Thermodynamic probability of various directions of complex reactions in oil refining processes. Chemical kinetics Probability of chemical reactions occurring

201. In which direction, under standard conditions, will the reaction N 2 (g) + O 2 (g) \u003d 2NO (g) proceed? Support your answer with calculations.

202. Calculate the change in the Gibbs energy of some reaction at 1000 K if ∆ r H° 298 = 131.3 kJ, and ∆ rS° 298 \u003d 133.6 J / K (influence of temperature T on ∆ H and ∆ S neglect).

203. Calculate ∆ rG° 298 systems PbO 2 + Pb = 2PbO based on ∆ r H° 298 and ∆ rS° 298 reactants. Determine if this reaction is possible.

204. Determine in which direction the reaction Fe 2 O 3 (c) + 3H 2 \u003d \u003d Fe (c) + 3H 2 O (g) will proceed spontaneously under standard conditions.

205. Calculate the change in the Gibbs energy and determine the possibility of reducing chromium (III) oxide with carbon at 1500 K according to the reaction Cr 2 O 3 (t) + 3C (t) \u003d 2Cr (t) + 3CO (g).

206. Tungsten is obtained by reduction of tungsten (IV) oxide with hydrogen. Determine the possibility of this reaction occurring at 500 and 1000 ° C according to the reaction WO 3 (t) + 3H 2 (g) \u003d W (t) + 3H 2 O (g).

207. Calculate the change in the Gibbs energy and determine the possibility of this reaction occurring under standard conditions CO (g) + H 2 O (l) \u003d CO 2 (g) + H 2 (g).

208. Calculate the change in the Gibbs energy and determine the possibility of the decomposition reaction of copper (II) oxide at 400 and 1500 K according to the reaction 4CuO (t) \u003d 2Cu 2 O (t) + O 2 (g).

209. Determine the temperature of the equiprobable reaction in the forward and reverse directions, if ∆ r H° 298 = = 38 kJ, and ∆ rS° 298 = 207 J/K.

210. Calculate ∆ rG° 298 and ∆ rG° 1000 for the reaction H 2 O (g) + + C (g) \u003d CO (g) + H 2 (g). How does temperature affect the thermodynamic probability of a process proceeding in the forward direction?

211. Which of the following reactions is thermodynamically more likely: 1) N 2 + O 2 \u003d 2NO or 2) N 2 + 2O 2 \u003d 2NO 2? Support your answer with calculations.

212. Determine the sign of ∆ rG° 298, without resorting to calculations, for the reaction CaO (t) + CO 2 (g) \u003d CaCO 3 (t), ∆ r H° 298 \u003d -178.1 kJ / mol. Explain the answer.

213. Determine the sign of ∆ rG° 298 for the process of assimilation of sucrose in the human body, which is reduced to its oxidation C 12 H 22 O 11 (s) + 12O 2 (g) \u003d 12CO 2 (g) + 11H 2 O (g).

214. Check if there is a threat that nitric oxide (I), used in medicine as a narcotic, will be oxidized by air oxygen to a very toxic nitric oxide (II) according to the reaction 2N 2 O (g) + O 2 (g) \u003d 4NO (G) .

215. Glycerin is one of the products of metabolism, which is finally converted in the body into CO 2 (g) and H 2 O (l). Calculate ∆ rG° 298 glycerol oxidation reaction if ∆ f G° 298 (C 3 H 8 O 3) = = 480 kJ / mol.

216. Calculate the change in the Gibbs energy for the photosynthesis reaction 6CO 2 (g) + 6H 2 O (g) \u003d C 6 H 12 O 6 (solution) + 6O 2 (g).

217. Determine the temperature at which ∆ rG° T \u003d 0, for the reaction H 2 O (g) + CO (g) \u003d CO 2 (g) + H 2 (g).

218. Calculate thermodynamic characteristics ∆ r H° 298 , ∆ rS° 298 , ∆ rG° 298 reactions 2NO (g) \u003d N 2 O 4 (g). Formulate a conclusion about the possibility of the reaction at temperatures 0; 25 and 100 °C, confirm it by calculation.

219. Is the reaction 3Fe 2 O 3 (c) + H 2 (g) \u003d 2Fe 3 O 4 (c) \u003d H 2 O (g) possible? Support your answer with calculations.

220. Calculate the change in the Gibbs energy of the reaction at 980 K if ∆ r H° 298 = 243.2 kJ, and ∆ rS° 298 = 195.6 J/K (influence of temperature on ∆ H and ∆ S neglect).

221. Calculate ∆ rG° 298 and ∆ rG° 1000 for reaction

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

How does temperature affect the thermodynamic probability of a process proceeding in the forward direction?

222. The interaction of calcium carbide with water is described by two equations:

a) CaC 2 + 2H 2 O \u003d CaCO 3 + C 2 H 2; b) CaC 2 + 5H 2 O \u003d CaCO 3 + 5H 2 + CO 2.

Which reaction is thermodynamically more preferable? Explain the calculation results.

223. Determine the direction of the spontaneous flow of the reaction SO 2 + 2H 2 \u003d S cr + 2H 2 O under standard conditions.

224. Calculate the change in the Gibbs energy of the reaction ZnS + 3 / 2O 2 \u003d ZnO + SO 2 at 298 and 500 K.

225. Determine the direction of spontaneous reaction

NH 4 Cl (c) + NaOH (c) \u003d NaCl (c) + H 2 O (g) + NH 3 (g).

under standard conditions

The rate of chemical reactions

226. Determine how many times the rate of a homogeneous gas reaction 4HCl + O 2 → 2H 2 O + 2Cl 2 will change if the total pressure in the system is increased by 3 times.

227. The reaction rate: 2NO + O 2 → 2NO 2 at NO and O 2 concentrations equal to 0.6 mol/dm 3 is 0.18 mol/(dm 3 ·min). Calculate the reaction rate constant.

228. How many times should the concentration of CO be increased in the system in order to increase the rate of the reaction 2CO → CO 2 + C (tv) by 4 times?

229. The reaction proceeds according to the equation N 2 + O 2 → 2NO. The initial concentrations of nitrogen and oxygen are 0.049 and 0.01 mol/dm 3 . Calculate the concentrations of substances when 0.005 mol NO is formed in the system.

230. The reaction between substances A and B proceeds according to the equation 2A + B \u003d C. The concentration of substance A is 6 mol / l, and substance B is 5 mol / l. The reaction rate constant is 0.5 l/(mol s). Calculate the reaction rate at the initial moment and at the moment when 45% of substance B remains in the reaction mixture.

231. How many degrees must the temperature be raised to increase the rate of the reaction by 90 times? The van't Hoff temperature coefficient is 2.7.

232. The temperature coefficient of the reaction rate of the decomposition of hydrogen iodine according to the reaction 2HI \u003d H 2 + I 2 is 2. Calculate the rate constant of this reaction at 684 K, if at 629 K the rate constant is 8.9 10 -5 l / (mol s).

233. Determine the temperature coefficient of the reaction rate if the reaction slowed down by 25 times when the temperature was lowered by 45 °.

234. Calculate at what temperature the reaction will end in 45 minutes, if at 293 K it takes 3 hours. Take the temperature coefficient of the reaction rate equal to 3.2.

235. Calculate the rate constant of the reaction at 680 K, if at 630 K the rate constant of this reaction is 8.9 -5 mol/(dm 3 s), and γ = 2.

236. The reaction rate constant at 9.4 °C is 2.37 min -1 , and at 14.4 °C it is 3.204 min -1 . Calculate the activation energy and temperature coefficient of the reaction rate.

237. Calculate how many degrees the temperature needs to be raised to increase the reaction rate by 50 and 100 times if the temperature coefficient of the reaction rate is 3.

238. At 393 K the reaction is completed in 18 min. After what time interval will this reaction end at 453 K if the temperature coefficient of the reaction rate is 3?

239. The initial concentrations of the reactants of the reaction CO + H 2 O (g) → CO 2 + H 2 were equal (mol / dm 3): \u003d 0.8; = 0.9; = 0.7; = 0.5. Determine the concentrations of all participants in the reaction after the concentration of hydrogen has increased by 10%.

240. The reaction between substances A and B is expressed by the equation A + 2B → C. The initial concentrations of the substance are: [A] = 0.03 mol/l; [B] = 0.05 mol/l. The reaction rate constant is 0.4. Determine the initial reaction rate and the reaction rate after some time, when the concentration of substance A decreases by 0.01 mol / l.

241. In the CO + Cl 2 = COCl 2 system, the concentration was increased from 0.03 to 0.12 mol/l, and the chlorine concentration was increased from 0.02 to 0.06 mol/l. By how much did the rate of the forward reaction increase?

242. How many times will the rate of the reaction 2A + B → A 2 B change if the concentration of substance A is doubled, and the concentration of substance B is reduced by 2 times?

243. What proportion (%) of novocaine will decompose in 10 days of its storage at 293 K, if at 313 K the rate constant of novocaine hydrolysis is 1 10 -5 day -1, and the activation energy of the reaction is 55.2 kJ / mol?

244. At 36 °C, the rate constant of the disintegration of penicillin is 6·10 -6 s -1 , and at 41 °С it is 1.2·10 -5 s -1 . Calculate the temperature coefficient of the reaction.

245. How many times will the rate of a reaction occurring at 298 K increase if the activation energy is reduced by 4 kJ/mol?

246. Calculate the temperature coefficient (γ) of the rate constant of the hydrogen peroxide decomposition reaction in the temperature range 25 °C - 55 °C at E a= 75.4 kJ/mol.

247. The decomposition of hydrogen peroxide with the formation of oxygen in a 0.045 M KOH solution at 22 ° C occurs as a first-order reaction with a half-life τ 1/2 = 584 min. Calculate the reaction rate at the initial time after mixing equal volumes of 0.090 M KOH solution and 0.042 M H 2 O 2 solution and the amount of hydrogen peroxide remaining in the solution after one hour.

248. With an increase in temperature by 27.8 °C, the reaction rate increased by 6.9 times. Calculate the temperature coefficient of the reaction rate and the activation energy of this reaction at 300 K.

249. For some first-order reaction, the half-life of a substance at 351 K is 411 minutes. The activation energy is 200 kJ/mol. Calculate how long it will take to decompose 75% of the initial amount of the substance at 402 K.

250. The rate constants of a certain reaction at 25 and 60 °C are 1.4 and 9.9 min -1, respectively. Calculate the rate constants of this reaction at 20 and 75°C.

Chemical equilibrium

251. The equilibrium constant of the reaction A + B = C + D is equal to one. Initial concentration [A] = 0.02 mol/l. How many percent of substance A undergoes transformation if the initial concentrations [B] = 0.02; 0.1; 0.2 mol/l?

252. The initial concentrations of nitrogen and hydrogen in the reaction mixture to obtain ammonia were 4 and 10 mol/DM 3 respectively. Calculate the equilibrium concentrations of the components in the mixture if 50% of the nitrogen has reacted by the time equilibrium is reached.

253. The reversible reaction proceeds according to the equation A + B ↔ C + D. The initial concentration of each of the substances in the mixture is 1 mol / l. After equilibrium is established, the concentration of component C is 1.5 mol/dm 3 . Calculate the equilibrium constant for this reaction.

254. Determine the initial concentrations of NO and O 2 and the equilibrium constant of the reversible reaction 2NO + O 2 ↔ 2NO 2 if the equilibrium is established at the following concentrations of reactants, mol / dm 3: = 0.12; = 0.48; = 0.24.

255. In what direction will the chemical equilibrium in the 2NO 2 ↔ NO + O 2 system shift if the equilibrium concentrations of each component are reduced by 3 times?

256. How many times will the equilibrium decrease partial pressure hydrogen during the reaction N 2 + 3H 2 ↔ 2 NH 3 if the nitrogen pressure is doubled?

257. In the system 2NO 2 ↔ N 2 O 4 at 60 ° C and standard pressure, equilibrium was established. How many times should the volume be reduced to double the pressure?

258. In which direction will the equilibrium shift as the temperature of the system increases:

1) COCl 2 ↔ CO + Cl 2; ∆ r H° 298 = -113 kJ/mol

2) 2СО ↔ CO 2 +С; ∆ r H° 298 = -171 kJ/mol

3) 2SO 3 ↔ 2SO 2 + O 2; ∆ r H° 298 = -192 kJ/mol.

Explain the answer.

259. The reaction AB (g) ↔ A (g) + B (g) proceeds in a closed vessel. The equilibrium constant of the reaction is 0.04, and the equilibrium concentration of substance B is 0.02 mol/L. Determine the initial concentration of substance AB. What percent of substance AB has decomposed?

260. When ammonia is oxidized with oxygen, the formation of nitrogen and various nitrogen oxides is possible. Write the reaction equation and discuss the effect of pressure on the equilibrium shift of reactions with the formation of: a) N 2 O; b) NO 2 .

261. In what direction will the equilibrium for the reversible reaction C (tv) + H 2 O (g) ↔ CO (g) + H 2 (g) shift when the volume of the system decreases by 2 times?

262. At a certain temperature, the equilibrium in the 2NO 2 ↔ 2NO + O 2 system was established at the following concentrations: = 0.006 mol/l; = 0.024 mol/l. Find the equilibrium constant of the reaction and the initial concentration of NO 2 .

263. The initial concentrations of carbon monoxide and water vapor are the same and equal to 0.1 mol/l. Calculate the equilibrium concentrations of CO, H 2 O and CO 2 in the system CO (g) + H 2 O (g) ↔ CO (g) + H 2 (g), if the equilibrium concentration of hydrogen turned out to be 0.06 mol / l, determine equilibrium constant.

264. The equilibrium constant of the reaction 3H 2 + N 2 \u003d 2NH 3 at a certain temperature is 2. How many moles of nitrogen should be introduced per 1 liter of a gas mixture in order to convert 75% of hydrogen into ammonia if the initial hydrogen concentration was 10 mol / l?

265. In the system 2NO (g) + O 2 (g) ↔ 2NO 2 (g), the equilibrium concentrations of substances are = 0.2 mol / l, = 0.3 mol / l, = 0.4 mol / l. Calculate the equilibrium constant and estimate the equilibrium position.

266. In a vessel with a capacity of 0.2 liters, 0.3 and 0.8 g of hydrogen and iodine were placed. After equilibrium was established, 0.7 g of HI was found in the vessel. Calculate the equilibrium constant of the reaction.

267. The initial concentrations of H 2 and I 2 are 0.6 and 1.6 mol/L, respectively. After equilibrium was established, the concentration of hydrogen iodine was 0.7 mol/l. Calculate the equilibrium concentrations of H 2 and I 2 and the equilibrium constant.

268. At a certain temperature, the equilibrium constant of the reaction 2NO + O 2 ↔ 2NO 2 is 2.5 mol -1 l and in an equilibrium gas mixture = 0.05 mol / l and = 0.04 mol / l. Calculate the initial concentrations of oxygen and NO.

269. Substances A and B in the amount of 3 and 4 mol, respectively, located in a vessel with a capacity of 2 liters, react according to the equation 5A + 3B \u003d A 5 B 3.

1.6 mol of substance A reacted. Determine the amount of substance B consumed and the product obtained. Calculate the equilibrium constant.

270. When studying the equilibrium of the reaction H 2 + I 2 = 2HI, it was found that at initial concentrations of H 2 and I 2 of 1 mol/l, the equilibrium concentration of HI is 1.56 mol/l. Calculate the equilibrium concentration of hydrogen iodine if the initial concentrations of H 2 and I 2 were 2 mol/L each.

271. When studying the equilibrium H 2 + I 2 = 2HI, it turned out that the equilibrium concentrations of H 2 , I 2 and HI are 4.2, respectively; 4.2; 1.6 mol/l. In another experiment, carried out at the same temperature, it was found that the equilibrium concentrations of I 2 and HI are 4.0 and 1.5 mol/l. Calculate the concentration of hydrogen in this experiment.

272. At a certain temperature in the equilibrium gas system SO 2 - O 2 - SO 3, the concentrations of substances amounted to 0.035, respectively; 0.15 and 0.065 mol/l. Calculate the equilibrium constant and the initial concentrations of substances, assuming that it is only oxygen and SO 2 .

273. In a vessel with a capacity of 8.5 liters, the equilibrium CO (g) + Cl 2 (g) \u003d COCl 2 (g) was established. The composition of the equilibrium mixture (g): CO - 11, Cl 2 - 38 and COCl 2 - 42. Calculate the equilibrium constant of the reaction.

274. How do the equilibrium shift and the equilibrium constant of the reaction H 2 (g) + Cl 2 (g) \u003d 2HCl (g), ∆ H < 0, the following factors: a) increase in the concentrations of H 2 , Cl 2 and HCl; b) increase in pressure by 3 times; c) an increase in temperature?

275. The initial concentrations of NO and Cl 2 in the homogeneous system 2NO + Cl 2 = 2NOCl are 0.5 and 0.2 mol/dm 3, respectively. Calculate the equilibrium constant if 35% NO has reacted by the time of equilibrium.

Using solutions of salts of manganese, iron, copper and zinc and a solution of sodium sulfide, precipitate the indicated sulfides in four test tubes, wash the precipitates with distilled water decantation method, and then add 2-3 ml of dilute sulfuric acid solution to each of the precipitates. What's happening? Compare the experimental data with the calculation results.

Experience 3. Choosing the direction of the reaction

The following interactions are possible between E 3+ and S 2– ions in an aqueous solution:

exchange interaction;

Mutually increasing hydrolysis;

Redox reaction, if the oxidation state (+3) of the element is not too stable and can drop to (+2):

2E 3+ + 3S 2– → E 2 S 3,

2E 3+ + 3S 2– + 6H 2 O → 2E (OH) 3 + 3H 2 S,

2E 3+ + 3S 2– → 2ES + S.

Using the data in Table 2, perform the necessary calculations and find out which of these options for the course of reactions is most likely from a thermodynamic point of view during the interaction of a sodium sulfide solution with salts of three-charged iron, aluminum, chromium, and bismuth cations.

table 2

Substance	∆ f G o, kJ/mol	Substance	∆ f G o, kJ/mol
Fe 3+ (solution)	– 10,53	FeS	– 100,8
Al 3+ (solution)	– 490,5	Bi 2 S 3	– 152,9
Cr 3+ (solution)	– 223,2	Al2S3	– 492,5
Bi 3+ (solution)	+ 91,9	Fe(OH)3	– 699,6
S 2 - (rr)	+ 85,40	Cr(OH)3	– 849,0
H 2 S	– 33,50	Bi(OH)3	– 580,3
H 2 O (l.)	– 237,23	Al(OH)3	– 1157,0
H 2 O (g.)	– 228,61

By what external signs in each specific case can one determine what kind of interaction has taken place?

Pour 1-2 ml of solutions of the indicated salts into three test tubes and add 1 ml of sodium sulfide solution. What is observed in each case?

Does the prediction match the experimental results?

Topic: CHEMICAL KINETICS. CATALYSIS. EQUILIBRIUM

Laboratory work№ 7

Chemical kinetics

Literature: 1. S. 104-112; 3. S. 65-68; 4. S. 61-64.

Goal of the work: study of the influence of the concentration of reacting substances and temperature on the rate of a chemical reaction.

Questions and exercises for self-preparation.

1. The subject of chemical kinetics. Define the rate of a chemical reaction. List the factors that affect the rate of a chemical reaction.

2. Give a mathematical expression for the rate of a chemical reaction. Explain why there is a minus sign in the mathematical expression for speed. How does the reaction rate depend on temperature?

3. Formulate the law of mass action. What is the physical meaning of the rate constant, and what factors affect its value?

Effect of temperature on the rate of a chemical reaction.

4. Activation energy. activated complex. Entropy of activation.

5. Kinetic classification of reactions. Molecularity and reaction order.

6. Catalysts and catalysis.

7. Homogeneous catalysis. Theory of intermediate compounds.

8. Reversible and irreversible processes. Conditions for the onset of chemical equilibrium. The constant of chemical equilibrium and factors affecting it.

9. Le Chatelier's principle. Shift in chemical equilibrium.

1. How will the reaction rate 2NO + O 2 = 2NO 2 change if the volume of the reaction vessel is reduced by 3 times?

2. Find the value of the reaction rate constant A + B \u003d AB, if at concentrations of substances A and B, equal to 0.5 and 0.1 mol / l, respectively, the reaction rate is 0.005 mol / l s.

3. Determine by how many degrees the temperature should be raised so that the reaction rate increases by 8 times if the temperature coefficient of the reaction rate is 2.

Equipment. Volumetric test tubes - 8 pcs. Pasteur pipette (5 ml) - 2 pcs. Chemical glass (100 ml). Bath water. Laboratory thermometer (100 0 С). Stopwatch (or metronome). Electric hob.

Reagents: Sodium thiosulfate 0.5% solution, sulfuric acid 0.5% solution, distilled water.

Introduction. Thermodynamic calculations make it possible to conclude that this process is possible, to choose the conditions for conducting a chemical reaction, to determine the equilibrium composition of the products, to calculate the theoretically achievable degrees of conversion of the initial substances and the yields of the products, as well as energy effects (heat of reaction, heat of change in the state of aggregation), which is necessary for compiling energy balances and determination of energy costs.

The most important concepts of thermodynamics are “heat of process” and “work”. The quantities characterizing the state of a thermodynamic system are called thermodynamic parameters. These include: temperature, pressure, specific volume, density, molar volume, specific internal energy. Quantities proportional to the mass (or amount of matter) of the considered thermodynamic system are called extensive; these are volume, internal energy, enthalpy, entropy. Intensive quantities do not depend on the mass of the thermodynamic system, and only they serve as thermodynamic parameters as states. These are temperature, pressure, and extensive quantities related to a unit of mass, volume or amount of a substance. Changing intensive parameters in order to accelerate chemical-technological processes is called intensification.

In exothermic reactions, the stock of internal energy of the initial substances (U 1) is greater than that of the resulting products (U 2). The difference ∆U = U 1 - U 2 is converted into the form of heat. On the contrary, in endothermic reactions, due to the absorption of a certain amount of heat, the internal energy of substances increases (U 2 > U 1). ∆U is expressed in J / mol or in technical calculations they are referred to 1 kg or 1 m 3 (for gases). The study of the thermal effects of reactions or states of aggregation, or mixing, dissolution is dealt with by the section of physical chemistry or chemical thermodynamics - thermochemistry. In thermochemical equations, the heat effect of the reaction is indicated. For example: C (graphite) + O 2 \u003d CO 2 + 393.77 kJ / mol. The heats of decomposition have the opposite sign. Tables are used to define them. According to D.P. Konovalov, the heat of combustion is determined from the ratio: Q burn = 204.2n + 44.4m + ∑x (kJ / mol), where n is the number of moles of oxygen required for the complete combustion of 1 mole of a given substance, m is the number of moles of water formed during the combustion of 1 mole of a substance, ∑x is a constant correction for a given homologous series. The more unlimiting, the more ∑x.

For acetylene hydrocarbons ∑x=213 kJ/mol. For ethylene hydrocarbons ∑x=87.9 kJ/mol. For saturated hydrocarbons ∑x=0. If the molecule of the compound has different functional groups and types of bonds, then the thermal characteristic is found by summation.

The thermal effect of a reaction is equal to the sum of the heats of formation of the reaction products minus the sum of the heats of formation of the initial substances, taking into account the number of moles of all substances participating in the reaction. For example, for the reaction general view: n 1 A + n 2 B \u003d n 3 C + n 4 D + Q x thermal effect: Q x \u003d (n 3 Q C arr + n 4 Q D arr) - (n 1 Q A arr + n 2 Q B arr)

The thermal effect of a reaction is equal to the sum of the heats of combustion of the starting substances minus the sum of the heats of combustion of the reaction products, taking into account the number of moles of all reactants. For the same general reaction:

Q x \u003d (n 1 Q A burn + n 2 Q B burn) - (n 3 Q C burn + n 4 Q D burn)

Probability the course of equilibrium reactions is determined by the thermodynamic equilibrium constant, which is determined by:

К р = e - ∆ G º/(RT) = e - ∆ H º/ RT ∙ e ∆ S º/ R Analysis of this expression shows that for endothermic reactions (Q< 0, ∆Hº > 0) with a decrease in entropy (∆Sº< 0) самопроизвольное протекание реакции невозможно так как – ∆G > 0. In the following, the thermodynamic approach to chemical reactions will be considered in more detail.

Lecture 4

Basic laws of thermodynamics. First law of thermodynamics. Heat capacity and enthalpy. Enthalpy of reaction. Enthalpy of formation of the compound. Enthalpy of combustion. Hess' law and the enthalpy of reaction.

First law of thermodynamics: the change in internal energy (∆E) of the system is equal to the work external forces(А′) plus the amount of transferred heat (Q): 1)∆Е=А′+Q; or (2nd type) 2)Q=∆E+A – the amount of heat transferred to the system (Q) is spent on changing its internal energy (∆E) and work (A) done by the system. This is one of the types of the law of conservation of energy. If the change in the state of the system is very small, then: dQ=dE+δA - such a record for small (δ) changes. For gas (ideal) δА=pdV. In the isochoric process δA=0, then δQ V =dE, since dE=C V dT, then δQ V =C V dT, where C V is the heat capacity at constant volume. In a small temperature range, the heat capacity is constant, therefore Q V =C V ∆T. From this equation, it is possible to determine the heat capacity of the system and the heat of the processes. C V - according to the Joule-Lenz law. In an isobaric process proceeding without useful work, given that p is constant and can be taken out of the bracket under the differential sign, i.e. δQ P =dE+pdV=d(E+pV)=dH, here H is the enthalpy of the system. Enthalpy is the sum of the internal energy (E) of the system and the product of pressure and volume. The amount of heat can be expressed in terms of isobaric heat capacity (С Р): δQ P =С Р dT, Q V =∆E(V = const) and Q P =∆H(p = const) - after generalization. It follows that the amount of heat received by the system is uniquely determined by a change in some state function (enthalpy) and depends only on the initial and final states of the system and does not depend on the form of the path along which the process developed. This provision underlies the consideration of the problem of the thermal effects of chemical reactions.

Thermal effect of the reaction is related to the change in the chemical variable quantity of heat, obtained by the system in which the chemical reaction took place and the reaction products took the temperature of the initial reagents (as a rule, Q V and Q P).

Reactions with negative thermal effect, i.e., with the release of heat into the environment, is called exothermic. Reactions with positive thermal effect, i.e., going with the absorption of heat from the environment, are called endothermic.

The stoichiometric reaction equation will be: (1) ∆H=∑b J H J - ∑a i H i or ∆H=∑y i H i ; j are symbols of products, i are symbols of reagents.

This position is called Hess' law: quantities Е i , H i are functions of the state of the system and, consequently, ∆H and ∆Е, and thus the thermal effects Q V and Q р (Q V =∆Е, Q р =∆H) depend only on what substances react under given conditions and what products are obtained, but do not depend on the path along which the chemical process took place (reaction mechanism).

In other words, the enthalpy of a chemical reaction is equal to the sum of the enthalpies of formation of the reaction components multiplied by the stoichiometric coefficients of the corresponding components, taken with a plus sign for products and with a minus sign for starting substances. Let's find as an example∆H for the reaction PCl 5 +4H 2 O=H 3 PO 4 +5HCl (2)

The tabular values ​​of the enthalpies of formation of the reaction component are, respectively, for PCl 5 - 463 kJ / mol, for water (liquid) - 286.2 kJ / mol, for H 3 PO 4 - 1288 kJ / mol, for HCl (gas) - 92.4 kJ /mol. Substituting these values ​​into the formula: Q V =∆E, we get:

∆H=-1288+5(-92.4)–(-463)–4(-286.2)=-142kJ/mol

For organic compounds, as well as for CO, it is easy to carry out the combustion process to CO 2 and H 2 O. The stoichiometric combustion equation organic compound composition C m H n O p is written as:

(3) C m H n O p + (p-m-n / 4) O 2 \u003d mCO 2 + n / 2 H 2 O

Therefore, the enthalpy of combustion according to (1) can be expressed in terms of the enthalpies of its formation and the formation of CO 2 and H 2 O:

∆H sg =m∆H CO 2 +n/2 ∆H H 2 O -∆H CmHnOp

Having determined the heat of combustion of the studied compound with the help of a calorimeter and knowing ∆H CO 2 and ∆H H 2 O , one can find the enthalpy of its formation.

Hess' law allows you to calculate the enthalpies of any reactions, if for each component of the reaction one of its thermodynamic characteristics is known - the enthalpy of formation of a compound from simple substances. The enthalpy of formation of a compound from simple substances is understood as ∆H of a reaction leading to the formation of one mole of a compound from elements taken in their typical states of aggregation and allotropic modifications.

Lecture 5

The second law of thermodynamics. Entropy. Gibbs function. Changes in the Gibbs function during chemical reactions. Equilibrium constant and Gibbs function. Thermodynamic estimation of the probability of a reaction.

The second law of thermodynamics called the statement that it is impossible to build a perpetual motion machine of the second kind. The law was obtained empirically and has two formulations equivalent to each other:

a) a process is impossible, the only result of which is the transformation of all the heat received from a certain body into work equivalent to it;

b) a process is impossible, the only result of which is the transfer of energy in the form of heat from a body that is less heated to a body that is hotter.

The function δQ/T is the total differential of some function S: dS=(δQ/T) arr (1) – this function S is called the entropy of the body.

Here Q and S are proportional to each other, that is, with an increase in (Q) (S) - increases, and vice versa. Equation (1) corresponds to an equilibrium (reversible) process. If the process is non-equilibrium, then the entropy increases, then (1) is transformed:

dS≥(δQ/T)(2) Thus, when nonequilibrium processes occur, the entropy of the system increases. If (2) is substituted into the first law of thermodynamics, we get: dE≤TdS-δA. It is customary to write it in the form: dE≤TdS-δA'-pdV, hence: δA'≤-dE+TdS-pdV, here pdV is the equilibrium expansion work, δA' is the useful work. Integration of both parts of this inequality for an isochoric-isothermal process leads to the inequality: A'V≤-∆E+T∆S(3). And integration for the isobaric-isothermal process (Т=const, p=const) leads to the inequality:

A’ P ≤ - ∆E+T∆S – p∆V=-∆H + T∆S (4)

The right parts (3 and 4) can be written as changes to some functions, respectively:

F=E-TS(5) and G=E-TS+pV; or G=H-TS (6)

F is the Helmholtz energy and G is the Gibbs energy, then (3 and 4) can be written as A’ V ≤-∆F (7) and A’ P ≤-∆G (8). The law of equality corresponds to an equilibrium process. In this case, the most useful work is done, that is, (A’ V) MAX = -∆F, and (A’ P) MAX = -∆G. F and G are respectively called isochoric-isothermal and isobaric-isothermal potentials.

Equilibrium of chemical reactions characterized by a process (thermodynamic) in which the system goes through a continuous series of equilibrium states. Each of these states is characterized by the invariability (in time) of thermodynamic parameters and the absence of matter and heat flows in the system. The equilibrium state is characterized by the dynamic nature of the equilibrium, that is, the equality of direct and reverse processes, the minimum value of the Gibbs energy and the Helmholtz energy (that is, dG=0 and d 2 G>0; dF=0 and d 2 F>0). In dynamic equilibrium, the rates of the forward and reverse reactions are the same. Equation must also be observed:

µ J dn J =0, where µ J =(ðG/ðn J) T , P , h =G J is the chemical potential of component J; n J is the amount of component J (mol). Great importanceµ J indicates the greater reactivity of the particles.

∆Gº=-RTLnK p(9)

Equation (9) is called the van't Haff isotherm equation. The value of ∆Gº in tables in the reference literature for many thousands of chemical compounds.

K p \u003d e - ∆ G º / (RT) \u003d e - ∆ H º / RT ∙ e ∆ S º / R (11). From (11) one can give a thermodynamic estimate of the probability of the reaction occurring. So, for exothermic reactions (∆Нº<0), протекающих с возрастанием энтропии, К р >1, and ∆G<0, то есть реакция протекает самопроизвольно. Для экзотермических реакций (∆Нº>0) with a decrease in entropy (∆Sº>0), the spontaneous flow of the process is impossible.

If ∆Hº and ∆Sº have the same sign, the thermodynamic probability of the process proceeding is determined by the specific values ​​of ∆Hº, ∆Sº and Tº.

Consider, using the example of the ammonia synthesis reaction, the combined effect of ∆Н o and ∆S o on the possibility of implementing the process:

For this reaction, ∆Н o 298 = -92.2 kJ / mol, ∆S o 298 = -198 J / (mol * K), T∆S o 298 = -59 kJ / mol, ∆G o 298 = -33, 2kJ/mol.

It can be seen from the given data that the change in entropy is negative and does not favor the reaction, but at the same time, the process is characterized by a large negative enthalpy effect ∆Hº, due to which the process is possible. As the temperature increases, the reaction, as shown by calorimetric data, becomes even more exothermic (at T=725K, ∆H=-113kJ/mol), but at negative value∆S about the increase in temperature significantly reduces the likelihood of the process.

I law of thermodynamics allows you to calculate the thermal effects of various processes, but does not provide information about the direction of the process.

For the processes occurring in nature, two driving forces are known:

1. The desire of the system to go into a state with the least amount of energy;

2. The desire of the system to achieve the most probable state, which is characterized by the maximum number independent particles.

The first factor is characterized by a change in enthalpy. The case under consideration must be accompanied by heat release, therefore, DH< 0.

The second factor is determined by temperature and change entropy.

Entropy (S)- thermodynamic function of the state of the system, which reflects the probability of the realization of one or another state of the system in the process of heat transfer.

Like energy, entropy is not among the experimentally determined quantities. In a reversible process occurring under isothermal conditions, the change in entropy can be calculated using the formula:

This means that with the irreversible course of the process, the entropy increases due to the transfer of part of the work into heat.

Thus, in reversible processes, the system performs the maximum possible work. In an irreversible process, the system always does less work.

The transition of lost work into heat is a feature of heat as a macroscopically disordered form of energy transfer. Hence the interpretation of entropy as a measure of disorder in the system arises:

As the disorder in the system increases, the entropy increases and, conversely, as the system is ordered, the entropy decreases.

So, in the process of evaporation of water, the entropy increases, in the process of crystallization of water, it decreases. In decomposition reactions, entropy increases, in compound reactions it decreases.

The physical meaning of entropy was established by statistical thermodynamics. According to the Boltzmann equation:

The direction of the spontaneous flow of the process depends on the ratio of the quantities on the left and right sides of the last expression.

If the process takes place under isobaric-isothermal conditions, then the overall driving force of the process is called Gibbs free energy or isobaric-isothermal potential (DG):

.

(15)

The value of DG allows you to determine the direction of the spontaneous flow of the process:

If DG< 0, то процесс самопроизвольно протекает в прямом направлении;

If DG > 0, then the process proceeds spontaneously in the opposite direction;

If DG=0, then the state is in equilibrium.

In living organisms, which are open systems, the main source of energy for many biological reactions - from protein biosynthesis and ion transport to muscle contraction and electrical activity of nerve cells - is ATP (adenosine-5¢-triphosphate).

Energy is released during the hydrolysis of ATP:

ATP + H 2 O ⇄ ADP + H 3 PO 4

where ADP is adenosine-5¢-diphosphate.

The DG 0 of this reaction is -30 kJ, so the process proceeds spontaneously in the forward direction.

Analysis of the ratio of enthalpy and entropy factors in the equation for calculating the isobaric-isothermal potential allows us to draw the following conclusions:

1. At low temperatures, the enthalpy factor prevails, and exothermic processes proceed spontaneously;

2. When high temperatures the entropy factor prevails, and processes proceed spontaneously, accompanied by an increase in entropy.

On the basis of the above material, it is possible to formulate II law of thermodynamics:

Under isobaric-isothermal conditions in an isolated system, those processes spontaneously occur that are accompanied by an increase in entropy.

Indeed, in an isolated system, heat transfer is impossible, therefore, DH = 0 and DG » -T×DS. This shows that if the value of DS is positive, then the value of DG is negative and, therefore, the process spontaneously proceeds in the forward direction.

Another formulation of the II law of thermodynamics:

An uncompensated transfer of heat from less heated bodies to more heated ones is impossible.

In chemical processes, changes in entropy and Gibbs energy are determined in accordance with the Hess law:

,	(16)
.	(17)

Reactions for which DG< 0 называют exergonic.

Reactions for which DG > 0 are called endergonic.

The value of DG of a chemical reaction can also be determined from the relationship:

DG = DH - T×DS.

In table. 1 shows the possibility (or impossibility) of a spontaneous reaction with various combinations of signs DH and DS.

Problem Solving Standards

1. Some reaction proceeds with a decrease in entropy. Determine the conditions under which the spontaneous occurrence of this reaction is possible.

The condition for the spontaneous occurrence of the reaction is a decrease in the Gibbs free energy, i.e. DG< 0. Изменение DG можно рассчитать по формуле:

Since the entropy decreases during the reaction (DS< 0), то энтропийный фактор препятствует самопроизвольному протеканию данной реакции. Таким образом, самопроизвольное протекание данной реакции может обеспечить только энтальпийный фактор. Для этого необходимо выполнение следующих условий:

1) DH< 0 (реакция экзотермическая);

2) (the process must proceed at low temperatures).

2. Endothermic decomposition reaction proceeds spontaneously. Estimate the change in enthalpy, entropy and Gibbs free energy.

1) Since the reaction is endothermic, then DH > 0.

2) In decomposition reactions, entropy increases, hence DS > 0.

3) Spontaneous occurrence of the reaction indicates that DG< 0.

3. Calculate the standard enthalpy of chemosynthesis occurring in Thiobacillus denitrificans bacteria:

6KNO 3 (solid) + 5S (solid) + 2CaCO 3 (solid) \u003d 3K 2 SO 4 (solid) + 2CaSO 4 (solid) + 2CO 2 (gas) + 3N 2 (gas)

according to the values ​​of the standard enthalpies of formation of substances:

Let us write the expression for the first consequence of the Hess law, taking into account the fact that the standard enthalpies of formation of sulfur and nitrogen are equal to zero:

\u003d (3 × K 2 SO 4 + 2 × CaSO 4 + 2 × CO 2) -

- (6× KNO 3 + 2× CaCO 3).

Let us substitute the values ​​of the standard enthalpies of formation of substances:

3×(-1438) + 2×(-1432) + 2×(-393.5) - (6×(-493) + 2×(-1207)).

2593 kJ.

Because< 0, то реакция экзотермическая.

4. Calculate the standard enthalpy of reaction:

2C 2 H 5 OH (liquid) \u003d C 2 H 5 OC 2 H 5 (liquid) + H 2 O (liquid)

according to the values ​​of the standard enthalpies of combustion of substances:

C 2 H 5 OH \u003d -1368 kJ / mol;

C 2 H 5 OC 2 H 5 \u003d -2727 kJ / mol.

Let us write the expression for the second consequence of the Hess law, taking into account the fact that the standard enthalpy of combustion of water (the highest oxide) is equal to zero:

2× C 2 H 5 OH - C 2 H 5 OC 2 H 5 .

Let us substitute the values ​​of the standard enthalpies of combustion of the substances involved in the reaction:

2×(-1368) - (-2727).

Consequences from Hess's law make it possible to calculate not only the standard enthalpies of reactions, but also the values ​​of the standard enthalpies of formation and combustion of substances from indirect data.

5. Determine the standard enthalpy of formation of carbon monoxide (II) according to the following data:

Equation (1) shows that the standard enthalpy change for this reaction corresponds to the standard enthalpy of formation of CO 2 .

Let us write the expression for the first corollary from the Hess law for reaction (2):

CO = CO 2 -.

Substitute the values ​​and get:

CO \u003d -293.5 - (-283) \u003d -110.5 kJ / mol.

This problem can be solved in another way.

Subtracting the second equation from the first equation, we get:

6. Calculate the standard entropy of the reaction:

CH 4 (gas) + Cl 2 (gas) \u003d CH 3 Cl (gas) + HCl (gas),

according to the values ​​of the standard entropies of substances:

We calculate the standard entropy of the reaction by the formula:

\u003d (CH 3 Cl + HCl) - (CH 4 + Cl 2).

234 + 187 - (186 + 223) = 12 J / (mol × K).

7. Calculate the standard Gibbs energy of the reaction:

C 2 H 5 OH (liquid) + H 2 O 2 (liquid) \u003d CH 3 COH (gas) + 2H 2 O (liquid)

according to the following data:

Determine whether the spontaneous occurrence of this reaction is possible under standard conditions.

We calculate the standard Gibbs energy of the reaction by the formula:

\u003d (CH 3 COH + 2 × H 2 O) - (C 2 H 5 OH + H 2 O 2).

Substituting tabular values, we get:

129 + 2 × (-237) - ((-175) + (-121) = -307 kJ / mol.

Because< 0, то самопроизвольное протекание данной реакции возможно.

C 6 H 12 O 6 (solid) + 6O 2 (gas) \u003d 6CO 2 (gas) + 6H 2 O (liquid).

known data:

We calculate the values ​​of the standard enthalpy and entropy of the reaction using the first corollary from the Hess law:

6 CO 2 + 6 H 2 O - C 6 H 12 O 6 - 6 O 2 \u003d

6×(-393.5) + 6×(-286) - (-1274.5) - 6×0 = -2803 kJ;

6 CO 2 + 6 H 2 O - C 6 H 12 O 6 - 6 O 2 \u003d

6×214 + 6×70 - 212 - 6×205 = 262 J/K = 0.262 kJ/K.

We find the standard Gibbs energy of the reaction from the relation:

T× = -2803 kJ - 298.15 K×0.262 kJ/K =

9. Calculate the standard Gibbs energy of the serum albumin hydration reaction at 25 0 С, for which DH 0 = -6.08 kJ/mol, DS 0 = -5.85 J/(mol×K). Estimate the contribution of the enthalpy and entropy factors.

We calculate the standard Gibbs energy of the reaction by the formula:

DG 0 = DH 0 - T×DS 0 .

Substituting the values, we get:

DG 0 \u003d -6.08 kJ / mol - 298 K × (-5.85 × 10 - 3) kJ / (mol × K) \u003d

4.34 kJ/mol.

In this case, the entropy factor prevents the reaction from proceeding, while the enthalpy factor favors it. A spontaneous reaction is possible if , i.e., at low temperatures.

10. Determine the temperature at which the trypsin denaturation reaction will spontaneously proceed, if = 283 kJ/mol, = 288 J/(mol×K).

The temperature at which both processes are equally probable can be found from the relation:

In this case, the enthalpy factor prevents the reaction from proceeding, while the entropy factor favors it. A spontaneous reaction is possible if:

Thus, the condition for the spontaneous occurrence of the process is T > 983 K.

Questions for self-control

1. What is a thermodynamic system? What types of thermodynamic systems do you know?

2. List the thermodynamic parameters known to you. Which of them are measurable? Which to the immeasurable?

3. What is a thermodynamic process? What are processes that occur when one of the parameters is constant called?

4. What processes are called exothermic? What are endothermic?

5. What processes are called reversible? Which are irreversible?

6. What is meant by the term "state of the system"? What are the states of the system?

7. What systems are studied by classical thermodynamics? State the first and second postulates of thermodynamics.

8. What variables are called state functions? List the state functions you know.

9. What is internal energy? Can internal energy be measured?

10. What is enthalpy? What is its dimension?

11. What is entropy? What is its dimension?

12. What is Gibbs free energy? How can it be calculated? What can be determined using this function?

13. What reactions are called exergonic? Which are endergonic?

14. Formulate the first law of thermodynamics. What is the equivalence of heat and work?

15. Formulate Hess's law and its consequences. What is the standard enthalpy of formation (combustion) of a substance?

16. Formulate the second law of thermodynamics. Under what condition does a process proceed spontaneously in an isolated system?

Variants of tasks for independent solution

Option number 1

4NH 3 (gas) + 5O 2 (gas) \u003d 4NO (gas) + 6H 2 O (gas),

Determine what type (exo- or endothermic) this reaction belongs to.

C 2 H 6 (gas) + H 2 (gas) \u003d 2CH 4 (gas),

3. Calculate the standard Gibbs energy of the b-lactoglobulin hydration reaction at 25 0 С, for which DH 0 = -6.75 kJ, DS 0 = -9.74 J/K. Estimate the contribution of the enthalpy and entropy factors.

Option number 2

1. Calculate the standard enthalpy of reaction:

2NO 2 (gas) + O 3 (gas) \u003d O 2 (gas) + N 2 O 5 (gas),

using the values ​​of the standard enthalpies of formation of substances:

Determine what type (exo- or endothermic) this reaction belongs to.

2. Calculate the standard enthalpy of reaction:

using the values ​​of the standard enthalpies of combustion of substances:

3. Calculate the standard Gibbs energy of the reaction of thermal denaturation of chymotrypsinogen at 50 0 С, for which DH 0 = 417 kJ, DS 0 = 1.32 J/K. Estimate the contribution of the enthalpy and entropy factors.

Option number 3

1. Calculate the standard enthalpy of the reaction of hydrogenation of benzene to cyclohexane in two ways, i.e., using the values ​​of the standard enthalpies of formation and combustion of substances:

Cu (solid) + ZnO (solid) = CuO (solid) + Zn (solid)

3. During the reduction of 12.7 g of copper (II) oxide with coal (with the formation of CO), 8.24 kJ of heat is absorbed. Determine the standard enthalpy of formation of CuO if CO = -111 kJ/mol.

Option number 4

1. Calculate the standard enthalpy of chemosynthesis occurring in the autotrophic bacteria Baglatoa and Thiothpix, by stages and in total:

2H 2 S (gas) + O 2 (gas) \u003d 2H 2 O (liquid) + 2S (solid);

2S (solid) + 3O 2 (gas) + 2H 2 O (liquid) \u003d 2H 2 SO 4 (liquid),

2. Calculate the standard enthalpy of reaction:

C 6 H 12 O 6 (solid) \u003d 2C 2 H 5 OH (liquid) + 2CO 2 (gas),

using the values ​​of the standard enthalpies of combustion of substances:

4HCl (gas) + O 2 (gas) \u003d 2Cl 2 (gas) + 2H 2 O (liquid)

known data:

Option number 5

1. Calculate the standard enthalpy of reaction:

2CH 3 Cl (gas) + 3O 2 (gas) \u003d 2CO 2 (gas) + 2H 2 O (liquid) + 2HCl (gas),

using the values ​​of the standard enthalpies of formation of substances:

Determine what type (exo- or endothermic) this reaction belongs to.

2. Calculate the standard enthalpy of reaction:

C 6 H 6 (liquid) + 3H 2 (gas) \u003d C 6 H 12 (liquid),

using the values ​​of the standard enthalpies of combustion of substances:

3. Calculate the standard Gibbs energy of the trypsin denaturation reaction at 50 0 С, for which DH 0 = 283 kJ, DS 0 = 288 J/K). Assess the possibility of the process proceeding in the forward direction.

Option number 6

1. Calculate the standard enthalpy of chemosynthesis in autotrophic bacteria Thiobacillus Thioparus:

5Na 2 S 2 O 3 × 5H 2 O (solid) + 7O 2 (gas) \u003d 5Na 2 SO 4 (solid) + 3H 2 SO 4 (well) + 2S (solid) + 22H 2 O (well .),

Determine what type (exo- or endothermic) this reaction belongs to.

2. Calculate the standard enthalpy of reaction:

C 6 H 5 NO 2 (liquid) + 3H 2 (gas) \u003d C 6 H 5 NH 2 (liquid) + 2H 2 O (liquid),

using the values ​​of the standard enthalpies of combustion of substances:

3. Evaluate the role of enthalpy and entropy factors for the reaction:

H 2 O 2 (liquid) + O 3 (gas) \u003d 2O 2 (gas) + H 2 O (liquid)

known data:

Determine the temperature at which the reaction will proceed spontaneously.

Option number 7

1. Calculate the standard enthalpy of formation of CH 3 OH from the following data:

CH 3 OH (liquid) + 1.5O 2 (gas) = ​​CO 2 (gas) + 2H 2 O (liquid) DH 0 = -726.5 kJ;

C (graphite) + O 2 (gas) \u003d CO 2 (gas) DH 0 \u003d -393.5 kJ;

H 2 (gas) + 0.5O 2 (gas) \u003d H 2 O (liquid) DH 0 \u003d -286 kJ.

2. Assess the possibility of a spontaneous reaction:

8Al (solid) + 3Fe 3 O 4 (solid) \u003d 9Fe (solid) + Al 2 O 3 (solid)

under standard conditions, if:

3. Calculate the value of DH 0 for possible reactions of glucose conversion:

1) C 6 H 12 O 6 (cr.) \u003d 2C 2 H 5 OH (liquid) + 2CO 2 (gas);

2) C 6 H 12 O 6 (cr.) + 6O 2 (gas) \u003d 6CO 2 (gas) + 6H 2 O (liquid).

known data:

Which of these reactions produces large quantity energy?

Option number 8

1. Calculate the standard enthalpy of formation of MgCO 3 from the following data:

MgO (solid) + CO 2 (gas) = ​​MgCO 3 (solid) +118 kJ;

C 2 H 6 (gas) + H 2 (gas) \u003d 2CH 4 (gas)

known data:

3. Which of the following oxides: CaO, FeO, CuO, PbO, FeO, Cr 2 O 3 can be reduced by aluminum to free metal at 298 K:

Option number 9

1. Calculate the standard enthalpy of formation of Ca 3 (PO 4) 2 from the following data:

3CaO (solid) + P 2 O 5 (solid) \u003d Ca 3 (PO 4) 2 (solid) DH 0 \u003d -739 kJ;

P 4 (solid) + 5O 2 (gas) \u003d 2P 2 O 5 (solid) DH 0 \u003d -2984 kJ;

Ca (solid) + 0.5O 2 (gas) \u003d CaO (solid) DH 0 \u003d -636 kJ.

2. Assess the possibility of a spontaneous reaction:

Fe 2 O 3 (solid) + 3CO (gas) \u003d 2Fe (solid) + 3CO 2 (gas)

under standard conditions, if:

3. Determine which of the listed oxides: CuO, PbO 2, ZnO, CaO, Al 2 O 3 can be reduced by hydrogen to a free metal at 298 K, if it is known:

Option number 10

1. Calculate the standard enthalpy of formation of ethanol from the following data:

DH 0 burned out C 2 H 5 OH \u003d -1368 kJ / mol;

C (graphite) + O 2 (gas) \u003d CO 2 (gas) + 393.5 kJ;

H 2 (gas) + O 2 (gas) \u003d H 2 O (liquid) +286 kJ.

2. Calculate the standard entropy of the reaction:

C 2 H 2 (gas) + 2H 2 (gas) \u003d C 2 H 6 (gas),

known data:

3. Calculate the amount of energy that will be released in the human body, who ate 2 pieces of sugar, 5 g each, assuming that the main way of sucrose metabolism is to oxidize it:

C 12 H 22 O 11 (solid) + 12O 2 (gas) = ​​12CO 2 (gas) + 11H 2 O (liquid) = -5651 kJ.

Option number 11

1. Calculate the standard enthalpy of formation C 2 H 4 from the following data:

C 2 H 4 (gas) + 3O 2 (gas) \u003d 2CO 2 (gas) + 2H 2 O (liquid) + 1323 kJ;

C (graphite) + O 2 (gas) \u003d CO 2 (gas) + 393.5 kJ;

H 2 (gas) + 0.5O 2 (gas) = ​​H 2 O (liquid) +286 kJ.

2. Without performing calculations, set the sign DS 0 of the following processes:

1) 2NH 3 (gas) \u003d N 2 (gas) + 3H 2 (gas);

2) CO 2 (cr.) \u003d CO 2 (gas);

3) 2NO (gas) + O 2 (gas) = ​​2NO 2 (gas).

3. Determine according to which reaction equation the decomposition of hydrogen peroxide will proceed under standard conditions:

1) H 2 O 2 (gas) \u003d H 2 (gas) + O 2 (gas);

2) H 2 O 2 (gas) \u003d H 2 O (liquid) + 0.5O 2 (gas),

Option number 12

1. Calculate the standard enthalpy of formation of ZnSO 4 from the following data:

2ZnS + 3O 2 = 2ZnO + SO 2 DH 0 = -890 kJ;

2SO 2 + O 2 \u003d 2SO 3 DH 0 \u003d -196 kJ;

H 2 O (solid) \u003d H 2 O (liquid),

H 2 O (liquid) \u003d H 2 O (gas),

H 2 O (solid) \u003d H 2 O (gas).

known data:

3. Calculate the amount of energy that will be released during the combustion of 10 g of benzene, according to the following data:

Option number 14

1. Calculate the standard enthalpy of formation of PCl 5 from the following data:

P 4 (solid) + 6Cl 2 (gas) = ​​4PCl 3 (gas) DH 0 = -1224 kJ;

PCl 3 (gas) + Cl 2 (gas) \u003d PCl 5 (gas) DH 0 \u003d -93 kJ.

2. Calculate the standard change in the Gibbs energy of carbon disulfide formation CS 2 from the following data:

CS 2 (liquid) + 3O 2 (gas) \u003d CO 2 (gas) + 2SO 2 (gas) DG 0 \u003d -930 kJ;

CO 2 \u003d -394 kJ / mol; SO 2 \u003d -300 kJ / mol.

3. Evaluate the role of enthalpy and entropy factors for the reaction:

CaCO 3 (solid) \u003d CaO (solid) + CO 2 (gas)

known data:

Determine the temperature at which the reaction will proceed spontaneously.

Option number 15

1. Calculate the thermal effect of the reaction of formation of crystalline hydrate CuSO 4 × 5H 2 O proceeding according to the equation:

CuSO 4 (solid) + 5H 2 O (liquid) \u003d CuSO 4 × 5H 2 O (solid),

One of the most important questions in chemistry is the question of the possibility of a chemical reaction. A quantitative criterion for the fundamental feasibility of a chemical reaction is, in particular, the characteristic function of the state of the system, called the Gibbs energy (G). Before proceeding to the consideration of this criterion, let us dwell on a number of definitions.

spontaneous processes. Spontaneous processes are processes that occur without energy supply from an external source. Many chemical processes are spontaneous, such as the dissolution of sugar in water, the oxidation of metals in air (corrosion), etc.

Reversible and irreversible processes. Many chemical reactions proceed in one direction until the reactants are completely exhausted. Such reactions are called chemically irreversible. An example is the interaction of sodium and water.

Other reactions proceed first in the forward direction, and then in the forward and reverse direction due to the interaction of the reaction products. As a result, a mixture is formed containing both the starting materials and the reaction products. Such reactions are called chemically reversible. As a result of a chemically reversible process, true (stable) chemical equilibrium, which is characterized by the following features:

1) in the absence of external influences, the state of the system remains unchanged indefinitely;

2) any change external conditions leads to a change in the state of the system;

3) the state of equilibrium does not depend on which side it is reached from.

An example of a system in true equilibrium is an equimolecular mixture

CO (g) + H 2 O (g) CO 2 (g) + H 2 (g).

Any change in temperature or other conditions causes a shift in equilibrium, i.e. change in the composition of the system.

In addition to true equilibria, apparent (false, hindered) equilibria are very often encountered, when the state of the system persists in time for a very long time, but a small impact on the system can lead to a strong change in its state. An example is a mixture of hydrogen and oxygen, which at room temperature in the absence of external influences can remain unchanged indefinitely. However, it is enough to introduce platinized asbestos (catalyst) into this mixture, as an energetic reaction will begin.

H 2 (g) + O 2 (g) \u003d H 2 O (g),

leading to complete exhaustion of the starting materials.

If the same catalyst is introduced under the same conditions into liquid water, then it is impossible to obtain the initial mixture.

Entropy. The state of any system can be characterized by the values ​​of directly measured parameters (p, T, etc.). This characteristic of the macrostate of the system. The state of the system can also be described by the characteristics of each particle of the system (atom, molecule): coordinate, vibration frequency, rotation frequency, etc. This characteristic of the microstate of the system. Systems consist of a very large number of particles, so one macrostate will correspond to a huge number of different microstates. This number is called the thermodynamic probability of the state and is denoted as W.

Thermodynamic probability is associated with another property of matter - entropy (S, J / (mol. K)) - Boltzmann formula

where R is the universal gas constant and N A is the Avogadro constant.

The physical meaning of entropy can be explained by the following thought experiment. Let an ideal crystal of some substance, such as sodium chloride, be cooled to absolute zero temperature. Under these conditions, the sodium and chlorine ions that make up the crystal become practically immobile, and this macroscopic state is characterized by a single microstate, i.e. W=1, and in accordance with (3.13) S=0. As the temperature rises, the ions begin to oscillate around the equilibrium positions in the crystal lattice, the number of microstates corresponding to one macrostate increases, and, consequently, S>0.

Thus, entropy is a measure of the disorder of the state of a system. The entropy of the system increases in all processes accompanied by a decrease in order (heating, dissolution, evaporation, decomposition reactions, etc.). Processes that occur with an increase in order (cooling, crystallization, compression, etc.) lead to a decrease in entropy.

Entropy is a function of state, but unlike most other thermodynamic functions, it is possible to experimentally determine the absolute value of the entropy of a substance. This possibility is based on the postulate of M. Planck, according to which at absolute zero, the entropy of an ideal crystal is zero(third law of thermodynamics).

The temperature dependence of the entropy of a substance is presented qualitatively in Fig. . 3.1.

On fig. 3.1 it can be seen that at a temperature equal to 0 K, the entropy of a substance is zero. With an increase in temperature, the entropy increases smoothly, and at the points of phase transitions, an abrupt increase in entropy takes place, determined by the relation

(3.14)

where Δ f.p S, Δ f.p H and T f.p are the changes in entropy, enthalpy and phase transition temperature, respectively.

The entropy of a substance B in the standard state is denoted as . For many substances, the absolute values ​​of the standard entropies are determined and are given in reference books.

Entropy, like internal energy and enthalpy, is a function of state, so the change in the entropy of a system in a process does not depend on its path and is determined only by the initial and final states of the system. The change in entropy during a chemical reaction (3.10) can be found as the difference between the sum of the entropies of the reaction products and the sum of the entropies of the starting materials:

The concept of entropy is used in one of the formulations second law of thermodynamics: in isolated systems, only processes occurring with an increase in entropy (ΔS>0) can spontaneously proceed. Isolated systems are systems that do not exchange with environment neither matter nor energy. Systems in which chemical processes take place do not belong to isolated systems, because they exchange energy with the environment (the thermal effect of the reaction), and in such systems processes can also occur with a decrease in entropy.

SO 2 (g) + 2H 2 S (g) \u003d 3S (t) + 2H 2 O (l), if the standard entropies of sulfur oxide (IV), hydrogen sulfide, sulfur and water are 248.1; 205.64; 31.88 and 69.96 J/(mol K), respectively.

Solution. Based on equation (3.15), we can write:

The entropy in this reaction decreases, which is associated with the formation of solid and liquid products from gaseous substances.

Example 3.8. Without making calculations, determine the sign of the change in entropy in the following reactions:

1) NH 4 NO 3 (c) \u003d N 2 O (g) + 2H 2 O (g),

2) 2H 2 (g) + O 2 (g) \u003d 2H 2 O (g),

3) 2H 2 (g) + O 2 (g) \u003d 2H 2 O (g).

Solution. In reaction (1), 1 mol of NH 4 NO 3 in the crystalline state forms 3 mol of gases, therefore, D r S 1 >0.

In reactions (2) and (3), both the total number of moles and the number of moles of gaseous substances decrease. Therefore, D r S 2<0 и D r S 3 <0. При этом уменьшение энтропии в реакции (3) больше, чем в реакции (2) , так как S о (H 2 O (ж)) < S о (H 2 O (г)).

Gibbs energy(isobaric-isothermal potential). In many cases, spontaneous processes in nature occur in the presence of a potential difference, for example, the difference in electrical potentials causes charge transfer, and the difference in gravitational potentials causes the body to fall. These processes end when the minimum potential is reached. The driving force of chemical processes occurring at constant pressure and temperature is the isobaric-isothermal potential, called Gibbs energy and denoted G. The change in the Gibbs energy in a chemical process is determined by the relation

ΔG = ΔH –TΔS, (3.16)

where ΔG is the change in the Gibbs energy of the chemical process; ΔH is the change in the enthalpy of the chemical process; ΔS is the change in the entropy of the chemical process; T is temperature, K.

Equation (3.16) can be represented in the following form:

∆H = ∆G + T∆S. (3.17)

The meaning of equation (3.17) is that part of the heat effect of the reaction is spent on doing work (ΔG), and part is dissipated into the environment (TΔS).

The Gibbs energy is a criterion for the fundamental possibility of a spontaneous reaction. If the Gibbs energy decreases during the reaction, then the process can proceed spontaneously under these conditions:

ΔG< 0. (3.18)

The process under these conditions is not feasible if

ΔG > 0. (3.19)

Expressions (3.18) and (3.19) simultaneously mean that the reverse reaction cannot (3.18) or can (3.19) proceed spontaneously.

The reaction is reversible, i.e. can flow in both forward and reverse directions, if

Equation (3.20) is a thermodynamic condition for chemical equilibrium.

Relations (3.18) – (3.20) are also applicable to phase equilibria, i.e. to cases when two phases (aggregate states) of the same substance are in equilibrium, for example, ice and liquid water.

Enthalpy and entropy factors. It follows from equations (3.16) and (3.18) that processes can proceed spontaneously (ΔG<0), если они сопровождаются уменьшением энтальпии (ΔH<0) и увеличением энтропии системы (ΔS>0). If the enthalpy of the system increases (ΔH>0), and the entropy decreases (ΔS<0), то такой процесс протекать не может (ΔG>0). With other signs of ΔS and ΔН, the fundamental possibility of the process proceeding is determined by the ratio of the enthalpy (ΔH) and entropy (ТΔS) factors.

If ΔН>0 and ΔS>0, i.e. Since the enthalpy component counteracts, and the entropy component favors the course of the process, the reaction can proceed spontaneously due to the entropy component, provided that |ΔH|<|TΔS|.

If the enthalpy component favors and the entropy counteracts the process, then the reaction can proceed spontaneously due to the enthalpy component, provided that |ΔH|>|TΔS|.

Effect of temperature on the direction of the reaction. Temperature affects the enthalpy and entropy components of the Gibbs energy, which may be accompanied by a change in the sign of the Gibbs energy of these reactions, and hence the direction of the reactions. For a rough estimate of the temperature at which the sign of the Gibbs energy changes, we can neglect the dependence of ΔН and ΔS on temperature. Then it follows from Eq. (3.16) that the sign of the Gibbs energy will change at the temperature

It is obvious that the sign change of the Gibbs energy with temperature change is possible only in two cases: 1) ΔН>0 and ΔS>0 and 2) ΔН<0 и ΔS<0.

The standard Gibbs energy of formation is the change in the Gibbs energy of the reaction of formation of 1 mol of a compound from simple substances that are stable under standard conditions. The Gibbs energy of the formation of simple substances is assumed to be zero. The standard Gibbs energies of the formation of substances can be found in the relevant reference books.

Gibbs energy of a chemical reaction. The Gibbs energy is a state function, i.e. its change in the process does not depend on the path of its flow, but is determined by the initial and final states of the system. Therefore, the Gibbs energy of the chemical reaction (3.10) can be calculated from the formula

Note that the conclusions about the fundamental possibility of the reaction proceeding in terms of Δ r G are applicable only to those conditions for which the change in the Gibbs energy of the reaction is calculated. If the conditions differ from the standard, then the equation can be used to find Δ r G van't Hoff isotherms, which for the reaction (3.10) between gases is written as

(3.23)

and between solutes

(3.24)

where are the partial pressures of the corresponding substances; c A, c B, c D , c E are the concentrations of the corresponding dissolved substances; a, b, d, e are the corresponding stoichiometric coefficients.

If the reactants are in the standard state, then equations (3.23) and (3.24) become the equation

Example 3.9. Determine the possibility of the reaction NH 3 (g) + HCl (g) \u003d NH 4 Cl (k) under standard conditions at a temperature of 298.15 K, using data on standard enthalpies of formation and entropies.

Solution. Based on the first corollary of the Hess law, we find the standard enthalpy of the reaction:

; the reaction is exothermic, therefore, the enthalpy component favors the reaction.

We calculate the change in the entropy of the reaction according to the equation

The reaction is accompanied by a decrease in entropy, which means that the entropy component counteracts the reaction.

We find the change in the Gibbs energy of the process according to equation (3.16):

Thus, this reaction can proceed spontaneously under standard conditions.

Example 3.10. Using data on standard enthalpies of formation and entropies, determine at what temperature equilibrium will occur in the system N 2 (g) + 3H 2 (g) \u003d 2NH 3 (g).

Solution. The equilibrium condition for the system is ΔG=0. To do this, using relation (3.21), we find the temperature at which ΔG=0. Calculate the standard enthalpy and entropy of the reaction:

The enthalpy component favors, and the entropy component opposes the reaction, which means that at a certain temperature it is possible to change the sign of the Gibbs energy, i.e. change the direction of the reaction.

The equilibrium condition will be written as follows:

∆G = ∆H –T∆S,

or, substituting numerical values, we get

0 \u003d - 92.38 - T (-198.3) 10 -3.

Therefore, the reaction will be in equilibrium at a temperature

TO.

Below this temperature, the reaction will proceed in the forward direction, and above this temperature, in the opposite direction.

Example 3.11. At a certain temperature T, the endothermic reaction A®B practically goes to completion. Determine: a) the sign D r S of the reaction; b) sign DG of the reaction B ® A at temperature T; c) the possibility of the reaction B ® A at low temperatures.

Solution. a) The spontaneous occurrence of the reaction A ® B indicates that DG<0. Поскольку DН>0, then from the equation
DG = DH - TDS implies that DS>0; for reverse reaction B ® A DS<0.

b) For the reaction A ® B DG<0. Следовательно, для обратной реакции при той же температуре DG>0.

c) The reaction A ® B is endothermic (DH<0), следовательно, обратная реакция В ® А экзотермическая. При низких температурах абсолютная величина члена TDS мала, так что знак DG определяется знаком DН. Следовательно, при достаточно низких температурах протекание реакции В ® А возможно.

Example 3.12. Calculate the value of the Gibbs energy and determine whether the reaction CO + Cl 2 ÛCOCl 2 is possible at a temperature of 700 K, if the equilibrium constant of the reaction at this temperature is 10.83 atm -1 and the partial pressures of all components are the same and equal to one.

Solution. The relationship D r G 0 and K r of the reaction A + B Û C + D is given by the isotherm equation (3.22)

Under standard conditions, when the partial pressure of each reactant is 1 atm, this ratio will take the form

Consequently, the reaction at T=700 K can proceed spontaneously in the forward direction.

Questions and tasks for self-study

1. Give the numerical values ​​of pressure and temperature in the international system of units, as well as in atmospheres, millimeters of mercury and degrees Celsius, corresponding to standard and normal conditions.

2. What condition do the state functions satisfy? What determines the change in the value of the state function in the process?

3. The constancy of what parameters characterizes isobaric-isothermal and isochoric-isothermal processes?

4. Formulate the first law of thermodynamics.

5. Under what conditions will the thermal effect of the process be: a) equal to the change in the enthalpy of this process; b) is equal to the change in the internal energy of the process?

6. The chemical reaction takes place in a sealed reactor. The change in which state function will determine the thermal effect of the reaction?

7. During a chemical reaction, the temperature of the system rises. Is this process exothermic or endothermic? What sign (+) or (-) does the enthalpy change of this process have?

8. Formulate Hess' law.

9. Define the term "standard enthalpy of formation of a substance."

10. What are the standard enthalpies of formation of molecular chlorine and stable at a temperature of 298 K modification of iron α-Fe?

11. The standard enthalpy of formation of white phosphorus is zero, and red - (-18.41) kJ / mol. Which of the allotropic modifications is more stable at a temperature of 25 o C?

12. Formulate the 1st corollary of Hess' law.

13. Define the concept of "standard enthalpy of combustion of a substance."

14. How are the standard enthalpy of formation of carbon dioxide and the standard enthalpy of combustion stable at T = 298 K modification of carbon - graphite?

15. Give 3 examples of spontaneous chemical processes.

16. List the signs of chemical (true) equilibrium.

17. Give examples of processes accompanied by: a) an increase in entropy; b) a decrease in entropy.

18. What sign should the change in the entropy of a spontaneously occurring reaction have if Δ r Н=0?

19. What sign should the change in the entropy of the reaction of thermal decomposition of calcium carbonate have? Why? Write the reaction equation.

20. What thermodynamic properties of the participants in the reaction do you need to know in order to resolve the issue of the possibility of a reaction?

21. An exothermic reaction between gases is accompanied by an increase in volume. What can be said about the possibility of such a reaction?

22. In which of the following cases is it possible to change the direction of the reaction with a change in temperature: a) DH<0, DS<0; б) DH>0, DS>0; c) DH<0, DS>0; d) DH>0, DS<0?

23. Find the standard enthalpy for the oxidation of gaseous sulfur(IV) oxide with oxygen to gaseous sulfur(VI) oxide. Standard enthalpies of formation SO 2 - (-297 kJ / mol) and SO 3 - (-395 kJ / mol).

Answer: -196 kJ.

24. Indicate the sign of the change in entropy in the following reactions:

a) CO (G) + H 2 (G) \u003d C (T) + H 2 O (G);

b) CO 2 (G) + C (T) \u003d 2CO (G);

c) FeO (T) + CO (G) \u003d Fe (T) + CO 2 (G);

d) H 2 O (W) \u003d H 2 O (G);

Answer: a) (-); b) (+); c)(~0); d) (+); e) (-).

25. Find the standard entropy of the reaction of oxidation of gaseous sulfur(IV) oxide with oxygen to gaseous sulfur(VI) oxide. Standard entropy of formation of SO 2 - (248 J / (mol K), SO 3 - (256 J / (mol K)), O 2 - (205 J / (mol K)).

Answer: -189 J/K.

26. Find the enthalpy of the reaction for the synthesis of benzene from acetylene, if the enthalpy of combustion of benzene is (-3302 kJ / mol), and acetylene - (-1300 kJ / mol).

Answer: - 598 kJ.

27. Find the standard Gibbs energy of the decomposition reaction of sodium bicarbonate. Is it possible for the reaction to proceed spontaneously under these conditions?

Answer: 30.88 kJ.

28. Find the standard Gibbs energy of the reaction 2Fe (T) + 3H 2 O (G) \u003d Fe 2 O 3 (T) + 3H 2 (G) (corrosion reactions of carbon steel with water vapor). Is it possible for the reaction to proceed spontaneously under these conditions?

Answer: -54.45 kJ.

29. At what temperature will chemical equilibrium occur in the system 2NO (g) + O 2 (g) Û 2NO 2 (g)?

Answer: 777 K.

30. Find the thermal effect of the evaporation process of 1 g of water (specific heat of evaporation) at a temperature of 298 K, if the standard enthalpy of formation of H 2 O (l) is (-285.84 kJ / mol), and gaseous - (-241.84 kJ /mol).

Answer: 2.44 kJ / g.

3.4. Tasks for current and intermediate controls

Section I

1. The process of formation of carbon dioxide during the combustion of graphite in oxygen can proceed in two ways:

I. 2C (g) + O 2 (g) \u003d 2CO (g); 2CO (g) + O 2 \u003d 2CO 2 (g), D r H ° \u003d -566 kJ.

II. C (gr) + O 2 (g) \u003d CO 2 (g), D r H ° \u003d -393 kJ.

Find D f H°(CO).

Answer: -110 kJ / mol.

2. Calculate the enthalpy of formation and enthalpy of combustion of carbon monoxide (CO) based on the following reactions:

I. 2C (g) + O 2 (g) \u003d 2CO (g), D r H ° \u003d -220 kJ.

II. 2CO (g) + O 2 (g) \u003d 2CO 2 (g), D r H ° \u003d -566 kJ.

Answer: -110 kJ/mol; -283 kJ/mol.

3. Find the standard enthalpy of formation of sodium sulfite from the thermochemical equation

4Na 2 SO 3 (cr) \u003d 3Na 2 SO 3 (cr) + Na 2 S (cr) - 181.1 kJ,

If kJ/mol and kJ/mol.

Answer: -1090 kJ / mol.

4. Find the standard enthalpy of combustion of methane based on the reaction CH 4 (g) + 2O 2 (g) \u003d CO 2 (g) + 2H 2 O (g), D r H ° \u003d -802 kJ.

Answer: -802 kJ / mol.

5. Predict whether it will be positive or negative

change in the entropy of the system in the reactions:

a) H 2 O (g) ® H 2 O (g) (at a temperature of 25 ° C);

b) CaCO 3 (t) ® CaO (t) + CO 2 (g);

c) N 2 (g) + 3H 2 (g) \u003d 2NH 3 (g);

d) N 2 (g) + O 2 (g) \u003d 2NO (g);

e) Ag + (solution) + Cl - (solution) = AgCl (t).

Give explanations without making calculations.

Answer: a) +; b) +; V) -; d) ~0; e) -.

6. Predict the sign of the DS system in each of the following

processes:

a) evaporation of 1 mol CCl 4(g);

b) Br 2(g) → Br 2(g);

c) precipitation of AgCl(t) by mixing NaCl(aq.) and AgNO 3 (aq.).

Give explanations.

Answer: a) +; b) -; V)-.

7. Using the tabular values ​​of the absolute values ​​of the entropies of substances under standard conditions (S °), compare the values ​​of the absolute entropies of substances at a temperature of 298 K in each of the following pairs:

a) O 2 (g) and O 3 (g);

b) C (diamond) and C (graphite);

c) NaCl (t) and MgCl 2(t).

Explain the reason for the difference in S° in each case.

8. Calculate D r S° for reactions

a) N 2 (g) + 3H 2 (g) \u003d 2NH 3 (g); b) 2SO 2 (g) + O 2 (g) \u003d 2SO 3 (g),

using tabular values ​​of the absolute entropies of substances under standard conditions.

Answer: a) -197.74 J/K; b) -188.06 J/K.

9. Using the tabular values ​​of the absolute en-

tropium (S°), calculate D r S° for the following processes:

a) CO (g) + 2H 2 (g) \u003d CH 3 OH (g);

b) 2HCl (g) + Br 2 (g) \u003d 2HBr (g) + Cl 2 (g);

c) 2NO 2 (g) = N 2 O 4 (g).

Does the sign of D r S° agree in each case with that which should be expected on the basis of qualitative representations? Explain answers.

Answer: a) -218.83 J/K; b) 94.15 J/K; c) -175.77 J/K.

10. The standard enthalpy of formation of CO (g) is -110.5 kJ/mol. The combustion of 2 mol CO (g) released 566 kJ of heat. Calculate

Answer: -393.5 kJ / mol.

11. Determine the amount of heat released when quenching 100 kg of lime with water: CaO (k) + H 2 O (l) \u003d Ca (OH) 2 (k), if the standard heats of formation CaO (k), H 2 O (l) , Ca(OH) 2(k) are respectively -635.14; -285.84; -986.2 kJ/mol.

Answer: -1165357.2 kJ.

12. Determine the enthalpy of decomposition of hydrogen peroxide (H 2 O 2) into water and oxygen using the data below:

SnCl 2 (p) + 2HCl (p) + H 2 O 2 (p) \u003d SnCl 4 (p) + 2H 2 O (l), D r H ° \u003d -393.3 kJ;

SnCl 2 (p) + 2HCl (p) + 1 / 2O 2 (g) \u003d SnCl 4 (p) + H 2 O (l), D r H ° \u003d -296.6 kJ.

Answer: - 96.7 kJ.

13. Calculate the amount of heat that is released in the production of 10 6 kg of ammonia per day, if

Answer: -2.7. 10 9 kJ.

14. Determine based on the following data:

P 4 (cr) + 6Cl 2 (g) \u003d 4PCl 3 (l), D r H ° \u003d -1272.0 kJ;

PCl 3 (g) + Cl 2 (g) \u003d PCl 5 (cr), D r H ° \u003d -137.2 kJ.

Answer: -455.2 kJ / mol.

15. Calculate the change in the enthalpy of reaction under standard conditions: H 2 (g) + 1 / 3O 3 (g) \u003d H 2 O (g), based on the following data:

2O 3 (g) \u003d 3O 2 (g), D r H ° \u003d -288.9 kJ,

kJ/mol.

Answer: -289.95 kJ.

16. Calculate the standard enthalpy of PbO formation reaction using the following data:

1) 2Pb (cr) + O 2 (g) \u003d 2PbO 2 (cr) - 553.2 kJ;

2) 2PbO 2 (cr) \u003d 2PbO (cr)) + O 2 (g) + 117.48 kJ.

Answer: -217.86 kJ / mol.

17. Calculate the standard enthalpy of the CuCl formation reaction using the following data:

1) CuCl 2 (cr) + Cu (cr) = 2 CuCl (cr) - 63.5 kJ;

2) Cu (cr) + Cl 2 (g) = CuCl 2 (cr) - 205.9 kJ.

Answer: 134.7 kJ / mol.

18. Calculate Δ f H° of methyl alcohol in the liquid state, knowing the following data:

H 2 (g) + 1 / 2O 2 (g) \u003d H 2 O (g), D r H ° \u003d -285.8 kJ;

C (gr) + O 2 (g) \u003d CO 2 (g), D r H ° \u003d -393.7 kJ;

CH 3 OH (l) + 3 / 2O 2 (g) \u003d CO 2 (g) + 2H 2 O (l), D r H ° \u003d -715.0 kJ.

Answer: -250.3 kJ / mol.

19. The standard enthalpies of combustion of benzene and acetylene are -3270 and -1302 kJ / mol, respectively. Determine D r H ° of the conversion of acetylene to benzene: 3C 2 H 2 (g) \u003d C 6 H 6 (g).

Answer: -636 kJ.

20. Determine the standard enthalpy of formation of iron oxide (III), if 146.8 kJ of heat was released during the oxidation of 20 g of iron.

Answer: -822 kJ / mol.

21. Calculate the amount of heat that is released when receiving 22.4 liters of ammonia (n.o.) if

N 2 (g) + 3H 2 (g) \u003d 2NH 3 (g), D r H ° \u003d -92 kJ.

Answer: -46 kJ.

22. Determine Δ f H° of ethylene using the following data.

C 2 H 4 (g) + 3O 2 (g) \u003d 2CO 2 (g) + 2H 2 O (g) -1323 kJ;

C (gr) + O 2 (g) \u003d CO 2 (g) -393.7 kJ;

H 2 (g) + 1 / 2O 2 (g) \u003d H 2 O (g) -241.8 kJ.

Answer: 52 kJ / mol.

23. Calculate the enthalpy of reaction F (g) + Li (g) \u003d F - (g) + Li + (g),

if F (g) + e \u003d F - (g) -322 kJ / mol;

Li (g) \u003d Li + (g) + e + 520 kJ / mol.

Answer: 198 kJ.

24. Calculate the standard enthalpy of the formation of Hg 2 Br 2 using the following data:

1) HgBr 2 (cr) + Hg (g) = Hg 2 Br 2 (cr) - 37.32 kJ;

2) HgBr 2 (cr) \u003d Hg (l) + Br 2 (l) + 169.45 kJ.

Answer: -206.77 kJ / mol.

25. Calculate the standard enthalpy of sodium bicarbonate formation using the following data:

2NaHCO 3 (cr) \u003d Na 2 CO 3 (cr) + CO 2 (g) + H 2 O (g) + 130.3 kJ,

If kJ/mol;

C (gr) + O 2 (g) \u003d CO 2 (g) - 393.7 kJ; H 2 (g) + 1 / 2O 2 (g) \u003d H 2 O (g) -241.8 kJ.

Answer: -947.4 kJ / mol.

26. Calculate the standard enthalpy of the formation of CaCO 3 (cr) using the following data:

Ca (OH) 2 (c) + CO 2 (g) \u003d CaCO 3 (cr) + 173.9 kJ;

C (gr) + O 2 (g) \u003d CO 2 (g) - 393.7 kJ;

kJ/mol.

Answer: -1206 kJ / mol.

27. Determine the standard enthalpy of formation of iron oxide (III), if during the reaction

2Fe + Al 2 O 3 \u003d Fe 2 O 3 + 2Al

for every 80 g of Fe 2 O 3, 426.5 kJ of heat is absorbed, kJ/mol.

Answer: -823 kJ / mol.

28. How much heat must be spent to obtain 11.2 kg of iron, if, in accordance with the thermochemical equation, FeO (t) + H 2 (g) \u003d Fe (t) + H 2 O (g) + 23 kJ.

Answer: 4600 kJ.

29. Find the heat of combustion of diamond if the standard heat of combustion of graphite is -393.51 kJ / mol, and the heat is

and the phase transition С(graphite) ® С(diamond) is

1.88 kJ/mol.

Answer: -395.39 kJ / mol.

30. How much heat is released when 1 kg of red phosphorus is converted to black phosphorus, if known,

that the standard enthalpies of formation of red and black phosphorus are -18.41 and -43.20 kJ/mol, respectively.

Answer: -800 kJ.

Section II

Calculate the standard change in the Gibbs energy of a chemical reaction at a temperature of 25 °C from the values ​​of the standard enthalpies of formation and absolute entropies of chemical compounds and establish the possibility of a spontaneous reaction:

1. 4NH 3g + 5O 2g = 4NO g + 6H 2 O g.

Answer: -955.24 kJ; reaction is possible.

2. SO 2g + 2H 2 S g \u003d 3S to + 2H 2 O well.

Answer: -107.25 kJ; reaction is possible.

3. 2H 2 S g + 3O 2g = 2H 2 O g + 2SO 2g.

Answer: -990.48 kJ; reaction is possible.

4. 2NO g + O 3g + H 2 O well \u003d 2HNO 3g.

Answer: - 260.94 kJ; reaction is possible.

5. 3Fe 2 O 3k + CO g \u003d 2Fe 3 O 4k + CO 2g.

Answer: - 64.51 kJ; reaction is possible.

6. 2CH 3 OH w + 3O 2g \u003d 4H 2 Og + 2CO 2g.

Answer: - 1370.46 kJ; reaction is possible.

7. CH 4g + 3CO 2g \u003d 4CO g + 2H 2 O g.

Answer: 228.13 kJ; reaction is not possible.

8. Fe 2 O 3k + 3CO g \u003d 2Fe k + 3CO 2g.

Answer: -31.3 kJ; reaction is possible.

9. C 2 H 4g + 3O 2g \u003d 2CO 2g + 2H 2 O g.

Answer: -1313.9 kJ; reaction is possible.

10. 4NH 3g + 3O 2g = 6H 2 O g + 2N 2g.

Answer: -1305.69 kJ; reaction is possible.

11. 4NO 2g + O 2g + 2H 2 O x = 4HNO 3g.

Answer: -55.08 kJ; reaction is possible.

12. 2HNO 3l + NO g = 3NO 2g + H 2 O l.

Answer: -7.71 kJ; reaction is possible.

13. 2C 2 H 2g + 5O 2g \u003d 4CO 2g + 2H 2 O g.

Answer: -2452.81 kJ; reaction is possible.

14. Fe 3 O 4k + 4H 2g \u003d 3Fe to + 4H 2 O g.

Answer: 99.7 kJ; reaction is not possible.

15. 2Fe 2 O 3k + 3C k \u003d 4Fe k + 3CO 2g.

Answer: 297.7 kJ; reaction is not possible.

16. Fe 3 O 4k + 4CO g \u003d 3Fe k + 4CO 2g.

Answer: -14.88 kJ; reaction is possible.

17. 2H 2 S g + O 2g \u003d 2H 2 O well + 2S c.

Answer: -407.4 kJ; reaction is possible.

18. Fe 2 O 3k + 3H 2g \u003d 2Fe to + 3H 2 O g.

Answer: 54.47 kJ; reaction is not possible.

Calculate the standard change in the Gibbs energy of a chemical reaction at a temperature of 25 °C from the values ​​of the standard enthalpies of formation and absolute entropies of chemical compounds and determine at what temperature equilibrium will occur in the system.

19. 4HCl g + O 2g ↔ 2Cl 2g + 2H 2 O f.

Answer: -93.1 kJ; ~552 K.

20. Cl 2g + 2HI g ↔ I 2c + 2HCl g.

Answer: -194.0 kJ; ~1632 K.

21. SO 2g + 2CO g ↔ 2CO 2g + S c.

Answer: -214.24 kJ; ~1462 K.

22. CH 4g + 2H 2 Og ↔ CO 2g + 4H 2g.

Answer: 113.8 kJ; ~959 K.

23. CO g + 3H 2g ↔ CH 4g + H 2 O g.

Answer: -142.36 kJ; ~ 963 K.

Calculate the change in the Gibbs energy of a chemical reaction at a temperature of 350 °C from the standard enthalpies of formation and absolute entropies of chemical compounds. Ignore the temperature dependence of D f H° and S°. Set the possibility of spontaneous reactions:

24. 2РН 3g + 4O 2g \u003d P 2 O 5k + 3H 2 O g.

Answer: 1910.47 kJ; reaction is possible.

25. Cl 2 g + SO 2 g + 2H 2 O w = H 2 SO 4 w + 2HCl g.

Answer: -80.0 kJ; reaction is possible.

26. P 2 O 5k + 5C k \u003d 2P k + 5CO g.

Answer: 860.0 kJ; reaction is not possible.

27. 2CO g + SO 2g \u003d S to + 2CO 2g.

Answer: -154.4 kJ; reaction is possible.

28. CO 2g + 4H 2g \u003d CH 4g + 2H 2 O g.

Answer: -57.9 kJ; reaction is possible.

29. NO g + O 3 g = O 2 g + NO 2 g.

Answer: -196.83 kJ; reaction is possible.

30. CH 4g + 2O 2g \u003d CO 2g + 2H 2 O g.

Answer: -798.8 kJ; reaction is possible.