesson

for medical students

Thermodynamics

Discipline: general chemistry.

Full-time education.

Lesson duration: 90 minutes.

The development was compiled by ass. Kulikova L.N.

Goals

formation of ideas about physical and chemical aspects as the most important biochemical processes in the body.

Initial knowledge of students:

The concept of the thermal effect of a chemical reaction from a school chemistry course. Exothermic and endothermic reactions.

The student must know: First law of thermodynamics. Concepts: enthalpy, entropy. Hess' law. Application of the first law of thermodynamics to biosystems. The second law of thermodynamics. Gibbs energy.

The student must be able to:calculate the standard enthalpy of a chemical reaction from the standard enthalpies of formation and combustion of chemical compounds, according to Hess's law, the entropy of a chemical reaction, the Gibbs energy.

Preparation plan:

1) Familiarize yourself with the questions to prepare for the lesson (Appendix 1)

2) Comprehend the brief theoretical material (Appendix 2) and the text of the lecture “Basic concepts of thermodynamics. First and second laws of thermodynamics.

3) If certain sections of the lecture are difficult to understand, then you need to refer to the following textbooks:

4) Answer questions for self-control (Appendix 3).

5) Run a training test and check the correct execution of the key (Appendix 4).

6) Solve the situational problem and compare the received answer with the standard (Appendix 5).

Annex 1: Questions to prepare for the lesson

1) Subject and methods of chemical thermodynamics. The relationship between the processes of metabolism and energy in the body. Chemical thermodynamics as a theoretical basis for bioenergetics.

2) Basic concepts of thermodynamics. Intensive and extensive parameters. State function. Internal energy. Work and heat are two forms of energy transfer.

3) Types of thermodynamic systems (isolated, closed, open).

4) Types of thermodynamic processes (isothermal, isobaric, isochoric).

5) The first law of thermodynamics.

6) Enthalpy. Standard enthalpy of formation of a substance, standard enthalpy of combustion of a substance. Standard enthalpy of reaction.

7) Hess' law.

8) Application of the first law of thermodynamics to biosystems.

9) The second law of thermodynamics. Reversible and irreversible processes in the thermodynamic sense. Entropy.

10) Gibbs energy. Forecasting the direction of spontaneous processes in isolated and closed systems; the role of enthalpy and entropy factors. Thermodynamic equilibrium conditions.

11) Standard Gibbs energy of formation of matter, standard Gibbs energy of biological oxidation of matter. Standard Gibbs energy of the reaction.

12) The concept of exergonic and endergonic processes occurring in the body. The principle of energy conjugation.

Annex 1: Brief theoretical material

Already at the end of the 18th century, it was known that human life is an interconnected process of chemical transformations (oxidation of food products, etc.) and energy metabolism in the body (A. Lavoisier, P. Laplace).

Chemical thermodynamics – This is a branch of physical chemistry that studies the interconversion of heat and energy during a chemical reaction.

Thermodynamics is based on a number of concepts: system, system state, system state parameters, system state functions, system internal energy, etc.

Thermodynamic system – this is a body or a group of bodies interacting with each other and separated from the environment by a real or imaginary interface.

Isolated system - is a system that does not exchange matter or energy with the environment.

Closed system - This is a system that does not exchange matter with the environment, but exchanges energy.

Open System - It is a system that exchanges both matter and energy with the environment.

An example of an open system is a living cell.

State of the systemis a set of system properties that make it possible to describe the system from the point of view of thermodynamics.

For example, to assess the state of the human body as a thermodynamic system, a doctor must evaluate some of its properties (temperature, pressure, concentration of biological fluids).

The physical properties that characterize the state of the system are called system state parameters.

The interaction of the system with the environment is noticeable by changing the parameters of the system.

Extensive Options – these are parameters that depend on the amount of substance in the system and are summed up when the systems are combined(volume, mass, energy, area, etc.).

Intensive Options – these are parameters that do not depend on the amount of substance and are aligned when systems are combined(temperature, pressure, concentration, density, surface tension).

The state parameters are linked by the state equation.

The transition of a system from one state to another with a change in at least one parameter is called thermodynamic process.

If the process takes place at constant pressure, it is called isobaric process. At a constant volume isochoric , at constant temperature - isothermal .

Status function- this is a characteristic of the system that cannot be directly measured, but is calculated through the state parameters. The value of the state function does not depend on how it is achieved, but only on the initial and final states of the system.

Internal energy is one such function.

Internal energy- the sum of all types of energies of motion and interaction of particles that make up the system.

In the 19th century, the German ship's doctor Mayer Yu.R. and the English scientist Joule D. showed that heat and work are capable of mutual transformations, being different ways of transferring energy.

Heat- a form of energy transfer by chaotic movement of microparticles.

Job- a form of energy transfer through the directed movement of the macrosystem as a whole.

Observing people in different climatic zones, Mayer concluded that the heat of combustion of food is used to maintain a constant body temperature and to perform muscular work. This observation formed the basis of the 1st law of thermodynamics.

First law of thermodynamics (first law of thermodynamics):
energy does not arise from nothing and does not disappear without a trace, but passes from one type of energy to another

or the increment in the internal energy of the system in some process is equal to the heat received by the system, plus the work done on the system.

∆U=Q+A

∆U - internal energy

Q - heat

A - work

Based on the 1st law of thermodynamics, which is a fundamental law of nature, simple calculations provide valuable information about the processes of metabolism and energy in the body.

Thermochemistryis a branch of thermodynamics that studies the heats of chemical reactions.

Hess' law:the heat of a chemical reaction proceeding at constant pressure or volume does not depend on the path of the process, but only on the initial and final states of the system.

If it is possible to obtain others from some substances in different ways, then the total thermal effect along the first path is equal to the total thermal effect along the second path. Hess's law makes it possible in practice to calculate the thermal effects of reactions that are difficult to monitor or take a long time. For example, the total heat of biological oxidation of food products in the body is equal to the heat of their direct combustion.

Enthalpyis a state function, the increment of which is equal to the thermal effect of the process occurring at constant pressure.

Methods for calculating the standard enthalpy of a chemical reaction

1) According to the standard enthalpies (heats) of formation of substances

Standard enthalpy (heat) of formation of a substanceis the thermal effect of the reaction of formation of 1 mol of a chemical compound from simple substances under standard conditions: T=289 K, P=1 atm=101325 Pa.

The heats of formation of simple substances are equal to zero.

ν i , ν j are stoichiometric coefficients in front of the corresponding substances in the reaction equation.

2) According to the standard enthalpies (heats) of combustion of substances

Standard enthalpy (heat) of combustion of a substanceis the thermal effect of the reaction of complete combustion of 1 mol of a chemical compound under standard conditions.

The heats of combustion of higher oxides (including CO 2 and H 2 O) are assumed to be zero.

The second law of thermodynamics establishes the possibility, direction and depth of a spontaneous process.

Spontaneous process- a process that proceeds without any external influences, and brings the system closer to a state of equilibrium.

Thermodynamically reversible process- a process that proceeds in the forward and reverse directions without changes in the system and in the environment, i.e. during the transition from the initial state to the final state, all intermediate states are in equilibrium.

In the presence of nonequilibrium intermediate states, the process is considered thermodynamically irreversible.

In nature, systems with minimal energy are stable. Then only exothermic processes should be spontaneous. But it's not. This means that there is another criterion for the spontaneous flow of the process - entropy (S).

Entropy - a measure of energy disorder in a system, a measure of chaos, a measure of the energy that dissipates in the form of heat and does not turn into work.

The second law of thermodynamics (second law of thermodynamics):

Processes occur spontaneously, leading to an increase in the total entropy of the system and the environment

∆S syst + ∆S medium ≥ 0 or ∆S ≥ 0

The physical meaning of entropy:

Entropy is the amount of energy dissipated by 1 mole of a substance per 1 degree.

Entropy is an extensive function. Entropy is proportional to mass, it refers to 1 mole or 1 gram of a substance. This means that the entropy of the system is equal to the sum of the entropies of its component parts:

S=

Entropy is a function of the state of the system. This means that it characterizes system, not a process. Its change depends only on the initial and final state of the system and does not depend on the transition path:

For a chemical reaction, the change in entropy: prod - ref

Similar information.

It is impossible to find the absolute values ​​of enthalpies and internal energies by thermodynamic methods, but only their changes can be determined. At the same time, in thermodynamic calculations of chemically reacting systems, it is convenient to use a single frame of reference. In this case, since the enthalpy and internal energy are related by the relation , it is sufficient to introduce a frame of reference for only one enthalpy. In addition, in order to compare and systematize the thermal effects of chemical reactions, which depend on the physical state of the reacting substances and on the conditions for the occurrence of XP, the concept of the standard state of matter is introduced. On the recommendation of the commission on thermodynamics of the International Union of Pure and Applied Chemistry (IUPAC) in 1975, the standard state is defined as follows:

“The standard state for gases is the state of a hypothetical ideal gas at a pressure of 1 physical atmosphere (101325 Pa). For liquids and solids, the standard state is the state of a pure liquid or, respectively, a pure crystalline substance at a pressure of 1 physical atmosphere. For substances in solutions, the standard state is taken to be the hypothetical state in which the enthalpy of a one-molar solution (1 mol of substance in 1 kg of solvent) would be equal to the enthalpy of the solution at infinite dilution. The properties of substances in standard states are denoted by the superscript 0. (A pure substance is a substance consisting of identical structural particles (atoms, molecules, etc.)).

This definition refers to the hypothetical states of a gas and a solute, since in real conditions the states of gases differ to a greater or lesser extent from the ideal, and the states of solutions differ from the ideal solution. Therefore, when using the thermodynamic properties of substances in standard states for real conditions, corrections are introduced for the deviation of these properties from real ones. If these deviations are small, then corrections can be omitted.

In handbooks, thermodynamic quantities are usually given under standard conditions: pressure R 0 =101325Pa and temperature T 0 =0K or T 0 \u003d 298.15K (25 0 C). When creating tables of total enthalpies of substances, their standard state at a temperature T 0 =0K or T 0 = 298.15K.

Substances, which are clean chemical elements in the most stable phase condition at R 0 \u003d 101325 Pa and the temperature of the reference point of enthalpies T 0, take the value enthalpy equal to zero: . (For example, for substances in the gaseous state: O 2, N 2, H 2, Cl 2, F 2, etc., for C (graphite) and metals (solid crystals)).

For chemical compounds(CO 2, H 2 O, etc.) and for substances that, being pure chemical elements, not in the most stable state(O, N, etc.) enthalpy at R 0 =101325Pa and T 0 not equal to zero: .

Enthalpy chemical compounds at R 0 and T 0 relies equal to the thermal effect of formation them from pure chemical elements with these parameters, i.e. . So, at T 0 \u003d 0K: and at T 0 \u003d 298.15K:.

The enthalpy of any substance at a temperature T will be equal to the amount of heat that must be supplied in the isobaric process so that from pure chemical elements at a temperature T 0 get a given substance and heat it from temperature T 0 to temperature T, i.e. the formula for calculating the enthalpy of any substance is:

, or in a more compact notation we have:

,

where the superscript "o" means that the substance is in the standard state at R 0 =101325Pa; is the enthalpy of formation of a substance at a temperature T 0 from pure chemical elements; = - excess enthalpy associated with the heat capacity of the substance, - total enthalpy, taking into account the enthalpy of formation of the substance.

For T 0 = 0:

,

For T= 298.15 K:

Scheme for calculating enthalpy at temperature T can be represented in the form

Here you will find examples of tasks for calculating such thermodynamic parameters as enthalpy, entropy, . Determining the possibility of spontaneous flow of the process, as well as compiling thermochemical equations.

Tasks for the section Fundamentals of thermodynamics with solutions

Task 1. Calculate the standard enthalpy and standard entropy of a chemical reaction. Determine in which direction at 298 °K (forward or reverse) the reaction will proceed. Calculate the temperature at which both directions of the reaction are equally likely.
Fe 2 O 3 (c) + 3H 2 \u003d 2Fe (c) + 3H 2 O (g)

Δ Hdistrict = ΣH 0 con– ΣH 0 ref kJ/mol

Using reference data standard enthalpies substances, we find:

Δ Hdistrict= 2 Δ H 0 Fe +3 Δ H 0 H2 O - Δ H 0 Fe2 O3 - 3 Δ H 0 H2 \u003d 2 0 + 3 (- 241.82) - (-822.16) - 3 0 \u003d 96.7 kJ / mol

Δ Sdistrict=Σ S 0 con– Σ S 0 ref J/(mol K)

Using reference data standard entropies substances, we find:

Δ Sdistrict= 2 Δ S 0 Fe + 3 Δ S 0 H2 O - Δ S 0 Fe2 O3 - 3 Δ S 0 H2 = 2 27.15 + 3 188.7 – 89.96 – 3 131 = 137.44 J/(mol K)

ΔG = Δ H – TΔS\u003d 96.7 - 298 137.44 / 1000 \u003d 55.75 kJ / mol

At T=298°K, ΔG> 0 - the reaction does not proceed spontaneously, i.e. the reaction will proceed in the opposite direction.

ΔG = Δ H – TΔS= 0, Then

T= — (ΔG – Δ H) / ∆S= — (0-96.7)/0.137 = 705.83K

At T = 705.83 K, the reaction will proceed with equal probability both in the forward and in the reverse direction.

Task 2. Calculate the Gibbs energy and determine the possibility of the reaction occurring at temperatures of 1000 and 3000 K.

The reaction equation for the combustion of liquid carbon disulfide is as follows:

CS 2 (l) + 3O 2 \u003d CO 2 + 2SO 2

We calculate the heat effect of the reaction by substituting the reference data of the standard enthalpies of substances into the expression:

Δ Hdistrict = ΣH 0 con– ΣH 0 ref kJ/mol

Δ Hdistrict= 2 Δ H 0 SO2+ Δ H 0 CO2- Δ H 0 CS2 - 3 Δ H 0 O2 \u003d 2 (-296.9) + 3 (- 393.5) - 87 - 3 0 \u003d -1075.1 kJ / mol

Those. during combustion 1 mole carbon disulfide is released 1075.1 kJ heat

and when burning x moles carbon disulfide is released 700 kJ heat

Let's find X:

x\u003d 700 1 / 1075.1 \u003d 0.65 mol

So, if 700 kJ of heat is released as a result of the reaction, then 0.65 mol of CS 2 will enter into the reaction

Task 4. Calculate the thermal effect of the reduction reaction of iron oxide (II) with hydrogen, based on the following thermochemical equations:
1. FeO (c) + CO (g) \u003d Fe (c) + CO 2 (g); ΔH 1 = -18.20 kJ;
2. CO (g) + 1 / 2O 2 (g) \u003d CO 2 (g) ΔH 2 \u003d -283.0 kJ;
3. H 2 (g) + ½ O 2 (g) \u003d H 2 O (g) ΔH 3 \u003d -241.83 kJ.

The reduction reaction of iron oxide (II) with hydrogen has the following form:

4. FeO (c) + H 2 (g) = Fe (c) + H 2 O (g)

To calculate the thermal effect of the reaction, it is necessary to apply , i.e. reaction 4. can be obtained by adding reactions 1. and 2. and subtracting reaction 1.:

Δ Hdistrict= Δ H 1 + Δ H 3 – Δ H 2 \u003d -18.2 - 241.3 + 283 \u003d 23 kJ

Thus, thermal effect of the reaction reduction of iron oxide (II) with hydrogen is equal to

Δ Hdistrict= 23 kJ

Task 5. The combustion reaction of benzene is expressed by the thermochemical equation:
C 6 H 6 (g) + 7½ O 2 (g) \u003d 6CO 2 (g) + 3H 2 O (g) - 3135.6 kJ.
Calculate the heat of formation of liquid benzene. Determine the calorific value of liquid benzene, provided that the standard conditions are the same as normal.

The thermal effect of the reaction is:

Δ Hdistrict = ΣH 0 con– ΣH 0 ref kJ/mol

In our case Δ Hdistrict= - 3135.6 kJ, we find the heat of formation of liquid benzene:

Δ Hdistrict= 6 Δ H 0 With O2 + 3 Δ H 0 H2 O - Δ H 0 C6 H6 - 7.5 Δ H 0 O2

-Δ H 0 C6 H6= Δ Hdistrict- 3 (-241.84) + 6 (- 393.51) - 7.5 0 \u003d - 3135.6 - 3 (-241.84) + 6 (- 393.51) - 7, 5 0 \u003d - 49.02 kJ / mol

Δ H 0 C6 H6 = 49.02 kJ/mol

Calorific value liquid benzene is calculated by the formula:

QT= Δ Hdistrict 1000 / M

M(benzene) = 78 g/mol

QT= - 3135.6 1000 / 78 = - 4.02 10 4 kJ / kg

Calorific value liquid benzene Q T = - 4.02 10 4 kJ / kg

Task 6. The oxidation reaction of ethyl alcohol is expressed by the equation:
C 2 H 5 OH (g) + 3.0 O 2 (g) \u003d 2CO 2 (g) + 3H 2 O (g).
Determine the heat of formation C 2 H 5 OH (l), knowing ΔH x.r. = - 1366.87 kJ. Write a thermochemical equation. Determine the molar heat of vaporization C 2 H 5 OH (g) → C 2 H 5 OH (g), if the heat of formation C 2 H 5 OH (g) is known, equal to –235.31 kJ mol -1.

Based on the given data, we write thermochemical equation:

C 2 H 5 OH (g) + 3O 2 (g) \u003d 2CO 2 (g) + 3H 2 O (g) + 1366.87 kJ

thermal effect reactions equals:

Δ Hdistrict = ΣH 0 con– ΣH 0 ref kJ/mol

In our case Δ Hdistrict= - 1366.87 kJ.

Using reference data heats of formation of substances, we find the heat of formation C 2 H 5 OH (l):

Δ Hdistrict= 2 Δ H 0 With O2 + 3 Δ H 0 H2 O - Δ H 0 C2 H5 OH(l) – 3 Δ H 0 O2

- 1366.87 \u003d 2 (-393.51) + 3 (-285.84)— Δ H 0 C2 H5 OH - 3 0

Δ H 0 C2H5OH(w)\u003d -277.36 kJ / mol

Δ H 0 C2 H5 OH(g) = Δ H 0 C2 H5 OH(l) + Δ H 0 vaporization

Δ H 0 vaporization = Δ H 0 C2 H5 OH(g) — Δ H 0 C2 H5 OH(l)

Δ H 0 vaporization\u003d - 235.31 + 277.36 \u003d 42.36 kJ / mol

We determined that the heat of formation of C 2 H 5 OH (l) is equal to

Δ H 0 C2H5OH(w)\u003d -277.36 kJ / mol

and the molar heat of vaporization C 2 H 5 OH (l) → C 2 H 5 OH (g) is

Δ H 0 vaporization= 42.36 kJ/mol

Task 7. How can one explain that under standard conditions, an exothermic reaction is impossible:
CO 2 (g) + H 2 (g) ↔ CO (g) + H 2 O (g)?
Calculate ΔG of this reaction. At what temperature does this reaction become spontaneous?

Calculate ΔG this reaction:

ΔG = Δ H – TΔS

To do this, we first define Δ HAnd ∆S reactions:

Δ Hdistrict = ΣH 0 con– ΣH 0 ref kJ/mol

Using reference data standard enthalpies substances, we find:

Δ Hdistrict= Δ H 0 H2 O(l) + Δ H 0 CO- Δ H 0 CO2 - Δ H 0 H2 \u003d -110.5 + (-285.8) - (393.5) - 0 \u003d -2.8 kJ / mol

Δ Sdistrict=Σ S 0 con– Σ S 0 ref J/(mol K)

Similarly, using reference data standard entropies substances, we find:

Δ Sdistrict= Δ S 0 H2 O(l) + Δ S 0 CO- Δ S 0 CO2 - Δ S 0 H2 \u003d 197.5 + 70.1 - 213.7 - 130.52 \u003d -76.6 J / (mol K)

Let's find Gibbs energy under standard conditions

ΔGdistrict= Δ H – TΔS\u003d -2.8 + 298 76.6 / 1000 \u003d 20 kJ / mol> 0,

so the reaction is spontaneous does not go.

Find at what temperatures this reaction becomes spontaneous.

In a state of equilibrium ΔGdistrict = 0 , Then

T = Δ H/ ∆S = -2.8 / (-76.6 1000) \u003d 36.6 K

Task 8. Calculating on the basis of tabular data ΔG and ΔS, determine the thermal effect of the reaction:
2 NO (g) + Cl 2 (g) ↔ 2 NOCl (g).

At constant temperature and pressure, the change Gibbs energy

ΔG = Δ H – TΔS

Based on the tabular data, we calculate ∆G and ∆S

ΔG 0 district= Σ ΔG 0 prod — Σ ΔG 0 ref

ΔGdistrict= 2 ΔG 0 NOCl(G) — 2 ΔG 0 NO(G) — ΔG 0 Cl2(d)

ΔGdistrict= 2 66,37 — 2 89,69 – 0 = — 40,64 kJ/mol

ΔGdistrict < 0 , so the reaction is spontaneous.

Δ Sdistrict=Σ S 0 con– Σ S 0 ref J/(mol K)

Δ Sdistrict = 2 ∆S 0 NOCl(G) — 2 ∆S 0 NO(G) — ∆S 0 Cl2(d)

Δ Sdistrict = 2 261,6 — 2 210,62 – 223,0 = -121,04 J/(mol K)

Let's find Δ H :

Δ H = ΔG + TΔS

Δ H = - 40.64 + 298 (-121.04 / 1000) \u003d - 76.7 kJ / mol

Thermal effect of the reaction Δ H = - 76.7 kJ / mol

Problem 9. With what will gaseous hydrogen chloride interact more intensively (per 1 mol): with aluminum or with tin? Give your answer by calculating ΔG 0 for both reactions. The reaction products are solid salt and gaseous hydrogen.

Calculate ∆G0 for the reaction of the interaction of gaseous hydrogen chloride (per 1 mol) with aluminum

2Al (t) + 6HCl (g) \u003d 2AlCl 3 (t) + 3H 2

ΔG 0 district= Σ ΔG 0 prod — Σ ΔG 0 ref kJ/mol

ΔG 0 districts1= 2 ΔG 0 AlCl 3 (t) + 3 ΔG 0 H2 — 2 ΔG 0 Al(t) — 6ΔG 0 HCl(G)

ΔG 0 districts1= 2 (-636.8) + 3 0 — 20 — 6(-95.27) = -701.98 kJ/mol

2 mol of Al(t) takes part in the reaction, then ΔGdistricts1 1 mole Al(t) is equal to

ΔG 0 district 1 \u003d -701.98 / 2 \u003d -350.99 kJ / mol

Calculate ∆G0 for the reaction of interaction of gaseous hydrogen chloride (per 1 mol) with tin:

Sn (t) + 2HCl (g) \u003d SnCl 2 (t) + H 2

ΔG 0 districts2 =ΔG 0 SnCl 2 (t) + ΔG 0 H2 — ΔG 0 Sn(t) — 2ΔG 0 HCl(G)

ΔG 0 district 2 \u003d -288.4 + 0- 0- 2 (-95.27) \u003d -97.86 kJ / mol

Both reactions have ΔG 0 <0 , therefore, they proceed spontaneously in the forward direction, but gaseous hydrogen chloride will interact more intensively with aluminum, because

ΔG 0 district 1˂ ΔG 0 district 2

Problem 10. Without resorting to calculations, determine which signs (>0,<0, ≅0) имеют ΔG, ΔH и ΔS для протекающей в прямом направлении реакции:
4 HBr (g) + O 2 (g) ↔ 2 H 2 O (g) + 2 Br 2 (g)
How will an increase in temperature affect the direction of a chemical reaction?

At constant temperature and pressure Gibbs energy change related to enthalpy and entropy by the expression:

ΔG = Δ H – TΔS

During chemical reactions, heat is absorbed or released into the environment. This heat exchange between a chemical reaction and the environment is called enthalpy, or H. However, it is impossible to measure enthalpy directly, so it is customary to calculate the change in ambient temperature (denoted ∆H). ∆H indicates that during a chemical reaction, heat is released into the environment (exothermic reaction) or heat is absorbed (endothermic reaction). The enthalpy is calculated as follows: ∆H = m x s x ∆T, where m is the mass of reactants, s is the heat capacity of the reaction product, ∆T is the change in temperature as a result of the reaction.

Steps

Solving enthalpy problems

Identify the reactants and products of the reaction. Any chemical reaction has reactants and reaction products. reaction product created as a result of the interaction of reagents. In other words, the reactants are the ingredients in the recipe, and the reaction product is the finished dish. To find the ∆H of a reaction, it is necessary to know the reactants and products of the reaction.

• For example, it is necessary to find the enthalpy of the reaction of formation of water from hydrogen and oxygen: 2H 2 (hydrogen) + O 2 (oxygen) → 2H 2 O (water). In this reaction H2 And O2- reagents, and H2O is the reaction product.

1. Determine the total mass of the reagents. Next, you need to calculate the mass of reagents. If you can't weigh them, then calculate the molecular weight to find the actual one. Molecular mass is a constant that can be found in Mendeleev's periodic table or other tables of molecules and compounds. Multiply the mass of each reactant by the number of moles.

   • In our example, the reactants hydrogen and oxygen have molecular weights of 2 g and 32 g, respectively. Since we are using 2 mol of hydrogen (the coefficient in the chemical reaction in front of H 2 hydrogen) and 1 mol of oxygen (the absence of the coefficient in front of O 2 means 1 mol), then the total mass of the reactants is calculated as follows:
     2 × (2 g) + 1 × (32 g) = 4 g + 32 g = 36 g

2. Determine the heat capacity of the product. Next, determine the heat capacity of the reaction product. Each molecule has a specific heat capacity, which is constant. Find this constant in the tables of a chemistry textbook. There are several units for measuring heat capacity; in our calculations we will use J/g°C.

   • Note that if there are multiple reaction products, you will need to calculate the heat capacity of each and then add them up to get the enthalpy of the entire reaction.
   • In our example, the product of the reaction is water, which has a specific heat 4.2 J/g°C.

3. Find the change in temperature. Now we will find ∆T - the temperature difference before and after the reaction. From the start temperature (T1) subtract the end temperature (T2). Most often, the Kelvin (K) scale is used in chemistry problems (although the Celsius scale (°C) will give the same result).

   • In our example, let's assume that the initial temperature of the reaction was 185 K, and after the reaction it became 95 K, so ∆T is calculated as follows:
     ∆T = T2 – T1 = 95 K - 185 K = -90K

4. Find the enthalpy using the formula ∆H = m x s x ∆T. If m is the mass of the reactants, s is the heat capacity of the reaction product, and ∆T is the change in temperature, then the enthalpy of the reaction can be calculated. Substitute the values ​​in the formula ∆H = m x s x ∆T and get the enthalpy. The result is calculated in Joules (J).

   • In our example, the enthalpy is calculated as follows:
     ∆H = (36 g) × (4.2 JK - 1 g - 1) × (-90 K) = -13608 J

5. Determine whether energy is released or absorbed during the reaction under consideration. One of the most common reasons to calculate ∆H in practice is to know if the reaction will be exothermic (releasing heat and reducing its own energy) or endothermic (absorbing heat from the environment and increasing its own energy). If the value of ∆H is positive, then the reaction is endothermic. If negative, then the reaction is exothermic. The greater the absolute value of ∆H, the more energy is released or absorbed. Be careful if you are going to do a practical experiment: during reactions with a high enthalpy value, a large release of energy can occur, and if it proceeds quickly, it can lead to an explosion.

   • In our example, the final result turned out to be -13608 J. There is a negative sign in front of the enthalpy value, which means that the reaction exothermic. Hot gases (in the form of steam) H 2 and O 2 must give off some heat in order to form a water molecule, that is, the reaction of formation of H 2 O is exothermic.

   Enthalpy Estimation

   1. Calculate the bond energy to estimate the enthalpy. Almost all chemical reactions lead to the breaking of some bonds and the formation of others. The energy from the reaction does not appear from nowhere and is not destroyed: it is the energy that is required to break or form these bonds. Therefore, the change in the enthalpy of the entire reaction can be estimated quite accurately by summing the energies of these bonds.

      Use the enthalpy of formation to estimate the enthalpy. The enthalpy of formation makes it possible to calculate ∆H by calculating the formation reactions of reactants and products. If the enthalpy of formation of reaction products and reactants is known, then you can estimate the enthalpy as a whole by addition, as in the case of energy discussed above.

   2. Don't forget about signs before enthalpy values. When calculating the enthalpy of formation, you reverse the formula for determining the enthalpy of reaction of a product, and the sign of the enthalpy should change. In other words, if you reverse the formula, then the sign of the enthalpy should change to the opposite.

      • In the example, note that the formation reaction for the product C 2 H 5 OH is written backwards. C 2 H 5 OH → 2C + 3H 2 + 0.5O 2, that is, C 2 H 5 OH decomposes, not synthesized. Therefore, the sign before the enthalpy in such a reaction is positive, 228 kJ/mol, although the enthalpy of formation of C 2 H 5 OH is -228 kJ/mol.

   Observation of enthalpy during the experiment

   1. Take a clean container and pour water into it. It is not difficult to see the principles of enthalpy in action - it is enough to conduct a simple experiment. It is necessary that the result of the experiment is not affected by foreign contaminants, so the container must be washed and sterilized. Scientists use special closed containers called calorimeters to measure enthalpy, but a glass beaker or flask is fine for you. Fill the container with clean tap water at room temperature. It is advisable to conduct the experiment in a cool room.

      • For the experiment, it is desirable to use a small container. We will be looking at the enthalpy of the reaction of water with Alka-Seltzer, so the less water used, the more obvious the change in temperature will be.