The probability of chemical reactions occurring. Entropy. Gibbs energy. Variants of tasks for independent solution
201. Under standard conditions, in what direction will the reaction N 2 (g) + O 2 (g) = 2NO (g) proceed? Support your answer with calculations.
202. Calculate the change in Gibbs energy of some reaction at 1000 K if ∆ r N° 298 = 131.3 kJ, and ∆ r S° 298 = 133.6 J/K (influenced by temperature T by ∆ N and ∆ S neglect).
203. Calculate ∆ r G° 298 system PbO 2 + Pb = 2PbO based on ∆ r N° 298 and ∆ r S° 298 reactants. Determine if this reaction is possible.
204. Determine in which direction the reaction Fe 2 O 3 (k) + 3H 2 = = Fe (k) + 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 (s) + 3C (s) = 2Cr (s) + 3CO (g).
206. Tungsten is produced by reducing tungsten(IV) oxide with hydrogen. Determine the possibility of this reaction occurring at 500 and 1000 °C by the reaction WO 3 (s) + 3H 2 (g) = W (s) + 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) = 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 (s) = 2Cu 2 O (s) + O 2 (g).
209. Determine the temperature of the equally probable reaction in the forward and reverse directions, if ∆ r N° 298 = = 38 kJ, and ∆ r S° 298 = 207 J/K.
210. Calculate ∆ r G° 298 and ∆ r G° 1000 for the reaction H 2 O (g) + + C (g) = CO (g) + H 2 (g). How does temperature affect the thermodynamic probability of a process occurring in the forward direction?
211. Which of the following reactions is thermodynamically more probable: 1) N 2 + O 2 = 2NO or 2) N 2 + 2O 2 = 2NO 2? Support your answer with calculations.
212. Determine the sign of ∆ r G° 298, without resorting to calculations, for the reaction CaO (s) + CO 2 (g) = CaCO 3 (s), ∆ r N° 298 = -178.1 kJ/mol. Explain your answer.
213. Determine the sign of ∆ r G° 298 for the process of assimilation of sucrose in the human body, which boils down to its oxidation C 12 H 22 O 11 (k) + 12 O 2 (g) = 12 CO 2 (g) + 11 H 2 O (l).
214. Check whether there is a threat that nitric oxide (I), used in medicine as a narcotic, will be oxidized by atmospheric oxygen to very toxic nitric oxide (II) according to the reaction 2N 2 O (g) + O 2 (g) = 4NO (G) .
215. Glycerol is one of the products of metabolism, which is finally converted in the body into CO 2 (g) and H 2 O (l). Calculate ∆ r G° 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 (l) = C 6 H 12 O 6 (solution) + 6O 2 (g).
217. Determine the temperature at which ∆ r G° T = 0, for the reaction H 2 O (g) + CO (g) = CO 2 (g) + H 2 (g).
218. Calculate the thermodynamic characteristics ∆ r N°298,∆ r S°298,∆ r G° 298 reaction 2NO (g) = N 2 O 4 (g). Formulate a conclusion about the possibility of the reaction occurring at temperatures of 0; 25 and 100 °C, confirm it by calculation.
219. Is the reaction 3Fe 2 O 3 (k) + H 2 (g) = 2Fe 3 O 4 (k) = H 2 O (g) possible? Support your answer with calculations.
220. Calculate the change in Gibbs energy of the reaction at 980 K if ∆ r N° 298 = 243.2 kJ, and ∆ r S° 298 = 195.6 J/K (influence of temperature on ∆ N and ∆ S neglect).
221. Calculate ∆ r G° 298 and ∆ r G° 1000 for reaction
Fe 2 O 3 (k) + 3CO (g) = 2 Fe (k) + 3CO 2 (g)
How does temperature affect the thermodynamic probability of a process occurring in the forward direction?
222. The interaction of calcium carbide with water is described by two equations:
a) CaC 2 + 2H 2 O = CaCO 3 + C 2 H 2; b) CaC 2 + 5H 2 O = CaCO 3 + 5H 2 + CO 2.
Which reaction is thermodynamically preferable? Explain the results of the calculation.
223. Determine the direction of spontaneous occurrence of the reaction SO 2 + 2H 2 = S cr + 2H 2 O under standard conditions.
224. Calculate the change in the Gibbs energy of the reaction ZnS +3/2O 2 = ZnO + SO 2 at 298 and 500 K.
225. Determine the direction of spontaneous reaction
NH 4 Cl (k) + NaOH (k) = NaCl (k) + H 2 O (g) + NH 3 (g).
under standard conditions
Speed chemical reactions
226. Determine how many times the rate of the 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 concentrations of NO and O 2 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 in the system be increased in order to increase the rate of the reaction 2CO → CO 2 + C (s) by 4 times?
229. The reaction follows 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 = C. The concentration of substance A is 6 mol/l, and that of 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 increased for the reaction rate to increase 90 times? The van't Hoff temperature coefficient is 2.7.
232. The temperature coefficient of the reaction rate for the decomposition of hydrogen iodide according to the reaction 2HI = H 2 + I 2 is equal to 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, when the temperature decreases by 45°, the reaction slows down 25 times.
234. Calculate at what temperature the reaction will complete in 45 minutes if at 293 K it takes 3 hours. Take the temperature coefficient of the reaction rate to be 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 by how many degrees the temperature must be increased 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 minutes. After what period of time will this reaction end at 453 K if the temperature coefficient of the reaction rate is 3?
239. The initial concentrations of the reactants in the reaction CO + H 2 O (g) → CO 2 + H 2 were equal (mol/dm 3): = 0.8; = 0.9; = 0.7; = 0.5. Determine the concentrations of all participants in the reaction after the hydrogen concentration increases 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 rate of the reaction and the rate of reaction after some time, when the concentration of substance A decreases by 0.01 mol/l.
241. In the system CO + Cl 2 = COCl 2, 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. How many times did the rate of the forward reaction increase?
242. How many times will the rate of reaction 2A + B → A 2 B change if the concentration of substance A is increased by 2 times, and the concentration of substance B is decreased by 2 times?
243. What fraction (%) of novocaine will decompose during 10 days of storage at 293 K, if at 313 K the rate constant of novocaine hydrolysis is equal to 1·10 -5 day -1, and the activation energy of the reaction is equal to 55.2 kJ/mol?
244. At 36 °C, the decay rate constant of penicillin is 6·10 -6 s -1 , and at 41 °C – 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 decreased by 4 kJ/mol?
246. Calculate the temperature coefficient (γ) of the rate constant for the decomposition reaction of hydrogen peroxide 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 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 6.9 times. Calculate the temperature coefficient of the reaction rate and the activation energy of this reaction at 300 K.
249. For a certain 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 substance at 402 K.
250. The rate constants of a certain reaction at 25 and 60 °C are equal to 1.4 and 9.9 min -1, respectively. Calculate the rate constants for this reaction at 20 and 75 °C.
Chemical equilibrium
251. The equilibrium constant of the reaction A + B = C + D is equal to unity. Initial concentration [A] = 0.02 mol/l. What percentage 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 produce 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 occurs.
253. The reversible reaction proceeds according to the equation A + B ↔ C + D. The initial concentration of each substance 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 equilibrium is established at the following concentrations of the reactants, mol/dm 3: = 0.12; = 0.48; = 0.24.
255. In what direction will it shift? chemical equilibrium in the system 2NO 2 ↔ NO + O 2, if the equilibrium concentrations of each component are reduced by 3 times?
256. How many times will the equilibrium partial pressure of hydrogen decrease 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 so that the pressure doubles?
258. In what direction will the equilibrium shift as the temperature of the systems increases:
1) COCl 2 ↔ CO + Cl 2 ; ∆ r N° 298 = -113 kJ/mol
2) 2СО ↔ CO 2 +С; ∆ r N° 298 = -171 kJ/mol
3) 2SO 3 ↔ 2SO 2 + O 2; ∆ r N° 298 = -192 kJ/mol.
Explain your answer.
259. In a closed vessel the reaction AB (g) ↔ A (g) + B (g) occurs. 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 percentage of substance AB has decomposed?
260. When ammonia is oxidized with oxygen, nitrogen and various nitrogen oxides can be formed. Write the reaction equation and discuss the effect of pressure on the shift in the equilibrium of reactions with the formation of: a) N 2 O; b) NO 2.
261. In what direction will the equilibrium shift for the reversible reaction C (s) + H 2 O (g) ↔ CO (g) + H 2 (g) when the volume of the system decreases by 2 times?
262. At a certain temperature, equilibrium in the system 2NO 2 ↔ 2NO + O 2 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 was equal to 0.06 mol/l, determine equilibrium constant.
264. The equilibrium constant of the reaction 3H 2 + N 2 = 2NH 3 at a certain temperature is 2. How many moles of nitrogen should be introduced per 1 liter of gas mixture 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. 0.3 and 0.8 g of hydrogen and iodine were placed in a vessel with a capacity of 0.2 liters. 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 iodide turned out to be 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 equal to 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 quantities of 3 and 4 moles, respectively, located in a vessel with a capacity of 2 liters, react according to the equation 5A + 3B = 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 iodide if the initial concentrations of H 2 and I 2 were 2 mol/l.
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 equal to 4.2, respectively; 4.2; 1.6 mol/l. In another experiment carried out at the same temperature, the equilibrium concentrations of I 2 and HI were found to be 4.0 and 1.5 mol/l. Calculate the hydrogen concentration in this experiment.
272. At a certain temperature in the equilibrium gas system SO 2 – O 2 – SO 3, the concentrations of substances were respectively 0.035; 0.15 and 0.065 mol/l. Calculate the equilibrium constant and initial concentrations of the 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) = COCl 2 (g) has been 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 reactions H 2 (g) + Cl 2 (g) = 2HCl (g) , ∆ affect the equilibrium shift and equilibrium constant H < 0, следующие факторы: а) увеличение концентраций Н 2 , Cl 2 и HCl; б) увеличение давления в 3 раза; в) повышение температуры?
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% of NO has reacted by the time of equilibrium.
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Thermodynamic probability of various directions of complex reactions in oil refining processes
In accordance with the Gibbs equation, the thermodynamic probability of a chemical reaction occurring is determined by the sign and magnitude of the change in the Gibbs free energy (isobaric-isothermal potential, free enthalpy). The change in Gibbs energy is related to the reaction equilibrium constant by the following formula:
chemical reaction thermodynamic oil refining
where Kr is the equilibrium constant,
K p = K 1 / K 2, (K 1 and K 2 are the rate constants of the forward and reverse reactions); - change in Gibbs energy;
R - gas constant;
T - temperature, K.
If K 1 > K 2 (i.e. the reaction goes towards the formation of products), then K p > 1 and ln K p > 0, i.e.< 0.
It follows from the equation that a negative value (at low temperatures T and pressure P) is a condition for the spontaneous occurrence of a chemical reaction. Moreover, the greater the absolute value of the negative value, the higher the probability of this reaction.
It is known that the value increases with increasing molecular weight hydrocarbons (except acetylene) and temperature. Consequently, high molecular weight hydrocarbons, which have a high potential for formation, are thermally less stable and more prone to decomposition reactions, especially when high temperatures.
To estimate the thermodynamic probability of a particular reaction, the magnitude of the change in free energy resulting from the reaction is used.
Free energy is the part internal energy system that can be turned into work. Reactions can be reversible or irreversible. Reversible reactions, which depending on the conditions go in one or the other direction, include:
1. Formation of the simplest hydrocarbons from elements and decomposition of hydrocarbons;
2. Hydrogenation of olefins - dehydrogenation of paraffins;
3. Hydrogenation of aromatics - dehydrogenation of six-membered naphthenes;
4. Condensation of aromatic hydrocarbons;
5. Isomerization.
Many reactions: cracking, coking, polymerization are irreversible.
For reversible reactions, any value of external conditions (temperature and pressure) corresponds to a certain state of the equilibrium system, characterized by a certain ratio of the amounts of starting substances and reaction products. This equilibrium state is estimated by the equilibrium constant.
The thermodynamic probability of any (including irreversible) reaction is determined by the sign of the change in the free energy of the reaction G (if the value of G is known for any isothermal reaction and if this value turns out to be positive, in in the indicated direction the reaction is thermodynamically impossible. If the value is negative, then the process can and does occur, even if at an immeasurably low speed.
Taking the values from the tables for the values of the free energies of formation from the elements of the initial and final substances G at any temperatures and using the difference to determine the change in the free energy of the reaction G for these temperatures, we obtain the coefficients A and B in the equation
G T = A + VT,
G = 0, then this is the temperature limit of the thermodynamic probability of the reaction: lower if ?G increases with increasing T, and upper if ?G increases with increasing temperature. For reversible reactions, the reaction also occurs at temperatures outside the thermodynamic probability, but with a depth less than that of the opposite reaction. However, for reversible reactions, the yields can be changed by changing the concentrations of the reactants.
The thermodynamic probability of a chemical reaction occurring is determined by the sign and magnitude of the change in the Gibbs free energy (isobaric-isothermal potential, free enthalpy). The change in Gibbs energy is related to the equilibrium constant of the reaction by the following formula:
ln К p = - ?G/(RT),
where K r is the equilibrium constant,
K r -- K 1 / K 2 y (K 1 and K 2 are rate constants of forward and reverse reactions);
G -- change in Gibbs energy;
R-- gas constant;
T—temperature, K.
If K 1 > K 2 (i.e. the reaction goes towards the formation of products), then Kp > 1 and lпКр > 0, i.e. ?G< 0.
It follows from the equation that a negative value of ?G (at low values of temperature T and pressure P) is a condition for the spontaneous occurrence of a chemical reaction. Moreover, the greater the absolute value of the negative value?G, the higher the probability of this reaction.
It is known that the ?G value increases with increasing molecular weight of hydrocarbons (except acetylene) and temperature. Consequently, high molecular weight hydrocarbons, which have a higher potential energy of formation?G, are thermally less stable and more prone to decomposition reactions, especially at high temperatures.
Industrial thermal processes are carried out, as a rule, under pressure and are accompanied by homogeneous or heterogeneous reactions.
In principle, any thermal processes of oil refining are accompanied by both endothermic reactions of dehydrogenation and decomposition of hydrocarbons, and exothermic reactions of synthesis, polymerization, condensation, etc. These types of reactions differ not only in the sign of thermal effects, but also in the temperature dependence of the Gibbs free energy values . For endothermic reactions of hydrocarbon decomposition, the ?G values decrease with increasing temperature, and for exothermic reactions they increase, i.e., decomposition reactions are thermodynamically high-temperature, and synthesis reactions are thermodynamically low-temperature. A similar conclusion follows from the Le Chatelier-Brown principle: an increase in temperature promotes the occurrence of endothermic reactions in the direction of the formation of products, and exothermic reactions in the opposite direction.
In the temperature range 300–1200 °C, in which most industrial oil refining processes are carried out, free enthalpy depends linearly on temperature:
In this equation, the value of coefficient b increases with increasing thermal effect of the reaction (for endothermic reactions b > 0, and for exothermic reactions b< 0). В реакциях с небольшим тепловым эффектом (например, изомеризации или гидрокрекинга) ?G мало зависит от температуры. В реакциях же со значительным тепловым эффектом (выделение или поглощение) эта зависимость заметно значительнее.
In accordance with the Le Chatelier-Brown principle, pressure has a significant influence on the value of the reaction rate constant. Its growth promotes the occurrence of reactions with a decrease in volume (mainly synthesis reactions). Low pressures accelerate decomposition reactions.
Predicting the probability of the formation of a particular decomposition product during thermal processes can also be based on thermodynamic data, in particular, on the values of binding energy between atoms in molecules. Thus, analysis of data on the free enthalpy of formation allows us to draw the following conclusions about the direction of hydrocarbon decomposition.
In alkane molecules, the bond breaking energy between the outermost carbon atom and hydrogen is highest in methane (431 kJ/mol), and it decreases as the number of carbon atoms increases to 4 and then becomes constant (at 394 kJ/mol).
In normal alkanes, the energy of breaking the bond between the hydrogen atoms and the carbon located inside the chain gradually decreases towards the middle of the chain (up to 360 kJ/mol).
The energy of abstraction of a hydrogen atom from a secondary, and especially from a tertiary carbon atom is somewhat less than from the primary.
In an alkene molecule, the energy of abstraction of a hydrogen atom from a carbon atom with double bond much more, and from a carbon atom in conjugation with a double bond, much lower than the energy of the C--H bond in alkanes.
In naphthenic rings, the strength of the C--H bond is the same as in the bonds of the secondary carbon atom with hydrogen in alkane molecules.
In molecules of benzene and alkylaromatic hydrocarbons, the bond energy between the carbon atom in the ring and hydrogen is comparable to the strength of the C--H bond in methane, and the energy of hydrogen abstraction from the carbon conjugated with the aromatic ring is significantly lower than the energy of the C--H bond in alkanes.
The energy of breaking the carbon-carbon bond in molecules of all classes of hydrocarbons is always lower than the energy of the C--H bond (by about 50 kJ/mol).
In alkane molecules, the length, chain structure and location of the bond to be broken affect the energy of carbon-carbon bond rupture qualitatively, similar to their influence on the strength of the C--H bond. Thus, the bond between the outermost carbon atoms weakens as the number of carbon atoms increases (from 360 for ethane to 335 kJ/mol for pentane and higher), and the bond between the internal carbon atoms weakens as the number of carbon atoms approaches (up to 310 kJ/mol ). For example, the energy of breaking the C--C bond in an n-octane molecule, depending on its location, changes as follows: 335; 322; 314; 310; 314; 322; 335 kJ/mol.
Bonds between primary carbon atoms are always stronger than C--C bonds in combinations with primary, secondary (Ct) and tertiary (Ctr) carbon atoms. The energy of carbon-carbon bond rupture (D c - c) decreases in the following sequence:
D C - C > D C - C tue > D C - C tr > D C tue- C tue > D C tue- C tr > D C tr- C tr.
10. In alkenes, carbon-carbon double bonds are much stronger (but less than 2 times) than C--C bonds in alkanes. Thus, the energy of breaking the C = C bond in ethylene is 500 kJ/mol. However, C--C bonds conjugated to a double bond (that is, located in the b-position to it) are much weaker than the C--C bonds in alkanes.
11. The energy of rupture of the carbon-carbon bond in the ring of cyclopentane (293 kJ/mol) and cyclohexane (310 kJ/mol) is slightly less than the C--C bond in the middle of the chain of normal hexane (318 kJ/mol).
12. In alkyl aromatic hydrocarbons, the carbon-carbon bond conjugated to the aromatic ring (C--C ap) is less strong than the C--C bond in alkanes. Conjugation to an aromatic ring reduces the strength of the carbon-carbon bond to approximately the same extent as conjugation to a double bond. Conjugation with multiple benzene rings reduces the strength of the C-C bond even further.
13. The energy of rupture (dissociation) of hydrogen atoms in a hydrogen molecule is slightly higher than the C--H bond in the most heat-resistant methane and amounts to 435 kJ/mol.
14. Durability C--S connection in mercaptans and S--S connection in disulfides is comparable to connection S--S in alkanes.
It is obvious that during the thermolysis of hydrocarbon raw materials, the weakest bonds will be broken first and products will be formed predominantly with lower free energy of formation. Thus, thermodynamic analysis makes it possible to predict the component composition and calculate the equilibrium concentrations of components in reaction products depending on the conditions of thermal as well as catalytic processes.
Schemes (I--IV) of chemical transformations of hydrocarbons during thermolysis are presented in Fig. 1.
Rice. 1. Schemes of hydrocarbon transformations during thermolysis of raw materials
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In exothermic reactions, the reserve of internal energy of the starting 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 it is referred to as 1 kg or 1 m 3 (for gases). The section deals with the study of the thermal effects of reactions or states of aggregation, or mixing, dissolution physical chemistry or chemical thermodynamics - thermochemistry. Thermochemical equations indicate the thermal effect of a reaction. For example: C (graphite) + O 2 = CO 2 +393.77 kJ/mol. The heats of decomposition have the opposite sign. Tables are used to determine them. According to D.P. Konovalov, the heat of combustion is determined from the relationship: Q combustion = 204.2n+44.4m+∑x (kJ/mol), where n is the number of moles of oxygen required for complete combustion of 1 mole of a given substance, m is the number of moles water formed during the combustion of 1 mole of a substance, ∑x is a correction constant for a given homologous series. The greater the uncertainty, the greater ∑x.
For hydrocarbons of the acetylene series ∑x=213 kJ/mol. For ethylene hydrocarbons ∑x=87.9 kJ/mol. For saturated hydrocarbons ∑x=0. If a molecule of a compound contains 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 starting substances, taking into account the number of moles of all substances participating in the reaction. For example, for a general reaction: n 1 A+n 2 B=n 3 C+n 4 D+Q x thermal effect: Q x =(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 reacting substances. For the same general reaction:
Q x =(n 1 Q A burn +n 2 Q B burn) – (n 3 Q C burn +n 4 Q D burn)
Probability the occurrence 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 From the analysis of this expression it is clear that for endothermic reactions (Q< 0, ∆ Hº > 0) with a decrease in entropy (∆Sº< 0) самопроизвольное протекание реакции невозможно так как – ∆G > 0. Subsequently, the thermodynamic approach to chemical reactions will be considered in more detail.
Lecture 4.
Basic laws of thermodynamics. The first law of thermodynamics. Heat capacity and enthalpy. Enthalpy of reaction. Enthalpy of formation of a compound. Enthalpy of combustion. Hess's law and enthalpy of reaction.
First law of thermodynamics: the change in internal energy (∆E) of the system is equal to the work of external forces (A′) plus the amount of transferred heat (Q): 1)∆E=A′+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 the work (A) performed by the system. This is one type of the law of conservation of energy. If the change in the state of the system is very small, then: dQ=dE+δA – this is the entry for small (δ) changes. For gas (ideal) δА=pdV. In an isochoric process δА=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 we can determine the heat capacity of the system and the heat of processes. C V – according to the Joule-Lenz law. In an isobaric process that occurs without performing useful work, taking into account 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 through 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 a certain state function (enthalpy) and depends only on the initial and final states of the system and does not depend on the shape of the path along which the process developed. This position underlies the consideration of the issue of thermal effects of chemical reactions.
Thermal effect of reaction– is related to a change in a chemical variable quantity of heat, obtained by a system in which a chemical reaction took place and the reaction products took on the temperature of the original reagents (usually Q V and Q P).
Reactions with negative thermal effect, i.e. with the release of heat in environment, are called exothermic. Reactions with positive thermal effect, i.e. occurring 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 – symbols of products, i – symbols of reagents.
This position is called Hess's law: quantities E i, H i are functions of the state of the system and, therefore, ∆H and ∆E, 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 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 tabulated values of the enthalpies of formation of the reaction components are equal, 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)=-142 kJ/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 equation for the combustion of an organic compound with the composition C m H n O p will be written as:
(3) C m H n O p +(р-m-n/4)O 2 =mCO 2 +n/2 H 2 O
Consequently, the enthalpy of combustion according to (1) can be expressed in terms of the enthalpy of its formation and the formation of CO 2 and H 2 O:
∆H сг =m∆H CO 2 +n/2 ∆H H 2 O -∆H CmHnOp
By determining the heat of combustion of the compound under study using a calorimeter and knowing ∆H CO 2 and ∆H H 2 O, you can find the enthalpy of its formation.
Hess's 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.
Second law of thermodynamics. Entropy. Gibbs function. Changes in the Gibbs function during chemical reactions. Equilibrium constant and Gibbs function. Thermodynamic assessment of the probability of a reaction occurring.
Second law of thermodynamics is called a statement that it is impossible to construct perpetual motion machine of the second kind. The law was obtained empirically and has two equivalent formulations:
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 less heated body to a more heated body.
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, as (Q) increases, (S) increases, and vice versa. Equation (1) corresponds to an equilibrium (reversible) process. If the process is nonequilibrium, then entropy increases, then (1) is transformed:
dS≥(δQ/T)(2) Thus, during nonequilibrium processes, the entropy of the system increases. If (2) is substituted into the first law of thermodynamics, we obtain: dE≤TdS-δA. It is usually written in the form: dE≤TdS-δA’-pdV, hence: δA’≤-dE+TdS-pdV, here pdV is the work of equilibrium expansion, δA’- useful work. Integrating both sides of this inequality for an isochoric-isothermal process leads to the inequality: A' V≤-∆E+T∆S(3). And integration for an isobaric-isothermal process (T=const, p=const) leads to the inequality:
A’ P ≤ - ∆E+T∆S – p∆V=-∆H + T∆S (4)
The right-hand sides (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 the equilibrium process. In this case, the maximum useful work is performed, that is, (A’ V) MAX =-∆F, and (A’ P) MAX =-∆G. F and G are called isochoric-isothermal and isobaric-isothermal potentials, respectively.
Equilibrium of chemical reactions characterized by a process (thermodynamic) in which the system passes through a continuous series of equilibrium states. Each of these states is characterized by the invariance (in time) of thermodynamic parameters and the absence of flows of matter and heat in the system. The equilibrium state is characterized by the dynamic nature of equilibrium, that is, the equality of the forward 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 forward and reverse reactions are the same. Equality must also be observed:
µ J dn J =0, where µ J =(ðG/ðn J) T , P , h =G J – chemical potential of component J; n J – amount of component J (mol). Great importanceµ J indicates greater reactivity of the particles.
∆Gº=-RTLnК р(9)
Equation (9) is called the Van't Haff isotherm equation. ∆Gº values in tables in reference literature for many thousands of chemical compounds.
К р = e - ∆ G º/(RT) = e - ∆ H º/ RT ∙ e ∆ S º/ R (11). From (11) we can give a thermodynamic estimate of the probability of the reaction occurring. Thus, 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 occurrence of the process is impossible.
If ∆Нº and ∆Sº have the same sign, the thermodynamic probability of the process occurring is determined specific values∆Нº, ∆Sº and Тº.
Let us consider, using the example of the ammonia synthesis reaction, the joint influence of ∆H o and ∆S o on the possibility of carrying out the process:
For this reaction, ∆H 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.
From the above data it is clear that the change in entropy is negative and does not favor the course of the reaction, but at the same time the process is characterized by a large negative enthalpy effect ∆Нº, due to which the process is possible. With increasing temperature, the reaction, as calorimetric data show, becomes even more exothermic (at T = 725 K, ∆H = -113 kJ/mol), but with a negative value of ∆S o, an increase in temperature very significantly reduces the probability of the process occurring.
(from the Greek kineticos driving) the science of the mechanisms of chemical reactions and the patterns of their occurrence over time. At 19 V. As a result of the development of the fundamentals of chemical thermodynamics, chemists learned to calculate the composition of an equilibrium mixture for reversible chemical reactions. In addition, based on simple calculations, it was possible, without conducting experiments, to draw a conclusion about the fundamental possibility or impossibility of a specific reaction occurring under given conditions. However, it is “possible in principle”reaction does not mean that it will go. For example, the reaction C + O 2 ® CO 2 from a thermodynamic point of view, it is very favorable, at least at temperatures below 1000° C (at higher temperatures the decomposition of CO molecules occurs 2 ), i.e. carbon and oxygen should (with almost 100% yield) turn into carbon dioxide. However, experience shows that a piece of coal can lie in the air for years, with free access to oxygen, without undergoing any changes. The same can be said about many other known reactions. For example, mixtures of hydrogen with chlorine or with oxygen can persist for a very long time without any signs of chemical reactions, although in both cases the reactions are thermodynamically favorable. This means that after reaching equilibrium in the stoichiometric mixture H 2 + Cl 2 Only hydrogen chloride should remain, and in the mixture 2H 2 + O 2 water only. Another example: acetylene gas is quite stable, although the reaction C 2 H 2 ® 2C + H 2 not only thermodynamically allowed, but also accompanied by a significant release of energy. Indeed, when high pressures, acetylene explodes, but under normal conditions it is quite stable.Thermodynamically allowed reactions like those considered can only occur under certain conditions. For example, after ignition, coal or sulfur spontaneously combines with oxygen; hydrogen reacts easily with chlorine when the temperature rises or when exposed to ultraviolet light; a mixture of hydrogen and oxygen (explosive gas) explodes when ignited or when a catalyst is added. Why do all these reactions require special influences such as heating, irradiation, and the action of catalysts? Chemical thermodynamics does not answer this question; the concept of time is absent in it. At the same time, for practical purposes it is very important to know whether a given reaction will take place in a second, in a year, or over many millennia.
Experience shows that the speed of different reactions can differ greatly. Many reactions occur almost instantly in aqueous solutions. Thus, when an excess of acid is added to an alkaline solution of crimson-colored phenolphthalein, the solution instantly becomes discolored, which means that the neutralization reaction, as well as the reaction of converting the colored form of the indicator into a colorless one, proceed very quickly. The oxidation reaction of an aqueous solution of potassium iodide with atmospheric oxygen proceeds much more slowly: the yellow color of the reaction product iodine appears only after a long time. Corrosion processes of iron and especially copper alloys, as well as many other processes, occur slowly.
Predicting the rate of a chemical reaction, as well as elucidating the dependence of this rate on the reaction conditions is one of the important tasks of chemical kinetics, a science that studies the patterns of reactions over time. No less important is the second task facing chemical kinetics - the study of the mechanism of chemical reactions, that is, the detailed path of transformation of starting substances into reaction products.
Speed reaction. The easiest way to determine the rate is for a reaction occurring between gaseous or liquid reagents in a homogeneous (homogeneous) mixture in a vessel of constant volume. In this case, the reaction rate is defined as the change in the concentration of any of the substances participating in the reaction (it can be the starting substance or the reaction product) per unit time. This definition can be written as a derivative: v = d c/d t, Where v reaction speed; t time, c concentration. This speed is easy to determine if there is experimental data on the dependence of the concentration of the substance on time. Using this data, you can construct a graph called a kinetic curve. Reaction speed in given point The kinetic curve is determined by the slope of the tangent at that point. Determining the slope of a tangent always involves some error. Most accurately determined starting speed reactions, since at first the kinetic curve is usually close to a straight line; this makes it easier to draw a tangent at the starting point of the curve.If time is measured in seconds, and concentration in moles per liter, then the reaction rate is measured in units of mol/(l
· With). Thus, the reaction rate does not depend on the volume of the reaction mixture: under the same conditions, it will be the same in a small test tube and in a large-scale reactor.Value d
t is always positive, whereas the sign of d c depends on how the concentration changes over time: it decreases (for starting substances) or increases (for reaction products). To ensure that the reaction rate always remains a positive value, in the case of starting substances a minus sign is placed in front of the derivative: v = d c/d t . If the reaction occurs in the gas phase, pressure is often used instead of the concentration of substances in the rate equation. If the gas is close to ideal, then the pressure R is related to concentration with a simple equation: p = cRT. During a reaction, different substances can be consumed and formed at different rates, according to the coefficients in the stoichiometric equation ( cm. STOICHIOMETRY), therefore, when determining the rate of a specific reaction, these coefficients should be taken into account. For example, in the synthesis reaction of ammonia 3H 2 + N 2 ® 2NH 3 Hydrogen is consumed 3 times faster than nitrogen, and ammonia accumulates 2 times faster than nitrogen is consumed. Therefore, the rate equation for this reaction is written as follows: v = 1/3 d p(H2)/d t= d p(N 2)/d t= +1/2d p(NH 3)/d t . In general, if the reaction is stoichiometric, i.e. proceeds exactly in accordance with the written equation: aA + b B ® cC + dD, its speed is determined as v = (1/a)d[A]/d t= (1/b)d[B]/d t= (1/c)d[C]/d t= (1/d)d[D]/d t (square brackets are used to indicate molar concentration substances). Thus, the rates for each substance are strictly related to each other and, having determined experimentally the rate for any participant in the reaction, it is easy to calculate it for any other substance.Most reactions used in industry are heterogeneous-catalytic. They occur at the interface between the solid catalyst and the gas or liquid phase. At the interface between two phases, reactions such as roasting of sulfides, dissolution of metals, oxides and carbonates in acids, and a number of other processes also occur. For such reactions, the rate also depends on the size of the interface, therefore the rate of a heterogeneous reaction is related not to a unit volume, but to a unit surface area. Measure the surface area on which the reaction occurs
It's not always easy.If a reaction occurs in a closed volume, then its speed in most cases is maximum at the initial moment of time (when the concentration of the starting substances is maximum), and then, as the starting reagents are converted into products and, accordingly, their concentration decreases, the reaction rate decreases. There are also reactions in which the rate increases with time. For example, if a copper plate is immersed in a solution of pure nitric acid, the reaction rate will increase over time, which is easy to observe visually. The processes of dissolution of aluminum in alkali solutions, oxidation of many organic compounds with oxygen, and a number of other processes also accelerate over time. The reasons for this acceleration may be different. For example, this may be due to the removal of a protective oxide film from the metal surface, or to the gradual heating of the reaction mixture, or to the accumulation of substances that accelerate the reaction (such reactions are called autocatalytic).
In industry, reactions are usually carried out by continuously feeding starting materials into the reactor and removing products. Under such conditions, it is possible to achieve a constant rate of chemical reaction. Photochemical reactions also proceed at a constant rate, provided that the incident light is completely absorbed ( cm. PHOTOCHEMICAL REACTIONS).
Limiting stage of the reaction. If a reaction is carried out through sequential stages (not necessarily all of them are chemical) and one of these stages requires much more time than the others, that is, it proceeds much more slowly, then this stage is called limiting. It is this slowest stage that determines the speed of the entire process. Let us consider as an example the catalytic reaction of ammonia oxidation. There are two possible limiting cases here.1. The flow of reactant molecules, ammonia and oxygen, to the surface of the catalyst (physical process) occurs much more slowly than the catalytic reaction itself on the surface. Then, to increase the rate of formation of the target product, nitrogen oxide, it is completely useless to increase the efficiency of the catalyst, but care must be taken to accelerate the access of the reagents to the surface.
2. The supply of reagents to the surface occurs much faster than the chemical reaction itself. This is where it makes sense to improve the catalyst, to select optimal conditions for the catalytic reaction, since the limiting stage in in this case is a catalytic reaction on the surface.
Collision theory. Historically, the first theory on the basis of which the rates of chemical reactions could be calculated was the collision theory. Obviously, in order for molecules to react, they must first collide. It follows that the reaction should proceed faster, the more often the molecules of the starting substances collide with each other. Therefore, every factor that affects the frequency of collisions between molecules will also affect the rate of reaction. Some important laws concerning collisions between molecules were obtained on the basis of the molecular kinetic theory of gases.In the gas phase, molecules move at high speeds (hundreds of meters per second) and very often collide with each other. The frequency of collisions is determined primarily by the number of particles per unit volume, that is, concentration (pressure). The frequency of collisions also depends on temperature (as it increases, molecules move faster) and on the size of molecules (large molecules collide with each other more often than small ones). However, concentration has a much stronger effect on collision frequency. At room temperature and atmospheric pressure, each medium-sized molecule experiences several billion collisions per second.
® C between two gaseous compounds A and B, assuming that a chemical reaction occurs whenever reactant molecules collide. Let there be a mixture of reagents A and B at equal concentrations in a liter flask at atmospheric pressure. There will be 6 in total in the flask· 10 23 /22.4 = 2.7 · 10 22 molecules, of which 1.35· 10 22 molecules of substance A and the same number of molecules of substance B. Let each molecule A experience 10 in 1 s 9 collisions with other molecules, of which half (5· 10 8 ) occurs in collisions with molecules B (collisions A + A do not lead to a reaction). Then in total 1.35 occur in the flask in 1 s· 10 22 · 5 · 10 8 ~ 7 · 10 30 collisions of molecules A and B. Obviously, if each of them led to a reaction, it would take place instantly. However, many reactions proceed quite slowly. From this we can conclude that only a tiny fraction of collisions between reactant molecules leads to interaction between them.To create a theory that would allow one to calculate the reaction rate based on the molecular kinetic theory of gases, it was necessary to be able to calculate total number collisions of molecules and the proportion of “active” collisions leading to reactions. It was also necessary to explain why the rate of most chemical reactions increases greatly with increasing temperature the speed of molecules and the frequency of collisions between them increase slightly with temperature proportionally
, that is, only 1.3 times with an increase in temperature from 293 K (20° C) up to 373 K (100 ° C), while the reaction rate can increase thousands of times.These problems were solved based on collision theory as follows. During collisions, molecules continuously exchange velocities and energies. Thus, as a result of a “successful” collision, a given molecule can noticeably increase its speed, while in an “unsuccessful” collision it can almost stop (a similar situation can be observed in the example of billiard balls). At normal atmospheric pressure, collisions, and therefore changes in speed, occur with each molecule billions of times per second. In this case, the velocities and energies of the molecules are largely averaged. If in this moment time to “recount” molecules with certain speeds in a given volume of gas, it turns out that a significant part of them have a speed close to the average. At the same time, many molecules have a speed less than the average, and some move at speeds greater than the average. As speed increases, the fraction of molecules having a given speed quickly decreases. According to collision theory, only those molecules that, when colliding, have a sufficiently high speed (and, therefore, a large supply of kinetic energy) react. This assumption was made in 1889 by a Swedish chemist Svante Arrhenius
. Activation energy. Arrhenius introduced into use by chemists the very important concept of activation energy ( E a ) this is the minimum energy that a molecule (or a pair of reacting molecules) must have in order to enter into a chemical reaction. Activation energy is usually measured in joules and is referred not to one molecule (this is a very small value), but to a mole of a substance and is expressed in units of J/mol or kJ/mol. If the energy of the colliding molecules is less than the activation energy, then the reaction will not take place, but if it is equal to or greater, then the molecules will react.Activation energies for different reactions are determined experimentally (from the dependence of the reaction rate on temperature). The activation energy can vary over a fairly wide range, from units to several hundred kJ/mol. For example, for the reaction 2NO
2 ® N 2 O 4 activation energy is close to zero for the 2H reaction 2 O 2 ® 2H 2 O + O 2 in aqueous solutions E a = 73 kJ/mol, for thermal decomposition of ethane into ethylene and hydrogen E a = 306 kJ/mol.The activation energy of most chemical reactions significantly exceeds the average kinetic energy of molecules, which at room temperature is only about 4 kJ/mol and even at a temperature of 1000
° C does not exceed 16 kJ/mol. Thus, in order to react, molecules usually must have a speed much greater than average. For example, in case E a = 200 kJ/mol colliding molecules of small molecular weight should have a speed of the order of 2.5 km/s (the activation energy is 25 times greater than the average energy of molecules at 20° WITH). And this is general rule: For most chemical reactions, the activation energy is significantly higher than the average kinetic energy of the molecules.The probability for a molecule to accumulate large energy as a result of a series of collisions is very small: such a process requires for it a colossal number of successive “successful” collisions, as a result of which the molecule only gains energy without losing it. Therefore, for many reactions, only a tiny fraction of molecules have sufficient energy to overcome the barrier. This share, in accordance with the Arrhenius theory, is determined by the formula:
a = e E a/ RT= 10 E a/2.3 RT~10 E a/19 T, where R = 8.31 J/(mol. TO). From the formula it follows that the proportion of molecules with energy E a , as well as the proportion of active collisions a , very strongly depends on both the activation energy and temperature. For example, for a reaction with E a = 200 kJ/mol at room temperature ( T~ 300 K) the fraction of active collisions is negligible: a = 10 200000/(19 , 300) ~ 10 35. And if every second 7 things happen in the vessel· 10 30 collisions of molecules A and B, it is clear that the reaction will not take place.If you double the absolute temperature, i.e. heat the mixture to 600 K (327 ° C); At the same time, the proportion of active collisions will increase sharply:
a = 10 200000/(19 , 600) ~ 4·10 18 . Thus, a 2-fold increase in temperature increased the proportion of active collisions by 4 10 17 once. Now every second out of a total of approximately 7 10 30 collisions will lead to a reaction 7 10 30 4 10 18 ~ 3 10 13 . A reaction in which every second 3 10 13 molecules (out of about 10 22 ), although very slowly, it still goes. Finally, at a temperature of 1000 K (727 ° C) a ~ 3·10 11 (out of every 30 billion collisions of a given reactant molecule, one will result in a reaction). This is already a lot, since in 1 s 7 10 30 3 10 11 = 2 10 20 molecules, and such a reaction will take place in a few minutes (taking into account the decrease in the frequency of collisions with a decrease in the concentration of reagents).Now it is clear why increasing the temperature can increase the rate of a reaction so much. The average speed (and energy) of molecules increases slightly with increasing temperature, but the proportion of “fast” (or “active”) molecules that have a sufficient speed of movement or sufficient vibrational energy for a reaction to occur increases sharply.
Calculation of the reaction rate, taking into account the total number of collisions and the fraction of active molecules (i.e., activation energy), often gives satisfactory agreement with experimental data. However, for many reactions the experimentally observed rate turns out to be less than that calculated by collision theory. This is explained by the fact that for a reaction to occur, the collision must be successful not only energetically, but also “geometrically,” that is, the molecules must be oriented in a certain way relative to each other at the moment of the collision. Thus, when calculating reaction rates using collision theory, in addition to the energy factor, the steric (spatial) factor for a given reaction is also taken into account.
Arrhenius equation. The dependence of the reaction rate on temperature is usually described by the Arrhenius equation, which in its simplest form can be written as v = v 0 a = v 0 e E a/ RT , Where v 0 the speed that the reaction would have at zero activation energy (in fact, this is the frequency of collisions per unit volume). Because the v 0 weakly depends on temperature, everything is determined by the second factor exponential: with increasing temperature this factor increases rapidly, and the faster the higher the activation energy E A. This dependence of the reaction rate on temperature is called the Arrhenius equation; it is one of the most important in chemical kinetics. To approximate the effect of temperature on the reaction rate, the so-called “van’t Hoff rule” is sometimes used ( cm. Van't Hoff's Rule).If a reaction obeys the Arrhenius equation, the logarithm of its rate (measured, for example, at the initial moment) should depend linearly on absolute temperature, that is, a graph of ln
v from 1/ T must be straightforward. The slope of this line equal to energy reaction activation. Using such a graph, you can predict what the reaction rate will be at a given temperature or at what temperature the reaction will proceed at a given speed. Several practical examples of using the Arrhenius equation.1. The packaging of a frozen product says that it can be stored on a refrigerator shelf (5° C) for 24 hours, in a freezer marked with one star (6° C) for a week, two stars (12° C) for a month. , and in a freezer with a *** symbol (which means the temperature in it is 18 ° C) 3 months. Assuming that the rate of product spoilage is inversely proportional to the guaranteed shelf life
t xp, in ln coordinates t хр , 1/ T we obtain, in accordance with the Arrhenius equation, a straight line. From it you can calculate the activation energy of biochemical reactions leading to spoilage of a given product (about 115 kJ/mol). From the same graph you can find out to what temperature the product must be cooled so that it can be stored, for example, 3 years; it turns out to be 29° C.2. Mountaineers know that in the mountains it is difficult to boil an egg, or in general any food that requires more or less long boiling. Qualitatively, the reason for this is clear: with a decrease in atmospheric pressure, the boiling point of water decreases. Using the Arrhenius equation, you can calculate how long it will take, for example, to hard boil an egg in Mexico City, located at an altitude of 2265 m, where the normal pressure is 580 mm Hg, and water at such a reduced pressure boils at 93 ° C The activation energy for the protein “folding” (denaturation) reaction was measured and turned out to be very large compared to many other chemical reactions - about 400 kJ/mol (it may differ slightly for different proteins). In this case, lowering the temperature from 100 to 93 ° C (that is, from 373 to 366 K) will slow down the reaction by 10
(400000/19)(1/366 1/373) = 11.8 times. This is why residents of the highlands prefer frying food to cooking: the temperature of a frying pan, unlike the temperature of a pan of boiling water, does not depend on atmospheric pressure.3. In a pressure cooker, food is cooked at increased pressure and, therefore, at an increased boiling point of water. It is known that in a regular saucepan, beef is cooked for 23 hours, and apple compote for 1015 minutes. Considering that both processes have similar activation energies (about 120 kJ/mol), we can use the Arrhenius equation to calculate that in a pressure cooker at 118°C the meat will cook for 2530 minutes, and the compote for only 2 minutes.
The Arrhenius equation is very important for the chemical industry. When an exothermic reaction occurs, the released thermal energy heats not only the environment, but also the reactants themselves. this may result in an undesirable rapid acceleration of the reaction. Calculating the change in reaction rate and heat release rate with increasing temperature allows us to avoid a thermal explosion ( cm. EXPLOSIVE SUBSTANCES).
Dependence of the reaction rate on the concentration of reagents. The rate of most reactions gradually decreases over time. This result is in good agreement with the collision theory: as the reaction proceeds, the concentrations of the starting substances fall, and the frequency of collisions between them decreases; Accordingly, the frequency of collisions of active molecules decreases. This leads to a decrease in the reaction rate. This is the essence of one of the basic laws of chemical kinetics: the rate of a chemical reaction is proportional to the concentration of reacting molecules. Mathematically, this can be written as the formula v = k[A][B], where k a constant called the reaction rate constant. The equation given is called the chemical reaction rate equation or kinetic equation. The rate constant for this reaction does not depend on the concentration of the reactants and on time, but it depends on temperature in accordance with the Arrhenius equation: k = k 0 e E a/ RT . The simplest speed equation v = k [A][B] is always true in the case when molecules (or other particles, for example, ions) A, colliding with molecules B, can directly transform into reaction products. Such reactions, occurring in one step (as chemists say, in one stage), are called elementary reactions. There are few such reactions. Most reactions (even seemingly simple ones like H 2 + I 2 ® 2HI) are not elementary, therefore, based on the stoichiometric equation of such a reaction, its kinetic equation cannot be written.The kinetic equation can be obtained in two ways: experimentally by measuring the dependence of the reaction rate on the concentration of each reagent separately, and theoretically if the detailed reaction mechanism is known. Most often (but not always) the kinetic equation has the form
v = k[A] x[B] y , Where x and y are called reaction orders for reactants A and B. These orders, in the general case, can be integer and fractional, positive and even negative. For example, the kinetic equation for the reaction of thermal decomposition of acetaldehyde CH 3 CHO ® CH 4 + CO has the form v = k 1,5 , i.e. the reaction is one and a half order. Sometimes a random coincidence of stoichiometric coefficients and reaction orders is possible. Thus, the experiment shows that the reaction H 2 + I 2 ® 2HI is first order in both hydrogen and iodine, that is, its kinetic equation has the form v = k(This is why this reaction was considered elementary for many decades, until its more complex mechanism was proven in 1967).If the kinetic equation is known, i.e. It is known how the reaction rate depends on the concentrations of the reactants at each moment of time, and the rate constant is known, then it is possible to calculate the time dependence of the concentrations of the reactants and reaction products, i.e. theoretically obtain all kinetic curves. For such calculations, methods are used higher mathematics or computer calculations, and they present no fundamental difficulties.
On the other hand, the experimentally obtained kinetic equation helps to judge the reaction mechanism, i.e. about a set of simple (elementary) reactions. Elucidation of reaction mechanisms is the most important task of chemical kinetics. This is a very difficult task, since the mechanism of even a seemingly simple reaction can include many elementary stages.
The use of kinetic methods to determine the reaction mechanism can be illustrated using the example of alkaline hydrolysis of alkyl halides to form alcohols: RX +
OH ® ROH + X . It was experimentally discovered that for R = CH 3, C 2 H 5 etc. and X = Cl, the reaction rate is directly proportional to the concentrations of the reactants, i.e. has the first order in the halide RX and the first in the alkali, and the kinetic equation has the form v = k 1 . In the case of tertiary alkyl iodides (R = (CH 3) 3 C, X = I) order in RX first, and in alkali zero: v = k 2 . In intermediate cases, for example, for isopropyl bromide (R = (CH 3) 2 CH, X = Br), the reaction is described by a more complex kinetic equation: v = k 1 + k 2 . Based on these kinetic data, the following conclusion was made about the mechanisms of such reactions.In the first case, the reaction occurs in one step, through direct collision of alcohol molecules with OH ions
(the so-called SN mechanism 2 ). In the second case, the reaction occurs in two stages. First stage slow dissociation of the alkyl iodide into two ions: R I ® R + + I . Second very fast reaction between ions: R+ + OH ® ROH. The rate of the total reaction depends only on the slow (limiting) stage, so it does not depend on the alkali concentration; hence zero order in alkali (SN mechanism 1 ). In the case of secondary alkyl bromides, both mechanisms occur simultaneously, so the kinetic equation is more complex.Ilya Leenson
LITERATURE History of the doctrine of chemical process . M., Nauka, 1981Leenson I.A. Chemical reactions. M., AST Astrel, 2002
Probability
Probability
flow
flow
chemical reactions.
chemical reactions.
Chemical speed
Chemical speed
reactions.
reactions.
Prepared by:
Prepared by:
chemistry teacher
chemistry teacher
1 qualifying
1 qualifying
categories
categories
Sagdieva M.S.
Sagdieva M.S.
Kazan 2017
Kazan 2017
Chemical speed
Chemical speed
reactions
reactions
Chemical kinetics studies the rate and
studies speed and
Chemical kinetics
mechanisms of chemical reactions
mechanisms of chemical reactions
Homogeneous and
Homogeneous and
heterogeneous systems
heterogeneous systems
Phase –
the totality of all
homogeneous parts of the system,
identical in composition and
everyone
chemical
physical
properties and delimited from
others
interface surface.
systems
And
parts
Homogeneous systems
consist of one phase.
Heterogeneous systems
Heterogeneous systems
The essence of chemical reactions comes down to the breaking of bonds in the initial
substances and the emergence of new bonds in reaction products. Wherein
the total number of atoms of each element before and after the reaction remains
permanent.
energy,
absorption
Since the formation of bonds occurs with release, and the breaking of bonds occurs
With
accompanied
energetic effects. Obviously, if the bonds being broken in the original
substances are less durable than those formed in the reaction products, then the energy
stands out, and vice versa. Energy is usually released and absorbed in the form
warmth.
chemical
That
reactions
The speed of chemical reactions is associated with ideas about the transformation
substances
V
industrial scale. The doctrine of the speeds and mechanisms of chemicals
reactions is called chemical kinetics.
efficiency of their production
economic
Also
A
The rate of a chemical reaction is the change
concentration of one of the reactants per unit time
with a constant volume of the system.
Let us consider in general terms the rate of reaction proceeding along
equation:
As substance A is consumed, the rate of the reaction decreases (as
shown in Fig. 4.1). It follows that the reaction rate can be
determined only for a certain period of time. Since the concentration
substance A at time T1 is measured by the value c1, and at time T2 -
value C2, then over a period of time the change in the concentration of the substance
will be ∆C = c2 - c1, from which the average reaction rate (i) will be determined:
Rate of chemical reactions
Rate of chemical reactions
(for homogeneous systems)
tVv
cV
A + B = D + G
A + B = D + G
cv
t
t = 10 s
C0 = 0.5 mol/l
C1 = 5 mol/l
v
5,05
10
45,0
mole
sl
Rate of chemical reactions
A + B = D + G
A + B = D + G
C0 = 2 mol/l
C1 = 0.5 mol/l
t = 10 s
v
25,0
10
15,0
mole
sl
cv
t
tSv
(for homogeneous systems)
v
25,0
10
15,0
mole
sl
(for heterogeneous systems)
Factors from which
speed depends
reactions
Nature of reactants
Nature of reactants
Concentration of substances in
Concentration of substances in
system
system
Surface area (for
Surface area (for
heterogeneous systems)
heterogeneous systems)
Temperature
Temperature
Availability of catalysts
Availability of catalysts
1. Effect of concentrations of reacting substances.
To be carried out chemical reaction substances A and B, their molecules
(particles) must collide. The more collisions, the faster it goes
reaction. The higher the concentration of reactants, the greater the number of collisions.
substances. Hence, based on extensive experimental material, a
the basic law of chemical kinetics, establishing the dependence of the rate
reactions depending on the concentration of reactants:
the rate of a chemical reaction is proportional to the product
concentrations of reacting substances.
For reaction (I), this law is expressed by the equation
v = kSA St,
where CA and Sv are the concentrations of substances A and B, mol/l; k - coefficient
proportionality, called the reaction rate constant. The basic Law
Chemical kinetics is often called the law of mass action.
2.Influence
temperature
Jacob Van'tHoff
Jacob Van'tHoff
(18521911)
(18521911)
Van't Hoff's rule
When the system heats up by 10 °C, the speed
reaction increases 2-4 times
- temperature coefficient
van't Hoff.
t
vv
10
0
100tvv
inform the molecules
inform the molecules
The energy you need
which one needs
(particles) reacting
(particles) reacting
substances to transform
substances to transform
them into active ones, called
called
them into active
activation energy
activation energy
EA – kJ/mol
a – kJ/mol
Using an example of a general reaction:
A2+ B2 = 2AB.
The ordinate axis characterizes the potential
energy of the system, x-axis - reaction progress:
initial state -* transition state
-* final state.
So that reactants A2 and B2
formed the reaction product AB, they must
overcome the energy barrier C (Fig. 4.2).
This requires activation energy Ed,
the value of which increases the energy of the system.
At
during the reaction from particles
reacting
is formed
intermediate unstable
grouping,
called a transition state or
activated complex (at point C),
the subsequent decay of which leads to
formation of the final product AB.
substances
this
The reaction mechanism can be represented by a diagram
If during the disintegration of the activated complex
more energy is released than is necessary for
activation of particles, then the reaction is exothermic.
