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Summary comments on mechanism For a reaction mechanism to be viable, two main conditions apply. 1. The sum of the elementary steps must lead to the overall balanced equation of the observed chemical reaction. 361
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Summary comments on mechanism For a reaction mechanism to be viable, two main conditions apply. 1. The sum of the elementary steps must lead to the overall balanced equation of the observed chemical reaction. 2. The slow step in the reaction mechanism should yield a rate law that corresponds with the rate law for the overall chemical reaction. 362
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Summary comments on mechanism For a reaction mechanism to be viable, two main conditions apply. 1. The sum of the elementary steps must lead to the overall balanced equation of the observed chemical reaction. 2. The slow step in the reaction mechanism should yield a rate law that corresponds with the rate law for the overall chemical reaction. For complicated situations, some manipulation may be necessary to prove this correspondence. 363
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A third condition also applies: 3. Each step in the mechanism should be chemically reasonable. 364
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A third condition also applies: 3. Each step in the mechanism should be chemically reasonable. This can be rather difficult to apply for a person with limited experience. 365
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A third condition also applies: 3. Each step in the mechanism should be chemically reasonable. This can be rather difficult to apply for a person with limited experience. The principal reason is that reaction mechanisms will often have highly reactive intermediate species present. And these would be difficult to conjecture for a person with limited background. 366
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An elementary step for a gas phase reaction that required three bodies to simultaneously strike each other would usually not be a good choice for an elementary step, particularly if the reaction took place under low concentration conditions. 367
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Catalysis 368
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Catalysis Catalyst: A substance that increases the rate of a reaction and can be recovered at the end of the reaction. 369
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Catalysis Catalyst: A substance that increases the rate of a reaction and can be recovered at the end of the reaction. Note that this definition does not exclude the possibility that the catalyst is directly involved in the chemistry. 370
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Catalysis Catalyst: A substance that increases the rate of a reaction and can be recovered at the end of the reaction. Note that this definition does not exclude the possibility that the catalyst is directly involved in the chemistry. If the catalyst is transformed in one step in the sequence, it must be regenerated in a subsequent step in the sequence. 371
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One way to speed up a reaction is to raise the temperature. This method can produce undesirable side effects. For example, at elevated temperatures the products formed may undergo other reactions. 372
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One way to speed up a reaction is to raise the temperature. This method can produce undesirable side effects. For example, at elevated temperatures the products formed may undergo other reactions. A catalyst accelerates a reaction without any need to change the temperature. 373
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One way to speed up a reaction is to raise the temperature. This method can produce undesirable side effects. For example, at elevated temperatures the products formed may undergo other reactions. A catalyst accelerates a reaction without any need to change the temperature. Regardless of their nature, all catalysts act in the same general manner. 374
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Consider the following two mechanisms: 1. No catalyst: A + B product (slow) (suppose it is a one step mechanism) 375
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Consider the following two mechanisms: 1. No catalyst: A + B product (slow) (suppose it is a one step mechanism) 2.Catalyst present: A + catalyst C (faster) B + C product + catalyst (faster) (suppose it is a two step mechanism) 376
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Impact of a catalyst on the energy profile 377
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Impact of a catalyst on the energy profile 378
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A catalyst lowers the overall activation energy required for a reaction by providing a completely different pathway for its progress. 380
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A catalyst lowers the overall activation energy required for a reaction by providing a completely different pathway for its progress. Consider A + B C + D (no catalyst) 381
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A catalyst lowers the overall activation energy required for a reaction by providing a completely different pathway for its progress. Consider A + B C + D (no catalyst) In the presence of a catalyst, A + B C + D 382
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A catalyst lowers the overall activation energy required for a reaction by providing a completely different pathway for its progress. Consider A + B C + D (no catalyst) In the presence of a catalyst, A + B C + D By definition: 383
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Note that the total energies of the reactants (A and B) and those of the products (C and D) are unchanged. 384
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Note that the total energies of the reactants (A and B) and those of the products (C and D) are unchanged. The important change is the lowering of the activation energy from E a to E a,c. 385
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Note that the total energies of the reactants (A and B) and those of the products (C and D) are unchanged. The important change is the lowering of the activation energy from E a to E a,c. The increase in the rate constant from k to k c can be understood by using the Arrhenius equation. 386
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Note that the total energies of the reactants (A and B) and those of the products (C and D) are unchanged. The important change is the lowering of the activation energy from E a to E a,c. The increase in the rate constant from k to k c can be understood by using the Arrhenius equation. 387
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Now take the ratio, so that: 388
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Now take the ratio, so that: hence 389
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Suppose the catalyst lowers the activation energy by 20. kJmol -1, that is, E a,c = E a - 20. (units kJ mol -1 ). In this case, assuming the reaction takes place at 25 o C, then 390
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Suppose the catalyst lowers the activation energy by 20. kJmol -1, that is, E a,c = E a - 20. (units kJ mol -1 ). In this case, assuming the reaction takes place at 25 o C, then 391
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Suppose the catalyst lowers the activation energy by 20. kJmol -1, that is, E a,c = E a - 20. (units kJ mol -1 ). In this case, assuming the reaction takes place at 25 o C, then 392
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Suppose the catalyst lowers the activation energy by 20. kJmol -1, that is, E a,c = E a - 20. (units kJ mol -1 ). In this case, assuming the reaction takes place at 25 o C, then that is. 393
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Suppose the catalyst lowers the activation energy by 20. kJmol -1, that is, E a,c = E a - 20. (units kJ mol -1 ). In this case, assuming the reaction takes place at 25 o C, then that is. The large increase in the rate constant is due to the exponential connection between k and E a. 394
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Heterogeneous Catalysis 395
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Heterogeneous Catalysis Heterogeneous catalysis: The catalyst is in a different phase than the reactants and products. 396
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Heterogeneous Catalysis Heterogeneous catalysis: The catalyst is in a different phase than the reactants and products. Usually, the catalyst is a solid, and the reactants and products are in the gas or liquid phase. 397
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Example: The synthesis of ammonia. N 2(g) + 3 H 2(g) 2 NH 3(g) 398
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Example: The synthesis of ammonia. N 2(g) + 3 H 2(g) 2 NH 3(g) The atmosphere provides a very cheap source for N 2. 399
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Example: The synthesis of ammonia. N 2(g) + 3 H 2(g) 2 NH 3(g) The atmosphere provides a very cheap source for N 2. Dihydrogen can be produced by passing steam over heated coal: H 2 O (g) + C (s) CO (g) + H 2(g) 400
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Example: The synthesis of ammonia. N 2(g) + 3 H 2(g) 2 NH 3(g) The atmosphere provides a very cheap source for N 2. Dihydrogen can be produced by passing steam over heated coal: H 2 O (g) + C (s) CO (g) + H 2(g) The formation of NH 3 is slow at room temperature. 401
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Example: The synthesis of ammonia. N 2(g) + 3 H 2(g) 2 NH 3(g) The atmosphere provides a very cheap source for N 2. Dihydrogen can be produced by passing steam over heated coal: H 2 O (g) + C (s) CO (g) + H 2(g) The formation of NH 3 is slow at room temperature. Raising the temperature does accelerate the reaction – but this also promotes the decomposition of NH 3 into N 2 and H 2. 402
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For an industrial process, require – 404
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For an industrial process, require – 1. An appreciable rate. 405
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For an industrial process, require – 1. An appreciable rate. 2. High yield. 406
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For an industrial process, require – 1. An appreciable rate. 2. High yield. The problem of preparing NH 3 using the above reaction with an appropriate catalyst was solved by Fritz Haber (1905). He discovered that iron plus a few per cent of oxides of potassium and aluminum catalyzes the reaction. 407
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In heterogeneous catalysis, the surface of the catalyst is usually the site of the reaction. 408
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In heterogeneous catalysis, the surface of the catalyst is usually the site of the reaction. There are two ways in which molecules may be attached to the surface of a solid. 409
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In heterogeneous catalysis, the surface of the catalyst is usually the site of the reaction. There are two ways in which molecules may be attached to the surface of a solid. Physical adsorption: Relatively weak intermolecular forces are responsible for holding the molecules on the surface. 410
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In heterogeneous catalysis, the surface of the catalyst is usually the site of the reaction. There are two ways in which molecules may be attached to the surface of a solid. Physical adsorption: Relatively weak intermolecular forces are responsible for holding the molecules on the surface. Physical adsorption usually plays no role or only a very minor role in heterogeneous catalysis. 411
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Chemical Adsorption: Involves the formation of covalent bonds between the molecules and the solid surface. 412
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Chemical Adsorption: Involves the formation of covalent bonds between the molecules and the solid surface. A very important consequence of chemical adsorption is that normal covalent bonds of the reactant molecules are weakened. 413
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One way to form NH 3 from N 2 and H 2 is to directly break the N N and H H bonds: 414
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One way to form NH 3 from N 2 and H 2 is to directly break the N N and H H bonds: N 2(g) N (g) + N (g) 415
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One way to form NH 3 from N 2 and H 2 is to directly break the N N and H H bonds: N 2(g) N (g) + N (g) H 2(g) H (g) + H (g) 416
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One way to form NH 3 from N 2 and H 2 is to directly break the N N and H H bonds: N 2(g) N (g) + N (g) direct bond breaking H 2(g) H (g) + H (g) 417
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One way to form NH 3 from N 2 and H 2 is to directly break the N N and H H bonds: N 2(g) N (g) + N (g) direct bond breaking H 2(g) H (g) + H (g) In the gas phase, this requires a large amount of energy (in the form of heat). This makes the process expensive on the industrial scale. Also, the NH 3 formed at these high temperatures would not be stable. 418
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With a catalyst: 419
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With a catalyst: N 2(g) N 2(chemisorbed) 420
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